$oluTloHs
1,1
Page 35
CLap
I
Vector Analysis
sot
,t.1.,1
Option (C) is correct.
M
-N
: ;,, ",,
:"
:=:,:"-_:z:::,;t;: ITa,l boul2a" ^, So, the unit vector in the direction of (M- N) is 73o,,*5ar*2o,, M-N -^ - jM:Tl-Wa,+5%+q -
:
sol. t.r.?
Option (B) is correct. Vector G at (-2,1,3)
G So,
0.92a,* 0.36a, + 0 Jayz.
:
: 4(- 2) (r) a, + 2 (2 + (_ z),) au _t S (J)2 a" :- 8a, *l2aui27a"
"":ft.:ffi
unit vector in the direction
:-
sol- {,r.3
of.
0.26a,i
G at
e
0.39a,
*
:
0.88o,
Option (B) is correct. cross product of two parallel vector fields is always zero since the angle between them is 0:0".
i.e.
AXB:Q loo o, o,l
It tt?-613l:o l"
Solving
sol.
1,,t.4
(- tz - Jp)ao 1-(Bo + 6)aa + @ - 2a)a" : it we have, 0:- 4 and a:- 2
Option (D) is correct. From the given field vector we have the component E, :9zf cos2r. So for the given condition E,: 0
9zf cos(2r)
We have,
:0
This condition met when,
Z:0
of,
a:0
or,
cos2r:0 9 2r:trl2 + r:r/4
Therefore the planes on which field component
Z:0 sol-
1.1.s
g
A:0
and
E, will be zero are
r: r/4
Option (D) is correct. FYom the given field vector we have the field components
Ev
:8zTsin2r
E" :2f sir,2r 9 A: 4z plane the in Now, A -- 42: 0
and So,
Eo:az(A2)sin2r:il2isin2r
: Er:
E"
Thus
2(4zl sin2r
:
32i sin2r
E"
Option (B) is correct. For the given condition
i.e.
-E: 0, we must have E,: Es: E,:0 9zf cos2r : Sz1sin2r: 2f sin2r :
This condition met in the Plane
Y:
0
$'
Option (D) is correct. of since, the options include spherical as rveli as cvlindrical representa'tion A, so, we will transform the vector in both the founs to check the result. The components of vector field / are A,: !, Ar:0 and '4' : Q as Now, we transform the vector componerrts in cylindrical systern
lA,l I cosd sino 0ll/
I
f;'l:l-;;; " ;;;;; fil[,,| ; ll,r,,l
:*
l;,1-l
i'
: @os$) (1) : cosp Ar:? sin@)(1) :-sind Ao
i
A,:o $
in cylindrical system is A(P'Q'') : cosfaP- sinPaa Hence, both the options (A) and (B) are incorrect' Again, we transform the vector components in spherical system
So, the vector field
'tt il
iltr
as
IA,l Isingcoso singsind cosg]['+".| I'lnl :lcos0t'o*p cosdsino - sindll'4,1
3
It
[o,l [
r
: A6 : Ao : A"
'4 .,{
ffi
So, the vector field
g
ol[1
]
: sindcosd (cos 0 cos @) ( 1) : cos I cos qt (.- sin P) (1) - - sin (':
(sindcos@) (1)
in spherical svstetrt
A(r,0,$): sol- 1.{,6
c{,s@
* *in o
sindcos@tt''
*
is
cos6cos?ao- sinda"
Option (B) is correct. We transform the given vector field in spherica'l system' Since the given vector tleld is F: IIa, (), F' -- O' The Cartesian components of the field are Fl, : 10, Fu: as So, the spherical components of vector field can be deterrnined I4l Isiudcoso sirrdsinp cot0|[{,1 I nl : lcosdcosd cosdsin o - sin0ll'fl,1 So' we
get
t.l 1""--,';;; cose
; : i3#::i
tllo
I
arld
F,r
:-
Now, for the given point
10sind
(z:3, g-: 2, z:_
Page 3?
1) we have
: /re)Tltf {_-tf : lu d : ran-, {ZLi: ,*"-,/ /fT(zF\ , \ (-1) l-- 105.5' d : tan_'(si;: ,""_,(i): 33.7. r..
Putting all the values in the rnatrix transformation, we have : 10sin(rOS.S")cos(33.2.) g 4 : Fo _- 10cos (105.5.) cos (83.7.) : _ 2.2 F+ :- 10sirr(33.7.) :_ 5.5 Therefbre, the vector field in spherical coordinate is tr- : F, a, * Foas * F*a5 : 80" 2.2a,e _ S.5a,o rlol
1
.1.9
Option (D) is correct. The given point is shown below
:
After 120' rotation looking down the axis the new co-ordinate axes (r,, y,, z') will be as shown below :
x, lJ
So, the rotation carries e axis
into E ; g_axis into
therefore the new co-ordinates of point g, : 7:
p are :
r
and
r into z.
q;
'u' :': :L:: a
:r:.a_:b i'e' (c' *. b) is the co-ordinates of point p in the transformed
system.
Chap
f
Vector Analysis
Page 38
Chap
$oL
t.'1"1$
1
Vector Analysis
Option (C) is correct. The given line r: 6, z:_ 2 is parallel to g-axis. So, the component of parallel to the given line is
Ar:(A'ar)ao :l(- 2a,* 2}ao* 4a") ' : (-2sirr@*20cosd)o, At Point P, Q:90o, so, Av:-2au {.{.'t{
anlo,
Option (C) is correct. The position vector can be defined as : R : fre,"-l yau* za"
R:nl7T7+7 So,
grad,R
: #o'*#",*Eo,
:iffi"'+16ft;7"'*Lffi-"' ra"* Aan* za" _ P ::-R --wT7i7
gol- {.1,t2
Option (A) is correct.
Y' A : fittl
+
ftQra
t + t) + St+'+
+
:0 +2r+2U :2(r+
sol. {,{.13
zu')
A)
Option (D) is correct. We have the vector field components as Fo: psinS , F6: P2z and F, : zcosQ I a" paa a,l V Now, X
'
:;l
:
f, [rri,\
i186,*" o - $
o
t'1", - i rl&'cos
@
- ft
+llftet, -
;,
: lt- zsin$ -
p'fo,-[0
: -|Qsin O t
p')
At point P (7,+,2)
VxF
:-r(2x :- 3ap*
so|- t,1.14
Option (D) is correct.
D
:
rJzo,r*
ar't
-
0]aa
(3pz
1+13)
-
+f,lzt
ao*(3x 1 x 2-o)a"
rzon* Wa"
v . D:fttu,)*ftoa+fi{w):o v
ou o,l
lu"
"" wl
xD:l* h :,1
f6n"i"ol",
pcos$la"
cos $) a"
6a"
lo,
r-
o';" ol"o
to the plane of triangle is given of vectors. so the unit vector perpendicular by
Page 40
Chai,
1
^:ffi#t Ia"%
Vector AnalYsis
Now,
ABx ag:l
a"]
18 -191
[- to 8 151 :[18 x 15-(-10 x 8)]4,-[zo x 15- (- 10) x (- 10)]a' +[zox8-(-10x18)]o, :350o" -200ao*340a" _ 350a,-20}ar*340a"
:@-
So,
: sol- {.{.{9
zO
0'664o'
-
0'379 ao
*
-
350%
- 2m"'+ 3a9e:--
7@I-
rooi'+
(r+o)'?
0'645 a"
Option (A) is correct. given by vector in the direction of vector AC is fi.u ""it ISa, 8a,* AC -!0a,1 aAc
.
:
WCI
-
1- ro;'1 1A)'+ (rs)'
:- 0.507 a,* 0.406o, * 0'761a" : - 0.514, -l0.4Lau* 0.76a"
is always perpendicular to the plane of Since the cross product of two vectors triangle which is perpendicular vectors. so, the unit vector in the plane of the
toACisgivenbycrossproductoftheurritvectorperpendicrrlartothe
pla,ne of the tria.ngle a,nd the
i.e. &p: 3(,L 1.1.20
unit vector aec'
anx a'ac:-0'5504' -0'832a'*0'077a"
Option (D) is correct' Unit vector in the direction of AB is given by aAB
20a,*l8ar-I9a"
:
@Jraf+1-to;'
:0.697 a,+
0'627 an
-
0'348a"
interior angle at A is defined A non unit vector in the direction ofbisector of AS
lloo, *
aec)
:
Lr10.697 o,
*
0'627 an
-
0'348a"
-
o'507
a'*
0'4064'
* 0'761o']
:
0.095a, * 0.516a, * 0.207 a" of interior angle at '4 is given So the unit vector in the direction of bisector
by
0.095o" * 0.516o0 * ou":@
: : sol- t"t.zt
O'2O7
a,
* 0.915a, * 0.367 a" 0.77 a,* 0.92a0 * 0.37 a"
0.168a,
Option (C) is correct.
ThevectorfieldFcanbewrittenincartesiansystemas
F(r,v,z):W:ffia,+ffi", F(r,y,z):#o,+ffw
(r: pcosQ,Y:
Ps|li,$)
The components
: !(.ordo. * sinda,.) p' of vector field F are
4:-p"p lcoso. 4,:
I
siu
d;
so the components of vector field
and.
F,:
Page
Chap 0
F in cylindrical system can be expressed
:::r
'q
lfi]
l]tfil
rr:)lcos'Q+sin'61:L
: f,lcos6(- sin/) * sin@cos@]: s F, :0
rr
field At the point P(p :2,
d
:
7r14,
F : +ao: s.
1.1.22
:
F(p,$,r)
So tlre vector
e:
1
poo
0.1)
0.ba,,
Option (A) is correct. Any vector field can be represented as the sum of its normal and tangent,ia! component to any surface as
A
:
At+_A,
where ,46 is tangential component and A,, is normal component to thr, srrrfa,ce r:20 at point P(20,150",330'). So, Ar, : T&r:20a. and
thererore'
"' :11,r?; cosgas :0.043oo
30t
1.1,?f,
*
*
t o,
100oa
Option (D) is correct. Any vector field can be represented as the sum of its normal and tangential components to any surface as
A:Ar+A, Here A1 and .4,, are the tangential and normal components to the conical surface Since the unit vector normal to the conical surface is ap. So,
An :-
8a,o
and therefore the tangential component to the cone is
Ar: A- An:-72a,*9aa 30L 1.1.:*
Option (C) is correct. Consider the unit vector perpendicular d : 150' is
b: b,a,* b6a6
to ,4 and tangent to the cone
(Tangential component to the cone will have Now the magnitude of unit vector is 1 So,
b7+b3:t
1
Vector Analysis
as
:[
41-
be: tl) ...(1)
Page 42
and the dot product of mutually perpendicular vectors is zero.
Chap
So,
1
Vector Analysis
A. b:0 -72b,* gb, : g
4:!u.
....(2)
So, from equation (1) and (2) we have
a3(r
+f;) : t
'':*,
u*:t
Therefore, A:f{S", *4ar) *61
'r-t.s$ Option (A) is correct. The separation vector trl can be defined as
R
and so,
: (r-
a)a"+
:
(y- b)ar*(z-
c)a"
n:M "
(+)
: *l@- o)' + fu- b)'+ (z- c)'\-'t2 a, +&l@
- a)'+ (v -
b)'+
(z
- ")'l-'/'o,
+&l@- a)'+ @ - b)'+ (z-
c1'1-'r' o"
: -!1ny'r, (* - o) - |1ny'r, (y - b) w - lgy,r2 (z ", __(r- a)a"*(y-b)q-f(z- c)a" :_& R3 Rslz as!.
{.{.rs
c)
a,
Option (B) is correct. The gradient of a scalar field at its maxima is zero. So at the top of hill
or,
Vh:0 (72r-ay+36)a,*(76y-4r- 56)o, : g
Therefore both the components
will be equal to zero
I2n-4y*36:0 l6y- 4r- 56 :0
I.e.
and,
Solving the two equations, we get,
. :x:'-2, U:3
Thus the top of the hill is located at 2 miles south ' miles east of the railway station. $oi-
l.t.t?
Option (B) is correct. Consider the position vector of point P is R : x&c* gan* za" So, the magnitude of R is
R: G+T+r' and unit vector in the direction of
a':!9@ " Ji+f+/
R
Therefore, the vector field .F at point
b-
is
P
is
(-2
miles north) and 3
,:#*:#(ffi7:*11vffffi] The divergence of the field
v''F'
F is given
Page 43
Chap
as
: tof*ffi;zy + 3a@ + +Lztr. *6;;;ryl : rclpfiqw - +@;#?d- + 6;fqw y\zA) J v(za) z(22) 1 B z(22) 3 -q ttr) 2@ii;4{/, 2@Ti;4l'- 1;' a S * Srt7, - 2(7+7T ye
I
(i+f + /) | :'o[#*#?#fu]
1o[*_*r] :o But at origin (r:0, U:0, z:0) _
expression for field zero else where.
jL
t.r.za
F
the position vector.fil:0 and so the blows up. Therefore, v . F is infinite at origin and
Option (B) is correct. For a vector function to be irrotational its curl must be zero. Now we check it for vector ,4.
VxA
:[&o-
ry
- ") - fu-
rz)fa. +ffia+ n
-
*a-
ra
-
z)la,
+l*c za - fta+ a)a"
: (- B *B) a,+ (1 - I) o"+(0 _ 0) &" :
0
So, vector A. is irrotational.
Again for a vector to be solenoidal its divergence must be zero. So we take the divergence of the vector ,4 as
v
: S{,* 4 +&eu1 +S@- ry -,)
.A
:1*o- 1:o
so, vector c is solenoidal Thus the vector A is both irrotational and solenoidal.
Note: since the curl of the gradient of a scal# field is zero. So, we can have directly the result
(-Vr:_:Vx (Vr:g
Vx A:VX Option (D) is correct. We have
Comparing
A
--v Y :-ffa,-#",-ffo"
it with the given
vector we get
:
ar:- @+ z) + f :-t - rz+ fi(y,z) dr u 6:3'- f :3Yz+f2(r,z) ^2
of :- ,- 3a a", -\ )+ r0z -_ @- - z) l:-
rz+3yz+ $"2 + fr(r,a)
In conclusion, from all the three results, we get
f :_+
_ rz*zyz+
{
I
Vector Analysis
Page 44
Chap
s{}L 1"*.30
1
Option (A) is correct. Consider that the vector
.A
is
in a" direction
as shown in the figure.
Vector Analysis
So we can
and
write the vectors in cartesian form
as
A:4a,
t::G:-ll''*to"(-o') _
_T*,
(A:4) (B:
2_,
3)
Now the resultant vector, So,
and
R :64,- 88 :3.22a,172a, R : {$8 +@ : 12.43units angle that R makes with z-axis is _ ," _ uuD 0-"^.-t13.22\:zr'
\AB)
So the graphical representation of vector
s$L,r"1.3"1
R
is
Option (D) is correct. We go through all the options to check the direction of the vector R for the corresponding directions of A, B ar'd C. Option (A)
since the direction of cross product is normal to the plane of vectors and determined by right hand rule. so B x c has the direction in which thumb indicates when the curl of the finger directs from Il to c. Thus B x c will be directed out of the paper and so we get direction of -a x (B x c) towarrl east. So the given direction of R is incorrect. In Option (B) : Direction of (B will be directed toward west.
x
is out of the paper so,
-fi|: A x (B x C)
In Option (C): Direction of (B x C) is into the paper so, will be directed toward north.
R: A x (B x. C)
C)
In Option (D) : Direction of (B x d) is into the paper so, R: A x (,B x C) will be directed toward south. so the given direction is correct.
3(
1.1.32
Option (D) is correct. As the vectors B and c are defined in cylindrical system. so, we transform the vector in cartesian form as below Given the vector field B : ap+ 04+3a,, the cylindrical components Bo:I, 86:1, B":3 So the cartesian components of vector B is
[E:1:[:r; lu"l B,
;ti iilfi]
: cosdBr- sin$Ba: cosd _ sin/ (Bo: Br: Bu : sinQBo* cosQBa : sind*cos@ (Bo: Br: B":P"-3
and so the vector field in cartesian system is
3:
So at the
wint (2,$,3)
(cos@
-
sin/)a" + (sin@
*
cosS)au
*
Ja"
11
Iy
Page 45
Chap
1
Vector Analysis
Page 46
Chap
I
now we transform
Vector Analysis
B -- a,l avl3a" the vector field C: "/2 ar+ Ja, in cartesian
system.
Co:J2, Cr:0, C":J
The cylindrical components,
So the cartesian components of vector
C
is
l3l:[:* "rr:[el C'
So,
: Ji
Cr: /2
cosd
Cr:/2sind
C":t :,/ C 2 cos da, + {i
and
So, at the
C"
noi"t
=
sin$au
*
3a"
(f,f,O)
c:
,/ 2
cos(+)",+
n,i"(T)n * to,
-- ax* au*3a,
So all the three vectors are same at
s6L l"{"35
their respective points.
Option (A) is correct. For checking whether a vector is perpendicular to a given vector or not we take their dot product as the dot product of the two mutually perpendicular vectors is always zero. Now we have ,4 * B : 4a,* 4au* 4a" So we take the dot product of (A + B) with the all given options to determine the perpendicular vector.
In option (A) (-+o,+ 4or) . (+a"+ 4ar+ 4a") :- 16 * 16 : 0 In Option (B) (4au*4a,). (4a,1-4ar*4a"):16+ 16 :J2 * In Option (C) (a"-l a")' (4a,+ 4ar* 4a") : 4 * 4: 8 * o so|-
"r-{.3d
Option (B) is correct. The given gradient is YV(r,Y, z) : 1.5t y/ a, So,we have the components as,
av -a.clr
* 0.5f /
au
*f
yza"
: l.Sty*
v : L|{:y/ +
f1(y,
z)
:0.5f i V :0.5f yl * fr(r,z)
:
0.513 yt
+
fr(y,
AV
-
oa
ff :fv' , : ry + f,(u,z):o.bf u* + l,(y,r) AV
Thus by comparing all the results we get,
V:O.bfg* $oL
,!.{"35
Option (B) is correct. Consider the given plane
z)
0
rlz :1
' So,
Wz-1:0 function, f :ryz-l gradient of function, Yf : yza,* rzou*
Page 47
Chap
1
Vector Analysis
rya"
Since gradient of the function of a plane is directed normal
to the plane scr
the normal vector to the plane at the point (2,4,+) is
yf :la,+\an+ea,,
Now consider (r,y,z) lies in the given surface r!z:r. so the tangential vector to the given surface at the point (2,4,f) is
r : (n - 2) a, r (y - +)q + (' - [)""
This vector will be perpendicular to V/. So, (") . (V/) : 0 (dot product of perpendicular vector)
|{. -
z) +
+@-
a)
+
r(,- *) :
o
2r*y*322:12 sot.
t.t.3c
Option (D) is correct. Let us consider a contour abcd, as shown in the figure.
As vector
A
has only a6 component so
its integral will not exist along
segments ab and cd, and so the contour integral for abcd is
Al
.
at
: (I+
.
I"* I^* L)d For bc segment, r: 1 and 0. 0 . +
at
:- rd0ae r: d, and 0 < 0 < T/2 dl : rdAae . :_ dl
and for da segment
so, A#
"
I=':(+){"ar)+ I=,_(i_)t :-(r-')(+) +(e-6)1trlzy :
te
As for the given contour C, 6 tends to zero
so,
. f d or:trnflI
.
dr
:I.
*t "-,
+
"-o)
,EC"-,+ e-o):+(r-
e-,)
I{OTE: l{osI of titc sllttlcnt,r r.lti a ntistak*: irere bv
."-&
r
".*'lo"if ]Wt"
"
Page 48
Chap
$id:l{".
d *"}!'
1
Option (B) is correct The divergence of unit vector a" is
Y . a. :
Vector Analysis
i*rttl:
$lz,):l
the divergence of unit vector oo is
Y.
Y
as:ffiStsin@):#:
and the divergence of unit vector ar is
Y. l;{.19-1.,1.3i1111111111111111t
ao:#&(1):o
Option (D) is correct. A vector can be expressed as the gradient of a scalar if it's curl is zero. Now we go through the options.
'
o,, clu o,
Option (A),
Curl of the ,recto, :
Option (B),
Curt of the vecror
Option (C),
Curl of [he vector :n]
:
ad; i, *' l: O 2iz 2"rYl :2Yz lao paq a"l t1l-:, ;l l* o i, p(#) 0l i0
"
o? ri ? t" i; do ?l:o P,'"io :
3sinO o'
lr So, $ilt."t.'tr,3$'
it
j
I
can be expressed as gradient of a scalar'
Option (D) is correct. Any vector for which rlivergence is zero can be expressed as the curl of another vector. For checking it we go through all the options'
In Option (A),
Divergence
:*{tto-
d} -&aa+&e)
: T- tr:O In Option (B). In Option (C),
: tr&(#):, Divergence : i*1r
Divergence
"y) * #,r", #(#):
:-\lzcoso)+-r-1 So all the vectors can be expressed as curl of another
^zsindcosd
0,,.
vector' i \
5i1t" t i,!i*
Option (C) is correct. \ for y) 0 i.e. above r-axis field will be directed towards *4" direction and\ will increase as we go far from the r-axis, since g-increases' For y ( 0 i"e. below r-axis, field will be directed tou'ards - o" direction and it's itrtensity will increases as we go away from the z-axis.
!i:;.
Option (D) is correct. Given the divergence of the vector field is zero i.e. V dA, , dAo _n
'l.l.t!l
'A:
-1. 1 ^ or oa -u AL - -94"
0r
and the curl of the vector field is zero,
dy
0
...(1)
-^\
t.e.
VX.4:0
Page 49
I a, a, a,l
Chap
li, *,,';l:o la,
i,
ol
_*".**o..(#_*)",:o 0A, ^-or
0A, _ n ^ -u oa
....(2)
(Since / is only the variable of r and g. So the differentiation with respect to z will be zero). Differentiating equation (2) with respect to r we get,
o^'4. %.Ai 0r0y :- o
ry-jrdl.\:o di 0y\0r )-
aA_ d l- 0A,1 _ A7-a a\-a;1:o
(from equation (1))
0rA, , }rAu _ #+fi:o Y2An
:
g
Again differentiating equation (2) with respect to y we get
{+-014.:s 0r0y Af
p-(a!\-0?4,:o
0r\ay ) at Lt--4a"1- a''4' : o dn\-i;)67
:
(from equation (1))
a2A- . A2A^ :+#:0 dxi
:{^ ' n" -0
rsot-
;
F. tsi F.:
iii 1
1.1.4*
Option (B) is correct. Given, vector position of P(r,y,z) R : ila, * ya, _l za"
: ^p 1f ,yj R":(t+f +iyt,
So, or
R
R"
R
: (t + t + *l21ra,* yo, + za"]
Now we take the divergence of the vector
v
. (R" n1:
as
*,@ + f + tyt, + &xr + f + ty,, + &Eo + f + *lt, : (i + f + *yn fu +,I+@ + f + /Y/,-||Q{ +(t + f
+
tf,ft+
t(t? +
t
+ tyr'
u[$(t +
x*+*:k***x**
f + ty,,-,](rr)
ft +,[g(i + f + /Y/'z-'f(2')
+2f +2*) : BIi" * nR" : (n * Z)R" = 3R" ++R"-r(2i
I
Vector Analysie
7-*-
soLuTloNs 1.2
Page 50
Chap
1
Vector Analysis
sol
t,2"1
Correct answer is
0.
As the direition of cross vector is normal to the plane. So, direction of B x C will be normal to the plane defined by the three vectors. Now the dot product of two mutually perpendicular vectors is always zero and since the direction of B x C will be perpendicular to the plane of vector A. So A ' (B x C) will be zero. $oL
1.2.2
Correct answer is - 1. Dot product of the two orthogonal vectors is always zero. l.e.
A' B:0 (4) (1) + (2k) (4) + (k) (- a) : o 4+8k-4h:0
4k:-4 k:-L soL {.2.3
Correct answer is 6.78
.
Since the two points are defined in different coordinate system so we represent
the point Q in Cartesian system
as
r : pcos$ - 4cos(- 50) :2.57 y : psind:4sin(- 50) :* 3.964 and z:2 So, the distance between the two points P(2,3, - 1) and QQ.57,-3.064,2) is given
as
IPQI: : 6.78 units $oL
1,2.4
Correct answer is 15. Since z-axis is normal to the surface z:5, so first of all we will find the angle between z-'axis and A which can be easily obtained from the figure shown below :
I
1.e.
A, 3 -f cos@-14:ffi:m d
:
cos
:74.g8" = 75' '/-L\ Jr34l \
3
Therefore, the angle between surface Srd- 1.2.5
z:
5 and vector .4 is (90" _ d)
:
Chap
Correct answer is 40.
I
Vector Analysis
In a cylindrical coordinate system Laplacian of a scalar field is defined
Y'f
Page 51
15. .
:i*(,H).#r#.#
: |ffofz'sin d * 8p) +
if
2pzsins -
as
a*
ftsinlcos d) +$QostnO * 6zcos2g)
:
|Qzsind+ 16p) - fienr"ir.| * 6lcos2d) + :16*6cos2/ -ffcoszO
6cos,O
At point P(S,r12,G)
Y'f :16+o -q+!q x (tol.
1.2.6
Correct answer is 0.b
1)
: 40
.
A vector field is called conservative (irrotationar) if its curl is zero.
VxM:0
l.e.
la,ayazl
I * y 2kr& *
l:o rcosrzl
zcosrz* I o" (0
-
0)
- an @os rz -
I
i
- cos rz + rzsin rz) + (2k - I) a" : 2k-7:0 k :+-- o.b.
rzsin rz
Of'
I I
g
i
sol 1.2.? Correct
answer is -0.b . For a scalar field to be harmonic,
Y'g:o }'_g_*d'g
_0,g _n
di'affa7-v
2(l + 2k)y sol- {,2.8
:
0 which results i" k: _i:
Correct answer is 1. For vector ,4 to be solenoidal its divergence must be zero.
i.e.
V./:0
*f, * +z) + fter - Bz) + $er* By - cz) : e 1*0-c:0 C:7 sol- t.2.s
Correct a,nswer is 0. Consider the differential displacement,
dl
so
:
dxa"* dyao*
dza"
f"o':(f"*)*+(f"au)u+(f"a")""
{ _ 0.5
For a contour the initial and final points are same. So, all the individual integrals described above will be zero. Therefore.
Page 52
Ctap
I
fat:o
Vector Analysis
Jc
,.2"10
Correct answer is 0. According to stoke's theorem.
so
f d.n:l(, xA).ds
. d,r: Since V x (VU) : 0 So
,"2"t1
f rvul
"(vu)l
.rrs
(curl of the gradient of a scalar field is always zero) the contour integral is zero.
Correct answer is 0. According to the divergence theorem surface integral of vector over a closed surface is equal to the volume integral of its divergence inside the region defined by closed surface. i.e.
fd.ds:f(v.A)d,u
v .A:s{zo4+&G,)+fiQ,u) :0*0*0
Now,
so 1"2.'!a
fl,
[n.as:o
J
Correct answer is 70.53
.
Consider that the cube has its edges on the r,y and z- axes respectively as shown in the figure. As the angle between any of the two body diagonals of the cube will be same so we determine the angle d between the diagonals OB and AC of. the cube.
Flom the figure we get the co-ordinates of points A, B and C B* (3,3,3) antl C- (3,3,0) 4- (0,0,3) OB : 3a, * 3ay l3a, So, the vector length,
and
ry;--. /
AC:3a,+3an-3a,
For determining the angle 0 between them. we take their dot product
as
:loBllAClcos? 9+9-9:(3v5)(3./e)cosd
page bB
is
Vector Analvsis
(OB) . (AC)
So, the angle formed between the diagonals
I : cos-j/l\: \J/ sot 4.?"t3 Correct
chapr
70.53'
answer is 32.
The circulation of
A
around the route is given by
,{d.m:(l.FI)d.at
where the route is broken into segments numbered 1st segrnent : (0,0,0) - (2,0,0) r changes from 0 to 2, 'y : 0, z : 0
so,
In.m:['rrl,t":3($)l':8
.11
Jo
-
2nd segment : (2,0,0)
tr:2,
I to 3 as described below
gr
(2,2,0)
changes from 0
to 2,
z:
0.
[d,. m: I'urrda:o 3rd segment : (2,2,0)
tr:2, a:2, So
-
.m:
(dI:
dya)
(itr:
dza")
(2,2,2)
z changes from 0 to
[,t
(dt:d,ra,)
2.
[' et ar: (3 x 4) z : 24 )')
total line integral will be :
[,] sol i.s.{"* Correct
- dt :
8
+o+
24:
32uints
answer is 32.
For the straight line from origin to the point (2, 2, 2) we have the relation between.the coordinates as
tr:y:z dr : dry: d,z
or,
and the line integral along straight line is given as
dl : d"ra,l dyar*
dza"
Therefore, the line integral of the vector field along the straight line is given AS
I a,. m: Izta,+ {ay,ay+ l#az : lzid,r+ {o}a,+ f z*a, : I i dr : (tr$) : tarl 6 12
:4X8:32uints
l'
sol "r.*.{$ Correct
answer is 0. For the closed path defined.
the line integral in forward path : 32 units the line integral in return path : -32 units. So, total integral in the closed path is :
J
A. dI:32-32:Ounits
Page 54
Chap
I
Correct answer is 6.25 . The circulation of .4 around the path
L can be given
as
f d.at:([*[*1.)o.*
Vector Analysis
where the route is broken into segments numbered 1 to 3 as shown in figure
below:
1"' segment ,
so,
(6:30", z:2,0 < p < 5) and dl:
[d.
2"d segment : (p
:
dpap
25 at: f'psin|dp : I'goo : T
z:2,
5,
30"
so, [d. m: Io*:o :180", :2, < 3'd segment, (Q
5
z
180")
p < 0)
(d:
and dl:
and
dl:
d,fao
dpap
. [n dI: fupsin|dp :o
(d:
Therefore, the circulation of vector field along the edge
I d' Correct
JT",
m
tr
3o')
180')
is
:ff+o*o : 1[ : 6.25units
is 0.5
.
Volume integral of the function is given by
, : III r
d,rd,yd,z
: [[
zo* a*aaa"
The surface of the tetrahedron will have a slope So, for a given value
integral will be
r* Y-l z :-I of y and z, r varies from 0 to (- 1 -yr(r-v-
z) and
r
JoO, :-7-g-z again for a given value of zt
y
")
ranges from 0
to
(- 1-
z). So y-integral will
be:
rt-l-z) Jre t - u -
z)dy
::[r-t 1 --z\r-fl-''a zta Vlo [\-
:(-1-r),-!#:eJ# \, :l+ ,+ {
Now there is only one remaining variable z thatranges from have the volume integral of the function as
v
: fo so*(i* ,*$)*
-
1
to 0. So we
:30[€ *i.fili_,
Page 55
Chap
=ao[o**-1*#] :30 x -#: *: o'u sot-
1.2.18
Correct answer is 12.5664 . The net outward flux through the closed cylindrical surface will be summation of the fluxes through the top(in a, direction), bottom(in - o, direction) and the curved surfaces(in oo direction) as shown in the figure.
Since, the vector field has no z-component so, the outward flux through the top and bottom surfaces will be zero. Therefore, the total outward flux through the closed cylindrical surface will be only due to the field component in ao direction(flux through the curved surfaces) which is given as
f n . ot : I:,ITf)bd,6d,z). 1 P'cos'|1" az1 -= l":f' l,S'"- o\'=t' )@ At p:r, o. d,s :9u-*,,0d4]ll'd,,] f a'4
o
:2XrX2:4tr:12.5664
tol
1.2,19
Correct answer is 25L.3274 . According to divergence theorem surface integral of a vector field over a closed surface is equal to the volume integral of its divergence inside the closed region:
i.e.
f n.as : [(o .A)du
Divergence-of vector
Ais
v.A : iS{oo,l*i&o,* *o, : i&Ofnp + 2psin2g)) +
i*$@"in2g)+
:8+4sin2d*2cos(2/)*6
: 8 + 4sin2 d * So bhe surface integral is
2cos2
$
-
2sin2
1
Vector Analysis
$* 6 :
16
firca
r Page 56
Chap
fd
1
. ds :
. I (v /) a,: lll
oe) pd,pd,gdz
:
Vector Ana$sis
rc [' pdp ['16 ['ar: Jo"Jo Jo : 80zr :251.3274
16
x2x T 2x
b-
NOTE: irrrtglal.e
$0L
{.4,2$
1-1ir
givcn lrtv:t{,r fir:i{i.
Correct answer is 7. According to stoke's theorem, line integral of a vector function along a closed path is equal to the surface integral of its curl over the surface defined by the closed path.
i.e.
{c. JrJ
m: f(v x c)ds
Curl of the vector field is
Vx
G:-6ta"
and the differential surface vector
dS : drdy(- a") So the line integral of the given vector field is
f c.m: Ir, x G)d,s:- ll -ara"au : I' I' o d,yd,r.t [' f'-' t aua, u
:
6
a
[' Jo
t rd,r* oJr[' ie - r) d,n
:^ltl'*alZ*t -"[4Jn'"1 3 -rnl' 4], : SoL
't"2"1t
u(+
- o)*'[(tf - f)-
tA
- i)l:
7 units
Correct answer is 2.3 . The relationship between cartesian and spherical co-ordinates is
,: ^ft 1f+7, r:
rsin9cosd,
rsind
: rEW
:
A: rsin9sinf
We put these values in the given expression of vector field as
n
/r +qllsn trn _Jrltaf r 2-n-to'*lq]
t+f 12-'' : #ry[(cos/ - sin@) a,* (cos6* sin@) an] ^t
: r[(cos d - sin /) o, * (cos d * sin /) or] Now we transform the vector field from cartesian system to spherical system lnl
Isindcos/ singsin@ cosdll41
cosdsin@ -'i"ollr,l l*l : lcosdcos/ sind cc,s@ 0l[4 j
F, : : Fe :
lFrl [ r(cos6 - sin/)(sindcos/)*
r(sin0sin/)(cos/+sin@)
rsin d r(cos Q
-
sin /)(cos 0 cos Q)* r(cos
@
*
sin @)cos d sin
/
: Fo :
rcos0
Page 57
-
r(cos Q
:T
i.e.
/)(-
sin
F : rsinla,*
sin @)
*
r(cos
rcos0a6a yq,
: f sin?dfrd,6a,
and the surface 5i is defined in the region
as
: [" Jo
:
^91
< d < 30', 0 < will be : &
1.2.23
Q
<2r
[tn" r'"in'0,1fr,16 "fi
f,'id I';"2gdn = *lr, :2.276 :2.3 3oL
is
r:2,0
So, surface integral through out the surface
I P' Js,
Chap
@
8
Correct answer is7.2552. The vector function in spherical form
a^s
*l
@t
r: 2)
calculated in previous question is
:
F : rsinfla,*
rcos9asa ra* The diff'erential surface r,rctor over the surface 5z is dS : rsin0dddra : and the surface ^9: is defined in the region 0 3 r < 2, 0 = 0 S 2r, 0 30" So, the surface integral of the field over the surface is : ^92
I r ' as : I' I" sot-
1.3"x3
:4o# :7.2sb2
rrinocosodfid.,
Correct answer is 42. Given the vector
A :3(Aa,*
field,
rao)
The differentiai line vector in the cartesian coordinate svstem is
dl :
so,
dro,,
I dyao* dza"
{d.s:fzydr+[s,au
y: ^/;12 put So. we r:2A' and dr:
The given curve is,
4Ed,y
in the line irrtegral tq[r,]?
I,a. m: f'ny,au+ ['ay,ay: 6x7:42units sot.
*.3.34
Correct answer is -6.2832 Given the vector
field
"
F :r+c9s22 a" f':so'
2cos2a
and the differential surface vector over the outer spherical surface is
dS : (lsin0d0d,$)a, (for r:2,0
. { r ot : I" I" (r#1rrsin0d0dft):set-
'!.t"e$
I
Vector Analysis
The differential surface vector over the surface d,S
@ + sin /)cos
(fr,O < d < 2n) 2n
:-
6.2812
Correct answer is 0.3927 . As the integral is to be determined in spherical volurne so, we transform the function in spherical system as,
2r :2rsin9cosd
Page 58
Chap
and so, we have the integral
f
Vector Analysis
I
: r l=,f,' l)'' ?rine"osg)(rsin0d.rd0dft) : z[['r a,]f^1"'' ;"oro] "i,.,eaalf : ,[t])Jg - yl:'rsingtrz
z,a,
:r*I"ft):2x#:+:o.Be2T r$t-
'1"?.**i
Correct answer is 5.141
.
contour integral of the field vector is evaruated in 3 segments as shown
below
(1)
In Segment
So, In segment
so, In segment
so, So
:
dl
0
dpa,
f,d. at : f\"o'dao : ("*o)[4], : t:2u,.it (2)
:
dt
pddaa
ld.at : L',to*l: 2[o];t' :71 3
It
.
dI
:-
at
:_
the contour integral is
Correct answer is _g. For the given contour
dl
so,
f'
oecosr/2)(d,p)
:o
drl: (z+n)units
0
pddao
2tr:r8tr
C2
d,I
:
_ pdeaa
. at :f",a, ['i,(odd):_2t
Therefore, the ratio of the contour integral is
fn.at rT_:,tgJ':_9 lt.n ?2")
J,:, a{1*,
t.2.2a
: b.141
e
f ,,+. at:[']{oab):ex
and for the contour
So,
:
03p<2,at6:rl2
dpap
f ,t ' at :l[+ I+ I]to . {.?"a?
0<6
Correct answer is _0.1667
.
0<032natp:l
l i I
i
i
The line integral (circulation) of force F around the closed path can be divided in four sections as shown below.
:
tI i
Page 59
Chap
I
For segment 1 we have,
So,
For segment 2 we have,
So,
For segment (3) we have
So,
Y:2:0 dI :- dxa,,
Ir
.m:
For segment 4 we have,
so,
I' re
0
:[-$]: :-+
I: Z:0 dl : dAas,
o
[r.n:f\-,,)(au):o tr:zrU:I dl : dna,l dzo,,
0
<
r<
1,0 < z<1
Ir. m:['ta,+1,'{-fyo" :
[+], -t,tL
:* - :-
Y:z'tr:\ d,I :- dyau- d,zo,
'
3
f r. u: ['{_o){_dil+ I efl?a,) :[gl:^*t+]'
:]++ :8
So, the net circulation of force .F a.round the closed path is
fr.n:(l*[*[*l)r."
:-$+o-3*8:-*:-0.1667
L t.2.2s
I
Vector Analysis
r!'-
Correct answer is 21.gg1 . The Iine integral (Circulation) of vector field .4 around the closed path can be divided into four segments as shown in figure below :
t-2lrr:/
The line integral (circulation) of force F around the closed path can be divided in four sections as shown below.
Page 59
Chap
1
Vector Analysis
\
!
U: z:0
For segment 1 we have,
0
dl:-dro,,
Ir
So'
.m:
For segment 2 we have,
fr.
So,
For segment (3) we have
I' re
tu)
tr: z:0 dl,:d,Uau,
0
m: [\-*)@e) : r:zrY:L d,l: d,ra"* dzo,,
:[+].-[,]i i:
For segment 4 we
have,
$'
t ir 6
n ft
so,
A
!
tl
't
F
r F
(
Itl
So,
0
<
r< I,0 < z
l, ::,;;t_
:t-':-3
o"o"
[r. a: l't-n){-ay1+ f'FflFa') :8 : : [4],.[4]. ]*] path J| around the the net circulation of f r-m:({*[*[*l)r." closed
force
:-|+
,.
I b
fr, u.zs
o
[r.m-['ia'+[,'?tryo,
So,
i,
:[-+]. :-+
o
- 3*3
is
:-*:-
0.1667
Correct answer is 21.991 . The line integral (Circulation) of vector field A around the closed path can be divided into four segments as shown in figure below :
For segment
1,
6:0, z:0
Page 60
Chap
1
Vector Analysis
d,I
so,
:
[d. il:l@sin6)d,p:s
For segrnent
P:2, z:O dl : pdbaa
2
so, [n. 41 : f" o'(odd) :8[dH : For segment 3 Q:r, z:o so,
6 : [-'prin|)(-0 For segment 4, p :1, z:0 d,I : pdfi(_ a6)
so,
l
d,pao,
[n.
It
.
at
0
-
a'
:
7T
***********
:2)
-2
o<4
: I" p,(-pdb) :-ld][ :-n
+ 8zr*
(P
(6: ")
f d,.at:([*F[+[)d.at 0
8?r
dp)
Therefore, the net circulation of the vector is
:
o
:
2|.9gl
sote,rrilox$'1 .3
Page 61
ehaF
I
Vector Analysis
L
13.'.r
Option (A) is correct. Given the vector field has the only component in aa direction and its magnitude is r so as r increa"se from origin to the infinity field lines will be larger and directed along ap as shown irr option (A).
I
l-3.2
Option (C) is correct. Options (A), (B), (D) are the properties of vector product' Now we check the relation defined in option (C). Since the triple cross product is not associative in general so, the given relation is incorrect' This inequality can be explained by considering vector A: B and C ,perpendicular to A as shown in the figure. According to right hand rule we determine that (B x C) points out of the page and so A x (B x C) points down that has magnitude ABC ' 'A But in L.H.S. of the relation, since : B So u'e have
and hence Therefore
(,4xIl):g (AxB)XC:0 (AxB)xC:0+Ax(BxC)
A:B Fti
i
a ii.
@xq
!i:
{'
:,.t:
Ax (Bx
C)
*i
hl 1i'
lr' :.1
ir fi
!.3.3 -.
Option (B) is correct. Consider the two vectors A and
B
as shown below.
$i
f.
i' fr,
lr
ti $!
; $ r As the angle between the two vectors is o. So component of vector A along B is ,4,
:
(cos cr)A
Page 62
Qhap
cosine of the a.ngle between the two vectors is defined as
I
cose
Vector Analysis
so, $gL
stlt
,t.3"4
1.3.5
: A,'JB A1
o,:(w)o t .ln :--Br
Option (D) is correct. \ Options (A), (B) and (C) are properties of V operator where as
:
Option (A) is correct. Consider V is a scalar field. So the gradient of the field is
yv
:***** or oy
oz
and the curl of the gradient of the field is
[o' q
o"l
6
El
vx(vn:l#fr#l l-a;
: (ffi ffi)*.(# -
#)".(ffi
ffi)"": o
So the curl of the gradient of any scala,r field is zero everywhere.
sor- r.3-6
Option (A) is correct.
sol
1.3"?
Option (A) is correct.
sol
1.3,8
Option (A) is correct.
$ot- {"3.9
Option (B) is correct.
$sL
1,3"{o
Option (C) is correct.
$0L 1"3.1{
Option (C) is correct.
$oL
Option (C) is correct.
a
1"3.12
***'k*******
1
I l
l
$oLUTIONS 1,4 3 I
Page63
chapr Vector Analysis
sol.
i.A.{
Option (A) is correct. Divergence of ,4 in spherical coordinates is given Y
.A
as
: $Sf* d,l : $fr{*r.') :4h+2\r*l (Given,V.,4:0t
=k(n*2)f-L:0 So, Of, sol-
{.4.2
n*2:0 n:-2
Option (C) is correct. Given, the vector, A : rUax+ r'ou Differential displacement along any path in the r-y plane is defined a-s dI : dra,* dAou (since, dz:Oi So, the line integral of the vector A along the closed square loop is giverr as
f d'
m
:
f"@uo,*
12 ar)
. (d,ra,*
d,yar)
:
$(rud,r+
: ['/f *a,+ ['f z,a*+ I'tor* I'tron : +l+- +].fl+ -*1.*p - 1l +$rr - at : 3, t.4.3
12 d,u)
r
Option (B) is correct.
Given,
V:YxA
...(i)
According to Stoke's theorem the line integral of a vector along a closed looir is equal to the surface integral of the curl of the vector for the loop.
i.e.
[d.
m: II"(, x A).
d,s
.
...(z)
where C is a closed path (contour) and 5c is the surface area of the loop. Flom equation (1) and (2) we get
t
1.4.4
fd.
m: !["v.
as
Option (D) is correct. The transformation of unit vector a4 in Cartesian coordinate system gives the result. where
/
at point
ao : (- sin/)a, * (cos/)an is angle formed with r-axis.
.4,
d :90'
B,
(try:-a, Q :90'*45' :
So, at Point
o,- :-lo,-
So, at point
C,
6
:-
a-3 135'
J2- 4o, J2 45"
b -'
b
So,
Page 64
Chap
I
Vector Analysis
:
or
ho,*h",
Q :0"
at point D, So'
Qd:Qv
0ption (B) is correct. Given, the solution of a Laplaces equation is
7: i.e. the field
I/
sinhrcoskgep"
satisfies Laplace's equation. So, we have
V2y
:0
azv,a2v,a2v r -r=--r
oft
dr
dlr
Now,
-u dz' d2v : uAf
'l
sinhrcoskgee"
a2v
#sinhrcosktte."
6h:d2v : Ti
p2sinhrcoskyep,
Putting all the values in equation (1), we get (sinh rcos kyee")(L - ti + p') : 0
t-#+p2:o
tC:/T+7 t*SYli
I
,1,1.uir,1;r
*&i- {.4-6
,,
tt".,tt ,',:
'
tt,"t,t'
,, . ,
,,r.'i;1 f
:ir: -:i.
i1:r,rj.';.ir1,i.
Option (C) is correct. The angle between two vector fields
.' : cos ttA'Bt \_Af
/
and
B
is defined
as
)
Given, electric field intensity at point P is
E :Ija,*
10a, *IAa, So, the angle formed between the field
e:cos rl E'n" \
\T'.ql/: "os'(ffi]:
Similarly, we get
//I _ - ^r_uun -.n. $0L
1"4,?
E and with r-axis (a,) is -'l 1q:\-"^"-r/-L\ cos
(6)
'/-l-\
\_/3)
Option (C) is correct. faplace equation is defined as v2V :A Now. rre consider the option (C) The scalar field is
v:+
So, the Laplacian of the field
v2v
I/
is given
as
01.dv\-7 Q'v *d'Y -uO -1 - , AV(Ai )-7 a6ld7101 _10 _LA /./_10\\rnt' _ : -T1 A7\ - FAt\'\-7 ,): V ))r
l.e. So,
v2v+o it
Page 65
dciebn't satisfy Laplace's equation.
Option (D) is correct. Laplacian of a scalar function is given v2V
: v
Chap
as
. (vy) :div(gradtr/)
i.e. The Laplacian of a scalar function is divergence of gradient of
sol.
1.4,3
I/.
Option (A) is correct. Given, the vector fielld,
A
: Jiaza,* f
zar+
So, the divergence of vector ,4 is
(f y -
2z) a"
V . A:6ryz-2+0 Therefore, it is neither divergence less and hor solenoidal Now, we determine the curl of vector as
Vx
A:O
since, the curl of the wector is zero so, it is irrotational (i.e., not rotational).
u_
1.4.10
Option (A) is correct. Laplacian of a scalar field
y in cylindrical coordinates v2V_LAI^AV\,IA2V,A2V * -
is given by
naol', )'VAF d7
Since, Laplace equation is defineil as
y2V So, we get
-.
rt l.l
t:
i"l
:0
i&(,#).#(#)*#=o
Option (D) is correct. The given curve is divided in three segments AB, So, the total integral is given as
BC
and
cD respectively.
I o' : (I"* I"* I")*
i t, ,.,
:
r"/2
Jo'-
nadao+
r-R
J"-"
at1-",)+
t'l,aaha,
:ERur_2Rq_Er"r :-zRa,r
il- *r, fi"l
Option (A) is correct. Given vector field,
1
Vector Analysis
A:2rcosdq
VO For the given contour we integrate the field in three intervals
as
v2v
l.e.
So, $1
t1'
it
+0
Page 65
dciebn't satis$r Laplace's equation.
Option (D) is correct. Laplacian of a scalar function is given
Chap
as
y2V: V . (vy) :div(sradl/)
i.e. The Laplacian of a scalar function is divergence of gradient of
L
r.4-9
z.
Option (A) is correct. Given, the vector field,
: Bfyza,* f
A
zau+
(f y _ 2z) a"
So, the divergence of vector ,4 is
V - A:GWz-2*
0
Therefor€, it is neither divergence less and hor solenoidal Now, we determine the curl of vector as
V X .24. :0 since, the curl ofthb vbctor is zero so,
r.
r.4..to
Option (A) is correct. Laplacian of a scalar fi'eld
i;,
i] i:
so, we
'1,
I
E
p.
*
wl* tr,
:0
get i&(,#).i(#)*#:o
Option (D) is correct. The given curve is divided in three segments AB, So, the total integral is given as
BC
and,
cD respectively.
I o' : (I"* I"* I")"
S*
l;i I l
ffi
: [)'' na4or+ !i-'at1- o,)+ t'r,na4o,
j::
!
,i
l
:IRur_2&ao_$n"r :- zfia!
i,
l
L \i:
i I
{I |,.rz
in cylindrical coordinates is given by
a7
y2V
"!?
il, rrrr
is irrotational (i.e., not rotational).
Since, Laplace equation is defined as
:
:
it
_10 1^0v1, tA,y*}ry -Vaol'r)*VdF*
vzv
i
y
Option (A) is conect. Given vector field,
I
Vector Analysis
A :2rcosda,
a
For the given contour we integrate the field in three intervals as
page66 Chap
. f n m: l,t.
I
vectorAnalvsis
:
d,ra,*
Io.
d,ra,
(,'{2r"os$)d'r+of\2rcosd)dr
-C--i|-
: : zltl' [Zlo -' sol i.4"i{
tuha.*
Io.
-(-"
.-
.
Option (B) is correct. B to be solenoidal
For a vector field
V
'B:0
It, -' B)rIu ::o ! n. as $oL
1.4.14
sol t.4.ts
Option (B) is correct. (o" an)+(anx a,)
"
: a"+(-a"):o
Option (A) is correct.
: o It gives the result that -F is irrotational (a -+ 2) div (r) : 6 (b) It gives the result that F is solenoidal' (b - 3) div Grad (d) : 0 (") (")
Curl (.F)
V . (V Q):O
Y'4:0
It
is Laplace
equation.
(d) As sor.
1"4.t6
@
(c div div(@)
:
-
1)
s
v '(v ' 6):0 is a scalar quality so its divergence is not
defined'
(d -.
)
Option (A) is correct. As by observing the given figure we conclude that the closed circular quadrant is in r-y plane and it's segments are , OP - dra, PQ :2d'6ao
QO
:
d,ra,
So, the closed looP integral is
f a. s:
['a,+ liho+ *
rk rk t< *<:k
*
X
* {<:f
[,'a,:z($):n
_==*
z1
=,
GHARTERS ELEGTROSTATIG FIELDS
INTRODUCTION Electrostatic fields are produced by charges at rest. The main objective of this chapter is to provide detailed concepts of electrostatics. They include: o Fundamental concept of electric charges and charge distributions
r o o r o o FC'
Coulomb's law and its applications
Electric field intensity due to various charge distributions Gauss's law and concept of Gaussian surface Potential functions and potential gradient Energy stored in electrostatic fields Electric Dipoles and dipole moment
ELECTRIG GHARGE
ti.
$;
Electric charge is a fundamental conserved property of some subatomic particles, which determines their electromagnetic interaction. The SI unit of electric charge is Coulomb. In electrostatic formulations charges have four types of idealisations, as described below.
F
rr $.
ir
i!
Point Charge Point charges are very small charges assumed to be of infinitesimally small volume, although they have finite volume considered as a single charge. For example, an electron is considered to be a point charge and has a charge of 1.6 x 10-1e Coulombs (C). The idealisation here is that the whole charge is e concentrated at a point. In generar, the point charges are denote{ by q. Q or
Line
Charga
/
This is a charge distribution in which the charge is distributed along a line like a filament, as shown in Figure 2.1(a). The chargg'per unit length along the line cha.rge is called line charge density. It is denoted by p1 and.defined as dQ ^ :_,,*A0 iLryoLT: di
pL
where AQ is small charge, and AZ is small hngth.
t.rt
Surface Charge
when a charge is confined to the zurface of a conductor, it is said to be surface charge distribution, as shown in Figure 2.1(b). The charge per unit area over the surface is called the surface charge density. It is denoted by p, and defined as
Page 6E
o,
Chap 2
: ]tvrils:
oo9
where AQ is small charge, and AS is small area.
Electrostatic Fields
++t+ + +P".*+++'+
.+ * ++++
(")
'
.*r
(c)
(b)
Fieure 2.1: Various Charge Distributions (a) Line Charge Distribution, (b) Surface Charge Distribution, (c) Volume Charge Distribution
2.2.4
Volume Charge when the charge is being distributed within a certain defined region, it is said to be the volume charge distribution, as shown in Figure 2.1(c). The charge per unit volume in the region is called volume charge density. It is denoted by p, and defined as
o,: hmS.:S .0AU dU
ao
where AQ is small charge, and Au is small volume.
2.3
GOULOilIB'S LAW According to Coulomb's law, the electrostatic force F between the two point cha,rges Qr and Q2 separated by a distance fi, is given by
,,- I QrQz- I Q'Qz ' - 4r€ R2 - 4n€o€' R2
where e is the absolute permittivity of medium, e " is the relative permittivity 12 of medium, and eo : 8'854 x 10 F/m is the permittivity in free space' In free space,
p
2,3.1
: #!#:(e
x
ro'g)
(';Y)
Vector Form of Coulomb's Law Consider the two point charges Q1 and Qz wlth separation distance -B as shown in Figure 2.2(a). The force exerted by Qt on Qz is
r,,
:4fuffo^o
where anrz is a unit vector directed from Q1 to Qt. If two charges have the position vectors 11 and 12, respectively, as shown in Figure 2'2(b); then the force acting on charge Qz due to charge Q1 is
p,,:9942-4 4neo]rr- rrlt
Similarlv, force exertedby Q, on
Q1 is
Fz,
:#^ry*21: QrQr(rt- rr) - _ F,, 4neslrl
Page 69
- ,rl'
Chap 2 Electrostatic Fi,elds
Y;,^-)ffi,, AQr&P
4
F",fnI
Origin i,.l
il,)
f igrl"r:2.2: Illustration of Coulomb's Law: (a) Force Representation in Vector Form, (b) Electrostatic Force in terms of Position Vectors
tft
Principle of Superposition
If there is a number of charges Qr, Qr, ..., Q,
placed
at points with position
vectors Tr, T,zt ...,7,,, respectively, then the resultant force F on a charge e located at point r is the vector sum of the forces exerted on Q by each of the charges Qr, Qr, ..., Qn.i.e.,
p _ QQ,(r- r,) - QQr(r- r') '
---T-T---T-
4treulr
- ,rl"
4ne olr
,:&2T# 2.4
- Tz
r,) - QQ,(rtreolr - r" l'
ELEGTRIG FIELD INTENSITY Electric field intensity is defined as the force per unit charge when placed in the field. If a point charge g placed in a field experiences a force .F then the electric field intensity in the region is defined as .E
: lim{ q-o Q
E:Lq
or simply
It
is seen that the electric field intensity is in the same direction as the force and is expressed in Newton per coulomb (N/c) or vort per meter (v/-).
2.4.L Electric Field Intensity due to a Point Charge The electric field intensity due to a point charge Q at a distance
,R
from the
charge is given by
Q E: - - nn*zan
If n point charges Qr,Qr,.......Q, be located at points
r1,T2,..........Tn the\, r is
using the principal of superposition the electric field intensity at point given as
* Qr@-rr) * ""' _r er@-r^) treolr- rrlt ' ATioV- r,"f 1 s, Qr(r- n) +"to?, fr{
B:,Qr(r-r') 4rcolr
- n l, '
-
Page 70
Chap 2 Electrostatic Fields
2.4.2
Electric Field Intensity due to a Line charge Distribution Consider a line charge distribution with charge density pr, as shown in Figure 2.3. The electrii field intensity due to the entire line charge is
, :#dI
*f ",(v/*)
...(2.1)
where r is the distance of point P from the small segment of line charge, r and o" is the unit vector along r directed towards point P.
f ig,u|e 2.ll : Electric Field.Intensity due to a small segmerrt of Line charge
Let us generalize equation (2.1) for some special cases' Electric Field Intensity due to a Finite Straight Line Charge consider the finite straight line charge ,48 shown in Figure 2.4. The net electric field intensity at point P due to the finite straight line charge is given try
E
: #;E(sinor -
sina2)a"+
ffi(ctsoz - cosor)ao"'(2'2)
E
}'igr:rcr 2.4, : Electric Field Intensity due
to a Finite Straight Line Charge
Electric Field Intensity due to an Infinite Straight Line Charge As a special case of the above expression, for an infinite line charge, point A is located at (z: @) and B at (z:-oo). So, (11
: Tl
and Qz: 0
Substituting these values in Eq (2.2), we get
E
: t#;R or
Electric Field Iutensity due to a Charged Circular Ring Consider the charged circular ring shown in Figure 2.5. The circular ring of
radius a carries a uniform charge pr,cf m and-is placed o,n the 4r-plane with axis the same aa the z-axis. Tfe net electric field intensity at point p due to the charged circular ring is given by pyaha" D
" --
r
2otr1"*;Y
li'igure 2,ir: Electric Field Intensity due to a Cha.rged Circular Ring
2-4.3 Electric Field Intensity due to surface charge Distribution Figure 2.6 shows a body with surface charges. Let p, be the surface charge density in Cfrnz. Then, the electric field intensity due to the entire surface
'charge is given by
F"lgrrrr'
':;^l *f".
...(2.3)
!.(i: Electric Field Intensity due to a surface charge Distribution
Electric Field Intensity due to an Infinite Sheet Charge For an infinite sheet of charge, the electric field intensity at any point defined
where
2.5
a"s
,:h^
p
is
...(2.4)
a, is a unit vector normal to the sheet and directed toward point P.
ELEGTRIC FLUX DENSITY As the electric field intensity is dependent on the medium, we define a new vector field D which is independent of the medium. This vector field is called electric flux density and given by
Page
7f
Cte
2
Electrostatic Fblds
D :9oE
Page 72
Electric flux density is also called electric displacement vector. in C/m2
Chap 2 Electroetitic Fields
It
is measured
2.5.1 Electric Flux The total electric flux passing through a surface S is given
t:Io.as
by
r
where dS is the surface area vector directed normal to the surface.
2.6
GAUSS'S LAW In Electrostatics, the Gauss's law states that the electric flux passing through any closed surface is equal to the total charge enclosed by that surface, i.e.
*--6o'ils:Q"o"r*"a ,Js This expression can be generalized in the following two forms: 1. Ihtegral Form: The Gauss's Law can be expressed. in integral form
as
fo.d,s:Ip.d,u where p" is the volume charge density.
2.
Differential Form: In differential form, the Gauss's Law is defined
as
Y'D--p, 2.6.1
Gaussian Surface
The closed surface to which Gauss' law is applied is known as Gaussian surface. The Gaussian surface must satisfy the following two conditions : 1. The D and E freld lines are normal to the chosen surface : This condition removes the dot product from the integral, leaving
Inas:
Js
2.
o.^"ro".,r
Thp D and E freld lines are constant over the surface : This condition means that D is independent of the position on the surface and hence can be removed from the integral, Ieaving
r{,T*Ji3"ffi
Following table provides some examples .r Gauss' law is applicable for electric field computation, with the corresponding Gaussian surface. Table 2.1: Gaussian surfaces for va,rious charge configurations
2.7
ELECTRIC
POTENTIAL
.,
.
The electric potential at a point is aennei'rs ihe *ork'done to bring a unit positive from. infinity to that point. The unit of electric potJntial is Joule per"9u: Coulomb (J/C) or Volt.
2.7.1
Potential Difference
:
The potential difference between two points .4 and B is the work done to bring a unit positive charge from poinf B to point ,4. It is defined as yn, :_ . ar
lh
where I,as represents the potential difference between the two points such
that B is the initiar point and ,4 is the final point. Fo'owing are sorre important points related to potential difference.
2.7.2
Potential Gradient The rate
of change of potential with respect to the distance is called the potential gradient. Electric fierd is equal to negative ofpotentiar gradient, i.e. This relatio" #;;;T["0 for the three coordinate systems 1. Cartesian"u,r coordinate system:
2
cyrindricar
".:;;,Lk
# :% "*
as
""1
3 sphericar ."",J;;[.ff; : i% "* # "'] n 2.7.3
:-[#w*l%
*.*u#*]
Equipotential Surfaces Equipotentiar surface is a surface with equal value of potentiar at every point on the surface- Following are some important properties of equipotentiar
surfaces:
Page Z3
.
Chry2
Electrostatic Fields
Page 74
3.
Chap 2
4;:l:$$,6s
Electrostatic Fields
2.4
:aiCI
No work is t"qoit"a to
r;{.eqi4$
{
,gtti:mn.
:,
niii*Ai*roS,':fur',4:$ehr$a& s,or.ar'6416rdth uniforml :'::'':: *.,w$.$:.fu *hi*p; :.':,: :::: line charge or a cylinder with unifom :
ENERGY STORED IN ELECTROSTATIG FIELD Electrostatic energy is defined as the energy required to establish the given charge distribution in space. The electrostatic energy in different charge distributions are determined below.
2.8.L Energy Stored in a Region with Discrete Charges consider a region with n point charges Q,, Q2,......,Q, located at points h,Pr,......,Pn respectively. If the total electrostatic potential at the points Pr,Pr,......,Pn be respectively V,V,...'.,V,. Then, the energy stored in the charge system is given bY
or 2,9.2
w
: r[Q,V+
w
:|fa,u
QzV*
.....,......'...
+ Q*v*f
Energy Stored in a Region with continuous charge Distribution If, instead of point charges, the region has a continuous charge distribution, the summation becomes integration. The electrostatic energy for line, surface and volume charge distributions are given below :
1.
Line charge distribution
w
2.
1f : iJrpsVdS
Volume charge distribution
w where pr,t pst region.
2.8.3
o,vat
Surface charge distribution
W
3.
:|l
1f : iJ,, p"vdu
pt alte the charge
density and V is the electric potential in the
Electrostatic Enerry in terms of Electric Field Intensity
with continuous charge distribution, assume that the electric field intensity is E, and the electric flux density is D. Then, the
In a certain
region
electrostatic energy stored in the system is given by
w:|l
o.na,
,
Flom the above expression, we define the electrostatic energy density in the
region as
u=#:*(il
Chap 2
"."0,)
:i@ . E):!u,R 2.9
Elecrnlc
DtpoLE
An electric dipole is formed when two point charges of equal magnitude but opposite sign are separated by a small distance. Figure 2.7 shows an electrir: dipole with charges *.q and separated by a small distance d,. .
-g
-q
*q
(u)
(b)
Figr,r ).7 : (a) Electric Dipole, (b) Illustration
2.9.1 Electric Dipole
of Electric Field due to a dipole
Moment
The dipole moment is defined as the product of the small charge g and the distance d between the charges. It is a vector quantity and denoLd by p, i.e.
p:qd
where d is the vector joining the negative charge to the positive charge. The line along the direction of dipole moment is called the axis of the dif,ole.
2.9.2 Electric Potential
due to a Dipole
Consider a point P located at a distance r from the dipole as shown irr Figure 2.7(b). The electrostatic potential at point p due to the dipole is given by
where
p is the dipole moment.
2.9.3 Electric Field Intensity due to a Dipole Electric fierd intensity at point p due to the dipole is obtained as
E:-YV
:
#r1f2cosoa,* sinr,ql *'krF*******x
Page 7b Electrostatic Fields
EXERCI$E 2.1
Page 76
Chap 2 Electrostatic Fields
MCQ 2,{.{
Two point cha,rges of 9 c and 36 c a,re located on z-axis at a separation of 3 m. A third point cha,rge g is placed on the r-axis at a dista,nce d from the 36 c cha^rge which makes the entire system in equilibrium. The value of q and d a,re
(b) -4C and 2m ,(D) -4C and 1m
(A) 4C a.nd Lm (C) 4C.and 2m iltcQ 2.1.2
consider that the point charges -5 nc and *2 nc are located at and (- b,0,8) respectively The net electric field inteniity at point
(- 4'0, (- 7,3, -
2) 1)
will be
(A) - I.004a"- l.284au+I.4a, (B) 1.004a, - 1..284an + L.4a, (C) - 7.004a, - l.284oo + I'4a, (D) + 1.004 a, * l.284au * r.4a, iltco
2,1.3
Which of the following
cha,rge
distribution produces the electric field intensity
E :2rUa,! 4gzar*6rza"Y lm (A) (B) (C) (D) incQ 2"1.4
MCO 2.1.5
?
?
infinite Iine charge of.2nClm along r-axis spherical shell of cha,rge density 3 nC/m3 plane sheet of charge density lnCfm2 at r-y plane field doesn't exist
An infinite line cha.rge of 1pc/m is located on the z-axis. Electric field due to the line charge at point (- 2, - 1,5) will be
(L) 2.4a,*
1.8a,
(C) -7.2a"
-
(B) 7.2a,*t4.4a, (D) -2a,- a,
3.64,
Electric'field intensity at any point (r,y,z) in free space is E . The electric flux density at the point (- 1,0,1) wiII be (B) e6a" (A) 0 (D) Atreoau (C) -eoa, Cornmon Data For Q. 6 and 7
: ia,+2ryan
:
Volume charge density in the free space in spherical coordinate system is given by
^:l!"'^'
0
r)
3m
MGQ
2.t.6
Net electric flux crossing the surface
(A) atr C (C) 2rC itcq
2.t.2
r: 1m is
Electric flux density at r
(il t"'c/^'
-
1
0
Electrostatic Fields
m is (B) a"C/m'
(Q) tra,Cfm2 lllcQ 2.t,8
ZI Ctap 2
Page
-r'c
(B) (D)
(D)
".
*
.
aC/m2
A point charge 8 c is located at the origin. The total electric flux r* A :2 m lying in the first oCtant is (A) lc (B) 4c (C) lclm (D) a c/m
the portibn of plane
crossing
:
ifcQ
2't's A uniform volume charge density nClmt is distributed
inside the region defined by a cylindrical surface of cross eectional radius a. The electric field intensity at a distance r (< a) from the cylindrical axis is proportional to
(A)
"
Q)# Commol Data For Q. 10 to 12
(B)
#
(D)
af
:
Charge density inside a hollow spherical shell of rad.ius origin is defined as
forr <
[0 o':14"1^t lf ",", scQ
2.1.{o
4m centered at
2
rurz for2l-r<4 \ 7'>
The Electric field intensity at any point in the region
(A) - ty /m (B) -4 y/^
(c)
r:
r<
2 will be
0
(D) zv /m i/rcQ
Itrco
2.{.ti
Electric field intensity at
(e)
shq
(c)
#q
r:
B
will
(s)
2't'12 If the region outside the spherical electric field intensity at r: b ? (a)
be
*q (o) *q shell is charge free then what will be the
#q
(B) \6n
dc0
ts
(c) 3[4"" McQ
2.t.rt In a certain
(D) region the electric flux density is
Volume charge density in the region
(A) 0Clm3
(c)
Tg
$'o+
c/m3
will
o:Y*+ffio,c/^,.
be
@)
(D
rylc/mg
#c/n'
Common Data For Q. 14 and 15 : In the entire free space electric potential is given by
Page 78
Chry
2
Electrostatic Fields
v Nqro& ?"'1.14
nx*& 3"1.1S
: rf t +th@ +zf +t*)
- 1) will be (B) 3'6o' lI!' ao- 35'6o'' (L) 7.7a,*22'8ar-Zt.ia"' (D) 2'2a, - t7'4au * 35'6a" (C) 3.64, - 17.4a, * 35.6a"'
Electric field at point P(3,2,
Electric flux density at point P will be (A) 31.44, * 1014, - 314.5a"PC1^" (B) 62.8a" *202an- 629a"tCfrlt2
(C) - 0.095a, - 0.304a, * 0.948a" nC f m2 (D) 7.14, | 22.8a'o - 71.!a" PC I ^' f{es
2"'1,1S
An electric dipole consists of two point
ch.arges
and - Q is magnitudel- Qis lying along r-axis such that * Q is ?I ::^dl? in ati:- d,l2.Electric field due to the dipole at any po\nt (r,0,$) spherical coordinate sYstem is given bY
E
:
'.Qd ul2cos'ga,*
Aireof
L
IS
(^)
##""
e)#",+ffion 2,'!,'l?
r4*& 2"*.1$
Me& 2.{"{$
lr/here
r >>d
e located at point
(0, y,0)
sindap]
The force applied by the dipole on a charge of +
!-,!*€3
of equal and opposite
1
@)m"" Q)##".
(- 1'0'0) Two equal point charges of *|nC each are located at points point charge of ana (t,b,o) iespectirref. What wiil be the position of third (0,1,0X at 0 +{i nC such ihat the net electric field E: (B) (0, - 1,0) (A) (- 1,0,0) (D) (0,3,0) (c) (3,0,0) ncf m' Plane 3r* 4y:O carries a uniform charge distribution with ps:2 The'electric field intensity at point (1'0,3) will be (B) 67'85o" * 90'48orV/m (A) - 67.8o, - 90.48anV/m (D) -3o, - AauY f m (C) 3a, * 4a,Y f m loop Electric field intensity at a distance 3 m above the center of a circular
ofradius4mlyinginthe4t-planeandcarryingauniformlinecharge *2nClm as shown in the figure is
(A) 21.72a,* (C) 10.86c, -t HCQ ?",!.26
HCQ 2,1.21
10.86o"
V/m
27.72a,V
(B) 10.86o, V/m (D) 72a,Y lm
lm
A dipole having
a moment
p
:
T€oa,C-m is located at origin in free space.
E,:
(A) a cone of angle b4.Z' (C) (") and (b) both :.{.aa
Chap 2
consider a point charge Q is located at the origin. Divergence of the electric flux density produced by the charge is (A) 0, at all points (B) +1, at all points (c) +1, at all points except origin (D) 0, at alr points except origin
the electric field produce due to the dipole is given by E then surface on which 0 bttt 8,, Eu + 0 will be
rce
Page 79
:
E" e,
*
Eo
a,
If
* E, a,
(B) a cone of angle 12b.8" (D) none of these
An infinite line charge * l nC/m is lying along entire z-axis. If the electrir potential at the point (1,2r/2,2) due to the line charge is zero then the electric potential at any point (p,S,z) will be (A) rs(r"(j))
f; vou
ro
rel
$r"(|)
('r
s+u'(*)
Common Data For Q. 28 and 24 : Electric field at any point (r,O,d) in free space is given by
I
.:
E:ffiyo,
HCQ 2",N.23
The electric potential will be maximum at
(A) infinity (C) at r:-2
Hcq
2't's4
(B) origin (D) r:* 2
Potential difference between the spherical surfaces
(A) 1/2 volt (C) l/s volt
r:0
and
r:2 will be
(B) l volt (D) 1/a voh
Cornmon Data For Q.25 to 27 A uniforinly charged solid sphere of radius .R has the total charge e. consicler the electric potential at a distance r from the centre of the ,fh"r" is v(r). z
HCe 2",!"?5
Electrostatic Fields
For
r
)
-8, plot of V(r) versus
r will be
v(r)
v(r)
(A)
(B)
./
r Page S0
Chap 2 Electrostatic Fields
(D)
iltcQ 2.{.26
With the increase in (A) increase (B) decrease (C) rgmain constant (D) be zero always
$co
If R:1m
2,1,27
and
r
None of these
potentibl V(r) inside the charged sphere will
Q:\c
then the total stored energy inside the sphere will
be
(A) 4.34 x 10e J (B) 6.75 x 10e J (C) 4.5 x 10e (D) 5.4 x 10e J !!,rcQ 2.1,?8
(0'0'0).a'nd Three point charges Q, -2Q and Q are located at (o'0'0)' (r'0'0) for point any at intensity fiLld (-a,0,6) respectively.lhe electric r>> 0 rs
(")"(Y)
(o)"(Y) G).(#) ilca
2,{.29
rB, and centered A volume charge is distributed throughout a sphere of radius intensity at a at the origin, with uniform density p,c lmt .The electric field distance r from the origin is outside the sphere(r > B) inside the sPhere(r < ,E)
(A)
*(#)*
(B)
*G)*
*(#)" *(#)-
(c)
*(#)" *$)"
*(#)" *(#)"
(D)
ri
MCA 2.'1.30
An infinite line charge of uniform density
pr, is situated along
The total electric freld flux crossing the portion of plane ! the first octant and bounded by the planes r: 0 and r:
@)*
e&
MCO 2,{.3{
(D) zero
the r-axis'
* z:1m
lying in
Im
(B)#
@)+
in the region Volume charge of uniform density 5 nC/m3 is distributed 5 m and radii of surfaces between two infinitely long, parallel cylindrical 2mandwiththeiru*urr"p*utedbydistanceofl-masshowninthefigure'
P4e
tr
CtQ Electrostatic
t
i'lGlk
The electric field intensity in the charge=free region inside the cylindrical surface of radius 2 m is (B) 5,65 x 1011V/m (A) 282.5a,Y lm (D) 1.77 x 1,0-12 V/m (C) 3.54a, mV/m
s"{.t2
A volume charge is distributed throughout a sphere of radius
"R
and centered
at the origin with uniform density p,Clmt.The electric potential at distance
r
a
from the origin is outside the sphere(r >
.R)
PoRl
(A)
Seoi
(B)
EEr
(c)
fuE
3eo
Seor
3P'Rt
(D)
€0fi
Comrnon Data.For Q. 33 and 34: Two infinite uniform sheets of cha,rge, each with density \Cf at y:a 1 and y:- L as shown in figure.
m2 ,
are located
:b c/m P":5 Qlm
sco 2.{.33 Electric field intensity at the origin will be (B) (A) 0
Q) ,
'
! I
t
f,co
?.'1"34
-f,o"vln
@)
fia,v lm
fia,vln
a test charge of 5 pC is placed at point (2,5,4) then the force applied by the sheets on test charge is (B) 2.5 x 10-'n N (A) 2.83 mN (D) 5.65 x 102 N (c) 2.83 N
If
/
Page E2
Chap 2
z"1.ss As we move away- frofrr the sheet charge U 1- 7, the electric field intensity will be
located
at y:- 1 in the
region
(A) linearly increasing (B) linearly decreasing (C) constant (D) zero
Electrostatic Fields
2.r,36
Consider a hollow sphere of radius .R centred at origin carries a uniform surface charge density pr. The electric field intensity at distance r from the center of the sphere is outside the sphere(r > R) inside the sphere(r S R) Pnl&Y (A) 0 eo\
(B)
rlrco 2.1.37
P" ctJ
o.
(c)
fiw
(D)
0
r / "-
Pn(&Y eo\
r/
o
0 o" ry o,
to
An air fiIled parallel plate capacitor is arranged such that the lower side of upper plate ca,rries surface charge density 2cf m2 and upper side of lower plate carries surface charge density -2Clm2 as shown in figure. The electric field intensity between the plates will be
+2 Clm
-2""
2a. B\ 'to
(C\ _LA,
La. D\ t't0
(^)
cu
xee
a.t.3s
In a certain region electric potential distribution is as shown in the figure.
v(m)
The corresponding plot of electric"fleld component E, will be
Page
t3
Ct4
2
Electrostatic Field3
(A)
-j
3
s(*)
e.t.3s Two electrons are moving with
equal velocities in opposite directions. A uniform electric field is applied along the direction of the motion of one of the electrons, so the electron gets accelerated while the electron moving in opposite direction gets decelerated. If the gain in the kinetic energy of accelerating electron is K.Esoh and the loss in Kinetic energy of decelerating electron is K.86," then the correct relation between them is
(A) K.Ec"n: K.Er.o", (B) K.EG'i") K.E1o"" (C) K.Ec"i,1K.E;ou (D) Can't be determined as initial velocities are not given {<>F:f,**+'t X+*,*
EXERGISK P-N
Page.84
Chap 2 Electrostatic Fields
Comrnon Data For Q. Four equal charges of side J 2
I
and 2
+2c
m in free space
:
are being placed at the corners
ofthe square of
in figure.
as shown
+2C
+2c.
1
,tT,
I
+2C+,[2m-*2C
ouES 2.2.1
m
The net force on a test charge
*
1
nC at the centre O of the square will be
N.
QUES 2,2.2
QUEA 2.2.3
ouEs 2,2,4
one of the four charges is being removed) what will be the magnitude of the net force (in Newton) on the test charge * 1 nC placed at the centre ?
If
The three point charges, each *5 nC, are located on the z-axis at z:L in free space. The electric field intensity at point P(0,0,3) will be N/m in o, direction. Charges
*0
and
*2Q
7,0,
1m. What will be the Q charge such that the net electric
are separated by a distance
distance (in meter) of point field intensity at P is zero. ?
P form *
Common Data For Q. 5 and 6
:
A uniform volume charge density of 2pCfm3 is spherical shell extending from r
:
2 cmto r
:
present throughout the
3 cm.
ouES 2.2.5
The total charge present throughout the spherical shell will be
QUES 2,2.6
For what value of
pC.
a (in cm) half of the total charge will be located in the
region2cmlr'-a? QUES 2,2.7
Electrons are moving randomly in a fixed region in free space. During a time interval ? the probability of finding an electron in a subregion of volume
1o 12 m3is 3o%.The volume charge density in the subregion for the time interval will be nC/m3.
Page E5
Chap 2 Electroetatig Fields
ouEs
Total stored charge on the cylindrical surface
x.R"€3
surface charge density p2zp,Cfm2 is
having
u,C.
Consider a triangular surface in the plane
Qt ffis 2.x,$
p:2,0
z:
0 as shown in the figure.
g
2r*y--5
If the triangular surface charge on it will be &r"**5
has charge density
x'2"'?* A circular disk of radius 5m has surface charge density ps:3r, where r ( < 5 m) is the distance of any point on the disk from its centre. The total charge stored on the disk is
a.s.'x
*
Coulomb.
Consider the electric field intensity in some region is found to be E
, in spherical coordinate system. What will in a sphere of radius 2 m, centered at origin &t"tx$
ps:3ry C/m2 then the total
Coulomb.
*"3.33 If electric flux density in a certain region
:
3f
a,Y lm be the total charge stored (nC) ?
is
D: Qf * 4z) a,*2ryarl4ra"Cl^' Thetotalchargeenclosedbythecube 0I rI2,0< U32,-l< z< t is Coulomb. &usr$
s"R.'t3 Two point charges * 1 pC and - 1 pC are being located at points (0,0,1) and (0,0, - 1) respectively. The net electric potential at point P(- 3,0, - 4) due to the two charges will be Volt. Common Data For Q. 14 and 15 : In the region of free space that includes the cubical volume 0 I , electric flux density is given by
r, y, z 1 L
D:faa,ly2ta"Cfm' &{Jss ;}"tr"'{}$ What is the
*r.lx$
a.R"t$ At
total flux (in Coulomb) Ieaving the closed surface of the cube
center of the cube, div D
:
cl-t
?
Comrnon Data For Q. f6 and 17
Page 86
Chap 2
:
In free space, flux charge density is given by
Electrostatic Fields
D:l\f onCfm2 r(
o'5m
lzlfunCfm2 r20.5m
quEs
2.2.16 Volume charge density at r:0.2m will
QUE$
2.2.1? Volume charge density at
be
cl^t cl^t
r:1m will be
p:ta,- au*2a"nC m is located at potential due to the dipole at point A(7,2,2) 6). The electric Volt.
2.2.ta An electric dipole having moment point B(0,1,
will ouE$
be
-
z.*.te A total charge 20nClis
being split into four equal charges spaced
at 90'
intervals around a circular loop of radius 5m.What will be the electric potential (in Volt) at the center of the loop ? eus$ 2.2.20 The work done in carrying a 2C charge from point A(1,I12,3) to the point 8(4,1,0) in the field .E:2ya,l2ranV/m along the curve y: /rl2 will
be
qUf;S
Joule.
2"2.21 In a certain region, the electric field intensity is given as -E : ra, - yarY f m . The amount of work done in moving a *2C charge along a circular arc in the region will be centred at origin from r: 1m to r: g:
#m
.Ioule. $ it I
Common Data For
q.22 and 23 :
Four equal charges of + 1nC is being ca,rried from infinity and placed at different corners of a square. Consider the side of the square is 1 m and the charges are being carried a,s one at a time.
lrs t i
2.2.22 How much energy (in nJ) does it require to bring in the last charge from infinity and place it in the fourth corner ?
i
QuEs
2.2.13 Total work done for assembling the whole configuration of four charges will
be OUES
n.I.
2.2.24 The electric field in a certain region is given by E : sindap+ (zI l) pcos$aa* psind,a"Y f m Work done in moving a 2 C charge from A(2,0' ,7) to 8(2,30',1) in the field
is
Joule.
eucs 2,2.2s Total work done in transferring two point charges tlpC and *2mC from infinity to the points A(- 3,6,0) and B(2,- 4,- 1) respectively is .Ioule.
eurs 2.2.26 Four point charges of 8 nC are placed at the eor"ngrs of a square of side L cm mJ. . The total potential energy stored in the system of charges is
Page 87
Cbap 2 Electrostatic Fields
Common Data For q,.27 and 28
:
The potential field in free space is expressed
as
v:lvryz Q{rEs
2,2"2? The total energy stored within the cube 1 1
x qu&s
10-13
r,
U, z 12 will be
J
2.e.28 The energy density at the centre of the cube will
be
x 10-13 J.
eur$ 2.2.2s The electric field intensity required to counter act the earth's gravitational force on an electron is
x
10-u
v/m.
2.?'3o Three point charges Q, kQ ar..d kQ are arranged as shown in figure. What will be the value of ,k for which the net electric field intensity at the point P(0, +, f) is zero ?
Qemmon Data For Q. 3L and 32 : Consider a total charge of 2 nC is distributed throughout a spherical volume of radius 3m. A small hole is drilled through the center of the spherical volume charge as shown in figure. The size of the hole is negligible compared to the size of the sphere.
eur$ 2.2.3'N If an electron is placed at one end ofthe hole and released from rest at l: 0 , what will be the distance (in meter) of the electron from center of sphere at t:1 psec. ? qu65 2.2.3? The frequency of the oscillation of point charge is
kHz.
7" Page
EE
euEs
Chap 2 Electrostatie Fields
2.2.33 A total charge of 9002r U,Q iq uniformly distributed over a circular disk of radius 6m The applied force on a 150pC charge located on the axis of disk N. and 4 m from it's center as shown in figure is
QUES 2.2,34
A charged sphere of radius 1m ca,rries a uniform charge density of 6 C/m3. A redistribution of the charge results in the density function given by
p,:k(3-f)clms where
r
is dista,nce of the point from center of the sphere. What is the value
of.k? auEs 2.2.35
A 50 pC point charge is located at the origin. The total electric flux passing through the hemispherical surface defined by r:48m, 0 < 0 < rl2 is pC.
euEs
2.2.36 Two identical uniform charges with pr,: 80 nC/m are located in free space at tr:0, U: -F 3 m. The force per unit length acting on the line at positive pN in a, direction. gr a.rising from the charge at negative y is
----
ouEs
2,2.3? Four 4 cm
1.2 nC point charge are located in free space at the corners of a square
on a side. The total potential energy stored is xxxx*******
----
pJ.
EXERCI$E;
2,;.3
P:se8e
pr."t"o.t"tiHra?
ucq
t'3"'!
Assertion (A) : Net electric field flux emanating from an arbitrary surface not enclosing a point charge is zero. Reason (R) : Electric field intensity at any point outside the uniformly charged sphere is always zero. (A) A and R both are true and R is correct explanation of A. (B) A and R both are true but R is not the correct explanation of A. (C) A is true but R is false.
(D) A is false but R is true. xcQ
2.3,4
Assertion (A) : No charge can be present in a uniform electric field. (R) : According to Gauss's law volume charge density in a region having electric field intensity E is given by po : €yE (A) A and R both are true and R is correct explanation of A. (B) A and R both are true but R is not the correct explanation of A. Reason
(C) A is true but R is false. (D) A is false but R is true.
rcq ?'3'3 A potential function V
satisfies Laplace's equation inside a certain region. In this region the potential function will have (A) a maxima only (B) 4 minima only (c) a maxima and a minima both (D) neither a maxima nor a
minima
xcQ
2.3-4
Coulomb's force is proportional to
(A) " (C) _L
'r
(B)
f
(D)
+ r
xcq 2.3"$ The proportionality constant in Coulomb's law has unit of (A) Farads (B) Farads/metre (C) Newton HcQ
2,3"$
The value of proportionality constant in Coulomb's law is (A) e x 10, (B) e x 10-'g
(C) 8.8b4 xcQ
2.3.7
(D) metre/Farad
x
(o)
10-i2
The unit of electric field is (A) Newton
,f"
rca 2'3'& If the direction of Coulomb's force on a unit
charge
electric field is
(A)
-o" o"
ro'g
(B) Coulomb/Newton (D) Coulomb/metre
(C) Newton/Coulomb
(c)
x
'
(B) a, (D) o"
is a,, the direction of
Page 90
M(:& 9.3"9
The unit of electric flux is
(B) Coulomb/metre (D) Weber/m2
(A) Coulomb (C) Weber
Chap 2 Electrostatic Fields
r!rl;d3
2.s.1o
wce
a.3"1{
The electric field on r-axis due to a line charge extending from
@#,p
@)
Q#
@h
oo is
#h
Potential at aII the points on the surface of a conductor is (B) not the same (A) the same
(D) infinity
(C) zero r${:rn 2.3,{?
Gradient of the potential and an equipotential surface (B) have opposite directions (A) have the same direction (D) have no directional relation (C) are orthogonal to each other
iltso 2,3.t3
The unit of electric moment is
(B) C-m (D) C-m'
(A) c/m
(c) ,1.r{:Q
t,3,1€
c/m'?
Point form of Gauss's law is
(A) V x (C) V x vs;c p"*,'t:
D: D:
Po
P,f
es
$t:e
2,3"ts
x D: x D:
P"
Q
(B) Coulomb/m (D) Tesla
Gauss's law is
I D.ds: Q fcl f od,s:Q
(^)
MS& 2.3,'17
(B) V (D) V
The unit of electric flux is
(A) Coulomb (C) Weber
Potential has the unit of (A) Joules/Coulomb
@)
f D.ds: Q
tol
Iod,s:
Q
(B) Joules (D) Joules/m2
(C) Joules/m3 M$& :"3"'lS
Electric flux lines (A) originate at (+)ve charge (B) originate at (-) ve charge (C) are closed loops (D) originate at (+)ve charge and also terminate at (+)ve charge
KfrQ 2,3.{9
The electric field in free space
(A) .-€o I
-co to
2
(B)
#
(D) q ' '60
(C) eoD **t<{
Ppge g1
-
.
Chap 2
Electrostatic Fields
b n t^'
Ih E IE I -In' Ib
If the electric field intensity is given by E: (ro,+yau* za,)volt/m, potential difference between X(2,0,0) and y(1,2,3) is (A) -r-r
(B) (C)
'olt -1volt *5 volt
\u., -ru
rcQ
2.4.2
EQ
2.4.3
vuru
There are three charges, which are given by e,,:lFC, er:2pC and Qt:3pc. The field due to each charge at a point p in free space is (a,+2a,u- &"), (an+3a") and (2a,- ar) newtons/coulomb. The total field at the point P due to all three charges is given by (A) 1.64" I 2.2o,u * 2.5a" newtons/coulomb (B) 0.3o, I 0.2au * 0.2s," newtons/coulomb (C) 3o, *2au* 2o, newtons/coulomb (D) 0.6o, -l 0.2 au * 0.5 a" newtons/coulomb Given that the electric flux density density at point (7,n14,3) is
rca
2"4.4
the
(A)
3
(c)
0.5
D:
zp(cos,il)
a"c/^r.
The charg.
(B) 1 (D) 0.5 o
An electric charge of Q coulombs is ]ocated at the origin. consider electric potentialT and electric field intensity E at any point (r,y,z).Then
(A) .E and V are both scalars (B) .E and V are both vectors (C) E is a scalar and I/ is a vector (D) ,E is a vector and I/ is a scalar rcQ
2.4,5
Assertion (A) : capacitance between two parallel plates of area ,A' each and distance of separation ,d' is e Af d for large Af d, ratio. Reason (R) : Fringing electric field can be neglected for large A/d ratio. (A) Both A and R are individually true and R is the.orr""-t explanation of
A
(B) Both A and R are individually true but R is not the correct explanatiorr
ofA
(C) A is true but R is false (D) A is false but R is true
lca
2"4.6
Assertion (A) : In solving boundary value probrems, the method of images is qsed. Reason (R) : By this technique, conducting surfaces can be removed from the solution domain.
(A)BothAandRareindividuallytrueandRisthecolrectexplanation
Page 92
A
Chap 2
(B)BothAandRareindividuallytruebutRisnotthecorrectexplanat
Electrostatic Fields
ofA (C) A is true but R is false (D) A is false but R is true MSQ 4"4,?
and opposite What will be the equipotential surfaces for a pair of equal charges ? (A) Spheres
(C) Non-concentric cYlinders {sfr$ 4.4"s
(B) Concentric cYlinders (D) None of the above
If the potential functions % and % satisfy Laplace's
equation within
closedregiorrandassumethesamevaluesonitssurface'thenwhichoft following is correct ? (A) K and Vz are identical
M&& 2"4"9
(B) y, is inversely proportional as V2 (C) Yt has the same direction as V (D)ylhasthesamemailnitudeas%buthasdifferentdirection electric field Assertion (A) : The expression E - - V I/' where E is the V is the potential is not valid for time varying fields' Reasou (R) : The curl of a gradient is identically zero' (A)BothAandRareindividuallytrueandRisthecorrectexplanation A.
(B)BothAandRareindividuallytruebutRisnotthecorrectexplanati of A.
(C) A is true but R is false (D) A is false but R is true xAe€ 2"4.ts
fficQ
caused What is the electric flux density (in pC/m'?) at a point (6' 4' - 5) 8? plane r: a at pC/m2 a uniform surface charge density of 60 (B) -604, (A) -304, (D) 60o, (C) 30o,
at concentric long conducting cylinders, the inner one is kept is what constant positive potential * % and the outer one is grounded. electric field in the space between the cylinders? (A) Uniform and directed radially outwards
2.4.,t,t of two
(B) Uniform and directed radially inwards (C) Non-uniform and directed radially outwards (D) Non-uniform and directed parallel to the axis of the cylinders lwcq
2.4.{c In a charge
free space,
the Poisson's equation results in which one of
following?
(A) Continuity equation (C) Laplace equation MCQ 2.4.13
Wr is the electrostatic energy stored
(B) Maxwell's equation (D) None of the above in a system of three equal point
arrangedinalinewith0.5mseparationbetweenthem.Ifl4lzisthe stored with 1m separation between them, then which one of the is correct
?
(A) W:0.5W (C) W:2W ICq
2.4"{4
2.4.,'5
Page g3
Chap 2
Equivalent surface about a point charge are in which one of the following forms ?
(A) (C)
rca
(B) W: W (D) W: 4W2
Spheres Cylinders
(B) planes (D) Cubes
consider the following statements regarding an erectrostatic field 1. It is irrotational
:
2. It is solensoidal 3. It is static only form a macroscopic view point. 4. work done in moving a charge in the field form one point to another is independent of the path of movement. Which of the statements given above are correct
(A) 1, 2 and 3 (C) Only 2 and
Ga
2.4.16
4
The potential (scalar) distribution is given as v: l}yn+20f .rf e6 is the permittivity of free space, what is the volume charge density p, at the point
(2,0) ? (A) -200e6 (C) 200e,
mo
2.4,1?
?
(B) 1, 2 and 4 (D) 1, 3 and 4
(B) _200/es (D) _240e11
The r-directed electric field E, having sinusoidal time variation eto, and in z-direction satisfies the equation y2E":0 under source free condition in a lossless medium. what is the sorution representing propagation in positive z-direction ? (A) 4: Eoe-k" (B) E,: Eoe+ik" (C) E": Eoe-ik' (D) E,: Eoe+h" space variation
-Q
2.4.18
An infinitely long uniform charge of density 30 nc lmis rocated at y - z, z : b ' The field intensity at (0, 6, 1) is rg:64.Tay-g6.Ba"v/m. what is the field intensity at (5,6,1) ?
(A) rc1
GQ
2,4.1S
E
(*ffi1'n
r"r
(#*#r)"
@(t##r.l'E
what is the magnetic dipole moment in Am2 for a square current roop having the vertices at the point ,4(10,0,0), B(0,10,0); C(_ 10,0,0) and D(:0,- 10,0) and with current 0.01A flowing in the sense AbCDA ? (A) (sf _za" (C) 4a, (D) 4(a,* a")
2a"
2'4"2a An electric
charge Q is placed in a dielectric medium. Which of the following quantities are independent of the dielectric constant e of the medium ? (A) Electric potential I/ and Electric field intensity E (B) Displacement density D and. Displacement ry' (C) Electric field intensity .E and Displacement density D
(D) Electric potential V and Displacement
ty'
Electrostatic Fields
r Page 94
rdcq
?.4.2"1
Chap 2
Two coaxial cylindrical sheets of charge are present in free space Ps : 5 C I m2 ' at' r:2 m and ps :- 2Clm2 at r:4 m. The displacement flux density D
at r:3m is (A) D: 5a,Cf m2 (C) D : l}l3a,Clm2
Electrostatic Fields
(B) D :2l3a,Clm' (D) D: 18f3a,Cfm2
rr{{:& 2"4"?2
An electric potential field is produced in air by point charge 1pC and 4 pC located at (-11,5) and (1,3, - 1) respectively. The energy stored in the field is (B) 5.14 mJ (A) 2.57 mJ (D) 12.50 mJ (C) 10.28 mJ
,tt*Q 2.4.23
A dipole produces an electric field intensity of 1mv/m at a distance of 2 km . The field intensity at a distance of 4 km will be (B) 0.75 mV/m (A) 1 mv/m (D) 0.25 mV/m (C) 0.50 mV/m
Mre z.d.a6 The energy stored per unit volume in an electric field (with usual notations) is given by (B) tl2eE (A) tl2elf (D) eE2 (C) Ll2eE'z r"lse
?"4..*$
A positive charge of Q coulomb is located at point ,4(0,0,3) and a negative charge of magnitude Q coulombs is located at point B(0,0, - 3). The electric field intensity at point C(4,0,0) is in the (A) negative r-direction (B) negative z-direction (C) positive r-direction (D) positive z-direction
r{*s
3"4"2$
The force between two points charges of 1 nC each with a 1 mm separation rn alr ls
(A) 9 x 10-3N (C) 9 x 10-eN ps*Q 3,4"X?
Gauss law relates the electric field intensity density po al a point as
(A) V x E:€op,
(C)VxE:p,leo rdeQ ?,4.1$
(B) I (D) I
x 10-6N x 1o-12N
E with the volume charge
(B) V . E:€opo (D) V'E:p,/€o
The electric field strength at any point at a distance r from the point charge q located in a homogeneous isotropic medium with dielectric constant e , is given by
(A)
E:ffio,
rc\ E:
,Q€
oa,
4nr"
(B)
E:
fD)
,E
f
nas"os}
: -t-o. 4trtr"
tQ
2.4.29
The vector statement of Gauss's a law,is
(tl f o. d.s:lo"o, tct
Eo
2.4,30
IIo. d,s: [c?n
(B)
In.as:f
Page gb
p,d,,
Chap 2
tol f"o . d.s: I o,o, apart. Now, if a glass srab is insertecl
Two charges are placed at a distance between them, then the force between the charge will (A) reduce to zero
(B) increase (C) decrease (D) not change
-o
2.4,3{
The following point charges are located in air
:
+0.008pC at (0,0)m +0.05pC at (3,0)m -0.009pC at (0,4)m The total electric flrrx over a sphere of bm radius with centre (o,o)
(A) 0.058 pC (n) b.o+o pc (C) 0.02e pC (D) 0.016 pC
rs
a
2.4.32
Electric flux through a surface area is the integral of the (A) normal component of the electric field over the area (B) parallel component of the electric field over the area (C) normal component of the magnetic field over the area (D) parallel component of the magnetic field over the area
-Q
2.4.33
Assertion (A) : The electric field around a positive charge is outward, Reason (R) : Gauss's law states that the differentiar of the-normal component of the outward electric fl,x density over a closed surface yields the positive charge enclosed.
(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A) (B) Both Assertion (A) and Reason (R) are individualry true but Reason (R) is not the correct explanation oi Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true
-a
2.4.34
Point charges of Q:2nC andQ2:JnC are located at a distance apart. with regard to this situation, which one of the following statements is not
correct ? (A) The force on the 3 nC charge is repulsive. (B) A charge of - 5 nc placed midway between e1 force.
and, ez
wirexperience no
(C) The forces Q1 and,ez are same in magnitude. (D) The forces on Ql and e, will depend on the medium in which they are placed.
Electrostatic Fields
Page 96
MSq 2.4"35
Chap 2 Electrostatic Fields
Which one of the'following is the correct statement Equi-potential lines and field lines (A) are parallel
?
(B) are anti-parallel (C) are orthogonal (D) bear no definite relationshiP tvtco 2,4.36
,!!e&
3.4.3r
ilrlQ *'4.38
point charges of -10 nc and 10 nc are located in free space at (- 1,0,0) m ? and (1,0,0j m respectively. What is the energy stored in the field (A) Zero (B) 450 nJ (C) -a50 nJ (D) - e00 nJ the electric A spherical balloon of radius a is charged' The energy density in is inflated field at point P shown in the figure given below is u' If the balloon at P ? density energy is the what to o radius b without altering its charge'
(*l
(A)
(*l
(")
(c)
,(*)
(D),
which one of the following statements is conservative
d.oes
not state that electrostatic field
?
(A) The curl of .E is ideutically zero (B) The potential difference between two points is zero (C) The electrostatic fietd is a gradient of a scalar potential (D) The work done in a closed path inside the field is zero flrcQ 2.4.39
Sphere of radius o
flux density at
rd*Q 2.{.4{l
r:
with a uniform charge density p,Cl^t shall have electric a, equal to
(A) $p,i,Clm2
(B) !p,i,Clm'z
(C) ap,i.,Cfm2
(D)
tP,ucl^'
Equipotential surfaces about a pair of equal and opposite Iinear charges exist in what form ? (A) Concentric sPheres (B) Concentric cYlinders (C) Non-concentric cYlinders (D) Planes
MCQ 9.4,4'l
For electrostatic fields in charge free'atmospherO, which one of the following is correct ?
E:0 andV . E:0 (B)VxE+}andV.E:0 (C)VxE:}andV.E+0 (D)V x E+oandV . E+0 (A) V x
3F on 3Q and 2F or 2e, then what is the force exerted on the point
charge Q
(A) (B)
r -r
?
(c) 5r (D)
2'4'43
-5r'
Which one of the following is the Poission's equation for a linear and isotropic but inhomogeneous medium ?
(.!J v2E:-EF
v . Y(eV):-p
(C) MCQ 2,4.44
z:
Plane
(B)
v . (evV):-p
(D)
V2V:-L
10 m carries surface charge density 20 ncf m2. what is the electric
field at the origin
?
(A) -70a"vlm (B) -182'a, vlm (C) 72tra"vlm (D) -3602ro, v/m MCQ 2,11.{5
Consider the following diagram
+e
i--,._
i'-i'-
---
:
?_-
di -- +
t
":,{
-----------
di l.t-t2/
------::f-,.'
..i'
-Qi'' The electric field E at a point P due to the presence of dipole as shown in the above diagram (considering distance r )) distance d) is proportional to
(A) llr (C) I/r3
MC6t :,,{.46
Cliap 2 Electrostdtid Fidlds
s"4'43 If the electric field established by three point charge e, ze and 3e exerts a force
Page.97
(B) rlr, (D) 7lr^
what is the value of total electric flux coming out of a closed surface (A) Zero (B) Equal to volume charge density (C) Equal to the total charge enclosed by the surface (D) Equal to the surface charge density
?
rT Page 98
MGq
2.4.4? A charge is unifdrmly-dfstributed
throughout the sphere of radius o. Taking
the potential hi infiirity as zeroT the potential at
Chap 2
rot-lJ#*
Electrostatic Fields
r:
b
< o is
@)-l:#d.
Q)-t#a,-l'ffia, @-t#d, mco
2.4.4s A potential field is given by V:3r2ytrue
3rz.
Which of the following is not
?
1A) .nt the point (1,0, - 1), V and the electric field E vanish (B) *'y: 1 is an equipotential plane in the 4t-plane
(C) The equipotential surface V:- 8 passes through the point P(2, - 1,4) (D) A unit vector normal to the equipotential surface V:- 8 at P is (- 0.832* 0.55s* 0.072) rvrce
2.4.4e
The relation between electric intensity ,8, voltage applied d between the plates of a parallel plate condenser is
(A) E: Vld (B) E:Vxd
(c) (D)
E:
vl@)'
E:Vx(A" *(*****t
X<>fi**
I/
and the distance
$ol.uTtol{s 2,{
Page 99
Chap 2 Electrostatic Fields
r
soL
t.1"1
Option (B) is correct. Since the two point charges are positive so the introduced third point charge must be negative as to make the entire system in equilibrium as shown below 3m
q
9C
-cgoc in equilibrium so the force between all the pair of
a3 the system must be charges will be equal
Fta (g)q (g)q
l.e.
G-df
: Fca: Ftc
-
_Gqs_Gq(e)
n---Try-
Solving the equation we get,
sol- 2.{"?
q:- 4C
and
d:2m
Option (A) is correct. Electric field intensity at any point P due to the two point charges e1 and Q2 is defined as
E:k(S+R,+ where,
R
8?^R,\
n, ,, ,;!!1f';;';'*?:;.1" 0,,", p rrom the rwo point
and
charges. So the net electric field due
g :9 x
10n
to the two given point charges is
x (-5) x 10-n[(-7*4)o,*(3-0
2)o,]
x 1Os x 2 x 10-e[(-7+5)a,+(3-0) ou+e 1 *g _-
: 30L
2.1.3
451- 3a,
*
3%
Ig"/'
1.4a,
-
!.284a,
*
-
a"f
*I8l-2a,
i
3)a,]
-l Ja, - 4a,f ,nt,,
L.004a"
Option (D) is correct. For an electric field to exist, the its curl must be zero. so, we check the existence of the given field vector first. Given the electric field intensity
E So,
:
2rya,* Ayzau l6rza"V f m la, aa az
YxE:Zl!lo, ou!:
lw
+ dz
2az 3rz
2f-2ya,
-
Szan
- ra,l*
0
Therefore, as the curl of the given electric neta is not equal
field does not exist.
to zero so, the
page
cbap
!00 ?
Efectrostatic
sol 2.1.4 Option (C) is correct. Electric field intensity in free space at a distance charge with charge density p7, is defined as
Fields
Given
10 6C/m
R:-2o,,-
2.{.6
7.2a,
-
3.6a0
kV/m
Option (B) is-correct. Electric flux density in a certain region for the given electric field intensity is defined as So
sot.
ao
, :0 i*!?,)e"#) --
2.1.$
line
: ,!*?;# ' pr.:lp,Cf m:1 x
so, $oL
R from an infinite
at the point
D: D:
(- 1,0,1)
eoE
: eo(i a,*2ryan)
eo(a,)
Option (A) is correct. According to Gauss law net outward electric flux from any closed surface is equal to the total charge enclosed by the volume
:
i.e.
th
or,
,r,:lo,d,u
:
Q"n"
I:, [:,
[, (#) rr,'", drd, 'Q)
:1x2x2tr:4rC sol.
2"{.?
Option (B) is correct. As we have already determined the total electric flux crossing the surface r:1m So, electric flux density D at r - 1m is evaluated as below:
Totalelectricflux
Sowehave
,1,:f,O.
A,S
("t,\V:4r)
dS :4r J{o.o,:l* D(4trl) :4v D : f.:1c/m2 l*
Thus soL
2.1.8
Option (A) is correct. As the point charge is located at origin. So flux due to it will be emanating from all the eight quadrants symmetrically. So the flux through the portion of plane r* E:2 m lying in first octant is Il8 of the total flux emanating from the charge located at origin. and from Gauss law, total flux: Q"," : 8 C So, flux through the surface
sol.
2.i.s
D : a,C/m2
r*
y
= 2 m is ai:$
Option (A) is correct. We construct a Gaussian surface at p
- r
as shown
:
t
"
in figure.
PAge
iOl
Chap 2 Electrostatic FieldE
So, according to Gauss law the total outward flux through the surface will be equal to the charge enclosed by it. l.e.
D(2nrh)
So,
D
:
p,(trf
h)
: p,r, r
from the cylindrical axis is
:
*: *(;) Eqr
Thus 5S$- g"{""${:!
1
(assume the height of the cylinder is h)
Therefore the electric field intensity at a distance
E
p:
Option (C) is correct. According to Gauss law the surface integral of the electric flux density over a closed surface is equal to the total charge enclosed inside the region defined bv closed surface.
i.e.
$o.as:e",,,
J
or
to ""'' J$n.as:+e"," As we have to evaluate E for r < 2 and since the charge r < 2 so Q",," :0
(sinceE:Z; €o ) density is zero for
(for r
<
2)
Therefore, JIn- as:lxo €o' E:0 s&L 9.t,1"!
Option (A) is correct. Again from Gauss law, we have the surface integral of electric field intensity over the Gaussian surface at r: 3 as 6n -'"' JIn.as:La"-"
{ n . as- :
J
E(4r x (3)') :
r_
[ '"-o,,rt
€oJ
: .€lrf oar- lf
{nt4a,
(#)tt,t"ndrdndft)
*l:[
I'" E(4nxg\:4"14(3-2) LU
n:#o. s$r-
g"*"€:e
Option (D) is correct. As calculated in the previous question, rve have the surface integral of the electric.field intensity over the Gaussian surface r: 5 as
fn
. as :
lf
r.o,:
*f
oo, +
r32
lf
tat*1au
+![
oau
4(r55
:ffiT'Fierds
E(4tr
Electrostatic
x (5)') :
E
sol-
2.t.13
*l:,I"
: +"41
(1Ur)
,:&*
I'"
(#)rr"t"0d'r.0d'Q)
!"^a,
Option (A) is correct. According to Gauss law the volume Charge density in a certain region is equal to the divergence of electric flux density in that region
P,:Y 'D
i.e.
:::tr.::ll:::;*"" - l- Q 1Sco20\+-J-41sin0s-in01 2f r"
)
r'
-0 sol-
2.{.'r4
Option (B) is correct. Electric field at any point is equal to the negative gradient of potential
i.e.
E
. --YV:-(!v+!v+4Y\ \dr" 0a''0r'l
:-lG t *7;ff r+zs -roz **)",+(zwt *a+#.*z)",
|,
r\
So, at the Point
sol 2.1.1s Option (A)
:3,U: 2,2:- 1) E : 3.6a' * l7.4an -
*p.t).*#rol",]
P(r
35.6a,Y f m
is correct.
Electric flux density in terms of field intensity is defined So' at point P(3,2'
sol.
z.t.t6
- 1),
as
D:eoE D : eo(3'6a,a71'4ar-35'6a") :31.4o,,* 101o, - 3l4.5a"pCl^'
Option (A) is correct. Electric force experienced by a point charge q located in the field
E
is
defined as
F:QE So, the force applied at the point charge
,
r'
*1C
located at (0,y,0) is
(q:+ 1C) : e)3%l2cos20a,lsindoa] ' ' treof
:
-44-1rin9o' Atresr
(- o,)]
(d
:
90",
ao:-
&",
r:
a)
- Qd^ o. -- 4trest' sor-
2.1.{?
Option (B) is correct. For d.etermining the position of the third charge, first of all we evaluate the total electric field at the given point c(0,1,0) due to the two point charges located at points 4(1,0,0) and B(-1,0,0) respectively as shown in figure.
fG Cte 2
Pete Electrctatic
-'2I
nC,
A(
'i+"" B(1,0,
1, o, 0)
0)
Electric field due to the charge located at point ,4 is
x to'(+) x
10
e'ffi
and the electric field due to charge at point
B
is
'':: rQ68T: I
e
ao)
ffi(a"*
D,: kaff&: e x 10'x (+) x 1o-e x : #f- a,t oy) so, Ett Ez:fi7{""+ q)+ :
4J-2-L(a,+
(- a,+ au) ('A + 1f
au)
9
2J2
-(ih, As the field is directed in a, direction so for making E:0 the third cha.rge of +^/2 nC must be placed on gr-axis at any point g> 1. Consider the position of the third charge is (0,y,0). So, electric field at point C due to
the third charge is.
*
_
e
x
10'g_x
(.1?)
x
1o-'g
@-1)'
(-o,):-d*o,
and since the total electric field must be zero So, we have Et,* Ez* Er
4n2'/2
as discussed above
3r
)
:0
,ef,?,uo,,:o
\y- r)' Q-7)':4ora:3,-1 1, so the point
will be located at
y:
3
i.e. Point P will have the coordinate (0,3,0) sol- 2.{.{8
Option (B) is correct. Electric field intensity at any point P due to the uniformly charged plane with charge density ps is defined as
: i;o" En" where a" is the unit vector normal to the plane directed toward point Since the unit vector normal to any plane /:0 is defined as
a' So, we have the
p
:+ Jl--lv/l
unit vector normal to the given charged plane
Zr*4y:e
AS
l4au ^ --r3a, 5 - Jg'+ 4' -*3a,r4a,
u|L_!----,:__j-
(f:
3r + 4y)
Fi:b
Since at
Page 10$
Chap 2
point (1,0,3)
E
Therefore,
Electroetatic Fields
/>
the positive value of o'' (2 x 1o n) l3o"*4411 (s:2nClm2) 2(ro-'g136r)\ b t
O, so' we take
:{ftw:
:ff{to,*
4an)
:67.85a"*90.48au $oL 2"{"tg
Y
lm
Option (B) is correct. Horizontal component of the electric field intensity will be cancelled due to the uniform distribution of charge in the circular loop. So the net electric field will have only the component in a, direction and defined a^s below :
n : ]6o,err) nt'7rna" = (e
x
10e)
:9 x , so|-
2.1.20
x
(2
x
10-'g)
x (2r x n)
Cir,y",
"U#3a":70.86a,Y1^
Option (D) is correct. Electric flgx density produced at a distance r from a point charge Q located at origin is defined as
o:ffi*
So, the divergence of the electric
v so
it
.
flux density
is
D:h*1r31:o
is 0 for all the points but at origin
(r:
0) its divergence can't
be
defined.
sol-
2.1.21
Option (C) is correct' p :Atreoa,C-m Given the moment point (r,9,@) produced due to an electric a,ny at intensity The electric field dipole lying along z-axis and having the dipole moment p in a" direction is defined as
E _ __2__
4neoru
B
1Z cos 0 a,
*sin
0oe)
: \1z"osla,+ sindae) r".
Now, given that the z-component of electric field is zero
(P:
Atteoa,C-m)
E"= E' . e,, .:0
l.e.
page
. a,) lsing(ae . a")]:0 \[2cos01a, ru'
Blectrosratic Fiolds
JJ2.o"'e-sin2dl:o r
r"'
'
2cos20
-
sin2d
f[t +:"o.zd] Thus
:54.7" or 0 :125.3' the conical surface of angle 0 :
sot.
?"*.*x
: 0
E,:
:o
o
0
Therefore electric field component E,
:6'
54.7"
or
125.3"
will
have the
Option (B) is correct. Electric field intensity produced at a distance p from an infinite line charge with charge density pr, is defined as
E : ^Ptzlf€n p and since the electric potential at point (I,r12,2) is zero so, the electric potential at point (p,d,z) will be equal to the integral of the electric fieid from point (7,r12,2) to the point to (p,6,r).
i.e.
, :_ I:,.,i,,j,r* :_ I,(#*)0, :l_hr"rali. v :2x
10
3
x e x 10eln(;)
: 18h(;) (pr.:*|nc)
f*&?&; ;t;i,,:itill. $ ,:tt,'; Z iiir,i
l]1.r1
]:tr,:i t,,;:ririr-t,L-ri.
sol n"r,z3 Option (B) is correct. Electric potential at any point for a given electric field
i.e.
v
E
is defined
as
at+c --[n. J
Now given the electric field intensity in spherical coordinate system
E
: -:r-a. (t' + 4)'-',
and since the differential displacement in the spherical system is given as
:
dra, * rdra6 So we have the electric potential d,l
* rsin0d$a,6
v:_[rr?.rr,udr_lc:nl J lt'+4)" f +4.+c dY : o At maxima'dr ,
1
@a+1ax2r:
o
Solving the equation we get,
At
r: o
d'.Y d"f
:-
So the electric potential
sot.
?"{.s4
r:
0 and
r:
oo
u"
will be maximum at origin.
Option (C) is correct. As calcrilated in the previous question, the electric potential at point (r,0, is
LOE
Chap 2
Q)
Y:-Jr+4.+C
Page 106
Chap
2
Electrostatic Fields
So
at
r:
and at
0, electric potential is
u:f,+
c
2 electric potential is
w:[+
c
r:
So potential difference between the two surfaces is
:
,,:(1+c) (**"; :{vort &ut- 2-t.25
Option (B) is correct. The charged sphere will be treated as a point charge for the field at any point outside the sphere. So, the electric field at distance r from the centre of the sphere will be :
I
"E,--Gq7Q
(For
the electric potential at the point will be
So
ve)
:- f'n . at
r)
-R)
:
(Taking oo as a reference Point)
:-hl:Ta,:-/a;l-91| ..Q
47f€o f
VQ)qI
So,
The graph
of.
V(r) will be as :
v(r)
${:}L
2.1"26 Option (B) is correct. z
+-
-'. \ * f\*," -x+ +..+ + r,'\'',+ *.* * ,"8\
'l,.*
f
t r *
\--:--1P ",n +O ' *:-v-
tI
+r
For determining the electric field inside the spherical region at distance r ( < R) from the centre of sphere we construct a Gaussian surface as shown in the figure. So the surface integral of the electric field over the Gaussian surface is given as
:!e",":*[(#)]
E@trf) So, the electric field
at a distance
Page 107
Chap 2
i
from the center is
, :#(g#)"
Electrostatic Fields
:&f,,*
(ror r
< rr)
Therefore the electric potential at the point P will be the line integral of the field intensity from infinity to the point P
i.e.
v(r):1lrh,.
where
d,r*
I'u. *l
-81
-
electric field outside the sphere as calculated irr
Ez
-
electric field inside the sphere
previous question.
,?)
:-lI. #90* I'(h#,)r,1
:&[+-#(ry)l So,
J- 2.1.2r
7(r)
decreases
with increase in r.
Option (D) is correct. The total stored energy inside a region having charge density po andpotential V is defined as
w,:|[
o,vd,,
As calculated in previous question the electric potential at any point inside
the sphere is
v(r):&lh-+,(ry)) : f";l*t, - tl1
:
1m,
e: 1c)
(r?:1m,
g:1c)
(n
Therefore the total energy stored inside the sphere is
,, :+I' (#)t^
"
$re
- r)j@rr dr)
:S" h"t['ttr-ra)dr :-
, =3 [l"'l' 16zr'€0[' 5 Jn
3 ,,4_3x9x10ex4 ^ 5-----lt-5:! * 10e: b.4 x 10:J -16?re.
,!J. z.1.za Option (A) is correct. The point charges can be represented as shown below.
a-2QA ' '
'
(-o,0,0) (0,0,0) (a,0,0)
>tr
So the electric field at point (r,0,0) will be directed along r-axis. Taking only magnitude we have the net electric field intensity at (u,0,0) as 2Q E'_ a A
" - lteilr-f - 4";?- G;GTfi
Fl _l
: #ey
Page 108
Chap 2 Electrostatic Fields
Since
*?*'(#l* I #*#Glr-!+'(*l-
r >> al neglecting higher
: *+* " #a['
powers
r(#l] -
:- 6Qa' _ r<(6Qo'\ 4tr,"t4 -'-\ F/
${}t- x,*,:$
l
* (#) we get
#
*
#al' - !
+'(#ll
Option (B) is correct. According to Gauss law the surface integral of electric field intensity over a Gaussian surface is defined as
In.as-rn -"-env-" r" So for the Gaussian surface outside the sphere at a distance r the centre of the sphere we have
E@nf)
:
"G#
(there is no charge outside the sphere)
Therefore at any point outside the sphere
will
be
"
(> n) the electric field intensity
E:p,(tr?t)o:p,lRr\^ "' - Eo\sv
)"'
and for the Gaussian surface inside the sphere at a distance
the center of the sphere
() ft) from
-4*ol we have
r(<
R) from
E@rrf):d*C Therefore at any point inside the sphere, the electric field intensity will be
':*+Po.:!(ft)o. ${lL 2"{-3{}
Option (C) is correct. The portion of the plane y I z: I m lying in the first octant bounded by the planes r:0 and r:7m has been shown in the figure through which we have to determine the total electric field flux.
l: l
rSril'
-tt, :r.r':.''l''' ':a)'.
:g*
zry1
""''-.
According to Gauss law the total outward flux through a closed surface is equal to the charge enclosed by it. ,p
l.e.
total electric field flux emanating flux from the line charge between 0 and r: 1m is
So the
r:
: f o . d.s: e"n"
r ds bE.
-o' J€oto60 -Q"o"-ot(t) and by symmetry, flux through the defined surface will be one fourth of the total electric field flux emanating from the defined portion . I
QE. dS the electric flux crossing the surface : -T:
i.e.
"
h
f{o?H:, l1. rrttli. lir::i in
sgi-
*",!"3{
l;11 !;'1r1.
r:;
rritirl !.lrlf
1.1u,tll,rrl r:li'rir:ir'1lu-r
fi rl"$ .f '
is :il-' .
r/$ rvhl!'tl:s ill.;i1 i,lix.t.ri(.
i'r.,ir.
Option (A) is correct. Consider a point P inside the cylindrical surface of 2m as shown in figure.
Now we make the use of superposition to evaluate the electric field at point P by considering the given charge distribution as the sum of two uniformly distributed cylindrical charges, one of radius 5 m and the other of radius 2 m, and such that the total charge in the hole is zero. Thus we obtain the net electric field at point P as En"t
: Et*
Ez
where .Er is the electric field intensity at point P due to the uniformly charged cylinder of radius 5 m that has the charge density (5 nc/m3), while .Ez is the electric field intensity at point P due to charged cylinder of radius 2 m that has the charge density (-S nC/m3) As calculated in MCQ.61 the electric field intensity at a distance r from the cylindrical axes having uniform charge density p, is n, P"T z--o - co4 So we have
and
Et:#;n1:L#", E, : #;R2: i#5",
So the net electric field
at point P
is
En"t:"#f"r_
Rr)
By the triangle law of vector
Rt- R, : C: So'
a,
En"t:5j#f*l :282.5a,,Y lm
(separation
:
1m)
Page 109
Chap 2 Electrostatic Fields
Page 110
Chap
$oL 2"*.32
2
Electrostatic Fields
Option (B) is coryrect. in Q'55' As we have calculated the electric field for the same distribution field of the integral line so we evaluate the electric potential by taking the intensity.
i.e.
V
:-[n'at [*{t).' :l*(#)''
torr<
R
forr>
R
is The electric potential at any point outside the sphere (r > R)
:- I *(#), : Q,R' :--geo - _p4t-tl' [-7j- - 3eo"
, :- ["
.
o,
and the electric potential at arry point inside the sphere
v
:-lfrn'
(r <
-E) is
l'n' atf :- I^ *#a'- l'fi(!)at at+
:-*l-+l:-#l+l:,.
:-#I-];lhl+-+l :#(+-+):hP'-t) soL 2,{.33
Option (A) is correct. is Electric field at any point due to infinite surface charge distribution defined as
where
t:h^ P,
+
surface charge densitY
o"+unitvectornormaltothesheetdirectedtowardthe point where field is to be determined' At origin electric field intensity due to sheet at g : a 1 it
,,: htw):_fi",
(w:-
and electric field intensity at origin due to sheet at
E_r _
9:-
f,a is origin at intensitY field So net E : E+t+ E-, :-t;u+fiu :0 $or* 2,'t,34
{;{",):
au)
1 is
(a^:
au)
Option (A) is correct. region As the test charge is placed at point (2,5,4)' So it will be in the y > + I for which electric field is given as E - E+1+ E-r
: {;{u)+ hei
(for both the sheet
a':
au)
_2x(SxtO-'g)- -----Zeo uo : Therefore the net force on the charge
F tol.
z.t.*s
will
bx10-e --- eo %
be
: QE: (5 x 10-6)(!r_1Q
j)*
Page 111
Chap 2 Electrostatic Fields
:
2.83
x
Option (C) is correct. since the electric field intensity due to a sheet charge is defined
1o-3N
as
n:fftu so
it
doesn't depend on the distance from the sheet and given
E
So,
tol 2.1.3e
:
E*r*
as
E_,
:ht-ai+htq) :-hou:-5j#o,
\"
it will be constant as we move away from the sheet.
Option (A) is correct. For any point inside the sphere when we draw a symmetrical spherical surface (Gaussian surface) then the charge enclosed is zero as all the charge is concentrated on the surface of the hollow sphere. So according to Gauss's law
rr,.
,oJ
dS
: J p,du:0
E :0 at any point
Therefore
inside the hollow sphere. now at any point outside the sphere at a distance r from the center when we draw a symmetrical closed surface(Gaussian surface) then the charge enclosed is Q"n"
:
P,(+trR2)
and according to Gauss's law
: esE(4iTR2) : I
,oJ
t'
d,S
Q"n"
p"(4trR2)
,:*(#)*
sot.
2.i.s?
Option (A) is correct. Electric field intensity at any point due to uniform surface charge distribution is defined as
where
n
:
&
--+
an
-r unit vector normal to the sheet directed toward the
,!fia surface charge density
point where field is to be determined. The electric field intensity due to the upper plate will be
nu
: At
o,)
(a": -
a")
(o*:
o")
and the field intensity due to lower plate will be
o,:-*e") So the net field between the plates is
E:Eu*Et
{ege 112 CIpn 2
: Ar-".y+[-2],t",)]
Etreltrostatlc Fields
s{3L k"1.38
: -
,
4 ,).- &, 7-a €o'9,,
Opti,rrr (A) is correct.
Electric lield intensitv at any point is equal to the negative gradient of electric potential at the point
E:-YV So, the y-component of the field is
Eo:-* oa Now, for the interval
-3 3 A 3-2,V :20(t+3)
En:-+:_ dy For the interval
-2 3 Y S-
1,
20Ylm V :20
g,,:-{:g ' dll <*1, V :-20t For the interval -7 3 y So,
E,
So,
For the interval
:-%:2oY
lm
I < y < 2;V :-20
So For the interval 2
Eu:O 3 a < 3, V : 20(r-
3)
dll " --dY:-2oY/m Therefore, the plot field componentr E, with respect to y for the intervals will be same as in option (A). 8,,
So,
r*L e.i.3$
defined
Option (B) is correct. Since the electrons are moving with equal but opposite velocities so assume that their velocities are *uoa, and -u64,. Now let the electric field is applied in o' direction
E:
t.e.
Eottr,
So the force applied on the electrons
will be
F:eE:_(1.6x10-1e).8 du :-(1.6 lt n.. rn-l9tn x 10-1e).8 -ffi therefore, change in the velocity
x 10-1'g)80d, a' x to-'n)p (rlt,\:_(1'6 : =\dt) rn m. of electron moving in * o, direction will change to z l[!P4t o, .u1 : o, :-.---_(t.o
So, the velocity
usax-Q'6
:[,0-
(1.6
Since velocity deceases so loss
K.Eu,,
x ro \e)Eodt 1".
in K.E.
is
: |*r|t - |^r? : (1.6 x 10-1e)E0d t-
Again the velocity of electron moving in
LO'6
x 10:'f ET@IY
- o, direction
will
change to
...(1)
,u2
--,uo(tr,-@z#Wo"
Pa_gc
ErectrostatrcFierds
:-[**@#@l* Since velocity increases, so Gain
'
K.Eco,n
Comparing eq (1) a,nd K.Econ
in K.E. is
: |*d -l*rfi :(1.6 x
)
eq.
lls
Chap 2
lo-re)Eodr*L@]#M
(2) we get
K.Er,o""
***********
t'
...(2)
soLuTloNs 2,2
Page 114
f,hap
2
Electrostatic Fields
soL
2"?.1
Correct answer is 0. from the centle. since all the charges are exactly equal and at same distance and so the so, the forces get cancelled by the diagonally opposite charges net force on the charge located at centre is F',r- 0N
sol
2.2.2
Correct answer is L8' treated as an since one of the four charges has been removed so, it will be due to the force the so additional -2c charge has been put on the corner, additional charge wiII be :
"
-lu(-z) I
x
(tl) x to' (r)-
:l-9 x LOs x 2 x L0-el :18N
is and so the net force experienced by the charge Iocated at center
F,"t--18+0:18N sol-
2.2,3
Correct answer is 19'0625 below, Flom the positions of the three point charges as shown in the figure will point charges the all to due intensity field we conclude that the electric be directed along 4,.
+5 nC
*5
nC
So the net electric field intensity produced
at the point P due to the three
point charges is
t :>#eaa
(where
ft is the distance
:&la+*+#*
of point P from the charge Q)
1a1'Y]'
(an: a")
: 5 x 1o-e x e x 1oe x [#+ t*L]"" :19.0625a" sol-
2"2,4
Correct answer is 0.414 " since, both the point charges are positive, so the point on the Iine joining the two charges as shown in figure'
P must be located
(1-r)
r rV-Tal{
Page 115
Chap 2
P
+1m-
Electrostatic Fields
Given the net electric field intensity at point
P is zero
:0
l.e. E.E Since, the direction of electric field intensity due to the two charges
so
will
be
opposite
l^"#l_l#etl:, 2f :(t_r)2 t+2r-|:o
-, + -/q-LA *ff:-l+J2
r:0.474
and
-
r:-2.414
As discussed aborb the point P must be located between the two charges, so we have the distance of point P from charge * Q as: r:0.4I4m
sol
2.2.s
,.''
Correct answer is 160. Given the volume charge density, So the
p,:2 jtC :2 X 10-o C total charge present throughout the shell is defined as the volume
integral of the charge density inside the region:
q
t.e.
:f :
o,d'u
I:,[,[i,,t' " 1o-6)(lsind
d'rdad'Q)
:f+n{zx1o-)"$1,,, : sot.
2"t"6
1.6
x
10-10
Correct answer is 2.6 . The charge located in the region
2
-
160 pC
cm
( r < o is
_Q_1 ^v-2-2x160:80PC Similarly as calculated in previous question we have
or or
80
x
q
:
80pc
:
10-,2
:f+nx2 x 106.+]or,
rherefore, , : sOL
2.2.?
Correct answer is
I
# I
P,du\
I:Lt:,,(2 [#*iS
x
10-6) (fsin0d,rd0dfr)
+ (o.oz;,]'/3 :2.becm
: 2.6 cm
-48.
Charge density in a certain region is defined as the charge per unit volume. Since the net charge in the subregion: B0% of. the electlonic charge So
the
charge density
: n%!3# _#x(-1.6x10-1e)
ht"
-
Page 116
4.8
x
10-8
:-
48 nC/m3
Chap 2 Electrostatic FieldE
soL
2.2,8
Correct answer is 25.1 Given the surface charge density ps: p2 z So the total charge distributed over the cylindrical surface is .
Q: I
o,as
: I), :8
[-,tr'
x
(d"S:
z) (P d'$ d'z)
pdfrdzj\
at
lili^"1r13.
P:2
:8XItZ":8zr:25.1p,C sol-
2.2.9
Correct answer is 6.5 . Given the surface charge density Ps :\ra Clm2 So,
total stored charge on the triangular surface is
. e : I p.d.s *" : [' [-"*t lzry\drd,y :6.59 J,:tJr=t J 's sol.
2.2.{O
Correct answer is 785.398 . Total stored charge on the disk is evaluated by taking surface integral of the charge density. i.e.
Q
:I
o,d's
: [\z,)(zn,a,)
: *[+], :25otr: 785.3e8 c aoL
2.2.11
Correct answer is 5.3
i
.
E:3fa, According to Gauss's Iaw the total charge stored in a closed surface is equal to the surface integral of its flux density over the closed surface. r.e.
Q,,,:f n-'as:qt'E'ds - ',l Gf d d,s : eoQf)(Anf) :toX 3X4ttx2a : 5.3 x lo-e - 5.3 nC
2.2.12
(f
as:4trfa,)
r:2rn
Correct answer is 16. According to Gauss law the volume Charge density in a certain region is equal to the divergence of electric flux density in that region.
P,:Y'D:2r
l.e. So total charge enclosed by the cube is
g:
I o"d'u : I' I' I-'{rr)(d,rd,vdz).
:4x2x2:16C 2.2.13
/
Correct answer is -578.9 Net electric potential due to two or more point charges is defined as : rz
' -\- L
I
4reoR
point P due to the two point
So, the electric potential at
t/-
Q,
Page 117
Q,
' - Ti;;ilr-
where
cha,rges is
Qhap 2 Electroetatic Fields
RyR2 are the distance of the point P Qt:l7p,C, Qr:- 1pC and -4r*R,
from the two point charges respectively. So, we
/ir :
have
,Rr
2"s,',4
:
u:H[#_f,]:-5z8ev
Thus sttt-
:
Correct answer is -0.1667
5.83
-- 4.24
.
The total flux leaving the closed surface is
: f O . aS
(
to surface) The closed cube has total eight surfaces but as the vector field has no component in a, direction so we have the integrals only through the four separate surfaces as shown in the figure ,b
d,S is normal vector
so, r: I I:,"=r,o_:o'* I'I:'_.-,0::.0,, *
f"I'-if
:- f' I'uaaar+ f f'iara, :-[4],1'u.
:-!, sol. s.t.d$
atY:o'
araz+
rt
['
rert
['tf atV=1,
a*a, IiSht
[dr].r'u
r+$x 1:-+:-0.16G7
Correct answer is 0.75 . Given the electric flux density
: iaa,l y'#anCl^' So, divD:V'D D
.
:
(*",*&",**"")' (iw,+f*a,)
V . D :l2rE+2iyl
: j + f,: o.rs
(center of the cube is located
*
(+,
L,L1l
Page 11E
sol.
2.2.{6
Chap 2
Correct answer is 4. Itom the given data v/e have the electric flux density at
Electrostatic Fields
r:0'2 m as
D:Sfa,nCfmz at any point is equal point, we ha,ve the volume so at that density flux of the to the d.ivergence charge densitY at r:0'2 m as According
to
Gauss law the volume charge density
Po:Y 'D
:$fter'fl):+ :20r :
2,2.17
x 5 x 4d
(r:
nClm}
0.2 m)
Correct answer is 0. Again from the given data we have the electric flux density a.t
D:2lfa,nCfm2 So, the volume charge density at, r:1m
r:1m
as
is
p.:Y.D-h*1r1tr-'
sol. 2,2"t8
Correct answer is 0.6 . Electric potential at a distance R from a dipole having moment p is defined as
,r-P'R
' - 4";F
so we have the potential at point .4 due to the dipole located at point B
t/ -
"-
as:
p'AB
4treolABl3
-(Ia,-
an+2a,)' (arj:u*8a,) x
10-e
:0.6V sol
2"2,19
Correct answer is 36. Since the charge is being split and placed on a circular loop so the distance of all the newly formed point charges from the center of the loop will be equal as shown in the figure.
+Ql4
+g/4
+Q/4
+Qla Therefore, the potential at the center of the loop will be
v
:4(H#) :
,n
x
loe)
'Pq+r!)
(Q
:
20 nC)
sL
2"2"20
Correct answer is - 15.5 . The work done in carrying a charge q from point ,4 to point B in the field
E
is defined as
w
-- t['n - at
Given the electric field intensity in the cartesian system
as
E :2ya,t2ra, and since the differential displacement in cartesian system is given as
dI : dra,* dyau* dza" So, the work done in carrying charge Q:*2C from point ,4(1,112,3) to the point B(4,1.0) is
w - - rl[^,rra, * [=,,?,or] The curve along which the charge is being carried is given
as
:G I we have w --zl[^z(/;E)ar+ l',z1zhdy] :- +l*vny+!ta'l|,) A
Therefore,
__
:saL
2"2.21
Correct answer is
r:2U2
TI -t 3 -'121 ^l7Jt
15.5J
1.
T
The work done in carrying a charge q from one point to other point in the field .E is defined as
w
--ol n.at
and since the differential displaceinent for the defined circular arc is dl : pd$aa as-obtained from the figure So, the work done
is
now we put r: pcosQ, the expression to get
W
W
- - 2J6:o [(;"" -
y:
- - z l"/n -zf
psinS
1J
-
Qddaa)
and o, . a6--sin$, ar. &6-
"inscos
cos@
in
(p:
t)
gdft
:-2xt[!sin(zQ)aft :*
Uau)
Electrostatic Fields
Page 120
saL
2"2.22
Chap 2
*1 nc
Electrosiatic Fields
1m
+1 nc
D
,1 m
{i*
B
*1
lm
nc
Lm
*1
nc
Considet the last cha,rge is'being placed at coiner D so the potential at D due
td the cha,rges placed tr.t the cornErd' A,B,'C is'' I L
':=*r;:ffif;-fiir /2)
\
:24.36volt potential is zero so the work done in at infinity As the charge from infinity tothe fourth corner is
ca,rrying the last
(s: 1nc)
W:eV:10-eX24.36 :24.36nJ soL
2,2.23
Correct answer is a8.72
.
Consider the first charge is being'placed at / so the potential at .4 will be zero as there is 1ro anJ chalge prggen! at. any of the corner and therefore the work done in carrying the first cha,rge is
w:0
now consider the second charge is being placed at will be only due to the charge at corner ,4 l.e.
, r/y, -
B
so the potential
at B
q
--
4T€oA
and therefore the work done in placing the second charge
at B
is
w2: qv2: r(#*)
:-1-*10-tt:9nJ I 47f€s and similarly the potential at the corner C corners A and B l.e.
will be due to the charges at
v,:@,'9=#(+.h)
therefore the work done in placing the third charge at C is
ws: qvs: r[#(t.#)]
:
(9
x
loe)
x ro-"/{=+ r\
\r'2
I
and the work done in placing the last charge at D has already been calculated
in previous question l.e.
Wr:24.36nJ
So the
total work done in a,ssembling the whole ctrnfrguration of four charges
is
9L
2.2.24
Chap 2
*
Electrostatic Fields
={'iY,!
# i,T;6 : 48 z2 nr
Correct a,nswer is 8. The work done in carrying a cha.rge g from point r4 to point B in the field -E is defined as
w
__tl,n. a
Q:2C
Given that
E:
sindop
* (z * l) pcos Sa6 * psinga" B have h: h:2 and zt: z2:1 so the
and since the given points .4 and
differential displacement in the cylindrical coordinate systep from A to B may be given as dI : pddao for0
w-
Therefore the work done is,
-,
* ty o cos6)bd6)
f
:-2 x ".tC (1+ 1) x (2)' x [sin@]lg' :_16r*:_r' sL
2.2-25
Correct answer is 1.604 . Consider the *LpC charge is tra,nsferred first, from infinity to the given point ,4(-3,6,0) so the work done for transferring the charge will be zero as there is no charge initially present. now the potential at point
B due to the charge at
. v:*#
.4 is
"tl
:gx10n-==4--9x103 -
JgiW+T
So the work done in transferring the charge
w:eav
tol.
2.2.26
:
(2
=
1.604 J
Correct answer is 0.312 8a-
*8
nc
8r=
x
1o-3)
.
9s=
1cm
*E
nc
,/2.m
*8
nc
1
cm
Page 121
4z: *8 nc
x
/t%
(q,r
:
1pc)
*2 mC at point Bis
(t#d)
(ga
:
2
mC)
page
122
The total potenJial energy stored in the system is given by
Chap2 ElectrostaticFields.
n
,r, :2lO"V W _l where q, is the charges ? ,n" four corners and % is the total electric potential at the corresponding corners' For the l"t corner :
Charge, 4r :8nC and potential, V: Vt* Vv* Vn
where Vrr,Vs, and %r are the potential at the 1"t corner due to the charges e2, Qsa\d qa resPectivelY
n:*l&*#6*u3t]
So,
(qz: qz: {a:8nc)
8x10-s[ 1 , 1 , 1l : --Ge, l0T1+ 0.u7, OnTl -1-
:7.944 x
104
V
Since all the charges are equal so the potential will be same at aII the corners and therefore the total potential energy stored in the system of the charges is
W
',{rr-
2.2.2?
:;x A(nVr) :2 x (8 x 10-'g) x (1'944 x 10) : 1
0'312 mJ
Correct answer is 9.68 Energy density in a certain region in free space having electric field intensity ,E ls defined as
uu:lesE.
E
and since the electric field is equal to the negative gradient of the potential
sowehave
B:-YV
:-\K""*%",*Na;",f
r a,+4q+]a""lv1* -1Ityr-' ' ' ,f "o ral*"i
So the energy density inside the cube
w,
:leo(E . E)
will
be
:ir,l# #. #l +
Therefore the total energJ stored in the cube is We
:
w:
J
wrdu
|uo
f' I'
[l#
+
;fu + fu]a'aua'
:zl' I'li)aw-#-#l:^** :\x3*#:9.68x10-13J ssl.
s.z.zs
Correct answer is 5.18 . As calculated in the above question energy density at any point inside the cube is
,,:!^1fu+#.#l
So, at the centre of the cube (1.5, 1.5, 1.5) the energy density is
,, : ]"[651fr
3-
2.2.25
(1.5)
Chap
l:s'rsx10-13J
e(n)
:
m"S(- a,) where e is the charge of an electron, rne is the mass of electron, acceleration due to gravity and a, is radial direction of earth. So, taking the magnitude only we have ttre required field intensity, gs (g
E =rneg:
2.2.30
Correct answer is 5.59
tr:t! ])f
g is
:b.s, x ro-,lvlm
.
Consider the electric field intensity produced at point p(0,1,1) due to the A, B and, c respectively as shJ*.r*in*/figure is E, , Es and.Ec' respectively.
charges located at points
kQ
'-l
.P(0,t14,r14)
A-Blm So the net electric field at Enet
point P is
: Ee* Ea*
Ec
and since the electric field intensity at any distance
Q is defined
so
as t : &fr
R from
:'hPrtfi+kartft.rartft] (!u*!*) (-i"*i"") :G^"I(+FilI*-
a point charge
En"'l
'^rffi.roffi.roffi (*" - 1"")
[(*1. (+r]'''
and since
En"t:0
(1)
so we have 3
" 1'uY'
aY-
4k
t(*l
Solving the equation we
sol
2.2.3r
+
get
in
t1
\4 n'''
-0
[(?i.Gll''' &
:
5.bg
Correct answer is 2.83 . As discussed earlier, the electric field at any point inside a charged solicl sphere is
' : *(5)",
r is the distance from center of the sphere and p, is the volume charge density given as where
2
Electrostatic Fields
Correct answer is 5.57 . The electric field to counter act the gravitational force must produce the same force as applied by gravity but in opposite direction.
i.e.
lq.
Page 123
A 2x10-e p,:T;fr:-;GF
Page 124
Chap 2
(Q:znC,R:3m)
:I.77 x 10-11C/m3
Electrostatic Fields
so the force acting on electron when the sphere is
it
is at a distance
F:eE
*"ffi: d,2
from the center of
(e is the charge of an electron)
(rn, is mass of an electron)
"XG) (_ 1.6 x 10-1rX1.72 X
r
r
E: let
10-11)
x 1o-31X8.85 x 1o-")
d2r
E-___(L.r7 x 1011)r
x5
,
#*(1.12x1011)r:o Solving the differential equation we have
r:
a'"o/(./t-r? x mtrt) + Arsi"(/t-rz x tbil t) ""(1)
where Ar and Az are constants'
Now, at
t:0, r:3m
So putting
as the electron is located
it in equation
(1) we get,
Ar
at one end of the hole.
:3
0 as the electron is released from rest' Az : O So putting it in equation (L) we get Thus the position of electron at any time t is
again at
t:0, #:
r=lcos(/ttzxrort) At sot.
2.2.32
t:1Psec
r':2.83m
Correct answer is 54.4 . As calculated in above question the position of the electron at any time
:
'r So,
2rf
2,?.33
is
g"or(r/tr? xloflt)
: JTl7Vffi
"nJ?Vffi :_ f :- -_--_'sol.
f
h o.:44
x
104
Hz :54.4KiH2
Correct answer is 9.44 '
Givei the total charge on the disk is Q :900tr pC : 9002r x 10-0 C a :6m radius of the disk is and since the charge has been distributed uniformly over the surface the small charge element dQ on the disk at a distance r from the center shown in figure is given as
de
--lQ\rq \37*" -
goon
x
1o-6
"(6Y
Qd,rdg)
so as
= 25 x l}-ortlrilg' The force applied by the charge.element dQ on at point P is d,F
:
x
(150
1o-6) dQ
_
the
Page 125
150pC charge located
x 70-6) dQ (l + to;
(150
As the disk has uniformly distributed charge so the horizontal component of the field is get cancelled and the net force will have the only component in a, direction and the net force by projection on z-axis is given as
F
p -.22.34
:zzo"l_ffi1:^:
e.44N
Correct answer is 2.5 . As the charge is redistributed so the total charge will remain same on the sphere.
Total charge before redistribution. Q,
:f
o,d'r:
(o
c/m3)($"(tI)
(p,:6Cl^')
8zr Coulomb and total charge after redistribution
Since So, we
have
an
:
['n1+tr)(Sf
_
ra)d,r
qrnh : +rk[f- 4l' -!]cJ olo:
or soL
2.2.35
2,2.36
I
I
Correct answer is 25. According to Gauss's law the total electric flux through any closed surface is equal to the total charge enclosed by the volume. Now consider the complete spherical surface defined by r: 4g m through which the total flux is equal to the point charge. So the total flux passing through the hemispherical surface will be half of the point charge. l
sol.
k :2.5
,!:8:
e.
50 ttC 2
:25p,C
Correct answer is 19.2 . The electric field intensity produced at a distance p from a line cha,rge of density pr, is defined as
Eot:2ffio0 where o, is unit vector directed toward point p arong p. so, the electric field acting on the line charge at y:3m due to the line charge located at
g:-3m
is
ffi" Therefore, the force
;:11? Ij?t
(Pr': 80 nc'a':
Chap 2 Electrostatic Fields
au'
P-
6
m)
exerted on the line charge rocated at
9:3m
Page 126
Chap 2 Electrostatic Fields
is
F:
: s0t
2,2.3?
11.
J,@rdz)(E) (80 x 10-'g)(240ar)
:
19.24v tr,N
Correct answer is 1.75 . The four charges located at the corners of square 4 cm has been shown in figure below : L.2 nc
L.2 nc
D
l"
1^
l"l
A
!.2 nc
1.2 nc
4cm
The net potential at the charge located at ,4 due to the other three charges
is
r''t "
,^:h(ffi+#*h) :9 x 1oe rv x to=- +6rtotr'2 x 1o-'g/.--!-+-]-+ Lu A x 1.2 :
10.8
x
4
730.92
\a
1 .\
Axto-")
102/z
+ -L\ \-'./z)
Volt
Similarly, the electric potential at all the corners will be
va : vc - vo: ve:730.92Yo1I net potential energy stored in the system is given the Therefore,
w
:llav :trQova* Qsvs* Qcvc+ qDvD) :Ir 4x(r.2 x 10-'g) x (730.e2) : 1.7b u,J ***xx***x<**
as
soLUTloN$ 2,3
Page 127 Chap 2 Electrostatic Fields
L
2-3.t
Option (C) is correct. According to Gauss law the total outward electric flux from a closed surface is equal to the charge enclosed by it
i.e.
,:fo.d,S:e"n"
so when the charge enclosed by the volume is zero then the net outwar
E
: -Q-a4nesrt
where r is distance of the point form center of sphere and a" is it's radial direction. So the electric field intensity at any point outside the charged sphere is not zero.
Therefore, Assertion(A) is true but Reason(R) is false.
;
'Option (A) is correct. According to Gauss's law
Po:eYE So when the field intensity is uniform
and
F 3.
yE :0 p, : 1YE:Q
So no charge can be present 2.3.3
in a uniform electric field.
Option (D) is correct. Laplace's equation for a scalar function V is defined
y2V:0
as
but at the point of maxima v 2 7 must have a negative value while at the point of minima V 2 7 must have a positive value. so the condition of maxima/minima doesn't satisfy the Laplace's equation, therefore the potential function will have neither a maxima nor a minima inside the defined region.
sol.
2.3,4
Option (D) is correct.
sol
?,3.5
Option (D) is correct.
sol
2,3.S
Option (A) is correct.
sol
2.3,7
Option (C) is correct.
sol.
2.3.9
Option (C) is correct.
sol.
2.3.9
Option (A) is correct.
ii
Page.128
Chap
2
Electrostatiq Fields
sol.
2"3.'lO
Opti'on, (A) tb Soft_Qftai{
\#r-
2.3,11
Optrion (A)'is correct.".
r
8SL 2.3.12
Option (C) is correcL ,'
${}L
2"3,'13
Option (B) is correct.
sol. 2.3-{4
Option (A) is correct.
sol.
2.3.15
Option (A) is correet,
sol.
2.3.1S
Option (B)is correat,
sol.
2.3.17
Optioq (A) is correet.
$(rL 2.3.18
Option (A) is correct.
soL 2.3.t9
Option (A) is eorrect.
,
*********+*
soluTloNs 2,4
Page 129 Chap 2 Electrostatic Fields
Option (C) is correct. Given, the electric field intensity, E : ra,* yar_l za" dI : a,drl ardyl a,dz So, the potential difference between point X and
u", :_ ["8 . d,t :
:-lg
f,,
rd,r+
y
is
!)'rar* 1'"a,
,',*41,'"*41,',1
: -i[r' - 1' + 02 - 22+ - 3r] : 02
J.
2.4.2
Option (C) is correct. Given the electric field vector at point and Q3 are respectively.
p
due
5
to the three charges er, ez
Et - a,*2au_ a,
Ez: a,, lBa, Es:2a"- ay So, the net field intensity at
point
p is
':il.';ii;"" toN. 2.{.3
Option (C) is correct. charge density at any point in terms of electric
D
is defined as
Pr:Y'D - zp(cos2d)a"Clm2 Po:Y 'D
So, we get
: :
sol
density
D
Since,
At point
fl'x
S[ro("or'6)o"l pcos2dcfms
, :1t;"orr(*) :t:o'5c/m3
(r,t,t),
:.4,4
Option (D) is correct. Electric field intensity .E is a vector quantity while the erectric potential V is a scalar quantity.
3(}L ?,4"5
Option (A) is correct. For an ideal capacitance the area of plates, ,4, is assumed very high ln comparison to the separation d between the plates. l.e.
#=*
so, the fringing effect at the plates edges can be neglected and thelefore' we get the capacitance between the parallel plates as
Page 130
Cbap 2
c:+
Electrostatic Fields
So
A and R both true and R is correct explanation of A'
sol.
2"4.6
Option (A) is correct. using method of images, the conducting surfaces are being replaced by the image of charge distribution which gives a system of charge distribution. so, in solving boundary value problems we can avoid solving Laplace's or Poission's equation and directly apply the method of images to solve it. Thus both A and R are true and R is correct explanation of A'
sol
2,4,7
Option (D) is correct. For.a pair of line charges equipotential surface exists where the normal distance from both the line charges are same. So, the plane surface between the two line charges will be equipotential. +
+
+ + + + r equipotential surface
This is the similar case to method of images. $oL
2.4.S
$oL 2.{.9
Option (A) is correct. According to uniqueness theorem : If a solution to Laplace's equation (a) be found that satisfies the boundary condition then the solution is unique' Here it is given that the potential functions y1 and v2 satisfy Laplace's equation within a closed region and has the same value at its boundary so both the functions are identical' Option (A) is correct' FYom Maxwell's equation we have
YxE:-# Y
xE:-ftlv r/)
(B: v x r4.)
v x(n+4a\:o .'\- dr I
Since, the curl of a gradient of a scalar field is identically zero. So' we get
E++A:-vv dt i.e. E * -v v in time varying field therefore A and R both are true and R is the correct explanation of A.
sol :d"{fi
Option (A) is correct. The surface charge density at plane r 8 is shown in the figure. =
Page 131
Chap 2 Electrostatic Fields
P(6, 4, -5)
The point P is located at (6,4, r: 8 pointing toward P is (trt AL' :-
- 5). So, the normal vector to the plane
Therefore, the electric flux density produced at point
P
is
D:to, :T(30L 2.4,{t
a,):-Joa,
Option (C) is correct. Consider the coaxial cylinder is located along z-axis. so at anv point between the two surfaces the electric field is given as
E --O,
:-&Vo,
(Since all other derivatives
will be zero)
Given that the inner surface is at potential I/s while the outer one is grounded so the region between the two surfaces will have a gradually decreasing potential and so, E will not be uniform and it is radially directed as calculated above (in o, direction). $L
2.4.12
Option (C) is correct, The Poisson's equation is defined
vrV where space
as
:-*
I/ is electric potential ar'd po is charge density. (p,: 0) we get the Poisson's equation as v2V
:0
So,
in charge free
which is Laplace equation.
Page 132
Chap 2 Electrostatic Fields
sol. 2.4.t3
Option (C) is correct. placed at a separation of 0'5 m Consider the three equal charges of Q C is as shown in figure below : 0.5 m
0.5 m
The net stored charge in the system of n charges is defined
_ w _!s-o, 2r,or
as
v, ,*
gr is one point chaigl and yA is the net electric potential at the point charge due to the other charges' Now,wehavethenetelectricpotentialatanyofthepointchargeQlocated
where
in the system
as
,,:+_^(&*#)
-a
'lf €o
given So, total energy stored in the system of charges is
as
w:r(+av):m
(1)
potential at Now, when the charges are separated by 1m then the electric any of the charge Q due to the other two charges is
h
w: +,,(1 *?): So, the stored energy
in the new svstem
is
w:B(+av):# FYom equation (1) and (2) we have Wz
z.4.ta
: O.5Wr or
W,
(2)
:2W
Option (A) is correct. Electric potential due to point charge is defined
as
IQ v : GqT So, for the equal distance r potential will be same i.e' equipotential
surface
about a Point charge is sPhere'
2.4.15
Option (D) is correct. An electrostatic field has its curl always equals to zeto' so electric field is irrotational. Statement 1 is correct' not solenoidal. is it Electric field divergence is not zero and so Statement 2 is correct' point' Electric field is static only from a macroscopic view Statement 3 is correct' work done in moving a charge in the electric field from one point to other is indePendent of the Path' Statement 4 is correct'
I-
2.4.16
Option (D) is correct. Given electric potential,
V
:
Page 13:l
Cbap 2
Ioua
+2oi
Electrstatic Flfds
FYom Poisson's equation we have
vrv
where,
:_foeo
7 -+ Electric potential P, --+ Charge density
(-:). 3t. #\toYa + 2or) : - !; l20r*120a:-ft P,:-240e0
h i'
f
(r:2, y:0)
p,:eo(I20x2+120x0)
il
I-
2.4.'!?
Option (C) is correct. Given, the wave equation in space for a propagating wave in z-direction is
Y,E,+PE,:O Now, from option (C) we have the electric field component
E'
:
Eoe
as
rk'
The Lapalacian of electric field is
: (_ lkf Ese-tu' Y2E, :- li Eoe lk":- E E' Y2E,
or, sol.
c.s."{,a
Y"E,+
t* 8,,
:
o
So,
it
satisfies the wave equation.
Option (A) is correct. Consider the infinitely long uniform charge density shown in.the figure. The electric field intensity produced at a distance p from an infinite line charge with density p, is defined as
E:#h
X
P (0,6, 1) and I (5,6,1) from the line charge will be same so, the field intensity produced. due to the infinite line at both the points P and Q will be same. Therefore, the field intensity at (5,6,1) is .8. Since, the normal distance vector of points
Page L34
sol 2.4"1s Option (A)
is cotrect Consider the square loop ABCD carrying current 0.1A as shown in figure.
Chap 2 Electmstatic Fields
B(0, 10,0)
c(-10, 0,0)
,4(10, 0, 0)
,(0, -10,
o)
The magnetic dipole moment is
:
IS where .I is current in the loop and III,
So,
?n
^9
is the area enclosed by loop.
: (o.olxrcnf :2
A-r'rr2
The direction of the magnetic dipole moment is determined by right hand rule.
m :2a"
i.e.
sol-
2,4,20
A.m2
Option (B) is correct. Electric flux density at a distance
r
from a point charge Q is defined
as
D:L^o, 4Tr* and the total flux through any defined surface is
: Io'
'$ as So, both the quantities has not the permittivity e in their expression. Therefore, D and Ty' are independent of permittivity e of the medium. sol- 2.4,2{
Option (C) is correct. According to Gauss's law, the total outward flux through a closed surface is equal to the charge enclosed inside it. dS : Q"n, Now, consider the height of cylinder is h. So, the cylindrical surface at p: 3 encloses the charge distribution (p, : 5 Al^") located aL p:2m.
i.e.
f D'
Therefore, we get
n(zn(3)h): OI' $or- 2.4.22
5
x 2r(2)h
o:*oo
Option (B) is correct. The electric potential produced by 1 pC at a distance (r x ro-u) gooo
v
:9
r,
,on
will be the energy of the
ast
W:QV : (4 x
1s-u;gQQq
is
-
rT
So, the potential energy stored in the field i.e.
r
: !9r j 10"
charges
where
r
is the distance between the charges given
as
Page 135
Chap 2
So,
J,
2A,23
1ry
:36 x-I0-3: 5.1b x 10-3 Joule
Option (D) is correct. Electric field intensity due to a dipole having moment from it is
Electrostatic Fields
p
at a distance r
O.\ r" p2
&--liri E" -QI 1 _GI n, : *mV/m |5|-
2.4.24
Option (C) is correct. Energy density (energy stored per unit volume) in an electric field is defined AS
,":|D.
E
:leot . B:leoE' $L
2.4.25
Option (B) is correct. The position of points A,
B
and C are shown below
,4(0, 0, 3)
since, position charge is placed at A and negative charge resultant field intensity at C is as shown below :
at B
so, their
A
Since, the forces fl - .s so the vertical component Fjy and .Sy are get cancelled while 4u and Fjs are get summed to provide the resultant field
in
- a" direction.
Page 136
sot-
2"4,26
Chap ?
Option (A) is correct. Given,
Electrostatic Fields
: Qr: 1nC : 10-e C r:1mm:103m
Charges,
Q,
Separation between charges, So, the force acting between the charges is
o
I
----,
kQtQ, f
-
o
x
toe(to-e)'z
-
:9x103N 30L
2"4.?7
Option (C) is correct. According to Gauss's law, the surface integral of flux density through a closed surface is equal to the charge enclosed inside the closed surface (volume integral of charge density)
i.e. fn-as:lp,d,u In differential form, the Gauss's law can be written
Y xD:p, V x E :* sol
2,4.28
2.4,23
2"4,30
esE)
: -J-a, 4trer"
o€-1 : -!-A4trf
Option (D) is correct. For according to Gauss's law the total outward electric flux through a closed surface is equal to the charge enclosed by the surface'
i.e. or. vrr sol.
(D:
Option (A) is correct. The electric field at a distance r from the point charge q located in a medium with permittivity e is defined as
E so|-
as
$n.as:Q","
J"
Io.as:Ip,,d, Jr'"-"
J"-
Option (C) is correct. The force between the two charges q1 and q2 placed in a medirim with permittivity e located at a distance r apart is defined as
7 qtq' n' '-4tre I
or
F.!
i.e. force is inversely proportional to permittivity of the medium. Since, glass has the permittivity greater than 1 (i.e. permittivity of free space) So, the force between the two charges will decreases as the glass is placed between the two charges. $oL
2.4.31
Option (B) is correct. According to Gauss's law the total electric flux through a closed surface is equal to the charge enclosed by it. Since, the sphere centred at origin and of radius 5 m encloses all the charges therefore, the total electric flux over the sphere is given as
,bu
-.2.4.32
3,
2.4.3*
: : :
Qrl- Qz-t Qt 0.008 + 0.05 - 0t009
Pte l37 CtA t Electrostatk
0.049 pC
Option (A) is correct. Electric flux thro'gh a surface area is the integral of the normal component; of electric field over the area. Option (C) is correct. The electric field due to a positive charge is directed away from outwards.)
it
(i.e.
According to Gauss's law the surface integral of normal component of flux density over a closed surface is equal to the charge enclosed inside it. So, A is true but R is false.
n
2,4"34
Option (B) is correct. Force between the two charges Q1 and e2 is defined as
F : QtQz *n n'' 4reuR2
when the charges are of same polarity then the force between them is repulsive. The electric force on both the charges will have same magnitude. As the expression of Force includes the term e (permittivity of the meclium)
it depends on the medium in which the charges are placed. so the statements (u), (") and (d) are correct while (b) is incorrect. so
2.4"3$ Option (C) is correct. Since the electric field is negative gradient of the electric potential so the field lines will be orthogonal to the equipotential lines (surface).
2.4.16 Option (C) is correct. Electric potential at - 10 nC due to 10 nC charge is t/,_ I Q
GqT
:gxtOnx 10-ll4 :
J2'+o+o
45
Volt
and so the energy stored is
W": QV : (_10 x 10-e) x :-450nJ 2.4.32
45
Option (D) is correct.
to Gauss's the outward electric flux density through any closed surface is equal to the charge enclosed by it. So electric field out side the spherical balloon doesn't change with the change in its radius and so the energy density at point P is wa for the inflated radius b of the balloon. According
2.4"3& Option (B) is correct. The curl of ,E is identically zero. l.e. So,
V it
x,E:0
is conservative.
The electrostatic field is a gradient of a scalar potential.
FiS
t t
l.e. So,
Page 13E
Chap 2 t
Electrostatic Fields
E:-YV V x,.E:0
(Conservative)
Work done in a closed path inside the field is zero
l
i.e.
i
i
In. at:o E:o
'ot
(Conservative)
So, (a), (c) and (d) satisfies that the field is conservative. As the potential difference between two points is not zero inside a field the statement (b) is incorrect. I i-
so,
;r-8.]r; option (A) is correct. Net outward electric flux through the spherical surface,
r:
a is
fo.ds:{:o,(t"d) : P:! no' p:ffa,C1m,
D(4na2)
1 :t,d.4r{3 Option (D) is correct. For a pair of equal and opposite linear chargeis the electric potential defined
is
a,s
IlYY
' -Ti€nrr-41ttrr,
and rz are the distances from the charges respectively. For the same
where
11
value
of 7 (equipotential surface) a plane can be defined exactly at the
centre point between them.
i :,
i:.,r...is Option (A)
is correct. charge free region characteristic
In a
(p,:0)
electrostatic field has the following
Y . E:?:O
and i{
r.
;.4".rs
V
x,E:0
(for static field)
Option (D) is correct. Consider the force experienced by Q is F1. Since, there is no any external applied field (or force) so, sum of all the forces in the system of charges will be zero. l.e.
or,
EF
3F+2F+
:0
Fr:0 R:-5F
i{ i. *.4.i!:} Option (B) is correct. Poissions law is deiived from Gauss's law as
Y'D:p For inhomogeneous medium e is variable and
so,
Y'(eE):p
v . ['(- v v)]: p V ' (uv V) :*p This is the Poissions law for inhomogenous medium.
J.
2-4.4
Option (D) is correct. Electric field intensity due to a infinite charged surface is defined
n
Page 189
as
Chap 2
: !*a^ zeo
Electrostatic Fields
where p, is surface charge density and a" is the unit vector normal to surface directed towards the point of interest. Given that, p, :20nCf m2:20 x 10-e Cf m2
and
an
:-
a,(Since the surface
z: I}m
th'
is above the origin).
So we have,
n:?4jff{?o") _20x10-ex9x
:l-
2.4./t5
4n
J6ltra"y /m
#(2cosoa,* sinda6)
So, for the given dipole,
d:
(for d <<
.R)
(, >>
d)
90'
Therefore,
p://-rt=7 t:##(o+oa)
i.e.
fa.
and
B
x
Option (C) is correct. Electric field intensity due to a short dipole having a very small separatiorr d, a* a distance .B from it is defined as
t :
b:
70e
1
t'
2.4.46
Option (C) is correct. According to Gauss' law the total outward flux from a closed surface is equal to the total charge enclosed by the surface.
2-447
Option (C) is correct. Electric {ield intensity at any.point
E
r
outside the sphere is defined
: --Q-a4nesI4
as
for
r)
and the field intensity inside the sphere is
F
':Cqffi* 0r :
-
So the electric potential
forr
:--!-0'
4trena3*
at any point
r:
b
< a is
a,
v:-le-at :--l' *"' (d,ra,)- fn'
Page 140 Chap 2 Electrostatic Fields
:- I" rdlL
*.6,;ss
^#ftu'-
(dra,)
I'L^neoo'd'
Option (A) is correct. Given, the electric Potential,
V
:Zty-
yz
of the Electric field intensity at any point is equal to the negative gradient Potential. E:-Y V i.e.
.
: -(Ary)a,-@t - z)an-?A)o" at (r: !, A:0' a:E:-4ay*0
So, electric field does not vanish
$*:-
?"4.q*
1)
at given point'
Option (A) is correct. to a consider iwo parallel plates separated by a distance d is connected voltagesourceV.So'thefieldintensitybetweentheplatesisdefinedas E' D-d -v
xx***x*{
_ at
*Qll_alxrH_n3
ELEGTRIC FIELD IN MATTER
INTRODUCTION The laws of electrostatics that we have formulated in the previous chapters, relate to phenomena where no bulk matter is present. The situation when matter is actually present in an electrostatic field is very different. This chapter concerned with the electric fields in matter includes the following
topics:
o o o o o o il'
Electric current density and continuity equation Electric field in conductors and dielectrics Electric boundary conditions for different media interfaces Capacitors: parallel plate, cylindrical, spherical Capacitance and energy stored in capacitors Poisson's and Laplace's equations, and their solutions
ELECTRIC CURRENT DENSITY
Electric current density is defined as the current at a given point through a unit normal area at that point. Current densities are vector quantities an6 it has the unit of Ampere/m2. It is represented by .r. It dan be classified in the following three types: 1. convection current Density: It is produced by a beam of erectrons flowing through an insulating medium. This does not obey ohm,s lau,. For example, current through a vacuum, liquid and so on is convection current. In general, the convection current density through a firament with charge density p, flowing at a rate u is given as J" : PoU 2. conduction current Density: It is produced. due to flow of erectrons in a, conductor. This obeys ohm's law. For example, current in a conductor like copper is conduction current. Conduction current density is defined AS
J:oE
3.
where a is the conductivity of the conductor, and E is the applied electric field. Displacement c'rrent Density: It flows as a result of time_varying electric field in a dielectric material. For example, current through a capacitor when a time-varying voltage is applied is displacement current. It is defined as the rate of change of erectric flux densitv with time, i.e.
t _aD_^AE
"a
-
0T
-'aT
Page 142
3.3
Chap 3
""*'J':::'=::::conductins
resion with vorumd charge densitv p., shown in Figure 3.1. If the current density through the closed surface ,5 the conducting region be J, then the continuity equation is defined as
Electric Field in Matter
ft
. as
:-*l
(Integral form)
o.o,
v'J:-4# Closed
surface,
S
o
Figtrre :.i.1 : Current through a closed Surface
3.4
(Differential form)
,S
ELECTRIC FIELD IN A DIELECTRIG TATERIAL When an electric field E is applied to a dielectric material, the electric flrrx density inside the material is given by
D:eoE*P where
P
...(3.1)
is the polarization of the dielectric material'
3.4.1 Electric Susceptibility Electric susceptibility is a measure of sensitivity of a given dielectric to electric field. Since, polarisation P of. a dielectric material varies directly as thenpplied electric field .E so) we can write Ort
PsE P - x.eoE
where eo is the permittivity of the free
...(3.2)
space and X, is the electric
susceptibility of the material.
3.4.2 DielectricConstant In a dielectric material, we define the electric flux density in terms of electric field intensity as ...(3.3) D -- eoe,E
D--eE ort where e is the permittivity of the dielectric material and e " is the dielectric constant also called the relative permittivity of the dielectric material. 3.4.3
Relation between Dielectric Constant and Electric Susceptibility FYom equations (3.1) and (3.2), we may express the electric flux density
as
D:eo(I+x")E Thus, by comparing the above expression to equation (3.3), we get the relation between dielectric constant and electric susceptibility as
e,
:1*X"
L43.1 Dielectric
.
Breakdown
Page 143
When the applied electric field in a dielectric is sufficiently large, it begins to pull'electrons completely out of molecules and the dielectric becorues condrrcting. When a dielectric becomes conducting, it is said thal d;ielectt1r. breakd,own has occurred.
tA.3.2 Dielectric Strength The dielectric strength of dielectric materials is defined ix the maximr:rrr electric field which a dielectric can tolerate or withstand without electrit.al breakdown. Once breakdown occurs, dielectric starts cond.ucting and 'rc, longer behaves as dielectric. The dielectric strength is measured in V7* o.
kV/cm.
IIB
ELECTRIC BOUNDARY CONDITIONS
If
an electric field exists in a region consisting of two different media, thgrr
the conditions that the field must satisfy at the interface betweerr th< two media are called electric boundary condit'ions. To define the electric boundary conditions, we decompose the electrical vector into two orthogoiracomponents as
E:Et+8, where -81 and En are the tangential and normal components of E to tjrt, media interface, respectively. Similarly, the electric flux density rnav irt decomposed as
where
Dt and D,
D:DtlD,
are the tangential and normal components of D to the media interface, respectively. Following are the electric boundary conditir,ns defined for different media interfaces:
L5.1
Dielectric-Dielectric Boundary Conditions Consider the two different dielectric media 1 and 2, characterisecl by thc permittivities e1 and s2, respectively, shown in Figure 3.2. Accorcli.g to boundary condition, the tangential component of the electric field is continuous at the boundary, i.e.
Et,: E* Qu:Pn tr
€z
It:';'t,i:i:::,t:]t:].-.:.''i.:l
l'igulc i].2 : Dieiectric-Dielectric Boundary Conditions
If no free charge is present at boundary
interface, then the boundan condition for the normal component is given by
.
Chap 3
Electric Field in Matter
D'n:
Chep $
Dzn
€zEzn
I
: Ps Boundary Conditions
I
or the
Page 144
: €t=rn
slrface charge density at boundary interface is ps, then the Wherr bouudary corrdition becomes (Dr,- Dr,)
nearic Field in Matter
3.5.2
Conductor-Dielectric
I I I
o - @,or the resistivity I is zero, (i.e. E:0). I ) a conductor inside is zero, p 0 and so' the field permittivity s, and medium 2 | Assume that medium 1 is a dielectric with is a conductor. Let E1 and E, respectively be the tangential and normal I conrponents of the electric field intensity in the dielectric at boundarY I
For a perfect conductor, the conductivity is infinite,
,::"._"
so, we denne
r?i:";_,y
condition for tangentiar comnonents as
Also, the boundary condition for normal component is defined
",: ?:
Conductor-Free Space Boundary
I
?
| i
Conditions
These boundary conditions will be identical as those for a conductordielectric boundary except that e will be replaced by e6 ' so that the boundary conditions for the tangential and normal cornponents become
Dt:O: Et, Dn: ps , n -Ps
and
3.6
'"
€o
CAPACITOR essentially consists of two conducting surfaces separated by a layer of insulating medium called Dielectric. The purpose of capacitor is to store the energy in Electrostatic Field. Figure 3.3 shows the capacitor with
A capacitor
three different shaPes. Metal Conducting
conducting spheres
Dielecftic Material (u)
I I
Drr:Ps
or, 3.5.3
as
(c,l
Fignle il.ll: (a) Parallel Plate'Capacitor' (b) Cylindrical Capacitor' (c) Spherical Capacitor
i l i
,,-
t t
t 6.1
Capacitance The property of capacitor to store the electricity is called its capacitance. suppose, we give e coulombs of charge to one of the two plates oi"up*ito, and if a potential difference of 7 volts is established between trr" t*i, tnu' its capacitance is defined as
t F
t a
c
:9:
p.Gnt-iqm,""*
The unit of capacitance is coulomb/volt which is also called Farad Capacitance of a parallel plate Capacitor Consider the two metallic plates of equal area ,g separated by a distance d. e be the permittivity of the dielectric medium between the parallel plates, then the capacitance of the parallel plate capacitor is given by
If
c:+
Capacitance of a Cylindrical Capacitor
consider a cylindrical capacitor of length z, inner radius o, and outer radius b' rt e be the permittivity of the dielectric medium between the cylinders, then the capacitance of the cylindrical capacitor is given by
L " _- rn@d 2re
t/1
Capacitance of a Spherical Capacitor
consider
a spherical capacitor
If e be the permittivity
with inner and outer radii, o and b respectively.
of the dielectric medium between the spheres, then the capacitance of the spherical capacitor is given by
C -
3.6.2
: -""\brru( . ob \ al
Energy Stored in a Capacitor The energy stored in a capacitor is equar to the work done to charge general, the stored energ"y in a capacitor is defined a,s Itr{to."d
: i"n
where is the energy in Joule, ''i**ain Volt. the voltage
3.7
it. In
c
is the capacitance in Fara.d, and
v
is
POISSON'S AND IAPLACE,S EQUAT consider a certain region with the volume charge density e. In this region, poisson,s equation is defineJas
p,
and.permittivity
,t, :_+
where
z
is the electric potential in the region. In the special case of (i.". p,:0), poisson'" reduces to "qultio., v2V :0
charge-free region
a
which is known as Laplace,s equation.
3.7.1
Uniqueness Theorem
According to uniqueness theorem, if a solution to Laprace's equation or Poisson's equation can be found that satisfies the boundary conditions, then the solution is unique.
Fage 145 Chap 3 Electric Eield in Matten
Page 146
Chap 3 Electric Field in Matter
*x*t
EXERCT$E 3.{
Page 147
art-^e Electric Field in Matter
-Common Data For Q. I and 2: In a certain region current density is given by
-
J :9a"-'
P
uc&
3"1'{
2o-sind
(n'+ t)
Total current crossing the plane
(A) oA
z:2 inthe a" direction
3'1.2
for p ( 4 will be
(B) 1.5mA (D) 20 A
(c) -32A ucq
a. A/m2
volume charge density in the region at a particular point (h,h,a) will be (A) non uniform (B) linearly increasing with time
(C) linearly decreasing with
time
(D) constant with respect to time
Comtttell Data For Q. 3 to 6 : In a cylindrical system, two perfectly conducting surfaces of length 2 m are located at p:3 and p:5cm. The total current passing radially outward through the medium between the cylinders is 6A. R'rcQ
Mcq
3'{"3 If a conducting material having conductivity o : 0.05 S/m is present for 3 ( r < Scm then the electric field intensity at p:4cm will be (A) 238.7a,Y1m (B) 150ooV/m (C) 318.3a0 V/m (D) 0 V/m 3"{.4
The voltage between the cylindrical surfaces will be (A) 4.88volt (B) 1.45volt
(C) 2.32 volt mcq
3"i"s
(c)
3"'''s
vott
o 0.50 0.813
@) 2.44a (D) s.13O
The total dissipated power in the conducting material will be
w (c) 2e.3w (A)
aI$Q
3
The resistance between the cylindrical surfaces will be
(A)
McQ
(D)
(B) 18 w (D) o.8W
175.7
3'{"? A solid wire of radius r and conductivity o1 has a jacket of material having conductivity o2. If the inner and outer radius of the jacket are r and .R respectively then the ratio of the current densities in the two materials will
(A) depend on r only (c) depend on both r and
-R
(B) depend on ,B only (D) independent of both
r
and
r?
rage Chap
ile 3
Electric Field in
Common Data For Q. 8 to 10 : The potential field in a slab of a dielectric material that has the relative permittivity €, : g lb is given by v : - 500y
Matter
.
Mco
3"1.8
Electric field intensity in the material will be (B) 500o, V/m (A) 50o, V/m (D) 0 (C) -500aoV/m
.lucQ
s"t.e
The electric flux density inside the material rvill be
tttcq
3.t.to
(A) 4.43 nCfm2 (C) s.85 tCfm2
(B) 3'544u wCl^' (D) 7.08o, nCfm2
Polarization of the material will be (A) 2.66o, nCfm2 (C) 5.31 x 10-12 aoCf rn2
(B) 7.08 nCf m2 (D) SarClm2
Cort mon Data For Q. L1 and 12:
Two perfect dielectrics with dielectric constant €n:2 and enz:S are defined in the region 7 (y> 0) and region z (a< 0) respectively. consider the electric field intensity in the 1"t region is given by
Et : 50a,* 2\au *tcQ
3.t.tl
The Flux charge density in the
70a"
2od region
kV/m will be
(A) 2.21a,* 0.35o, - 0.44a,PCl*' (B) 2.2L a,* 0.35o0 - 0'44a, nC f m2 (C) 2.27a,* 0.88o, -0.44a"nCf m2 (D) 0.44, 1,0.O7ar- 0.084, nCf m2 ilcQ 3,1.'12 The energy density in the 2"d region will be (B) 118 mJfm3 (A) 66.37 mJfrns (D) 59 mJ/m3 (c) 472 x 106 J/m3 Mce
3.1.13 An infinite plane dielectric slab of thickness d and having permittivity €.:4€o occupies the region 0 < z< d.If a uniform electric field ,E - Eya" is applied in the free space then the electric flux density(Di") and electric field intensity(-8,) inside the dielectric slab will be respectively (A.)
*""
esEna" and !a"
and
(C) 4Esa,
(B) esB6a" and !a' (D) esEoa" and 4Eoa,
ncq 3,t"{4 The energy stored in
an electric field made up of two fields ,Er and Ez is Wn"t where as the energies stored in individual fields Et and E2 are 1[/t and Wz respectively so the correct relation between the energies is
(A)
W: W+ W,
(c)w>w+w
wtce
3.,t.ri
(B)
W: JW'w,
(D)w
When a neutral dielectric is being polarized in an electric field then the total bound charge of the dielectric will be
zero (c) negative
(A)
(B) Positive (D) depends on nature of dielectric
rcQ
3"{"rs A cvlindrical
wire of length / and cross sectional radius r is formed of a material with conductivitv 106(fr-)-t. If the total conductance of the wire is 106(f)) -1 then the correct relation between / and r is
(l) ,: ,/ I
(C)
2m:
I
rca 3'{"t? Medium between the two conducting
@)
,: tE
(D)
r:
I
parallel sheets of a capacitor has the
permittivity s and conductivity a. The time constant of the capacitor will be
(A) q
rB)
(C)
(D) lloe
"o
oe
q
Common Data For Q. 18 to 20 : In spherical coordinate system, the current density in a certain region is given by
J :L"-rc"'
rco 3.'!"'ts At t:1ms,
a, Af m2
how much current is crossing the surface
75.03A (c) 0.37 A (A)
rcQ
3'{"*s At a particular
,:6?
(B) 27.7 A (D) 2.77 A
time l, the charge density p,(r,t) at any point in the region to. (Assume pa + e as t , @)
is directly proportional
(A)
r
(c) 1,
'r'
xcq $.'r.2{:} The velocity of charge density at (A) 6o, m/s (B) 1000o, m/s (C) 0.6 x 10-3a"m/s (D) 600o, m/s Common Data For
Q.2l
r:0.6
and,22
(B)
l
@)
f
m will be
:
Two uniform infinite line charges of 5pc/m each are located at r:0, g: 1 and t: d, y: 2 respectively. Consider the surface g: 0 is a perfect conductor that has the zero potential.
Ga 3.1.2"{ Electric potential at point p(_ I,-2,0) will be (A) (C)
rca 3.1.s?
1.2volt +0.2volt
Electric field at the point
(B) _0.2volt (D) _0.04volt
p will be
(A) 0.12a, - 0.003o, V/m (B) 0.12o, - 0.086o, V/m (C) 723a, - I8.9arV f m (D) 0.02aa, - 0.086o, V/m
page 149
cbapB EbctricFieldinMatt€tr
chap
3
Erectric Fierd in
ta For Q' 23 and 24 A sphere carries a polarization
Matter uce s.,r.23
P(r): 2ra, where r is the distance
| | ;:,:t:::"ctric
:"::::
sphere. The plot of
'^'
:
E
fierd component in the radial direction with respect to r will be
",,
ruce
l
3.1.24 If the radius of the sphere is a then the electric field (A)
- trar
i",ia" tn"
I t
I I
'"'
cL,", 7\'r' lx.
from the
",,
",,
outside the snhere
(B) 8zio3 (D) -8zro3
I
1
I
willl
|
Data For e.25 to 27 z | A short cylinder of ra.dius r and length .L carries a uniform nolarization P{ parallel to its axis as shown in the figure'
::""
ffiL P (Uniform)
I
I
GE1
3.1.ru
Total bound charge by the cylinder will be
(A) 2P coulomb (C) 0 coulomb
Ge 3.t"4{t If L:2r
Page 151
(B), P coulomb
(D)
-P
Chap 3
coulomb
then the electric field lines of the cylinder will be
Electric Field in Matter
as
, ,, ,.F rl
iil '.'ril
(c) , l { lr
\'\ ''.: l-
..
McQ
3.{,2?
The lines of flux charge density will be
(A)
as
J 1
i
Gl l-.*-*l
| .;', -'{ (D)
r3
An infinite plane conducting slab carries uniformly distributed
surface
it's surface. If the sum of the charge densities on the two surfaces is p""Cfm2 then the surface charge densities on the two surfaces will be (A) P,"f2, p""f2 (B) 2p*, - p* (c) 0, p,, (D) None of these charges on both of
McQ
3"{'2$ Two infinite plane parallel surface charges Psrtt the figure.
conducting slabs carry uniformly distributed and Pnz on all the four surfaces as shown in
Pstzt Psrr
Ps't
Page 152
Chap 3 Electric Field in Matter Psn
Pnt
Which of the following gives the correct relation between the charge densities
(A) Rrt : (C) R"rt :
Paz, Pstz:
Pszr
pszt:
ps22
psr2t
?
(B) P"tt : Pszzt Pstz:- Pszr (D) Ar, : - P"22. Psr2: - Ps2i
Common Data For Q. 30 and 31 : The plane surfaces tr:0, t:1, a:0 and 9:1 form the borrndaries of conductors extending away from the region between them as shown below.
between the surfacds is given by 5ry volts then the surface charge density on the surface I
If the electrostatic potential in the Iegion
3.1.30 r:
0 is
(A) -5e6y (C) -5eo(r+ y) tucQ 3.{.3{
9:0is
(A) -5ess (C) -seo(r+ u)
MCQ 3.r.32
(B) -5e6r (D) 5eo(rs) (B) -5eor (D) Sesry
Two infinitely long coaxial, hollow cylindrical conductols of inner radii 2 m and 5m respectively and outer radii 3m and 6m, respectively as shown in the figure, carry uniformly distributed surface charges on all four of their surfaces.
If net surface charge per unit length is 10 c/m and 6 c/m for the inner and outer conductor respectively then the surface charge densities on the four surface will be Surface p:2m
(A) (B)
5l3tr
(c) i
b
rl"
(D)
&3
rr.rs
Plane
o
z:0
p:5m -Iln 0 21" 7ln
p:6m alSr 4l3r
-2ln 0
a
surface charge layer with the charge density in figure. If the electric field intensity in the region 0 is Ez:2a,*3au-2a"Yfm then the field intensity Elin the region defines
ps:2nCfm2
z(
P:3m 5l3tr -lln -Il" -Iln
0
as shown
z)0willbe
1,, Ps:2
"'7" .." z:0
Plane
I
nc/m2
(A) 220a,*279a0-2a" (B) 2a"*3an*224a" (C) 222a"4-227ar+2a, (D) 2a,l3a,o*226a"
bo-*oo Data For Q. 34 and 35 : An infinite plane dielectric slab of 1m thickness is placed in free space such that it occupies the region 0 < y < lm as shown in the figure.
Dielectric slab has the non uniform permittivity defined
-_ ;o
4eo
as
for0
$+ar
3't'34 If a uniform electric field E :
4asY lm is applied in free space then bound surface charge densities on the surface g: 0 and y: 1 will be
y:0 O -3eo 3eo - 5eo
at (A) (B)
(c) (D)
at gr: _3eo o o 8eo
_'
'_ -
t//-
l
Page 153
Chrp
3
Electric Field in Matter
Page 154
,i{",*
3'1"3$ As we move from the surface g : 0 toward the
surface will be density charge polarization volume dielectric slab,
Chap 3
A
- 1 inside
the
(A) linearly increasing (B) linearly decreasing (C) Constant (D) zero at all points
Electric Field in Matter
.s."r.:*,a
In a spherical coordinate system the region a < r < b is
occupied
dielectric material. A point charge Q is situated at the origin. that the electric field intensity inside the dielectric is given by
n:ffio,
It
for
by
a
is found
a1 r1
b
The relative permittivity of the dielectric will be
(N
(b'z
@)
(a'
lf) lf)
(c) (f l3) (D) (a'lb')
r:rt i"l r?
Two perfectly conducting, infinite plane parallel sheets separated by
a
uniformly distributed surface charges of equal and opposite densities p, and - ps' respectively. The medium between the sheets is filled by a dieleliric of non uniform permittivity which varies linearly from a value of €1 rr€&r one plate to value of e2 near the second plate. The potential difference between the two sheets will be 1a1 ,Eoa ru) ' '€z-€t
distance d" carry
reP;tt"(?)
(c) r:{.13 3"$"3fi
(o)
#th(?)
p'h(?)
A parallel plate capacitor has two layers of dielectrics with permittivities 6r: 3Eo and ez :2€o as shown in the figure.
1m
3m
If the total voltage drop in the capacitor is 9 Volt then the voltage drop in 1"t and 2"d dielectric region will be respectively (A) ++ Volt, ff Volt (B) 3Volt, 6Volt (C) f+Volt, lf volt (D) 6Volt, 3Volt
I 3.t.3s A dielectric slab is inserted
in the mediurn between two plates of a capacitor
as shown in the figure
Page 155
Chap 3 Electric Field in Matter
Dielectric slab
The capacitance across the capacitor will remain constant (A) if the slab is moved rightward or leftward (B) if the slab is pulled outward of the capacitor
(C) (A) and (B) both (D) none of these
3.t.4o A potential field in free space is given as V:4}cos?sin$V Point P(r:2,0 : rf 3,$ : r12) lies on a conducting surface. The equation of the conducting surface is
(A) 32cosdsin$: 'f (C) 16cosdsins:'f
(B) 16cos@sinl: (D) 32cos@sin0:
********,t<**
f I
Page156 QhaF 3 Electric Field in
Matter
EXERGI$E 3.2
! --
&ei€$
3.a.,{ A certain current density at any point (p,Q,z) in cylindrical coordinates is given by J : t}e"(farta")Afm2. The total current passing the plane Ampere' z:0,0 3 p 3 2 in the a" direction is
----
Q#*ts
**j€$
3.3.x Given the current density in a certain region J : rcos20a, * f sin0a6 - f ^, Al^'.The total current crossing the Ampere' surface defined by 0:90',,0
The current density in a cylindrical wire of radius 16 mm placed along the z-axis is J: $a" Af m2. What will be the total current (in Ampere) flowing through the wire ?
Common Data For Q. 4 and 5 : Atomic hydrogen contains 5.5 x 101e atom/cm3 at a certain temperature and pressure. If an electric field of 40 kV/m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7.1 x 10-16 m.
*ss$
s.4.4
The polarization due to the induced dipole will
be
nCf m2'
el,rs 3.x.5 What will be the dielectric constant of the atomic hydrogen
?
&{Js$
3.?.s
The dielectric constant of the material in which the electric flux density is double of the polarization is
&&JKs
3.a"?
An electric dipole is being placed in an electric field intensity 1.5a, - a,Y f m If the moment of the dipole be p-- 2a'*3avC m then energy of the Joule. dipole will be
----
Common Data For
i.
A and O
:
A lead bar of square cross section has a hole ofradius 0.5 cm bored along its length as shown in the figure.
-1 l cml
3cm
_l 3cm
(Conductivity of lead
:
5
x
106(ftm)-1)
If the length of the lead bar is 8 m then the resistance between the square will be mQ.
ends of the bar
Page 157
Chap
B
Electric Field in Matter
3"2.s If the hole in the lead bar is completely filled with copper, resistance (in pA) of the composite bar
what will be the
?
(Resistivity of copper
:
I.T2
x
10-8
f)m)
Cornmon Data For Q. 10 and ll : A capacitor is formed by two concentric conducting spherical shells of radii a : 1 cm and b : 2 cm centered at origin. Interior of the spherical capacitor is a perfect dielectric with s,:4.
3.2.10 What is the capacitance (in pF) of the capacitor
?
F r,e.rr If a portion of dielectric is removed from the capacitor such that 6, : 1 for , + < 0 < zr and €, : 4 for the rest of the portion, then the capacitance of the composite capacitor will be
pF.
zs
3.2.12 Two conducting
Bs
3'2'13 Two coaxial conducting cylinders of radius 4 cm and
'"
surfaces are present at r:0 and r:5mm and the space between them are filled by dielectrics such that €n:2.5 for 0 < r( 1mm and e,'z: 4 for I < r < 3 mm rest of the region is air filled. The capacitance per square meter of surface area will be nFf m2
g cm is lying along z -axis. The region between the cylinders contains a layer of d.ielectric from P:4cmto p* 6cm with €,:4.If the length of cylinders is 1m then the capacitance of the configuration will be ____ pF.
3'2"14 A parallel plate capacitor is quarter filled with
a dielectric (e" : 3) as shown in the figure. The capacitance of the capacitor will be pF.
A:10
d:4
cm2
rnm
3'2"15 A thin rod of certain cross sectional area extends along the gr-axis from a:0m to y:5m. If the polarization of the rod is along it's length and is given by Pr: 2U2 * 3, what will be the total bound charge of the rod ? s
3'2.ts A neutral atom of polarizability o is situated at a distance 1 m from a point charge f nC. If the force of attraction between them is fl, then $ : N
Common Data For Q. 17 and 18 :
Page 158
Chap 3
ThetwodipolesPl,P2withdipolemoment2nC-mandgnC-mrespectively in figure' are placed at 1m distance apart as shown
Electric Field in Matter
euss 3.2.{? The torque on Pz due to R will be
----
*u€$ 3.2.{B The torque on R due to Pz wiII be
----
Common Data For Q' 19 to 21
pN
- m'
pN
- m'
:
Athicksphericalshellismadeofdielectricmaterialwithapolarization
P(r)
:lwrrClm'
where
r
is the distance from i
the inner radius 2 m 3.?.1e If the spherical shell is. centred at origin and has
outerradius6mthentheelectricfieldintensityatr:l-mwillbe--
v/*.
cu€$ 3.2"20 What will be the electric field at
{lu&$
s.2.zt If at r:
b n},
r:7 m ?
the electric field intensity is
.E
: &9
thervalue of k is
Ql}frs3.2'2?Asphericalconductorofra,diuslmcarriesacharge3mC.Itissur
outtoradius2m,byalineardielectricmaterialofdielectric :::;, ;;;;*; ;;'th" fig*"' what will be the enersv (in Joule) configuration
of
?
with
di
material sphere of radius llJi m is ma'de of dielectric is embedded i 0'6nO/m3 density charge free constant €,:2.If a uniform Volt. it then the potential at the centre of the sphere will be --'
&r.,€s 3,2.23
A
ouEs
Aparallelplatecapacitorisfilledwithanonuniformdielectriccharacteri plate in meter' If ir" ,):- ifi+ iOOi,,l where a is the distance from one them is 10 between separation surface area of the plates is 0'2 m2 and be PF' then the capacitance of the capacitor will
3.2"24
----
t2'25 A two wire transmission line consists of two perfectly
conducting cylinders. each having a radius of 0.2cm, separated by a centre to centre distancc gt 2cm. The medium surrounding the wires has relative permittivity €,-- 2. If a 100 V source is connected between the wires then the stored charge rrer
unit length on each wire will be
nC/m.
3J.26 A tank
is filled with dielectric oil of susceptibility y": 1. Two long coa-rial cylindrical metal tubes of radii lmm and 3 mm stand vertically in tht' tank as shown in the figure. The outer tube is grounded and inner ong is maintained at 2 kv potential. To what height (in pm) does the oil risc iu the space between the tubes 7 (mass density of
oil
:
0.01gm/crn")
FEs 3'2'27 A conducting spherical shell of inner radii 2 m and outer radii 3 m ca.rri, uniformly distributed surface charge on it's inner and outer surfaces. If the net surface charge is 9 C for the conducting spherical shell then, the surfa
3'*"28 An infinite plane dielectric slab with relative permittivity r.: 5 occupies the region r > 0. If a uniform electric field .E : r}q,,v/m is applied iir tlr; region z < 0 (free space) then the polarization inside the dielectric will bc ke
luEs
6a,Cf m2 such that value of ,k is
3'2'2s Two perfectly conducting, infinite plane parallel
sheets separated bv
m cany uniformly distributed surface charges of equal ancl opposite densities *bnO/m2 and -5 nCfrn2 respectively. If the medirrrn between two plates is a dielectric of uniform permittivitv e : 4€o then tlrr,' a distance
2
potential difference between the two plates will be lu€s
kv.
3'2'30 Two perfectly conducting, infinite plane parallel sheet separated by
a
distance 0.5 cm carry uniformly distributed surface charges of equal ;in,l opposite densities. If the potential difference between the two plates is 5 k\: and the medium between the plates is free space, what will be the chargg densities (in pC) on the plates ?
Page 159
Chap 3 'Electric
Field in Matter
Pgge 160
Cbrp
auEs
3.2.3{
I
Dbctric Field ia Matter
A steel wire has a r-adius of 2 mm and a conductivity of.2 x',; steel wire has a.rr- alurrinium (o : 3.8 x 10? s/m) coating of 2 mm thickness I The total cuuent carried by this hybrid conductor be 80 A. The curred I x 105 A/m2' density in steel .I,, is I
--.--
I &:q"$ffits 3"?":??.
The medium between two perfectly conducting infinite plane parallel sheetsl consists of two dielectric slabs of thickness 1 m and 2 m having lermittivities I 4:2€o and ez:4eo respectively as shown in the figure'
I
Infinite
Infinite sheet +
+ + + 0.6 nClm2
-
0.6 nC/m2
+ + + + +
1m
If the conducting
2m
sheets ca,rry uniformly distributed surface charges of equal
and opposite densities 0.6 nc/m2 and - 0.6 nc f m2 respectively, what will be the potential difference (in Volt) between the sheets ?
{<**+r(*<{<***
{<
I
EXERCTSE 3.3
?bge 16r CUep
S
dbctric Field in tr,iatte,r
xco
3.3.{
The electric field in the three regions as shown in the figure a"re respectively Er, E, and .Es and all the boundary surfaces are charge free.
If
e1
:
es
*
€2,then the correct relation between the eledtfic fibld is
(A) E, + E2+
(B) E
E3
(C)E=Ez:Et a
rco
3.3.2
ICA
3.3"3
The unit of poldrization in dielectric is
The taplacian opbrator,
(B) c/m (D) C-m'? V2
(A) has unit of m2 (C) has unit of 1/m2
I rco
3,3,4
Laplace's equation has (A) two solutions
(C) no solution
rcQ
3.3.5
Ez
(D)El=Ez*Es
(A) c/m' (C) C/m3
t
: Et*
(B) is a vector operator (D) has no unit (B) infinite solutions (D) only one solution
The surface charge density in a good dielectric is
(A) zero (C) infinity
(B) p" (D) -p"
ICO
3.3.6
If e": 2 for a dielectric medium, its electric susceptibility (A) 1 (B) 2 (c) 3 (D) 2es
rcQ
3.3.7
Example of non-polar type of dielectric is (A) water (B) hydrochloric acid (C) sulphur dioxide (D) oxygen
HCO 3.3.a
Example of polar type of dielectric is
(A) oxygen (C) hydrogen teQ
is
(B) water (D) nitrogen
3.3.e If the voltage applied across a capacitor is increased, the capacita,nce (A) increases (C) remains constant
(B) decreases (D) becomes infinity
value
162 chap 3
page
Electric Field in
nilcq
3.3.10 If electric suscepiibility.of (A) (C)
Matter rucQ
3,3.{{
3"3.rt
dielectric is 4, its relative permittivity is
(B) 4 (D) 2
g
Poisson's equation is
(A) V' (C) V' tutce
€t
5
(B) V'Y:- P,f e (D) V'Y:- P,f e
e V:- P, Y:
p,f
Boundary condition for the normal component of dielectric is
E,r: B* (C) E*r: ?t,,
(.F')
ilrco
3"3.t3
The boundary condition on .O is
(A) o" x (E (C) Er:6, McQ
- Er):o
3"3"14 The electric field just
E
on the boundary of a
(B) E"r -
fi,2:
(D) E"r:
o
Ps
(B) '(E'- Er):o "" (D) none of these
above a conductor is always
(A)normaltothesurface(B)tangentialtosource (D) (C) zero McQ
3.3.{5
The normal components of D are (A) continuous across a dielectric boundary (B) discontinuous across a dielectric boundary
(C) zero (D) * x**x
EXERGI$E 3.4
hC! Elechic
rco
nca
3"4""!
3"4"2
A
parallel plate air-filled capacitor has plate area of 10-a m2 and plate separation of 10-3m. It is connected to a 0.5 V,3.6 GHz source. The magnitude of the displacement current is (e : frfO-'g n/m;
(A)
10 mA
(c)
10 A
(B) 100 mA (D) 1.5e mA
t has the electrical permittivity ar: 1.5e0 farad/m and occupies the region to the left of r: 0 plane. Medium 2 has the electrical permittivity €z: 2.5€o faradf m and occupies the region to the right of r : 0 plane. If 4 in medium 1 is ,Er : (2a, - 3au * la") voltf m, then fi in medium 2 is (A) (2.0o, - 7.5art2.5a") volt/m (B) (2.0a, - 2$an+ 0.6a,) volt/m Medium
(C) [2.0o, (D) (2.04,
* 3.0ar*
-
l.Oa") volt/m
2.0a,* 0.6o") volt/m
MCA 3.4,3
The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with e.: 80eo. The surface charge densitv on the conductor is (e : i# P/*) (A) o C/m2 (B) 2 Clm2 (C) 1.8 x 70-11 Cf mz (D) 1.41 x 10 e C/mz
MCO 3.4,4
The space between the plates of a parallel-plate capacitor of capacitance C is filled with three dielectric slabs of identical size as shown in the figure. If dielectric constants &r€ 61, 6z and e3, the new capacitance is
Me& 3.d"$
(A)
g
(C)
(er
r"l
@i#:tr
-t ez* ez)C
If the potent ial, V : 4r -12 V, the electric field is (A) 6 v/m . (B) 2Ylm (C) av lm (D) -4a,Y lm
raE
Cfrt
FiSink
Page 164
rtrcQ 3.4.e
Chap 0
Two dielectric rireditr,'with permittivities 3 and tharge-free boundary as shown in figure below :
,/E
ut" separated by a
Electric Field in Matter
The electric field intensity in media 1 at point Pr has magnitude Er and makes an angle or : 60o with the normal. The direction of the electric field intensity at point P2,a2 is
(A),*-,(4E')
(B) 45'
(c)
(D) 30'
"",-'(4E')
MCQ 3"4.7
Assertiou (A) : Under static conditions, the surface of conductor is an equipotential surface. Reason (R) : The tangential component of electric field on conductor surface is zero. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A) (B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true
MCO $.4.8
Along
1
metrethickdielectric (e
:
3eo)
(A) (C) l,rcQ 3.4"s
4eoa, 36soo,
< r < 5 is 6e,.The polarization
slab occupyingthe region 0
placed perpendicularly in a uniform electric field Pr inside the dielectric is
.Oo
:
(B) 8eso, (D) Zero
The flux and potential functions due to a line charge and due to two concentric circular conductors are of the following form : (A) Concentric circular equipotential lines and straight radial flux lines.
(B) Concentric circular flux lines and straight equipotential lines (C) Equipotentials due to the are concentric cylinders and "hurg"are straight lines. equipotentials due to two conductors (D) Equipotentials due to line charge are straight flat surfaces and those due to two conductors are concentric cylinders. ilrcQ 3.4.10
There are two conducting plates of sizes 1m x lmand2mX 2m. Ratio of the capacitance of the second one with respect to that of the first one is (A) 4 (B) 2 (c) 112 (D) rl4
-Q
3.4.{{
Consider the following
:
In a parallel plate capacitor, let the chalge be held constant while
Page 165
the
dielectric material is replaced by a different dielectric. Consider 1. Stored energy 2. Electric field intensity.
3.
Capacitance
Which of these changes (A) l only (C) 2and3only
?
(B)land2onIy (D) 1, 2 and
3
3cl 3.4"!2
By what name is the equation V . J: O frequently known (A) Poisson's equation (B) Laplace's equation (C) Continuity equation for steady currents (D) Displacement equation
-a 3.4"'!3
Method of images is applicable to which fields (A) Electrostatic fields only
?
?
(B) Electrodynamic fields only (C) Neither electrostatic fields nor electrodynamic fields (D) Both electrostatic fields and electrodynamic fields
DQ 3.4"{4 What is the unit of measurement of surface or sheet resistivity? (A) Ohm/metre (C) Ohm/sq. meter
(B) Ohm metre (D) ohm
GO 3,4.{5
Which one of the following statements is correct ? On a conducting surface boundarv, electric field lines are (A) always tangential (B) always normal (Q) neither tangential nor normal (D) at an angle depending on the field intensity
*o
which one of the following is correct ? As frequency
3.4"{6
increases, the surface
resistance of a metal
(A) decreases (B) increases (C) remains unchanged (D) varies in an unpredictable manner
rca
3.4.1?
Application of the method of images to a boundary value problem in electrostatics involves which one of the following ? (A) Introduction of an additional distribution of charges and removal of a set of conducting surfaces (B) Introduction of an additional distribution of charge and an additional set of conducting surfaces
(C) Removal of a charge distribution and introduction of an additional set of conducting surfaces (D) Removal of a charge distribution as well as a set of conducting surfaces
Chap 3 Electric Field in Matter
Page 166
i/lco
3.4",rs
Assertion (A)
: Potential
everywhere on a conducting surface of infinite
Chap 3
extent is zero.
Electric Field in Matter
Reason (R) : Displacement density on a conducting surfdce is normal to the surface.
(A) (B) (C) (D) [,t*s
3,4-*$
Both A and R are true and R is the correct explanation of A Both A and R are true but R is NOT the correct explanation of A A is true but R is false A is false but R is true
A parallel plate capacitor of 5pF capacitor has a charge of 0.1pC on its plates. What is the energy stored in the capacitor
rscg
3"4.20
A
charge
(D)
l
pJ
of 1 Coulomb is placed near a grounded conducting plate at
distance of 1m. What is the force between them
(A)
#.N
tcl rela ifico
3-4"21
?
-(B)1pJ
(A) 1mJ (C) 1 nJ
*
tut
a
?
sh *
(D) 4zes N
The capacitance of a parallel plate capacitor is given by sf where A is the area of each plates. Considering fringing field, under which one of the following conditions is the above expression valid ? (A) is tending towards zero @ is tending toward.s infinity
+
1c;
fisr
+
(D)4i. ''d€r€o
1
!$cQ 3.4.22
What is the expression for capacitance of a solid infinitely conducting solid sphere of radius '.R in free space ? (A) 2reoR (B) AtreoR (C) 8aeo.E (D) 0.52'es,R
fr.ltcQ 3.4"23
A point charge of + 10 pC placed at a distance of 5 cm from the centre of a conducting grounded sphere of radius 2cm is shown in the diagram given below:
What is the total induced charge on the conducting sphere ? (A) 10 pC (B) a pc (c) 5 pc (D) 12.5 pC I$Cq 3.4.24
For an electric field E- Essinwt, what is the phase difference between the conduction current and the displacement current ?
(B) 45' (D) 180'
33.25 An infinitely long line charge of uniform charge density prclm
.
is situated parallel to and at a distance from the grounded infinite plane conductor. This field problem can be solved by which one of the following ? (A) By conformal transformation (B) BV method of images (C) BV Laplace's equation (D) By Poisson's equation
3.4'15 An air condenser
of capacitance of 0.00b pF is connected to a d.c. supply of Volts, disconnected and then immersed in oil with a dielectric constant of 2.5. Energy stored in the capacitor before and after immersion, respectively 500
IS
(A) (B) (C) (D) rGo
x 10-4 J and 250 x 250 x 10-4 J and 500 x 625 x 10-4 J and 250 x 250 x 10-4 J and 625 x 500
10-4 J 10-4 J 10-4 J 10-4 J
\
3'4.2? A 3 pF capacitor is charged
by a constant current of 2 pA for 6 seconds. The voltage across the capacitor to the end of charging will tre
(A)3v (c)
rcQ
3.4.28
(B)4v (D)ev
6v
Consider the following statements
:
A parallel plane capacitor is filled with a dielectric of relative permittivity r'r &nd connected to a d.c. voltage of I/ volts. If the dielectric is changed to another with relative permittivity s,r :2et, keeping the voltage constant, then 1. the electric field intensity E within the capacitor doubles. 2. the displacement flux density D doubles
3. 4.
the charge Q on the plates is reduced to half. the energy stored in the capacitor is doubled. Select the correct answer using the codes given below (A) 1 and 2 (B) 2 and 3 (C) 2 and 4 (D) 3 and a
HCQ 3.{,29
A coil of resistance 5o and inductance 0.4H is connected to a bOv d.c. supply. The energy stored in the field is (A)
10 joules
(C) 40 joules HCO 3.4.30
:
(B) 20 joules (D) 80 joules
The normal components of electric flux density across a dielectric-dielectric boundary
(A) (B) (C) (D)
are discontinuous are continuous depend on the magnitude of the surface charge density depend on electric field intensity
Page 167
Chap 3 Electric Field in
Matts
Pt"-!6E- ,
ctai
t
"
,
Ebctric'field in Matter
Mcqi.3'4i3r.. Qopqide,r the fuflqwrnglpt&fegleFrts in conneetion with boundary relations of , r el.gctric field : 1,. ,,, :l-,., ,, . ., 1. In a single medium electric field is continuous., - i 2. The tangential components axe the same on both sidd of a boundary between two dielectrics.
3. 4.
The ta,ngential electric field at the boundary of g, dieJe.ctric and a current carrying conductors with finite conductivity is zero. Normal components of the flux density is continuous across the chargefree boundary between two dielectrics.
Which of these statements is/are correct ?", (A) l only (B) 1, 2 and 3 (C) 1, 2 alnd 4 (D) 3 and 4 only , MCO 3,4,32
The capacitance of an insulated conducting sphere of radius .R in vacuum is
' ,
(L) 2tresR (c) 4;eoB2
lvlcQ
(.8),4neoR
(D) 4tresfR
3.4.33 A parallel
plate air capiacitor ca,rries a cha,rge Q at its maximum withstand If the capacitor is half filled with an insulating slab of dielectric constant 4 as shown in the figure given below, what are the maximum withstand voltage and the cha4ge.qn the Capaeitor at this voltage, gespectively ? voltage tr/.
(B) 4V,2.5Q (D) Vl4,Q MCQ 3.4.34
when an infinite charged conducting plate is placed between two infinite conducting grounded surfaces as shown in the figure given below, what would be the ratio of the surface densities A and p on the two sides of the plate ?
t
-__/ t
(A)
(c)
Charged plate
(dr+ t)
(d'+
Ld.
o,/^'
t1
(B)
(d"+ t) (dr+ t)
(D)
h d,r
].*.
t
(A) l9 - eoD+ P' (B) D: eo(g* P) (C) D: eoE* P
(D)E:D+eoP
F
f. *". h
r 3437 :
't F
The polarization in a solid dielectrie ts,related,to'thd electric field .E a"nd the electric flux density D according to which on of the following equations ?
Image theory is applicable to problems involving
(A) electrostatic field only (B) magnetostatic field only (C) both electrostatic and magnetostatic fields (D) neither electrostatic nor magnetostatic field six capacitors of different capacita.nce s c1, c2, cr, co, G and c6 are connected in series. Cr) Cz) Ct) Ce) Cs > Co. What is the total capacitance almost equal to ? (A) (B) c. (c) c4 (D) G
c,
e.ela
:
Two extensive homogeneous isotropic dierectrics'meet on a plane z: 0. For z) 0, €r= 4 and for z 1 0, €a=2. A uniform electric field exists at z> 0 as Er : 5o, - 2q* 3a, kw/m. What is the value of. E," in the region z < 0 ?
I
i
l
(A) 3o (B) 5o,-2ao (C) 6a, (D) o'-q
flux density D = L Coulomb/m2. The slab is uniformly polarized. What is the polarization P of the slab in Coulomb/m2 ? (A) 0.8 (B) 1.2
(c)
4
(D) 6
ts lr'eo Which one of the following
glves the approximate value of the capacitance between two spheres, whose separation is very much larger than their radii
R? (A) hr/esR (C) 2neo/R
I r'c"c't
(B) 2reoil (D) aneolR
Assertion (A) : For steady current in an arbitrary conductor, the current density is solenoidal (R) : The reciprocal of the resista,nce is the conductivity. Fpj" (A) Both A a,nd R are true and R is the correct explanation of A (B) Both A and R are true but R is Nor the correct explanation of A (C) A is true but R is false (D) A is false but R is true
Page 169
Chap 3 Electric Field in Mdtter
Page 170
";$(:Q S.€.d3
Cbap 3
Assertion (A) : Displacement current can have only a.c components. It is generated by a change in electric flux. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is Nor the correct explanation of A (C) A is true but R is false
Reason (R) :
Electric Field in Matter
(D) A is false but R is true .ii *
3.4".13
A plane slab of dielectric having dielectric constant 5, placed normal to a uniform field with a flux density of 2 cfm2, is uniformly polarized. The
polarization of the slab is (A) 0.a Clm'
(B) 1.6 C/m'? (C) 2.0 clm'z (D) 6.a Clm'z IL&
3"4.d.4
;{.& 3.4.4$
Ohm's law in point form in the field theory can be expressed
(A) v: (B) J: (C) J: (D) ,? :
as
Rr
Elo oE ptlA
A medium behaves like dielectric when the (A) displacement current is just equal to the conduction current (B) displacement current is less than the conduction current
(c)
displacement current is much greater than the conduction current (D) displacement current is almost negligible r,*{.&.}"S"dS
A copper wire carries a conduction current of 1.0A at b0Hz. For copper wire e : €oi |.tr: lh, o: 5.8 x 10mho/m. what is the displacement current in the wire ? (A) 2.8 x 10 A (B) 4.8 x 10-11 A (c) 1A (D) It cannot be calculated with the given data
llil{r* 3"4"i}?
Assertion (A) : when there is no charge in the interior of a conductor the electric field intensity is infinite. Reason (R) : As per Gauss's law the total outward electric fl,x through any closed surface constituted inside the conductor must vanish. (A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is Nor the correct explanation of A (C) A is true but R is false (D) A is false but R is true f,tfi&
3"4"d}.*
A point charge * I is brought near a €orner of two right angle conducting planes which are at zero potential as shown in the girren figure. which one of the following configurations describes the tQtal effect of the charges for calculating the actual field in the first
quadrant\_--,
page 171 Chap
B
Electric Field in Matter
(A)
(B)
F H
I
(c)
(D)
$
r h i
h*" i
The electric field across a dielectric-air interface is shown in the given figure. The surface charge density on the interface is
I
€:1
--t
i ":n E:26.,
(A) -4e, (C) -2es Q
(B) -3ee (D) -to
3'4"50 when air
pocket is trapped inside a dielectric of relative permittivity ,b,, for a givenapplied voltage across the dielectric, the ratio of stress in the air pocket to that in the dielectric is equal to
(A) r/5
(e)s (D) 5_1
(c)1+b x***>t**>t**{<
I
soLUTloNs 3.{
Page 172
Chap 3 Electric Field in Matterr
sol' 3.{.{
Option (A) is correct. For a given current density, the total current that passes through a given surface is defined as
r:It.as
where dS is the differenjtial surfaee area having the direction normal to the surface.
fl$ : pd,pdftaz for the plane z: 2 Therefore the total current crossing the plane z:2, p < 4 is So we
have
r
:(ff",-ffi""){oo'00"") : f" ( l-2}=sin|' -
./a=oJp=o\ c. +
t
)@aoao)
:-ll=,ftoaolll?t"oaol :0A sol.
3.1.2
Option (D) is correct. Flom the equation of continuity we have the relation between the volume charge density, p" and the current density, .I as
9*:-v 0t
.J
Given the current density,
t:Too-ffio"N^' So, we have the components
Therefore,
Jo:
T,Jr:
0 and
J,:- (ffi
*:-li#;@r,)+I#.*]
:\tr&*'.*(#y)l -0
So, volume charge density
sol. 3.t.3
will be constant with respect to time.
Option (A) is correct. Given the current 1: 6 A is flowing radially outward (in ao direction) through the medium between the cylinders. So the current density in the medium between the cylinders is
t:h+:ffiq : $a" 27tp
(I:2ml
Af m,
'
For a given current density in a certain medium having conductivity o, the
electric field intensity is defined
as
Page 173
E_J _lf 3 -t D-o-o\fi-,1
Chap 3 Electric Field in Matter
(p: 4 x Lo-z m1 o :0.05 S/m)
3
2rx4xl0'
x
0.05
:238.73Y lm Option (A) is correct. Voltage between the cylindrical surfaces is defined as the line integral of the electric field between the two surfaces
i.e.
v:-[n.m
Now the electric field int6nsity in the medium between the .two cylindrical surfaces as calculated a.nd
:
in previous question
is
": *(h"r)
the differential displacement between the two'cylindrical surfaces is dl
dpap
So the voltage between the cylindrical surfaces is
, :- I:::i,(#r",). {aoo,):-#'(*) :-
4.88 volt So, the voltage between them will be 4.88 volt.
Option (A) is correct. As we have already calculated the voltage between the two cylindrical surfaces and the current flowing radially outward in the medium between the surfaces is given in the question. So the resistance between the cylindrical surface can be evaluated directly as
r F
R
L
L
3.r.6
3.1.7
:Y:#:
0'813o
Option (C) is correct. Since voltage between the cylindrical surfaces is and cuirent flowing in the medium is So, Power dissipated in the medium is p : VI: (4.88) x 6:29.2gwatt
(Y:
4.88
I/ :
4.88 volt
volt,l: 6 A)
1:6A
Option (D) is correct. Consider a constant voltage is applied across the ends of the wire so, t6e electric field intensity throughout the wire cross section will be constant.
i.e. where
E :4:o1
J2 02
.Ir is the current density in the material having conductivity o1.
is the current density in the material having conductivity So,the ratio of the current density is "I2
Jr Jz--Qoz i.e.
n
3.1.8
it will
be independent of both rand
.R.
Option (B) is correct. Electric field intensity is defined as the negative gradient of the potential
o2.
E
Page 174
Chap 3 Electric Field in Matter
:-YV
:-(#**%"*#"") :
$aL
3.{"9
500a,,
V/m
Option (D) is correct. For a given electric field intensity E in a material having relative permittivity e,, the electric flux density is defined as :
" : g*1r 8b x 10-12) x (booan)
e,:815
:7.}8aynCf'm2
l
${3t- 3"t.,lo
Option (A) is correct. For an applied electric field intensity ,6 in a material having relative permittivity e', the polarization of the material is defined as P : eo(€,- t)E
$$r- 3"t"1{
l
l
=[H:il:i];'j{$::"'
Option (B) is correct. since the two regions is being separated by the plane y : 0, so the tangential and normal component of the electric field to the plane 9: 0 are given as Ert :50a'- 70a"
Eu =
20an
FYom the boundary condition, the tangential component of electric field
will
be uniform. I.e.
Ezt
: Eu- 50a,- 10a"
and the normal component of the field is nonuniform and given as
E2E2n: E1E6 Ern
- Z"-:!1zoo4:
e",
the electric field intensity in the second region is Ez : Eztl Ezn : (504, - 10a,) * (Sau) : 50a, * 8ao - \\a,kY f m Therefore the electric flux density in the region 2 is
So
Dz: €re€oEz : 5 x 8.85 x 10-12(50o, ]-San- 104,) x 103 : 2.2I a" -f 0 .35 ou - 0 .44a" pC I ^' $ot-
3.1.12
Option (D) is correct. Energy density in the region having electric field intensity Ez is defined as W, :le*eoEz' Er, where the relative permittivity of the medium is e,r As calculated in previous question the electric field intensity is Ez :50a"* 8a, - I}a"kY f m So the energy density in the region 2 is
*":i,x F
:
Y.
lar-rs
5
x
(s)r+ (10),] x
ee[(bo)r+
106
Page 178
.
59 mJ/m3
Option (B) is correct. As the dielectric slab occupies the region 0 < z< d and the field intensitv in the free space is in * a, direction so, the field will be normal to tht boundary of plane dielectric slab. so from the boundary condition the field normal to the surface are relatecl as
€Ei, :6o9
Therefore, L
3.1.i4
E^
: &Eoa": *a"
Din
:
.e
Ein
= +eo!
o,:
(s
:|a :$uo
|
:
41111
€sE()a,
Option (C) is correct. Total energy stored in a region having electric field is given
w
as
(E . E)du
. f,tn,* az) (8,+ n,)au
(E: fi-r
E2;
: !^ [ {al + E3 + 2r,, . E,)d, : l^ | nla, +|r, fidu + eo(n, . E")d,, | | : w+ r,f..rs
wr+
[
eo@,'
E2)du
Option (A) is correct. Consider a neutral dielectric is placed in an electric field.E, due to whic6 the dielectric gets polarized with polarization p, the bound surface chargc density of the dierectric be prs and the bound volume charge density a" p* So the total bound charge by the dielectric is given as Qbound:
frorras+
[
oo,o,
Since for a given polarization P of a dielectric material, the bound surfircc charge density over the surface oi material is defined as
Pps:P'an where
a'
is the unit vector normal to the surface directed outward. while the bound vorume charge density inside the material is defined as
ppr.:_ V . p
so we have,
Qbouna
: frt, . a,)d,s - v |
. e a,
:fv.ds- lv.va,
But according to the divergence theorem
$p. ds: J,fv
I vr
Therefore,
Qoou,a
:0
. pdu
Chap 3
Electric Field in Matter
(d,sa,:6s1
pist
rzo
Qt'h
s
sol.
3-,1-{6
olribn (nlt$,mlum Given thd conduAii"itv
Etirc{ric FieH in }iaitdr
"f
a = lo€(Om)-t 5r :19619;-1
-"t"it"t,
and cond'ubtance of the wit'e, Sin'ce the cohtluctance
So we
have,
1oo
of a Wife of 'length I having cross sectional aida S
: Jdf"t
(5:
:.8
, -! sol. 3.{,17
n
Optioh (A) is correct. Ad the medium between capacitor plates is conducting so
it
carries
aS well as capaditine property. Considei thd platds'dre separdtbd' liy a dista^nce d and the surface area pbtes ib ,9 as dhbwn in the ffgure.
resi$tive
d
So the
total resistance of the medium between plates
ib
R:4 OD
arrd capacita,nce of the capacitor.is
u:_T ^ g,S Therefore the time constant of the capacitor will be
, : RC:3 soL
3.1.18
Option (B) is correct. For a given current density, the total current that passes through a surface is defined as
t:ft.w J
where dS is the differential surface area having the direction normal to surface. Since the current density is independent of d and
the current
d so we can have di
I - J. S- J(4trf a,) - !-"-to'noP :4trx(6)'xlx ato"to* : 4n x 6 x e-r: 24tre-L: 27.7 A
.t
i.te
Option (C) is correct. of continuity we have the relation between the volume charge density, p, and the current density, J as FYom the equation
Q*:-v dt
.J
and since the curent density have only the mmponent in
aet -_ t -
0t
4
direction so we bale,
o ,.
S6rt'tJ')
0p, :-7A;\' _ _ l 0 ts1
t"--ro,r\)
E
Integrating both sides we get,
p,e,t) where
/(r)
:- I i"-""t dt+ f(r)
is the function independent of time.
P,(r,t)
:$-"-tot *
.,1'1
co P,(r,t) :0 put the given condition in the equation to get
Now for t -+ So, we
3.1.20
therefore
p"(r,t)
-
i.e.
p,(r,t)
.h
/(r):0
1Q1"-to" rt
Option (D) is correct. The velocity of eharge density can be defined as the ratio of current density to the charge density in the region
(1/')"-to"to' :h:W:ro3ran u:L So, at r:0.6m, o : 103 x 0.6a, : SSQa"m/s
i'e' 3.1.21
Option (B) is correct. The given problem can be solved easily by using image theory as the conducting surface g: 0 can be repiaced by the equipotential surface in the same plane g:0 and image of line charges (p', :- 5pC/m at r:0, A :- L and r :0, A - - 2) as shown in the figure
o]:
5 pclm
Pl,: -5 P..l^
Pr: +
+ + +
+ + + +
5 Pcf m
Pr:
5 Pc/m
Page 177
Chap 3 Electric Field in Mritter
r178 QhaF 3 Page
Electric Field in
The work done to p?rly a unit positive charge from a point located at a distance a from the line charge with charge density n, to another point located at a distance b from the line charge is defined as
Matter
Uu:-&n(L\ z'^Eo \u I and since the surface
g: 0 has zero potential, so the potential at point P
will be equal to the work done in moving a unit positive charge from plane y: 0 to the point P. So the potential at point p will be
the
v,:_2fun(!) where a is the distance of the surface
g:
0 from the line charges while b is
the distance of point P from the line charges.
so, scl.
3.{.t2
ve
:-b \=L!-'lzt_r(*)_ r(+)*,"(+).r(+)] t :- O.r';;;
Option (A) is correct. Electric field at a distance .fi| from a line charge having uniform charge density p, is defined as
P:=Pt - 2neo 4
172
so the net electric field intensity produced at the point line charges discussed in previous question is given as
p
due to the four
,:Eh#
where .fi| is the distance of point P from the line charges Therefore, E : :-pt [(- 1' -2'0) - (0,1,0) * (- 1, - 2,0)
--r,r6,[1tr 1-BOT- *_ (- 1, - 2,0) - (0, - 1,0) _ (- 1, - 2,0) - (0, - 2,0) l(- 1, - 1,0) l' l(- 1,0,0) I' (t# : u;#l_s*!) _ * s+E . q+al
:
O.\2a,
-
0.0032a,
:
0.12a,
-
0.008a,
(0,2,0)
V/m
sol 3.{.23 Option (C) is correct. For a given polarization P inside a material, the bound volume cha,rge density inside the material is defined
as
pr, __v . p Since the polarization of the sphere is p(r):2yq,, So the bound volume charge density inside the sphere is pp,
:-v . p(r) :_-hf;fn2t :-+ x 6f :- 6
Therefore the electric field intensity inside the sphere at a distance the center is given by
: I P"xtnf E-,1 " - 4reoQ'!"n-7
: Ho,:-ffio,:-(*]ir", so the radial component of the'electric field inside the sphere is
D_
Dr
--
2^I
^ c0
r from
$ p
IB
F
h
linearlv decreasing with a.roo"
F-" r
fj';}
br"" F
Ir
(-*)
with respect to
r as shown
Chap 3
B
Electric Field in Matter
F,
i-
f r
B $
F p
t h
h F'
[,r. t
Page 179
3.1.?4 Option (C) is correct. For a given polarization P of a material, the bound surface charge density over the surface of material is defined as
t F
PPs:P'o'n so the bound surface charge density over the spherical surface is prt : P(r)' a,
:2r:2a
So,
(at the spherical total bound surface charge over the sphere is Qps
:2a x
4tra2
(a": surface r:
a,) a)
:8ra3
and the bound volume charge density inside the sphere as calculated above is
Pro:-6 So,
total bound volume charge inside the sphere is Q,,
: o,,(tra'): (- 6) x (+"0):-
stra'
Therefore the total bound charge in the sphere is Qboud : Qrs* Qr, :8ra3 - 8zro3 : 0 According to Gauss law the outward electric field flux through a closed surface is equal to the charge enclosed by the surface and since the total bound charge for any point outside the sphere is zero So, the electric fielcl intensity at any point outside the sphere is .E: 0.
3.'t.as
Option (C) is correct. For a given polarization P of a material, the surface charge density over the surface of material is defined as
Ps: P' where
an
a, is the unit vector normal to the
surface directed outward of the material. while the volume charge density inside the material is defined as
P'--V
'P
Since the the cylinder has uniform polarization So, volume charge density inside the sphere is
p.
P'-- v'P:0
and the surface charge density over the top and bottom surface of the cylinder is
Ps:P'an:tP
(+P at top surface and -p at bottom
surface)
7
Pase 180
So the
Chap,3
total bound charge by the cylinder is Qbouna:
Electric Field in Matter
Qs*
Q,
: l, oras + | sol.
,,,0,
:I+
P(rrf)
- rQrf)l* o :
o
3.1,2S
Option (A) is correct. As calculated above th" .,rolrr-" charge density inside the cylinder is zero while the surface charge density at top and bottom surfaces are respectively *P and -P, so the cylinder can be considered as the two circular plates (top and bottom surface) separated by a distance .L. since the separation between the plates is larger than the cross sectional radius (L : 2r) so the fringing field(electric field) will exist directed from the upper plate towards the lower plate.
sol. 3.t.27
Option (C) is correct. The electric flux lines will be the same as the electric field intensity outside the cylinder but as the volume charge density is zero pu: 0 inside the cylinder n ' ds:0 and therefore the flux lines will be continuous.
$oL 3.1.28
Option (A) is correct. consider the charge densities of the two surface of the slab is p,1cfm2 p*Clm'as shown in the figure. As the sum of the charge densities is p",Cfm2 so we have
"" f
Pt*
P"2
: Psl
and,
"'(1)
and since the electric field intensity inside the conducting slab must be zero Sot
Et* Ez :0 where ,81 is field inside slab due to charge density slab due
to
...(2) p"1 and
Ez is field inside
p,2
2
ti:+:*::r*:+'r+' +r*.r+ +. +. + +
+.+ri+.+r'+:+ *,+: +:+
+..i+..+
As the electric field intensity at any point
p
due to the uniformly charged
plane with charge density p5 is defined as
E:#o, where
o'
is the unit vector normar to the plane directed toward point
So we have,
: He E, : Hu" E,
F}om equation (2)
a
o")
Putting the result in equation (1) we get
Prt: Pr2-
PL
:-
a,)
(a,:
o,)
(a,
#Fo,\-t n!!a.:o tc\ ' -c0' PsI
:
Ps2
p
Option (B) is correct. As the slabs are conducting so net electric field inside the slab must be zero. and since the electric field intensity at any point P due to the uniformly charged plane with charge density pe is defined as
t : #,",, where
a, is the unit vector normal to the plane directed toward point P
So, the net electric field intensity inside slab 1 is
Hta")rffia"*Ho"+ffia":o (o,,
:-
a" for pnt while
4.,
:
a" for rest of the charge densities)
- P*t* P"n* P,zr* Przz :0
... ( 1)
and the net electric field intensity inside slab 2 is
ffie ""7+ffit a"1+ffi? a)+ffia":o (on: o. for pnz while o,, : -
Solving
3"$"3{i
-
-
Ps2t
and
P"12
9s12
*
Przz
:
for rest of the charge densities) ...
O
(2)
eq (1) and eq (2) we get, PsI1'
s8L
Ps11
o,z
:
Ps22
: -
P"zr
Option (A) is correct. As all the four surfaces form the boundaries of the conductors extending away from the region between them so, the medium outside the defined region is conductor and sb the field intensity outside the region will be zero. Now the electric potential in the non conducting region is given as V :5rA So the electric field intensity in the region is E :- YV :- bya,- brao From the conductor-free space boundary condition we have the surface charge density on the boundary surface defined as P"
:
€oE'
where E, is the normal component of the electric field intensity in the free space.
So, the surface charge density on the surface
&: (- 5y) (the "":1"i::,:ponent eo
__
oclg
En:
-
r:
0 is
5y for the surface r
:
0)
Page 181 Electrlc triold
ir
s Mritter
th"p
Page 182
$ol
3,'t"3,1
Chap 3 Electric Field in Matter
Option (B) is correct. Again as discussed in above question, the surface charge density on the surface g: 0 will be given by p" : €sEn and since the field component normal to surface .r/: 0 is
En:-5r So, the surface charge density on the surface
P, :$oL
3.,t-32
g:0
is
Seor
Option (A) is correct. Flom the symmetry associated with the charge distribution the electric field must be radially directed. Then choosing Gaussian surfaces which are cylinders having the same axis (p: 0) as the conductors and of length l, we get
(2rpI)Eo:O
for
p<2m
for
p<2m
(Since there is no charge enclosed by the Gaussian surface)
Thus
Ep
:0
Now, since the field inside the conductor 2 < p < 3m is zero; there cannot be any charge on the surface p:2m.
Ps:0
1.e.
at
P:2m
and all the charge associated with the inner conductor resides on the surface
P:3m.
10C/m: b n pt:ffi rlclm2
1.e.
atp-$*
Proceeding further we have
2nprEo: where
I is length
So
*(ro
for5< p<6m
Clm)r
of the cylinder.
,:#oo
for5
This the field produced by the inner conductor but the fact is that the field inside the conductor 5 < p < 6m is zero that gives
[prLr:r- : eolEL,r:r' (-
and
:'.(r:ftO",)' (- :-ffi:_,Lclm' ",)
[p,L,o:u*
:*rr.; sol. 3,t,33
or)
:r;r,,:,zn(s)]
Option (B) is correct. FYom the boundary condition for the charge carrying interface, the tangential component of electric field on either side of the surface will be same.
Et;:
1.e.
Ezt
while the normal components are related
E6-
E,,
:
as
P1 €g
the field intensity in the region z ( 0 is Ez - 2a"*3au-2a" So the tangential component, E,21 :2a"| now
a,s
3ao
I r,
tI i
component, Ezn :_ 2a" Therefore the field components in region (z > 0) are
t
and the normal
I t
and
Eu : Ezt:
Eu _
F
2a"
*
hlFl
crr!
ElectrfoFHdLn
3av
n^+*:[_z+#*1ftfa,:224a,
So the net field intensity in the
r
Et
I
,"gio, z > 0 is Eul E6 : 2a,* Jau *
:
I
: 3.1.34
224a,"
Option (B) is correct. As the dielectric slab occupies the region 0 < y < 1m and the erectric field in the free space is directed along o, so, the field will be normal to both the boundary surfaces g: 0 and g: 1. So from the boundary condition the field normal to the interface of dielectrics are related a^s €86
:5oB
(where -o, is the fierd inside the dielectric)
E,:AG+vYE:9{@o,) so the polarization
t;ji#kectric
p : €Er_ €oEt : t, -
ffi
since
e: . 4eo $+yY
=
is
: (6p -,,)tr + ur a,
€oE,
,.,;,L1;s.;{H:
Now for a given a dierectric material, the surface charge density over the surface of dielectric is defined
PPs: P'
an
where an is the unit vector normal to the surface directed outward of the dielectric. So, the bound surface charge density
at y _ 0 is
[p*L,,=o:P'(-ar) (o^: - au) :14 - (r + l)le,(_ 1) :_(4 _ 1)ro 3eo -U:0 and the surface charge density at y:1 m is [P*Lu=r: P ' (ar)
:
[a
-
(1
+ yI]so(l)
:
(4_
4)eo :
(o":
or)
s
r- 3.1.35 Option (A) is correct. As calculated in previous question, polarization inside the dielectric is
p :[4 _
(t + s)]eoa, since for a given polarization p inside a material, the bound vorume charge density inside the material is defined as
Ppr:- V ' P
So
the volume charge density inside the dielectric is
e,
:-&14 - (r + yfle,
so when we move linearly increasing.""*:
;tt
'r!', :1m,
the votume charge density wiil be
i i
I I
Page.lM. Chap Q Electric Field in Matter
sol.
3.{.36
i
Option (A) is correct. As the charge is'being located at origin so the field intensity due to it will be in radial direction and normal to the surface of the dielectric material. Therefore the flux density will be uniform(as from boundary condition) and at any point r inside the dielectric flux density will be
: -Q-a4rt'
D Now
it
is given that electric field intensity at any point inside the dielectric is
n:ffiu
and since in a medium of permittivity e
:
D
:
e,so the
flux density is defined
as
€,eoE
So for the given field we have
ffi",:r,"(ffi")
Fd
^b2 sot.
3.{.37
r
Option (C) is correct. The electric field between the plates carrying charge densities is defined as
*p,
and
-p*
E:+ where e is the permittivity of the medium between the plates. Now consider that near the plate 1 permittivity is e1 and near the plate 2. permittivity is e2. So at any distance r from plate 1 permittivity is given by E
: €t*(T).(Since
So the field intensity
the permittivity is linearly increasing)
at any point in the medium will be
P-
D-
Pn
e*(T)r
Therefore the potential difference between the plates will be
v:foro
sol.
3.t.3s
Ps o, e,+(,+)_
Option (A) is correct. The capacitor of a parallel plate capacitor is defined
c:+
So, the capacitance
in 1't dielectric region will
Cr:+:+
and the capacitance
in
2nd
Therefore the voltage drop V, -= et,
as
be
dielectric region
in 1't dielectric region is
Cz v 'r-"7;,V
(where
I/
is total voltage vo dropi
:ffiffi(ovotp; and
similarly,
Y,
PtG
:]&rvort
Ct4t
::9t-u' - eJ4' - SeJiJ-ee (o) : f{ 3eo'5
Option (A) is correct. Consider the dielectric slab is of thickness f and dr, width in the medium as shown in the figure.
h
vort
are the remaining
Dielectric slab sfi.i*ffiffis##H#j; .IS$ffi.
,$w 'ffi
d,
Now the capacitance of the whole configuration will be considered as the three capacitors (capacitance in the three regions) connected in series as shown in the figure
c1
c2
c3
C,:+, tr:#
So,
The equivalent capacitance, is defined
1 1 1 1 e;:q-6*q:
and G -dt -qE
as
t u-s
.@t+da) ---;r
Since t; (&+ dn) will be constant although if the dielectric slab is moved leftward or rightward so the equivalent capacitance will be constant. But if the slab is pulled outward then the capacitance will change as the effective surface area of the capacitance due to dielectric slab changes.
t 3.1.40 Option (C) is correct. Given, the potential field in free space
: $coslsing So, the potential at point P (r:2,0:t,O:$) V
v,
:
ffi.o'(+)""(6)
is Siven as
: 2.5 vort
Now, as the conducting surface is equipotential, so, the potential at any point on the conducting surface will be equal to the potential at point P. l.e.
or
V : Vp:2.5Volt 49cosdsind : 2.5 I"
l6cosdsind
: I
This is the equation of the conducting surface.
>k*********>k
fl5
Electric Field in
Mattc
soLUTloN$ 3,2
Page L86
Chap 3 Electric Field in Matter
Correct answer is 125.664 . For a givel current density, the total curr€nt that passes through a given surface is defined as
t: Jft.
as
where dS is the differential surface area having the direction rormal to the surface.
dS : pdpd$a, for the plane z:0 Therefore, the current passing the plane z:0r 0 < p So we have
t:
[:,[:,ltoe'(da,+
:10
f2t
Jo
o")f
'
<
2 is
(PdPd'\o")
f2
Joe"
odzdf
- ,o I f,'oaoao :'[+]:,,xtotl :
10
x 2 x 2r
:
(z: 0)
40rA
:
125.664A
Correct answer is 1.5708 . For a given current density, the total current that passes through a given surface is defined as
t: J[t.
as
where d$ is the differential surface a.rea having the direction normal to the surface. So, we
have
Thereforei 0
<
4$
:
(rsind d,$)Gr)aa for the surface
the total current crossing the surface
r< lm is
t : [ ("o"' 0o, + f
l'
of'of
:lol?">< -$ilt.. 3"2"3
0: 90' d : 90",0 < Q < hr.
sinlae -
rz o,o)
. (rsin0d,gd,rae)
,in'odidr
[+1,
at
o:90"
:2n xi:+:1.5708A
Correct answer is 5.026 . For a given current density, the total current flowing through a cross section is defined as
r: [t- as where dS is the differenjtial cross sectional area vector having the direction normal to the cross section. So we
have
ffi : pdpdba,
(since the cylindrical wire is lying along z-axis)
Therefore the total current flowing through the wire (cross section) is
r: (?") . _ :
Page 1E7
@aod,Qa"1
Chap 3
/50\,^. Jo=o lr=,\ftYoaoao1 116xto-3 sz"
Electric Field in Matter
:;: x t#f ;-iJr,i :5 026A Correct answer is 6.25 . Since hydrogen atom contains a single electron (-ve charge) and a single proton (+ve charge). So, the dipole moment due to one atom of the hydrogen
will
be
p
: qd
where g is electronic charge and d is effective
length i.e. So'
L.6 x 10-1e C and d,:7.7 x 10-16 m p :(1.6 x 10-1e) x (7.1 x 10-16) the polarization in a material is defined as the dipole moment per
e:
and since unit volume.
ie
P:
Therefore
npwhere n is the number of atoms per unit volume.
":ll: il;:::H:,#'
p :(5.5 x 1025) x (1.6 x 10-1e x 7.1 x :6.25 x 10-e Cf m2 :6.25nCf m2
So,
Correct answer is 1.0177
10-16)
.
When an electric field E is applied to a material with dielectric constant €, then the polarization of the material is defined as
P : eo(e,- l) E So'
^tl ar
x 10-e :1.7655 x x40x10 ":1*0.0177:1.0177
'-
enE€o.u 8.85
6.25
x
10-
10-2
Correct answer is 2.
D :2P + P: Dl2 lf the polarization of a dielectric material placed in an electric field E is P
Given
, then the electric flux density in the material is defined as
D:eoE+P
:
or
eoE* Dl2
D:2eoE
(1)
and since the relation between the electric field, .E and flux densitv, D inside a dielectric material with dielectric constant e" is defined as
D:
eoe,E
So, comparing the result
with equation (1) we get,
Correct answer is 6. Energy on a dipole with moment
p in an electric field
Wn:-p. E :- (- 2a"f-Jay) . (7.ba,- a,)
:*(_3_3):.61
e": ). .E is defined as
Page 188
suL 3.3"&
Chap 3 Electric Field in Matter
Correct answer is 1.95 . Resistance of a conductor area is ^9
of length
R:JOS
where
I
and having uniform cross
o is the conductivitv of the conduc
Given the conductivity, the length of the conductor,
o I
side of the square cross section, and radius of the bored hole, So, the net cross sectional area is
o
,s:
: b x 106(Om) I :8m
:3cm r:0.5cm
area of square cross section(bar)
or
S
- area of circular cross section(hole) : a2_rf: (3)r_zr(0.5)r:(9 _i)cm,
The total resistance between the square ends is given
: R:-lo OD : $$L 3"r"$
1.948
x
(5
x
106)
10-3f,)
:
, [(r-i)
as
x ro"1
1.95mf)
Correct answer is 924.6 . The two materials of composite bar will behave like two wires of -R1,(resistance due to lead) and -R6'(resistance due to copper) connected parallel. As frorn the previous question we have the resistance due to the lead is l?; : 1'948mO and since the area of the cross section filled with copper is equal to the of the cross section defined by hole so we have Sn -4:4
Cross sectional area
cm2
I :8m
Length of the bar
"'_7-
and conductivity of the copper,
ur--
resistivity of
tle
copper
1
- 1J2t So the resistance due
Rc rhererore th"
lo.E
to copper is
: -l-:
8
*i'd;i;J.L;t-1Hm
;":
R:Rcllft,:q##+P :
s{}*- 3.2.ts
924.62
x
Correct answer is 8.9 . Given the radii of spherical shell
a
: 1cm:
10-6
e:
:
-"
g24.6p,e
as
0.01 m
b:2cm:0.02m The capacitance of a spherical capacitor having inner and outer radii a b respectively is defined as
n _ 4r€,€o _ 4r x 4 x 8.85 x 1 1i 11
la-a/
onT - onz
10-12
:
9.9 pF
F 3.2.1{ :
Correct ansifi/er is 7.26 . Since the dielectric has been removed from the portion defined bV so the composite capacitor will have the dielectric fillecl only in \
(f < d < ") fth portion
of the total capacitor and so the configuration can be treated as the two capacitors connected in parallel with each other. The capacitance of the portion carrying air(e" : 1) as the medium between the spherical shells 1 /1 4z'x 8.85 x 10-12 : 4x t'r: 47 ..x 4treo i 11 -a-6 onT - 0.-02
&
i
:
0.56
x
Page 189
Chap 3 Electric Field in Matter
10-12- 0.56 pF
The capacitance of the portion carrying dielectric(e": 4) as the medium between the spherical shells
c,:f;x
c
where C is the capacitance if no any portion of dielectric was removed already calculated in previous question. So we have
"r:tx Therefore the equivalent
8.9
x
10
12
:6.7 x
10
12
as
:6.7pF
capacitance of the composite capacitor is,
C"o: Ct+ C2:0.56+ 6.7 :T.26pF 3
3a.t2
Correct answer is 3.05
.
z(mm)
I Capacitance of a parallel plate capacitor is defined
/'1
as
e^9 d,
where
is the surface area of the parallel plates d is the separation between the plates
^9
Here, the three different regions
will be treated
as the three capacitors
connected in series as shown below
ct
c2
c3
So the capacitance of the region 1 is
Capacitance of the region 2 is Capacitance of the region 3 is
: ffi:2oooeo's c3 : o3#: 5ooeoS Cz
Therefore the equivalent capacitance of the whole configuration is
Page 190
1 1 1 1 lr 1 1 lt e,: q*6* 4 :;;sl25o-o+2ooo*-soo/
Chap 3 Electric Field in Matter
So,
C"c
:3.45 x
102es,9
The capacitance per square meter of surface area will be C"o'
s6L
3,1,13
:
QSo
:
3.45
x
102e6
:
3.05 nF
f
rn2
Correct answer is 143. Capacitance between the two cylindrical surfaces is defined as
" - t"(b6 Where
J
-
length of,the cylinder
0
--+
inner radius of the cylinder
b -+ outer radius of the cylinder Since, the medium between the conducting cylinders includes the die
layer(e,:4) from p:4cm to p - 6cm ahd air(e": 1) from p: to p- 8cm, so the configuration can be treated as the two connected in series. Now for the dielectric layer
(e,:
4) from p
:
4 cm
to p -
6
cm,
6
c
is
.n _2reoe,I _ 8zree "'-ln(614) -ln(1.5) and for the air medium (r" 1) from p: =
(l: 6 cm
to
p:8cm,
1
capacitancei
So, the equivalent capacitance of the configuration is evaluated as
111 e; : q:6 C"q
s$L
3.2.'14
:
rn(1.5) -r, h(aft) : -€?i€o -Zr€o
L43PF
Correct answer is 2.76 . The equivalent arrangement of the capacitor can be drawn in form of ci as below
For which the capacitances are calculated as below n _ eoSl2_gaq _ so x 10 x 10-4 _€o ., vrdl2 d - 4x10-3 -4
n _e,esSf2 _e"eoS
"'---d,lr-T--T _Seo
- _l I Cr -
n _ ao X 10 x 10-a _ eo "'-,_esSf2 d - 2x4xro3 -8 Therefore the equivalent
$acitance of the capacitor
is
€n Sen
c"c: Lz15
cc*ffe: f + #k T-T -T
Page 191
:?:2.T6pF
Chap 3 Electric Field in Matter
Correct answer is 0,
For a given polarization P inside a material, the bound surface chargt: density over the surface of material is defined as
PPs: P ' an where o, is the unit vector normal to the surface directed outward. while the bound volume charge density inside the material is defined
as
Ppo:-V ' P
Since the component of polarization of rod along g-axis is pu: 2y2+ 3. Sc,. the polarization of the material is p : (zt + 3) or. and the charge densiti. on the surface of the rod is po, : P . o,n
At y:0(top surface) At y: 5(bottom surface)
po:(2f*3)au. (-o,):-g po : (2f *3) a,. (aJ: bg
and since the polarization has no radial component so no charge will be stored on its curvilinear surface and so the total bound surface charge orr the surface of the rod is
ooror: p51,S+ pnS(S is the cross sectional I :-3,5*53^9:50^9
Qos:
areai
Now, the bound volume charge density inside the material is
pp,:- V . P --v . (zf +B)oo:-+a
I t
so the total bound volume charge stored inside the material will be
i
Qr,
:I :
So, Total bound charge Qboand:
t
3.2.15
oo,d,u
I\- ay) sd,y :- +{$l', -sos Qs-t
Q':50,S- 50S:0
Correct answer is 2.
0:n
r:1 m
z--\
o:anc "9
E
Electric field produced by the point charge at a distance
E: +1t€o ,t fr
r
is
Qoo-
'
,
So, the induced dipole moment in the neutral atom due .E produced by the point charge will be
to the electric fiekl
P:aE:=9!-oA 4T€sf and since the electric field intensity produced due to a dipole having momerrt p at a distance r from the dipole is defined as
Eaip:
ffi1r"ood,+
sinoaol
where d is the angle formed between the distance vector
moment p
r and dipole
So the field.produced by the induced dipole at the point charge is
Pqge 192 Qbap 3 EQetric Field in Matter
D
.t| 99 \ 2p _'\+treof ) :tAreorn tresf -
'
2aq (Areof
(0
rs
: r
bs shown in the
Therefore the force experienced by the point charge due to the field by induced dipole is
F
:
qEarp
:r"(hf
#
:2"(+ x 1o-e x e x ro'f x
4:2N a
So' $sa- 3.2.{7
Correct answer is 0.16
r"*
S:
.
Electric field intensity produced due to a dipole having moment distance r from the dipole is defined as Earp
:
ffi1,
0
a,
+
sir. 0 ae)
"o, is the angle formed between the distance vector
where d r and moment p So the electric field intensity produced due to dipole Pr at Pz is
n,:ffio,:]ffio,
Therefore the torque on Pz due
(o:r
to Pr is
l-p2xEt Taking the magnitude only we have the torque on
' ssl-
3,2"18
:::'
Correct answer is 0.324
P2 is
sin e0 "
:::;
f : :;],(,H)
.
Electric field intensity produced due to a dipole having moment p, distance r from the dipole is defined as Ear.p
:
ffi
1, "o"
o
a,
+
al,
si'. 0 aa)
where 0 is the angle formed between the distance vector r and dipd moment p So the electric field intensity produce due to dipole Pz at Pr is __g X 1ol x 2o' Ec :_ P, u2
-
4t
=2n._ fi72a'
:-|ffi
Therefore the torque on P1 due
to
l
,r:{
.
Pz is
T:prXEz Considering the magnitude only we have the torque on P1 is
y:
:2 x 10-e x /-9i::10-'g x 2\ \ 47160 I :3.24 x 10-z N-m : 0.324 pN,m
sot-
3.2"19
p1%2sin0
(0
:
rl7
Correct answer is 0. Since the spherical shell is of inner radius r : 2 m so region inside the spha will have no polarization and therefore the total charge enclosed inside tl shell for r
<2m will be zero.
Le.
:0
Qur"
According to Gauss law the total outwa,rd electric flux from a closed surface is equal to the charge enclosed by the surface and since the total enclosed charge for r< 2m is zero so the electric field intensitv at r:lm will be
P,age 1p3 Electurtc Fleld
Chap 3 in Mattq
zero.
l' 3-2.20 Correct answer is 0. i Since the total bound charge by a polarized neutral dielectric is zero as I discussed earlier. So for any point outside the spherical shell the total enclosed charge(bouna chargl) will be zero and as discussed in the previous i to Gauss law the electric field intensity outside the spherical shell will be zero. question, according
i'
So, for the surface
r: 7
at
a,ny point
:0 E :0
e"n"
is
Therefore the electric field intensity
l. 3.2.2{ Correct answer is -1.
As we have to find electric field at r: 5 m so we determine first the charge enclosed by the surface r : 5 m which will be equal to the sum of the volume charge stored in the region 2 < r < 5m and the surface cha,rge stored. at
r:2m.
Since for a given polarization P of a dielectric material, the bound volume charge density inside the material is defined as
.Ppu:-V ' P So the bound votrume charge density inside the dielectric defined region 2 ( r { 6m will be
Pp.:-v so the
. P(r)
:-#*(n*)
total bound volume charge in the region 2
e*: | 0,,d, : I:,-l :-ZOn[rll:-
in the
:_i
< r < 5 m is
x 4nf dr
60zr
Now for a given polarization P inside a dielectric material, the bound surface charge density over the surface of dielectric is defined as
PPs: P' an where a' is the unit vector normal to the surface pointing outward of the material. So the bound surface charge density pos
at r:2 m is
= P(r)' (-
"") charge over the surface Therefore the total bound surface
e*
:-* x 4trf :-$, 4n x 22 :-
(a^:- q) r:
2
m is
(for spherical surface
S: &rf) r:2m
40r
total enclosed charge by the surface r: b m is Q"n" : Qoo+ Qo":- 60zr - 40r :- 700tr So the electric field intensity at r: 5 m will be, So, the
E:h"%*:h*
=#h*
:-$*
Since, from the given problem, we have the electric field
at
r:
...(1)
b
as
//
" E:Lo to
Page 194
Chap 3 Electric Field in Matter
Thus, by comparing equations (1) and (2), we get
k:-l sol.
3,2,22
Correct answer is 27. Since the electric field intensity
at any point inside a conductor is zero, so the electric flux density at a distance r from the center of spherical conductor can be given as
r
D:l[0,Q
lfi7w' r> 1m
where Q : 3 mC is the total charge carried by the conductor. and since the dielectric material surrounding the spherical conductor permittivity €,:3, so the electric field intensity at a distance r from center of the sphere is
r( 1m l
2m
total energy of the configuration is
,,:+lo.
Ed,u
: Ll[',a, * f
&)G*r){+"r a,). f (#)(#e)e"n Lffi*{!f'ga,++I- i*} g{#l+1,.*t-+ :#{*.}**} l;l (3x10-3)2xgx10e 2
:2.7 x sol
3.t,23
L04
0,4
x#
J :27kJ
Correct answer is 9.
The electric potential at the centre of sphere will be equal to the work done to carry a unit charge from infinity to the centre of the sphere ( Iine integral of the electric field intensity from infinity to the center of sphere) i.e.
v:_[oe.
at
p,:0.6nCf
Since the sphere has uniform charge density
so the electric field intensity at a distance can be given as
I
o".,
E:l l#an, .\
I
p,Rt -
16""
r
m3 embedded in it from the center of the sphere
rlR r)R
where B is the radius of the sphere i.e. r? :
1
-F a/ 1f ^
So, the potential at the centre 9f .sphere
V:
:- I''"
E 3
r! *
:-#(
F
rti
will
be
Page l9b
, (where differential displacement is
d,l:
d,ra,)
t tt
#fu(h)'a'- [,',ffa' hl "[-i],: -#h[+]:,,
:&:#+r^_
:
5 x 0.6 x
10_
e
xex
: 9 volt
i :
1oe
(e,:2)
go:o.6ncf
lot. 3.2"*d Correct answer is 45.1 . consider the surface chi,rge density on the parailer plates is electric flux density between the plates is defined as
*
m3
ps so the
D : pron where a, is the unit vector normal to the surface of plates directed from one plate toward the other plate. Since permittivity changes from layer to layer, but the field is normal to the surface so electric flux densitv D will be uniform throughout the plate separation as from boundary condition. so the electric field intensity at any point between the parallel plates is
D : E: €o€,
Ps&n
2es(1
e,:2(I
+ 100o)
+
100a'z)
Therefore the voltage between the plates can be evaluated by taking the line
integral of electric field from one plate to the other plate
, :- I'
.
: *hI
"
:- I::Gtrftu').
ra"t
(dI:
d,a)
a-#.a (the direction or o is along a") :f,xfr>. #[,."-'(#)]i' :ftxhtt-ol :ffi Now charge stored at the parallel plates is
Q
:
:
br)(E
where ,S is surface area of the plates
p' X (0'2)
So, the capacitance of the capacitor is evaluated as
S:0.2m2
n_Q _p"x(0.2) 16e^ v \p"T) /80es : 4.bI x 10 rr : 45.1pF 7r
ssL
3.x"a$
Correct answer is 3.64 For the two wire transmission line consists of the cylinders of radius b and separated by a distance 2h (centre to centre), the capacitance per unit length between them is defined as .
: ---Jl9cosh-l (h/ u) Herc, 2h: 2 cm and b : 0.2 cm C'
Chap 3 Electric Field in Matter
: {;,(*1 "^+ffi*fi:ffi+#*
r
t
- lon . m
F
Prgc.
Ctlp
lqf
Eloc'trh Fleld
b.Matter
7tx2x8.85x10-12 :3.64 x 10 11 F/m (e,:21
C':
So'
I
So the charge per
cosh-'(1/0.2) unit length on each wire will be,
(%
'=i:i::^n"ilo 30L
3.2.26
Correct answer is 41.1
:
100
V)
.
Consider the oil rises to a height h in the space between the tubes. So, the capacitance of the tube carrying oil partially will be treated as two capacitors connected in parallel. Since the capacitance between the two cylindrical surfaces is defined as
"
1t:-
Where
In(bla)
I -+ Iength of the cylinder
o --r inner radius of the cylinder b - outer radius of the cylinder So the capacitance of the portion carrying oil (1": 1) as the medium between the cylindrical surfaces is n _ ..._2tre,eoh ""'-ln(3fl)
treoh
(":x"+1:2)
-ln(3)
and the capacitance of the portion carrying air(e,: 1) as the medi'n between the cylindrical surfaces is
- h) /1 _2neo(7 _ _,"Cn,
uanr
Therefore the equivalent capacitance of the tube carrying oil to the height
his C : Coal
Ca;,r
Since the energy stored
0 + h) :27r^-ol;l3f
in a capacitor is defined
W : |CW
where
as
7 is the applied voltage to the capacitor
So the net upward force due to the capacitance is given by
o _ dWt _7 rr2dc _l
'' -_aF_2,
m-2,
172
2tr€s
ET3)
and net downward force on the oil due to gravity
F : nlg :
(0.01gm/cm3)
x
will
r(b2
-
be az)h
xg : 0.01 gm/cm3
mass density
:ffi Since
x
r(9-
1)
x
1o-6
x hx s :o.o8trhl
in equilibrium both the upward and downward forces are equal
So,
0.08trhg
0.08r'hx(e.8)
:Lr'l{(;l :|xQx
h:+*
"_Xtr-"i-
(z x 103)2x2x8.85x10-12
:4.11 x
:
103)'zx
41.1u,m
0.08x9.8xIn(3) 10-5 m
Correct answer is 0.0796. From the symmetry associated with the charge distribution the electric field must be radialrv directed. As, there is no charge enclosed
r:2m
by the surface
so we get
hie
t0?
Ctrp s ENechlc
FlihlhMette
E,:0
forr<2m Now from the conductor-free space boundary condition we have the surface charge density on the boundary surface defined P"
where space.
:
as
€oE,
E, is the normal component of the electric fierd intensity in the free
r:2m Ps : E':0
So the charge density
at
Therefore the total charge
is given
wil
as
is
be concentrated over the outer surface which
A : s : t C/m, ^ : 4;F ^, : d* fi
ps2
0.0796C/m2
Correct answer is g. As the dielectric slab occupies the region r ) 0 and the electric field in the free space is directed along o" so, the field will be normal to the boundary surface, z: 0 of the dielectric slab. so from the boundary condition the field normar to the interface of dielectrics are related a^s €,€sE;
E'
: eoE
(where ^8, is the field inside the dielectric)
: *:Y:2o'
(e":
So, the polarization inside the dielectric is P : (e - eo)Eo : (5uo _ eo)Eo: Since, from the given problem, we have
P: So, we get
sol-
3.2.29
lce
g€s
5)
a,
oa,
ft:8
Correct answer is 2g3. Consider the parallel sheets arrangement as shown in the figure. + +
5 nC/m2 +
- 5 nC/m2
+ +
l
I
Electric fierd intensity at any point with charge density ps is defined as
p
due
to the uniformly charged
prane
u:#o^ where a, is the unit vector normar to the plane directed toward -* wv"wru point l, e is the permittivity
of the
medium.
p
and
t=r
So
Page 19E
the field interrsity inside the dielectric due to the left sheet will be
Qhap 3
(w:
o,:5_x]91p,1
Electric Field in Matter
au)
and again the field intensity inside the dielectric due to right sheet will be
n,:-j4_J{{(-o,) :
*#:"
so the net field intensity inside the dielectric
E
:
Et*
(a": will
an)
be
E2:L9o,
Since the field intensity is unifor"m inside the dielectric So potential difference
between the plates will be directly given as
V
:
E
x
(distance between the plates)
-------i=' --5x10-n.,., +ts0
:2.824 x
3.2"s0
102
Volt
:
283
kV
(e
Correct answer is 8.85 . Assume that the surface charge densities on the plates is + field intensity between the plates will be
Pm so
:
4eo)
the electric
E:Z and the potential difference between the plates will be given by V : E x (Distance between plates)
5
x 103:(e) x (o.s x
1o-'?)
Therefore the surface charge density is (8.85 x 10 ''?)I15
Pt:6-'''
3.2.3{
x
t03)
:8.85p,C
Correct answer is 1.10 . Since, the wire is coated with aluminum So,the configuration can be treated as the two resistance connected in parallel and therefore, the field potential will be same across both the material or we can say that the field intensity will be same inside both material. l.e.
where
ort where
: ErE"r E"t
J*
Eot
Field intensity in steel Field intensity in aluminum.
:Ja
O31
Oat
..f1 --+
current density in steel current density in aluminum
Jor
-
o', -+ conductivity of steel ool + corductivity of aluminum So, we get,
J,t_o"t- 2x106 Jot - oot - 3.8 X 107 --119 Ja : 19J't
Now, the total current through the wire is given
...(1) as,
I : J"r(nat)* J"t(nb2 - na') where
o -+ cross sectional radius of inner surface (steel wire)
b -+ cross sectional radius ofouter surface (with Since, thickness of coating is
coating)
: 2 x 70-3 b : a* 7: (2 x 10-3)+(2 x 10-3): (4 x 10-3) t
So, Therefore, we get,
80: J"tr(4 x 10-6) + J"tfn(to x 10-6)_a.(a x to{)l 80: J"ttr(A x 10-b) +rcJ"rlr(tz x 10-6)l (from eq.(1)) *, : ,rtTlb= : 1.10 x tol Afm2
t
b
l,1232 t
or, So,
Correct answer is 67.8
.
As calculated in previous question the electric field between the two dielectrics having surface charge densities p" and -p.s is
E:+
where e is the permittivity of the medium between the sheets. So electric field
and electric field in slab 2
Since
: *: h n, : ?: h
in slab 1 is
A,
is
the electric field between the sheets is uniform so the potential will be V :EE x (distance) : ,',(1m)+ rr(zm)
difference between the plates
:fi{D+h@:*
_
0.6 x 10-e 8'85 x 10=rt ****r<>t*X*'1.*
:
67'8
Volt
page 199 Erectri.
r.ru r"*oli.l
Plse ?99 Chap
sol.urloNs 3,3
-,
l'
Hectic Fie$ lq.Mattq ''-i,,;
'.r,,
sol
3.3"r
Option (B) is correct. According to boundary condition the tangential components of electric field a,re uniform l.e.
fl1 : E2r- fit,
...( 1)
but the normal component of electric fields are non uniform and defined as €rEn: €zEzn:4Esn (Given) e1 e3 Since -
So,
Enn:Er,*
Er^
...(2)
and as the net electric field is given by E : Ett E, (sum of tangential and normal component) Therefore by combining the results of eq (1) and (2) we get
Et: Es* Ez
i
ir i
t
fl
sol-
3.3.2
Option (A) is correct.
sol.
3.3.3
Option (C) is correct.
3()L
3-3"4
Option (D) is correct.
sol-
3.3,5
Option (A) is correct.
$oL 3.3.6
Option (A) is correct.
$or-
3.3.?
Option (D) is correct.
$oL 3,3.8
Option (B) is correct.
sol.
3.3.9
Option (C) is correct.
sol.
3.3.'ro
Option (A) is correct.
soL
3"3"11
Option (B) is correct.
sol.
3.3.'12
Option (C) is correct.
EOL 3.3.{3
Option (A) is correct.
sol- 3"3,{4
Option (A) is correct.
sol.
Option (A) is correct.
3.3.15
soluTloNs 3,4
Page 201
,
Chais
nnctrtc Fteld in Matter
Option (D) is correct. The capacitance of a parallel plate capacitor is defined c : fu-A: 8.85 x 1o*12 x lg-a :8'85
:-ff*-
x
..
as
. \rl
4*
,
1o-r3
!
\,*\
The charge storedpn the capacitor is
Q:CV : 8.85 X 10-13 :4.427 X 10-13 Therefore, the displacement current in one cycle
I:8:fQ :
4.427
x
10-13
x 8.6 x 10e:
(/:
1.b9 mA
Option (C) is correct. The electric field of the EM wave in medium 1 is given
3.6 GHz)
as
Et:2a"-Zou*\a" since the interface lies in the r: 0 plane so, the tangentiar
and normal
components of the field intensity in medium 1 are Ett :- 3au* a,, and Eh: lq,,
Flom the boundary condition, tangentiar component of electric fierd is uniform. so, we get the tangential component of the field intensity
medium 2
in
as
Ezt : Eu:- Saul ay Again from the boundary condition the for normal component of electric flux density are uniform l.e.
or
Dn: e1E1n
:
Dz, €2E2n
So, we get
1.5eo2a, :2.5eoE2n
or
Ern
- {a,:7.2a,
Thus, the net electric field intensity in the medium 2 is Ez : Eztl E2n :- Bau* a"* I.2a,
l'
3"4"3
Option (D) is correct. The surface charge density on a conductor is equal to the electric at its boundary.
ie l, 3.44
":i;'i;:o:'r:r-,,X2:141
fl'x (e
x to-scfmz
density
-
80e,)
Option (B) is correct.
The configuration shown in the figure can be considered as the three capacitors connected in parallel as shown below
.l
Page 202
Chap 3 Electric Field in Matter
Now, consider the distance between the two plate is d and the total surface area of the plates is S. So, for the three individual capacitors the surface area is S/3 and the separation is d. Therefore, we get'
n eo61(s/3) ", -- d
-
"
eoez(!l3) d
n vt:--d
sos3(s/3)
-
Since, the three capacitance are in parallel So, the equivalent capacitance is
C"s: Ct* Czl
:
Ct
r0sr(^9/3), €o€2(S13), 60s3(s/3)
d
T-A---T
: (#X..€t") : (u,*?*")" it
.?"6,$
(r
: #)
Option (D) is correct. The electric field is equal to the negative gradient of electric potential at the point.
E:-YV
l.e.
Given, electric potential
:4r*2 E --4a,Yfm V
So, .*1.
*.4"$
Option (B) is correct. The angle formed by the electric field vector in two mediums are related
tanor
as
gr
tana2 -- e2 So, for the given field vectors we have,
tan60":: 3 lano.2 J Z :1 oz : tan-1(1) :45'
tarta2
or *,39 3"4.7
Option (A) is correct. The tangential component of electric field on conducting surface is zero (since the surface conducts current) so, under static condition we have,
or
-V
V : E:0 V : constant
i.e. the conducting surface is equipotential. So, (A) and (R) both true and R is correct explanation of A.
f..f.8
Option (A) is correct. since, the electric field is incident normal to the slab. so, the electric field intensity(E,) inside the slab is given as
€Ei :6ofio
Page 208
.
Chap 3
Electric Field in Matter
,r:#:ro" Therefore, the polarization inside the slab is given as Pt: EoX"Et where X" is electric susceptibility defined &s X" : €, _ I
P,
lt*e
:
eo(3
-
L
So, we have
l)Eo :4€se,,
Option (A) is correct.
to both the line charge and concentric circular conductors, the equipotential surfaces are circular (cylinder) i.e. concentric equipotential Due
lines.
The flux lines due to both the configurations (line charge and concentric circular conductors) are in straight radial direction. I
er,ls
Option (A) is correct. Capacitance of 1"' plate is given
as
n -€St-e(txt) "t- d -----{-:a The capacitance of
2od
€
plate is
n "z--€Sz-e(2x2)
4e
d -----tr-:T
So, the ratio of capacitances is
' an
11
fr:+ Option (D) is correct. Consider the dielectric material with permittivity e1 is replaced by a dielectric material with permittivity e2. The capacitance of parallel plate capacitor is defined as
c:+
i.e. the capacitance depends on the permittivity of the medium and so, due to the replacement of the material between the plates the capacitance changes.
Now, the charge is kept constant l.e. OT,
Qt: CtV:
Q, CzVz
so, due to the change in capacitance voltage on the capacitor changes and therefore the electric field intensity between the prates changes. The stored energy in the capacitance is defined as
w:g2C As total stored charge Q is kept constant while capacitance changes so, the stored energy in the capacitance also changes.
Thus, all the three given quantities changes due to the replacement of material between the plates.
li
soL
Page 204
3"rt"t2
Chap 3
Option (C) is correct. According to continuity equation we have
v.J:-Qp 0t
Electric Field in Matter
As for electrostatic fien
*:0
so, we get
V.J:0 sal-
3"4.{3
$sL
3.4.'14 Option (C) is correct.
$0L
3.4.15
Option (B) is correct. Since a conducting surface is equipotential so no electric field exists tangential to the surface and therefore the electric field lines normal to a conducting surface boundary.
ssl.
3,rl,"lS
Option (B) is correct' Surface resistance of a metal is defined
Option (A) is correct. Electrostatic fields onlY. Surface or sheet resistivity is defined as resistance per unit surface area' the unit of surface resistivity is Ohm/sq' meter'
as
R"={#:rc
i
,l
so, as frequency
I I
I
(/)
increases the surface resistance increases.
sgl-.3,4,17
Option (A) is correct. when we determine force using method of images then in this method, cond,ucting surface is being removed and an additional distribution of is being introcluced symmetrical to the existing charge distribution.
sOL 3.4"{8
Option (B) is correct. The conducting surface is equipotential and since the potential at infinity zero so, the potential every where on a conducting surface of infinite rs zefo.
Since the conducting surface is equipotential so displacement density on conducting surface is normal to the surface' So A and R both true but R is not correct explanation of A' $01
3.4.1e
Option (A) is correct.
Capacitance, C:5PF:5
x
10-2F
Charge on caPacitance, 6 Q :O.IPC : 0'1 x 10 C The energy stored in the capacitor is defined as (0.1 x to-6f *" :4 - 2x 5 x 10-12 :lmJ - 2C _
3.d.zo
Option (C) is correct. Consider the charge of 1C is placed near a grounded conducting plate at r distance of 1m as shown in figure.
I
t
,lC
1C
t
a
t
F
t t
+
I
i' '*l
.
Chap
-----i-----1
E
2m
C
a
r r
using image of the charge we have one negative charge opposite side of the plate at the same distance as shown in the figure and the force between them is
i k &
i
-1 - 16t-1* "' r:G)iJ- 4"-e'f C;W:
F
'Hr
$
r I
Negative sign indicates that the direction of force is attractive.
r
Fringing field has been shown below in the figure
l.*'
Option (B) is correct. Ilinging field
I
i
r t
i
ftinging field
The capacitance of a parallel plate capacitor is given
as
_ eoe, C-'''A il
It is valid only
when the fringing is not taken into account. Now; the fringing field can be ignored only when the separation d between the plates is much less than the plate dimensions. so, for the fringing field taken under consideration, Af d, is tending towards infinity.
J
3J'22
Option (B) is correct. The capacitance of a solid infinitely conducting sphere is defined.
C:
as
treoR.
where .R is radius of the solid sphere.
l'3.4.?3
Option (A) is correct. The electric potential produced by a point charge Q at the a distance r from it is defined as
tl-.4 v where u is permittu"t*n*'io"medium. so, the electric potential produced by the point charge *10pC at the centre of the sphere is
A ' - Gar
rl-
t
Electric Field in Matter
F
t t
P4e 2[E
1oxlo-6 an6o@t10T
(Given
r:5cm)
,,1
As the surface of sphere is grounded so' the total voltage on the capacitor will be equal to the potential at its centre as calculated aboreNow, the capacitance of the isolated sphere is defined as
Page 206
Chap 3 Electric Field in Matter
C :Atrea where a is the radius of the sphere. Therefore, the induced charge stored
the sphere is given
as
ena: CV _ (z x
:4x $0L
3"4.24
:t+r*r)fff$ ro-'?)
x
x
(10
1o-6)
(Given
(r x ro-)
a:
2
10-6C:4p,C
Option (C) is correct. E : Eosinat Given electric field The conduction current is defined as J" : o(E): ogosinat where o is conductivity and ,O is electric field intensity' and the displacement current density is
, : dD -AE E:, E : eno(wcosut) :
,la
So
sol
3.4"25
the phase difference between J. and
ernosi"($ "/a
- rt)
is 90''
Option (B) is correct. Method of images are used for the charge distribution at a distance from grounded plane conductor.
sol-
3,4.*S
Option (C) is correct. Given
C :0.005pF:5 x 10-eF V :500V
Capacitance of condenser,
Supply voltage, €, Permittivity of oil, befQre immersion is in condenser stored So, the energy
:2.5
w:+cV:Ixsx :6.25 x
(immersed
to-'gx(soof
10-4 J
After immersing the condenser in oil the capacitance changes while the charge remains same.
i.e.
Qafterinmersi,on
:
Qb"fur"i**"rrion
:2.5 x
: (5 X 10-'gX500)
10-6 Coulomb
The capacitance of the cond.enser after immersion is
: €,C : (2.5X5 x 10-'g) :1.25 x 10-8 F Therefore, the stored energy in the condenser immersed in oil is Coft",im
. i
ssL
3,4,2?
"r"ion
d :J?-+*:2.5 w:^o 2C1uft"ri**",sion) 2(1.25 X 10-")
Option (B) is correct. Given,
x
1o 4J
Capacitance,
:3pF:3 x 10-6F I :2p,A:2 X 10-6 A
C
Current,
Page 20?
Chap J Electric Field in Matter
Charging time, t:6sec So, the total charge stored on capacitor is Q : Charge transferred
: It :
(2
x
10_6X6)
Therefore, the voltage across the charged capacitor is _(2 x 10-6)(6)
_a -e -
:4
-;10=-
Volt
Option (C) is correct. Given, the total charge on capacitor : V (1) Electric field between the plates will be given
as
E:-YV which is independent of permittivity of the material filled in capacitor so .O
(2) The
will be constant
displacement flux dehsity inside the capacitor is given as
D:eE (3)
As ,E is constant while permittivity is doubled so The charge stored on the plates is given as where
D will
also be doublecl.
Q:CV
I/ is constant but capacitance c will be doubled as it
proportional to the permittivity given
is directh,
as
g:rt -ad So, the charge on plates
will be get doubled. As discussed already, the capacitance will get doubled. Therefore, the statements 2 and 4 are correct.
()
Option (B) is correct. since, resistance doesn't store any energy. so, the energy stored in the is only due to inductance and given as
c'il
w:|r'P where ,L is the inductance and .I is the current flowing in the circuit. At the fully charged condition, inductor is short circuit and therefore. current through the circuit is
I So, the energy stored
w tol
3.4.30
:X: T: toa
in the field (in the inductor) is
: |{0.+)(1of :
2o Joules
Option (C) is correct.
The normal component of erectric flux density (D) across a dielectric-
dielectric boundary is given
as
D1n- D2n: p" where p, is the surface charge density at the interface.
Option (D) is coirect The equivalent capacitahte bf series cpnnected capacitance has the value less than the smallest capbcitance here the smallest capacitance is ca so the total capacitance is less then Co
CrlCu C*= Cu sot.
p.4.ss
Option (C) is correct. Given, the electric field intensity in rnedium
1.
Et:5a,-2o4*3a" Since, the mediuril interface lies in plane
z:
0.
So; we get the field compdnents as
and
Eit:\ar-2an En :3a"
Now, From the boundary condition for electrii field we have
Eu: Ezt 4E1n :6rB;, So, the field coi,nponents in medium 2 are
Ezt: Eu:5o;-2ou
Ern:ZE*:go" Therefore, the net electric field intensity in medium 2 is givdn as
Ez : Ezt* E2n : 5a, - 2ar* 6a" So; the z-component of the field intensity.in rnedium 2 is
:6a;
Er,
3"4.3e Option (A) ls correct. Electric flux density, D : lC/mz Relative permittivity, Cr =O Since, the normal .componerlt of flux density is uniform at the boundary surface of two medium so, the flux density insid'e the slab is
D:LCfmz Therefore, the polarization of the slab is given
P
3.4.4a
as
=(+)o :tX 1 :0.8
Option (B) is correct. The capacitance of a isolated spherical capacitor of radius
C:
-R
is defined
as
treoR
Since the two spheres are identical and separated by a distance very much larger then R. So, it can be assumed as the series combination of crprcitu,rr"es.
Therefore, the net capacitance between two spheres is given l.e.
3.4.4{
c:##;:ffiW#:2,reo'
as
Option (C) is correct. For steady current in an arbitrary conductor the current density is given
r_I "-A
as
Page 209
Chap 3 Electric Field in Matter
and since 1 is constarrt So, J is constant and therefore V x J:0 is true' So, the current density is solenoidal' i'e' Assertion (A) false' The reciprocal of resistivity is conductivity' i'e' Reason (R) is
Page 210
Qhap 3 Electric Field in Matter
sol-
3.4,42
Option (A) is correct. Sinceo the displacement current density is defined
as
to:#
So,itisgeneratedbyachangeinelectricflrrxandthereforethedi is current has only A'C. components as derivative of D'C' components A' i.e. A and R both are true and $ is correct explanation of sol-
3.rt.43 Option (B) is correct. €,
Dielectric constant,
:5
D :2Clm2 FInx density, So, the polarization of the medium is given as
l\l : * x 2 :7.6Clmz p - :(€' a \€rl-
80L 3,4.44
Option (C) is correct. as below The ohm's law in point form in field theory is expressed (For constant V:RI
u:fin
wherelislengthintegraland,4isthecrosssectionalarea.So,weget
D:PJ
E:Lo l.e.
sol.
3.4.45
J:oE
Option (C) is correct. Displqcement current density is defined
as
Jo:rQ -" At and the conduction current density is defined as
J":oE
for a dielectric e must be larger while conductivity must tend to zero. So, we get
Ja
))
J"
i.e. displacement current is much greater than conduction
sol.
3"4.46 Option (B) is correct. Conduction current,
/.
:1A
Operating frequencY,
f :50H2
Medium PermittivitY'
e -
PermeabilitY
current'
c0
lt : lto o : 5.8 x
10 mho/m ConductivitY' densit5 The ratio of conduction current density to the displacement current
is
J"o
Jo-a€
I
I"IA *_
Io/A
o
(,4 is cross sectional area)
Q€
ra : ffr":
,"r5
tl6*(1) :
Chap 3
+.a
x
1o-11A
sot
3.4"4?
Option (D) is correct. when there is no charge in the interior of a conductor, the electric field intensity is zero according to Gauss's law the total outward flux through a closed surface is equal to the charge enclosed. Now if any'charge is introduced inside a closed conducting surface then an electric field will be setup and the field exerting a force on the charges a.nd making them move to the conducting surface. so all the charges inside a conductor is distributed over its surface. Therefore the outward flux through any closed surface constructed inside the conductor must vanish. A is false but R is true.
sol-
3.4"d,A
Option (D) is correct. when the method of images is used for a system consisting of a point charge between two semi infinite conducting planes inclined at an angle /, the no. of images is given by
'nr:/369"-r\ \@ I
Here the angle between conducting planes is So,
N:3
/:
gg'.
and since all the images lie a. a circle so we have the image charges as shown in figure.
sol
3.4,49
Option (B) is correct. Consider the two dielectric regions as shown below. Et: a'
Er:2a,
Since the field is normal fields are,
Region
et:
1
1eo
Region 2
e2:2es
to the interface so, the normal components of the
En : 7- and E2n: From boundary condition we have
)
Page 211 Electric Field in Matter
€rEu-
Page 212
€zEzn
:
P"
(where p, is surface charge density on the interface)'
Chap 3 Electric Field in Matter
G,X1) -(zeo)(z)
: p"
P,:-3€o $ct
3.4"s0
Option (B) is correct. The'stress is called the force per unit area which is directly proportional to the electric field intensity and electric field intensity is inversely proportional to the permittivity of dielectric material. l.e. So, ratio of stress is
noL€
:s :.rllo 4:rl?o llSes Ez - lle *****(*{<**t<*
CHAPTER 4 MAGNETOSTATIG FIELDS
Magnetostatic fields are produced by the moving charges, (i.e., charges that are moving with constant velocity) or constant current flow. In this chapter we shall study the subject of magnetism that includes following topics: o Concept of magnetic flux, magnetic flux density, and magnetic field intensity.
o
Biot-Savart law, which defines the magnetic field at a point due to a differential current element.
r o
Ampere's circuit law, which defines the magnetic field in a loop.
o 42
Magnetic field intensity due to various current distributions: straight line current, square current loop, solenoid, etc. Scalar and vector magnetic potential.
MAGNETIG FIELD CONCEPT Steady magnetic fields are also called static magnetic fields or magnetostatic fields. The two opposite ends of a magnet are called its poles. If a magnet is floated freely, one pole will point towards the north pole and is called the north pole of the magnet, denoted by N. The other pole is the south pole, denoted by 5, as shown in Figure 4.1. Following are some important,terms related to magnetic field.
4.2.1
Magnetic Flux Magnetic flux is the group of magnetic field lines emitted outward from the north pole of a magnet, as shown in Figure 4.7.It is measured in Weber and is denoted as @.
I
\ I
Fignle 4..1.: Magnetic Flux Lines from a Magnet
Page 2L4
4.2.2
Qhsp 4 Magpetostatic Fields
Maguetic Flux DensitY Magnetic flux density is the amount of magnetic flux per unit area of a section? perpendicular to the direction of magnetic flux. It is a vector quantity and also known as the magnetic induction. The unit of magnetic flux density is weber per squared metre (wb/m') or Tesla (?). It is denoted by B.MathematicallY'
u:
*@^
the where djD is a small amount of magnetic flux through small area d^9 of to normal vector unit is the a, and section perpendicular to magnetic flux
thesurfacearea.Theaboveequationmaybealsoexpressedas
o:In'as Js
i.e. the magnetic flux through any surface is the surface integral of the normal comPonent of B'
4.2.8
Magnetic Field IntensitY The degree to which a magnetic field can magnetise a material is called Magnetic field intensity or Magnetising force. It is a vector quantity and d.enotedbyll.ItsunitisNewtonperWeber(N/wb)orAmperepermetre
(A/*).
4.2.4
Relation between Maguetic Field Intensity (rr) and Magnetic Flux Density (B) The magnetic field intensity is related to the magnetic flux density as
B : p,H: FIF,H
where, pl is the permeability of the medium, ps: 41r x 10-7H/m is the permeability of free space, and. p, is the relative permeability of the medium'
4.3
BIOT.SAVART'S LAW
when electric current flows through a conductor, it produces a magnetic field. Biot-savart's law gives the magnetic field intensity produced due to a current element. According to the Biot-savart's law, the magnetic field intensity d.f1 produced at a point P due to a differential current element 1d-t, shown in Figure 4.2, is given bY dH
__
u##s
As we know, the magnitude of cross product (I d.L x a6) is equal to so, in vector form the magnetic field intensity can be given as
(l
d,l sin
a)
d'H:t#
_ IdL x_R 4nR"
where o6 is the unit vector along the distance vector tr|. Hence, the total magnetic field intensity produced by the current carrying conductor is
H:|ry# fIdLx R :J,.-ffi
. .(4.1)
Page 215
Chap 4 Magnetostatic Fields
F'igurc
.tr,.2:
Illustration of Biot-Savart's Law
Direction of Magnetic Field Intensity As the magnetic field intensity is the cross prod.uct of Id,L and .fi| so, its direction can be determined by either "Right-hand rule,' or *Right handed screw rulett.
1.
Right-Hand Rule : The direction of dH can be determined by the righthand rule with the right-hand thumb pointing in the direction of the curtent, the right-hand fingers encircling the wire in the direction of dll as shown in Figure 4.3(a).
2.
Right Handed screw Rule : The direction of d,H can be determined also by the right handed screw rule, with the screw place along the wire and pointing in the direction of current flow, the direction of rotation of the screw in the direction of. d,H as shown in Figure 4.3(b).
dH
Irigurr" '1 i} : Determination of Direction of Magnetic Field Intensity using (a) Right Hand Rule, (b) Right Handed Screw Rule
13.2
Conventional Representation of (H) or Crurent (1) The direction of the magnetic field intensity (11) or current (1) is represented by a small circle with dot O if fI or 1 is out of page or by a small circle with cross sign a if 11 or ,I is intci the page as shown in Figure 4.4. H( or4 is out
d l,r )
II(or
1) is in
a (1,)
l'igur:e 4.4 : conventional Representation of
a
(or
r) (a) out ofthe
page (b) Into the page
Page 216"
"
'
4.4
AtlPEREtSjCIRCUlf;AI,:'l*tlr' ' r.,
Chap 4
'
,
!.
l
,
Magnetostatic Fields
, ;
,; ,,;
'
Ampere's circuital lau states that the line integral of the magnetic field intensity around any cloSed,,nq!\.is Eqp{ to the {irect curlF+t" enclosed,,by the path. The closed path oh which Ampere's law is applied i's known as Amperian Path or Amperitin'tbop: F6llotvirtg are the two matherriatical forrns of Ampeie's circuital law:
Integral Form of Amperets Circuital Law
If the total current enclosed by a closed loop .t ,be ,I as shown in Figure 4.5, then from Ampere's circuital law tLe hirc integral of magnetic field intensity I[ around the closed loop .t'is equal to 1, ire. ' , " r1 r'r'
[n. at:t
Jr.
Il.igrrrr.:,1.i.l:IllustrationofAmpere'sCircuitalLaw.
Differential Fsrm of Amperets Circuital Law In differential form Ampere's circuital law is defined
as
YxH:J i.e. the curl of the magnetic field'"intensiby"(.Ef) is equal to the current density (J) pJ the point in space. ff at agy point, no current density exists, then the curl of the magnetic field is zero at that point. 4.5
MAGNETIC FIELD INTEIIISITY DUE TO VARIOUS CURRENT IilsTRtBUT|ONS
Similar to the different charge distribution discussed in previous chapters, we can have three types of current density distribution given as: 1. Line current density, 1, given in Ampere, 2. Surface current density, I{, given in Ampbre per meter (A/m), and 3. Volume current density, J, given in Ampere per squared metre (A/m'z) These current densities are related to each other as
IdL=KdS=Jdu Thus, in terms of the distributed current sources, equation (4.1) becomes
H:II4#
,:ly# H:lw
(Line current) (Surface current) (Volume current)
Now, we obtain a more generalised explUnion.Sf ,some typical ,€urrent distributions.
Page 217
Chap 4 Magnetoetatic Fields
Iegnetic Fteld Intensity due to a Stralg[t Line Current Consider a straight current carrying filamentary conductor of finite length ,4.B located along the z-axis as shown in Figure 4.6. The current flows from point ,4 to point B. The magnetic field intensity produced due to the straight line current is given by
I, ,}, : fr(roscz - cosal)c6
[email protected])
where or and a2 tuto the angles subtended at point P by lower end ,4 and upper end B respectively.
I(into the page)
Figrrrc
"1.1i:
Magnetic Field Intensity due to a Straight Line Current
Magnetic Field Intensity due to an Infinite Litre Current A special case to the above expression is found when the conductor is infinite in length. For this ca.se, point A is at (0,0, - oo) while B is at (0,0,"o). So, the angle subtended by its lower a,nd upper ends respectively becomes i (}r : l80o 0z :0o Therefore, equation (4.2) reduces to
and
rt',r-- I ^ wrrd This expression can also be derive using Ampere's circuital law.
Magnetic Field Intensity due to a Square Current Carr
'yrng
Loop
in the z: 0 plane and carrying in the anti-clockwise directionas shown in Figure 4.7. The net magnetic field internity at the origin due to the square current carrying loop Consider a square loop of side 2a located
a current
r
is given by
H
:fr1o. 7ra
Page 218
Chap 4 Magnetostatic Fields
2a l;
igirr*r
.1,.7:
Magnetic Field Intensity due to Square Current Carrying Loop
1.5.4 Magnetic Field Intensity due to a Solenoid Consider a short solenoid of length ,t and radius a as shown in Figure 4. . It has n turns per metre of its length and it carries a current 1. The section of solenoid is shown in the Figure 4.8(b). Magnetic field intensity point P due to the solenoid is given by
n : !posa2-
(a)
cosc,ll&,
(b)
Figurr,r,l.li; (a) Solenoid with n T\rrns per Unit Length, (b) Cross Section of Solenoid
4.5.5
Magnetic Field Intensity due to an Infinite Sheet of Current Consider a thin infinite current carrying conductor plane having a uniform surface current density ,K. The magnetic field intensity due to the infinite sheet of current is given by
H:;Kxa" 1
where a, is the unit vector normal to the surface directed towards the point of interest.
4,6
MAGNETIC POTENTIAL
Just like an electric potential, we can define a potential associated with magnetostatic field. In fact, the magnetic potentials are of two types:
1. 2.
Magnetic Scalar Potential, and Magnetic Vector Potential
Scalar Potential
Page 219
Chap 4
From Ampere's law we know
Magnetostatic Fields
YxH:J If the current density J is zero in
V
some region of space, then we have
x.EI:0
Since the curl of magnetic field intensity is zero, so we can write the magnetic field .EI as the gradient of scalar quantity as
H:-YV*
'..(4.3)
where, 7- is called the magneti,c scalar potential.lts unit is Ampere. Equation (4.3) can be expressed in integral form as
(v^)no:- Juf'n .
at,
Following are some important points related to magnetic scalar potential:
s:r,,rscalai
tlb 63i,
,lofiir*eryte.
Magnetic Vector Potential FYom law
of
conservation
of magnetic flux density, we know that
the
divergence of flux density is zero. i.e.,
V.B:0 Since, the divergence of the curl of any vector is zero. So, vector expressed as the curl of another vector field, (say ,4), i.e.
B
can be
B:Y xA The vector field ,4 so defined is called the "uector magnet'ic potent'ial".Its unit is weber per meter (Wb/m). Magnetic vector potential satisfies the Poisson's equation, i.e. y2
A
:_
LtoJ
***********
EXERCT$E 4.1
Page 220
Chap 4 Maguetostatic Fields
M3* 4.t.{
Magnetic field intensity .E[ exists inside a certain closed spherical The value of V ' .EI will be (A) 0 at each point inside the sphere.
(B) 0 at the center of the sphere only. (C) 0 at the outer surface of the sphere only. (D) Can't be determined as .E[ is not given MeQ 4.1.2
A circular loop of radius a, centered at origin a"nd lying in the rg carries current 1 as shown in the figure.
The magnetic field intensity a the centre of the loop will be
(A)
*"" Q) *"" trac& 4.1"3
@)
-*""
@)
+",
In the free space a semicircular loop of radius a carries a current 1. will be the magnitude of magnetic field intensity at the centre of the (A) ,,4 @)+
I
Q)*
@+
Co'nrnon Data For Q. 4 and 5 : A long cylindrical wire of cross sectional radius distributed over its outer surface. ilca
4.1"4
.R carries a steady
Magnetic field intensity inside the wire at a distance
r(<
.R)
current
from it's
will be (A) non uniform (B) zero (C) uniform and depends on r only (D) uniform and depends on both r and R
axes
*{{:a 4-1.5
The magnetic flux density outside the wire at a distance r(> .R) from it center axes will be proportional to
(B) Ilr (D) rlR
(A) " (C) rlR
Page 221
Chap 4 Magnetostatic Fields
Two infinite current carrying sheets are placed parallel to each other in free space such that they carry current in the opposite direction with the same surface current density. The magnetic flux density in the space between the sheets will be (A) zero
(B) constant (C) Iinea,rly increasing from one sheet to other (D) none of these I
F.
1..r.7 i
l.al-8
In a spherical co-ordinate syslem magnetic vector potential at point (r,0,4) is given as A : l2cos|on. The magnetic flux density at point (3,0,2r) will be (B) 0 (A) 4aa (D) 36oa (c) 4a
An infinite plane current sheet lying in the plane g: 0 carries a linear current density K: Ka," Alm. The magnetic field intensity above (y > 0) and below (g < O) the plane will be (A)
g>0
y<0
f",
-Ta,
K
K
K
(B)
(c) (D)
2e'
io,
-2Ka" K
2Ka,
K
q%
-Tav
Comrnon Data For Q. I and 10 : In a Cartesian system, vector magnetic potential at a point (r,y,r) is defined as
: 2t ya, + 2f rao - Sryza"wb/m I +t.e The magnetic flux density at point (1,-2,- 5) will be (B) 40o" * 80a, * 6a, wbfmz (A) 40a" *6a,"wbfml (D) 80a" - 6a"wbfm2 (C) -40a, - 804, - 6a"wbf m2 I rlt.'ro The total magnetic flux through the surface z:4,0 < r S 1, -I < U < 4 A
will
be
(A) 20 wb (C) a0wb
'
'
(B) - 10/3 wb (D) 130/3 wb
, 41.11 The current density that would produce the A:2aa in cylindrical coordinates is
magnetic vector potential
(A) '
-Lo, PnP'
re)#",
Q)
h",
@)h",
F Page 222
mcQ
{1"{.'12 Magnetic field intensity produced due to a current.souxce is given
Chap 4
g :
Magnetostatic Fields
(zcos ay) o,y + (z *
The current density over the (A) (o, - 0,y- a")Afm2
rz
plal;re
eu)
will
as
a,,
be
(B) -o, * &v- &" (C) -2a,t au-2a, (D) o, * ar* a,, &{eo
,s"x'13 Assertion (A) : In a source free region, magnetic field intensity expressed as a gra.dient of scalar function.
Reason (R) : Current density for a given magnetic field intensity is AS
J:YxH (A) (B) (C) (D) lxen
A and R both are true and R is correct explanation of A. A and R both are true and R is not the correct explanation of A. A is true but R is false. R is true but A is false.
4.'l-'14 An electron beam of radius a travelling in o, direction, the current is given
as
J :2(r\ - 1\", al
Fc
The magnetic field intensity at the surface of the beam will be
(A)
n6rQ 4.{"{S
$"'
(n)
#",
Q)4",
@)
*",
If there is a current filament on the r-axis carrying (A) 0.1(a"-2a)Afm (B) 1.76a"- 1..62au Af m (C) (- !.077a,* I.62a,n) Alm (D) -0.1(2o"- ou)
\
\ r$*fi {.{"{6
A in a, (4,2,2) ?
4.4
then what will be the magnetic field intensity at point
d
In the plane z : 0 a disk of radius 3 *, centered at origin carries a ^/ surface charge density pr:2C/^'. If the disk rotates about the z
an angular velocity w:2rad/s then the magnetic field intensity point P(0,0,1) will be (A) a" Alm (B) 2a" Alm (C) au Alm (D) 2a, Alm
Common Data For Q. fZ and 18 : In a Cartesian system two parallel current sheets of surface current Kt:3a,A-r/m and Kz:-3a,Afm are located at r:2m and r:respectiv'ely. The net vector and scalar potential due to tfre sheets are at a point P(I,2,5).
+r.r?
Consider the scalar potential at any'point (r,U,r) in the region between the two planar sheets is V*. The plot of I/- versus g will be
Page 223
Chap 4 Magnetostatic Fields
I/_(A)
: -{ g(m)
V^(A)
s(m)
E! .,1.18
The vector potential at origin will be
(B) -3paa"Wb/m (D) -3wb/m
(A) Spaa,Wb/m
(c)
0
Common Data For Q. 19 and 20 : A uniformlv charged solid sphere of radius r is spinning with angular velocit i' a:4radls about the z-axis. The sphere is centered at origin and carrics a total charge 5 C which is uniformly distributed over it's volume.
Ee
/r,{.19
The plot of magnetic dipole moment of the sphere, m(r) versus the ratlir of the sphere, r will be
"(*)
r(m)
r(m)
r(m)
rn(r)(A-m2)
s
PVge 224
$cq
4.1.?o
Chap 4 Magnretostatic Fields
:,,
:;Q
.4.1.21
The anrage.Enm$ig$dd intensity wtthi4'the sphere will be
(N
#w
(B)
fi",
Q
7",
(D
#""
o
d
rect4ugglar cail, lyipg.ip t[re plane r+7g-1.52-3.5 carries a quch that the qragretic nom.ent of the coil is dirocted away frorn origin. If the area of the rectangular coil is 0.1 m2 then the magnetic of the coil will be (A) -0.2c" - 0.2sn* 0,3.q A-mz 7A
(B) ?c" + EA - 6p" !r-lrr' (C) 1.4a, + 4.2a0 - 2.1o" A,-mz (D) 0.2c" * 0.6a, : 0.&c, .A,-m2 r*cQ 4.{.22
Vector magnetic poterrtial
:
(6y
-
2z)
a"*
in a
certain region
The electric curront density at any point (4 (A) (- 8o, * zau* 6q") Nmt (B) (3c, * o") Af m2
(c) 4",t.23
y, z)
will be
A-/m2
Magnetizing force at arry point. P on z-axis due to a eemi infinite element plaped along poeitive o-axis is Il. If one more similar cu element is placed along positive y-axis then the resulta,nt magnetizing at the point P will be
(N
HlJ'
(B)
J'
H
(c) 2H @) t*tQ
4"1.24
is
0
(D) fr(8o" *2an-6o") Mgo
of free space
4xzon
A
-nH
.[-shqped fi'lanrentary wire with semi infinite long legs making 90' at origin atrd lying in y-z plarre as stpwn iir the figure.
an
If the current flowlng in the whe is I = 4 A thetr the magnetic Ilux at (2 m,0,0) will be (A) -2 x 10*7(c,+ a,)Wb/m'? (B) 2 x 10-t(o,+o,)Wb/m' (C) -4 x 10-?(a,* a)Wb/m2 (D) a x lO-'(a"+a,)Wb/m,
Common Data For Q. 25 and 26 : An infinitely long straight conductor of cylindrical cross section and of radius .R carries a current 1, which is uniformly distributed over the conductor cross section.
If the cond uctor is located along z-axis then the magnetic flux density at a distance p ( < -R) from the cylindrical axis will be
@)#"
@)#"-
Q)#",
ro)#"r
Magnetic flux density at a distance p(>,R) from the cylindrical axis will be proportional to (A) 1 (B) ',p 'p' +
(c) 4.1"2v
(D) p'
p
The magnitude of the magnetic field intensity produced at center of a square loop of side a carrying current 1 is
(A\
rc\ '
2/' I TA
-JJ2ra
tst
hI
(D)
EI
'
4,1"28 For the
7T O,
' Ifo,
single turn loop of current shown in the figure the magnetic field intensity at the center point P of the semi circular portion will be
1:8
A
1m
,-l
P
1'2m
(A) 5.8 A/m (B) 5.8 A/m (C) 3.8 A/m (D) 3.8 A/m
outward inward outward inward
***rk**{<***<*
Page 225
Chap 4 Magnetostatic Fields
EXERGISH 4"2
Page 226
Chap 4 Magnetostatic Fields
Meo 4,2"{
A conducting filament carries a current 5 A from origin to a point . Magnetic field intensity at point (3, 4, 0) due to the filament curr€d be
wbf m2
MCQ 4.2"s
A circular conducting loop of radius 2m, centered at origin in the z: 0 carries a current of 4 A in the a6 direction. What will be the field intensity at origin in a, direction ?
$0&
A long straight wire placed along z-axis carries a current of 1: 10A ir *4" direction. The magnetic flux density at a distance p:5cm fron wire will be x 10-5 wbf m2
4"2.3
mcQ 4.2.4
For the currents and the closed path shown in the figure
Ampere.
, Jd H . dl:
L 12:10 A
o MCQ 4,2.5
Two infinitely Iong wires separated by a distance 5m, carry currents opposite direction as shown in the figure. If 1: 8 A, then the magnetic intensity at point P is A/m in o, direction.
f
1m
&
.
P>
01
5m
MStt
4"?.G
Two point charges Q1 and Q2 are located at (0,0,0) and (1,1,1) respectivelj A current of 16 A flows from the point charge Q,. to Q, along a straight win
connected between them. What will be the value $ n. dt (in A/m) "f {he around the closed path formed by the triangle having vertices (1,0,0). (0,1,0) and (0,0,1) ?
Common Data For Q. 7 and 8 : An infinite current sheet with uniform current density K:20a,, A/m is located in the plane z:2. The magnctic field intensity at origin will be
Magnetic field intensitv at point (2, direction.
itcs
4"?.s
-1, 5) will
A/m in o, direction. be
____
:
An infinite current sheet with uniform surface current density K is located at z:0 as shown in figure.
4'x'{ii}
ao
In the free space two cylindrical surfaces p :0.5 cm and p : 0.25 cm carries the uniform surface current densities 2a, Af m and - 0.8o, A/m respectively and a current filament on the entire z-axis carries a current of 14 mA in the *o, direction. The surface current density on the cylindrical surface at p:8cm. which will make the net magnetic field H:0 for p> 8cm will be A/m in a. direction. Common Data For Q. 10 and 1L
HcQ
Afm in
:
4a, Alm
Magnetic flux density at any point above the current sheet (z > 0) will be
x
pocr,, n'b/m2
uca 4.x"1!'i The vector magnetic potential at
z:-
2 will be
x poarwbf m2.
rcQ ,4.*.52 In the free space, magnetic field intensity at any point b,6,r) is given by H:2p2 aa A/m. The current density at p :2 m will be Afrf in a" direction. HCQ 4"*"1l;3
Magnetic field intensity produced at a distance p from an infinite cylindrical wire located along entire z-axis is 3pae A/m. The current density within the conductor will be A/m2 in o, direction
HCA r*,R,{d
A circular loop of wire with radius r?:0.5m is located in plane tr:0, centered at origin. If the loop carries a current I : 7 A flowing in clockwise as viewed from negative r-axis then, its magnetic dipole moment will be A-mz in a, direction.
MC& d"*"'rs
In the free spaceT the positive z-axis carries a filamentary current of 10 A in the -o, direction. Magnetic field intensity at a point (0,2,3) due to the
.
Page227 Chap 4
Magnetostatic Fields
Prge 22E
filamentary current will be
A/m in a, direction.
Ctap 4 Magnctctatic Fields
iltca
4"e"'!6
A filamentary conductor is formed into an equilateral triangle of side 2 m that carries a current of 4 A as shown in figure. The magnetic field intensity at the center of the triangle will be A/m in o. direction. s(m)
r(*)
trrcQ 4,2",t?
A current sheet .I(:4a,, A/m flows in the region -2 < z ( 2 in the plane r: 0. Magnetic field intensity at point P(3,0,0) due to the current sheet will be A/m in a,, direction.
afrcQ 4"2"1S
A square conducting loop of side 1 m carries a steady current of 2 A. Magnetic flux density at the center of the square loop will be X 10-6 wbfm2
Me& 4.2"{S
A filamentary conductor is formed into a \oop ABCD as shown in figure. If it carries a current of 3.2 A what is the magnitude of magnetic field intensity (in A/m) at point P ?
tvtc{ 4,x.2{}
The magnetic field intensity at point P due configurations shown in figure will be
I:4 MCQ 4"?"41
to the steady current
Al^.
A
In the plane z:5m a thin ring of radius, o:3m is placed such that z -axis passes through it's center. If the ring carries a current of 50 mA in o6 direction then the magnetic field intensity at point (0, 0, 1) will be mA/m in a, direction.
Ga 4.2.23 An infinite solenoid (infinite in both direction) consists of 1000 turns per unit length wrapped around a cylindrical tube. If the solenoid carries a current of 4 mA what will be the magnetic field intensity along its axis
?
Common Data For Q. 23 to 25 : The two long coaxial solenoids of radius o and b carry current 1: 3 mA but in opposite directions. Solenoids are placed along y-axis as shown in figure. The inner solenoid has 2000 turns per unit length and outer solenoid has 1000 turns per unit length.
4'*"fr3 Magnetic field intensity inside the inner
solenoid
will
Al^
be
directed along or.
ICQ
4,X";t6
The magnetic field intensity in the region between the two solenoids will be A/m directed along o,.
4.2""X&
What will be the magnetic field (in A/m) outside the outer solenoid
4,*.}{{
A long cylindrical wire lying along z-axis carries a total current 1o : 5 mA as shown in the figure. The current density inside the wire at a distance p from it's axis is given by J s P. If the cross sectional radius of the wire is 2cm, what will be the magnetic flux density (in nwb/m') ut p:1cm ?
l
11,:s
rca
,4"3"a?
*l
Magnetic field intensity is given in a certain region
as
":P*io,*3#tau-ffia,Afm
What is the total current (in Ampere) passing through the surface ,l3 A ( 4m, 3 3 z< 4m in a, direction ? HCQ 4"4.i?*
?
fr:2m
A phonograph record of radius 1 m carries a uniform surface charge density ps:20C1^'.If it is rotating with an angular velocity cu : 0.1rad/s; what will be the magnetic dipole moment (in A - m2)
Page 229
Chap 4 Magnetoetatic Fields
Common Data For Q. 29 and 30 : A circular toroid with a rectangular cross section of heiglrt h: 10 m, carries a current 1:10A flowing in 105 turns of closely wound wire around it as shown in figure. The inner ancl outer radii of toroid are a: 1m and b : 2m
Page 230
Chap 4 Magnetostatic Fields
respectively.
ib
I
t
I to, I
4.*"3$
What will be the total magnetic flux (in weber) across the circular toroid
4.X.3n
If the magnetic flux is found by multiplying the cross sectional area by the flux density at the mean radius then what will be the percentage of error ?
eee& 4.*.35
An infinitely long straight wire carrying current 5 A and a square loop of side 2 m are coplanar as shown in the figure. The distance between side AB of square loop and the straight wire is 4 m. What will be the total magnetic flux (in pwb) crossing through the rectangular loop 7
?
C +
l2 rn
m square loop is lying in r-E plane such that one of it's side is parallel to 3r-axis and the centre of the loop is 0.3 m away from the y-axis. How much Ampere current must flow through the entire y-axis for which the magnetic flux through the loop is 5 x 10 5 Tesla rnz ?
lvlLu {.&.Jd
A
!r{*Q 4"2"33
Consider a filamentary wire is bent to form a square loop of side 2 m lying in the r-y plane as shown in the figure. If the current flowing in the wire
is
0.5
1: 1 A then the magnetic flux density at the center of the loop will be xlO-7 a"Wb/m2
2m
An infinitely long straight wire carrying a current 20 A and a circular loop of wire carrying a current 1 are coplanar as shown in the figure. The radius of the circular loop is 10 cm and the distance of the centre of the loop from the straight wire is 1m. If the net magnetic field intensity at the centre of the loop is zero then the current 1: Ampere.
+2.33
Two perfect conducting infinite parallel sheets separated by a distance 2 m carry uniformly distributed surface currents with equal and opposite densities 4o,, and- 4a, respectively as shown in figure. The medium between the two sheets is free space. The magnetic flux between the sheets per unit length along the direction of current will be kp,sarWbfm such that the value of k is
t<*
t(*t(* ** ***
Page 231
ehaF 4 Magaetostatic Fields
hlDrt8
HXFRCI$E 4.3
qri'
rfj+rr.tr.;c
1lie16s
MC(l
,4,"3"t
Assertion (A) : For a static magnetic field the total number of flux lines entering a given region is equal to the totar no. of flux lines reaving the
region. Reason (R) : An isolated magnetic charge doesn't exist. (A) Both A and R one true and R is the correct expranation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false.
(D) A is false but R is true. ItcQ
4.3.?
Match List-I with List-II and select the correct answer using the codes given below. (Notations have their usual meaning).
List-I
List-II
a. b.
Amperets
c. d.
Gauss's
law
Conservative nature of
l. Y . D: magnetic field
law
B.
Non existence of magnetic monopol
Codes
Z.
e
4.
ft. V
p,
as:
In.
at
'B:0
f n.at:o
:
abcd (A)3142 (B)2143 (c)2413 (D)3421 MCQ 4.3,t
The source which doesn't cause a magnetic field is (A) A charged disk rotating at uniforl speed
(B) An accelerated charge (C) A charged sphere spinning along it's axis (D) A permanent magnet rllrc& 4.3.4
The correct configuration that represents magnetic flux lines of a magnetic dipole is
(A,
tAl rVl
,",@l@
,",@)l@ ls
Page 233
,',@)l(@
The correct configuration that represents current I and magnetic fieid intensity .6I is
I aa
I oo
o a ,no o
(c)
oo
aa
I oo
I aa
(n)o o H8
oo
A
aa
In a certain region consider the magnetic vector potential is C and the current density is J. Which of the following is the correct relation between
J
and
A?
(B) V'z-A : p4J (D) V'?A :- p'J
(A)VA:J (C) V x A:-FJ ,o.
a
If the magnetic field, (A) 1.6r'o,p wb/m2 (C)
,3"8
7.6np,
wb/m
{(o r H).
(A) zero
(c)
/
H:
4a,, A/m, flux density in free space is
(B) I6na"pwb/mz (D) 160no, wb/m
ds (B)
1"""
1o1
f n.as
The magnetic field in a an ideal conductor is
(A) zero (C) finite
(B) infinite (D) the same
The unit of scalar magnetic potential is (A) Ampere (B) Volt
(C) Amp/m
V
x
(D) Volt/m
VZ-is
(A) zero (C) J
(B) V2V* (D)
v.v%
a,s
its outside field
Chap 4 Magnetostatic Fields
4.)".r* v xAis
Page 234
n
Chap 4
(A)
Magnetostatic Fields
(c) ./.
(B) B (D) 0
n".l,.,':i Torque has the unit of (B) N/m (D) N
(A) N-m (C) N'm'? '{.;1.{,$
{a.asi,
(B) I (D) J
(A) zero
(c) :;
*t. {."}"': i
11
Absolute permeability of free space is
($ ar x 10-7 A/m Q)
atr
x
10-7
(B) aa'x (D) azr x
F/m xx
r
*x t<* *i(*
10-7
H/m
70-7 Hf m2
mKmKcilsK 4.4
Page 235
Chap 4 Magnetostatic Fields
Common Data For Q. I and 2 : An infinitely long uniform solid wire of radius of density .I The magnetic field at a distance to
r
c,
carries a uniform dc current
from the center of the wire is proportional
(A) rforrlaandlf 12forr> a (C) rforr( aandlfrforr) a
(B) 0forr( oandTf rforr> a (D) 0for r< ound'41r'forr> a
A hole of radius
b(b < a) is now drilled along the length of the wire distance d from the center of the wire as shown below.
at
a
The magnetic field inside the hole is (A) uniform and depends only on d
(B) uniforrn and depends only on b (C) uniform and depends on both b and d (D) non uniform Two infinitely long wires carrying current are as shown in the figure below. one wire is in the y - z prane and parallel to the 3r axis. The other wire is in the r - g plane and parallel to the r axis. which components of the resulting magnetic field are non-zero at the origin ?
(A) ,, g, z components (C) y,, components
(B) components ",y (D) r,z components
':
'?o
i.'. i.'L . '
'G'Page 236
d'P
&,{s*4'4^4Afluxofl.2mWbeXeItSinamagnethavingacross-Sectionof30cm2 flux densitY in tesla is (A) 4
4
Magnetostatic Fields
(c) w**
2.5
The magnetic flux densitY
4"4"$
related
as
consicler
B
and the vector, magnetic Potential4
A:Vx B (D)A:Y'B
(B)
(A) B:Vx A (C) B:Y ' A
m*c& 4"4,$
(B) 0.4 (D) 40
electrostatic and the following statements relating to the
magnetostatic field
: body is of charges on an isolated conducting distribution relative The 1. of the bodv' Jependent on the total charge surface is zero' 2. The magnetic flux through any closed is/are correct ? Which of the above statements (B) 1 onIY iLj x"t,rr* 1 nor (D) Both 1 and 2 (C) 2 only
2
Thelineintegralofthevectorpotential.4aroundtheboundaryofasurface S-r"pr"r"rrtt1tni"n one of the following? (A) nfux through the surface S (B) Flux densitY in the surface S (C) Magnetic field intensitY (D) Current densitY
f$3*& 4"4"?
&.**qd'.*'&Aninfinitelylongstraightconductorlocatedalongz-axiscarriesacurrentr g plane field at any point P in the rin the *ve z-clirection' The magnetic is in which direction?
(A) In the Positive z-direction (B) In the negative z-direction radial line OP (in r (C) In the direction perpendicular to-the joining the origin O to the Point P (D) Along the radial l\ne OP M*e
- g plane)
of 5 cm radius' What A 5 A current enters a right circular cylinder
4.4.9
is
end surface? linear surface current density at the
(A) (50/zr) A/m (B) (100/zr) A/m (C) (1000/zr)A/* (D) (2000/zr)A/f!$*&
potential due to an infinitesimally is the value of the magnetic vector it ? evaluated at infinite distance from small
d..&.rs what
.
""p'""t "t"*""t'
(A) InfinitY (B) UnitY (C) Zero depending on the strength of th (D) Any number between zero and infinity cur.rent element
h
a.n.rr
what is the magnetic field intensity vector .E[ between two parallel sheets with separation 'd' along z-axis both sheets carrying surface current
K:
K,,a,v ?
Current density
(4, ir cylindrical coordinate system
J (p,d,
r)
:
fr,rof orr. i:l
3
is given as:
:i:;
where a, is the unit vector along z-coordinate axis. In the region, a < r < , what is the expression for the magnitude of magnetic fielci intensity (r1)
(q
"'l
#o'-
1C;Jo(43p
o3)
Ja-
b ?
a'1
@1
fi6'+
@l
*rp'- dl
n
4.4.'t3
which one of the following concepts is used to find the expression of radiated .E and 11 field due to a magnetic current element ? (A) Concept of vector magnetic potential (B) Concept of scalar electric potential (C) Concept of scalar magnetic potential (D) Concept of vector electric potential
n
4.4,,t{
The circulation of ,E[ around the closed contour
0 (c) 4l (A)
c,
shown in the figure is
(B) 2l (D) 6l
e 4.4.1s The unit of magnetic flux density is (A) gauss (B) tesla (C) bohr (D) weber/sec Q
4'4'{$
The magnetic flux density created by an infinitely long conductor carrying a current I at a radial distance _R is
(^)#
(C\
P'I
^ '"' 2iF
Chap 4 Magnetostatic Fields
(A) -k,a, (B) +kao.u (C) -k,a, (D) Zero E 4J.12
Page 237
@)# ,n, \"/ -4rR2l
r
'l1
Page 238
nAse 4"{.r?
II
Match List I with List
and select the correct answer using the codes
below the lists.
Chap 4 Magnetostatic Fields
List-II
List-I
a. b. c. d.
1. 2. 3. 4.
Work Electric field strength Magnetic flux Magnetic field strength
Codes
Ampere/metre Weber
Volt/metre Joule
:
abcd (A)4327 (B)1324 (c) 4231 (D)r234
14CQ 4.4, 1E
A long straight wire magnetic field 1 Am
1
carries
a current
1
: 1 A. At what distance
?
(A) 1.5e m (B) 0.15e m (C) 0.015e m (D) 0.00159 m e{&Q 4.4,'tti}
How much current must flow
field lmAm (A) 1.0 mA (C) 2.0 mA ffis&
4"4"4e
1
in a loop radius 1m to
Produce a
?
(B) 1.5 mA (D) 2.5 mA
Assertion (A) : Knowing magnetic vector potential density B at the point can be obtained.
A at a point, th
Reason(R):V',4:0.
(A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true and R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
r*f& 4,4"t{
The magnetic vector potential
1. B: Y x A
3.
o:l#
A
obeys which equations
2.
Y2A:-
Select the correct answer using the code given below (B) 2 and 3 (A) 1 and 2
(C) 1 and 3 ! I
I
itbl
t-
s8s&
d,,,;l..j|?
?
p4J
:
(D) 1, 2 and 3
A long straight wire carries a current 1:10 A. At what distance is th 1 magnetic field 11: 1 Am ? (A) (C)
1.1e m 1.5e m
(B) 1.3e m (D) 1.7e m
what
is the magnetic field due to an
infinite linear current carrying conductr,r
?
Chap 4
:#ofu (c) H: # ol^ (A)
Page 239
(B)
11
p
s:*ol^
(D) fI
Magnetostatic Fields
:*otrr
Equation V-. B : 0 is based on (A) Gauss's Law (B) Lenz's Law (C) Ampere's Law (D) Continuity Equation Plane
y:
0 carries a uniform current density 30a"mAf m. At (1,20,
what is the magnetic field intensity (A) - 15a,lrrA/m (B) 15,a" mA/m (C) 18.85o, mA/m (D) 2\a"mA/m
GQ 4.4.*6 Which vector
(A) (C)
-
2
?
one of the following is not the valid expression for magnetostatic lrei.i
B
?
B:v.A v . B:0
(B) B: V x,4 (D) V x B: poJ
4..1'R? Which one of the following statements is correct ? Superconductors a,, popularly used for (A) generating very strong magnetic field (B) reducing i2r? losses (C) generating electrostatic field (D) generating regions free from magnetic field
-Q 4.4.c*
.\\
Assertion (l) : I A . d,S :0 w]rere, B : magnetic flux density, d,S: vector with dire6tion normal to surface elements d,S. Reason (R) : Ttrbes of magnetic flux have no sources or sinks. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false (D) A is false but R is true
\
rcQ 4"4"4$ Plane definedby z: 0 carry surface current density 2a, Af m. The magnetrc intensity' Hr' inthe two regions -a < z< 0 and 0 < z
(A) on and -a, (C) o" and - o,
(B)
-*r
- an and a, (DJ - a, and e,,
ir'
t1 I I
sol-uTlol{$
Page 240
Chap 4 I I
4.1
Magnetostatic Fields
L
s{}L
4"*"':
Option (A) is correct. Since the field intensity exists in a closed surface and lines of field intensity makes a closed curve so the flux lines leaving the spherical surface equal the total flux entering the surface and So the net flux
*:{u.d,s:o According to divergence- theorem
fa.as:f(r.B)d,u
I
o: Jfv
Since volume of the sphere
I
V . H :0 at all points
or I
will have certain finite value
so,
V.B:0
I
I
. Bdu
inside the sphere
:'',i 4.1.2 Option (A) is correct. The magnetic field intensity produced due to a small current element /dl
I
defined as
I
d,rr where
__
Idl x
?n
4rR2
dl is the differential line vector and oa is the unit vector di
towards the point where field is to be determined. So for the circular carrying loop we have
I
I
dI
:
aR:-
addao QP
Therefore the magnetic field intensity produced at the centre of the circ
I
loop is
Iadba' x ao) rr : Ja:o [" -(4trd
t I
: ffilol\^ (o") : fia"
t
t
s$I-
4.,t"3
i
Alm
Option (C) is correct. As calculated in previous question the magnetic field intensity produced the centre of the current carrying circular loop is
H:*
t
will produce the field intensity half produced the field intensity by complete circular loop. So by symmetry the semicircular loop
i.e. ${}L 4.1"4
I
$*-
Field intensity at the centre of semicircular loop
:Lr:
*
Option (B) is correct. Since current in the wire is distributed over the outer surface so net enc current, 1"n". for any Amperian loop inside the wire will be zero. and as from Ampere's circuital law we have
] F
So.
F
rt
f n. f n.
or
b
d,r:r"n"
d,t:o .E[
:0
I
;-
4-1.s
a
F
b
f
r
Page 241
(1""":0) for r<.R
Option (B) is correct. consider the cylindrical wire is lying along z -axis as shown in the figure. As the current 1 is distributed over the outer surface of the cylinder so for an Amperian loop at a distance r(> ,?) from the centre axis, enclosed current is equal to the total current flowing in the wire.
i i
Now from Ampere's circuital law we have, I
J B' dI: B(2rr)
It
4.1.6
:
thl"n" PnI
or
B:
or
B.+
u^I
(1"
": I)
Tnror
Option (B) is correct. consider one of the sheet carries the current density Kr. So, the other sheet will have the current density -I{r. Magnetic flux density produced at any point p due to a current sheet is defined as
B:ffxxw where r{ is current density of the sheet and an is the unit vector normal to the sheet directed towards point p. so for any point in the space between the sheets normal vector will be opposite in direction for the two sheets as shown in figure i,e. 0,n2 :anl
Chap 4 Magnetostatic Fields
Therefore, the resultant magnetic flux density between the two sheets will be
Page 242
ch"p
4
Magnetostatic Fields
a : flKt
X
@^
at any point
* (- r(r) x (-
a' is unit
vector normal to the surface, and density. So the cross product will be a constant. Since
sol. 4.t,7
ort
"
o^
(at (3,0,2r))
Option (B) is correct. Consider the current sheet shown in the figure.
defined as
P
due
to a current sheet is
g:!xxa^
where -I( is, current density of the sheet and the sheet directed towards point P.
4.t.s
an,
I(r is given current
: USuq ao :4ao
Magnetic field intensity produced at any point
$oL
x
B:Y XA: V x(l2cosdae) :l${rrr"oso)aa:l439d
4.1.4
paKl
Option (A) is correct. The magnetic flux density at any point is equal to the curl of magnet vector potential ,4 at the point. l.e.
sol.
:
o,')]
the
a, is the unit vector normal to
so, for s >
0
n :!(xa") x (or) :-|xo,
and for y <
0
n :|(xa,) x (-
c.r)
(K:
: *o" (K:
Ka"
A.f
m,
Ka, Lf m,
a,:-
Option (B) is correct. Magnetic flux density is defined as the curl of vector magnetic potential l.e.
B:Y xA az I la,a o1r a a -l 0, dy dz -l
I
lzty zf
I
-8ruzl : (- 8rz -0) a, + (0 + 8yz) a, + (2f So the net magnetic flux density at (1, - 2, - 5) is B :40a,* 80au *6a"wbf m2
4.t.{o
a,:
Option (C) is correct. Total magnetic flux through a given surface 5 is defined
- 2t) a"
as
ao)
an
o:Ia.as
Page 243
Js
where d^9 is the differential surface lvector having direction normal to the surface So, for the given surface
49
z: 4, 0 I r I 1,
1-
<
:
3r
< 4 we have
(drdy)a" and as calculated in previous question we have B : (- 8rz- 0) o.,+ (0 +8yz) ar+ (2f - 2*) a, Therefore, the total magnetic flux through the given surface is
*:
I'. ,L,rrt-2i)(drdy)
:2xtl^fau-zxs['ta, : r[4]_, -rr[$],:2xS-.tf :40wb sol-
4"{",t,1
Option (D) is correct. The magnetic flux density at any point is equal to the curl of the vector magnetic potential at the point
B:Y xA :f,fiodr)o, :|ffeao, :7o"
l'€'t
The current density
J
J
in terms of magnetic flux density B is defined
:L(v P4'
as
x B)
:h[-*G)]":fua This current density would produce the required vector potential. sol- 4"{.r2
Option (B) is correct. The current density for a given magnetic field intensity .EI is defined
J:Vx,6[
H : (zcosay)au*(z-t
Given
So
o, vrI/:l I *
:l- *, or,
av arn
""s
ay
)
a,
- ? &t, *
: - cos aAa, * : J V X H:J -- a,* aa-
4.1"{3
eu)a,
a,l
*i
l(z+es) zcosay 0l au
e\)
-
au
+
r-z
(*, *, ou - ft 1, +,, ))l *
eu o,"
cosayeryJ-
Therefore the current density in the
sol
as
ou-
ev
a"
plane is
a, Af m2
(a:0
in r-z plane)
Option (A) is correct. In a source free region current density, J: 0. The current density at any point is. equal to the curl of magnetic field intensity .EI.
i.e.
J:YXH
Chap 4 Magnetostatic Fields
rft 1
(J:0)
VxII:O
or
Page 244
and since the curl of,a given vector field is zero so gradient of a scalar field
Chap 4 Magnetostatic Fields
sol- 4.{.{4
can be expressed as the
H:Yf
i.e. So
it
A and R both are true and R is correct explanation of A.
Option (A) is correct. As the beam is travelling in o" direction so the field intensity produced by it wilt be in aa direction and using Ampere's circuital law at it's surface we have
H5(2ra)
: I*"
H6(2r a)
:
I' r(t - f,)z" pap
H6(2na):n"l+_*l:^ H6(2na)
or sol. 4-t.{5
:ry
":to,
Option (A) is correct. According to Biot-savart law, magnetic field intensity at any point P due to the current element ldl is defined as
n:IIq4P J 4rR'
where
R
is the vector distance of point
P from the current
element.
I:4.44
Now the current element carries a current of' 4'4 So we have,
A in * a' direction'
P: (4a,,*2ar*3a") - (ra,) (Since on
r-axis E- and z-component will be zero)
R: (4- r)a,*2ao*3a" R
and .
I
,itI b I
il L.
Idl
: r@_,jf+t'+5z : Jirir+29 :4.4d,ra,
(filament lies from
Therefore the magnetic field intensity is (4.4a,) x l(+ - r) a, + 2g,! + 3o.1 _
H -- - J-[** *
ar e?
- 8, + 291',' 4'4 t2o.- u*r) [*- ,,_ sid* zolt/,,,, - 4n \o-, 3o,,)J_* + 1f
r:-cx)
dr
to
r:
@)
:#eo"_s",y[u*S$yz].-
Page 245
Chap 4 Magnetostatic iields
: ffi{Zo" - 3or) 4.4
:
0.1077 a,
-
0.762an
:0.1a"
-
0.2a, Af m
ALTERNATIVE METHOD:
According to Ampere's circuital law, the line integral of magnetic field intensity .Ef around a closed path is equal to the net current enclosed by the path. Since we have to determine the magnetic field intensity at point (4, 2, 3) so we construct a circular loop around the infinite current element that passes
though the point (4,2,3) as shown in the figure. Now from Ampere's circuital law we have,
sH. dr:r",, (2nr)
H
:
(I^"
+.+
4.4
2tr x J73 Now direction of the magnetic field intensity is defined
where
a4: o.tx aP or + unit vector in the direction
r: t/!3
=
4.4 A)
from figure.
as
of flow of current
unit vector normal to the line current directed toward the point. -t So we have, a6:a,x l(4a" 2au t 3a,) - @",)l (4- 4)'+22 +82 @p-+
:(,,x@#A Therefore the magnetic field intensity at the point (4,2,3) is
Lr
-
: :
4.4
(2a,
zrt/ts
- 3ao) Jts
^4L{2o"-Jau) - 0.2o" Alm
O.Ia"
Page 246
sol.
4.{.'ls
Option (A) is correct'
Chap 4 Magnetostatic Fields
Sincetheuniformlychargeddiskisrotatingwithanangularvelocity
u:2ndls
about the z-axis so we have the current density K : p"x (angula.rvelocity) : p"(up) : 2
x2x
p
K :47aa
or
AccordingtoBiot-savartlaw,magneticfieldintensityatanypointPdueto the current sheet elemenf KdS is defined as
[4!Exgt s :- J" 4trR2
where
R
is the vector distance of point P from the current element'
Now from the figure we have
R : a"-
or and
So the magnetic field
Pis
P&p
R: rE+V or:4 ^/I+p-due to a small current element intlnsity
d,H:eW:
On integrating the above over @ around the
complete with us leaving components get ca,ncelled by symmetry,
H(z): h['" Jo f
-Kd^9
at point
4p(ar* pa") 4tr (p' + l)t/' circle, the o,
'
!""ru@anat) atr@'+ 'n{ l)
:rIn (f,fua* :21{v.:l#-l:
I
""
:rl*#lo,:o,At^
$0L rt,1.t7
Option (C) is correct. fit" *ug*tic field intensity produced at any point P due to an infinite as sheet carrying uniform current density K is defined
u :l1x x a) toward the point P' where a, is the unit vector normal to the sheet directed
Sothefieldintensityproducedbetweenthetwosheetsduetothesheet Kr:3a" located at tr:2m is (o' :- c,) gr:!1to,) x (- a,) :-lao A,lm
and the field intensity produced between the,two sheets due to the sheet I{z:- 34, located at r:- 2m is
n, :i?Ba,) x (o") :-|a,
(o,:
A/m
o,)
Therefore the net magnetic field intensity produced at any point between the two sheets is
H : Hrl H2:-3ou Since the magnetic field intensity
at any point is the equal to the negative gradient of scalar potential at the point
H :-Y V* So for the field H-- 3o, in the region between t.e.
the two current carrying
sheets, we have
a, (the field -3au -- 4%: dy
V*:3yl
or
Putting
V*:0
has a single component
where
Cr
for point P(1,2,5) (given), we have
0:3x Cr:-6
or Thus, and the graph of
in o, direction)
Q is constant
(2)+Cr
V^:(By-6)A I/-
versus y
will be as plotted below
s(m)
L
4.1,18
Option (B) is correct. The magnetic flux density at any point is equal to the curl of the vector magnetic potential at the point l'€'t
B:Y xA
(1)
B : hH :- SpnavWb/m, (calculated in previous question) As the magnetic flux density is in ao d.irection so A is expected to be z
Since
-directed. Therefore from eq (1) we have
-44' - ^ or. or Putting
A":
A; :3por* Cz O at point P(1,2,5) (given), 0
or So,
'2,. Ol.[O
:3pn]-
we have
Cz
:- 3po A,:\pr(r- 1):-3p" Cz
Thus, the magnetic vector potential at origin is
A --Spaa"Wbfm
at origin (0,0,0)
Page 247
Chap 4 Magnetostatic Fields
Page 24E
sol. 4.{.19
Chap 4 Maguetostatic Fields
Option (C) is correct, Magnetic dipole moment of a spherical shell of radius charge density p, is given as
* : {prrrn
where
a.,
r
having surface
is angular velocitr-
total charge of 5 C is distributed over the volume of the sphere the magnetic dipole moment of the sphere is given as Since the
m(r):
so.
(p,:p,dr\
f*fo,oOrr
where p, is uniformly distributed volume charge density of the sphere Therefore, we have
m(r)
:*""8 :!Q,t
(": #\
1 - "^ (s) x (+) x f :4f A'-m2 (Q:5c, ti:4 -5
rad/s)
I i i
rn(r)(A-m2)
r(m)
sot
4"r.20
Option (B) is correct. The average magnetic field intensity over a sphere of radius r, due to steady currents within the sphere is defined as
l2x4f I2m :G--Frr no,":G-;F
:-z7Tr
(m:4f
y
As the sphere is spinning about the z-axis so, the produced magnetic field will be in o, direction as determined by right hand rule. Thus, we have Hoo"
sot- 4,{,21
:1a" fir
Option (D) is correct. Magnetic dipole moment of a conducting loop carrying current / is defined as: rn : ISa, where ,S is the area enclosed by the conducting loop and 4,, is normal vector
to the surface.
So we have
m
:
(7) (0.1)a"
Now the given plane is
t*3Y-1'52:3'5 For which we have the function
f : r+3Y-1'52 and the normal unit vector to the plane is,
,':ff:
So the magnetic dipole moment of the coil is
(I
:
7 A,
^9:
0.1 m2)
na
:
(o.T@!ll+; t'59)
:
0.2a,x* 0.64,
-
Page 249
Chap 4 Magnetostatic Fields
0.3a, A-m2
Option (C) is correct. The magnetic field intensity, (^tr) in the terms of magnetic vector potentiiil.
(A) is defined
as
H:L(v Pt '
x A)
:*to x (6y- 2z)a,+4rzaol
:
frl-
a",
-
2au
+
6a,f
Since the electric current density at any point is equal to the curl of magnetic field intensity at that point.
J:YxH
l.e.
So, we have the electric current density in the free space as
:
J :.4"i.2i
Yx
fr[-a",
- 2artBa"l :
g
Option (B) is correct.
consider the point P on z-axis is (0,0,h) and current flowing in the current element is 1 in a, direction. since the magnetic field intensity at any point P due to a current element I is defined as
Hf :4fofcosar- cosol]46
where
p --+ distance of point p from the current element. 01
+
at p. by the upper end of the element at p.
angle subtended by the lower end of the element
02 -'+ angle subtended
So for the given current element along positive e-axis we have
0r :90o 0z Therefore,
:0o
n:ffia6
Now the directicin of magnetic field intensity is defined
(p:
h)
as
a'o:Q4xaP where al is unit vector along the line current and oo is the unit vector normal to the line current directed toward the point p.
So,
Page 250
Chap 4
Magddatic
ea
:
e" X
e,":-
6o
Therefore magnetizing force is
Fblds
H
or
I I
F.
:he
o')
H:h
r
Now consider the current flowing in the current element introduced i the positive gr-axis is 1 in o, direction. so, the magnetic field inte produced at point P due to the current element along the positive gr-a
i i
t I
ru r
:fi["osc2-
cosatfa4
:
cos
T
4;;[cos O' -
90"]a,4
(p
:
h,r]1
:
gQo, 42
=
: ho,
(a6: a, X a" = magnetic field intensity produced at point p dr
Therefore the resultant both the current elements will be
or,
Hn"t
:
Hn"t
: fin
*nC
ao* a,)
Thus, from equation (1) we have
H'"t ${3;.4.1"?,1
: /i H
Option (A) is correct. Consider the flux density at the given point due to semi infinite wire alr y-axis is .El1 and the flrix density due to wire along z-axis is 82. The magnetic flux density B produced at any point P due to a straight n carrying current ,[ is defined as
a
: ffiposo2- cosal]a4
'o.l:11/2
where
p - distance of point p from the straight wire.
+ a2 + a1
angle subtended by the lower end of the wire at p. angle subtended by the upper end of the wire at p.
and the direction of the magnetic flux density is given as
q is unit
a'6:otxdP
where vector along the line current and a, is the unit vec normal to the line current directed toward the point P. So, we have
:2m ad : av X (a"):P
a"
(ar:
o,r,
Q,o:
,
f:
1
: a, X a":-
Page 250
So,
Chap 4
Therefore magnetizing force is
a6
Magnetostatic Fields
n : fn6t-
H:h
4,
",)
Now consider the current flowing in the current. element introduced the positive y-axis is 1 in an direction. So, the magnetic field i produced at point P due to the current element along the positive g-axis
n:*l.ose2-cosarla6 4TPt
: ffi1cos}' - cos90']o6 :ffio.
(p
:
h,or
:
9oo, oz
: o'l
X a": ql Therefore the resultant magnetic field intensity produced at point P due to both the current elements will be
: ffitHo"t : hrn
Hn"t
or.
(or:
on
au* a,)
Thus, from equation (1) we have
H'*:/2H $*,,4"1"f;"*
Option (A) is correct. Consider the flux density at the given point due to semi infinite wire aloug y-axis is Br and the flux density due to wire along z-axis is Bz. The magnetic flux density B produced at any point P due to a straight wire carrying current I is defined as
A
:
ffiposa2 -
cosal]a,p
(2,0,0)
where
p + distance of point P from the straight wire.
+ angle subtended a2 + angle subtended a1
by the lower end of the wire at P. by the upper end of the wire at P.
and the direction of the magnetic flux density is given as
o'4:Q4xQ'P where or is unit vector along the line current and ao is the unit vector normal to the line current directed toward the point P. So, we have
P:2m a6 : Qy X (a"):-
a"
(at: anrap: a,)
: E, az: 0
at
Therefore the magnetic flux density produced at point infinite wire along gr-axis is
t' :
similarly
*)
Page 251
due to the semi
Chap 4 Magnetostatic Fields
(as y tends
cos$)1- o4
p
to
:-#o"
ffi(coso we have the magnetic fl'x density prod.uced at point p due to
semi infinite wire along z-axis as
Ar:-ffiU Thus, the net magnetic flux density produced at point -shaped filamentary wire is
a
:-ffa_ :-
4.1,26 :
p
due to the tr
2n&o"
2(ar* o") x
1o-7
wb/m2
Option (A) is correct Using ampere's circuital law we have f
l B' dl:
.
lhI"n"
As the conductor carries current .I which is uniformly distribute over the conductor cross section so, the current density indde the conductor is
l:4
we construct an Amperian loop of radius p(< R) inside the cylindrical wire for which the enclosed current is
t*:(#)trt : \#)
and since the current is flowing along z-axis so using right hand rule we get the direction of magnetic flux density along *a6. Thus, from Amperes circuital law, we have (86)(2rp) : 7","
or E-
4.1-26
'-:#(#) B-PIIP^ -
2nfr*o
Option (A) is correct. Similarly as calculated above we construct a,n Amperian loop of radius p(> R) outside the cylinder for which the entire current flowingin the wire will be enclosed. i.e. I"r" : f and from Ampere's circuital law we get, B6(2trP) : p4I
Bo:
So tu- 4.1.27
fia
PaI
w 1
p
Option (A) is correct. Consider one half side of the square loop to determine the magnetic field
r
';'
I $ p
'ilage 252
intensity at the centre O as shown in the figure.
Uhap 4 i' f aguetostatic Fields
,l
i
o
0i al2
il
I h
The magnetic field intensity f/ produced at any point wire carrying current I is defined as
P
due
to a strail
n:fiposa2-cosa1] where
p + distance of point P from the straight wire. ar + angle subtended by the lower end of the wire at P. az angle subtended by the upper end of 'thefire at P.
'
So we have
P:al2 ar:7t/2 az : n/4
and
Therefore the magnetic field intensity produced at centre O due to the h side of the square loop is
I t -r --^7r\-: Hr- 2"Gn\"oZ - "usZ/
I
m
As all the eight half sides produces same field intensity at the centre of I loop so, net field intensity produced at the center due to the complete sqri loop is Hnet
sot- 4.{.2a
:g/-J-\ :2qra I \J2ra)
Option (D) is correct. For the shown current loop we divide the loop in two segments as shown figure Segment
1
I:8
A
Ill
m
1m
;-J
P
I s.emuot z
2m Now the field intensity due to segment (1) (Semicircular loop) at point can be given directly as calculated in Que.60 l.e.
n,
: *
or
F'
: Zfu :2
where a is radius of semicircular lo
Alm
(o:
1r
again for determining the field intensity due to segment (2) we consider as the half portion of a complete square loop of side 2 m and since the fir intensity due to a completer square loop of side a carrying current 1 can
b-.
directly given from previous question. l.e.
Page 2r;r:
H_ zJzt 7fa
so the field intensity due
H,
ChaP 'r Magnetostatic FielJ.
to the half portion of square loop is
/i I :IH: zTa
s,:/z-G)-4/, " ,r(2)
(1:
7r
8
A,
a:2m)
As determined by right hand rule the direction of field intensity produced at point P due to the two segments will be same (inward) therefore, the net magnetic field intensity produced at point p will be Hn"r
-
Ht* Hz : ,
*****:***{<**
* nf :
3.8
A/m
inward.
Tt'
1
soLuTloNs 4,2
Page 254
Chap 4 Magnetostatic Fields
sol
4.2.,t
r0 I
Magnetic field intensity
at any point P due to a filamentary current
defined as
i
j
u:fi1cosa2-cos01]tr4 where
p + distance of point p from the current filament. or ' arrgle subtended by the lower end of the element ut p-[l '! 02 + angle subtended by the upper end of the element at p-] l
Flom the figure we have
P:/fa4:5 ot = IEl2 + Cosgr:
and
cos'c
0
: -L:P Js'iIE tt
Now we Put these values to get,
H:#(+3-9", :#o* :0.074aowbf
(1:5i m2 l
$oL
4.2"2
Correct answer is 1. According to Biot-savart law, magnetic field intensity at any point p drnl the current element ldl is defined as
n:IldlxF J 4trR" where -E is the vector distance of point Here current is flowing in o6 direction
p from the current
element.
So the small current element
Idl : Ip64o, :4 x 2dbao : gddao and since the magnetic field to be determined at center of the loop so r have
R
:2m
,l
(radius
: 2l
I
and
dn :- &p (pointing towards origiri Therefore the magnetic field intensity at origin is
t
n : [* (8deq):_t_ : a" Afm
/t2,3
a,)
:
f,; ,*_!_uo" : #[e]:0.
cm
According to Ampere's circuitar law, the line integrar of magnetic fier,r intensity 11 around a crosed path is equal to the net current enclosed by thtr path. since we have to determine the magnetic field intensity due to the infirril r: line current at p = 5cm so we construct a circular loop around the jine current as shown in the figure. Now from Ampere's circuital law we have
.
or rhererore
J'8.
dI:Fof","
B(2rp): /a X 10
(1*"
*" n'*
5
=
l() A)
cm as
l:ii-1"ff-ry'x""1':'l'' :4 X 10:;"wb/m2 Z' x S x-iOf
L
4.2"4
Correct answer is _20. According to Ampere's circuital the contour integrar of magnetic intensity in a closed path is equal.law to the current by the path.
i.e.
fi:i
"rr"tor"a
$8. dI:1"," "right
Now using hand rure, we obtain the direction of the magnetic fiertl intensity in the loop as it will be opposite to the direction of ,t.
so,
$n. m:-r"n":-2oA
(10A is not'inside the loof. So it won,t be considered.)
4.2.5
Correct answer is _1.5915 . The magnetic field intensity produced at a distance straight wire carrying current 1 is defined
H:+
p froman infinitery l,rrg
as
As determined by right hand rule, the direction of magnetic field intensily will be same(in _ a, direction) due to both th" magnetic fierd intensity due to both"rrr"ntir;;;;";;.,ri _r,,. the current carrying wirr,.; ;,rlHnet
H:Hr*Hz ^\, I ---I2r@\-/ av)+ ,'4it
au)
Cbap 4 Magnetostatic Fields
Correct answer is 4.
P:5
Page 2bb
t
:-q$* :-ln:-
Page 256
Chap 4 Magnetostatic Fields
i"
?,6
(r:
1.5e15av
Correct ansli'er is 6. Since the current flows from Q and terminates at Q2 and the charge located at the surface ofthe contour so the actual current is not enclos the closed path and the circulation of the field is given as
fn . m: pnllaf"," : *lf ' ds + | an, ' llo1"o"
M"h
J
and
asl
"" where .Ei is the electric field intensity produced by charge field intensity produced by charge
so,
Q2.
froL,":#[''(g).,,(g)]
Q1
while
.Ez
is
:+#*+#
As the current flows from Qr and terminates at Q2 so the rate of change the net charges is given as
-#:#:16A Therefore from equation (1) we have the enclosed displacement current
r
lIoL^":f t- rol +f {ro) : oa Thus, the circulation of magnetic flux density around the closed loop is
Ia'a:/ro(6)
J
..i
: 6po Wb/m d,l :6
LIn. J lto
'
t'
I.
5$L
4"1.7
ln.
d,l
Correct answer is
10.
:6
Magnetic field intensity at any point P due to an infinite current carrying sheet is defined as
n:!xxu
where K is the current density and o" is the unit vector normal to the current sheet directed toward the point P. Since we have to determine the magnetic field intensity at origin so from the figure we have
lol.
0* :- O, Therefore the magnetic field intensity at the origin is
Page 25?
n :${zou) x (- a"):toe'A/m
Magnetoctatie FtcHe
Cbap 4
(K = 20a,)
4,2"8
Magnetic field intensity at any point P due to an infinite current carrying sheet is defined as
6 :|xx
o,
where ,fi( is the current density of the infinite sheet and a" is the unit vector norpal to the cuirent sheet dF ctgd,to.i;af0 the point P. Since we have to deterlnirre tbe medhetic field:intensity ut point (2, -1, 5) which is above the plane sheet ai shown in figUie, so we have,
Arr:l
A"
Therefore the magnetic field intensity at the point (2,
n :|Poo,x iol.
4.?,t'
a)
-l,
b) is
:-$eu :-loouAlm
(K:4Qd,)
Correct ans\4/er is -0.13 . Consider the current density at p=8cp'is / drectgd along Now the magnetic field for p > 8 cm nqst bb zero.
ta:.
l.e.
(for p'!.8 cm)
I[
=0
So flom Ampere's circuital lp,w wo hbve
In'm-l",":o
I
Since for the region p > 8 cm the A.qperian loop wilt have all thd currs:rt distributions enclosed inside it. i.e.
I"n"
: l4x
10-3
+2 x (2Tx 0.s x :6.43 x
10-2)
-0.g x
(2zr
"
10-2+
x
x 10-2) +J(ZrxSx10-'z)
0.25
/(tdn'x l0a)
So we have
6.43 or or
x 1o-2+
,I(16n
x
10-1'+"0 d
a
:--
J':-
(1"""
6:43 x 10-! 16n x 10-2 0.13o,
A/m
= 0)
Page 258
sol. 4.2.ro
Chap 4 Magnetostatic Fields
Correct answer is -2. For determining the magnetic field at any point above the plane z: 0. draw a rectangular Amperian loop parallel to the 37-z plane and an equal distance above and below the surface as shown in the figure. Flom Ampere's circuital law,
m:thl"n"
Ie.
Since the infinite current sheet is located in the plane z: 0 so, the -component of the magnetic flux density will be cancelled due to and in the closed Amperian loop the integral will be only along y-axis. we have
: 231 :
B(21)
pal","
paKl
(1"n":
As determined by right hand rule, the magnetic flux density above the plarl z: 0 will be in - a, direction. So we have the flux density above the currc{ sheet as
A :-^l
a
&y
=-21taouwbfm2
ALTER}IATIVE iIETHOD: The magnetic flux density produced at any point carrying uniform current density I{ is defined as
(K:4Alnl
P due to an infinite shed
n -luo(K x a,)
where a" is the unit vector normal to the sheet directed toward the point P. So, magnetic flux density at any point above the current sheet I{ :4a" k
a :|n@a,) sol.
4.2,11
Correct answer is
(a":
x (o") :*2ttnaswbfm2
o,l
-4.
Magnetic flux density at a certain point is equal to the curl of magnetfu vector potential at the point. r.e.
B:Y xA
So from the above determined value of magnetic flux density
B
we have,
V x .4 :-2poauwbfm2 (l) Since ,4 is parallel to I{ so the vector potential ,K will depend only on z. Hence, we have
A
:
A(z) a"
Flom equation (1) we have,
I o, a, a,l _2pna,:l * *l
00 -2pnar:'-uilO, lA(z)
or
A(z)
Page 259
h
Che 4 Magnetctatic Fi,ldg
| I
:2pr,
A :2pnza,
So,
Therefore the vector magnetic potential
A 3,4.?,.12
Correct answer is
--
at
z:-2
is
4paa.wbfrn
12.
current density at any point in a magnetic field is defined as the curl of magnetic field intensity at the point.
J:Y xH Since the magnetic field intensity in the free space is given as
H
:2p2
ao
Therefore the current density is
t :iryA",:|firzp,)a" :6Pa":l2a,A/m2 ltol
4.!",13
(p:2m)
Correct answer is 6. Given that the cylindrical wire located along z-axis produces a magnetic field intensity, H : 3pa6.
so, applying the differential form of Arnpere's circuital low we have the current density with in the conductor
as
J:YxH a"l .lao pa6 a," .la,a pa6 _Ilplan a , ao a E do a;al_11 l:716 ' a
A,l I o 3p, 0 : |,fttttla,: 6a" Al^' ,lAo
sol
4"2"14
pA6
Correct answer is 5.5 . Magnetic dipole moment of a conducting loop carrying current 1 is defined
as:
m:
IS
where ,5 is the area enclosed by the conducting loop. So we have
m :7 x (zr x 0.52) : 5.5 (I : T A,.R : 0.5 m) The direction of the moment is determined by right hand rule as when the curl of fingers lies along the direction of current,then the thumb indicates the direction of moment
So, sol
4,2.,r5
m :5.5a,
A-m?
Correct answer is 0.73 . Since the current is flowing in the So,
Idl
: l|dz(-
a")
-
a, direction
Pace 100 c,fnlP
4
,,
Ma6mtort*le Flelds
defined as
n:f,l"osa2-costrr]a" - distance of point P from the current filament. or + angle subtended by the lower end of the element at P. oz + angle subtended by the upper end of the element at P.
where p
Now from the figure we have,
oI
P:2 az:r-0 Cos02 : cos(n -
0)
:-cosg:--+:,/2'
0r
and or
COS01
:0 : COS0:
n:
So,
+
3'-
3"15
(angle subtended by end
z: *)
1
fiPosa2 -
cosar]o4
:u!ml'-(-#)]",:9 Now the direction of magnetic field intensitv is defined
x (r+
h)"
as
'a4:atxaP where or is unit vector along the line current and an is the unit vector normal to the line current directed toward the point P. aa : (- a,) x (a,): a" So we have Therefore,
':#('+#)(".) : 0.73a, m
. sol
4,2,{6
A,f
Correct answer is 2.8648 . As the magnetic field intensity at the center of the triangle produced by all the three sides will be exactly equal so we consider only one side lying along c-axis that carries 4 A current flowing in + a, direction a,s shown in the figure. Now the magnetic field intensity at any point P due to a filamentary current / is defined as
n:fiPose2-cosa1]o4 where
p+
distance of point P from the current filament.
01
J
02 +
angle subtended by the lower end of the element at p. angle subtended by the upper end of the element at p.
c(m)
c(m)
From the figure we have
tan3o.
and
:t * e:h
(11
:,rT
0z
:
-
rf
6:+ + cos01 :
30o =e cos02
"orf
lt :- _T
- g9g$g.: /5
so the magnetic field intensity produced by one side of the triangle at centre
of the triangle is
H' :
az
;!J-[cos -', ,' ,ti
-
cos ar] a6
3:- /51{-a r ITt2,1il^ ]u,:ior
Now the direction of magnetic fierd intensity is determined
as
Q'6:0'1 XQ'P
where ar is unit vector along the line current and, ao is the unit vector normal to the line current directed toward the point p. and since the line current is along r_axis so we have A6 : A, X Ay: &z i,t (At= A, Op= Ov) Therefore the net magnetic field intensity due to all the three sides of triangie
is
H : BHt: s x (*)o :9a, A/m
(ao:
o,)
:2.9649a" Alm s{tl 4"*,{7
Correct answer is -0.75 . According to Biot-savart law, magnetic field intensity at any point p due to the current sheet element Kd,S is defined as
n : J"frqf_l:es 4rR.2 R is the vector distance of point p from ihe current element. Nou'we consider a point (0,y,2) on the current carrying sheet, from which we have the vector distance of point (3,0,0) where
R:(3a,*\an*0o,)-(0a,+ aay* za"):(3s, - yq_ a, : 3aL-J!+J? 3e' :-!.9- nsz
"/3"+f+l
-
^/g+ur+/
za.)
h1:fr f&cldE
'trpr
Page 262
Therefore the magnetic field intensity due to the current sheet is
Chap 4 Magnetostatic Eields
4(- za, - 3q,,) J":-rJ,:-* 4Rg + yp + \wdadz
_ : f, f-
we note that the r component is anti symmetric in z about the origin (odd parity). since the limits are symmetric, the integral of the r component over z is zero. So we are left with
':I:f
+^1sffi4qauaz
I:le; s)rtrv;71. :a,z :-** I,hd* :- n9o,l*,."-'(6)]1,
- - + ""
:-? , 7f
:sol.
4.2,{A
Correct answer is 2.26
(2)
0.7ba"
x
(0.5e) o,
Alm
.
C
'o
I-2
a
A
0.1
1m
As all the four sides of current carrying square loop produces the same magnetic field at the center so we consider only the line current AB for which we determine the magnetic field intensity at the center. Now the magnetic field intensity at any point P due to a filamentary cuuent .I is defined as
where
HI : 4fifcosa, - coscl]o6 p-+ distance of point p from the current filament. or
+
angle subtended by the lower end of the filament at
a2
+
angle subtended by the upper end of the filament at
P. P. Flom the figure, we have
p: i,^, 1
So
CLz=
45" and d,r=
180'- 45.
the magnetic field inten'sity at the centre o due to the line current
n':
ffi1"os02
-
cosar]
: GIG/D[cos45' - cos (180" - 45')]
t 2: :f;
-- 1\x J2
1\
n,'.,
A-B is
and the magnetic flux density produced by the line current .4.B is
Bt: :
lhHr :4tr X tO.7 X fr1f 5.66
x
10-7 wbf m2
Therefore the net magnetic flux density due to the complete square roop will be four times of .B1
i.e.
B:4Bt:4X :'2.26 x
(5.66
x
10-?)
10-6 wbfm2
Correct answer is 0.2 . According to Biot-savart law, magnetic fierd intensity at any point p due to the current element ldl is defined as
, : JI IdIx an 4trR2 where -R is the vector distance of point p from the current element. As the cross product of two parallel lines is always zero so the straight segments will produce no field at p. Therefore the net magnetic field produced at point P will be only due to the trvo circular section. i.e. H : Hco-f Hde
tE
:[["''
@ffi-l]*,:,. +[1"* wffi:trsrl,,:,_ :
I"''/%ao- ["''fuao :fr *['-+] (t)" :0.2 Alm "
ALTERNATIVE METHOD:
The magnetic field intensity produced at the center of a circular loop of radius .R carrying current 1 is defined as
g:J-2R
and since the straight line wilr not produce any field at point p so due to the two quarter circles having current in opposite direction, magnetic field at the center will be
,:i[*-*)
where
o -+ inner radius b -+ outer ra.dius
,:+[#_#]:o2Atm
f,iffi q{ff,
rUagtretrr*tlt
m P{G
e04
$oL 4.2.20
Chap d Ittagnstotatic FioLls
Correct answer is 0,82 . The magnetic field intensity at any point current 1 is defined as
.
p
due
to an infinite fi
H:* z7t p
p from the infinite current filament. Now the two semi infinite lines will be in combination treated as a si infinite line for which magnetic field intensity at point p will be where p is the distance of point
:#
H,
(.R is the length of
point P from line current
4 - nxz _t -i
(I:4A,R:2
As the magnetic field intensity produced at the center of a circular loop radius .R carrying current f is defined as
H:#
so magnetic field produced at point P due to the semi circular segment is
nr:t";E:+
Therefore net magnetic field intensity produced at point
p is
H:Ht+Hz_**i : 0.82 Alm $AL 4.2.21
Correct answer is 1.8
.
Magnetic field intensity produced at any point loop carrying curre4t -I is defined as
'E
p on the axis of the
: 2(f, ^It tflzrz^,+
^,
where ft, is the distance of point the radius of the circular loop.
P from the centre of circular loop and
p
Flom the figure we have
p:3m
and
h:5-1:4m
and using.right hand rule we conclude that the magnetic field intensity is directed along -Fa,.So the magnetic field intensity produced at point p is
H_
50
x
1o-3(3)2
+ 421e/z 9x50x10-3 :ffi@":l'8o'mA/m 2(32
Correct answer is 4. Assume the cylindrical tube is of radius a for which we have to determine the magnetic field intensity at the axis of solenoid. Now we consider a small ring (small section of solenoid) of the width d,z at adistance z frompoint p lying on the a:ris of the solenoid as shown in the figure.
i
The total current flowing in the loop of the ring will be d,I : nld,z where n is the no of turns per unit length since magnetic field intensity produced at an-y point p on the axis of the circular loop carrying current f is defined as
':T#w
where h is the distance of point
P from the centre of circular loop and p is the radius of the circular loop. So we have the magnetic field intensit;r due to the ring as
(p: o,h:
z)
Florn the figure we have
z:acoto+dz*_ffin and
sind:g:-L:+ r Ja2+l
1 -sin3d (o,+/yr'e3
The total magnetic field intensity produced at pdint
p
due to the solenoid is
u:ll--ffffi:+l:,ff-s,;-_odn) :-9|'::"odo :$("oro - coszr) :nI:1000X4X10-3:4Alm i 4.2.*3 Correct answer is -3. r
(n:1000,1:4mA) '
As calculated in the previous question the magnetic field intensity inside a long solenoid carrying current .I'is defined as
H
: nI
where n is no. of turns per unit length and since using right hand rule we conclude ihat the direction of magnetic field intensity will be right wards (+ or) due to outer solenoid and left wards ((due to inner solenoid. So the'resultant magnetic field intensity "r)) inside the inner produced solenoid will be
H : Ht+
H2 : nrl(- o) * n2loo where n1 arld na ate the no. of turns per unit length of the inner and outer solenoids
So
respectively. ' H -- (B x to{)(2000)a"+ : 3 x 10-3(- 1oo0)a, :-
.
(3
x
Bou
10-3)(1000)a
A/m
Page 265
Chap 4 Magnetostatic Fields
r,' 1
Page 266
sol
4-2.24
Chap 4 Maguetostatic Fields
Correct answer is 3. Since no any magnetic field is produced at any point out side a solenoid in the region between the two solenoids field will be produced only due the outer solenoid.
H:
l.e.
rulay
:1000x3x10-.or:lqAlm sol.
4.2,25
Correct a,nswer is 0. Since no any magnetic field is produced at any point out side a so, at any point outside the outer solenoid, the net magnetic field produced due to the two solenoids will be zero.
soL
4,2.26
Correct answer is 12.5
.
Since the current density inside the wire is given by
Jqp
J:kp
So we have,
where ,t is a constanL
and the total current flowing in the wire is given by
h: IJ'as J"
or
5x1o-3:[To'[,i"Aoo 5
x
10-s
-hrk(2 \
(/o
:
5
mA)
l0-2)g
u:*xtx#:ffix,03
So we have
Now for the Amperian loop at
p:1cm
enclosed current is
h":lJ.as :
fl x 1o-2
I kp(2rp) dp
JP=o
:(# x 1d) 'r"[+].
,
:+r+x1o-3:fixfi-s
So from Ampere's circuital law we have
$a.
Jt
m:thr"n"
B(2np)
:#
x
10-3
Therefore the magnetic flux density
B
at
p: lcm is
:#r rd$rT
x 4n x
ro-7
:1.25 X 10-8
: sQL 4.2.27
l2.5nWb/m2
Correct answer is -259. Current density at any point in a magnetic field is defined as the curl of magnetic field intensity at the point.
J:Y xH
l.e. So the current density component
in o, direction
is
J,:(V
I4,
:w-# "
Page 267
Chap 4 Magnetostatic Eblds
:-( : -\G;1Y -. 't,,o+6lz\t + ox' z
)ar
Therefore the total current passing through the surface x =
,3Sz<4mis
:-I^f
(#++z+z)dudz
:- I^l#+z+,afn,az ::4,?"28
m,
t < y< 4m
r:lJ,.ds :- I:,t,(&+otz)d'ud'z
aol-
2
259
Correct answer is 1.5708
(d,S:
dyd,za")
(r:2m)
+T2z)dz I^ (Zo
A
.
Magnetic dipole moment of a conducting loop carrying current 1 is defined AS:
m:
IS
where S is the area enclosed by the conducting loop. So for a ring of radius r, magnetic dipole moment
m: I(trf). Now as the charged disk(charge d.ensity, fu:20C1^')!4g!atinS witn angular velocity tl:0.1rad/s so, the current in the loop is givEn as d,I
:
p,
ard,r
Therefore the magnetic dipole moment is
*: :
f 'rl
a4nf1
l:o(rr"dr)(rf) : pru^ f d,, [' :20 xo.t t "[fl,
: t:1'5708 A-m2
r I
Page 268
&oL 4.2.25
Chap 4 Magnetostatic Fields
Correct answer is 1.39 . Magnetic flux density across the toroid at a distance
r
from it's center is
defined as
a:ffiao 1-
N
Total no. of turns Current flowing in the toroid across the toroid is given by the surface integral flux magnetic So, the total of the flux density
where
o: In'
l.e.
as
where d,S is differential surface area vectot. Consider a width dr of toroid at a dista,nce
r from it's center as shown in
figure Centre axis I
a
I
'i
H
;t+ffi
Irr+lln
total magnetic flux across the toroid
So we have the
o:
2tr
sol. {.2,30
(d,S:
I:-,^(#a,)@ara,1 4nx7O-7x105x10x
:
as
1.39
10
'(?)
(N:
105,
/:
hd,ra6) 10
A)
Wb
Correct answer is -4.31 . As determined in previous question the magnetic flux density across the toroid at a distance r from it's center is
a:ffiao So at the mean radius,
'--5!:1'5m B:hll67f o^
we have,
(r:
1.5
m)
Therefore the total magnetic flux is
o'
:I
"
.
d,s
: I:,(#
*){na,"o)
105 x 10 x 101-12 :Fl',l1 - 4zr x 10-7 x
(dS:
hdra5)
(l[:105, 1:10A)
:1.33Wb Thus, the percentage of error is Tnerror
:P4
a
x
100%
(d:
1.39wb as calculated above)
: E};#a x Tooro : - 4.3r%
Correct answer is 0.81 . The magnetic fl,x density produced at a distance p from a straight wire ca^rrying current .I is defined as
B:PI Now consider a strip dp of thesquare roop at a distance p from the straight wire as shown "iri$o in the figure.
Total magnetic flux crossing the strip is d{^: B(2dp)
(area of sftip
:ffiear) So,
the flux crossing the-complete square
:2dp)
loop is
":',,ikedp)
: Sl"d2
:e.#(t"f) --
Jn,
4.2.sa
g.11
x
10-7 Weber
:
0.g1!r,Wb
Correct answer is 20g.5 . As calculated in previous question the total fl'x crossing through the square Ioop due to the straight conducting element .
is
,!* : f'
"ffi{t
ao)
.where 1 is the current carried by the conductor, tr is the side of the square loop and a,b are the distance of the two sides of square loop from the conductor. So we
have
: 0.5 m o:0.3-g[:0.0bm
tr
b:0.3+ry:0.b5m Thus,
,b-
: I::",H@ sap) : ffU"olg;iE :
fflnny
Therefore the current that produces the net flux
t:^&t)x5x1o-5 = 208.5 A
{)*: 5 x 1o-5 Tm2 is
pagel Ch+r Magnetostatic
Fieb
Page 270
fist.
4"2.33
Correct answer is 5.6569
.
We consider only the half side of the loop to determine the flux density at the center as shown in the figure.
Chap 4 Magnetostatic Fields
The magnetic flux density B produced at any point P due to a straight carrying current I is defined as
3:/;1cos02-coso1]tl,, 41T
PL
p - distance of point P from the straight wire. or + angle subtended by the lower end of the wire at Poz + angle subtended by the upper end of the wire at Pand the direction of the magnetic flux density is given as where
a6: a1X aP where a; is unit vectolalong the line current and ao is the unit vector normal to the line current directed toward the point P. Therefore, the magnetic flux density produced at centre o due to the half side of the square loop is nr:]';1"os02-cos01){r6 ' 4trPt where Thus,
P: trn
o,
",
: E and coso2 : +V: VI +r
#
\,/ z
: Go#$D (#_ o)", (ao: aux (-0"): o,) :t!:o.Wb/m'? J2
As all the half sides of the loop will produce the same magnetic flux densitlat the centre so, the net magnetic flux density produced at the centre due to whole square loop will be
B sor-
4"2,34.
:8Bt :4J2 x !o-7 a"Wb/m'z :5.6569 x I0-7 o,"Wb/m2
Correct answer is 0.6366 . Using right hand rule we conclude that the field intensity produced at centre of the loop by the loop wire and the straight wire are opposing each other. so, the field intensity at the centre of the loop will be zero if Hrir" :
Hloop
...(1)
where H.i," is the field intensity produced at the center of loop due to the straight wire and H6* \s the field intensity produced at the center of loop due to the current in the circular loop. Since the magnetic field intensity produced at a distance p from an infinitely
]
f
long straight wire carrying current
H:+
So we
have
is defined
as
Page 271
Chap 4 lidagnetostatic Fields
H*,"
!Q:1q - =J--: n 2r(I)-2tr-
(I:20A,p:1m)
and as calculated earlier the field intensity produced by circular loop at its center is
Hr* : *,
Huop:rGd*F)
oft So
where o is the radius of the loop
-$:sr
(o
:
10 cm)
putting the values in eq. (1) we get 10
-.t
T -ut
I : Z: 7f
Thus, Dr.4,2.35
0.6366
A
Correct answer is -8. The flux density due to infinite current carrying sheet is defined
as
s:Sxxw r(
is surface current density and a, is unit vector normal to the surface directed toward the point where flux density is to be determined. So, for the sheet in z:0 plane, where
B, and for the sheet in
: t(4o,) x (",) :-
2pnav
(w: a")
z:2 m plane
ar: ttao')
x (- a,) :-2poan
(o":-
Therefore, the net flux density between the sheets is
B : Bt+ B2:- 4na, Thus the magnetic flux per unit length along the direction of current is
r^
l,
;ffiff; =t_
*****+*****
between the prates).
o")
tl.
soluTtoNs 4.3
Pag€272 Chap 4 Magnetostatic Fields
$0L
4"3.1
Option (A) is correct. is not possible to have an isolated magnetic poles (or magnetic If we desire to have an isolated magnetic dipole by dividing a magnetic successively into two, we end up with pieces each having north and poles. So an isolated magnetic charge doesn't exist. That's why the tota,l flux through a closed surface in a magnetic field
It
be zero.
i.e.
r
I
B'
dS
:0
or more cle-ar, we can write that for a static magnetic field the total of flux lines entering a given region is equal to the total number of flux leaving the region. So, (A) and (R) are both true and R is correct explanation of A. $sL
4,3,2
Option (C) is correct.
sot-
4"3,3
Option (B) is correct. The Magnetic fieid are caused only by current carrying elements aqd
gr
as
R AtrR3
Since an accelerated Qlectron doesn't form any current element(Idl) so it not a source of magnetic field.
sol.
4.3-4
Option (C) is correct. Accordirlg to right hand rule if the thufnb poipts in the direction of out or inw#d cufrent then rest of the fingers will curl along the direction magnetic flux lines, This condition is satisfied by the configuration in option (C),
sot.
4.3.$
Option (C) is correct. According to right ha,rd rule if the thumb points in the direction of then rest of the fingprs will curl along the direction of magnetic field li This condition iq satlsfied by the configuration shown in option (C).
sol
4.3.6
Option (D) is corrept. Since the magpetic flux density is defined as
and
B:Y xA Vx.B=pnJ
Now using the vector identity, we have
V x (V x .1t): V(v .
or,
A)- vzA Vx.B=V(V.A)-V'zA poJ: V(V . A)-Y'A
ort As the vector pciteritial is always divergence free so we get, y2A:_ poJ
lE
,L3.7
Optioh (A) h corffitr
L
/L3.8
Option (A) is correct.
aot
4.3.9
Option (A) is correct.
I_ d3.lo
Option (A) is correct.
$L
4.3.1{
Option (A) is correct.
t_
4_3.{2
Optiotr (B) is cprrect.
Jt
/t.3,t3
Option (A) is correct.
43.14
Option (A) is correct.
4.3.15
Option (B) is correct.
-. trL
rIrl *r ttlqn*e**nd
***rt*!t*ttt**
soLUTloN$ 4,4
Page 274
Chap 4 Magnetoatatic Fields
sol. 4.4.t
Option (C) is correct. For r) a,
I"ncro""d,
f u,
: (lra2) J
d,l:1"n"1o""4
H(2rr) : (tra2) J Io ttLr - in,
uol,
l.e.
For
r<
(ra2)
,t
forr>a
f_
o,
So,
$
n. m: H(2trr)
I"nctosed
:# +
H:^J'u 2ra' 3(}L 4"4.2
I":
Option (B) is correct. Assume that the cross section of the wire lies in the figure below
Hqr, r-y
forr(r
plarre as shown
ir
:
Since, the hole is drilled along the length of wire. So,
it
can be assumed tha
the drilled portion carriers current density of. - J. Now, for the wire without hole, magnetic field intensity at point P will b given as
H6(2rR)
: JQrR') =)
Ho,
: *
Since, point O is at origin and the cross section of the wire located in rplane. So, in vector form the field intensity due to the current carrying wir without considering the hole is given as
Hr: {@a,+ yau)
(1)
Page 275
Again, only due to the hole magnetic field intensity at point P will be given
Magnetostatic Fields
as
(Haz)(2rr)
:- J(trr')
n*:=* Again, if we take O' at origin then in vector form
n,
: j@,a,+ u,an)
(2)
r' and g'denotes point'P'in the new co-ordinate system. Now the relation between two co-ordinate system will be where
' So, putting
r:fi'*d
it into equation
and A:g' (2) we have
Hr:+Kr_Aa,+yayl Therefore, the net magnetic field intensity at point Hn"t
P
is
: Htt Hz: {do,
i.e. the magnetic field inside the hole will depend only on d.
tot
4,4"3
Option (D) is correct. Due to 1 A current wire in r-y plane, magnetic field be at origin will be in r direction as determined by right hand rule. Due to 1 A current wire in y-z plane, magnetic field be at origin will be in z direction as determined by right hand rule. Thus, r and z-component is non-zero at origin.
tol.
4.11.4
Option (B) is correct. The total flux, Cross sectional area, So, the flux density is given as
iD:L2mWb :1.2 X 10-3Wb ,4 : 30 cm' :30 x 10-a m2
D O 1.2x10-3 IJ:=;:*:0.4TeSla A 30 x 10-. soL 4.4.5
Option (A) is correct. The relation between magnetic flux density
B
and vector potential ,4 is
given as
B:Y xA sol
4.4.S
Option (D) is correct. For an isolated body the charge is distributed over its region which depends on the total change and the curvature of the body. Thus Statement 1 is correct Since the magnetic flux lines form loop so the total magnetic flux through any closed surface is zero. Thus Statement 2 is correct.
sol.
4-d.?
Option (A) is correct. The magnetic flux density in terms of vector potential is defined
B:VX,4
In.as:f(oxA)ds
as
Chap 4
o-fd"at
Pryo.{o
cFsl
u*irbutc
i.e. the line integral of v6ctor potential A around the boundary of a 5 is equal to the flux through the surface S.
.
Fhla
sot
4.4.8
Option (C) is correct. the figure. consider the current element along z-axis as shown in
of magnetic field directing normal Using right hand rule we get the direction to radial line OP.
sol.
4.4.9
Option (A) is correct' Forthegivencircularcylinder,considerthesurfacecurrentdensityisK.So. is given as the totJ current / through the cylinder
K(2rr) where
r
is radius of circular cylinder'
So,
so|. {.4,{O
: I
K:
*
: c,f
ro-\:5Q
A/"'
Option (C) is correct' r, -----^--+ element small current Magnetic vector potential of an infinitesimally
ii
defrned as
A:
l':
T##
Given that 'R where .R is the distance from current element'
So 3ol
+.n.r't
oo
A:0
Option (D) is correct. Considerthetwopa^rallelsheetsareseparatedbyadistance.d'asshownil the frgqre below
The two sheets carries surface currents
K:
Kuay
Pa[{p277
At any point between them the magnetic field intensity is given
u :i,*
X
(a,,*
a,t)
r ',r i,
4.4"1a
and ad :
g :|Kuo, x (-a"+
So,
trL
a,
az
a,)
Option (C) is correct For the given current distribution, the current enclosed inside the cylindrical surface of radius p for a< p < 6 is
L,"
: l,'(t,$)e"oao)
:\f{o'-':)
and the magnetic field intensitv is given as
In. at :!f;{t
-
H(2rp) H
5L
4"4"13
Jo(p'-
- --iF p
"l
ot)
Option (A) is correct. The radiated .E and .EI field are determined by following steps (1) Determine magnetic field intensity .EI from the expression
fl : p,H: Y X A (2) then determine
V
E
from the expression
x.EI:rdP -' 6t
So, the concept of vector magnetic potential is used E and -EI field.
to find the expression
of radiated
tol.
4-4".t4
Option (C) is correct. Using right hand rule, we conclude that the direction of field intensity is same as determined for the two correct elements 31 and 21 whiie it ig opposite for the current element 1. Therefore, from thd ampere's circilital law, we get the circulation of .E[ around the closed contour a.s ,{
J sol
4.4-15
n .m:
lencro,ed
"'**-
:21+BI- I : 4I
Option (B) is correct. The unit of magnetic flux density
sol.
4.4,16
(B) is Tesla or Wb/m2
Option (A) is correct. From Ampere's circuital law, the circulation of magnetic field intensity in a closed path is equal to the current enclosed by the path
i.e.
f
I'H'd,I:1","
So, for the 6urrent 1 the circulation at a radial distance .R is given as
H(2trR): I
u :4a
1
Mqulrfi }ttb
where o,u is the normal vector to the upper plate and a,r is normal vector to the lower plate both directs toward the point between them
anu:-
Cr+
as
; :..'
1
Page 27E
Therefore, the magnetic flux density at the radial distance
Chap 4
E
is
_ P4tt B-"^H-PyI - tiE
Magnetostatic Fields
sol-
4,4,17
80L 4.4,18
Option (A) is correct. Unit of work is Joule. Unit of electric field strength (.8) is volt/meter. Unit of magnetic flux is Weber, Unit of magnetic field strength is Ampere/meter. So, inthe matchlist we get, A - 4, B' 3, C Option (A) is correct. Magnetic field intensity at a distance current I
r
2,D' I'
from a long straight wire carrying
is defined as
H:-I - 2rr 1:=1- 2rr r : +* :1.59m so|.
4.4.19
Option (C) is correct. Consider the current flowing in the loop is 1 and since the magnetic field intensity is maximum at the centre of the loop given as
H:J -2r r
is radius of the loop. So, the current that must flow in the loop to produce the magnetic field 11: 1mA/m is
where
I:2rH:2X1X1:2mA 30L 4,4.20
Option (B) is correct. Magnetic flux density B in terms of vector potential A is defined
as
B:Y xA
A also we know V ' A: 0 but it is not the explanation of Assertion (A). i.e. A and R both are true but R is not the correct explanation of A. So,
sol- 4.4.2r
B
can be easily obtained from
Option (D) is correct. (1) Magnetic flux density in terms of vector potential is given
as
B:Y xA (2)
Poission's equation for magnetic vector potential is
:_ pnJ potential for a line current is defined (3) Magnetic vector y2
A
as
[P^nt A-_J_GE So, all the statements are correct.
sol
4,4,22
Option (C) is correct. Magnetic field intensity due to a long straight wire carrying current 1 at distance r from it is defined as
H:*
\L--
a
1
::!q zTr
Page 279
'Chap4
Magnetostatic tr'ields
r:*:1.59m Option (B) is correct. Magnetic field intensity due to an infinite linear current carrying conductor is defined as
f n.
d,t:r"," H(%rr): t 3 H:*
Option (A) is correct. The net outward magnetic flux through a closed surface is always zero as magnetic flux lines has no source or sinks.
f n.
t.e.
[*'-"
as
:o
(1)
Now, from Gauss's law we have
It,
a,
.B)d,u:f
a.
as:o
(2)
So, comparing the equation (1) and (2) we get
V.B:0 ao{. 4.4.25
Option (A) is correct. Given, the plane g: 0 carries a uniform current density B0a, mA/m and since the point ,4 is located at (r,20, so, unit vector normar to the -2) current sheet is A":AY Therefore, the magnetic field intensity
, :L*
x
K:30a" mA/m) toL
4.4.26
an
:!{soo") x (ar) __
t5&,mA/m
(
Option (A) is correct. The magnetic fl'x density at an;r point is curl of the magnetic vector potential at that point. Le.
B:Y xA
Flom the Maxwell's equation, the divergence of magnetic
fl'x
density is
zeto. i.e.
V.B:0
Again from the Maxwell's equation, the curl of the magnetic field intensity is equal to the current density.
i.e. or,
VxH:J VXB:pnJ
The expression given in option (A) is incorrect t.e.
sol.
4.4.27
(B:
paH)
B+V.A,
Option (A) is correct. superconductors are popularly used for generating very strong magnetic field.
I
rY I L
fl
hr" rao chry4'
so!.
4-4.24
ttft69{p{*h llclb
Option (A) is carrgct. As tb€ maguetic flu:i lines have no source or sinks i.e. it forms a loop. So thc total outWird flux through a clmed surface is zero.
i.e.
I
ll
{
sol.
4.4.29
{
J
B.
ds
:o
Option (A) is correct. The magnetic field intensity due a surface current density .K is defined
g =!xx
as
o^
Where c^ is unit normal vector to the current carrying surface directed toward the point of interest. K :2o,., Girren that, and since the surface cqrrying cunent is in plane z: 0.
for.-a < z
F. 1
iI
f
i'i
Fbr 0'<
tl
and
z1s,
^ On::
g,
nr:!(zo,) x (- @") : g"
Or:
nr:l(za,) ********;i:f
a,
*
x (",)
--
av
HAqTER 5 MAGNETIC FIETDS IN MATTER
txTRODUCTtol{
ffi"#:il::jililfTff
o t . o
;I1rJ""1-
to provide detaled concepts of magnetic
Magnetic f6rce on moving charge, current element, and between two current elements. Lorentz force equation that describes the net force on a moving charge in presence of both electrio and magnetic forces. Magnetic dipole moment and magnetic torque. Magnetic boundary conditions that relates va'v ta.ngential --.--v the v.-rsE and normal components of magnetic field.
o Magnetic of materials: magnetization, magnetic: . susceptibility,characteristic permeability. o Types of magnetic materials: paramagnetic, diamagnetic, ferromagnetic. o Energy stored in magnetic field and inductors. 3 Analogy between electlic and magnetic circuits.
IT
MAGNETIG FORGES Magnetic
exerted by a magnetic field on charged particles, current i:1::current loops. These t"**r*
[r]""r'uru discussed in folrowing
$:ffi::'and
L2.l
point Charge in Magnetic Field If we atow a charge to move e with a verocity u in the presence
Force on a Moving
l;'o' ",
then the magnetic force
F^:
.ff
exertei
o'ti"
of a magnetic
particle is given "n"rged
Q(ux B)
..(5.
1) Lorentz Force Equation since' the electric force p on a stationary -r or moving erectric charge Q in an electric field E is given by
.
F"_QE
so' from equations is.r;
ffi;tL**:}fr?in
*a
the presence or
(5.2), the. total force experienced [}? electric as weu magnetic
utr,'ir,"
;-;;"
F:F"*F^
:
QE+ Qfux B\
This equation is known
* i)*"tz
y'orce equation The comparison of the
two forces (electric and magnetic forces) experienced by a moving charge been summarized in the table below:
Page 282
Chap 5 Maguetic Fields in Matter
Table 5.1: Comparison between Electric Force and Magnetic Force
5.2.2
Force on a Differential Current Element
in Magnetic Field
The differential magnetic force experienced by various differential elements are given below.
r^:
[.tat, x n
(Line current
,^:l*xBds
(Surface current)
r^:ftxBda
(Volume current)
IdL is the line current element, KdS is surface current element, Jdr is volume current element, and .FL is the magnetic force exerted on thc respective elements in presence of magnetic field B. where
5.2.3
Force on a Straight Current Carrying Conductor in Magnetic Field Flom above discussion, we can say that if a conductor of length .L a current ,[ is situated in uniform magnetic field B, it would experience force given by OI,
F*:ILxB F- : BILsin9
where d is the angle between the vectors
5.2.4
B
ar,Ld
L.
Magnetic Force Between Two Cunent plements Consider the two differential current,elements ltd,Lt and lzdLz separated by a distance r. The magnetic force between the two current elements is given by
[dl'x(4'xo') F:FlIz[ an Jr,Jr,-------This equation is also called Ampere's force law.
5.2.5
Magnetic Force Between Two Current Carrying Wires Consider the two straight, long and parallel wires separated by a distance R as shown in Figure 5.1. Wire(1) carries current 1r and wire(2) carries current Iz. The force exerted by wire(l) at wire(2) is given by
p,:4*(-
oJ
Similarly, the magnetic force exerted by wire(2) at wire(l) is given by
E
P'
-
EyI'I' ^ -riTq
Thus we conclude that there is a force of attraction between the two wires carrying currents in the same direction, i.e.
l4l: +#:141 4:_ 4
and
,L-, R
!'igure i.1: Magnetic Force between Two Current Carrying Wires
F
TAGNETIC DIPOLE
A magnetic dipole is created when a negative magnetic charge positive magnetic charge - Q^ and, a Q* arepraced ai a smal ,Jo.r"iio' l. A bar magnet or a sma'filamentary loop carrying a current is known as a magnetic dipole. Figure 5.2 shows a bar magnet anj its equivalent
Magnetic Dipole Moment of Bar Magnet
"urr*
toop.
The magnetic dipole moment of the bar magnet is a vector quantity directed - e- to + e,". rt is denoted l, _-""1'gu"r.-*"'""'
from
m:
Q^l
Bar Magnet Surface
area,
^9
Figiire 5 2 : Dipole Moment of a Bar Magnet.and its Equivalent current Loop
Magnetic Dipole Moment of Current Loop If a magnetic dipole is rinked to a current loop with area s and carrying 1 as shown in Figure 5.2' then that ioop magnetic dipole
il ;
ffi#i
where,
a'
is the
"#;i:;ii,3r
r" the plane of the roop and its direction
page 28J MagLeiic Fields
b
Chap b Matter
is determined by the right-hand rule.
Page 284
Chap 5
Magnetic Dipole Moment of Current Carrying Coil If a coil has N turns (i.e., N loops), the net magnetic moment of the coil is given by
Magnetic Fields in Matter
m:NIS 5.4
IUIAGNETIG TORQUE
When a current loop is placed parallel to a magnetic field, forces act on the Ioop, that tend to rotate it. The tangential forces times a radial distance at which it acts is called the torque or mechanical moment, on the loop Mathematicallv,
T:d,xF where F is the force acting on the loop and d is the moment-arm (i.e., radial distance of the force from axis of rotation). Torque is expressed in Newtor metres. It is a vector directed along the axis of rotation of the loop.
5i.4.1 Torqrie in Terms of Magnetic Dipole Moment as a vector rn, with direction normal to the plane of the loop; the torque relation can be expressed in a more general form using vector product. i.e.,
If the magnetic moment is regarded
.l-nxxB where, T : Torque on loop (Nm) rn : rnan: magnetic moment of loop (Attt') B : flux density, Wb/m' 5.5
MAGNETIZATION IN ilIATERIALS Magnetization is defined as the amount of magnetic moment per unit volume. It is also called magnetic polarisation density. It is a vector quantilr and denoted AV (m) Its unit is Ampere per metre (l/*). If there are N atoms in a given volume Ao with magnetic moments rrt4,TTI4,Trt4,....rrL5 respectivelv, then' magnetization is given by
M 5.5.1
: Jir.(#i-,)
Magnetic Susceptibility In a linear material, magnetization is directly proportional to field intensity. i.e
M gH M -- x*H
or, where
X- is the
magnetic susceptibility of the medium. The magnetb
susceptibi,li,ty of a magnetic material is a measure of the degree of magnetization
of a material in response to an applied magnetic field.
5.5.2
Relation between Magnetic Field Inten^sity and Magnetic Flux Density
In a magnetic material, magnetic field intensity
3:
magnetic flux density is expressed in terms of
as
p,o(H* M)
: FoF,H :
:
P,H
u^(t+x^)n
where,
coil b
Page 2Eb
:
poF, is called permeability of the medium, expressed in Henry per metre (H/m), Ho : 4tr x 10-7H/m is the permeability of free space, known as absolute permeability,
F
tl, :(I*X*):fi
dimensionless.
)tl tb€ *ance
ir tt" t
relative permeability of the medium,
it
is
Classification of Magnetic Materials
loop-
Depending upon the values
of the magnetic susceptibility (x^) or the relative permeability (/"), -ugtretic materials are broadly classified into three groups as 1. Paramagnetic materials 2. Diamagnetic materials, and
3.
Ferromagnetic materials.
Paramagnetics and diamagnetics are linear magnetic materials whereas ferromagnetics are non linear materials. The characteristics of these magnetic materials are given in Table 5.2.
to n€ral nal
|.G
MAGNETOSTATIG BOUNDARY GONDITIONS Magnetic boundary conditions are the conditions that B or .E[ or M field must satisfy at the boundary between two different magnetic media. In the following sections, the boundary relationships are described separately for the field components normal to the bound.ary and tangentiai to the boundary.
unit
ntitr E are
.-?ll-r
15.f
Boundary condition for the normal components Consider the normal components of magnetic field shown in Figure 5.3. The two different magnetic media 1 and 2 are characterised by the permeabilities p4 and,6l2 respectively' Flom the boundary condition, the normal components of magnetic field are related as
Bn:
Bzn
In terms of the field intensity, the boundary condition can be written
lhiln:
as
p,zHzn
Thus, the normal component of B is continuous, but normal component of .EI is discontinuous at the boundary surface.
L'igurc ii.i): Magnetic Boundary Conditions
Chap b Magnetic Fields in Matter
Thble 5.2: Classification of Magnetic Materials
Page 2E6
Chap 5 Magnetic Fields in Matter
q.N. Paramagnetic 1.
Diamagnetic
The atoms or
molecules
have a
permanent magnetic dipole moment.
Ferromagnetic
The magnetic material Ferromagnetic materiak does not have permanent are made up of smafl patches called magnetk
magnetic dipoles.
domains. 2.
In the absence of
any
applied external magnetic
field, the
permanent
The presence of an external field Bs will induce magnetic dipole moments in the atoms or
magnetic dipoles in a paramagnetic material molecules. However,
An externally applied field Bo will tend to Iine up those magnetk dipoles parallel to thc
external field. The stron6 are randomly aligned induced magnetic dipoles interaction betweel and thus do not have any are anti-parallel to Bo, neighbouring atomk magnetization (M:0) leading to a magnetization dipole moments causes r and thus, the average M and average field much stronger alignmeil magnetic field (B;) is also anti-parallel to 116, and of the magnetic dipoler zero. therefore, a reduction in than in paramagnetk the total magnetic field materials. strensth
.).
these
The enhancement of thc
When placed in an external field (fr), the dipoles
applied external field can be considerable, with the total magnetic field inside a ferromagnet 103 or 1d times greater than tbe applied field.
a torque that tends to align rn with -E[, thereby producing a net magnetization (M) parallel to H . Since 81 is parallel to H, it will tend to enhance the field. The magnetization (M) is not only in the same direction * (II), but also linearly proportional to it. experience
4.
5.
Paramagnetism
is
temperature dependent. 6.
7.
lJ,)t,X^>0
p,,
1I,x*<0
X^) 0,F,) I
(Itt paramagnetics X^ (In diamagnetics X- is is usually of the order of usually of the order of 10-6 to 10-3.) - 1o-5 to - 1o-s) Examples: Air, platinum, Examples: Copper, Examples: Iron, steel gold, silver, lqad, silicon, nickel, cobalt etc. tungsten, potassium, aluminium, chromium, palladium, copper sulphate, manganese, etc.
diamond,
bismuth,
antimony, mercury, tin, zinc, alcohol, hydrogen, nitrogen, water, etc.
Boundary Condition for the Thngential Components Now' we assume that the boundary carries a surface current density k normal to the plane as shown in the figure. According to boundary condition, the ta^ngential components of magnetic field in the two media are related as (H"- Hr,) : K
or
(E -E"\: n \/rt Fzl
In vector form the boundary condition can be given as (Hr- H) X a,tz: K where, a'rz is the unit vector normal to the boundary directed from medium 1 to medium 2. If the media are not conductors, then the boundary is free of current, i.e., K: 0; and then,
Hrt:
Hzt
Eu:Pn
and
Fr
l-tz
Law of Refraction for Magnetic Field For a boundary interface with no surface current, we define the law of refraction as !u,n-d,
lrt
:
tan?z Fz
where d1 and 02 are the angles formed by the field components with the boundary interface, and pr,1, r-r2 are the magnetic permeabilities of the tvno different media.
TAGNETIG ENERGY
Iir order to establish a magnetic field around a coil, energy is required but no energy is needed to maintain it. Just as energy is stored in electrostatic field, energy is also stored in magnetic field of inductor. Energy Stored in a Coil Consider a coil with self-inductance .L. If the current flowing through coil is 1, then the magnetic energy stored in the coil is given by
w*:|tF
Energy Density in a Magnetic Field In a magnetic field with flux density B, the stored magnetic energy density is given by
,*:!1n . ny where
-Er is the magnetic fierd intensity in the region. The total magnetic energy stored in a region is obtained by taking the volume integral of the energy density, i.e.
w^: 3.8
f **d,,: I*@
.
H)d,u
MAGNETIC CIRCUIT If all the magnetic fluxes associated with a particular distribution of currents, is confined to a well defined path, then that can be considered as analogous to an electric circuit in steady state and refer as magnetic ci,rcu,it. tr'igure S.4
hrr
Chlr
Magnetic Fields
ir - -
T'
I Page 2EE
Chap 5
ciicuit fhiih is analogous to electric circuit. between electric a,nd magnetic circuits is given in Table 5.3. shows a magnetic
Magnetic Fields in Matter
Figure
l'-l..1: Analogr
betwen Elatric md Magnetic Circuit.
lhble 5.3: Analogr between Electric and Magnetic Circuits
*
rgpetiE :.1;tiii.,:.,,,
i ::a'::::? :::;,. I t4t.:;a : a.Li:a :t: :iar::.:.|
::.:is: j95:::i...i
ffi*
r!:}:irir]li!i;.s1,it
t:....:t::.::,,',:; 1:t
:.,::..:
..t
:ii?a*-,21 .,aa...;..,riiiit!.:.li::::i::i:li
ii:
W* ffiffiiff;
.',ri.'r;::il:li{i;!t4:.1-,!!
:t::l.l:;
r$
*********rt:1.
EXERCTSE 5,{
Pacs Magnetic
An electron is moving in the combined fields E:0.7a,- 0.2ar* 0.3a,kV/m and B:- 3a, f2an- a,Tesla. If the velocity of the electron at t:0 is y(0) : (200a" 300an 400a,)m/s then the acceleration of the electron at t: 0 will be (charge on electron, e: 1.6 x 10-1e C1 mass of electron,rn" : g.1 x 10-31 kg)
(A) (B) (C) (D)
x 1013(1.la, * 7.4a, - 0.ba") mfs2 2.1 x 104(a, * au- a,)mf s2 3.5 x 1013(6a, t6ar- a")mfs2 3.19 x 10 17(6o" *6ar- a")mf s2 1.75
DQ
s,1.2
A
lca
5"1.3
consider two current loops Q and cz carrying current d and .I2, separated by a distance ,8. If the force experienced by the current loop G due to the current loop G is -F, then the force experienced by current loop G due to the current loop G will be (A) (B) .F
current element of 2m length placed along z-axis carries a current of 1: 3 mA in the * o, direction. If a uniform magnetic flux density of B : o,,* 3o, wb/m2 is present in the space then what wiII be the force on the current element in the presence of the magnetic flux density ? (A) 6a, - 18a, mN (B) - 18a, * 6o, mN (C) 18a, - 6o, mN (D) - 1.8a, * 6ao mN
-r (c) - r(*)
(D)
\+,)
A rectangular coil of area 1m2 carrying a current df 5 A lies in the plane 2r * 6y - 3z :4. such that magnetic rnoment is directed away from origin. If the coil is surrounded by a uniform magnetic field B =6a"*4au*5a, wb/m2 then the torque on the coil will be (A) 3o, - 20an - 20a, N-m (B) 30a" - 20a, - 20a, N-m (C) 2la, - 74au - 14a, N-m (D) 6a. - 4an- 4a. N-m ICA
5"'t.$
A circular current loop of radius 1 m is located in the centered at origin. What
of magnetic field B : is flowing in the loop ?
(A) 2otr(2a,- a,) (C) &r(a,* ao) sco
5"1.6
prane
z- 0 and,
will be the torque acting on the loop in presence 4e, - 4ou - 2a"wbf m2 , if. a uniform current of 10 A (B) 40tr(a,+ ar) (D) 40tr(a, - au)
Magnetic flux density inside a magnetic material is B. If the the permeability : 3/o then the vector magnetization of the material will be
of the material is p
(A)
#
@)ffi
rc)
fn
D\ ?A ' Jgo
$e
6bap 6
ficldr ln lfidf€r
Page 29O
iltc{l
s.1.2
Chap 5 Magnetic Fields in Matter
t$cQ 5.1.8
A portion of B--H curve for a ferromagnetic material can be by the analytical expression 3 : pakH . The magnetization vector the material is (A) (6,k- t)a (B) kH (D) (k - r)H (c) (r+ t)a
M
A large piece of magnetic material carries a uniform magnetization M magnetic field intensity H, inside it. The magnetic flux density inside material is given by
Bs: Pa(Hsl Il/1) If a small spherical cavity is hollowed out of the material then the field intensity
H
at the center of the cavity will be
(A) 2rro (C) rr,
magnetic
@)
-ry
H0++
(D) rr.
-
+
Common Data For Q. 9 and 10 : A nonuniform magnetic field B inside a medium with magnetic suscepti X^:2 is given as B : Aza,Tesla ilcQ
lllcQ
5.'t.9
s,t.to
Bound current density inside the medium will be
(a) $a,.0./m'
(n) 3fo4, {/m'
Q)
(D)
fia,Alm'
ffio,A/m'
Total current density inside the medium will be
(^)
h", tl^'
@)
(D) 4paan Alm2
Q) fia,Alm' Common Data For Q. 11 and 12
f,a,Alm'
:
The two homogenous, linear and isotropic medium is defined in a Cartesian system such that medium 1 with relative permeability Hn: 7 is located in
the region g < 0 and medium 2 with relative permeability region y > 0.
pa: 6 is in the
s.l,{l
The magnetic field intensity in the 1"t medium is.EIr:9a,i76ou-10o,. What will be the magnetic field intensity in the 2"d medium ? (A) 9a, - 78.67ar-lI\a" Af m (B) 9o, | 2.63a0 - l\a" Afm (C) 9o, * 18.67 ao - 10a" A.f m (D) 18.67a" - 9ao -t 70a" Afm
5.1.12
Magnetic flux density in medium 2 will be (A) (6.8o, - 74.1au * 7.5a") x 10-5 wbf m2
(B) (6.8o, 1-I4.7a,r-7.5a") x 10-5 wbfm2 (C) (14.1a, - 6.8a, *7.5a") x 10-5 wbf m2 (D) (5ao, + I77oo- 60a")wbf m2
I
:
4a" * 3 a, Wb f m2 The magnetic flux density in the regio n z 0 is given as B 0 carries a surface current density K 4ay A/m; then the .If the plane magnetic flux density in the region z > 0 will be
z:
:
(A) 44, + 3(1 + un)a"Wbf m2 (B) 4o, * p4au*3a, (C) (ao, + 3a,)(1 + po) wb/m'? (D) 4o, lApnaulSa,Wbf m2
rq s.t.{s An infinite plane magnetic material slab of thickness d and relative permeability p" occupies the region 0< r< d. An uniform magnetic field B : Boa, is applied in free space (outside the magnetic material). The field intensity Hin and flux density Br inside the material will be respectively (A) Wp^Bo and p,,Bs @) J-r",t^and Bo
(C) t""Bo ""a
ft
(D)
lce s."i.ts In the two different are (81, 11r) and
ff
and. p,,Bo
mediums of permeability 1,tr and p2, the magnetic fields as shown in the figure.
(Br, Hr) respectively
If the interface carries no current then-the correct relation for the angle d1 and
dz is
(A) Blcos& : Bzcosgz (C) Brsindr : B6in0z
(B) Ilrcos4 : Hzcosflz (D) Both (B) and (C)
$"i"1s In a three layer medium shown in the figure bglow,
Magnetic flux impinges at an angle dr on the interface between regions 1 and 2. The permeability of three regions a"te [ht p2 and tr,lr. So the angle of emergence da will be independent of
(A) pt and (C) All h,
pz both Fa and Fz
(B) pa only (D) p1 only
Page 291
Chap 5 Magnetic Fields in Matter
u--'
$.1.17 The magnetic circuit
Pqsg 202 Ma6Dstic Dields ln
shown
in the figure has .l[ turns of
coil.
a,nalog for the magnetic circuit shown in the figure is
CbrD 6
Mrtter
H,B
HcGl
(A) r+
,u;
(c)
(D)
h
,5,1.18 The coil of the magnetic circuit 100
ro
NIo
shown in figure has 100 turns.
turns
E,B
Which of the following is correct electrical analog for the magnetic circuit
lR2 Rr -.r*O 8u
R7
1000/
0
(c)
tl6
,R3
1000r
lR1
10001
l
In the free space the magnetic flux density B points in the a, direction and electric field .E points in the a, direction as shown in the figure. If a charged particle at rest is released from the origin, then what path will it follow ?
$,"r"*o
A point charge *2 C of mass rt : 6 kg is injected with a velocity uo: 2aum/s into the region y > 0, where the magnetic field is given by B : 3o,wbfm2 . If the point charge is located at origin at the time of injection then in the region y > 0 the point charge will follow (A) a circular path centered at (0,0, - 2) (B) an elliptical path centered at origin (C) a circular path centered at (1,2,0) (C) a parabolic path passing through origin
$.1.x"? Two filamentary currents of -54, and 5o"A are located along the lines U : 0, z : - Im and A : 0, z : lm respectively. If the vector force per unit length exerted on the third filamentary current of 10a,A located at y= k, z: 0 be F then the plot of ,F versus k will be
10
e(*)
e(m)
Pryc
Mqntic
Fioldr
203
thrp 5 h Ithttctr
Page 294
Chap 5 Magnetic Fields in Matter
5
tt(^)
5.t.22 A current filament placed on r-axis carries a current 1:104 in *4, direction. If a conducting current strip having surface current density K:3a,A/m is located in the plane y:0 between z:0.5 and z:1.5m then what will be the force per unit meter on the filament exerted by the strip ? (A) 6.64" pN/m (C) 6a, pN/m
,i
Common Data For Q. 23 and 24 : A thick slab extending from U:- a to density J: Joa, trtcQ 5.1.23
(B) 6.6o, pN/m (D) 0
y:1o
carries a uniform current
Plot of magnetizing factor 11 at any point in the space (inside or outside slab) versus 9 will be
s"1-24 If a magnetic dipole of moment rrl:nr4ar is placed at the origin then the force exerted on it due to the slab will be (A) oN (B) mapaloya" (C) mapalsa" (D) - rnstuJ(.)!o,,
l
l
Common Data For Q. 25 and 26 : A long circular cylinder placed along z-axis carries
M:
s.{.zs
bp2
a
magnetization
J at any point inside the cylinder
is proportional
ao.
The volume current density
to
(A) (C)
(B) tlp (D) p"
p psin@
5.1.t6
The plot of the magnetic flux density B inside the cylinder versus p will be
Ge s.r.2?
Magnetization of a long circular cylinder is M along it's axis. Which of the following gives the correct pattern of magnetic field lines (B).
M
M
u
(c)
(D)
{t
\\
rcq $.{"zs
Magnetic flux density
B
inside a sphere that carries a uniform magnetization
M will be 0
(n) *p"iw
Q) @y
@) ?p,wt
(A)
P4r15 CtT5 Magnetic Fields in
I\[&
Paf"
Mcq
106
Chap 5
IUlgrlth
Sieldr in Matter
5.{.29
A short cylinder placed along z-axis carries a "frozen-in" uniform magnetizatim M in * a, direction. If length of the cylinder is equal to its cross sectional diameter then pattern of its surface current density .I( will be as
l
K
IM
(B)
K (D)
IM 11
til iirl
ti,l
ii,
ri
l
lii; iir! iiii
5.{.30 A short cylinder of length equals to it's diameter carries a magnetization M as shown in the figure.
uniform
L:2a
The correct sketch for the magnetic field intensity
-If
inside the cylinder is
5-1.31 An infinitely
long straight wire of radius o, carries a uniform current 1. The energy stored per unit length in the internal magnetic field will be (A) uniform and depends on I only
(B) non uniform (C) uniform and depends on a only (D) uniform and depends on both 1 and
o
s.1.32 A
mass spectrograph is a device for separating charged particles havirig different masses. Consider two particles of same charges Q but different masses m and 2rn injected into the region of a uniform field B with a velocity o normal to the magnetic field as shown in the figure. When the particles will be releasing out of the spectrograph the separation betweel them will be
o
s,,,,..]O
O,
Q
tJ
E
(B)ffi
rnu
(D) 0
(A) 2mu
(c) i , ;
i
t !
i N
!
?
I
w
Common Data For Q. 33 and 34 : Consider a conducting filamentary wire of length l meter arrd mass 0.3 kg oriented in east-west direction, situated in the earth's magnetic field at the magnetic equator. (Assume the magnetic field at equator has a value of 0.6 x 10-a wb/m2 arrtl directed northward)
I i
s'1.33 fa I I t r
I i
i
f I f !
t
r
!
The current that required to counteract the ear-th's gravitational force orr the wire must flow from
(A) west to east (B) east to west (C) any of (A) and (B) (D) none of these
I
I I
t-a s'1'34 ;
t ?,
I I L
i
i
lI lI
What will be the magnitude of thc currenl flowing in the wire the gravitational force ? (A) 4e kA (B) 24.5 kA
(c)
e8
kA
(D) 4.e kA
|Do s.1.3s B-.Er curve for a ferromagnetic lr f
,,
r E'
Ir I
material is given as B : 2paHH. what will be the work done per unit volume in magnetizing the material from zero to a certain value 86 :2poH3 ?
g) +p,4
tD ap^#
(c) ap,H\
@2+
lH $ E
l
b
I I
&.
r r I
as to counteract
Page 297
Chap
5
Magnetic Fields in Matter
s.1.36
Page 29E
Electron beams are injected nolmally to the plane edge of
field
Chap 5
.EI
:
a
uniform
Ho&, as shown in figure.
Magnetic Fielde itr Matter
The path of the electrons ejected out of the field will be in (B) - o, direction (A) +a, direction (n) (O - a,) direction (C) (O* o) direction
ileQ
$,1.37
Two perfectly conducting, infinite plane parallel sheets separated by e distance d, cany uniformly distributed surface currents with equal and opposite densities
.I( and -
K
respectively, The medium between the two
plates is a magnetic material of non uniform permeability which varies iinearly from a value of [17 neat one plate to a value of pq neat the second plate. What will be the magnetic flux between the current sheets per unit length along the direction of flow of the current ?
(o)
(s) (p'+ w)Kd
e+!')*o
P)(ry)Kd
(c) (fr + fi)xa
Comrnon Data For Q. 38 and 39 : The magnetic field intensity inside an infinite plane magnetic material slab is given as.Ef : 4a"!2ao. The permeabilitv of the magnetic material is;r :2114
s.t.3n If the magnetic material slab occupies the region 0
(n) (-aO *2a,)
and (44,
-2o')
(C) (Aa,+ 4ar) and (2a,- 4a') !!!eq
s.{,39
J* will be (B) 4a,*2au
The magnetization volume current density
(A) 0 (C) 8a"* tvrcQ s,l.40
(B) (-2a,+ 4q) and (2a"- aa,l (D) (2o, + 4q) and (-2a,+ 4o,l
4au
(D)
-4a,-2a,
Two infinitely long straight wire and6 third wire of length I are pa"rallel to each other located as shown in the figure.
Infinitely long wire carries a current l while the wire of length / shown at the top carries a current 2I . The magnitude of the force by the top wire is "*pJri".r""d
@)
+
ltrl @)# (B)
Q)#
Medium 1 comprising the region z > 0 is a magnetic material with permeability r^:4ltn where as the medium 2, comprising the region z< 0 is a magnetic material with permeability p2: zpn. Magnetic fl1x density in medium 1 is given by .
Br : (0.4a,* 0.ga, + a,)Wb/m2
If the boundary z: 0 between the iwo media carries a surface current of density .K given by
K :[email protected],_o.aau)Alm then the magnetic flux density in medium 2 will be
(A) (0.8o" + au+ a,)Wb1m2 (B) (- q,,*0.8aa- a,)Wbfm2 (C) (r" * 0.8a, * a")Wb/m2 (D) (o, * 0.8o,) Wb/m,
x***x*x>Fx**
page 299 Chap b Magnetic Fields in Matter
HXERCTSH 5.2
Page 300
Chap 5 Magnetic Fields in Matter
&ua$
5.2,1
An electron beam is passed through a uniform crossed electric magnetic fields .E :7\aaV/m and B:3a"wb/m2 (.8 and B are mut perpendicular and both of them perpendicular to the beam). If the passes the field without any deflection, what will be the velocity (in m of the beam
?
Common Data For Q. 2 and 3:
In the free
with surface current densit are located in the plarre z:0, z:7
space three uniform current sheets
Kt:4ar, Kz:-2a,,, Ks:-2a, z: - 7 respectively. &tiH$
5.?^3 Net magnetic field intensity produced between the sheets A1m in o, direction. and z: 1 will be
qrjtr$
5.4"3 If a conducting filament located along the line y:0, z:0.2m carries ? current in I a, direction then what will be the force per unit length on
gus$
5"2.4
it will be
x
proa,
located at
N/m
Magnetic flux density inside a medium is So"mwb/m2. If the permeability of the medium is 2.3, what will be the magnetization (in A; inside the medium ?
QL'r$ 5"2"$
A magnetic material of relative permeability F, : 4/r is placed in a field having strength H:2p2 aoA/m. The magnetization of the material p: 2 will be A/m in a, direction.
euss
5.2"s
A metallic bar of cross sectional area 2m2 is placed in a magnetizing H : 70 A/m. If the field causes a total magnetic flux of Q : 4.2 mWb in bar then the susceptibility of the bar will be
suxs
5"2.?
An infinite circular cylinder is located along z-axis that carries a uni magnetization M:0.7a, Afrn. The magnetic flux density due to it i the cylinder will be x 10-7 a5.
*ux$
5,2.s
Magnetic flux lines are passing from a nickel material to the free space. the incident of the flux line makes an angle at: 75" to the normal of tha boundary in the nickel side as shown in figure then what will be the angh a2 (in degrees) with normal of the flux when it comes out in free space ? (relative permeability of Nickel : 6fi)l
Page 301
Chap 5 Magnetic Fields in Matter
atrEs s.;t"s
Two infinite plane conducting sheets are located in the plane z: 0 and z:2m.The medium between the plates is a magnetic material of uniform permeability p : 4pn.If in the region between the plates a uniform magnetic flux density is defined as B: (3a,+4a)x 10-3Wb/m2, what will be the magnetic energy stored per unit area (in J l^r) of the plates ?
l'€s $"!.'!o A
conducting wire is bent to form a circular loop of mean radius 50 cm . If cross sectional radius of the wire is o, such that o << 50cm then the internal inductance of the loop will be nH.
q'Es s'a'{"t A 200 turns of a coil is wound over a magnetic core of length 15 cm that has the relative permeability of 150. The current that must flow through the coil to produce 0.4 Tesla of flux density in the core is Ampere. Common Data For
e.
L2 and LB
:
A filamentary conductor is formed into a rectangle such that it's corners lies on points P(1,1,0), 0(1,3,0), -R(4,3,0), S(4,1,0). An infinite straight wire lying on entire r-axis carries a current of 5 A in a, direction.
3Es 5'2"'t? If the filamentary conductor carries a current of 3 A flowing in a, direction * from Q to rt then the force exerted by wire on thc side Qi? of rectangle will be
____
orpN.
c',Es 5'2"t3 The total force exerted on the conducting loop by the straight wire will be avFN'
QEs 5'?'14 A conducting current strip of 2 m length is located in the plane r: 0 between 9 : 1 and u : z. rf surface current density of the strip is .r(: 6o, A/m the1 the force exerted on it by a current filament placed on z-axis that carries 1 current 1:10A in *a, direction will be pN in au direction. Common Data For Q. lb and 16 : A conducting rod of square cross section of side 2 cm carries a unifbrm magnetization M: 4 Alm along it's axis. Length of the rod is .t )) 2 cm.
ItEs 5"2.{5 If the rod is bent around it into a complete circular ring then density inside the circular ring will be x 1,r,swbf m2
magnetic flux
rut i.if'
ir t
Page 302
Chap 5 Magnetic Fields in Matter
s*Js$
$"2.16 Assume that there remains a narrow gap of width 0.1mm between the o{ the rod when it is formed into a circular ring. The net magnetic X 10-7 wbf m2 density at the center of the gap will be
strg$ s,2"{?
Mutual inductance between an infinite current filament placed along in and rectangular coil of 1500 turns placed in r-y plane as shown
will
mH.
be
(6, 1, 0)
QUrC 5,*"',18
&sn$ 5"9.',s
QUS$ 5.2.20
2 m wic A planar transmission line consists of two conducting plates of in * placed along r-z plane such that the current in one plate is flowing p the If both direction. a, irt iirection. while in the other it is flowing carries4Acurrentandthereisaverysmallseparationbetweenthemt x l-to force of repulsion per meter between the two plates will be
Averylongsolenoidhaving20,000turnspermeter.Thecoreofsolenoid and it carr formed of iion. If the cross sectional area of solenoid is 0'04 m2 per meter (J/ stored enelgy the be will a current .I:100mA then what iron, of (relative Permeability P.": 100) in it's field ? i A rigid loop of wire in the form of a square is hung by pivoting one of
abt side along the r-axis as shown in the figure. The loop is free to swing it's pivoted side without friction. The mass of the wire is }'2kglm carries a current 2A. If the wire is situated in a uniform magnetic
B:1.g6Wb/m,
then the angle (in radian) by which the loop swings
the vertical is
QUf;S 5.2.21
The medium between the two infinite plane parallel sheets current densities 4a" and, -4a'Afm, consists of two magnetic
carr
m and 2 m having permeabilities pr1 : 2pa and p"2: respectively as shown in the figure' The magnetic flux per unit len between the current sheets along the direction of flow of current will slabs of thickness
L
X p'oaowb/m.
!L--.'
K=-
page 303 Chap 5 Magnetic Fields in Matter
4a,
aiQa 9,@aa 2m
oooooeo K:4a"
Lm
5'2'22 Two infinite
plane conducting sheets lying in the plane r: 0 and z: 5 cm carry surface current densities * 20 mA/m au a,nd -zo ^Al*a, respectivery. If the medium between the plates is a magnetic material of uniform
permeability rr: 2pn then what will be the energy stored per unit area (in J /^") of the plates ?
5'2'23 A square
loop of a conductor lying in the yz plane is bisected by an infinitely long straight wire carrying current 2 A a,sshown in the figure. If the current in the square loop is 4 A then the force experienced by
the loop will be
pN in a, direction.
,
lEs
5'2'24 A certain region z < 0 comprises a magnetic iiredium with permeability lt':l\pn' The magnetic flux density in free space (z> o) *.r.o-ur urrgt" d, with the interface whereas in medium 2 flux density makes an angle d2 as shown in the figure. If. then whai will be the urrlgrrt* deflection (0r_ 0r) in degrees ',2:l.2au*0.ga, ?
Z:0
q'Es 5'2'25 The coil of a magnetic circuit has 50 turns. The core of the circuit has a relative permeability of 600 and length of the core is 0.6m. What must be the core cross section (in cm2) of the magnetic circuit so that the coil may have a 0.2 mH inductance ?
a
a
F
Common Data For Q. 26 and,27 Consider the magnetic circuit shown in figure
Page 304
z
Chap 5 Magnetic Fields in Matter
The cross sectional area of the section on which coil is wound is ,9r where all the rest of the section has the cross sectional area ^92. Magnetic core the permeability 1t : 7000p14.
:
:
cm' then the total reluctance of the circuit will
*i,ts$ 5"?.26
If
c{rss 5.2.2?
If the number of turn of the coil is 100 then the equivalent
,9r
5 cmz
of the coil
is
and
,92
10
self
i
mH.
Common Data For Q. 28 and 29 : A System of three coils on an ideal core is shown in figure below. The sectional area of all the segments of the core is ,S: 100 cm2.
2cm
1fi : 500 then what will be the self inductance (in mH) of the coil /y'' turns ?
$cr€s 5"3.28
If
$rrfis
If lfr:
5"2,e9
250 then the self inductance of the coil Nz
be
mH.
Common Data For Q. 30 and 31 : A system of three coils on an ideal core that has two air gaps is shown the figure. All the segments of core has the uniform cross sectional 2000 mm2.
Page 30b
Chap b Magdetic fieias in Matter
&:250 turn
4:500
turn
QUES
5':"3o what will be the mutual:lnductance (in mH)
QUE*
s"2.3* The mutual inductance between l/2 and lI, wili be
auE$
$.2.3p The magnetization curvd for an iron alloy is approximately given by
a If
Il
alloy
incredses from 0
is
: !n+
between Nr and /fz
?
mH.
H2prwblm2
to 210 A/m, the energy stored per unit volume in the
MJ/rn3.
**+*t<******<
EXSRC|$E 5.3
Page 306
Chap 5 Magnetic Fields in Matter
n{co
5,3.t1
Path of a charged particle A that enters in a uniform magnetic field (pointing into the page) is shown in.the figure'
aaaa
Path of charged pa.rticle
8A8A
The deflection in the path of the particle shows that the particle is (B) negatively charged (A) positive charged (D) can't be determined (C) uncharged MCQ 5,3.2
Unit of a magnetic point charge is (A) Ampere meter (C) Ampere meter square
(B) coulomb meter (D) doesn't exit
rvrcQ 5.3.3
Which of the following statements is correct for a current free interface between two different magnetic media ? (A) Normal component of magnetic field intensity will be continuous. (B) Tangential component of magnetic flux density will be continuous. (C) Magnetic scalar potential will be same in both the medium. (D) None of th"r"
i,tco
List I shows the type of magnetic materials and List-II shows their criterions. Match List I with List II and select the correct ansv/er using the codes given below : (Notations have their usual meaning)
5,3.4
List-II
List-I
a. b. c. d.
Diamagnetic Non-magnetic Paramagnetic
Codes
(A) (B)
(c)
(D)
1. 2. 3. 4.
Ferromagnetic
:
a
b
c
d
2
3
1
4
4
3
1
2
4
1
3
2
1
3
4
2
X.*:0, F,:I
X^)0, F,2
SI 0, p,,)>
X^<0,
X*))
7
p,,
7
XXHRffiilffiM S-4
Page 30E
Chap 5 Magpetic Fielde
ir
Mptter
MC& 5.4"1
A current sheet J :
A/rn lies on the dielectric interface r: with e"r :5, lt,t: 1 in R,egion-1 (r < 0) and €,2:2t lr,z:2 in Region-2 (" > 0). If the magnetic field in Region-l J r:0- is ,E[r : 3o, * 304, A/rn the magnetic field in Region-2 at r: 0- is 10q,,
between two dielectric meclia
x>0 (Region-2) : e,z: 2, V,,:2
t
+
l_ I
.I
x<0 (Region-1)
;
s"r:
5
(A) Hr: 7.5o,, * 30a, (B) Hr: 3a' * 30ao -
* l\a, Af m 10a" Af m
(C) Hr: Lba, * 40a, Af nt (D) II, : 3a", * 30o, * I]a, Af MCA 5.4.2
nr
A bar magnet rnade of steel has a tnagrretic nroment of 2.5 A-rn2 and a rnass of 6.6 x 10-3kg. If the density of steel is 7.9 x 103kg/m3, the intensity of magnetization is
(B) 3 x 106A/rrr (D) 8.2 x 106 A/m
(A) 8.3 x I0-7 Af nI (C) 6.3 x 10-7 A/m tlltcQ 5.4.3
If the current
magnetic field of
(A) -2.0a"mN (C) -2.04, N l|ltcQ $,4.{
H:5a. lp
Match List I with List below the lists :
List
a. b. c. d.
Magnetic flux Reluctance
Permeability :
abcd (A) 3421 (B) 1243 (c)3247 (D) t423
x
in a
(D) 2.0a. N
II
and select the correct ilnswer using the code gil'en
I
MMF
Codes
10 aa, Arnp-m is placed A/m, the for<-.e otr the current element is (B) 2.04, mN
element represented b1' 4
List
1. 2. 3. 4.
II
Conductivity Electric crrrrent Eil,IF R.esistance
ttc{i
$..i{. $
consider the following statenents as$ociated'nith boundary conditions
between two rnedia: 1" Normal component of B is continuous at the surface of discontinuity. 2. Normal cornponent of D may or may not be continuous.
Wrich of the.staternent(s) given above is/are correct? (A) 1 onlv (B) 2 onlv (C) Both 1 arrd 2 (D) Neither I rror 2 HCix s"4*s I
Magrretic currellt is cornposed of which of the following (A) Only condur:tion cornponent
?
(B) Onlv displacement cornponent (C) Bottr conduction and displacement components (D) Neither corrduction cornponent nor displacement component urc& $,4.?
wt'ch
one of the following is the correct expression for torque on a loop in rnagnetic field B ? (Here M is the loop moment)
r:v. B (C)":MxB (A)
MS*
$,;*"{1
Nlatr:h List I with List below the lists :
(B)
":M.B (D)":BxM II
and select the correct answer using the code given
List-I
a. b. c. d.
Line charge Magnetic flux density Displacernentcurrent Power flriw
Codes
List-II 1.
Maxwell
2.
Poynting vector
3.
Biot-Savart's law
4.
Gauss's law
:
aircd (A) 1243
(B) 43r2 (c) i342 (D)4213 [,t68
$"4.$
MC{} S"4,9*
What cloes the expression lJ . A represent (A) Elt't't ric ('rr(,rg.v derrsit.y (B) Magnetic errergy density (C) Power dersity (D) Radiation resistarrce
?
Two thi'parallel wires are carrying current along the same direction. The lbrce experienced by one due to the other is (A) Parailel to ihe lines
(B) Perpendicular to the lirres and attractive (C) Perpendicular to the lines and repulsive (D) Zero
Pnge
i09
Chap 5 Mnglietic Fftjtdl iri Matter
hr{O ft5 lltdc
iltGQ 5,4.11
ffiinMatter
ilCO 5.4.'t2
A
boundary sepaxatdsitwo,rnagnetic materials of permeability p1 and A, . The magnetic field'iector in pr is rrr with a normal component H6 anr! tangential component f4r while that in pq is H2 with a normal componem H,,2 and a tangential component r{2. Then the derived conditions wourd bc (A) f/' - H2andHn: Hn (B) //rt - Htza\dFrHa: FzHnz (C) flt: HzatdlhH,*: FzHnz (D) //t : Hz,Hn- H12and [hHa: paHnz
The dependence of
B (flux density) on H
(magnetic field intensity) for
different types of material is Paramagnetic Ferromagnetic
'-,eDiamagnetic
rvrco 5.4.13
Statement I : Pola.rization is due to the application of an electric field to dielectric materials. Statement II : When the dipoles are created, the dielectric is said to be polarized or in a state of polarization.
(A) Both Statement (1) and Statement (2) are individuallv true and Statement (2) is the correct explanation of Statement (1) (B) Both Statement (1) and Statement (2) are individually true but Statement (2) is not the correct explanation of Statement (1) (C) Statement (1) is true but Statement (2) is false (D) Statement (1) is false but Statement (2) is true
ilca
l
5.4.14 I I
The following equation is not valid for magneto-static field in inhorhogenous magnetic materials
(A) V . B:0, (C) V x A: B
f,,tco 5.4,{5
(B) V . H:o (D) V x H:J
j
{
Assertion (A) : Superconductors cannot be used as coils for production of strong magnetic fields. Reason (R) : Superconductivity in a wire may be destroyed if the current in the wire exceeds a critical value. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A)
(B) Bolh Assertion (A) and Reaso.n,.(R), are ,individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true 5,4"tS
A conductor 2 metre long lies along the z-axis with a current of 10 A in o, direction. If the magnetic field is B:0.05a" T, the force on the conductor is (A) 4.0a, N (B) 1.0a, N (C) 1.0a"N
5'4'4v
(D) 3.0aN
The force on a charge moving with velocity o under the influence of electric and magnetic fields is given by which one of the following ?
(A) q(r+Bxu) (B) q(.8+ ax H) (Q q@+ax E)
(D)q(.8+axB) s'4'18 If a very flexible wire is laid out in the shape of a hairpin with its two ends secured, what shape will the wire tend to assume if a current is passed through it ? (A) (B) (C) (D)
I
5.4.ts
Parabolic
Straight line Circle Ellipse
Consider the following : Lorentz force F : e(a x ts) where e,u and, B are respectively the charge of the particle, velocity of the particle and flux density of uniform magnetic field. which one of the foilowing statements is not correct ? (A) Acceleration is normal to the plane containing the particle path and B (B) If the direction of the particle path is normal to B, the acceleration is maximum
(c) If the particle is at rest, the field will deflect the particle (D) If the particle path is in the same direction of B, there will be no acceleration
5.4'2s What is the force on a unit charge moving with velocity u in presence of electric field E and magnetic field B ? (A) E-a.B I
u. B (C).8+Bxa (D) .E+ ax B (B) ,u+
HCQ 5.4"2t
l
vtrhat is the force experienced per unit length by a cond.uctor carrying 5 A current in positive z-direction and placed in a magnetic field B : (Bo" + aau) ? (A) 15a, * 20a, N/m
(B) -20a,* 15au N/m (C) 20a,- 15a, N/m (D) -20a,- 20a,Nf m
P4r 3fr
ct? Magnetic Fields' in
ltfie
5
Page 312
$cq
5.4.22
Chap 5
boundary'"'d
Which one of the followjrig formulae is not correct for the bound two rnagnetic mat€rials
9",
?Y
(A) B^: I (B) B, : ^/ B,r+ I (C) fl, : Ha* Hr I (D) a,zt X (H, - Ilir) : I{ where a,zt is a unit vector normal to the interface and directed from region 2 to region f
Magnetic fields in Matter
Bo
i i
i
ilc{t
5.4"23
Interface of two regions of two magnetic materials is current-free. The region 1, for which relative permeability lrt:2 is defined by z< 0' and region 2'
0 has Fa:7.If 81:7'2a,10'8ar* 0'4a'T; then IIz (A) r/rlo[0.0 o,,* 0.4o,y+ 0.4a"]Llm
z>
is
(B) tlp4lt.za,*0.8as+0.8o.]A/m (C) tlprlt.za,*0.4ay*0.4a"f Alm (D)
I
I
I
I
I
1/k[0.6a,*0.4ay+0.8o,]A/m
i
I
itca
r.4"24
If ,4 and J
are the vector potential and current density vectors associated
with a coil, then I n ' t (A) flux-linkage '
du has the units
(C) energy
XXXXXT(t<*(***
of
(B) Power (D) inductance
I
]
$Ol-UTlOl{S"5;'*
Page 313
Chap 5 &Iagnetic fields'ini i'1lttder
Option (C) is correct. For a moving cha,rge Q in the presence of both electric and magnetic fielcis. the total force on the charge is given by
p
time
J
glB+ (u x B)]
E --+ electri'c field o - velocity of the charged particle B - magnetic flux density
where
So, at
:
:
0 total force applied on the electron is
F(o): Nowwe
y(0)
have
,
e[E+(V(o) x B)]
I B :(200a,-300o*- 400a,) x (-i3.a,*2a,-r 1100o, *I400a, - 500o"
a,i
therefore the applied force on the electron is
F(0)
:
m"a(0)
(1.6
:
x
1.6
10-1e)[(0.la,
x
(f(0) : o(o;
:
-0.2a,*0.3o,) x
103
*11004, *1400a, - |r00ru.1 o,'+ + - 200)oo* (300 - 500) a ] m"a(0), where o(0) is a,cceleration of electron at t -- 0)
10-l'g[(100
(1400
1100)
i9+ ' "-s a) ---\.--, *6ao_ -./ 10-,,x 2oo(6o,, : 3.5 x 1013(6a, * 6a, - a")mf s2 19 9.1 1 x
Option (B) is correct. Force F applied on a current element density B is defined as
in the presence of magnetic ilux
F:I(LxB) .I
where
tr
--+
current flowing in the element --> vector length of current element in the direction of
current flowing
F :3 x l0-3f2a" x (o,+3a)] :6 x 1o-3[or-3o,] -- 18o,*6aomN
So,
I rol
5.1.3
Option (A) is correct. The magnitude of the force experienced by either of the loops will be sa,rne' but the direction will be opposite. So the force expfrienced by G due to C2 will be -.F.
3()L 5"t.j1
Option (B) is correct. Magnetic dipole moment of a coil carrying current .I and having area .9 is given by
m: where
ISa,.
o, is normal vector to the surface of the loop.
Since the coil is lying in the plane
2r*
69
- 3z:4
so the
unit vector normal
to the plane of the coil is given
Page 314
as.
Qhap 5
(f:2r*6Y-
So,
Magnetic l'ields in Matter
Therefore the magnetic dipole moment of the coil is
rn
: (5)G)p9'l+ ]e) _ 5(2a,+
6oo 7
B
As the torque a magnetic field is defined as
(1:5A,S:1
* 3a")
on the loop having magnetic moment
T:r'z.xB So the torque on the given coil is
,'-t_[5(2a"+ : ${}L
5"'1.5
304,
-
7-
6_au
20au
-
3a")
I x (oo, l 204, N-m
*
4ou
+
ba")
Option (B) is correct. Magnetic dipole moment of a coil of area ,9 carrying current 1 is defined
rn : ISan where a, is the unit vector normal to the surface of the loop. and since from the given data we have
1 S
:10A : rf :7r
X
(I)':
n
(normal vector to the surface z: I So the magnetic moment of the circular current loop lying in the plane z: an
: az
rS
m : IUra" Now the torque on an element having magnetic moment of magnetic flux density B is defined as
T:mX
m in the presenoe
B
Therefore, the torque acting on the circular loop is
| : (I\ra") x (4a" - 4ao - 2a") (B: : I}tr(4ay* 4a") : 40r(a,* ay) s$i-
s.1"6
4a,,
Option (D) is correct. Given the permeability, p: 3/o and magnetic flux density So the field intensity inside the material will be
i H,:#:&
-
4a,
-
2o,)
: B
i
Since the magnetization of a magnetic material is defined hs
lw
:E-
H
114
where B and So we get
5{}L 5"r,?
fil
are the flux density and field intensity inside the material.
BB28 M: Pn JUo
JUn
Option (D) is correct. As the magnetic flux density and magnetic field intensity inside a magnetic material are related as
ri
B:hIhH
,,
so, comparing it with given expression for magnetic flux density we get relative permeability as
th.
Page
Jlb
Chap 5 Magnetic Fields in Matter
!':k:k-I Therefore, the magnetization vector inside the material is given as
M:(w_t)H:(k_l).tl
Option (B) is correct. For the spherical cavity of magnetization Bcavuu
M, the flux density is given
\ by
:?^*
since the cavity is hollowed. so not magnetic flux density at the center of
cavity is
Bo"t:
Bo- B"o-tv
- Bo-?^*
and so the net magnetic field intensity at the center of cavity is
Hn"t
: *",",:
:
#[", -?^*] hl^*'* pnM-?^rl
:[**#],
(Bs:
Pa(Hs+ M))
Option (B) is correct. In a magnetic medium the magnetic field intensity and magnetic flux density are related as
B: n(I*x)H So the magnetic flux density inside the medium is
n:6#n:ffio"
(B:
4za,T,
y*:2)
Now the magnetization of a magnetic medium having magnetic field intensity .E[ is given as
M:X^H :q(lz\-\3n)-' o=:'z JPo
The bound current density inside a medium having magnetization given
M
is
B
is
a^s
Jt:Y xM l
l 1
:VX(ffi^):firo1^, 1
aoL 5.r.{0
Option (A) is correct Total current density inside a medium having magnetic flux density given as
"-m':h[w]"
(B:4za,,T)
l
Page 316
Chap
$oL
$"1.1't
Option (C) is correct. Magnetic field interrsity in 1't medium is given
5
o
Magnetic Fields in Matter
lfle
respectively the tangential and normal =ni,',\':,:'-'0"" to the boundary interface in rnedium intensity of the magnetic field Flom boundary condition we have H,,: H,, where H17 and
and
a,rQ
1.
lhHz,,: FtHu
respectively the tangential and nornlal component magnetic field intensity in medium 2. so we get the components in medi
where Hzt arld H2n
a.re
2as
Hy-9a,-I}a" Hzn: "*rt"
and
.
:4H,, p,2pn
:
n
6
(i6@J
:
t8.67 aa
Therefore, the net magnetic field intensity in medium 2 is
Hz: Hzt* Hzn : 9a,1-18.67 ao - l\a" Af m $sL
5.1."14 Option (B) is correct. Magnetic flux density in any medium in terms of magnetic field intensity defined as
B:FH where
p is the permeability of the medium.
medium 2 is given
So, the magnetic flux densilv
as
t' ::3?rl-Y?rY;x (ea.
: sol
5,{,*3
(6.8o"
I
*
!4.1a0- 7.5a,) x
18.6T an
- roa") (t"o:
10 5 wb/m2
Option (D) is correct.
The magnetic flux density in region z ( 0 is given
B :4a, l3a"Wbf
as
m2
Now we consider the flux density in region 1 is Br. So, we have
Br : 4a,* 3a" Therefore the tangential component Brr and noimal component rnagnetic flux density in region 1 are
Brt
and
Page 317
Br, of the
:4o''
86 = 3a,"
From the boundary condition the tangential and normal components of magnetic flux density in two mediums are related as
Bn: Bz' Brr- Brr: pnK where B21 and B2n ore respectively the tangential and normal components of the magnetic flux density in region 2 and K is the current density at the boundary interface.
get
and So, we
Bzn
u"
: Bn:3a"
(8":3a') (Brr:4a,, K:4ao Alm)
jIZ,,:f:T,,
Therefore the net flux density in region 2 (z > 0) is
Bz
s.i.14
r-
: Bzt* Bzn - 4a,*
Apnau-f 3a,
Option (D) is correct. As the surface boundary of the slab is parallel ro yz-plane so the given rnagnetic flux density will be tangential to the surface. 1.e.
Bro
and
llt'o
:
Bo
Lr -Bro-Bn ---- ltn Pn
Since the tangential component of magnetic field intensity is uniform at the boundary of the magnetic material so, magnetic field intensity inside the materiai is
il
F
i
H;n : l{ro-
r
Btlo
P'o
i
Therefore, the flux density inside the material is
Bt,:
i
:L
5.1"{5
ltHon
: f,^#
:
F,Bo
Option (D) is correct. From boundary condition the normal component of flux density is uniform at boundary
: Brsin4 :
i.e.
Brn
Bz, Bzsinqz
and the tangential component of field intensity is uniform
i.e. s"1.t$
Hu: H* -lricosdr : Hzcosqz
Option (B) is correct. Relation between d1 and
d2 at boundary of region (1) and region (2) p4tanfi - p2tan1z and at the interface between region (2) and region (3) is
pqtan9s
:
p4tan9ai So, combining the two eq. we get,
Thus,
da
patanfi - p4tan9a will be independent of pa only.
since
Chgp
E
Magnetic Fields in Matter
as
d2:
Bt
$tll.
Chap 5
Option (A) is.correct. Flom the analogy between electrical and magnetic
Magnetic Fields in Matter
following relations,
Page 31E
5.1.17
.f
we have
- 7 (voltage) -' 1 (current) @ (magnetic flux) R (Reluctance) - -R (Resistance)
(magnetomotive force)
Now, magnetomotive force,
f,:NIo and so, the electrical analog of the magnetic circuit is
sol- 5.{.14
Option (A) is correct. For drawing the electrical analog replace the coil by a source ( force) and each section of the core by a reluctance. In the shown material there are 9 sections so we draw the reluctance for each of them we get the magnetomotive force as
.f
:10001
So the equivalent circuit is
$oL 5.1.19
Option (A) is correct. Consider the particle carries a total cha,rge Q. Since for a moving charge Q in the presence of both electric and fields, the total force on the charge is given by F = QIE+ (u x B)l .E - electric field where o - velocity of the cha,rged particle B --+ magnetic flux density So initially the magnetic force on the particle will be zero as the is released at rest (o: 0). Therefore the electric field will accelerate particle in y-direction and as it picks up speed (consider the velocity a: kou,,k is very small) a magnetic force develops which will be given by
F:axB since the magnetic field is in a, direction while the beam has the velocity i a, direction so the magnetic force will be in o, (au x a") direction.
Therefore the magnetic force will pull the charged particle around to th right and as the magnetic force will be always perpendicula,r to both th li
velocity of particle and electric field. so the pa,rticle will initially goes up in the gr-direction and theh followitrg a path lowers down towards the r "otrie -axis.
Option (A) is correct. For a moving charge Q in the presence of both electric and magnetic fieicls, the total force on the charge is given by p glE+ (u x B)l where .E --+ electric field u --+ velocity of the charged paiticle Il - magnetic flux density since initially the velocity of the charge (at the time of injection) is
:
uo:2aum/s and for the region y > 0 magnetic flux density is B: 3a, wb/m2. will be no any velocity component in -l a, direction caused by the field (since the magnetic field is in a, direction). so there
a, :0 So we consider the velocity of the point charge in the region y ) 0 at a particular time I as a : uaay* u"a" Therefore we have the force applied by the field on the charge particle at time f as I.e.
-fft
"
So, we get
and
*#
F
:
""]
: Ql- 3u,a" *
Ql(urau
I
u"a,) x(3a")l 3a
"
ool
du,,
38 dt - mu"
+:-3Q atm
u.,
From the two relations we have
#*(#1,":o u, :,4,cos(
# r) *r,.i"(f;
r)
where ,41 and 81 are constants. and since at t:0, u":0 (since charge was injected with a velocity direction) Putting the condition in the expression we get 1,:0 and so we
have ,, : B,.ir(. ,r r): Brsinl
in o,
e:2C, m:6kg
Page 319
Chap b Magnetic Fields in Matter
:. ' 'r 1
tsage 320
Chap 5 Magnetic Fields in Matter
(0, 0, 1)
(0,
&,
o),1
rrro A
(0, 0, -1)
Net magnetic flux density arising from the two current filaments 5o, A at the location of third filament is given by
B:Bt*Bz
-5o.
an
(r
wlrere Br and Bz are the magnetic flux density produced by the currer filaments 5o, and -5o' respectively. Since the magnetic flux densil producetl at a distance p from a straight wire carrying current 1 is define AS
B
tt.af : ffioo
Page"321
and the'direction of the magnbtic,flux.,denSity is given as
a4:
o4
X
Chp
o'P
where ol is unit vector along the line curreirt and oo is the unit vector normal to the line cuirent directed to;ward the point p. so, the magnetic flux density produced by the currejnt filament 5o, is
B':';dfr6[""(tr#)] : *#qfta"+
a')
sinrilarly the magnetic fl'x density produced by the current filament
a':;frffi1[r-"'l "(ffi)] :
^#F,(
ko"*
(-
5o,) is
a'Y)
Therefore from equation (1), weget the net mh,gnetic flux density experienced
by the third filamentary current of
10 o,
A
as
: *ffi4.;fka,* au- ka,+ ao)
a
_
5p, b.^, 2r(l + k2)\--at
:
bu 4177540
As the force experienced by a curreiit element
flux density B is de{ined as dF :141*
ldl in the presence of magnetic
3
where 1 is the current flowing in the elernent and dJ is the differential vector length of the current element in the direction of flow of current. Force per unit rneter length experienced. by the third filament is
r:
['_ottoa,dr)x#na,
:u*f,*fu*:ffiuN
or,
r :6f6r'u
Thus, the graph between
10 sol- 5"f,22
1A:an-)
F
and ,t will be as shown in the figure below
:
&(-)
Option (B) is correct. Consider the strip is formed carrying curcent Kdz.
of many adjacent strips of width dz
each
b
'tr,tald*ic.riaas in Metter
Page 322
Chap 5 Magnetic Fields in Matter
Since the magnetic flux density produced at a distance wire carrying current 1 is defined as
p from
B:!"1 - 2rp So the magnetic flux density produced by each differential
strip
dB:tt"\Kd')' - --frl -u
is
Q
(Using right hand rule we get the direction of the magnetic flux along ar) Therefore the net magnetic flux density produced by the strip on the filament is
(K:3 Al
I'_:,Wa,:ffn(t*)" "::6.6 x IA-'auwbfm2 As the force experienced by a current element
flux density B is defined
ldl in the presence of
as
dF : Idlx B where 1 is the current flowing in the element and dl is the differential length of the current element in the direction of flow of current. So the exerted on the filament per unit length is
p
: JI tilx n : ft Ood,*a") x (6.6 x 10-7au) :6.6a,p,N/m Jr:n'
sol.
5.1,23
Option (A) is correct. Consider a rectangular Amperian loop of dimension (I) slab as shown in the figure below. Amperian loop
x
(2y)
---'l As from the Amperes circuital law, we have
Page B2B
I H. dl:1"," I
Chap b
So for the Amperian loop inside the slab we get
Magnetic Fields in Matter
HeA:eyx[)(Js)
for -a < U < a (Net magnetic field intensity along the edge 2g7 will be cancelled due to
symmetry) Therefore the magnetic field intensity (magnetizing factor)
inside the slab is
H:
or
at any point
JoUa"
n:Jolyl
(for
,
o)
lul< and the magnetic field intensity (magnetizing factor) at any point outside
the slab is
Thus, the plot of
5.1.24
H:Joa Il versus y will be as shown below
(for
lyl>
o)
Option (A) is correct. Force on any dipole having moment rn due to a magnetic flux density B is
defined
a^s
F:V(*.8) Since the magnetic moment of the dipole is given as ,ITt,
:
mnLa (1) and as calculated in previous question the magnetic field intensity produced due to the slab is
H:
Joga" So we get the magnetic flux density produced due
B - paH:
to the slab
as
trhJoAa"
(2)
Therefore from equation (1) and (2) we get
m. B:0 Thus the force acting on the dipole is
F:0 5.{.25
Option (A) is correct. Magnetic flux density inside a magnetic material is defined
B
B : pn(H+ Il)
as
and M will be in same direction inside the cylinder. Now as the magnetic field lines are circular so outside the cylinder make a loop. Thus, the magnetic field lines will be as shown below So,
it will
PsBe'324
Chop 5 Megr6ie. FioldE iri Ma*ter
sol- s,{,26
sol
5,1-27
As calculated above the volume current densitv inside the cylinder
J : liltaSo, we can get the flux densitv by Arnperer's circuital law
is
| I
as
!;,,.::I,';u
:
|
'#ot*'/: is
I
Now the enclosed current in the loop
I
l",":[J.rts : ['
I
fTr'"r)lpdpdrt)
I
:2tr xrrl(]:, : I}trf So, the magnetic flux densitv inside the cylin<,ler is
B:#1,,":ltl-r,np'
(I :
10trp3)
Thus the plot of magnetic flux derrsity B versus p is as shown below
sol.
5.4.28
Option (D) is correct.
Let the
magnetized sphere
magnetization be
be of radius r, ceutered at origin and the
M in o, direction
I
as shown in figure.
PsEe.0g5
Ohap 5 Magnuiic Fieldc in Matter
Volume current density inside a material is equal to the curl of magnetization
M
J:Y xM
i.e.
So the volunte current density inside the cylinder is
J:Y
x(L[a"):g
and siuce the surface current density in terms of magnetization is defined as K -- M x a,, where o,, is unit vector normal to the surface. So the surface current density on the sphere is
K:(Ma) x (o") : Msin9aa
(a": w) ...(1)
Now, corrsider a rotating spherical shell of uniform surface charge derrsity a . thzit corresponds to a srtrfirce current density at any point (r,0,Q). So we have
K : oaRsin9aa .,., - angular velocity of spherical -R - radius of the sphere.
whcre
...(2) shell across z-axis
and the magnetic flux density produced inside the rotating spherical shell is defined as
B
:lpoaR
...(3)
Comparing the eq.(1) and eq.(2) we get
M:
oaH,
Putting this value in eq.(3) we get the magnetic flux density for the rnagnetized sphere as
n
:!u,M
(M:
ouil)
Optiorr (B) is correct. The surface current density of a material in terms of its magnetization is defined as
K : M x a",
where
a, is unit vector normal to the surface.
So, the surface current density of the cylinder is
6:(Ma,)x(a,)-Mo4
(M : Ma,,an:
ap)
Therefore the surface current density is directed along a6 as shown in option
(B).
F"Tt
Page 326
sol. 5.{.30
Chap 5 Magretic Pields in Matter
Option (B) is correct. Since the magnetic flux density inside a magnetic material is defined : pa(H+ ItD
as
So, we have the magnetic field intensity inside the material as
n :LBM P4 and outside the material the magnetic field intensity is
II:LB P4 So the field lines outside the material
will be same
as for
the case of
flux density shown earlier. Whereas inside the material the direction magnetic field intensity will be opposite to the direction of magneti Thus the sketch of the field intensity will be same as shown in the (B).
sol 5.{.3t Option (A) is correct. Consider the wire is lying along z-axis. So at any point inside the wire ( distance p < a ftom it's axis) magnetic field intensity will be determined
f n. dr: r"n" H(2np): r(#) OI,
(Ampere's circuital
(for Amperian loop of radius
n:ffioo
The direction of the magnetic field intensity is determined using right rule.
Now the stored energy in the magnetic field 11 is defined
w^
as
: ftu"fi a,
in the internal magnetic filed per unit length (over unit length in z-direction) will be So the stored energy
I:, I:", I:,## Pd'Pds'lIz # 'Therefore, the energy per unit length depends only on / and is uniform I
:
:
the uniform current. sol- 5.{.32
Option (A) is correct. Consider the path followed by the two particles are the curvatures havii radii r, and rz as shown in figure. So at balanced condition centrifugal fcr will be equal to magnetic force.
f, i
[i I
k
r#
p ffi
t
ptr
:{ it
Ir
k l
t
d
Therefore for the first charged particles
+:
Bqu
) r':ffi
and So
PPI T2 :
Bqu
'" ) y,-p_Bq
Page 327
Chap 5
the distance between the two particles at releasing end is
d
Magnetic Fields in Matter
- 2rz_?":r(W)_r(ffi):W
5.1.33 Option (A) is correct. The wire is oriented in east-west direction and magnetic field is directed northward as shown in the figure.
I r I
I
I I
Since the direction of gravitational force will be into the paper(toward the earth) so for counteracting the gravitational force, applied force must be outward. Now the force experienced by a current element ldl in a magnetic field B is
r:f
Qat)x a As the magnetic field B is directed toward north therefore, using right hand rule for cross vector we conclude that for producing the outward force current must flow from west to east as shown in the figure below.
L3(,L 5.1,34
Option (A) is correct. consider the current flowing in the wire is 1. so the magnetic force applied by the field Bo on the wire is F^ : ILBo where .D is length of the wire At balanced condition the magnetic force will be equal to the gravitational force:
F^:
mg
where rn is the mass of the wire and g is acceleration due to gravity. So comparing the two results we get the current flowing in the vrire as
r-m9 ,_LBO Since
Therefore
tol
5,1.35
Bo
:0.6 x
I-
10-4Wb/m2, m (0.3) x e.8
: O;los , tol
Option (B) is correct. Given the B-H curve for the material,
:0.3 kg and -t : lm
49
kA
(S: 9.8 m/s)
'
Page 32E
Chap.5
j,, .,,,"8."*ieuaHlff
The work done per unit volume in m4gnetizing a material frorn 0 to has non uniform permeability is defined as
Magnetic Fields in Matter
.
,trm
-Bo
thet
: fth . aa Jo
Now for determining dB . we can express
B :2pnlf
dH
where a;r is the unit vector in direction of .EI.
So,
!,.
#:4paHa11 :4PnH
and
,^: Js -^l 3ln [u"H.(ap^H):+^141'":!P!!{3 3
sol-
5.{"36
Option (B) is correct. As discussed earlier, the path of electron will be parallel to the input beem but in opposite direction. so the ejected electrons will be flowing in the q direction.
sol
s"1.37
Option (A) is correct. As the permeability of the medium varies from p4, to 1t2 linearly. so at any distance z from one of tire plate near to which permeability is p1, the permeability is given
as
(ttz-
l-r:Ft+T,
tt't\
(u
The magnetic flux density between the two parallel,sheets carrying equal and opposite current densities is defined as
B:FK where K is the magnitude of the current density of the sheets. Therefore the flux per unit length between the two sheets is
!,Jo: foAa,
:
where d is the separation between the two sheets.
: *l'lr,*@;@,)0, + #(4)]:. : *(*+*1
(from equation (1))
liouxd,z
: Kfu, sol 5.1.3s Option (B) is correct.
z
o
Given the field intensity inside the slab is
H :4a,*2aa So the magnetic flux density inside the slab is given as
B : FH where ,u is the permeability :2pr(4a,* 2ar) Therefore the magnetization of the material is
of the material.
0t:2po)
M:E-H P4 : 8o" * 4a,u - (4a, * 2ar) : 4a, * 2ay Now the magnetization surface current density at the surfaces of a magnetic material is defined as
K^-Mxa" where o" is the unit vector normal to the surface directed outward of the
material So, at
z:
0 magnetizatibri purface current density is x (_ o", fK,^),r"=o
,: y :
and at
z
:
so, t39
4av
-
Page 829
(an: - a,)
Chap
5
Maguetic Fields in Mattsr
2a"
d, the magnetization surface current density is
Lr(^],r"=o:Mx(q") -(4a,*2au) x (o") :-4ay*2a,
(a,,: a,)
Option (A) is correct. As calculated in the previous question the magnetization vector of the
material is
M:4a,j-2q The magnetization'volume current density inside a magnetic material equal to the curl of magnetization., l.e.
rs
J-:Y xM
Therefore the magnetization volume current density inside the slab is
la- a,
a"l
l^:l{ ,i il:o lao
l+ | 2 ol
|
Option (A) is correct. As @ shows the direction into the,paper while o shows the direction out of the paper. So the wire of length / carries current 2I lhat flows out of th. paper.
The Magnetic field intensity produced at a distance straight wire carrying current 1 is defined as
l{-
p frorn an infinite
-wI
so the magnetic field intensity produced at the top wire due to the infiniter wire carrying current inward is 11^,
: --J-2n(J2I)
(p:izt';
H?
and the magnetic field intensity at top wire due to the infinite wire carryi'g current outward is H42
::-+;
-'#[!i"l!"sity at the wire of rength r is Hy : (H6* Har)coso
k:
lT
t,)
Therefore the resultan
:-2!:-
2tr1Jzt1" -L ur2 r.
I - 2rl
since the force exerted on'a current element ld,t by a magnetic field
i l
fl
:
is
ll1a 3t0
Ctq
defined as
"'
5
dF =(pIr(IdD So the force experienced by the wire of length I is
Magnetic Fields in Matter
r
:0,H,)(21)t
: u(fi)eU
:+
sol s"'r"4'r Option (C) is correct.
'
since the boundary surface of the two medium is z:0, so the nor component 81, and tangential component Bu of magnetic flux densitr medium 1 are
: a" Bt :0.4a,*
Brn
and
0.8o,
As the normal component of magnetic flux density is uniform at the bound
of two medium so, the normal component of magnetic flux density in medium 2 is
Brn: B1.n: a" Now for determining tangential component of field in medium 2, we I calculate tangential component of magnetic field intensity in mediur which is given as Hu :4!
where pr is the permeability of mediun
Ft'
:fitr.no"*0.8o,)
:Vt#t
0a:4
Again from the boundary condition the tangential component of magn, field intensity in the two mediums are related as
a, x (H1t- Hr,)
: K
where Hzt and H11 are the tangential components of magnetic field intenr in medium 2 and medium 1 respectively, -r{ is the surface current densit5 the boundary interface of the two mediums and o, is the unit vector non to the boundary interface. So we have ""
"
: o, 0.4 e,) fi{0., (# - Hu,)ou - (X - Ho,)o, : [email protected], - 0.4a,)
l!19#2!'
Comparing the
r
-
(Hz* a, + Hu,
a,)f
and gr-components we get
Hru:T*#:E
and
Hztu
:H *#:#
Therefore the tangential component of magnetic field intensity in medi
2is
Hr,:ff",+ff", and the tangential component of magnetic flux density in medium 2 is
Bzt : kzHzt: a' * 0.8ov Thus the net magnetic flux density in medium 2 is
Bz: Bzr* Bzn: +{<**1.*t< t
a,J-0.8a,a* q"
SOLUTION$,5.2
Page BBI
.
Clrap 5,
Magnetic Fields in Matter
tL
5"2,,!
Correct answer is 5. For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by
where
F
:
.E
+
e[E+ (u x B)] electric field
o - velocity of the charge
B
-+ magnetic flux density
Since the electron beam follows its path without any deflection so the net force applied by the field will be zero l.e.
Q[E+(o x B)]
:6
ISau*uX3a":Q As the electric field is directed along a, and magnetic field is directed along &u so the velocity of beam will be in a" direction (perpendicular to both of the field).
V : ka" 1Bg,":0 lSan-3kao:g
Consider the velocity of the beam is So we have 15ao*ka,
' So, the velocity of the beam
L s.2.2
k:f:5*/,
will be 5 m/s along the r_axis.
Correct answer is -2. The magnetic field intensity produced at any point in the free space will be the vector sum of the field intensity produced by all the current sheets. Since, the magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density I{ is defined as
n :|(x
x
a,)
where a" is the unit vector normal to the sheet directed toward the point p . so in the region 0 < z < 1 magnetic field intensity due to f{z and I(s will be cancelled as the unit normal vector to the two sheets will be opposite to each other.
Therefore in this region magnetic field intensity will be produced only due to the current density Kt: 4a, which is given as
n :*Krx an:!t+o,) x (a,) :_
2%
(o,:
o,)
Alm
Correct answer is - 14. As the conducting filament is located along the line g: 0, z:0.2m which is in the region 0 < z < 1m, so, the net magnetic field intensity produced on the conducting filament by the current sheets is
H--
2a"
Alm
(as determined in previous question)
Chap
B = UgH:-2Pna,t
oft
Page 332
Now the force experienied by a current element magnetic flux densitY B is defined as
5
Magnetic Fields in Matter
IdI in the presence
dF : IdIx F wherelisthecurrentflowingintheelementanddtisthedifferential length of the current element in the direction of flow of current' conducting filament is So ior"e per unit length experienced by the (I:7 A, dl: (-2prao)
#:7a,x :- lLtno,,Nlm
s{31-
5.2,4
Correct answer is 2249. field i In a magnetlc medium the magnetization in terms of magnetic is defined
as
M:X*H where
1-
is magnetic susceptibility given as
(relative permeability, pr'': \n, : lt - 1: 1.3 flux densiqv and since the magnetic field intensity in terms of magnetic given as
H:1:ffi :
So
sol
5.2,5
(B:\a,"mwb/
b
4zr
\_Ig-t ^ =o, x 10-' x 2.3
:
lTJ}a, Af m
the magnetization inside the medium is
M : X*H :2249 Alm
Correct answer is 2.19 Magnetic flux density
.
in a medium in terms of magnetic fiIed intensity b
defined as
B:pH:1,l,pnH : (lr)$n x 10-1 Qt ' :32 x 70-7 p2 a6
(p'':
"o)
4f
r
'
H:2f
ao
Alml
Againthemagneticfluxdensityinsideamagnetizingmaterialisdefinedas 6 : p^(U+ M) where M is the magnetization of the material' So' we have
* :f,-,
:lT#-'tf"r : r,,l+_ ,]",
M sot
5.2,6
p:2
M:
2.19o0
Correct answer is 22.87 Given
A/m
'
:70 Alm
H Magnetic field intensitY' @ :4.2 mWb Total magnetic flux in the bar, S:2m2 Cross sectional area of bar, in the bar So we have the magnetic flux density I
I
B
_F__2_-_, :9 - 4'2 >t'10-'
,,
Page 383
Chap 5
:
2.t mwl;lmz since tfre magnetic field intensfty and magnetic flux density are related
Magnetic Fields in Matter
as
B = tu(L*xiH 2.1,x 10-3 (4n y i0-?) (1 + .r._) (20) = .
r
So, we
have
r
(t+xi:ift#fi
I
!,
y _/3x10-5 _f\ ^n-\4,rxloi-L) : (23'87 - r) :22'87
trL
5.3.?
correct answer is g.g
.
volume current density inside a material is equal to the curl of magnetization
M
i.e.
J:YXM
So the volume current density inside the cylinder is
J:Y x(0.2a"):O
1M:Q.Ta"Afm)
and since the surface current density in terms of magnetization is defined as K : M x an where o, is unit vector normal to the surface. So the surface current density of the cylinder is 7S
:
(0.7a")
x o, :0.7a0
(M:0.Ta"Afm, a,:
ao)
Therefore the current flowing in cylinder is just similar to a solenoid and the field inte'sity produced due to a solenoid at any point inside it is given as
B
: paK: panl
where n, is the no' of turns per unit rength of the sorenoid current flowing in the solenoid.
a'd 1 is the
Thus, the magnetic flux density inside the cylinder is (direction is determine,l
by right hand rule)
B pL
$.2.8
:
0.7
pna,
:
8.8
x
10-7
a,
(K
:0.7)
Correct answer is 0.3b6 From snells law we have the relation between the incidence and refracted angle of magnetic flux lines as
tanat _ tana, where
trr'1
prr 1t*
and pt2 ate relative permeability of the two medium.
tan /b tano2
Page 334
Chap 5 Magnetic Fields in Matter
600
tan /b
tano2 Q2
600
_.^_-1[tan75"1 _ udr
: sol.
5,2,9
(relative permeability of air
1
t-T0fJ
0.356'
Correct answer is 5. The magnetic stored energy per unit volume of the plate for a given flux density (uniform permeability) is defined as
:t"- "
wm
Given
B :(Ja,+4au)x 1o-3wb/m,
So we have,
H
- L-(V#t)x
:L,r.
and therefore
u
:;[#] :2.49
1o-3A/m
x10-6:#4,lx+F
Jlm3
now the separation between the plates is given as d:2m Thus magnetic energy stored per unit area of the plate is
W*lA:u*xd -(2.49)X2:5Jlm2 sol-
5.2"10
Correct answer is 157.1 . Internal inductance of a loop of radius tn
r
is defined as
: ffi(arr1 :4tr x lo-7 x
2rrX 50 x
: $oL 5.2.{t
(r:
10-2
50
157.1nH
Correct answer is 1.6 Given that,
.
B :0.47 N :200 I : lScm:
the magnetic flux density, no. of turns of coil, length of magnetic core, permeability of the core, So, current required
15
X
to produce the given magnetic field
10-2
is
'-BI " -iN (1501a)(200)
$oL 5.2.12
Correct answer is -3. The magnetic flux density produced at a distance p from an infinitely straight wire carrying current 1 is defined as
B:lt - 2np So the magnetic flux density produced by the straight wire at side Qfi of loop is (direction of magnetic flux density is determined by right hand rde)]
Bqp:
ffi""
(P:
page 33F
(1r:5A)
Magnetic Fields in Matter
Chap
:15 o. o7f 8(1,3,0)
3)
a(4,3,0)
1r:5 A Force experienced by a current erement
density
B
is defined
as
:
dF
J41*
ldr in thepresence of magnetic fl'x
6
where 1 is the current flowing in the element and dl is the differential vector length of the current element in the direction of flow of current. So the force exerted by wire on the side of the square
er?
loop is
Feo:l)^ho'xBqn i
where 'I2 is the current flowing in the square roop as shown in the figure. so, we get
Fe,
:
" (W) rl(- a,)
In ,{saro,1
:*v-
(Iz:
3
A,
dl:
dra")
_-5x4zrx10-7x3_ --2i-aY :- 3 x 10-6 arN -- B ou pN tol. 5.2.!3
Correct answer is 6. Total force on the loop will be the vector sum of the forces applied by the straight wire on alr the sides of the loop. The forces on sides pe and .R,S will be equal and opposite due to ,y-*Ltry and so we have FPQ* Fns :0 Therefore the total force exerted on the conducting loop by the straight wire
is
: Fgp* Fsp (1) where Fqa and Fro are the forces exerted. by the straight wire on the sides Q.R and .9P of the conducting loop respectively. As calculated in previous question we have Ftotal
Fgp :_ 3 x 10-oa, N similarly we get the force exerted by the wire on the side ^9p of the loop
Fsr:
f"rrarx
as
B,sp
where Bsp is the magnetic flux density produced by the wire on the side . So, we get.
sp
E
tu-ffio,
Page 336
Chap
5
Magnetic Fields in Matter
:*n" Fs,
(1t
: I^'t(- d,ua,) x ffo,
(Ir:
3 A,
:'i
dI: -
:9 x 10-6arN Thus, from equation (1), the total force exerted by the straight wire on conducting loop is Ftot"l
tt*L
5"2.1&
:-3 x 10-6an* I x :6 x 10-6arN
10-64,
Correct answer is -26.4 . Consider the strip as made up of many adacent'strips of width dg. carrying current, Kdy
I x:oo"
Since the magnetic
t1^
flux density produced at a distance p from a str
/ is defined B_ el 2np
wire carrying current
as
So the magnetic flux density produced at distance 3r from the filament located along z-axis as shown in the figure will be
"
:
c
Hf- o,) (Direction is determined using right hand
:-#A-10r*' zTra
As the force experienced by a current element
flux density B is defined
ldl in the presence of
as
dF:Id,lxB and since the length of strip is of strip is given by
l:
2
m so, the force exerted on the width
dF:l(KdixB Therefore the net force exerted on the strip is
r : [' ,tz) (6a") x (-H *)ty :-Y*1t"'1' :-
26.4a, pN
Q
= 2 m, K
:
5.2-t s
Correct a.nswer is 4.
L€t the circular ring being placed such that magnetization M is iri direction and So, we
have
the ring is centereil at origin.
page B3Z aa Mag;nbtic Fields
M :4aa
Circular Ring
As the surface current density of a materiar in terms of its malnetization is dcfined as
K : M x qn
where
a, is unit vector normal.to the sutface.
So the surface current density of the ring is 4ao X ao)
6
:
(-
4q" (Iu[ = 4aa, a,:, ap) and since the volume current density insiJe a material is equal to the curl of magnetization M
J:YxM
l.e.
So the volume current density inside the ring is
J:Vx(4aa):g
(M:'Aaa)
Now from Ampere's circuital law we have
{a. u:t'^l. l,
Pw ' Pnf
-'
and for determining the fierd inside the circurar ring, the current present on the inner surface of ring will be considered only. So we get
(B)(arfi: wegetrp)
Therefore the magnetic flux density inside the circular ring is
B
:
(p*) (4)
:
(K: a Alm)
aprwb/m2
ALTERNATIVE METHOD: Magnetic flux density inside a magnetic material is defined
B:thM
and since the magnetization of the rod is directly the magnetic flux density inside the
B :4pnWb/m2
M: 4A/m ring
as
a^s
so, we can have
h
Chap b Matter
Page 338
Chap 5 Magnetic Fields in Matter
5(}L 5,2.t6
Correct answer is 50.04 . As calculated above for the complete circular ring, magnetic flux inside the ring is
B:
pnaawbf m2
(Magnetic flux density will be directed along the assumed d magnetization)
Now we calculate ttre flux density contributed by the gap at its centre it was the complete ring. The gap has its cross section in form of a loop as shown in figure below
a:2
cm
o,
I As calculated in previous question the surface current density of the
K :4
ri.g
Alm
and since the width of the gap(square loop) is ut sb, net current in the is
I : Kw:4w Now the magnetic flux density at any point P due to a filamentary 1 is defined as
n:f,poso2-coso1]a,
where
p - distance of point P from the current filament.
+ angle subtended oz + angle subtended
by the lower end of the filament at Pby the upper end of the filament at PSo the flux density at center of the square loop produced due to one side the loop is
or
B"q,:^f&)"(h)
(P:1cIrI,Q1 :135'
Summing the flux density produced due to all the four sides of loop, we total magnetic flux density produced by the square loop as
B"q:4"(#ffi):/iu!') *n' -
/Tpo x
(4)
(o.r T
x
1o-3)
x 1o'
(tu:
0.1
:{}!u^or
Page 339
Therefore at the centre of the gap the'net magnetic flux density
will reduce by this amount of the flux density. Thus at the centre of the gap the net magnetic flux density at the centre of the loop will be
u*'
-i;!'u"-,^ : ^(n
- +/i x ro-)
:50.04 x
s.2.1r
Correct answer is 0.33
10-7 wbfm2
.
The magnetic flux density produced at any point filamentary current 1 is defined as
p
due
to an infinite
B:!^I27Tp p from the infinite current filament. Now consider a small area d,s of the coil located at a distance r from where p is the d.istance of point
the current filament. The magnetic flux density produced on current filament along y-axis is
B:#:
it
due to the
(p: ,)
since the flux density will be normal to the surface of the coil as determined by right hand rule therefore, the total magnetic flux passing through the
coil is
.,t
^
: I a . d,s :
: I:,1:,(#)@,a0 Snz
As the mutual inductance in terms of total magnetic
M:+
fl*
,lr* is defined
as
where 1-+ current flowing in the element that produces the magnetic flux. N-r Total no. of turns of the coil that experiences the magnetic flux. Thus the mutual ind'ctance between the current filament and the loop is
,:ry(#rn3)
5.2.18
:o.33mH
N:1500
Correct answer is 4. since the two conducting plates of width w:2m carry a uniform current of I: 4 A each so, the surface current density of each plate is
K:L:t:2Alm
Now consider the first plate carrying current in *a" direction is located at g: 0 and the second plate carrying current in - a" direction is rocated at U : d, where d is a very small separation between the plates. since the magnetic field intensity produced at any poinf l due to an infinite sheet carrying uniform current density I( is defined as
n :|(x
x
a,)
where o" is the unit vector normal to the sheet directed toward the point P. So, the magnetic field intensity prod.uced at the second plate due to the
Chap b Magnetic Fiel& in Matter
first plate is
Pale,S{0 Chap 5
Hn
Mbgn€tlc tr'telds in Matter
* a,) :- |,tro-
Now the force per meter, exerted on the
F" :
I' IiKz
x
B,z)
ds
s,' 2"'1
(Kr:2a*-'
plate due to the 1"t nlate will
be
I
where .f(2- rurrent density of the 2"r plate B12J rrl&gn€tic flux clensity produce
It I
F : I' I?-2o,) x (poHtz)d,ad'z (Kr:-fa"tn'r: prHr) |
So,
: I' I'r. 2a,)'x(- sno) dvd"z :41tua,
As the force applierl by first plate on the 2nd plate is in a., direction so it is a repulsive force. Therefore the reprrlsive force between the plates is 4pn$.?"{s
I
Correct answer is
10.05
n : 20,000 turns/meter turns, Relative permeability, lL, : 100 ,S : 0.04 m2 cross sectional a,rea, 1 : 100 x 10 3 A Current in the solenoid. So, its self inductance will be, L' : ltnp,n,zs : (4n x 10 ;) x (100) x (20.000)'z x (0.04) :2.011 x 103 Therefore the energy stored per unit length in tlte field is
No. of
I I
I I I I | I I I |
W',, -'!L'f :lr 2'011 x 103 x 10 ' :10'05J/m
soL . 5.2.20
Correct ans!\,'er is 0.785'1 consider the square loop has sicle a. Now, when the loop is situated in the field B:1.96wb/m'. suppose it swings with an angle o. So in the nes position the torque must be zero. Gravitational forces acting on all the sides of loop will be down wards ancl the force drrc to magnetic field will be in horizontal direction as shown in the ligure'
al2
omg So, in balanced conclition.from the shown figure we have a ms sin a(a)
+
2
a rns sin
"($) tanfY
: rc, _B_ 2mg
! I I I ! ] I
a
s.?"*1
:
tan
t(t):
rf 4=0.78s4
FaEe 341
Correct answex is -40. At any point in between the two parallel shuts the net magnetic flux density produced by the two sheets is given as
B:BtiBz where 81 is the flux density produced by the lower sheet and, 82 is the flux
density produced by upper sheet. Now the rnagnetic flux derrsity produced at poirrt having current density ,fi( is defined as
B
:
#K x
p
due to a plane sheet
a,
p
where o, is the unit vector normal to the sheet and directed toward point . So, the flux density produced by lower sheet is
B, :5(+o,) x
o,
(K:4a,, an:
a")
: - 4a,, &, - -
o,)
and the flux density produccd by the lower sheet is
B,' So tliel net rnagnetic
: t?
4a,)
x (- a,)
(K
flux density produced in the region between the two
sheets is
B
: +(4&,) x :-
a"+$F+a") x (-o,)
4p,au
where p, is the permeabilitv of the medium. Therefore the flux density in region 1 is Br-"qru,r1
o,r:
--
4p,1
: -
4p,z et,
-
8pn
a,
1tr:
and the flux densitv in region 2 is Br"grorr2
So the net flrrx per
:-
16 pa
a,
(
p,
:
2p
4tto)
unit length in the region between the two sheets is (B**,.1) (width of region l) + (8,"n,o,,2) (width of region
! : : (- 8p, a,)( ) + (- 16 p1 a,)(2) :* 40pa1a, Wblm
)
2)
1
$'*"aa
Correct auswer is 2\.IB2T . Considelr the sheets its showrr in figrue that having the surface current densities -1-20 mA/rua, and - 20 tu.Alma, *20arrli'Af m 9r6i*ri4.*r:
:
: z/, t, .,' 5 cnl 't
-20a mA/nt So the field intensity between the plates
will be given
as
Chnp 5 Magnetic Fields in
Mdter
H:HrlHz
Page 342
where .EIr is the field intensity produced by the sheet located at r: .EIz is the field intensity produced by the sheet located at r: 5 cm. Now the magnetic field intensity produced at point P due to a plare having cuuent density .fi( is defined as
Chap 5 Magnetic Fields in Matter
s:|Xxa^ where o" is the unit vector normal to the sheet directed toward point P.
the magnetic field intensity in the region between the plates is
n :LK.X
.-;-'-
a,,r*
lxrx *ffi
:LQo x
10-3ou)
:-
10-3o,
20
x
x
o*
(o")
+!?zo x 10-3ar) x (-a)
and magnetic flux density in the region between the sheets is
B : pH - _ 40tn x
I0-3
a"
Qr:
Therefore the stored magnetic energy per unit volume in the region is
,*:LH.B-${soorcx10-6) : 400 x 4tr x 70-7 x 10-6 : I60tr J lm} Since the separation between plates is
d:5cm.
So, stored energy per
area between the plates is W^l
so|.
5.2.23
A
: u)^ x 4 : (1602r) x (0.05) :8r J lm' :25.1327 J lmz
Correct answer is 6.4 . Consider the square loop is of side 2a as shown in the figure
Since the sides BC and AD crosses the straight wire so no force will be experienced by the sides, while the flux density produced by the straigl* wire at sides AB arrd CD will be equal in magnitude. Now the magnetic flux density produced at a distance p from a straight wire carrying current .I is defined as
B:lt - 2rp *
r i i
the magnetic flux density produced by the straight wire at the two sides of the loop is So
:fr.il:ffi B:y\2):p" Since the force exerted on a current element defined as
/r1r:2A'P:ol IdI by a magnetic field B
is
4p:(tdt)xB
I
Therefore the force experienced by side
F :14(2a)",lxfffio,l t
Page 343
AB of length 2o is
Chap 5
: f{_",)
(1:
4
A)
Magnetic Fields in Matter
Similarly force experienced by side CD is
4 :la(2a)(-
t
")1"[#t-".)] Yg,l
Thus the net force experienced by the loop is
i
F:
;
Ft+ F2
:Yer)
:16x 4xI0-7a, : 6.4a, Jr,N irl
s"x"s4
Correct answer is b0.6 . According to Snell's law the permeability of two mediums are related patanfi - p,tan9z
as
tanh _l\pn
tan01 :
tan?z
p.n
l5tan02
...(1)
Now, the given flux density in medium 2 is
Bz:7.2a,0*0.8a, So the normal and tangential component of the magnetic flux density in medium 2 is Bzn
:0.8a"
Bzt
:
and
l.2au
FYom the figure we have
tanlz
or
0z
- k:#:3 : tan t (2lB)
from equation (1)
tandt : L5tan02 tandr : 16 d : tan-1(10) Thus the angular deflection is or- oz: tan-t(10) : 50.6'
5.2.2t
- tanl(2lz)
Correct answer is 358098.62 . For calculating total reluctance of the circuit, we have to draw the electrical analog of the circuit. In the given magnetic circuit, there are total six section for which six reluctance has been drawn below.
For a given cross sectional area and length. of the core ^9
I
reluctance is
'
l
I
defined as
Paqe $44
R.:J-
Chap 5 Mag4etic lields
ir
Matter
I.LD
Where p is permeability of the medium in core So, we have
5 x 10-2 R;: '-1rooos";(5x10- ) lRz
5
:
x
10-2
x
(1000a)(10
6
lRt:
x
10-2
x
(1000po)(10
(D :
14
^'
,o_a;
x
10-4)
10-2
(loooto)(lotlo-o)
1
1oP"
* -
:
Rs:Rs: Ufu 4x10-2 Ra:
[ooofiFotlo-5:
1
20A, .)
50p,0
7
5o/h
1
xtt^
since all the reluctance are connected in series so total reluctance of magnetic circuit is
.'::T;:#I#.J# : #^:
sol.
5,2.2S
Correct answer is 27.9
d^.*^
358098'62
.
For a given reluctance R.
of.
a magnetic circuit, the self, inductance is
AS
.N u_(R
,
"f _-
Then,
Where
N is no. of turns of
(1ooy \-.^/
(o,:
@lZ\p")
don
:2.79 x 10-2 :27.9rr'H $oL $,2.t7
Correct answer is 0.64 Give that
:50 l:0.6m
N
No. of turns of coil, Length of the core, Relative permeability, Inductance of the
p, :600 L : O.2mH : 0.2 x
coil,
10-3
H
So, the cross sectional area of core is
o- Ll - (o'2 x 1o-3Xo'6) o: u*:-(6oorrJGof: -
sol.
5"2.2$
Correct answer is 62.8
6.366
x
10-5 m2
0.64 cm2
Since the core is ideal so it's reluctance will be zero and so the electrical analog for the magnetic circuit will be as shown below'
1 Page'3{5
'
Chap 5
Magnetic Fields in Matter
The reluctanc€ R1, Rz and R3 is produced by the air gap.
h _ 4x10' 4) 'rr _ &oSr p6(100 x 10 (D 2x10' 2 (D ,ar
......:-
-
"t -
roltoo
,1
&,
t rol:
,R,: *-2-\n-2 --2^ pr{,(100x10-{)-ph So, the
total reluctance seen by coil 1[
R,
: !lr+ Rr llR,
:!+a:5 llnt ltt,
,
and the self inductance of coil
L,
is
: #;:
lt'o
will
be
62.8 rnH
5.?"*s 'Correct answer is 23.6 The total reluctance of the magnetic circuit as seen from the coil 1/2 is
R.,
: (R,l :r t 4-
R,)+ R, ) - 11 1 "11 2t I
(as calculated abov' \
\lt"" lu,)' ya, 4,2 70 :3/h-/r":3r
,
Therefore the self irrductance of the coil 1[r is
Ni Qsor : , - -R'r: Lr: CoF^)
5.s"*8
23.6 mH
Correct answer is 78.54 . Since the coil ifr and N: are directly connected through ideal core so entire
flux produced b;' lI, will link with 1f1. The electrical analog of the magnetic circuit is shown below where thc reluctance R1 and fl2 are the reluctance due to air gap.
So, the reluctance seen by coil
R.,: -L-:"'* prS
,AI2
is
--9.!C) ga(2000 x i-J - -- ir, 2
Consider the current flowing in coil Nz is ,iz. So, the total flux produced by
Page 346
Nz zz is
Chap
500i" :250pnw pltt^-) Since the entire flux wiII link with lfr So mutual induction between 16 is
,
5
qh
Magnetic Fields in Matter
N,'i, : fr:
M: Lrz:ry:@#9:28.54mH s&L
5.4.3t
Correct answer is 0. As the coil Nr and Nz are directly connected through an ideal core so entil flux will produced by 16 will link with y'fi and so flux linked with Ng wiil t zero. Therefore the mutual inductance between Ns and Nz is zero.
srlr
5"2.34
Correct answer is 6.2
.
Given, the expression for magnetization curve,
a
: ln + IP p"wblmz
The energy stored per unit volume in a magnetic material having magnetic flux density is defined as
.*:
ffu . an
Jn:o
Since, magnetic field intensity varied from 0 r27O
'*: Since,
JuHodB
#:!+zn
So, putting it in equation we get,
,* : I?,r(!+zn)an H2 , 2H312'o :[T*-Flo I
:6.18 x 106J/m3 :6.2MJlm3 ***
>1.
{.
*
rl.
****
to
210
A/m
So, we have
linea
-1
$olurloNs
5.3
Page 147
Chap 5 Magnetic Fields ia Matter
Option (A) is correct. Force applied by a magnetic field
B
on a moving charge with velocity o is
defined as
F:uX
B
Since the direction of velocity u and B are perpend.icular to each other as obtained from the shown figure so the resultant force will be perpendicular to both of them. i.e. the force on the moving charged particle will be in upward direction. and as the particle is also deflected in upward direction with the applied force so it gives the conclusion that the particle will be positively charged.
x,
5-3.2
Option (D) is correct. since a magnet bar must have south and north pole i.e. a single pole charge can't exist. So, a magnetic point charge doesn,t exit.
!t
5.3.3
Option (C) is correct. The boundary condition for the current interface holds the following results. (1) normal component of magnetic flux density is continuous. l.e.
(2) Tangential
Bn: Brn component of magnetic field intensity is continuous.
Hrt: Hzt So, (A) and (B) are wrong statement. Now, we check the statement (C). consider the rnagnetic field intensity in 1"t medium is .EI1 and magnetic field intensity in 2"d medium is .Er2. so, it's tangential component will be equal i.e. Htt: Hzt (tangential component) Since scalar magnetic potential difference is defined as the line integral of r.e.
magnetic field intensity
i.e.
u- w: f n. dl: r
and since there is no cuirent density at bound.ary. So, we have U- U:0 or V : V i.e. magnetic scalar potential same in both medium.
will
be
tol.
5,3.4
Option (B) is correct.
soL
5,3.5
Option (C) is correct. A diamagnetic material carries even no. of electrons inside it's atom. Number of electron in carbon atom is six. which is even so it is a diamagnetic material rest of the material having odd no. of electrons.
sol.
5.3.6
Option (A) is correct. A paramagnetic material have an odd no. of electrons and since atomic no. of Al is 13, which is odd. So, it is a paramagnetic material.
i I
$
{
J
So,
Page 34E
Chap 5 Magnetic Fields in Matter
A and R both true and A is correct explanation of R'
s4)t. *"3"?
Option (A) is correct.
s*il
$""3"S
Option (B) is correct.
s*i-
s.3"$
Option (A) is correct.
&*1- $,s.'r$
Option (A) is correct.
$$*-
s"3,'N$
Option (B) is correct.
s{}i-
$"3"''1fr
Optit-,n (C) is correct.
Volumecurrentrlensityirrsi
i.e.
J:YxM
andthesurfacecurrentdensityinterrnsofmagnetizationisdefinedas
K:Mxa, wherea,,isuuitvectorrrormaltothesurface.Considertlrecylirrderis along z-axis so,
(tr,,:
Ap and
M :
MA"
is Therefore the volume current density inside the cylinder
J:Yx(Ma"):g
(M is not the function of
and the surface current density of the cylinder is
K : Ma,X ao:1t'1or Sothecurrentflowingincylinderisjustsiurilartoasolenoidandthefi it is zero' Thus intensity produced due to a solenoid at any point outside as have the magnetic field intensity outside the cylinder 'Efou1ti6"
$*i-
$"3,13
: 0
Option (B) is correct. Forceappliedonamovingchargeinthepresenceofelectricandmagnetk field is defined as F : F"* Fn,: S@+ a x B) where.{andF-aretheelectricarrdmagneticforcesappliedonthecharges it is clear that the rroving charge experiences both the electric and magnetk (in direction of forces. The electric force is applied in a uniform clirectiou force is magnetic the electric field) i.e. it is an accelerating force while, of, velocity field and applied in the normal direction of both the magnetic both the options the charged particle i.e. it is a deflecting force. Therefore, A' of are correct but R is not the correct explanation x*x***r<****
s$LarTIsKs s"4
Page 349
Chap 5 Magnetic Fields in Matter
Option (A) is correct. From boundary condition we have the following relation between the magnetic field intensity in the two mediums :
PrHn : FaHz, (Hr- Hr) X aa12: K
and I[
(1) (2)
and H2 are the magnetic field intensity in the two mediumst an12 is the unit vector normal to the interface of the mediums directed from medium 1 to medium 2 and K is the surface current density at the interface of the two rnediums. where
Norv. the rna,gnetic field intensity in rnedium 1 is
Ht :3a,*
30a,
As the interface lies in the plane
Htn
A/m
u:
0 so, we have
- 3a'
From equation (1), the normal component of the field intensity in medium 2 is given as
: ft:
Hzn
t.so,
Therefore, the net magnetic field intensity in medium 2 can be considered as
Hz :1.5a,* Aar* Bo. where ,4 arrd B are the constants. So, from equation (2) we have [(34, + 30an) - (7.5a,* Aao* Ba")f x a, : 1go, [1.5o, * (30 - A) a, - Ba"f x a" - I\a, 0 - (30 - A) o"- Bo, - I\a, Comparing the components in the two sides we get
30-,4:0e/:30 -B - 10 + 6:-
and
(3)
10
Putting these values in equation (3) we get the magnetic field intensity in medium 2
as
:1.5a,*
Hz
sol
$"4"*.
30oo
-
70a, Af m
Option (B) is correct. Given,
the magnetic mornent m, :2.5 A-rn2 Nlass of magnet, mass:6.6x103kg density of steel, density : 7.9 x 103 kg/m3 So, the net volume of the magnet bar is mass _ 6.6 x 10 3
" -density
:
-
7.9
x
105
0.835 x 10 m3 Now. the magnetization of the magnet is defined as the magnetic moment per unit volume sor we get magnetization of the rnagnet bar as 6
2.5 a, m tvt: r:0s55"10-
Page 350
Chap 5
:3x106A/m
Magnetic Fields in Matter
sol-
5"4.3
Option (A) is correct. Given,
Magnetic field intensity,
rr :5a" A/^ lt
Id,l:4 x 10-aorA-m Current element, So, the magnetic flux density is given as B : FH :Sa"Alm Since, the force exerted on a current element 1dl placed
B
is defined
So,
in a magnetic
as
r:(rdt)x
B
putting all the valdes we get,
p :(4 x 10-aar) x (5a,) :-2x10=3a,N:'-2o,mN $0L 5.4.4
So, for the given match
$cL
5.4,5
$oL s.4.5
-
3, B
'
2,
C'
4, D
'
7.
Option (C) is correct. At the surface of discontinuity (interface between two medium) the component of magnetic flux density are related as B'n: Btn i.e. normal component of magnetic flux density is uniform at the surface discontinuity. Statement 1 is correct At the boundary interface between two mediums, the normal component the electric flux density is related as i.e. disconti Dzn- Dn: P, where p^ is surface charge density at the interface. If the interface is free (pr:0) then, the equation changes to i'e' con Drn : Drn So, the normal component of flux density at the surface of discontinuity or may not be continuous. Option (C) is correct. Magnetic current is composed components.
sol* 5.4.7
list we get, A
Option (C) is correct.
of both
displacement and cond
Torque exerted on a loop with dipole moment defined as
M in a magnetic fiekl B
is
,
T:MxB as defined below
B:-J-_----. 4rR"
(b*3)
Displacement current is determined by using maxwel's equation as Y x H : J.* Ja where -Ia is displacement current density (c _+ 1) Time average power flow in a field wave is determined by poynting vector as
:!n,
x n,
using Gauss's law line charge distribution can be -L 5d.9
(d-2) determined.
Option (B) is correct. Magnetic energy density in a magnetic field is defined
,^:!t
1
Wire
as
:
2
The force exerted due to the wire 2 at wire 1 is given
'
(a -+ a)
.A
Option (B) is correct. Consider the two wires carrying current as shown below
Wire
as
F:(rdt)x(.B)
where 1dl is the small current element of the wire 1 and Il is magnetic flux density produced by wire 2 at wire 1. As determined by right ha;d rule the magnetic flux density produced due to wire 2 at wire 1 is out of the paper. which will be towards wire 2. In the similar way the force due to wire 1 at wire 2 will be toward wire 1 i.e. attractive and perpendicular to the wire.
sol
5.4.'tl
Option (B) is correct. Flom the boundary condition for magnetic field, we have the followirrg derived condition as
and sol.
5,4.12
PtHa:
Hn:
fuH,,2
Hn
Option (C) is correct. The magnetic flux density B and magnetic fierd intensity with permeability p are related as
B
- p,H: F,lhH
rl
in a medium
Now, for the different magnetic materiar relative permeability p, are listed
below:
Chap 5
Magnetic Fields in Matter
Option (B) is correct. Biot savart's law gives the magnetic fl*x density p, I tat* n
Poo"
Page 351
Page.352
Free space (var-'uurn)
,lf,r-L
Chrip 5 Mapietic Fields in Matter
Diamagnetic
pt,'
{ 1 F,2 7' F' )'2 1
Paramn.gnetic
Ferromagnetic
So, the
IJ-fI
1
curve for the respective material has been shown
(depending on their slopes p). Ferromagnetic OParamagnetic .Vacrnrm
I
B Diamagnetic
H&.r:rr.
$",{,"13 Option (A) is correct. When a dielectric material is placed in an electric field then the electri dipoles are created in it. This phenomenon is called polarization of th dielectric rnaterial. So, we conclude that both the statement are correct and statement (2) correct explanation of (1).
${:i-
ir
$.4.,rjr Option (B) is correct. For an inhomogeneous magnetic material, magnetic permeability is a variabh and so, it has some finite gradient. Now, from Maxwell's equation we kntr
V'B:0 B:FH v.B:V,(pH) 0:V.p,+V.H
Since, So,
In the above equation
V ' p have some finite
v.H+o
$$1. $"4.r$
Option (C) is correct. Force on a current element
' !9*t
s"4"{g
value therefore,
(in inhomogeneous mediuml
ldl
kept in a magnetic field
B
is defined
as
F : Jt{ Id,Ix B
:
[(10)(2)a,] x [o.obo,]
:
1.oas N
Option (C) is correct. Magnetic energy density in a magnetic field is defined
as
,r, :72J ' A wliere $$L
$"4.'t?
s{3L $,4"'tS
J
is the current density and ,4 is the magnetic vector potential.
Option (D) is correct. The force on a moving charge q with the velocity tr in a region having magnetic field B and electric field .E is defined aus p : q(84- a y.. B) Option (B) is correct. The currents in the hairpin shaped wire flows as shown in the figure.
ffi:H
exerted on a roop with dipole moment
M in amagnetic fierd B is
T:MxB 5.4.8
,
tatx
-p,l
n
Displacement current
VxH
:
i, d"t**;r"d by using maxwe',s equation as J.l^Ja where .Ia is displacement current
(b
-_>
density (c _ . Time average power flow in a fierd *urr" t, a"rurmined by poynting vector
Pou":|n"xn"
oL
5.4.9
1) as
(d -+ 2)
as
A
Option (B) is correct. Consider the two wires carrying current as shown below
Wire I
Wire
:
2
The force exerted due to the wire 2 at wire 1 is given
F:(rdr)x(B)
s.4.rt
J)
using Gauss's law rine charge distribution can be determined. (a -+ 4) Option (B) is correct. Magnetic energy density in a magnetic field is defined
,,_:|J. 0L 5.4,{o
Chap b
Magnetic Fietds iu Matter
Option (B) is correct. Biot savart's law gives the magnetic flux density as defined below
u
pagie Bb1
as
where ,dt is the smalr current element of the wire 1 and B is magnetic flux density produced by wire 2 at wire 1. As determined by right hand rule the magnetic flux density produced due to wire 2 il;; I i, ort of the paper. which w'l be towards wire 2. In the sim'ar *"y t;; force due to wire 1 a,t wire 2 will be toward wire 1 i.e. attractive and perpendicular to the wire. Option (B) is correct.
condirion for magnetic fietd, we have the fouowing
fi:ftJi:,1J,,::dary PtH6: p2f!* and Hn: Hn 5.4.t2
Option (C) is correct. The magnetic flux density B and magnetic field intensi ty with permeability p are related as
ffik:'the H
differeBn;::r-r2fl"ri'r
H ina medium
relative permeabilitv p. are risted
J
Page 352
Free space (vacuurn)
,Fr-t
CLip
Diamagnetic
P',Sl
Paramll.enetic
F,Z\
5
Malnetic Fields in Matter
Ferrornagnetic
So, the
1
11,')'>
l]-fl curve for the
L
respective material has been shown
(depending on their slopes p). !'erromagnetic
"
-"uParamagnetic
.Vacuum Diamagnetic
*.,i!r. $.4"'!3
${3;- $.4"44
Option (A) is correct. When a dielectric material is placed in an electric field then the electrb dipoles are created in it. This phenomenon is called polarization of tbe dielectric naterial. So. we conclude that both the statement are correct and statement (2) b correct explanation of (1). Option (B) is correct. For an inhomogeneous magnetic material, magnetic permeability is a variabh and so, it has some finite gradient. Now, from Maxwell's equation we knor
'B:0 B:FH v.B:1.(p,H) 0:V'p,+V'H V
Since, So,
In the above equation V
v.,Et+0 .$$k. s"4.'t$
' p have some finite value
Option (C) is correct. Force on a current element IdI kept in a magnetic field B is defined
:fo',Jr',rirt"rx [o.osa.] *#L
S",{"'tS
wliere $.4,,!?
:
J
as
A
is the current density and
A
is the magnetic vector potential.
Option (D) is correct. The force on a moving charge q with the velocity magnetic field B and electric field .E is defined a^s
o in a region having
P:r1(E*axB) $*L
5,d1,'{8
as
1.oav N
Option (C) is correct. Magnetic energy density in a magnetic field is defined
'r'r,:lrJ' s{}t
therefore,
(in inhomogeneous medium)
Option (B) is correct. The currents in thc hairpin shaped wire flows as shown in the figure.
i
Hairpin
Page
BEB
Cbap 5 Magnetic Fields itr Matter
As the direction of current are opposite so the force acting between them is repulsive, and So it tend to a straight line.
5.4.1s
Option (C) is correct. Given, the Lorentz ibrce equation,
F : e(ax B) is at rest then u :
If the particle 0 and particle due to the magnetic field.
3"4.3*
so there
will be no any deflection in
Option (D) is correct. Force acting on a small point charge'g moving in an
p : qE* q(u+ B)
EM
wave is defined as
So,forg:1 ! t
F:E*axB s.4"*'t Option (B) is correct. I
.
' tol-
Given,
Current flowing in the conductor, .I:5A Magnetic flux density, B :Ba"* ay since. the force experienced by a current carrying element ldl placed in a magnetic field B is defined as
d7;,:(Idl)x B As the current flowing in a, direction so, we have
dI : dla" and the force experienced by the conductor is
dF :(Sdla") x (Ja" *4au) Therefore, the force per unit length experienced by the conductor
is
dF
u sol.
5.4.22
:'_'#:.io**,*
Option (B) is correct. From the boundary condition for magnetic field we have the following relation : Normal component of magnetic flux density is continuous
Ba:
l.e.
B,tz
Any field vector is the sum of its normal and tangential component to any
surface
Hr: Ha*
i.e.
Ha
when the interface between two medium carries a uniform current r( then the tangential c,mpo*ent of magnetic fierd intensity is not uniform.
Hs-Hp:ff
l'e.
Ot,
(1,p1
X (Ht- Hr): K
Page 354
Chap
5
Magnetic Fields in
But, Matter
s.,..
B,
s.rr.z3 option (A)
*
J
B"z* Bn
is correct.
Given, the magnetic flux density in medium 1 is
Br : 1.2a"*
0'8ao
*
0.4a,
and the interface lies in the plane z: 0' So, the tangential and normal components of magnetic flux density in tbc
two mediums are resPectivelY
:
Bu:7'2a"* Bn:0'4a,
and
0'8a,
Now, from the boundary condition of current free interface, we have tbe following relations betweea the components of field in two mediums.
Brn:
Brn
: E" F:PD\
EIL
a,nd
I
Therefore, we get the field components in medium 2 as
B2n: Bn:
and \
Bzt
:
O'4a"
""(#)
:!g.zo,*0.8o,):(0.6o'+0.aan)
Thus, the net magnetic flux density in region 2 is
Bz:
Bzn* Bzt
:0.6o' *0'4ar*0.4a" So, the magnetic field intensity in region 2 is
,, : #: $oL
5.4'24
fi{o.uo,*0.4a,*0.4a")
Option (C) is correct. Energy stored in a magnetic field is defined
w.:![n'
So, f J
a'
ta,
Jd,u hasthe units of energY.
**X*rk***t
as
Alm
gHAPTER 6
TIME VARYING FIELDS AND MAX}UEIL EQUATIONS
3t
INTRODUGTIOlI very populalald ther are known as Electromagnetic HnYi:::::oT::" rhe *"i" til" ;"pt";";;';;;;,;:"il?fi:1
*:li:.::irioas. "f equations. background and concepts on _,1h Maxwell's
o
Faraday's law of electromagnetic induction for three different
|:: "::ltl-,^^*l_crl*i: -
.2
o r
uaa)eD: cases:
field, moving conductor with static magnetic
,"ar"r"t^*r;;;Jrfi;
moonafi^ ft^ll magnetic field.
o
rent
They include:
Lenz's law which gives direction of the induced current
associated with magnetic flux change. Concept of self and mutual inductance
in the loop
Maxwell's equations for static and time varying fierds in free space and conductive media in differential and integrai form
FARADAYS LAW OF ELECTROUAGNETIG INDUGTION According to Faraday's law of electromagnetic induction, emf induced in a conductor is equal to the rate of flux lint
v^t:#:-#[a.as
...(6.1) the total magnetic flux through the closed roop, B is the magnetic flux density through the loop and .9 is the surface area of the loop. If the where
@ is
closed path is taken by a.n becomes
Y*:-
.r[-turn filamentary conductor, the induced emf
NH
1.2.1 Integral Form of Faraday's Law we know that the induced emf in the closed loop can be written in terms of electric field as
I/emr:fra.at
...(6.2)
From equations (6.1) and (6.2), we get
f,n.
ar,
:_#!,"
. o,
This equation is termed as the integral form of Faraday,s law.
...(6.3)
Page 356
6.2.2 Differential Forrn of Faraday's Law
Chap 6 TiEe Vary'ug Fields rnd
Applying Stoke's i;heorem to c'cluation i6.li), we obtain
Maxwell Equatioos
It,
"
E)'ds
--!t[a.
as
Thus, equatirrg the integrands in above equa,tion, we get
V r: E :-d^P dt This is the differential form of Faradav's 6.3
law-.
LENZ'S LAW extrrration is ciue to Lenz's larv wlfch states that the direction of ermf inclucerl {)pposes tire carrse prorirrcing it. To understancl tbe Lenz's law, con-sider the two conducting lo
The negative sign in F-ararliry
rt-:
i t,,.."ori,,g
lr (")
. I]r;,11;,1 1; .l.
.B
I D,,.r.uring B
r"-J-\ '-*"{*1 (b)
Deternrinationc,fDirectiouof lntlr.rcedCurrerrtin aLoopaccortlingtoLenz's B in Upward Direction
(a) B lrr Upward l)ircctiou Increasing with Tirne (b) with Time
l{E Yt{€ FSL&*|!?V AF
r
Ta detdrrnine the polarity of induced emf (direction of induced we may follow the steps given heiow. Step',1r Obt;ain the direction of'magnetic flru density thrirugh the In both tlre Figurcs 6.1(a) (b) tlrL rnirguet,r field i's upward. Step 2:
Deduce whether the field is increasing or decreasing with along its directiou. In Figure ti.1(a). the magnetic field
upward is increa-sing', wtrerc;us in Figurc 6.1ib), the"rrn fickl directed upward is decreasing with time. Step'3: For increasing fieid aesign tire dirtxtion of induced current the loop such that it produces the field opposite'to th6 gi ,rnagn€tic lield clirection. Whereas for decreasing field Escign direcrion of irrcluccrl cnrroirt irr the lcop such that it prodtrces field in the same direction that of the given magnetic field. Figure 6,1(a), rrsing riglrt hand rule ue conihrde that any flowing in clockwise direction in the ioop will cause a m field directed downward and hence, opposes,the increase in (i.e" opposcs the ficld that carrscs it). Simiiarlv in Figure 6.1 using riglrt hand rule, we conclude that a,nv current flowing qnti-cklckwise directior in thc loop wili cause a magnctic dire.t'ted rrpward and iience, opposes the decrease in flux opposes the field that carrses it). ,
the polarity of intluced ernf in the loop corresponding to thc obtainccl rlirection of indrrced current.
*S.iS:
aA
MOTIONAL AND TRANSFORMER EMFS According to Faraday's law, tbr a flrrx thro'gh a loop, there wilr 'ariation be i'clnr:e
6.4.L Stationary Loop in a Time Varying Magnetic Field FoI a stationarv ioop located emf in the loop i-s given b.y i,,,,,
in
ti,
tirne va,rying rnagnetic field, the induced
: .[ n",tn-
J,o,#. ot
This emf is induced by the tirne_varyirrg currerrt (producing time_varying rnag'etit: fieltl; irr a, statiorrar,v loop is called transfor
", "*j.
6.4.2 Moving
Loop in Static }\Iagnetic Field whc'n a condrrcting krup is *rr,vi.g irr a static fierltl. a,n ernf is induced in the loop. Tlris i'tlucecl errrf is <:alled, rn,ol:icnnl .ntf antd given bv V.,,i
: Jti n,,,. rtl : [ir> Bl. Jt' r
{tL
where u is the velor:ity of loop in rnagnetic field. Using Stoke's theorem in above equatiorr, wr_, gct
y x 8,*: V x (ux 6.4.3
B)
Moving Loop in Time Varying Magnetic Field Ihis is the gerreral ca,se of in
is
: f,"n . at :_ I,#. d,s+ {.{, ^ B) . dL .ds+ffurB\.dL v,,,= J.fl.I " - | dt. %u,r
or.
!L\-
using st.ke's thettt:rn,
t";:H;;["
as
tn,,
"tilJ,i'lo,i*ruoo
i' differential
form
v xE=-O,f+Vx(uxB) 6.5
II{DUGTANGE
Arr inductatrce is
the iner:tial property of a cilcuit caused by a1 incluced reverse voltage that opposes thc flow of current when a voltage is applied. A circuit or a part of circuit that has inducta,nce is called an inductor.-A device can have either self ind'ctance or rmrt'al inductance.
6.5.1
Self Inductance
consider
a circuit .arrvi*g a varving currr:nt which prod'ces
rnagnetic I'ieid r,vhich in
the change in
fl'x.
trlr*
varying
prr:duces irr
Page
iil
Chap 6 Tlme Varyiry fielde ana Maxwell Equetiols
Page 35E
Cha[' 6 Time Varying Fields and Maxwell Equations
the change in flux is produced by the circuit itself. This phenomena is called self-induction and the property of the circuit to produce self-induction b known as self inductance. Self Inductance of a Coil Suppose a coil with N number of turns carrying current ,I. Let the curred induces the total magnetic flux @ passing through the loop of the coil. Thus, we have
NIDcl I
: LI ,NO o: I
or
NID
or where
.t is a constant of proportionality known as self inductance.
Expression for Induced EMF in terms of Self Inductance a variable current i is introduced in the circuit, then magnetic flux linked with the circuit also varies depending on the current. So, the self-inductance
If
of the circuit can be written
L,
as
: ft
diD
...(6.4)
Since, the cha,nge in flux through the coil induces an emf in the coil given dO r/ V"^r:-ft
\r
...(6-51
So, from equations (6.4) and (6.5), we get
V^r:- L# 6.5.2 Mutual
Inductance
Mutual inductance is the ability of one inductor to induce an emf acr6 another inductor placed very close to it. Consider two coils carrying curreil I and -Iz as shown in Figure 6.2. Let 82 be the magnetic flux density produced due to the current 12 and $ be the cross sectional area of coil 1So, the magnetic flux due to Bu will link with the coil 1, that is, it will pasr through the surface &. Total magnetic flux produced by coil 2 that p* through coil 1 is called mutual flux and given as
Qrr: I Br. dS 1
to current
12,
a,s
the ratio of the flux linkage on coil
i.e.
Mn
-
Nt9" h
where Nr is the number turns in coil 1. Similarly, the mutual induciance M21 is defined as the ratio of flux linkage on coil 2 (produced by current in coil 1) to current i, i.e.
Mz,
N'9"
--j-
The unit of mutual inductance is Henry (H). If the medium surrounding the circuits is linear, then
Mtz:
Mzr
j
]
Jg,
We define the mutual inductance Mp
i
Page Bb9
Chap 6 Time Varying Fielde and Maxwell Equations
Circuit
Circuit
1
I;'ii;u'' 6.2 : Mutual Inductance
2
between Two Current Carrying Coils
Expression for Induced EMF in terns of Mutual Inductance If a variable c'rrent f2 is introduced in coil 2 then, the magnetic flux linked with coil 1 also varies depending on current rz.so, the mutual inductance can be given as
Mr,
:
d!,"
...(6.6)
axz
The change in the magnetic flux linked with coil 1 induces an emf in coil
given as
(V^r\:-d9t'
1
...(6.7)
So, from equations (6.6) and (6.2) we get
(%-r}
: - rrr#
This is the induced emf in coil 1 produced by the current i2 similarly, the induced emf in the coil 2 due to a varying current 1 is given as
(V"-r),
I 6.6
in
coir 2.
in the coil
: - Mrr#
Mru1ry6aa's EQUATTONS The set of four equations which have become known as Maxwel|s equations are those which are developed in the earlier chapters and associated with them the name of other investigators. These describe the sources "qoutiorrs and the field vectors in the broad fields to electrostatics, magnetostatics and electro-magnetic induction.
6.6.1
Maxwell's Equations for Time Varying Fields The four Maxwell's equation include Faraday's law, Ampere,s circuital law, Gauss's law, and conservation of magnetic flux. There is no guideline for giving numbers to the various Maxwell's equations. However, it is customary
to call the Maxwell,s equation derived from Maxwell's equation.
Faraday,s law as the first
Maxwellts First Equation : Faradayrs Law The electromotive force around a closed path is equal to the time derivative of the magnetic displacement through urry s,rrface bounded by the path.
Y
or
xE:-ry dt
frn. at :_!r[ n . as
(Differential form) (rntegral form)
T'"' I
I
!
kl
Page 360
I
Chap 6
Maxwell's Second Equation: Modified Amperets Circuital law
The magnetomotive force around a closed path is equal to the cond plus the time derivative of the electric displacement through any bounded by the path. i.e.
Time Varyiry Fields and Maxwell Equetionr
F I
n
V x.EI
f n'
ar,
: t*#
(Differential
: l(t+@ur)' *
(Integral
Maxwellts Third Equation : Gauss's Law for Electric Field
The total electric displacement through any closed surface volurne is equal to the total charge within the volume. i.e.,
Y'D:p, ort
enclosing
(Differential (Integral
fo.4s:lp,d,u
This is the Gauss' law for static electric fields. Maxwell's Fourth Equatiou : Gauss's Law for Magnetic Field The net magnetic flux emerging through any closed surface is zero. In words, the magnetic flux lines do not originate and end anywhere, but continuous. i.e.,
B
V
f,a.
Of,
-0
(Differential (Integral
dS
This is the Gauss' law for static magnetic fields, which confirms the existence of magnetic monopole. Table 6.1 summarizes the l\{ax equation for time varying fields. Table 6.1: Maxwell's Equation for Time Varying Field
ffiriti$',&l
:?
..p469ii$$.,:;,,,,try
in*isi .i ;!!l;.li.l.:,J ;:..: -i.ir' j t,::.ijrr"irr,r.,r!1ii r:.,
,olp.tttgmi.bgnetii ''',',
:;:llidffi{s6
r,;::i,r::;;,:rr) :i r,t"r '
r-
v.t tr; ri4&;'t ir:r,,
All.I :,::i*
l
i:
i-','.:. j ..! -jl
l:.r.iir r:
::r:i,.
:rr'1
:.
r,' 1'1
.1],:ttttttttttttttt;a:''a,:t
l::dgl
l
le;.,;:*€$.ff,:
i4i r '.'. ,i .,i..' {i}.ir11:lru!rr:r,r...i..r.4 r'i . ..:,,: :r:r " i:..r,....t :r,.!ii,::.. r: :,. r._:r,,r; i ;rit i: i" 1:ai ;a u:r i:.!,Jl.jtfi t!.r'!,i ,,.-:.......aa
:.4
:.' \ t
a, :.'.., l:..f-.'
4,'.. .t 1'
. !,
t.i. 1ri,;:: 1!i :ii!:t:rt.,l i,!;'ii:,-!?-'i.,-. . ;-.:::"::-a..:::._i -:-... |: r . ...1,
6.6.2
vell's Equations for Static Fields Maxwell For static fields, all the field terms which have time derivatives are zero.
aB _,,
nt -v ^-r ano
0D : u -At -n
Therefore, for a static field the four Maxwell's equations described reduces to the following form.
\
q
ab
Thble 6.2: Maxweil's equation for static field
:s;N.:,
Chap 6
Iffiatial,Ibrm
L,.
6
Jt
,
,,v x.fi,.:,,ir,
r..:..'
, ,,.,,
'
y. ,
i,.r.*
g t ar, *a
F,erad'ayls ; .'lqw
Jt
., -,lrl
$,n,
Js -.
:
..,,, ,, .of
electromaffretic induction
In.an:{t.n
,.'
* ,V: ;'o t...'
Page 361
..._t.
X-+"{4,au,
Modified Ampere's circuital law Causs' Iaw of 'Eloetiixtatics
s,
l:
h
r
t
Maxwell's Equations in Phasor Form Inatime-varyingfield,thefieldquantities E(r^y,z,t),D(r,A,z,t),B(r.y,z,t) ) H(r,y,z,t), J(r,A,z,t) and po(r,y,z,t) can be represented in their respective
'
phasor forms as below:
:P"e{E"e''} :pte{D,e''} B: R"{8,"-'} H : R*{II,"-'} E D
,
...(6.8a''
...(6.8b) ...(6.8c) ..(6.8d)
and :":Y"l:.I,!j
illi]
where E,rD,,B",H".J, and pu" a,te, the phasor forms of respective fiel,:i quantities. Using these relations, we can directly obtain the phasor form ol Maxwell's equations as described below. Maxwell's First Equation: Faraday's Law
In time varying field, first Maxwell's equation is written
as
v x.E:-%|
.(oe)
Now, from equation (6.8a) we can obtain
YxE
:V
X P.,e{E,e,'}: Re{V
x
E"eP'}
and using equation (6.8c) we get
#
:$n"1n,
".')
: R"{i,-o, ".'}
Substituting the two results in equation (6.9) we get
Re{v x 8""''}
Hence, or,
: - n"{jrn"*'I
VxE,--jwB, . $ n". an :- ja "JsI a" as Jt
(Differentialfornt) (Integral forrn)
Maxwell's Second Equation: Modified Ampere's Circuital Law
In time varying field, second Maxwell's equation is written
v x .Er :
t*#
...(6.10)
From equation (6.8d) we can obtain
Vx
.F/
:V
X Re{,Er"e''}
as
: Re{v x
H"eYrl
Time Varygg Fields and Maxwell Equations
Page 362
Chap 6 Time Varying Fields and Maxwell Equations
From equation (6.8e), we have
J : Re{J"err} and using equation (6.gb) we get
H : $n" 1o
"
",'j
:
Re{j, D
"
"to,}
Substituting these results in equation (6.10) we get Re{v x H""tor} :Re{J"er, * juD"ept} Hence, V x trI" : J"+ jaD,
or,
. frn, dL :
I
e,+ i,tD") .
(Differential fo
d,s
(Integral fo
Maxwell rell's Third Equation: Gauss's Law for Electric Field In tirne varying field, third Maxwell's equation is written
Y.D:p,
as
...(6.
}}om equation (6.8b) we can obtain
v . D:v.Re{D""to'}:Re{v. D"".'} and from equation (6.8f) we have
p,:Pue{p*er'} Substituting these two results in equation (6.11) we get
Hence,
or,
Re{v . D""r,} :R"{p*"to,) v . D, : po,
f
o" .
(Differential for
aS : p*d,u f
(Integral for
Maxwell's ell's Fourth Equation: Gauss's Law for Magnetic Field
v'B:o
...(6.r
F}om equation (6.8c) we can obtain
v . B :v . Re{B,"to,}:Re{v. 8""*,}
Substituting it in equation (6.12) we get Re{V' B,ePt} -0
Hence,
V.B,:Q
or. ,
(Differentialfon
: v ds:o 6a,.- uD Isus
(Integral fon Table 6.3 summarizes the Maxwell's equations in phasor form. Table 6.3: Maxwell,s Equations in phasor Form
,,tj--:
9: t.: r'i
.r$*:ff;*i,g& , :.r;..:.:li'.:l
e :
.-..
4' i'
r ,l.l::ii
l.a,.i:.ll:al
f a"
. ar=
. f{.r"+ tnD,) as
iag*{
ItMi66d',,::, ;.:. circuiAf l*+. ,,
EOUATIONS IN FREE SPAGE
Page 363
For electromagnetic fields, free space is characterised by the following parameters:
Relative permittivitY, €, : 1 Relative permeabilitY, 1tr,: !
1. 2. 3. 4. 5.
J:0 Po:0
Volume charge densitY' As we have already obtained the four Maxwell's equations for tirnevarying fields, static fields, and harmonic fields; these equations can be easily written for the free space by just replacing the variables to their respective values in free space'
Maxwell's Equations for Time Varyrng Fields in trYee Space By substituting the parameters, J: 0 and p,:Lin the Maxwell's equations given in Table 6.1, we get the Maxwell's equation for time-varying fields in free space as summarized below: Thble 6.4r Maxwell's Equations for Time Varying Fields in Free Space
Bi&redal&q&.'
L:t','.
*..x',b';:#
tffi-,:;l
$[am., i#'r'rr
feliilrystta'tr'
2, 3.
r.r;r.r-ri Li-,ii..-,,-r,i:,ri... :r;i..._:ir:rri,rrrr.i.
v,' D;.0
dn. dg=o
d&riddl l&!e,'of Ebbtrmt&tiis
'
Js
@d,lryq
4;,
,,{4
ri# Mqgqrtwtatip
e
,q{,: .qigneiic
nono.pote)
J7.2
Maxwell's Equations for Static Fields in trbee Space Substituting the pa,rameters, J: 0 and pu : 0 in the Maxwell's equatiorr given in Table 6.2, we get the Maxwell's equation for static fields in fre,: space as summarized below. Thble 6.5: Maxwell's Equations for Static Fields in Flee Space
* $t $i :,1:l::::::a:!;l:l[':::rir,tir::
S.X.
il*.hi&;i.,,.::i I : .i:t:ai:::
,1:
, $:,
tlli;,|:,,,i'::;;' :',1,r,.
i.'.
F.,*, :f,fi,*&
.,. .:tr: j,.
4;
;::::.,
:.:,:.:'
:t::l*..' :..,:,,:.li:ll.rit
tit:itilii
i
l
r:qri:'X:',,&,,+tS l
.:,]$l::l',,iilr,lr.rg
,V..r
jir:0,|:ii-:i.tii
:t,,.:.-rtit.
g'*
:
O
Time Varying Fields and Maxwell Equations
l i I l
I
ConductivitY, o :0 Conduction current densitv'
s.N.
ehap 6
.,.:r:..,.:u
*€:t
*irixdts;}}*sr
Page 364
Thus, all the four Maxwell's equation vanishes for static fields
Chap 6 Time Varyiug Fields and MaxweII F,quations
space.
6'7'3
Maxwell's Equations for Time Harmonic Fields in Fbee space Again, substituting the parameters,.f:0 and po:Q
in
the
equations given in Table 6.3, we get the Maxwell's equation for time fields in free space as summarized below.
Ihble 6.6 : Maxwell,s Equations for TimeHarmonic Fields in Flee Space
*'t<********x
i
l.
J'
kH
EXERGISE 6.{
Page 365
Chap 6 Time Varying Fields and Maxwell Equations
A perfect conducting sphere of radius r is such that it's net charge resides on the surface. At any time f o magnetic field B(r, t) inside the sphere will be (A) 0 (B) uniform, independent of r (C) uniform, independent of f (D) uniform, independent of both r and f
ln
e.r.:
t
A straight conductor ob of length I lying in the ry plane is rotating about the centre o at an angular velocity t.r as shown in the figure' @
&
@
@
@
@
@
@ @ @ 1,/
@
@@&@@
@
@@@@&
@
@ @
(jp
lG
If
@s@@@@@ a magnetic field
B
w
is present in the space directed along a, then which of
the following statement is correct ? (A) V"b is positive (B) %, is negative ', (C) %" is positive (D) l/a" is zero
tQ
6"r.s
Assertion (A) : A small piece of bar magnet takes several seconds to emerge at bottom when it is dropped down a vertical aluminum pipe where as an identical unmagnetized piece takes a fraction of second to reach the bottom. Reason (R) : When the bar magget is dropped inside a conducting pipe, force exerted on the magnet by induced eddy current is in upward direction. (A) Both A and R are true and R is correct explanation of A. (B) Both A arid R a,re true but R is not the cortect explanation of A'
(C) A is true but R is false. (D) A is false but R is true. Self inductance of a long solenoid having proportional to
(A)
" (c)
,,?
n turns per unit length will
(B) 1.ln (D) tln'
be
Page 366
&qc&
$"1"5 A wire with resistance .R is looped
on a solenoid as shown
Chap 6 Time Varying Fielde and Maxwell Equatiofs
''
If a constant
current is flowing in the solenoid then the induced flowing in the loop with resistance r? will be (A) non uniform (B) constant (C) zero tdlC& $"1.S
(D) none of these
A long straight wire carries a current
locos(arf). If the current along a coaxial conducting tube ofradius r as shown in figure then field and electric field inside the tube will be respectively.
(A) radial, longitudinal (C) circumferential, radial f'{C& S"'!"?
,I:
(B) circumferential, longitudinrl (D) Iongitudinal, circumferentii
Assertion (A) : TWo coils are wound around a cylindrical core such that primary coil has Nr turns and the secondary coils has N2 turns as showr figure. Ifthe same flux passes through every turn ofboth coils then the of emf induced in the two coils is
Vtz - Nz %-rr I[t i---..--..-.---
Reason (R) : In a primitive transformer, by choosing the appropriate no t turns, any desired secondary emf can be obtained I (A) Both A and R are true and R is correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of (C) A is true but R is false. (D) A is false but R is true.
t.
l.
A'
I
I
l
Page 366
t!t*&
s"{.5 A wire with resistance ft
is looped on a solenoid as shown in figure.
Chap 6 Time Varying Fields and Maxwell Equations
If a constant
current is flowing in the solenoid then the induced flowing in the loop with resistance -E will be (B) constant (A) non uniform (D) none of these (C) zero r!:€& $,1.6
A long straight wire carries a current 1:.Iocos(cut). If the current
along a coaxial conducting tube of radius r as shown in figure then field and electric field inside the tube will be respectively.
(A) radial, Iongitudinal (C) circumferential, radial erc& 6"1"7
(B) circumferential, longit (D) longitudinal, circum
Assertion (A) : Two coils are wound around a cylindrical core such that primary coil has Nr turns and the secondary coils has Nz turns as shown figure. Ifthe same flux passes through every turn ofboth coils then the of emf induced in the two coils is
Vatz %^tr
Nz
- n[ :-*-o Secondarv coil (N, turns')
Primary coil (N, turns)
+
tr-----''
.a
rl:.i,.ii..r:.r" .i..{:i:,.,::. . i,ri:.ir r,., : rj.U
t:
I
I
j:::r
Reason (R) : In a primitive transformer, by choosing the appropriate noturns, any desired secondary emf can be obtained. (A) Both A and R are true and R is correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
l.
t,
lqQ
5.1.rt
In b non magnetic medium electric field E : Eocosut is applied. If the permittivity of medium is e and the conductivity is o then the ratio of the amplitudes of the conduction current density and displ.acement current density will be (A) pa lae (B) olwe (C) opalue (D) uelo
Fa
6",!.s
In a medium, the permittivity is a function of position such that E = O. tf the volume charge density inside the medium is zero then V . ,o i. roughlv equal to (A) e,E (B) -e.E (c) 0 (D) -Ye. E
:a
6.1"{O
In free space, the electric field intensity at any point (r,0,$) in
spherical
coordinate system is given by
"
sindcos(wt_ =-7ae
kr)
^
The phasor form of magnetic field intensity in the free space will be bsin_Q ' A\' (tp,gr e-jh a^Y
(B)
-H
(C) @!n
(D)
&EPd e jb a,
(
"-in'
o
e-ir'a,
Common Data For Q. Ll and 12 : A conducting wire is formed into a square loop of side 2 m. A very long straight wire carrying a current 1:30A is located at a distance 3m frorn the square loop as shown in figure.
/:30
A
a:2
Ee 6."t.tt
Eo 5"r.r2
m
If the loop is pulled away from the straight wire at a velocity of 5 m/s then the induced e.m.f. in the loop after 0.6 sec will be (A) 5 pvolt (B) 2.5 pvolt (C) 25 pvolt (D) 5 mvolt If the loop is pulled downward in the parallel direction to the straight wire, that distance between the loop and wire is always 3 m then the induced e.m.f. in the loop at any time I will be (A) linearly increasing with f (B) always 0 (C) linearly decreasing with I (D) always constant but not zero. such
ErQ 6.1,13
Two voltmeters ,4 and B with internal resistances .Ra and -Rs respectively is connected to the diametrically opposite points of a long solenoid as shown il figure. current in the solenoid is increasing linearly with time. The correct relation between the voltmeter's reading V1 and Vs will be
Page 367
Chap
O
Time Varying Fields and Maxwell Equations
Page 36E
Chap 6
Tlne Varying Fields and Maxrell Equations
(A) Vo:
(B)
1rt
sv::H
yA--va
Q)h:-h
Common Data For Q. 14 and 15 : Two Jrarallel conducting rails are beirrg placed at a separation of 5 m with a resistance B : 10 f) connected across it's one end. A conducting bar slider frictionlessly on the rails with a velocity of 4mf s away from the resistance as shown in the figure.
(}9
o @
c,
A
If
a uniform magnetic field
B:
Tesla pointing out of the page fills entire region then the current 1 flowing in the bar will be 2
(B) *40A (D) -4A
(A) oA (c) 4A fft{:s 6,1.t5
The force exerted by magnetic field on the sliding bar will be (A) 4 N, opposes it's motion
(B) 40N, opposes it's rnotion (C) 40N, in the direction of it's motion (D) 0 ri*&
6.,!.16
Two small resistor of 250 o each is connected through a perfectly conducting filament such that it forms a square loop lying in n-g plane as shown in the figure. Ma,gnetic flux density passing through the loop is given as
fJ
aaa aB8
aaa
:--
7.5cos(120a"t,.-
30')a"
The induced current /(f) in the loop will be (A) 0.02sin (72hrt - 30') (B) 2.8
(C) -5.7sin (l2hrt 6.1.17
-
x
Page 360
103sin(120zrt- 30")
(D) 5.7sin(120zrt
30')
-
30')
A
rectangular loop of self inductance tr is placed nea,r a very long wire carrying current i1 as shown in'figure (a). If ir be the rectangular pulse of current as showir in figure (b) .then the plot of the induced current iz in the loop versus time t will be (assume the time constarrt of the loop, r > LIR)
(a)
f,co 5-t.19
Two parallel conducting rails is placed in a varying magnetic field B
:
0.2 cos ufa,. A conducting bar oscillates on the rails such that it's position :0.5(1 - cosr..,f ) m. If one end of the rails are terminated in a resistance .R:5O, then the current f flowing in the rails will be
is given by A
x
0.5 m
0
&@@ @@@ @@@ @@@
@@a
a&sB
8@@ @@@
R
v
Chap 6
Tin€ Varying Fieldr and Maxwell Equations
(A) 0.01c,.'sin0/t(1+2cosut) (B) -0.01arsina,r(1*2cosc,.'t) (C) 0.01c..'cosr.,'t(1+ 2sinat) (D) 0.05c,.,sinu.'(1+ 2sir.''t)
Page 370
Chap 6 Time Varying Fields and Maxwell Equations
ffeq s.1.1$ Electric flux density in a medium
(e.:
10,
F,:2)
is given
as
D :1.33sin(s x to\- o.2r)arp.Clm' Magnetic field intensity in the medium
\ *t\"
(A) 1o-5sin(3 x to8t - 0.2r)a, Alm (B) 2sin(3 x 108, -0.2r)a,Alm (C) -asin(3 x 108, -0.2r)auA,lm (D) asin(3 x 108t -0.2r)auAlm McQ
_-_-_-_
6.{.2$ A current filament located on the e-axis in free space with in the interval -0.1 < r< 0.1m carries current 1(0:8f A in a, direction. If the retarded
vector potential at point P(0,0,2) be ,4(f) then the plot of ,4(l) versus time
will be A(nwb/m)
A(nwb/m)
.4(nwb/m)
1(nwb/m)
Common Data For Q. 2l and,22 In a region of electric and magnetic fields .E and B, respectively, the force experienced by a test charge qC are given as follows for three different z
velocities.
m/sec ar
Force, N q(ay+ a")
e!
qal
az
q(2ar+ a")
Velocity
6"1.21 What will be the magnetic field B in the region ? (A) (C)
{ l'
h
o, o,
(B) o, - ou (D) a, - a"
6.1.2? What will be electric field (A) a,- a" (C) a,1- a"
6'l'e3
,E in the region
?
Page 371
(B) au (D) o, *
a"
a"-
Chap 6 Time Varying Fields and Maxwell Equations
o,,
Inanon-conductingmedium (o:0, I,t,: €,: 1), the retarded potentials are given as V : y(r- ct)volt and ,4 : A(+ - t)a,Wblm where c is velocity of waves in free space. The field (electric and magnetic) inside the medium satisfies Maxwell's equation if (A) J:0 only (B) p,: o only (C) .r: p,: o (D) Can't be possible
6.1.24 In Cartesian coordinates
magnetic field is given by B --2f ra". A square loop of side 2 m is lying in ry plane and parallel to the g-axis. Now, the loop is moving in that plane with a velocity n:2a, as shown in the figure.
What will be the circulation of the induced electric field around the loop (B) q (A)
(c)
4-#9-
"r
a*q
(D)q?)
Cornrnon Data For Q. 25 to 27 : \ In a cylindrical coordinate system, magnetic field is given by
forp< m [0 B:lZsinata" for4< p15m forp>5m [0 rflcg 6,'t.?5
The induced electric field in the region
(A)
0
p 14m will be (s) uff@"r
(C) -2coswta6 MCQ 5.1.26
The induced electric field at p
(A)
:
4.5
s'{"27
x*a",
13;
-
m is
0
.raaff!& Mcq
(o)
l7oggsc'rt
(P1 -I7uc'-osut
The induced electric field in the region p
) 5 m is
(A) -f;c.,cosuta4
@)
(C) -gpcosata6
-_gffe!", @) 9effet",
?
$cq
Page 372
Chap 6
Tine Jqryiry Fi€ld8 ara Maxi€ll Equations
6.r"s8
Magnetic flux density, B:O.rta, Tesla threads only the loop abcd lying in the plane ry as shown in the figure.
@t d
+ 40
aaa aaaB aaa
I
l_ I
Consider the three voltrneters V1, V2 and V3, connected across the resistance in the sarne rgr plarre. If the area of the loop abcd is 1m2 then the voltmeter readings are
u
V2
V,
(A) 66.7 mV (B) 33.3 mV
33.3 mV
66.7 mV
66.7 mV
(c)
33.3 mV
66.7 mV
66.7 mV
33.3
(D)
33.3
mV
66.7 mV
66.7 rnV
mV
Cotttmon Data For Q. 2g and J0 : A square wire loop of resistance ft rotated at an angular velocity ru in the uniform magnetic field B: 5av mWb f m, as shown in the figure.
a
ilcq 6't"rs If the angular velocity, a:2tadl will be (A) 2sindpV/m (C) 4cosd rNlm n{3Q
then the induced e.m.f. in the loop
(B) 2cosd rN lm (D) 4sindpV/m
6'1'30 If resistance R: 4omQ then the current flowing in the square ' (A) 0.2sin0mA (B) o.tsindmA (C) 0.1cosdmA
mcq
sec
loop will be
(D) 0.5sind mA
6'1"31 In a certain region magnetic flux density is given as B:
Bssinwt ar. A
rectangular loop of wire is defined in the region with it's one corner at origi., and one side along z_axis as shown in the figure.
ti
\ Page S73
Chap 6 Time Varying Fieldg and Maxwell Equations.
If the loop rotates at an angular velocity
c,., (same as the angular frequency of magnetic field) then the maximum value of induced e.m.f in the loop will
be
(B) zBosu (D) 4BgSw
(A) !B,)Su
(c) &s,
Corrmon Data For Q. 32 and 33 : Consider ihe figure shown below. Let B: 10cos120ntWb/m2 and assume that the magnetic field produced by i(t) is negligible
ooo o,ao o.,/ o
EA 6,{.32
o
The value of u,* is
(A) * 118.43 cos 120zrt V (C) -118.43sin t20nt V
3e
6,t"s3
(B) 118. 3cos120zrt V (D) 118.43sin720nt Y
The value of z(t) is
(A) -0.a7cos120nf A (C) -0. 7sin120nt A
(B) 0.47cosI2Urt A (D) 0.47sinl2hrt A *,F
\
r(** {.***t('f
:l
EXERGISE 6"2
Page 374
Chap 6 Time Var 'yrng Fields and Maxwell Equations
QUES 6.?,t
A small conducting loop is released from rest with in a vertical evacuated cylinder. What is the voltage induced (in mV) in the falling loop ? (Assume eaxth magnetic field : 10 6 T at a constant angle of 10" below the horizontal)
6.t.2 A square
loop of side 1m is located in the plane r: 0 as shown in figure. A non-uniform magnetic flux density through it is given as B : 4lf a,,The emf induced in the loop at time f : 2 sec will be Volt.
Qt^rgs
6.2'3 A very long straight wire carrying
a current 1: 5 A is placed at a distance of 2 m from a square loop as shown. If the side of the square loop is lm then the total flux passing through the square loop will be x 10-7 wb
a:1m
l,'
quEs 6.2.4
In a medium where no D.C. field is present, the conduction current density at any point is given as ./a : 20cos(1.5 x 108i)o, Al^'. Electric flux density in the medium will be Drsin(l.5 X 1084 aunCf m2 such that Do: __**
ouEs
A conducting medium has permittivity, €:4€s and conductivity, o:L.74 x 108s/m. The ratio of magnitude of displacement current and conduction current in the medium at 50GHz will be X 10-8.
6,2,5
rsrEs 6.2.s
In a certain region magnetic flux density is given as B: 0.Lta,Wp/m'?. An electric loop with resistance 2 O and 4 O is lying in r-y plane as Shown in the figure. If the area of the loop is 1m2 then, the voltage drop v1 across the 2
O resistance
is
mV.
Page 375
chrp
a 88+oaAA a -'4-* a a aaaaa a a aaaaa a V, +11-a a a a a a a a
I
l
5
I I
k
b
f*
: crEs
6.2.7
f p t
magnetic core of uniform cross section having two coils (Primary and secondarv) wound on it as shown in figure. The no. of turns of primary coil is 5000 and no. of turns of secondary coil is 3000. If a voltage source of 12 volt is connected across the primary coil then what will be the voltage (in
A
Volt) across the secondary coil ?
%:12 Volt
[ *="
u.u"u
Magnetic field intensity in free space is given
r[ lt eirEs
6.2.3
:0.1cos(15r'y)sin(6n'
as
x 10et- br)a"Af n
satisfies lVlaxwell's equation when I b l:
Two parallel conducting rails are being placed at a separation of 2 m as shown in figure. One end of the rail is being connected through a resistor ,R : i0 0 and the other end is kept open. A metal bar slides frictionlessly on the rails at a speed of 5 m/s away from the resistor. If the magnetic flux density B : 0.1Wb/mz pointing out of the page fills entire region then the Ampere. current 1 flowing in the resistor will be
R:10
O
l,:eucs
6.a"to
An infinitely long straight wire with a closed switch ,5 carries a uniform current I:4A as shown in figure. A square loop of side a:2rn and resistance
6
Time Varying Fields and Maxwell Equations
\------/'
distanee 2m from the wire. Now at a.ny time tthe switch is open so the current .[ drops to zero. what will be the totd cha"rge (in nC) that passes through a corner of the square loop after t: to ?
R-l0jq,lqpated.a$,p
Page 376
Chap 6
Tine Varying Fields ard Maxwell Dquations
a:2
QUE$
m
6.2.t1 A circular loop of radius 5m carries a current I:2A. If another small circular loop of radius lmm lies a distance 12 m above the large circula,r the common axis as shown in figure then total flux through the small loop
will
be
12m
Qus$
6.2.{2 A non magnetic'medium at frequency ,f:1.6 x 108H2 has permittivity
:
resistivity p : 0.77 Q - m. What will be the ratio of amplitudes of conduction current to the displacement current ? e
QUE$ 6"2"t3
54eo and
In a certain region a test charge is moving with an angular velocity 2raaf sec along a circular path of radius 2 m centred at origin in the r-y plane. If the magnetic flux density in the region isB: 2a,"wbf m2 then the electric field viewed by an observer moving with the test charge is V/m in o, direction. l
Common Data For Q. 13 and 14
:
In a nbn uniform magnetic field B:8ta,Tesla, two pa.rallel rails with a separation of 20 cm a,nd connected with a voltmeter at it's one end is located in r-y pla.ne as shown in figure. The Position of the bar which is sliding on the rails is given
as
r : t(r+0.4f)
]
l
lEs
e.e.ra
What will be the voltmeter reading (in volt) at f :0.4sec
?
Page 377
''t :i
Chap 6
l , \rar:i
Time Varying Flelds and Maxwell Equations
cm
J,
-s
T
6.2"'t5 What will be the voltmeter reading (in volt) a,t r: !2 cm
?
I
I
E
6.2-{E In a non conducting medium
(o:
0) magnetic field intensity at any point
.tf : cos(1010t - bx)o" A,lm. The permittivity of the medium is e:O.I2nF/m and permeability of the medium is p:3 x 10-5H/m. D.C. field is not present in medium. Field satisfies Maxwell's equation, if I 6 l: is given by
r i
:.
lF
o.z.r:
Electric field in free space in given .E
I
It
:
as
5sin(10zry)cos(6zr
satisfies Maxwell's equation for
I
b
x
Los
-
bn)a,
l: I
V
Fs 5'2.18 8 A current is flowing along a straight wire from a point charge situated at the origin to infinity and passing through the point (2,2,2). The circulation of the magnetic field intensity around the closed path formed by the triangle having the vertices (2,0,0), (O,Z,O) and (0,0,2) is equal to ____ Ampere.
{}s ,
6.2.{e A 50 turn rectangular loop of area 64 cm2 rotates at 60 revolution seconds in a magnetic field
B
:
per
z77twblm2 directed normal to the axis of rotation. what is the rms value of the induced voltage (in volt) ? 0.25 sin
******trl.*'k*
');tx'
EXERCI$E 6.3
Page 378
Chap 6 Time Varying Fielde and Maxwell Equations
.d(,e &"3"{
Match List I with List II and select the correct answer using the codes gi below (Notations have their usual meaning)
List-I
a b c
List-II
Ampere's circuital law
L. Y . D:po
Faraday's law
2. V'B:0 3. y x E:-A-B^
Gauss's law
dt
d
Non existence of isolated magneticharge
Codes
4. V
X
. , AD .E[: t*-at
:
abcd
(A)432t (B) 4r32 (c) 2374 (D)4312 F"{eG S"3"!
Magneto static fields is caused by (A) stationary charges
(B) steady_currents (C) time varying currents (D) none of these 8l$c& 6"3"3
Let a be magnetic vector potential and E be electric field intensity certain time in a time varying EM field. The correct relation between and ,4 is
$rcs 6"3"{
(A)
E:-#
g1
p: 0! dt
A closed surface 5 defines the boundary line of magnetic medium such t the field intensity inside it is B. Total outward magnetic flux through closed surface
(A)B.s (C) Bx s
[*]e{} 6,3.5
A:-# (D) A:# @)
will
be
(B) 0 (D) none of these
The total magnetic flux through a conducting loop having electric E: 0 inside it will be (A) 0 (B) constant (C) varying with time only (D) varying with time and area of the surface both
3e
!e
6.3.S
6.3.?
A cylindrical wire of a large cross section made of super conductor carries current 1. The current in the superconductor will be confined. (A) inside the wire (B) to the axis of cylindrical wire (C) to the surface of the wire (D) none of these
If B;
denotes the magnetic
flux density
increasing
with time and
a
Ba
denotes the magnetic flux density decreasing with time then which of the configuration is correct for the induced current 1 in the stationary loop ?
I
b. 6.3.i ; Fr
A circular loop is rotating about z-axis in a magnetic field The total induced voltage in the loop is caused by (A) tansformer emf (B) motion emf. (C) Combination of (A) and (B) (D) none of these
B:
,
6'3.e
E 6.3.rs
For static magnetic field,
(A) V x B: (C) V ' fl:
P
P'oJ
Displacement currerit density is
(B)
(A) D (q aDl at
tQ 6.3.t1
(B) V x B-- pJ (D) V X B:0
Q) arl at
The time varying electric field is
(A)
E:-YV
(c) E :-vv-
/
B
-Q 6.3.12 A field can exist if it satisfies (A) Gauss's law (B) Faraday's law (C) Coulomb's law (D) All Maxwell's equations
(B) E:-VV- A (D) E:-YV-D
Bocosat&u.
tt (lry3
h6a
Tine Varying FitLlt
ad
Maxwell Equatbns
Page 380
uce
6,3.{3
Cbap 6 Time Varying Fiel& and Maxwell Equatiors
Manwell's equations give the relations between (A) differerit fields
(B) different souices (C) different boundary conditions (D) none of these ilrcc
o.3.t4 If .E is a vector, then V ' V X E is
o
(A) (C) does not
(B) 1 (D) none of these
exist
Mcq
6.3.15
The Maxwell's equation ,
r$ce
6.3.16
For free space, (A) o: o"
Y ' B:
0 is due to (B) (A) B (D) none of these (C) non-existence of a mono pole
:'pH
(C) J +
0
B:I
(D) none of these
nca 6.3.1? 'For time va,rying EM fields
(A)VxH:J (C). V x E :0
(B)Vx-f,[:b+J (D) none of these
***********
EXERGISE 6.4
Page 3El
Chap 6 Time Var 'yrng Fields end Maxwell Equations
n
64''
fi,ffn";:"- ;13,;1"ffii"ffr#;i,*#
B
: B'(vti
"' -
7+r".)
[Hint : The algebra is trivial in cylindrical coordinates.] (A) -r: @)
#(r*t=-),r*
(C)
n
6.A"2
J:
0,r
*
o
r:-#(,fu),," (D) r: #(r#),r*
0
?
(A)V .E:O,VxB:0,(B)V
.E:0,V.8:0 (D)VxE:O,V.8:0
(C)VxE:0, VxB:0 6.4"3
If c
is closed curve enclosing a surface s, then magnetic field intensity the current density J an{the electric flux density D are related by
fr1t*#). * Q) ff,H' d,s: [1t*#). " (^)
F_ F sr
t
rt Ir. t t fr.
IIH.d,s:
The unit of V (A) Ampere
x
cr"5
The Ma:cwell equation (A) Ampere's law
(C) Faraday's
I
r^c.5
H,
dl: #,1r*#1. ot d,t:
il(r*%?).
*
-EI is
(B) Ampere fmeter (D) Ampere-meter
(C) Ampere fmeterz
I
o
For static electric and magnetic fields in an inhomogeneous soruce-free medium, which of the following represents the correct form, of Maxwell's equations
to
o
Yx
law
H:
J
*4 "
ir based on (g) Gauss'law (D) Coulomb's law
A loop is rotating about they y-a:
:
Bocos
(
ut
*
Q) a"
(A) zero (B) due to rotation only (C) due to transformer action only (D) due to both rotation and transformer action I
5.4.7
The credit of defining the following current is due to Maxwell. (A) Conduction current (B) Drift current
(C) Displacement current Q 6.4.8
(D) Diffusion current
A varying magnetic flux linking a coil is given by
:
A:LlJ\f .If
the emf induced is 9 V, then the value of ) is. (A) zero (B) 1Wb/s2 (C) -1wb/s' (D) swb/s' f
3 s,
at time
Page 382
Assuming that each loop is stationary and time varying magnetic field , induces current ,I, which of the configurations in the figures are correct
MSe 8.4.9
QhaF 6 Time Varying Fields and Maxwell Equations '
1.
Increasing
Decreasing
B
B
o o
2.
B
B
4.
(B)land3only
(A) 1, 2,3 and 4 (C)2and4only *tEQ e.4.t0
o o Increasing
Decreasing
3.
?
(D)3and4only
Assertion (A) : For time varying field the relation E - - V V is inadeq Reason (R) : Fa,raday's law states that for time varying field V x E: 0 (A) Both Assertion (A) and Reason (R) are individually true and (R) is the correct explanation of Assertion (A) (B) Both Assertion (A) and Reason (R) are individually true but (R) is not the correct explanation of Assertion (A)
(C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true .v'!e6, S"d.{
t
Who
developed
magnetic field (A) Gauss (C) Hertz fdcQ 8.4,{2
t?rcc
6.4.13
(B) Faraday (D) Maxwell
A single turn loop is situated in air, with a uniform magnetic field normal its plane. The area of the loop is 5 m2 and the rate of charge of flux \s 2'Nblrnzls. What is the emf appearing at the terminals of the loop ? (B) -2v (A) -5v (D) -1ov (c) -0.4v Which of the following equations results from the circuital form of Ampere's law
(A) V x
(c) &cQ 6,4.{4
the concept of time varying electric field producing
?
E:-#
v . D:
P
v . B:o (D)vx H: J+#
(B)
Assertion (A) : Capacita,nce of a solid conducting spherical body of r
o is given by Atresa in free space.
----!
Reason (R)
: V X trl: ja€E+ J
(A) Both A and R are individually true and R is the correct explanation of A.
(B) Both A and R are individually true but R is not the correct explanation
Page
BEB
Chap 6 Time Varying Fields and Maxwell Equations
of A.
(C) A is true but R is false (D) A is false but R is true
6'4-t5
Two conducting thin coils X and Y (identical except for a thin cut in coil Y) are placed in a uniform magnetic field which is Jecreasing at a consta,nr rate. If the plane of the coils is perpendicular to the field lines, which of the following statement is correct ? As a result, emf is induced in (A) both the coils (B) coil I/ only (C) coil X only (D) none of the two coils
6'4'tc
Assertion (A) : Time varying electric fierd produces magnetic fields. Reason (R) : Time varying magnetic fierd produces erectric fierds. (A) Both A and R are true and R is the correct expranation of A
(B) Both A and R are true but R is (C) A is true but R is false (D) A is false but R is true
6'4'{?
Nor
'f
A
Match List I (Electromagnetic Law) with List II (Different Form) and select the correct answer using the code given below the lists :
List-I
t
List-II
a. b.
Ampere's law
1.
Y.D:po
Faraday's law
2'
v . J:-qdt
c.
Gauss law
3' Vx.E[: J+Q dt
d.
Current
4'
Codes
(A) (B) (c) (D) rGQ 6.a.r8
the correct explanation
Y
x
E:-dP dt
:
a
b
c
1
2
3
3
4 4
1
2
1
3
2
3
2
1
4
d 4
Two metal rings 1 and
2 are placed in a uniform magnetic fierd which is decreasing with time with their planes perpendicular to the field. If the rings
are identical except that ring 2 has a thin air gap in following statements is correct ? (A) No e.m.f is induced in ring 1
(B) An e.m.f is induced in both the rings (C) Equal Joule heating occurs in both the rings (D) Joule heating does not occur in either ring.
it, which one of the
I
t Page 38d
fficq
6.4.te Which'one of the following radiation
Chap 6 i
Time Varying Fields and Maxwell Equations
Maxwell's' equatious glires the basic
? .:i
V x -E[: ADI At) v x E: ABI Atl Y ' D: oI (C) v . D:'ol
(A)
i
Y x E:- 0Bl 0t1 (B)v. D--aB'l atl
(D)v'B-p "vxH:(aDl
t|llc{l 6.4-21
(A)
v x H:#*,
(B)
v x E:A#
(c)
v . D:
(D)
V.
Match List i (Maxwell eciuation) with List correct answer : List
II
I
at)l
Mrtr $"4"2o Which one of the following is NOT a correct Ma:
P
-l
?
B:o
(Description) and select the
I
a. f a. ai:o b. fo. as:ln,du
c. f n.at:-l#.ot d' n'at:14'orl4'ot f
List
II
1.
The mmf aiound a closed path is eciual to the conduction current plus the time derlvative of the electric displacement eurrent through any surface boundeitl by the ilath.
2.
The emf a,round a closed path is equal to the time d.erivative is equal to the time derivatirre of the magnetic displacerhent through any surface bounded by the path.
3.
The total electric displacement through the surface enclosing a volume is equal to total charge within the volume
4.
The net magnetic flux emerging through any closed surface is zero.
Codes
:
abcd (A) (B)
(c) (D)
nics
6,4.22
1324 432r 4231 r234
The equation of continuity defines the relation between (A) electric lield a"nd magnetic field
(B) electric field and charge density (C) Ilux dehsity and charge density (D) current density and charge density
s"d$"ar What is the generalized Maxwell's equation V space
(C) V 0.4"24
x
H:#
Magnetic field intensity is 'EI density J Almz ?
:
3a,
*
x
H:
J"
(D) V
x
H:
D
A circular loop placed perpendicula,r to a uniform sinusoidal magnetic field of frequency c.rr is revolved about an axis through its diameter at an angular velocity a2 radf sec (uz < ur) as shown in the figure below. What are the frequencies for the e.m.f induced in the loop ?
+
+ + + + + + + +
+ + + + + + +
+
!
(B) ayu2l uzanduz (D) arr - trr2 and at* az
(A) ,t andaz (C) az,ar- c.2andu2 Hcq
s.4,26 Which one of the following
(A)V x
H:(o+ire)E
is not a Maxwell's equation
6.4.2?
Consider the following three equations
tol
f," 'ds:o
:
x.E:-# 2.Y xH:J+# 1.
V
3' V'B:0 Which of the above appear in Maxwell's equations ? (B) 1 and 2 (A) 1, 2 and 3 (D) 1 and 3 (C) 2 and 3 i,Ico
6.4.28 In free space, if p,:0, (A) (B) (C) (D)
the Poissonts equation becomes Maxwell's divergence equation V ' B:0 Laplacian equation
YzV:0
Kirchhoff's voltage equation None of the above
?
(B)r:Q(D+uxB)
tl f"r-dl:f t.as+ffr.as
Hca
* 2ra" A'f m' What is the current
(B) -7a" (D) I2o"
(C) 3a,
EV:0
Page 3Eb
Chap 6
Tine Varying Fiel& and
(B) V
Taan
(L) -2a, IGQ 6.jl.S$
for the free
?
(A) V x .E[:0
lca
x -E[: t"*#
Maxwell Equitions
Page 386
rucQ 6"4"29
Chap 6 Time Varyiag Fields and Maxwell Equations
A straight current carrying conductor and two conducting loops A are shovm in the figure given below. what are the induced current
two loops
and IN
?
(A) Anticlockwise in A and clockwise in B (B) Clockwise in A and anticlockwise in B (C) Clockwise both in A and B (D) Anticlockwise both in A and B McQ
6"''1'3t' Which one of the following equations is not Maxwell's equation for a static electromagnetic field in a linear homogeneous medium ? (A) V .
lllcQ
6"4'3{
B:0
(B) V
tl f"t.
dl: kr
Match List below :
I with List II
List
D:6
x
:
(D) V"4
pyJ
and select the correct answer using the codes given
I
List
II
a
Continuity equation
1. vx H: J+#
b
Ampere's law
2.
,dD dt
c
Displacement current
3.
Y x E:-A-B dt
d
Faraday's law
Codes
:
(A) (B)
a 4 4
b
c
3
2
1
1
2
3
2
3
2
1
4 4
3
(c)
(D) McQ
4.
Yx
J:-Q!n dt
d
1
6'4'32 The
magnetic flux through each turn of a 100 turn coil is (13 - 2f) milliwebers where f is in seconds. The induced e.m.f at r: 2 s is
(A)
1v
(c)
0.4
v
-1v
(B) (D) -0.4
v -\-
rca
6,4.33
Match List I (Type of field denoted by A) with List select the correct answer using the codes given'below : List
II
II 1. V 'A:0 VxA*0 List
I
a
A static electric fietd in a charge free region
b
A static electric field in a charged region
c
A steady magnetic field in a current carrying conductor
d
A time-varying electric field in a charged medium with time-varving magnetic field
Codes
(Behaviour) and
2. v.A+0 V xA:0 3. v.A+0 VxA.*0 4. V 'A:0 VxA:0
:
abcd
(A)4231 (B)4213 (c)2431 (D)2413 rco
6.4.34
Which one of the following pairs is not correctly matched D ' ds: {v (A) Gauss Theorem , f"
(B) Gauss's Law
fo.
:
(C) Coulomb's Law
(D) Stoke's Theorem
rcQ
6,4.35
rcl 6.4.36
-
' a":{oa,
Ddu
db^
dt
fe.m:f,$x{)'ds
:
Y x E:- (aBldt) .-ou:-&f B .
Maxwell equation
i^t/; tGQ
rr v-
:
?
dI
is represented in integral form
f Ex at:-$f B.dt
o,:-&lB'
rst
f E'
rnl
f Exat:-ftf B'dI
as
d's
Two conducting coils 1 and 2 (identical except that 2 is split) are placed in figure' a uniform magnetic field which decreases at a constant rate as in the If the planes of the coils are perpendicular to the field lines, the following statements are made
1. 2.
:
an e.m.f is induced in the split coil e.m.fs are induced in both coils
2
Page 3E7 ChaP 6 Time Valying Fields and
Marwell Equations
3. 4.
Page 38E
Cbap 6 Time Varyiag Fields aud Maxwell Equations
equal Joule heating occurs in hoth coils Joule heating does not occur in any coil
Which of the above statetnents is/are true ? (A) 1 and 4 (B) 2 and 4 (C) 3 only (D) 2 only MCq 5"4"3?
For linear isotropic rnaterials, both .E and .EI have the time dependence and regions of interest are free of charge. The value of V x .EI is given
(A) oE (C) oE+ mcc
6,4,38
E
Which of the following equa,tions is/are not Maxwell's equations(s)
(A) V . I
(C)
I
jwe
(B) jueE (D) oE * j,.oe E
v
J:-*
(B)
-E:*#
v . D:
folf
?
pu
H.rlr:l("r*,#).*
{ I
i t
Select the correct ans$'er using the codes given below
(A)2and4 (C) 1 and 3
.
l
:
(B)lalone (D) 1 and 4
.l
i
t{so
s.4"3s
Assertion (A) : The relationship between l\Iagnetic Vector potential ,4 and the current density J in free space is
V x (V x,4)
: hJ
For a magnetic field in free space due to a dc or slowly varying current
Y2A:-hJ
,1
Rea,son
(R) : For rnagnetic field
V',4:0.
I ! i :
(A) (B) (C) (D) MCO 6"4,4$
clue
to
dc
fo
or slowly varying curreri
Both A and R are true and R is the correct explanation of A Both A and R, are true but R, is NOT the correct explanation of A A is true but R is false A is false but R is true
Given that
V x .E[: l+Qdt
Assertion (A) : In the equation, the ad
(C) A is true but R is false (D) A is false but R is true HCo 6.4.4{
1
A circular loop is rotating a,bout the g-axis as a cliameter in a magnetic field 6 : Bosincrta,Wbf m2. The induced emf in the loop is (A) due to transformer emf only (B) due to motional emf only (C) due to a combination of transformer and motional emf (D) zero
which are placed Consider coils C1, Cz, Cr and G (shown in the given figures) by the coils produced in the time-varying el""tric field ,E(t) and electric field : C'r,C', and Cin c.rrying time varving current 1(t) respectively E(t)
2.
1.
Coil planes are orthogonal
Time va.rying electric field E(t) parallel to the Plane of coil Ct
'al
(r)
\--/
OO.
3.
4.
Coil planes are orthogonal
Co-Pianer coils
The electric field lv'ill induce an emf in the coils
rcQ
6.4'cs
(A) C' and Cz (C) Cr and C3
(B) G and G (D) G and Ct
Match List I (Law/quantity) with List select the correct answer :
II
List
law
Gauss's
b.
Ampere's
larr
2.
c.
Faraday's
law
3.
d.
Poyrrting
vector
4.
1'
c.
(A) (B)
(c)
(D)
:
a
b
c
d
1
2
4
3
3
5
2
1
5
2
3
4
1
1
3
II Y'D:P
List
I
a.
Codes
(\ilathematical expression) and
,
{<******{<***
Yx
E:-0#
P:ExH P:q(E*uxB) V x ,E[:
J"*#
Page
3t9
ChnP 6 Time Varying Fields and
Maxlell Equhtlons
soLUTloNs 6.{
Page 390
Chap 6 Time Varying Fields atrd Maxwell Equations
30L 6,{.t
Option (C) is correct. FYom Faraday's law) the relation between electric field and magnetic field is
v x E :-a-B^ 0t Since the electric field inside a conducting sphere is zero. l.e.
E:0
So the rate of change
in magnetic flux density will be
AB
6:-(v Therefore sol- 6"{.2
B(r,t) will
x@:s
be uniform inside the sphere and independent of time.
Option (A) is correct. Electric field intensity experienced by the moving conductor ob in the presence of magnetic field B is given as E: aXB where u is the velocity of the conductor. So, electric field will be directed from b to o as determined by right hand rule for the cross vector. Therefore, the voltage difference between the two ends of the conductor is given as
vt:-
fun'
m
JO
Thus, the positive terminal of voltage will be a and 50L
6,1,3
Voo
will be positive.
Option (A) is correct. Consider a magnet bar being dropped inside a pipe as shown in figure.
Pipe Falling magnet
Ring of pipe
I in the magnet flows counter clockwise (viewed from above) as shown in figure. So near the ends of pipe, it's field points upward. A ring of pipe below the magnet experiences an increasing upward flux as the magnet approaches and hence by Lenz's law a current will be induced in it such as to produce downward flux. Thus, 4,a must flow clockwise which is opposite to the current in the magnet. Suppose the current
I
lt-
Since opposite currents repel each other so, the force exerted on the magnet
due to the induced current is directed upward. Meanwhile a ring above the magnet experiences a decreasing upward flux; so itts induced current parallel to 1 and it attracts magnet upward. And flux through the rings next to the magnet bar is constant. So no current is induced in them. Thus, for all we can say that the force exerted by the eddy current (induced current according to Lenz's law) on the magnet is in upward direction which causes the delay to reach the bottom. Whereas in the cases of unmagnetized bar no induced current is formed. So it reaches in fraction of time.
Thus, A and R both true and R is correct explanation of A.
sol 6.t.4
Option (C) is correct. The magnetic flux density inside a solenoid of n turns per unit length carrying current 1 is defined as 6 : panl Let the length of solenoid be / and its cross sectional radius be r. So, the total magnetic flux through the solenoid is O : (pnnl) (trf) (nt) (1) Since the total magnetic flux through a coil having inductance .t and carrying current .I is given as
iD:LI So comparing
it with equation (1) we get, L : por*I#I
and as for a given solenoid, radius
r
and length / is constant therefore
Lqn2 n 6.1,s
Option (C) is correct. The magnetic flux density inside the solenoid is defined
where So the
as
B - p4nl ??
+
1
--+
no. of turns per unit length current flowing in it.
total magnetic flux through the solenoid is
o:IB.d,S:(panl)Qrar) J where
o --+ radius of solenoid Induced emf in a loop placed in a magnetic field is defined
as
tr dO V"^f :-E where @ is the total magnetic flux passing through the loop. Since the resistance .R is looped over the solenoid so total flux through the loop will be equal to the total flux through the solenoid and therefore the induced emf in the loop of resistance will be
Vmr:-Ta2p-"# Since current 1 flowing in the solenoid is constant so, the induced emf is
%-r:0 and therefore the induced current in the loop
sot.
c.'t.G
Option (B) is correct.
will be zero.
Page 391
Chap 6 Time varyiug Fields and Marwell Equatione
Page 392
Chap 6
*-l#ilTi:"#lT:',rffTffilt*i,r
frne'Varying fielda and Mar*ell Equations
$oL 5,{,7
wh'e the erectric'-
{
Option (B) is correct. I In Assertion (A) the magnetic flux through each turn of both coils are "qud l So, the net magnetic flux through the two coils are respectively I
A: NriD
and
i\z
l
: NziF
where @ is the magnetic flux through a single loop of either coil and ,A/r. trfz are the total no. of turns of the two.coils respectively Therefore the induced emf in the two coils are diD,
tf
Yemfl
yernr2
--E--
:_
_
do" T:_
n, dib trlzl
^, do lur6
Thus, the ratio of the induced emf in the two loops are
V^tz
Nz
%,nir- ifr
Now, in Reason (R) : a primitive transformer is similar to the cylinder core ca,rrying wound coils. It is the device in which by choosing the appropriate no. of turns, any desired secondary emf can be obtained. So, both the statements are correct
80L
6"'1.8
but R is not the explanation of A.
Option (B) is correct. Electric flux density in the medium is given
D
:
eE
as
: €Eocosat
(E:
Escoswt)
Therefore the displacement current density in the medium is
h : #:-aeEosinut and the conduction current density in the medium is
J" : oE: oEocOsut So, the ratio of amplitudes of conduction current density and displacement current density is
W-- u€" lJ"l
$gL 6.1.9
Option (C) is correct. Given the volume charge density p,: So, from Maxwell's equation we have
Q
Y'D:p, V.D:0
(1)
Now, the electric flux density in a medium is defined as D : eE (where e is the permittivity of the medium) So, putting it in equation (1) we get,
or, and since Therefore,
:0 E. (v e) *e (V . E) :0 Y9=o+Ve=o V'
(e.E)
V.lq=0
(given)
I I
f
i
L 6.{.{*
Option (A) is correct. Given the electric field intensity in time domain
Page 393
Cbap 6
as
!)o, 6 :l-1"6_(yt:T So, the electric field intensity
E, :V0
and
h
Tine Varying Ftelds atd Maxwell Equations
in phasor form is given
"*i*"
as
*
: (- ltelwo "-ir' or
Y x E, :lf7@,")qo
Therefore, from Maxwell's equation we get the magnetic field intensity as
t
g":-!-lA:
lAro
6.{.11
h.
k
v!-"i*o r
dTo
Option (B) is correct. Magnetic fl1x density produced at a distance p frctm a long straight wire carrying current 1 is defined as
I
B
B
: !ta^ z1T p.
where a6 is the direction of flux density as determined by right hand rule. So, the magnetic flux density produced by the straight conducting wire linking ihrough the loop is normal to the surface of the loop' Now consider a strip of width dp of the square loop at distance p from the wire for which t[e total magnetic flux linking throrrgh the square loop is
I t t
f,
I II
given as
t
o:fa.as
E'
I
Js
I
: # [r'.'f,toor) -T'"\- p ) --P.IaulP*a\
t
F
(area of the square loop is
t]s:
atl,fi')
The induced emf clue to the change in flux (when pulled away) is given
I
v"mt
i
Therefore, Given
:-# :-g
as
r1[*(']')l
v^,:-#G"#-i't#) # :
r"lo"itY of looP
:
5
m/s
and since the loop is currently located at 3m distance from the straight wire, so after 0.6 sec it will be at (o --+ velocity of the loop) P :3+ (0'6) x u'
:3*0.6X5:6m
So,
%*r
:_po x [3!) x 2[trri _f rsll
:25 x 10 ? volt : sol. 6,t,"t2
(a:
2 m,
1:
30
A)
2.5 pvolt
Option (B) is correct. Since total magnetic fl1x through the loop depends on the distance from the straight wire and the distance is constant. So the flux linking through the Ioop will be constant, if it is pulied parallel to the straight wire. Therefore the induced emf in the loop is
V^r:-# :0
(@ is constant)
Page 394
${}L 6"*.{3
Chap 6 Time Varying Fields aad Maxwell Equations
Option (D) is correct. Total magnetic flux through the solenoid is given as E : panl where n, is the no. of turns per unit length of solenoid and
/
is the current
flowing in the solenoid. Since the solenoid carries current
that is increasing linearly with time
Iqt
t.e.
So the net magnetic flux through the solenoid
will be
iDqt
itr:kt
where&isaconstant. ort resistances ,Ba, .Re is loop consisting Therefore the emf induced in the
,r ,enr __
dO
dt
V^t:- k and the currert through .Rr and ,Bz will be r_k__HTE ,na Now according to Lenz's law the induced current I in a loop flows such as to produce a magnetic field that opposes the change in .El(d)i.e. the induced current in the loop will be opposite to the direction of current in solenoid (in anticlockwise direction).
So,
Vt: IinaRe:-#+fu
and
r D -t kRn -\ Va-- tindtlB - \E;+E )
Thus, the ratio of voltmeter readings is
Vr
Va---Re Rs $oi* s.t.{4
Option (D) is correct. Induced emf in the conducting loop formed by rail, bar and the resistor is given by
rr v" r :_
dO
ilt
where O is total magnetic flux passing through the loop. The bar is located at a distance r from the resistor at time magnetic flux passing through the loop at time t is O rails
: IB. J
d,S
:Blr
where
l. So the total
I is separation between the
Now the induced emf in a loop placed in magnetic field is defined V*t :-Edo
as
is the total magnetic flux passing through the loop. Therefore the induced emf in the square loop is where
@
V^,
:-*t(Bln)
:-
Since from the given figure, we have
l:5m B :27
BI#
(iD:Blr)
and
drf dt
-
velocity of bar
:
Amls
So, induced emf is V.mr
:-
(2) (5) (4)
: - 40 volt
Tine
I:ry1:-#:-4A
Option (B) is correct. As obtained in the previous question the current flowing in the sliding bar is
I:-4A
Now we consider magnetic field acts in a, direction and current in thrr sliding bar is flowing in * a, direction as shown in the figure.
Therefore, the force exerted on the bar is
r : I rmX B : [\_ +aro") x (2a,) i.e. rhe force exerteo:."tiiltJli;#l3T motion of the sliding bar. h, ar.ro i q
in
opposite direction
to
the
Option (C) is correct. Given the magnetic flux density through the square loop is B : 7.5cos (I2\trt_ 30.) o, So the total magnetic flux passing through the loop will be as
o:{a. Js
: [-7.5cos(l20rt- JO")a,](r x 1)(- a,) : 7.5cos(120zrf - 30')
Now, the induced emf in the squa.re loop is given by V.
,
:-#:
7.5
x
1202'sin(120a.t- 30')
The polarity of induced emf (according to Lenz's law) will be such that induced current in the loop will be in opposite direction to the current 1(i) shown in the figure. So we have
IU):-W -:x.
6.,r.1?
7'5 l!204'sin(120zrt
VaryingF*d nqft
Maxwell
Therefore the current in the bar loop will be
- g0") (R:2s0*
2b0
:
b00
o)
5.7sin(12hrt- 80")
Option (A) is correct. consider the mutual inductance between the rectangular loop and straight
Page 396
wire be
M.
Chap 6
So applying
KVL in the rectangular loop we get,
M#: tffi+ nb
Tino Varying Fielde and Maxwell Eqrntions
...(i)
Now from the shown figure (b), the current flowing in the straight wire b given as
h:
Itu(t)
- Iru(t- 11 (/r is amplitude
or, ffi : tro1t1- t,6(t- t1 So,atf:0 ff:t, and Solving
MI1
it
:
t
ffi
+
of the current) (2)
nA
(from equation (1))
we get
i,
: ff Ire-t*rtt'
Again in equation (2) at
for0
t: 7 we have
__i dt -- t'
dh
and Solving
-MIL:tff+nU it
(from equation
(l))
we get 'io
:-4Le-@/L)$-n -rv
for
L
t>
T
Thus, the current in the rectangular loop is
[ {r,e-@/L)t t, _._l L2 -1
0<
l-ft'"-'o'uu-n t>
t
Plotting i2 versus f we get '2(t)
sol- 6.t,{8
Option (A) is correct Total magnetic flux passing through the loop formed by the resistance, bar and the rails is given as:
o:la-as
So, the induced
I
- B . S : [0.2cos ata,]. [o.f (t - g)a,] | : 0.1[1 - 0.5(1 - cosrut)]coswt (y:0.5(1 - cos*rt]{ : 0.05cosc..,t(1 + cosot) : 0.05(cosc.r t* cos2ut) | emf in the loop is I
V^r:-#
and as determined by Lenz's law, the induced current will be nowins
{ { 1
r
opposite direction to the current
i.
so the.current
i
o:-Vu!' :-I / d@t R- --F-\-E)
'
: S[-
,
:_
6.t.te
,
,
in wt
-
in the roop will be
Page
.
30i
Chap 6
Time Varying Fields ard
2w cos wtsin
Maxvell Equations
wtf
0.01c^,,sina..rt(1+2coswt)
Option (D) is correct. Given the electric flux density in the medium is D = 1.B3sin(a x rO8l _0.2r)arp.C/m, So, the electric field intensity
in the medium is given
E:Q€
where e is the permittivity of the medium
: D _ E --€,eo
oft
Now, rrom maxweu,s
as
1.33
x
t0-6sin(3
@or(e.:10)
;rl":;"t$Jf"g
t
108'
-
x t08t-0.2r), 0'2r)
.^\
ao
y xE:-#
or,
+:-y xE dt :--!4"
AE,,
dx
:- (- 0.2) x (1.b x 10a)cos(3 x 108, _ 0.2r)au :3 x 103cos(B x 108t -0.2r)a,
Integrating both sides, we get the magnetic flux density in the medium x loscos(3 x 1o8r_ o.2r)au
: I, " :fi-f$,tn(3 x rosr -0.2x)au
as
a
a
:
10-5sin(3 x 108, _ 0.2r)a, Tesla Therefore the magnetic field intensity in the medium is
Thus
H:B-P
B :
ll":2
lt, pn
II :4sin(3 x
108,
-
0.2r)oo Alm
Option (B) is correct. P(0, 0, 2)
fir
The magnetic vector potential for a direct current flowing in a filament is
given as
A
: [ ffio.a,
Here current 1(t) flowing
in the firament
shown
in figure is varying with
I'
time
Page 398
as
I(t) :311'
Chap 6 Time Varying Fields and Maxwel! Equations '
So, the retarded vector potential
at the point P will be given
n:f4ffio'*
as
where -R is the distance of any point on the filamentary current from P as shown in the figure and c is the velocity of waves in free space. So, we have
R:/;\4andc:3x108m/s
o:[)j_,,@wo".o,
Therefore,
:Wlli:ffia'-
l,)ia"]
:8 x 1o-7r[rn(r+ JATn)f :,-ffit"]5,
I
t-
:8 x rc-tr1^( o'r+{4'u \-0.* +,/ 4.01
I
\-0.1
I
I
x
10-15
:8 x 1o-8t- 0.53 x 1o-15 A : (80f - 5.3 x 10-7)a" nWb/m or, t : 6.6 X 10-e : 6.6 nsec So, when A: 0 A *-5.3 x l0-?nWb/m andwhen f :0
I
l
(U
Flom equation (1) it is clear that, A will be linearly increasing with resped to time. Therefore the plot of A versus f is ,4(nwb/m)
15.3 x 10
I{OTE
7
r
'.lirn.r: var'.virrg potrrrt.ial is
$or- 6,{-21
tistallr calkrl thc lr,:inrdi:d pi,tutT'ial
Option (A) is correct. The force experienced by a test cha,rge g in presence of both electric .E and magnetic field B in the region will be evaluated bv using force equation as
p : q(E*ax B) putting the given three forces and their corresponding velocities in equation we get the following relations q(ar* a") : q(E+ a,x B) So,
: q(E* oo x B) q(2ar* a") : q(E* a" x B) nsry
Subtracting.equation (2) from (1) we get
a" : (a,- an) x B and subtracting equation (1) from (3) we get au : (&,- ar) X B
B: B,a,* Buau* B"a,in eq (4) to get a" : Bve"_ B"&v* Bra"_ B"o,
Now we substitute So, comparing the
r,y
page 899 Chap 6
and,z components of the two sides we get
Time Varying Fields and Maxwell Equatiom
:1 and B" :0 Again by substituting B: B,a,* Brau* B,a" ineq (5), we get ay : B,&s- Bue,_ Boa"a B"au B,+
BU
so, comparing the
r,y
and.z components of the two sides we get
B,+8,:r
and
By
as calculated above
:0
B,:O,therefore B,:L
Thus, the magnetic flux density in the region is
B : a,Wblm2
(8,:1, Bu: B":0\
Option (C) is correct A' rul"uiuiud i;;;"rs So,
putting it
q]restior the magnetic flux densitv in the region is t : aaWb/m2 in Lorentz force equation we get
F:q(EavxB)
ort
q(%+ a,)
:
q(Ey a, X a,)
Therefore, the electric field intensity in the medium is
: q* a"yfm
E
30L 6.1.23
Option (C) is correct. Given
potential, and retarded vector potential, Retarded scalar
: y(r_ ct)volt A : U(t _ ,)o,Wb/m V
Now the magnetic flux density in the medium is given
B:VxC
:-*o":
as
(r-f)o,r""ru
(1)
So, the magnetic field intensity in the medium is
H:BPo
(pa is the permittivity of the medium)
: *(r-
f,)o" 11^
(2)
and the electric field intensity in the medium is given as
E so, the erectric
__o
r_#
:(ct- r)au o,*;;1;; jih;lff j ,,:' D : eoE
Now we determine equation.
(a) or'
(ee is the
rou";"!?; :)#?#"rd
Y.D:p, pa:y.[ro("r-x)ar]
(B)
permittivity of the medium)
to satisry au the
r"",nn**",!1]
(from equation (a))
-0
Page 400
-
Chirp
e
It
means the field satisfies lVlaxwell's equation
(b)
Ti$e Varying Fields atrd Maxwell Equatiors
if p,:9.
V.B:o
Now, v.B-v.[(,-f)",] So,
it
(from equation
(l
already, satisfies Maxwell's equation
(")
vx.EI-t*Qatr
Now.
v x .tr :-%4;",:ho,:
t[fro,
(from equation
and from equation (4) we have
dD : €ncay: tE; -At
(Since in free space
J fi"
c:
Putting the two results in Maxwell's equation, we get the condition
1.,
J:0
u
iI .i
J 'I
ii
(d)
Vx.O:-#
Now
VxE:Af;a,:-a, q. :o, ot
it already
satisfies Maxwell's equation. Thus, by combining all the results we get the two required conditions as 0 and p, 0 for the field to satisfu Max.well's equatiorr. So,
J:
se:L
s"1.s4
:
Option (A) is correct. Given the magnetic flux density through the loop is
B:-2f ra. So the
total magnetic flux passing through the loop is given
* : I u . ds : I'-' I'-' (-1",).
(-
as
d,rd.ya,)
: (zi" "Ir)t
l:4h(#) Therefore, the circulation of induced electric field in the loop f n. m:-@d, :-*l^"(#)l 4 d (r+2\ :__ -- (qElal-\ " / \r ) 4r r 2drt : -tTZ(? ,Il ) __ : 8 ror__ : 16
s$L
s.*"as
is
tdr
'1'44"r r'@+'
\A:
Option (A) is correct. As the magnetic flux density for p<4 is B:0 so, the total flux through the closed loop defined by p:4m is
o:Ig.d,s:o J So, the induced electric field circulatir.rn for the region p
{n.m__di| _n il" a
I
4
\
m is given
as
E:0
or, 6.t.26
forp<4m
Option (B) is correct. As the magnetic field for the region p14m and p) 5m is zero so the distribution of magnetic nr* a"rrrity a^s shown in figure below. we get
'm @-t-J
{^ao \w
At any distance p f.o.* origin in the region 4 < p <5 m, the circuration of
induced electric field is given as
$n. Jc
OI,
n:-dq dt --ddt (l'' *) :-$1z"inrtQrp, nar)l -
.:_2arcoswtQrp2_I6n) :- 2ucosut("t _ rcn) F _ 2(pt_ 16)c..,cosc.rf
E(2rp)
t
L
So, the induced electric nufa irrt"i3i
E
:-t
Kl.s)t
p:
ty at
- 16)r..,cos
4.5
m is
cr.,f
:-fiwcoswt n
6.7.27
Option (B) is correct. For the region P)5m the magnetic flux density is 0 and so the total magnetic flux passing through the closed loop defined by p :5 m is
a:l'". :o+
so, the circuration
is
ds
_[^a.n+|,n.
f\zrinwt)a".
d,S
-t"yrl:
nagnetic flux density
"r;11:::lt"r1r
as
{or
l8zrsinc"'r
any loop in the region p
f n' m:-#
' E(2rp):-$(tSnsinwt)
:_
.a
lgrwcosut
So, the induced electric field intensity
,:Jfftlo,
in the region p > 5m is
)
5
m
page 401 Chap 6 Time Vmying t'tslds
ad
Maxrrcll Equatirons
Page 402 Chap 6
Tine Varying Fields and Maxwell Equations
soL
6"1"29
Option (D) is correct. The distribution of magnetic flux density and the resistance in the circuit are same as given in section A .(Q. 31) so, as calculated in the question, the two voltage drops in the loop due to magnetic flux density B :0.1t az are 14 : 33.3 mV and V :66.67 mV : 66.7 mV Now % (voltmeter) which is directly connected to terminal cd is in paralld to both V and I{. It must be kept in mind that the loop formed by voltmete V3 and resistance 2 O also carries the magnetic flux density crossing through it. So, in this loop the induced emf will be produced which will be same as the field produced in loop abcd at the enclosed fluxes will be same. Therefore as calculated above induced emf in the loop of % is V^t : 100 mV According to lenz's law it's polarity will be opposite to 73 and so
-V^r:V+vs
or, sol.
6"1.29
%:100-33.3:66.7mV
Option (D) is correct. ThelatludE\emf in a closed loop is defined as /\ rr . 'emt --dQ dt where @ is the total magnetic flux passing through the square loop At any time f , angle between B and dS is g since B is in oo direction the total magnetic flux passing through the square loop is
so
o:Ia.as J
: (B)(S)cos0 : (b x 10-3)(20 x 10-3 x 20 x 10-3)cosd' :2 x 10-6cosd
Therefore the induced emf in the loop is
t,v. f :_ dO ilt
:-2
x ro-6$(cosd)
:2 x l}-Bsinlffi and So,
$oL 8.r.30
as
a: d0
V^r
: :
:2rad/
angular velocity (2
x
sec
lO-o)sin0(2)
4 X 10-6sin0
Y
fm
:
4sind pV/m
Option (B) is correct. As calculated in previous question the induced emf in the closed square loc;r is
V^t
:
So the induced current
I_
4sin0
p,Y f
m
in the loop is
W R
where ,R is the resistance in the loop-
_ 4sin0 x
(ft
40 x 10-3 0.1sind mA
: scl. 6.{.3{
10_-o
:_# :-
no
40 mO)
Chap 6
as
:-$Unorina;r) (,gcosdl s
ftlsin
(0:
i,,'tcos a'rtl
o.'t)
--
BoScos2ut Thus, the maximum value of induced emf is
lV"6l: sol. 6.{,32
BoSu
Option (C) is correct. e.m.f. induced in the loop due to the magnetic flux density is given
aa
as
$1to"os l2}rt)(trp2)
UT.: ':-r(to x T!
10-'?f
:
x
(1202r)(-10sin120zrt)
L2zjsinl2}nt
As determined by Lenz's law the polarity of induced e.m.f will be such that b is at positive terminal with respect to o. l.e.
or
sol
6.{,33
Vo: Vt: L2tf sinl2}nt Vt :- I2fisinl2}nt : - 118.43sin 120zrt Volt
Option (D) is correct. As calculated in previous question, the voltage induced in the loop is
Vt
:-
12z2sin720trt
Therefore, the current flowing in the loop is given
IU):-h: :
0.47
sinl2hrt
rI t
***X*rl.****.*
Page 403 Time Varying Fields and Maxwell Equations
\ Option (C) is correct. The total magnetic flux through the square loop is given , : f t . d,S :(&sinc.,t)(S)cosd So, the induced emf in the loop is
V*
:
as
$oLr,sTx$rus s.2
Page 404
Chap 6 Time Varyfuig Fieldi and Maxwelt Equa^tions
sol.
6,2.1
Correct answer is 0. As the conducting loop is falling freely So, the flux through loop will constant. Therefore, the voltage induced in the loop will be zero.
$oL
6"2.2
Correct answer is -4. The magnetic flux density passing through the loop is given
B:4ff
as
a,,
Since the flux density is directed normal to the plane z: 0 so the totd magnetic flux passing through the square loop located in the plane r: 0 i
, : I u . d.s : [=,[),fnor)
d,yd,z
:r
@s
Induced emf in a loop placed in magnetic field is defined
:
(dyd,z)
+!
as
V"^t:-Eda where @ is the total magnetic flux passing through the loop.. So the induced emf in the square loop is
v"*
:-#:-
(@=fl
2t
t:2sec the induced %",i :- 4 volt
Therefore at tirne
sol
6"2.3
ernf is
Correct answer is 4.05 Magnetic flux density produced at a distance carrying current 1 is defined as
p from a long straight
fl '= unl fio, where o6 is the direction of flux density as determined by right hand rule. the flux density produced by straight wire at a distance p from it is
(a"
is unit vector normal to the ":Ho, Therefore the total magnet flux passing through the loop is
*:
Iu. ds : I'"'#oo,
where dp is width of the strip of loop at a distance p from the straight Thus,
, : I'(#)y : #^(t): :
sol-
6"2,4
(2
x
10-1 (5)ln (1.5)
:
gpr,,1r
4.05
x
10-7
s)
Wb
Correct answer is 133.3 . The displacement current density in a rnedium is equal to the rate of in electric flux density in the medium. Jo
:-dtap^
Since the displacernent current density in the medium is given as
.Ia
:
x
20cos(1.b
1o8r)a,
Page 40b
Chap 6 Time Varying Flelds o.d Maxwell Eqrntions
A/^,
So, the electric flux density in the medium is
n: f-loat+c As there is
'o
DC
(C*
constant)
,;/#::1if ;"'fij1,?f*l,"-" getc:0 and thus,
, _: 20sin(1.5 x 108r). 1;;10*-Jo,:1.33 x 10-7sin(t.b x 108t)a, : 133.3sin(1.b x 108t)onnCfm2
Since, from the given problem we have the flux density
D
:
D6sin(1.b
Do
:
133'3
x
1o8t)ornCfm,
So, we get
:
IOL
$.2.5
Correct answer is 9.75 . The ratio of magnitudes of displacement current to conduction current in any medium having perurittivity e and condrrctivity o is given as
l-&!q"."$9!L9qII9!t | _
I
Conduction current j
..,e
-7
where ur is ttre angular trequency of the current in the medium. Given frequency, / : 50 GHz
Permittivity, € -4es:4x 8.85X 10-12 Conductivity, o:1.I4 x 108s/m So, a :2trf:2tr x 50 X 10e: 1002r x 10e
Therefore, the ratio of magnitudes of displacemelt current to the conduction
current is I
Ial
ll;l--
F
6,2.$
_mq11
x
l0n_X_4_X&!5;1 10-" :9.75 1.14
x
108
x
10-8
Correct answer is 33.3 . Given magnetic flux density through the square loop is
B :O.lta, Wb/m, So,
total magnetic flrrx passing through the loop iD : B.
d,9
:(0.1rX1):
0.11
The induced emf (voltage) in the loop is given
V^,
:-ff,:
-
is
as
0.1Volt
As determined by Lenz's law the polarity of induced emf will be such that
V+
V':-
%*r
Therefore, the voltage drop in the 2 O resistance is
Vr: I 2 \'
lT+z)e
sol
6,e,7
Correct answer is 7.2 Voltage,
r/ t I %*r) :
f
: sr'3 mv
.
V:- Nr#
where @ is total magnetic flux passing through it.
Again
Page 406
Qhnp 6
Vr:-
Nr#
Since both the coil are in same magnetic field so, change in flux
Time Varying Fields aud Maxwell Equations
will be sane
for both the coil. Comparing the equations (1) and (2) we get
V _M V-W, ,, : u#: (12)9333 :7.2vott
sot
6.2.8
Correct answer is
41.6
In phasor form the magnetic field intensity can be written
II, :
o.1cos(1"5zry)e-
jb'a"
Af
!
{ I I I I I
I
as
I
m
Similar as determined in MCQ 42 using Maxwell's equation we Set the
relation
I I 1
(r5nf+b'--l#-' +
So, sol-
6.2,9
f
b:*41.6ndfm
lbl-41.6rad/m
Correct answer is 0.01 . Induced emf. in the conducting loop formed by rail, bar and the resistor is given by V"-f
:-E do
total magnetic flux passing through the loop. Consider the bar be located at a distance r from the resistor at time the total magnetic flux passing through the loop at time t is
where
@ is
,:lt.ds:Btr
(area of the loop is
Now the induced emf in a loop placed in magnetic field is defined do V" :-E
l.
So
^9: lr)
as
f
where @ is the total magnetic flux passing through the loop. Therefore the induced emf in the square loop is
,:-fitau) :-
y"
Bl dr
(O:BLrl
a
Since from the given figure, we have
and
I:2m and B :0.1Wb/m2 drf dt : velocity of ba,r : 5 m/s
So, induced emf
is
-
V^r
:-
(0.1) (2) (S)
:-
I I
(15rf + b2 : Groeo r..r :6zr X 10e Here so, (1b?r'I+ 6z : /6r' x 1Qe12
b2:l7bi
I
1't
o11
According to Lenz's law the induced current ,I in a loop flows such as to produce magnetic field that opposes the change in B(l). As the bar move away from the resistor the change in magnetic field will be out of the page so the induced current will be in the same direction of f shown in figure. Thus, the current in the loop is
r- V^t (-ti ,:--r:-'10':0.01A 6.2"1c
Correct answer is 2ZT. Magnetic fl'x density produced at a distance carrying current 1 is defined a^s
(ft:
Chap 6
p from a long straight wire
t:Ho,
where aa is the direction of flux density as determined by right hand rule. since the direction of magnetic fl'x density produced at the roop is normal to the surface of the loop So, total nu* p*rirrg tfr.ough tlr" loop is given by
*:
-
.s :
I"'
I:,(H)r"*l
PoIa fa dp -Ztr p
(ds:
ad,p)
l,
:$!,mz:$mp1 The current flowing in the loop is
So,
V"
flooo
a.\d.induced e.m.f. is Zu6.
r.: Ir"""R:-4 tvuP'e dt
ffos:-f,r'12yff where
e
is the total charge passing through a corner of square loop.
# :- **r"@# :_#tne)
d,e
(
R:4e)
dr
Therefore the total charge passing through a corner of square roop is
q :-ffime1foar
: -ffitn1z1(o - 4)
:uffEmp. :2.77 x
Dt 6.a.r{
10-z
C :2TT nC
Correct answer is 44.9 . since the radius of sman circular loop is negligible in comparison to the ra'dius of the large loop. so, the flux density aitr"rn the small loop will be constant and equal to the flux on the axis ofthe loops.
So,
pal R2 B ---2-@17y""
where
r?
-
radius of large loop
z
--+
distance between the loops
:
5
ry1
:
12
m
(5)'
o,,:ffia, tr,rrS*, : ^:*# " tii,o"si, \iJ"rr*"tt
Therefore, the total flux passing
': I "'
d,S :
page 402
10O)
il;i,
ffi " "f wherer is radius of small circular loop. _25x4trxIO7 -=-6f* x z(10-3f :44.9fWu
Time
litying
Fields and Maxwell Bquations
sol. 6.2.t2
Page 40E
Correctans\r,er'is'2i7.
,
,
Electric field in any medium is equal to the voltage drop per uni
Chap 6 Time Varying Fields and Maxwell Equatiors ,
i.e.
E
:Y -d
where
y
-+ potential difference between two points.
d
-
,
distance between the two points.
The voltage difference between any two points in the medium is V : Vocos2rft So the conduction current density in the medium is given as (o --' conductivitY of tl J" : oE
: E (p ' V -Vocos2trft --pdpd OT,
resistivity of
tl
(v\.
IJ"l:h
and displacement current density in the medium is given as
,,
ort
: # :,fr :,&l*P9],u: : *l-r"ftsin(22rfi)l
vscos2nft)
rrr_2rfeVo l"dl-- -T
Therefore, the ratio of amplitudes of .conduction current and displacement current in the medium is
v') lba lr_ V4 -- @lQ"kv')
lr,l 't'' : n x :2.7
sol.
6.2.{3
-
1
2rIep 1
1t.e x 10) x (54 x 8.85 x
10-1'?)
x
0.77
Correct answer is 8. Let the test charge be q coulomb So the force presence of experienced by the test cha,rge in the presence of magnetic field is
F:
q(a
x 4)_
and the force experienced c6n b4written in terms of the electric field intensity as
F:QE Where .6 is field viewed by observer moving with test charge.
Putting it in Eq. (i)
qE:q(uxB) E:(upa6)x(2a,) where ar is angular velocity and p is radius of circular loop.
:(2)(2)(2)a, :8a,Y lm
6.2,14
Correct answer is -0.35 . As shown in frgure the bar is sliding away from origin' Now when the bar is located at a distance dc from the voltmeter, then, the vector area of the loop formed by rail and the bar is 45 : (20 x 10-'z) (d*) o"
total magnetic flux passing through, the loop is
So, the
Page 409
o:In.as : ['
Chap 6
o") (20
tat
time Varyig Fields and
x ro-2 dra")
Marwell Equations
_ r.o[(r + o.a4]' 3
Therefore, the induced e.m.f. in the loop is given as
V^, 7"-r
:-#:-+
x 3(r+ o.4ff x (t+r.2f)
:-
1.0[(0.4)+(0.4fF x [1+ (1.2)(0.4),] (t: 0.4 sec) 0.35 volt Since the voltmeter is connected in same manner as the direction of induced
:-
emf (determined by Lenz's law). So the voltmeter reading will be
:
V
F
6.2"15
Ve^r:
-
0.35
volt
Correct answer is -2J.4
Since the position of ba,r is give as
r: So for the position r:
t
+0.4f)
l2cm we have
0.12: t(t+0.4f) f : 0.1193 sec
of,
I
t(L
As calculated in previous question, the induced emf in the loop at a particular
time f is
I/"-,
:
_(1.6)[r +
F
So,
t
at l:0.1193sec, z"-r
r l
o.aff (1+ r.2f)
since the vortmeter
:-
1.6[(0.
1
1es)
+ o.1{-sr.{tf f [r + (1.2)(0. 1 1e3f ]
t,l;l;1?3i1;#:lXJ*
the direction or induced
emf as determined by Lenz's law. Therefore, the voltmeter reading at r: 12 cm will be V : V^r:- 23.4 mvolt
;
l
6,2.15
Correct answer is +600. Given the magnetic field intensity in the medium is Ir = cos(lolot- br)a" Af m Now from the Maxwell,s equation, we have
:# # :-ffi":-
V x.EI
or,
bsin(10,0t- br)o,
O=
since no
where C is aconstant. I-bsin(1010r -br)dt+ C D.c. field is present in the medium so, we get c 0 and therefore, --
o:
c /mz fmcos(lo'or and the electric field intensity in the med.ium is given
E:2 r :7 :
br) an
012 xjfrFl'rdcos(101o'
-
as
br)a,o
@:O.r2nF/m)
Again From the Maxwell's equation
Page 410
Chap 6
Y
Tine Varying Fields and
or,
Maxwell Equatirons
xE:-# #:-V x[#"ot1ro"'-br)a,f :-$sin(r
So, the magnetic flux density
"
o'o'-
bn)a,
in the medium is
:- I #sin(1010' :
br)a,itt
C#lomcos(10'or
- bt)a"
We can also determine the value of magnetic flux density as
(U :
B:FH : (3 x 10-5)cos(101ot- br)a,
(2)
Comparing the results of equation (1) and (2) we get,
, ,b' 10'".^:
(1.2)
x
3
x
10-5
b2:3.6x105 b
6.2.{?
:*
600rad/m
Correct answer is 54.4L4 . Given the electric field in time domain
as
.E :
Comparing
it
5sin(1ozrg)cos(6n x Los - br)a, general equation for electric field intensity given as with the
B : Eocos(ut- pr)a" r.r:6zrX10e
We get,
Now in phasor form, the electric field intensity is
aA: 5sin(10r'g)e-jb a, Flom Maxwellfuqrrhtion we get the magnetic field intensity H"" ----!-*(o "1u114
(1) as
t B"):*l*",-* hl**-*"1
:--r^luo""os(102'y)e-'o'o,+ j5bsin(10zry)or]"-t* Again from Maxwell's equation we have the electric field intensity
n,:fi6(v
x rr,)
:hl*_+\""
as
:6ft4fnb) (- ib) sin( tory) s-tb' + (50n) (10zr)sin (10 rv) e-ib'la"
:
ffiltt2+
Soo,f]sin rorye-ib'a,
Comparing this rezult with equation (1) we get
-J--tsu'+ 5ool)' : b dPreo' b2 + tOO# : uj p.eo or, b2+toof :(62-x roef x b2
+t00# :400tf b2:300#
d-T
(w:6r x 10e,'/a,r':+)
b:-F ^/S}}nrad/m lDl:/300 r:54.474radfm
So, EL 6,2..!8
Correct answer is 7. Let the point change located at origin be of the page through the closed triangurar
page 4l
Time Varying Fields and Maxwell Equations
e and the current 1 is flowing out iath as shown in the figure.
(0, 2, 0)
As the current 1 flows away from the point charge along the wire, the net charge at origin will change with increasing time and given as
#=-,
so the electric field intensity will also vary through the surface and for the varying field circulation of magnetic field intensity around the triangular loop is defined as
f H . dt :[kfa*\U"].". I
where [1"[".- is the actua( flor/of ctra.ge car]ed encrosed conduction current and [16[," is the current due to the varying field called enclosed displacement current which is given as fIoL,"
: #[a,o.
d,s
: $f o . as
(1)
From symmetry the totar electric flux passing through the triangular surface is
ln.as:$
so'
fIoL,"
whereas
: #(g):a#:-{ LI"L,":
(from equation (1))
I
so' the net circulation of the magnetic field intensity around the
triangular loop is
f"n'
(1: 8 A)
Correct answer is 21.33 . As calculated in previous question the maximum induced voltage rotating loop is given as
lVtl:30*
From the given data, we have Bo
S
:0.25 Wb/m,
:64cm':
closed
il:f16f,,"+fl"L,"
:-{+r:$1s):za t 6.2.'rc
64
x
I
Chap 6
10-a m2
in the
Page 412
Chap 6 Time Varying Fields and Maxwell Equations
and is covered)
, -.60 x Ztr :377 rad'f sec (In one revolution 2zr radian I ".
So, the r.m.s. value of the induced voltage is
[/"-r]^^,
:
:
h"r*
ftlY"-r | 1,o.zsx64x1o-4x377)
- rE\
:0.4265 Since the loop has 50 turns so net induced voltage will be 50 times the calculated value. l.e.
:50 x (0.4265) [v"-L-" L fi,n,s
:
21.33 volt ****'F******
o
soLUTtONs 6,3
'
page 413
Chap6
Tine l&rying Fields and
'
Maxwell Equatious
tol.
6,3.1
Option (D) is correct.
tol.
6,3"2
Option (B) is correct. The line integral of magnetic fierd intensity along a crosed loop is equal to the current enclosed by it. .f r.e.
lH.dl:1"n"
so, for the'constant current, magnetic field intensity will be constant i.e. magnetostatic field is caused by steady currents.
Ilr,
s.r.r
r
Option (A) is correct. Flom Faraday's law the electric field intensity in a time varying field \
defined
h
as
Vx
F
-E
F.
where
I;# \)
field.
:
B
is
is magnetic flux density in the EM
and since the magnetied'x density is equal to the curr of magnetic vector potential l.e. So,
lt p
B:Y xA
putting it in equation (1), we get
y xE:_$tv r,a)
or
V X.E -v - Y ,\-l
Therefore, lol
6.3.4
E
:-#
0 r\ \-aT^)
Option (B) is correct. Since total magnetic flux throrrgh a surface S is defined
o:In.as
Flom Maxwell's equation
it
zeto
'. I
as
is known that curl of magnetic flux density is
V.B:0
fn
. as :
If,
.
B)d,u:o
(Stokes Theorem)
Thus, net outwards flux will be zi:ro for a closed surface. sol.
6.3.5
Option (B) is correct. trlom the integral form of Faraday's law we have the relation between the electric field intensity and net magnetic flux through a closed loop as
----\
da 9E'. dt:-ff r
since electric field intensity.is 'ero (E:0) inside the conducting loop. so, the rate of cha^nge in net magneiic fl,x through the closed loop is t
d@
Page 414
-v a -n
Chap 6
i.e. @ is constant and doesn't vary with time.
Time Varying Fields and Maxwell Equations
s$L
{!"3,S
Option (C) is correct. A superconductor material carries zero magnetic field and zero electric inside it. B :0 and .E: 0 l.e. Now from Ampere-Maxwell equation we have the relation between magnetic flux density and electric field intensity as
y xB:pallp"€rT So,
J
:0
(B:0, E:
since the net current density inside the superconductor is zero so all current must be confined at the surface of the wire.
sol 6.3.? Option (C) is correct. According to Lenz's law the induced current 1 in a loop flows produce a magnetic field that opposes the change in B(t)'
such as
Now the configuration shown in option (A) and (B) for increasing flux B;, the change in flux is in same direction to Br as well as the current flowing in the loop produces magnetic field in the same direction so it not follow the Lenz's law.
For the configuration shown in option (D)' * the flux Ba is well with time so the change in flux is in opposite direction lo 86 field produces magnetic the loop the current 1 flowing in the direction so it also does not follow the Lenz's law. For the configtrration shown in option (C), the flux density 116 is with time so the change in flux is in opposite direction to 116 but the c 1 flowing in the loop produces magnetic field in the same direction to (opposite to the direction of change in flux density). Therefore this is correct configuration. $oL
s"*.s
Option (C) is correct. Induced emf in a conducting loop is given by do where @ is total magnetic flux V" t:-Zl through the loop. Since, the magnetic field is non-uniform so the change in flux will be it and the induced emf due to it is called transformer emf.
by
Again the field is in a, direction and the loop is rotating about z-axis s flux through the loop will also vary due to the motion of the loop. Thb causes the emf which is called motion emf. Thus, total induced voltage ir the rotating loop is caused by the combination of both the transformer and motion emf.
$0L s,3.$
Option (B) is correct.
g$L
6"3.10
Option (C) is correct.
soL
6"3"1t
Option (B) is correct.
_-__r_
r I
I
I p
sor o.r.rz
Option (D) is correct.
-. ..'.',
Option (A) is correct.
ror. s.r.rc
I ..r.,, [ -. ror- e.s.rs I ror- o.e.rz I
page 4tb Chap 6 Time Varying Fields and Maxwell Equations
Option (A) is correct. Option (C) is correct. Option (B) is correct. Option (B) is correct. *>t********x
r $
r
i
t
a
r I a
t F
.G.
Page 416 Chap 6 Time Varytng Fiel& and : Maxwell Equatious
sol. 5.4.{
Option (C) is correct. Given, the magnetic flux density in air
B:
as
h(ataa-ffi")
'(1)
Now, we tra,nsform the expression in cylindrical system, substituting
and
r':rcosf and g:rsind oo : cosbzn- sin/46 o'v : sindo"* cos$oa
So, we get
B:
Boa6
Therefote, the.magnetic field intensity in air is given
as
Boo, ,which is constant = L: Pn I'to So, the current density of the field is
H
J:Yx.E[:0 30L 6.4,2
(since I1 is consta.r*)
Option (D) is correct. Maxwell equations for an EM wave is given
as
V'B:0 Y ' E:fu€ Y
xD:-ry ot
v x .Er :#*, So, for static electric magnetic fields
V.B:0 V.E:pu/€ V
YxH:J gol.
6.4.3
(#:rl
x.E:0
(#:o
i
Option (D) is correct.
''AD vxrr : t*at
) sol-
6.4.4
f[to " H1.
as --
Il,tr.oz;.
Maxwell Eq
lntegral
as
fH.dr:ll1r*#).*
Stokes
Option (C) is correct. Flom Maxwells equations we have
v Thus,
Vx
xfir:#*,
.EI has
unit of current density
J
(i.e',
Al^')
sol
6.4.5
Option (A) is correct. This equation is based on Ampereis t#",is from Ampere,s circuitar law we
have
$n. Jt
or,
as
Applying Stoke,s theorem we get'
/tv
I
><.rr).ds
V Then,
it
=lt.as
x.EI=J
is modified using'continuity equation
Vx.EI:J+#
h."" . tol-
as
Option (D) is correct. when a moving circuit is put in a time varying magnetic field induced nt" tyTimponents. one due to time variation of magnetic flux densityemf B and ot|€f\ue to the motion of circuit in the field.
()
i
6.4"?
Option (e) is correct. Fbom maxwell equation we have
Vx-6t:r*#
The term sol.
6.4"8
S
a"no", displacement burrent.
Option (C) is correct. Emf induced in a loop carrying a time varying magnetic flux
U^l:-# g
__
9
: -,\l
@
is defined as
d
/_1 r*s\ a\3"'i
at time, f : 3s, we have e =_)(BI ,\ :- lWb/s2 sol.
6,4.9
4lZ
Chap 6 Time Varyhg Fields and Maxwell Equations
a:/encreed
f,n. u=J"r.
Page
Option (B) is correct. According to Lenz's law the induced emf (or induced current) in a roop flows such as to produce a magnetic field that opposed the change in B. The direction of the magnetic field produced by the current is determined by right hand rule. Now, in figure (1), p directed upwa.rded increases with time where as the field produced by cufrent -I is downward so, it obey's the Lenz's law. In figure (2), B directed upward is decreasing with time whereas the field produced by current 1 is downwards (i.e. additive to the change in B) so, it doesn't obey Lenz,s law. In figure (3), B directed upward is decreasing with time where as current -I produces the field directed upwards (i.e. opposite to the change in B) so, it also obeys Lenz's law.
In figure (4), B directed upwa,rd is increasing with time whereas current 1 produces field directed upward (i.e. additive to the change in B) so, it
doesn't obey Lenz's law. Thus, the configulafion 1 and 3 are correct.
Page 41E
Chap 6 Time Varying I'ields and Maxwell Equations
sol
6.4.10
Option (C) is correct. Faraday's law states that for time varying field,
vxE:-AP At Since, the curl of gradient of a scalar function is always zero
v x(vv):o
t.e.
so, the expression for the field,
E
E:- v I/ must include some other terms b
--O r-#
i.e. A is true but R is false'
6"4.{t
Option (B) is correct. Faraday develops the concept of time varying electric field producing e magnetic field. The law he gave related to the theory is known as Faradayb law.
d
a.4..2
Option (D) is correct. Given, the area of loop
,9:5m2 Rate of change of flux density,
ry:zwb/m2lS dt So, the emf in the loop is
. 6.4"{3
%*r
:-$[a'
as
: (5x-2) :-
10
v
Option (D) is correct. The modified Maxwell's differential equation.
V x,EI
: r*#
This equation is derived from Ampere's circuital law which is given
as
f H' dI:1"," I
f t, "'H\'
ds
: lns
YxH:J
$oL
6.4."14
Option (B) is correct. Electric potential of an isolated sphere is defined as C : treoa The Maxwell's equation in phasor form is written as
(free
V x ^EI: jteE*oE: jueE*J
So
A and R both
are true individually
but R is not the correct
of A.
sol.
6.4.1s
Option (A) is correct. a coil is placed in a time varying magnetic field then the e.m.f. will in coil. So here in both the coil e.m.f. will be induced.
If
tol
6.{.16
30L 6.4.{?
Option (B) is correct. Both the statements are individually correct,but R is not explanation of A. Option (B) is correct.
Ampere'slaw
V
Faraday'law
Vx.E:#
x,tI:J+ODdt
(a-3)
(b-a) (c'1) (d-2)
Y -D:pu V .J :-U
Current continuity 6.11.18
dt
Option (B) is correct. since, the magnetic field perpendicular to the plane of the ring is decreasing with time so, according to Faraday's law emf induced in both the ring is v"-r
: _ftf a.
as
Thereforeqmf will be induced in both the rings. 6,4.tS
/\
Option ($) i$ correct. The Basikea of radiation is given by the two Maxwell,s equation
:# y xE:-#
V x.EI
; tol. 6.4.20
Option (B) is correct. The correct Maxwell's equation are
Vx-EI:t*# V x.O :-# tol
6.4.2{
V . D:
p
V 'B:0
Option (B) is correct. In List I
a.
f
fB.dS:O
The surface integial of magnetic flux density over the closed surface is zero or in other words, net outwa,rd magnetic flux through any closed surface is zero. (a - 4)
b.
fn.as:Ip,du
Total outward electric flux through any closed surface is equal to the charge enclosed in the region. (b-+ 3)
c
/ fn.m:-l#ot of
i.e. The line integral the electric field-intensity around a closed path is equal to the surface integral of the time derivative of magnetic flux density
(c-2)
d
Chap 6 Time Varying fields and Maxwell Equations
Gauss law
30L
Page 419
f n.ot:l(#+r)aa
i.e. The line integral of magnetic field intensity around a closed path is equal
to the surface integral of sum of the current density and time derivative of electric flux density. (d-+ 1)
soL
Page 420
6"4.tr2
Chap 6 Tiqe Varyiry Fields and
Option (D) is correct The continuity equation is given
as
Y . J:--p,
Maxwell Equatibns
i.e. so!-
6.4.23
it
(J)
relates current density
and charge density
p,,.
Option (C) is correct. Given Maxwell's equation is
' AD Vx.EI - ,1,+ dt For free space, conductivity, o:0 J":6fi:g
and
so,
Therefore, we have the generalized eqttation
V x.EI :A^D -0t sc|.
6"4.24
d
Option (A) is correct. Given the magnetic field intensity,
H:3a,*7yan*2ra, So from Ampere's circuital law we have
J:YxH la" a,. a.l t' ooo
I d"
: $oL 6,4.?5
l3
dp
.,.
7y 2ri
a,(0)
-
a,(2
-
0)+ a,(0)
:-20,
Option (A) is correct. The emf in the loop will be induced due to motion of the loop as well as t variation in magnetic field given as
%*r:-fffi-as+t'@xe)at So, the frequencies for the induced e.m.f. in the loop is
80L
6"4,2S
c,.,r
and
c,r.
Option (B) is correct.
p:
Q(E + a
x B) is Lorentz fbrce equatiorr.
sol.
6.4.27
Option (A) is correct. All of the given expressions are Nlaxweil's equation.
sol.
6"4.28
Option (B) is correct. Poission's equation for an electric field is given as
v, V
:-*
I
where, I/ is the electric pltential at thc point and p,, is the volume charge' density in the region. So, for po: Q we get.
Y2V :0 Which is Laplacian equation. $oL
6.4.2e
i
I Option (A) is correct. I The direction of magnetic flux due to the current 'i' in the conductor hl determined by right hand rule. So, we get the flux through ,4 is pointing i"t"l the paper while the flux through B is pointing out of the paper. According to Lenz's law the induced e.m.f. opposes the flux that causes iL !
1
1
so again by using right hand rule we get the direction of induced e.m.f. anticlockwise in A and clockwise in B.
Page 421
Chap 6
$.,1"*s Option (D) i_s correct. y2 ! :_ p,J
Time Vap 'ymg Fields atrd Mai:s,ell Equations
This is the wave equation for static electromagnetic field. i.e. It is not \,[axwell's equation. 6"4"3.t
Option (B) is correct. Continuity equation
VxJ
:_q' dt
Ampere's law
Vx,EI
:
(b-1)
+o-P dt
*At J _aD
Displacement current
Faradav'k* "/
VX,E
t/ Option (Bfn correct. l
Induced emf in a coil of
t/ Yt|t
I
(a-+ 4)
lf
(c-2)
AB dt
(d-3)
turns is defined
as
nrdib --- r\ (l,l
where @ is flux linking the coil. So, we get
V^r
:*
roo#J(3
-
2q
- 1oo(31 - 2) - 100(3(2F - 2):-
1000
mv
(at
t:
2
s)
-1V 8.4"3n Option (B) is correct. A static electric fiekl in a charge free region is defined
and
V ..8:0 V xE:0
as
(a-a)
A static electric field in a charged region have
v'E:Q:t+0
and
(b -+ 2)
Vx.E:O
A steady magnctic field in a crrrrent carrying conductor have
V.B:O
VxB:traJ*0 A time varying electric fiekl in
field
have
-'
a charged medium
VxE:-ry+0 dt v.E:fu+0 so*
6.4.$d
(c -+ 1) with time varying magnetic
(d'+
3)
Option (C) is correct.
, :-'lLtdt
t.'
It
is Faraday's law that states that the change in flux through any loop
induces e.m.f. in the loop.
Page 422
$oL
6"4"35
Chap 6
Option (B) is correct. From stokes theorem, we have
Time Varying Fields and Maxwell Equations
r
[{r"E)'ds:fn'at Given, the Maxwell's equation
vxE:_(aBl
aq
Puttlpg this expression in equation (1) we get,
f n.m:-ftf"a.as sol
6,4.3S
3(}L 6,4.37
Option (D) is correct. Since, the flux linking through both the coil is varying with time so, induced in both the coils. Since, the loop 2 is split so, no current flows in it and so joule heati not occur in coil 2 while the joule heating occurs in closed loop 1 as flows in it. Therefore, only statement 2 is correct. Option (C) is correct. The electric field intensity is
E
: Eoetot
where -Es is indePendent
So, from Maxwell's equation we have
V x.EI
: t*# :
30L
6"4.38
sot- 6,4,39
oE
*
l
e(1't)Eoe''
:
oE
I
jaeE
Option (C) is correct. Equation (1) and (3) are not the Maxwell's equation. Option (A) is correct. Flom the Maxwell's equation for a static field (DC) we have
VxB:pnJ v x(v xA):pal v (v . A)- v'A : p\J V
For static field (DC),
'A:0
Y2A :- FnJ therefore we have, So, both A and R are true and R is correct explanation of A' $oL
6"4.40
Option (A) is correct. For a static field, Maxwells equation is defined
as
YxH--J and since divergence of the curl is zero
v.(vx,EI) :s
f.e.
V'J:0 but in the time varying field, from continuity equation (conservation of charges)
v . J :-o!, il +o So, an additional term is included in the Maxwell's equation. t.e.
V x.EI
: r*#
\
),/ where :
ff "
displacement current density which is a necessary term.
Chap 6
Therefore A and R both are true and R is correct elphnation of A.
tol
6,4,41
Option (C) is correct. since, the circular loop is rotating about the y-axis as a diameter and the flux lines is directed in a, direction. so, due to rotation magnetic flux changes and as the flux density is function of time so, the magnetic flux also varies w.r.t time and therefore the induced e.m.f. in the loop is due to a combination of transformer and motional e.m.f. both.
tol
6"4.d2
Option (A) is correct. For any loop to have an induced e.m.f., magnetic flux lines must link with the coil. observing all the given figures we conclude that loop e and G carries the flux lines through it and so both the loop will have an induced e.m.f. Option (C) is correct. Gauss's law
Y.D:p :1.+0P^
Ampere's law
Vx
Faraday's law
YxE:-ABdt P:ExH
Poynting vector
.EI
*,1.x****+***
dt\
Page'C2B
(a-1) (b-5) (c-2) (d-3)
Time Varying Fields and Maxwell Equations
n
CHAPYER 7 ELEGTROMAG
a ITTRODUCTION
In this chapter, we
concentrate
on the fundamental properties
of
electromagnetic waves as they propagate in various types of media, and their interaction with matter. Following are the topics included in the chapter:
'
o r . o o r r o o
Definition of EM waves
Wave equations and solutions
Propagation characteristics Waves
in conductors and dielectrics
Poynting vector and average power consideration in EM wave Polarisation: Linear, Circular, and elliptical polarization Normal and oblique incidence of EM waves Reflection and transmission coeffrcients Brewster angle
ELECTROTAGNETIC WAVES In general, wave is a carrier of energy or information and is a function of time as well as space. As far as we are concerned, a wave means Electromagnetic
wave (or simply EM wave). some examples of EM waves are radio, radar bearns and TV signals. Maxwell predicated the existence of EM waves and established it through his weil-known Maxwell's equations.
General Wave Equation for Electromagnetic Waves Consider a linear, homogeneous and isotropic medium. The three-dimensional equation or Helmholtz equationin an absorbing medium or lossy ".l"lor\uu dielectric medium is defined as
p,#- po#:o v2H- p,#- po#:o v2D-
...(7.1) ...(7.2)
Now, we will consider the modifications in the wave equations for different cases.
|2.2
Wave Equation for Perfect Dielectric Medium In a perfect dielectric medium, the conductivity is zero, i.e. it in equations (7.1) and (7.2) we get
y2E
: rt#
o:0.
substituting
1{
ETI C WAVES
-T
Page 426
A2 II YzH== p€_aT
and
Chap 7
These are the wave equations for perfect dielectric medium.
Electromagoetic Waves
7.2.3
Wave Equation for Free Space We have already studied the parameters of free space which are as
1. 2. 3.
Relative permittivityt Relative permeability,
€,:I F,:7
e:es) (i.e. p: trto)
(i.e.
Conductivity, o :0 Substituting these parameters in equations (7.1) and (7.2) we get A2E
== Lt o€o-Af A2 II YzH= poeo-AT Y2E
and
=
These are the wave equations for free space
7.2.4
Wave Equation for Time-Harmonic Fields
The standard form of wave equations for time harmonic fields (in form) are defined as
vtEr- tos:o v'.[lr- "f Hs:0
and
where
7 is a complex
consta.nt called the propagation constant.
Propagation Constant For a medium with permittivity e, permeability tt, and conductivity a. propagation constant is given by
1:{fuilo+w)
Propagation constant is expressed
in per meter (m-1). It can be
defined as
''l
:
o+ jP
where o is the attenuation constant, and p is the phase constant as describd below.
7.3
1.
Attenuation constant (o): The real part of the propagation constad is defined as the attenuation constant (a). The attenuation constax defines the rate at which the fields of the wave are attenuated as th wave propagates. Its unit is Neper per metre.
2.
Phase constant (B): The imaginary part of the propagation constant i , defined as the phase consta"tt (B). The phase constanf,defines the rat at which the phase changes as the wave propagates. Its unit is radia per metre.
UNIFORM PLANE WAVES
The uniform plane wave has two important words in it; uniform and planr L. Plane: The term plane means that the electric and magnetic field vecto: both lie in a plane and all such planes are parallel. In addition, tl phase of the wave is constant over a plane.
2. '
Uniform: The word uniform means, the amplitude and phase of vecto E and -E[ are constant over the planes.
In Figure 7.1, we have chosen the eiectric field to be in r-direction and the magnetic {ield will turn out to be orthogonal to r-direction and so is directed in the g-direction. The field vectors ,E and .E[ are in the r-g plane flane waves. The word uniform, also means that the field vectorr of r and g in these planes. Thus, the vector fields E and oly be functions of. z and time, i.e. E : E,(z,t) a, ...(7.3) H : Hr(z,t) au ...(7.4) le associated with these field vectors propagates in the z-directions
hopagation of a Uniform Plane Wave the time factor e,t into above equations, we get the instantaneous the field components as E (2, t)
:
Rnl(E o e-'" e'') a,f
H (2, t)
:
Re[(flo e-'"
"*')
ou)
E6 and flo of the field components are related as
Hn:fu "n rhere
r7 is i,ntrinsi,c i,mpedanced.the medium. Following axe some important, points related to uniform plane waves.
r,.t:il*,:trtq$${ri
.1;r:rl
!i:,,t,t;:ir:.
'bfe''?gM
Page 427 Chap ? Electromagnetic Waves
Page 42E
Chap 7 Electromagaetic Wave
7.4
WAVE PROPAGAfION IN LOSSY DIELEGTRICS
'
7.4.1
Consider a uniform plane wave travelling in a medium where the conductivity is nonzero, (o * 0), that is, a lossy medium. Let us study the various characteristics of the plane waves in the medium.
Propagation Constant in Lossy Dielectrics We have already defined the propagation constant as
and
l:61+ip
Solving the above two equations, we obtain the attenuation constant (c) and phase constant p as ...(7.5) and
B
: r\l +
...(7.6)
Thus, c and p a,re both nonzero for the medium and hence, thepropagation constant, 7 includes both imaginary and real parts.
7.4.2
Solution of Uniform Plane Wave Equations in Lossy Dielectrics We have the field components of the wave propagating along o, direction as
:
Re[(Eo e-'"
H(z,t):
Re[(I/oe-'"
E (2, t)
a.nd
o'l
...(7.7)
"'')ql
...(7.8)
"to')
: a+ j0 in equation (7.7), we get E(z,t) - Re[Eee-"' eiPt*Pz)ar7 \Q,t) : Eoe-o"cos(ut- Bz)a,
Substituting 1
or
Simila"rly, the magnetic fie1d component is obtained as
: II(z,t) : H(z,t)
or
Re[I/oe-"'eiPt-8") asl Hoe-o"cos(wt- Bz)a,
Here, Ese-o' is the instantaneous amplitude of electric field, f/6 e-"' is the insta.ntaneous amplitude oilnagnetic field and (4,'t Bz) isthe instantaneous phase angle of the wave.
-
Velocity of Wave Propagation in Lossy Dielectrics
'
.
The velocity at which a fixed point on the wave trlielling is called velocity of wave propagation. Let us consider a fixed point P on the wave ,4 sin (u.r t gz) -
'
'
i.e.
wt- Bz:
constant
As the wave advances with time, point P moves along * a, direction. Differentiating both sides with respect to time, we get
,_g#:o dza E:
B-
,o:ff
...(7.e)
where uo gives the velocity of fixed point
P on the
wave which is called the
velocity of wave propagation
Wavelength of Propagating Wave The wavelength of the propagating wave can be given as ), : upT where uo is the velocity of wave propagation, and ? is the time period of the wave. So, substituting equation (7.9) we get
or 1.4.5 Intrinsic lrnpedance Consider
^:(fr), ^:4p
:(4f),
a uniform plane wave travelling in * a" direction with
field
components
E"
and
- Ese-l" a, H": Hoe-l'av
...(7.10) ..(7.11)
The ratio EolHo is called intrinsic impedance a,nd dpnoted by 4, i.e. ,1
E"
-Ho
Solving equations (7.10) and (7.11) using Maxwell's equation, we get
or
n-i'tt:-L ' 't J jwp(" + rul€)
- | t--Fu Gifr,
n =
"'(7'12)
Thus, the intrinsic impedance is a complex quantity. Its unit is ohm (O). can be also expressed as
,t :lrtl/!_;ln!4. with
and
.4.6
,/nE l'l:ldFr tan2lr:L
Loss Thngent For a time-harmonic field, we have the Maxwell's equation, V x II" :(o* jue)E"
It
.
Page 429
Chap 7 Electromagnetic Waves
Page 430
Chap 7 Electromagnetic \ilavcs
where
t"
=
i,rl, - t#1",
-
ite"E"
is the complex permittivity given by,
,
,f e":elI-il d€] r e":(€'-jr")
where (e') is the real part and (e") is the imaginary part of the permittivity. The ratio of. (e") to (e') is the ratio of the magnitude of conduction current density to the magnitude of the displacement c density. This ratio is defined as the loss tangent or loss angle of the medi i.e.
o ld:|,: | €' ' a€-,loE,l,lweE"l
lJ"ooao"tionl
:tang
lJo*l,*"^.o, where tand is the loss tangent and I is the loss angle. The Ioss tangent a measure of how lossy a medium is. For a lossy dielectric, loss tangent I
b
the order of unity.
7.5
WAVE PROPAGATION IN LOSSLESS DIELEGTRICS lossless dielectric is one in which o << u€. So, for lossless dielectrics medium parameters may be considered as
A
o=
0,
t:
eg€7
F
: Folt,
The propagation characteristics of the lossless dielectrics are obtained follows:
7.5.1 Attenuation Constant Flom equation (7.5), we have the attenuation constant
+l/'.#-'l
Substituting o: 0 in the above equation we get the attenuation constant a lossless dielectric medium as
o
7.5.\
:0
n
(7.13) ...(7.13) ...
Phase Constant
Flom equation (7.6), we have the phase constant
+hE;#.'l
Vl
$
Substituting o: 0 in the above equation we get the phase constant in a lossless dielectric medium as
g
7.5.3
: rr/G
...(7.14)
PropagationConstant Propagation constant in a medium is defined by
'y:a+ jp Substituting a and B from equations (7.13) and (7.14) , we get the propagatim constant in the lossless dielectric medium as
j : ja{G
Velocity of Wave Propagation
Page 431
Chap 7
The velocity of wave propagation in a medium is given by
Electromagnetic Waves
oo:fr Substituting 73 from equation (7 .L4), we get the velocity of wave propagation in the lossless dielectric medium as wp
a1 -
@"/
lt€
t/
U,e
?.5.5 Intrinsiclrnpedance Fbom equation (7.12), we have the intrinsic impedance
n: 't t-@(o + jue) ^/
Substituting o: 0 in the above equation we get the intrinsic impedance in a lossless dielectric medium, as
,t-l ttr e
7.5.6
Field Components of Uniform Plane W'ave in Lossless Dielectric Since, the field components of uniform plane wave in ,lossy dielectric is giverr
by
and
E(z,t) H(z,t)
: :
Eoe-o" cos(wt
-
Bz)a"
Hoe-o"cos(ut- 0z)A So, by substituting a : 0 in above equaiions, we get the
in lossless dielectric
field. components
as
: g : B
E6cos(ut- Bz)a, t/6cos(wt_ By)a,
where Eo and Ilo a,re related as
Eo
-^- fO fr:n:\/E Thus, we see that for a lossless dielectric medium, the wave propagates without any attenuation, and the electric and magnetic fields are in phase with each other.
7.6
WAVE pROpAGATtOt{
tl{
PERFECT qONDUCTORS
A
perfect, or good conductor, is bne in which o>> a€. So, for good conductors the medium parameters may be considered as d: oo r c-€s' F:FolL" The propagation characteristics of good conductors are obtained as follows :
7.6.1 Attenuation Constant Ftom equation (7.5), we have the attenuation constant
) (r€) so we get toFa \/ --r-
Here, the condition is, o
...(7.15)
Page 432
7.6.2
Phase Constant
Chap 7
FYom equation (7.6), we have the phase.constant
Electromagnetic Waves
o
Again, applying in a
phase constant
2 ae in above
equation, we
as
,.j (7 16il
7.6.3
PropagationConstant Propagation constant in a medium is dg{ined
jp
'y:a+ . 7.6.4
as
Substituting o and B from equations (7.15) and (7.16), we get propagation constant in the good conductors as
t{ i
"t:'/ry+l/ry
I
Velocity of Wave Propagation The velocity of wave propagation in a medium is given by
,,:Bu Substituting
B from equation (7.16), we get the velocity of wan
propagation in a good conductor
",-a-@ "r - ,S .;i'.e*;-,
as
,,1 p,o
7.6.5 Intrinsiclmpedance Flom equation (7.12), we have the intrinsic impedance
Since, o
2 we,i.".T (
1. So, we ca,n
write
j
t+ffct Hen\tre
intrinsic_imp
_f
"OffiLn
n:\/
I a good conductor reduces is oUtainea
I
I
7.6.6 $kin Effect
I
When an AC current is applied to the conductors, it shows skin etrect- { is the tendency of an alternating electric current to distribute itself wttl{ a conduc\or in suc\ a way \hat \tre cunen\ densi\y is \\re \argest o"o q surface of the conductor, while decreasing at greater depths. Mathematiclt skin effect is expressed by skin depth' Skin
I I
Depth
is defined as the depth in which the magnitude of the wave is attenu:{ to e-1(- 37%) of its original value. The electric field of a propagating { in a good conductor in a, direction is given by I
It
E
-
Ese-o'cos(ut-
Pz)a"
I
l
where Eoe-o" is the instantaneous amplitude of the field. So, after travelling a distance z:6, the field reduces to
E
: Eoe-"6
...(7.LT)
Now, from definition of skin depth, the electric field of uniform plane wave after traveling a distance 6 (skin depth) reduces to e-1 of its original vahre. i.e.
:
E
...(7.18)
Eoe-r
So, from equations (7.17) and (7.18) we get
c6:1 u:a 't
or
Substituting equations (7.15), we get
^-1-- ! L2uu'o
Thus, in a good conductor, the depth of penetration decreases with increasing frequency (a,).
7.7
WAVE PROPAGATION
11{
FREE SPACE
In free space, we have the parameters
O:0,
F:
T}1e propagation characteristics
7.7.1
FO
in free space are obtained as follows.
Attenrration Constant FYom equation (7.5), we have the attenuation constant
in a medium ...(7.1e)
Substituting o: 0 in the above equation we get the attenuation constant in a lossless dielectric medium as
a:0
7.7.2
Phase Constant From equation (7.6), we have the phase constant in a medium ...(7.20)
Substituting o:0, t : 6o and phase constant in free sPace as
g : u,/
F: llo in the above
equation we get the
ttoeo
7.7"3 Propagation Constant Propagation constant in a medium is defined by 'y
:
o+
i[3
substituting a and. p from equations (7.19) and (7.20), we get the propagation constant in the free sPace as .y : ju^/_p*q
Page 433
Chap 7 Electromagnetic Waves
Page 434
Chap 7
7.7.4
Electromagnetic'tilaves
Velocity of Wave propagation The velocity of wave propagation is given by
,o:fr Substituting p from equation (2.20),we get the velocity of wave in the free space as Q1
__-.::
.y
a^/
7.7.5 Intrinsic
:'
poeo J poeo
108m/s
Impedance
Flom equation (7.I2), we have the intrinsic impedance
n: E@_
substituring Jf'r): po io the above equation, we set 1: 0, in,:r:," intrinsic impedance free space as
TF; ,10:JTi:7ZOn=377Q
7'7'6
Field components of Uniform plane wave in Fbee space we have field components of uniform plane wave in lossy dielectric as E (z.t) : Eoe-o' cos(at 7z)a, and II(z,t) : Hoe-o,cos(ut_ Az\a, free space attenuation ctnst"nt,'a":0. so, the fierd components l]Te^lt reduces to
6: .EI :
Eocos(ut_ pz)a,, flocos(wt_ py)a, These are the field components of ,roifor* prane wave propagating in free space) where E6 and IIs are related as Eo : t4n H-o
_^rlo:
4
'/ the Thus, we see that in free space) wave propagates without any attenuation and the electric and magnetic fields it inJr" with each other. 7.8
"r"
POWER COHSIDERATION
II{ ELECTROMAGNETIC WAVES
One of the important characteristics of electromagnetic wave is that it can transfer energ"y from one point to another. The law of conservation of electromagn etic fi:W is mathematicallv expressed
ffTiltr,#:H*** 7.8.1
Poynting's Theorem According to Poynting's.theolem, at any point in an erectrornagnetic field. the power per unit area is described by a ,recto. termed as poynting vector which is basically a curl of erectric field intensity vector and magnetic fierd intensity vector. i.e.
F:ExH The magnitude of Poynting vector (p) is the power flow per it points along the direction of wave propagation vector. Ils unit area, and unit is watt per
squared meter (W/m'). For time harmonic field,. the phasor of vector is defined as
P,: E, x II,*
Poynting
Page 435
...(7.2L)
Electromagneticwaves
Chap 7
where the asterisk (*) sign in the magnetic field vector represents the complex
conjugate of the vector. This is the instantaneous power in uniform plane wave also called the compler Poynti'ng uector.
'?.8.2
Average Power Flow in Uniform Plane Waves
To determine the time-average power flow in a uniform plane wave, integrate equation (7.21) over the time-period 7. i.e'
we
Poo:#[, r"0' po,:+l?t"xH,+)dt
or
This is equivalent to
g",:!n*lE, x f/"x] P,, is the time-average power density vector in a uniform plane wave which is expressed in Watt per squared meter (W/m2). The total time-average power passing through a surface ,S is given by
P.,:f P",'dS Let us generalize this expression for lossy and lossless dielectric media. Average Poynting Vector in Lossless Dielectric Medinrn
For a uniform plane wave propagating in a lossless medium in o" direction, we have the field components
E"
H""rl:
and
Eoe-iB'a'
E!"-io"
ou
Hence, the time-average Poynting vector in the field is
p^,:fRe[E,x H"*f 1
: $n"[ta" -p,a,) * (]r'o,a)l: fr",
Similarly, we can also obtain the average Poynting vector magnetic field as
D foo:
in terms
of
-!4t 2 a,
Average Poynting Vector in Lossy Dielectric Medium
Let us consider a uniform plane wave in a lossy dieleefrc medium. The phasor fields written in the vector form are E, : Eo€-o" e-i?" A, H" : Se-"2 So, the average
" l'tl "-ioz "-fr'^ vector in the uniform plane wave is
?*t*
P",:fRe(E, x II,*)
: ]n"[ta"-o" e-ie" a,)
" (ffiu"'"tu' "n,",)]
Page 436
: ffi"-r", a, "oslg,,)
Chap 7 Electromagnetic Waves
The average power in the \4rave contains the faetor e.'2"' . rvhich saJ's d the power is being dissipated in the merdirirn. as thc w-a\€ passes through
7.9
WAVE POLARIZATION
wave polarization is the electric fir,:lti vector orientation as a function time. at a fixecl point in space. rhe three important types of polari are (1) Linear, (2) Elliptical, a,nd (3) Circular polarization.
7.9.1
Linear Polarization
A wave is said to ber linea,rl.,, irolarised if the elcctric fielcl rernains along straight line as a function of tinre at, sorrre poirrt in the rnediurn. For the propagation irr * o.. directicln, the iinear polarized wave r,vould in have its electric field phasor exprcssr:d as B, : (86u,*
E,1t1ar)e.-
"'p-tit"
where E6 and E*r are constant arnplituries along z and g. As the e; arrd -components of eiectric field are in sa,me phase. so it is a lin,early pola waae' rn general, for tire rn'ave traveling betvi'een diff'erent media, li polarization may be of lbllowing two tl"pes:
1.
Parallel Polarisation:
lt i-\ defined as the
polarisatiorr
in which t
electric fieid of the rvave is pa,rallel to the pla,rre of incide'ce" par polarisation is a,iso called vertical polarisation,
2-
Perpendicular Polarisation: It is clefined as the polarisation irr whicl the electric field of t,he wave is yrer:pcndicrrlar to the plane of incidencePerpendicular polarisat,ion is alsr.r cralleti h.c,rlzontai polarisation.
7.9.2 Elliptical
Polarization
In elliptical polarized wave. the tip of the, electric fiel
E, : (E,s.a' * Eat{''
ar)r,:
rt:
,
i3:
where p is thc pilast t-liffert'rrcc bctw'ccri tlrc .r and y-comporjcnts of elect field rector such dhat 0 < r! < rl2.
7.9.3 Circular Polarization A circularly polarisecl-wave is oire in which the tip of the electric vector .E traces a c'irclt: a,s tinie virries, This is the special case of ellipti polarizerl wave. Here, the arnplitude, of :r and g-components of electric vector are equai. (i.e. 81, -- E,p : ^&'e) and ph,l-se clifference between them i x12 (i.e. 6:* nl2). Thuq, eiectric fielci phas.-rr of circrrlar polarizecl wave
in general form as E" : (Eno,,* Elte'r'r/2 11r)e "' ,-t"' In time .domain, we tlrav express thc electric fiekJ vector of
expressed
polarizcd we\/e
as
E(t) : Eneosurta,f
E6cos(r^,'l
+
nf 2)a,,
Circular polarization rnay be of the followirrg two types:
Chap 7
1. Right-hand Circular Polarization
When E, lealls E, by r 12 (i.e. tlt : * T 12), we obtain the r and 3t-components of electric fie.'ld vector as A, - Escoswt and E, : Eosin",rt The instantarleous angle that .6 rnakes with the z-axis is given by
d
g
: tan '{*\: \,
oL ,,
This indica,tes that electric field vector (E) rotates at a uniform rate with an angular velocity iu irr a counterclockwise direction as shown in F igure 7 .2(a). \\'hen tlie lingers of the right hand follow the direction of the rotatiorr of ,8, tlie thuuib points to the direction of propagation of the wave. Tlrus, it is a ri,qkt-harr'd ot Txtsitirte c'irctilurly polari,zed waue.
(") l:r,-t:rr','
:..i
(b)
Ilhistratbn of Circular Polariza,tion (a) Right-hnnd Circular Polarization, (b)
Lcft,-hand Circular I'olarizat
ior
r
2. Left-hand Circular Polarization
Wherr E, lags 8,, by rf 2 (i.". ,p : rl2), we obtain the r and y-components of electric field vector ol. circular polarized wave as E,: Iloelswf and En :- Eos\nut The iristantaneous angle tllat -E rnakes with the z-axis is given !y
g
'
:
rari
'{l:\:-rt \t',!
This indicateg that elcctric field vector (E) rotates at a uniform rate with an angular vekrc,it]' ar ill a clockwise rlirectiorr as showrr in Figure 7.2(b). When thc fingers of ther left hand follow the direction of the rotation of .O, tLre thurnb points i,o the direction of propagation of the wave. Thus, it is a left-h,and or n.egati.ue cir
1.
"s
R€H!EM&#R
For a linearly polarized wave, the, r'and::y-1c!tp{loncnt- d*t$$.Sdd... are in same phase.
2. For an elliptical
:
r and;g-comibd t$dktdg|. O < 6 I rl2.
polarized wave, the
field have the phase difference
3.
For a circular polarized wave, the r a.ud gr-components of"electric field :t':' have the equal amplitude (i.e. Ea - E ; &1's&d:i.p&d{i, hretween them
Page 437
is n12 (i.e.
d:tnl2);
Electromagletic Wavea
Page 438
Chap 7 Electromagnetic Waves
7.1O
REFLECTTON
A
REFRACTION OF UNIFORM PLAI{E WAVES
when a plane wave propagating in a homogeneous medium encounters interface with a different medium, a portion of the wave is reflected from interface while the remainder of the wave is transmitted. Depending the manner in which the uniform plane wave is incident on the boundar.rl two types of incidences may occur: 1. Normal Incidence: When a uniform plane wave is incident normally to the boundary between the media, it is known as normal incidence. 2. Oblique incidence: When a uniform plane wave is incident obliquel-v (making an angle other than g0") to the boundary between the two media, it is known as oblique incidence. In the following section, we will consider these two types of incidences ai various media interfaces.
7.11 NORMAL INCIDENCE OF UNIFORilI
PLANE WAVE AT THE INTERFACC
BETWEEN TWO DICLECTRIGS Consider a plane wave propagation along the t o, direction incident normaftr on the boundary z:0 between medium I (r< 0) characterized by o1, €y Fr a,nd medium 2(z > 0) characterized by o2, €zt ltzt as shown in Figure Z.&
I"lgur:c 7.i): Normal Incidence
of uniform Plane wave at the Interface
between
Dielectrics
In general, the field components of incident, reflected, and transmitted waves are expressed as given in table below. Table 7.1: Representation of Field Components of Incident, Reflected, and Tlansmitted Wave lnterface between Two Dielectrics
fr
I
E,Jt\ Eo8" & \ -/- = t---
r .-- -l-
-
:j!
w,;)'
l
I
I I r
p.n.r Reflection I
r
and Thansmission Coefficients
The ratio E,nf En is called reflect'ion coeffici,ent denoted by
I
-f
and given by
_rlz-\t I _- E,o Enu - \z* rlr
t I
f
Page 439
The ratio Er l E* is called transmiss'ion coefficient denoted by
r
and
given by
E,' -E*-
t I
t
2nz
,lr+rh
r
Ft-tt.z Standing Wave Ratio I i I
i I I I
The total electric field intensity in medium 1 is the sum of the electric fields of the incident and reflected waves, i.e. Er,(z) : Ei"(z) + 8,,(z) ,-/-. The ratio of the -a*imu- amplitude to the minimum of the total elQctri| \ field , E1 | is called stand,'ing waue ratio, i.e'
l.a'l
1+l-rl
r lt!qq!! . -1-lfl "-laL.
i
|
' lmrn
i
:-
-.ji,,..u.rr:ir .::.rir,,..i::r
f;'arer
1.
jr::r,rrlr:tiilii. ;:t
dit$e
a:rt
::|f...
:a
i-rr..: ::::..ri,irti,ut.rl
,,
2.
3.
4.
7.12 NORMAL INCIDENGE OF UNIFORM PLANE WAVE ON A
PERFEGT
CONDUGTOR
consider that a uniform plane electromagnetic wave) propagating along * a" direction in a perfect dielectric medium (ot : 0) of permeability pr and permittivity e, is incident normally on the surface of a perfect flat conductor ( 02: oo). The conductor surface is taken at z:0 pla,ne, as shown in Figure 7.4. Mediurn
(o':
f
ifrur
o)
1
2 fr, Medium (ar: oo)
1.1: Normal Incidence of Uniform Plane Wave on a Perfect Conductor
Chap 7 Electromagnetic T[avee
Since the depth of penetration for a perfect conductor is zeto, the
Page 440
*ac
will not travel beyond the z: 0 plane. In general, the fiqld components of incident and reflected waves for the interface are expressed as given in table
Chap 7 Electromagnetic Waves
below. Table 7.2: Representation of Field Components of Incident and Tlansmitted Wave for Interface between Dielectric and Perfect Conductor
id
H*(z) = H,oeil'" (- q\ : -
Eo "it'"
o,
7.12.1 Reflection and Thansmission Coeflicients Notice from Figure 7.4tlnat the wave is totally reflected. So, the reflection coefficient is given by
E'o
'' - E,o--_ ' Since, there is no transmitted wave (i'e' E61: 0)' So' the
transmission
coefficient is zero. i.e.
r:0 7.12.2 Standing Wave Ratio Since, the wave is totally reflected in this case with reflection coefficient lf I : f . So, the standing wave ratio is infinite' i'e' g:oo
7.13 OBLIQUE II{CIDENCE OF UNIFORT PLANE WAVE AT THE INTERFACE BETWEEN TWO DIELECTRICS
/)l\
l_/
For oblique incidence of uniform plane wave at the interface between two dielectrics, we consider the following two cases: 1. Parallel Polarization 2. Perpendicular polarization
7.L3.1 Parallel Polarization Figure 7.5 shows the oblique incidence of a uniform plane wave (parallel polarized) where the .D field lies in the rz-plane) the plane of incidence. Both the mediums are lossless. Reflection and Tlansmission Coefficients for Parallel Polarization The reflection and transmission coefficients for the parallel polarised wave are defined as
and
n Ea, r|:fr:
rlzcosflt
''' - FP En-
rl2cos01* 4rcosd,
11
-
rncos9t
qrco;6t+nlc;sdt 2rl2cosot -
Bfewster Angle for Parallel Polarized Wave The incident angle at whicir there is no reflection (i.e. E"o: 0) is called
i3.-
Brewster arrgle. For the para,llel polarized wa\€ propagating through lossless mediums, the Brewster angle dal is expressed as sin2
ds,
: : L,---azlLuL9 I -\t1le2f
z:0 l:'irlitir:
I i;: Oblique
Lossless I)ielectrics
Inciderice of Parallel Polarized Wave at the Interface between Two
13.2 Perpendicular Polarization corrsider the oblique incide'ce of a perpenclicular polarized wave shown in Figure 7.6. Both the mediurns are lossless and the .E field is perpendicuiar to the rz-plane, the plarre of iricidence. Medium 2(€",p")
rk \:,:1.
fl
.,6 Z:0 I:'iliiir'<' ?.ii: Obiique Incidence of Perpendicular Polarized Wave at the Interface between Two Lossless Dielectrics
Reflection and rlansmission coeffrcients for perpendicular polarization The reflection and transmission coefficients for perpendicular polarized wave are given by
,.8 and
E6 Ett
_ r12cos0i- r1pos06 - ryrcosf3 rjtcos9, E,D _ 2r12cos0,
Ei-
arcor9,ar"";6-
Page iWl
Chap 7 Electronagnetic Waves
Paee 442 Chap 7 Electromagnetic Waves
W'ave Brewster Angle for Perpendicular Polarized
Fortheperpendicularpolarizedwaveplopagatingthroughlosslessmediums. the Brewster angle' ds' is expressed as sin2du,
: I-
u,Gzf p,zet
I-GJNT
7'|4oBLIouEINGIDENGEoFuNIFoRtPLANEwAvEoNAPERFEGT CONDUGTOR
Considerthatauniformplaneelectromagneticwave'propagatingalong p dielectric medium (or: 0) of permeability i:", Jo**ton in ea perfect surface of a perfect flat and permittivity , ls incident obliquely on the polarization and Aguirr, we will consider the two cases: parallel
i"TuI"L,
PerPendicular Polarization' I
7.14.1 Parrillel Polarisation
/
,i
7.7 shows the obliqle.ingidSn1 :: ',o"?1!il-",Y'::^i.:?,"-",:" " reflected wales lie in the ",*,rr" perfect conductor. Electric field of the incident and
rz-p|ane,theplaneofincidence.Asthedepthofpenetrationforaperfect
,"r"a""toriszero,thewavewillnottravelbeyondthez:0plane'So,the reflectioncoefficientofparallelpolarisationisobtainedas r rr:
f,:_
I..iglrr'e7.7:obliquelncidenceofParallelPolarized.WaveonthePerfectConductor
7.t4.2 Perpendicular Polarisation Inthiscase,theelectricfieldisperpendiculartotheplaneofincidence. Consideraperpendicularpolarisedwaveincidentonaperfectconductoras shown in Figure 7'8' is zero, the since, tie depth of penetration for a perfect conductor
wavewillnottravelbeyondthez:0plane.So,thereflectioncoefficientof perpendicular polarisation is obtained
,r_$:_1 - En-
as
--Tl
Page 448
Chap Z Elettromagnetic Waves
lligurc 7'ti; oblique Incidence of Perpendicular Polarized Wave on the perfect Conductor
******,k*+**
0
xxHKCtSH
Page 444 Chap 7
?.{
|
Electromagnetic Waves
! f,sc&
?.r.t
propagating wave is given a certain medium electric field intensity of a
In
{3r)anY
rhe erectric n"rd ph"fl,9'i1;""#;ffi1?:(A) Eoe-1"*'B)'a,nY l:at j1)'aaY (R) jAoe-e* ltr_ (C) - jE,)e-1"+jo)" avY lm (D) -i&os("-j1')aoY lm f",*sc*
?,d"2
lm
b;-
I I I I
plane Eosin(z)cos(ct)o, represents the electric fiell of a
E: wave in free sPace' Assertion (A) :
|
,.
i j I
Reason(n):Aplanewave/propagatingwithvelocitytloin*o,direction: must satisfY the
equation
i
:o tJ dl dt - "'r',a]!,
f$
of A' (A) Both A and R are true ancl R is the correct explanation
(B)BothAandRaretrrrebutRisnottlrecorrectexplanationofA. (C) A is true but R is false (D) A is false but R is true r$*&
?.*.3
A propagating wave in free space has magnetic field intensity .E[ :0.1cos(10e1 - aE)a"Alm : lcm at time' What will be the electric field intensity of the wave at I t
)
:
0.1ns
?
(A) -37.7 a"Y lm (B)-37.6o, V/m (C) -1e'84,Y l^ (D) 37.6o" V/m ir,r{&
r..!.4
space is given Electric field intensity of linearly polarized plane wave in free field magnetic of form phasor by E :(Sar- 6o')cos(r"'t-502)Y lm' The
intensity of the wave will be
(A) -r7o(5a,* 6ar)e rt" v l* ^-c50,
(C) (5o, -6a,)e
^
r$'!s{3 ?,"!"$
Yln
(B) (5a" (D)
- oa")ffvp
-(5o"'+aa.,1effvp"
EM wave In a perfect conductor (resistivity' p = 0) magnetic field of any (A) lags electric field bY 90' (B) leads electric field bY 45" (C) lags electric field bY 45' (D) will be in ptra,se with electric field
-l!
Commsl Data For Q. 6 and 7: An electromagnetic wave travels in free space with the electric field component
E" *r*s
McQ
:
(5a"* l0a")e-la'-'"'t
V
lm
?.'t.s
what will be the phasor form bf magnetic field intensity of the wave (A) -29.66e j(' 2")tr.L/m (B) -5"/5 e-i14x 2z)mA/m (C) 29.66e't4r-2'ztmA/m (D) -29.66e 44'-2'\ A/m
?'{'7
What will be the time average power density of the electromagnetic wave
?"i'li
?
(A) -3o, * ^/i ar- 2a" (B) -2^/i a,l {i ao- a" (C) 3o, -,/S au+2a, (D) 2,/i a, - ,/i a, + a"
*rcs
Common Data For Q. g and l0 : In free space an electric field intensity vector is given by E :100cos(at_ pz)a" where r,,, and p are constants.
?.'r's If P: -#: I Pneo -.
what will be the magnetic flux density vector
Il
?
(#:3xro+8m/s)\
(A) 3 x l01ocos(cur _ gr)a, (B) 3.JB x 10-7cos(c..,t - |r)on (C) B x 108cos(a.,r - gr)o, _
lnl us x 1o 6cos(u., t- Bz)a, ucQ
?"{.{r}
The poynting vector of the E_M field will be
(il
n^
pr)o"
fficos2(at(C) f04 fik cosz(r,.,t - 0r)o" HcE
7't"*'t
?
A propagating wave has the phasor form of its electric field intensity 4efined as E" :.(-2^/5 0,+ r/5 ao- o,,)2-!'r"(-t'aliv-22)v/m. The wave is rinearly polarized along the direction of
l,
(B) [email protected](wt- pr)o, (D)
(ut _ [32)a, +cos, \/ po
The electric field associated with a sinusoidallyaime varying electromagnetic field is given by E : Ssinrrsin(2r X 108 _ ,[i rz)orV li The time average stored energy density in the electric field is
(A) ,4aeosin2rr (c) lflsnzzrr
-'-.-.
(B)
f
(D)
ffsin2rr
sinrnr
Chap 7 f, lssfr'gmagnetic Waves
(A) (o6b.9o" - 331.6a")W l^, (B) (1a8.9a" - T4.rSa")W/-, (C) (- 331.6 a" * 66s.9a)W l*' (D) (- 7 4.lS a, * 148.9 a,) W l^, Hc$
Page 44b
[ae
aa6
MCA
7,',1"',12
Chap 7 Electromagnetic Waves
Electric field associated with a sinusoidally time varying blectromagr field is given by .E :10sin(ry)sin(6tr x 108' -'/lrt)a'V/rn' What field ? be the time average stored energy density in the magnetic
1l; 1p1zs * (B)
5osin2a-r)
#(25 * 5osin2zrr)
(c) 14q-!d(2s + trosin'?zrr)
Q) fuQ; * ile6
7.{.13
McQ
?.{.'14
( )
5osin2z'r)
on the surfr An electromagnetic wave propagating in free space is incident field electric the of of a dielectric medium (Pn, 4Eo)'If the magnitude field electric incident wave is Eo then'what will be the magnitude of the the reflected wave ? (B) -Eoi3 (A) -2Eol3 (D) - E, (c) Eol2
is given by Magnetic field intensity of a propagating wave in free space
I[ :
0.3cos (ut
ff the total time period of the wave be time,
-
?
BY)a" A'lm
then the plot of
I/ versus y
r: $ will be
(B)
MCo ?.1.{t
plane wave is propagating with a velocity of 7 '5 x 107 m/s fi a lossless medium having relative permeability 1t',:4'8' The electric phasor of the wave is given bY E" : geflst o"Y lm What will be the magnetic field intensity of the wave ? (A) 11.05cos(9.54 x 106r+ 0.3r)an mA/m (B) 22.13cos(9.54 x 106r+ 0.3r)ao mA/m
A uniform
(C) 22.13cos(9.5a (D) 11.05cos(2.25
x x
106r+ 0.Zz)umA/m 107r+ }-3r)armA/m
ilcq 7"'r'{6 I' uniform
prane.
wave t. oroolrl*]ng at a veloc-lty of z x roz mf sin dielectric such that the eleciric a perfect and magnetic neias or the wave are given bv E(r, t) : 900 cos (5 x 106n _ pr) y f a,
,
lm
ff (A) "::',x#'T'*'il';'l 1.70,2.6s
(c) itcE
?.t""r?
HCQ ?,{.rs
1.20,
1.58
!'";"'"T I ; #^Y/'', n" m edi um wi,, i,, : (B) B'4, 5.32 (D) 2.37, 2.6s
":T
An electromagnetic *.," in free space in _ o, direction i. propagating with a frequencv o and prt*" u"gj" ,ulr. rir" EM wave is polarized i. * o' direction. If the ampritude or lt"ctri" field of tt *uu" is ao tnen tt magnetic field of the wave will e " be
ral
#"*(wt+fr)au
(B)
(c)
-#"os(wt+?r)*
(o)
4s.a'cos
(wt*ucr)ao
ffcos(c,.,
t_i,)u
what will be the electric fierd of a prane wave polarized parallel to trre :r_: plane and propagating in free .p."" io it direction
(1'1'1)' that has the ampritud< (A) Eocosl, t +,
\!]W) + ft-(r +, . 4](T) -
;
(B) z'"o'[,
t
" ui;;iou"rr",
(c)'Eocos[, t _
ft;t,
from origin to the
p<.rint
i.r with zert.r phase a'gre.,
f(r +, -
4](*#) (D) Eo"osl,t + f(r +,. 4]("#) HCQ ?,r"ts
il:"lr: j*1:"::"L:l|_"i^":?:*s_is
done for a microwave experiment ri,
lj"trT:tT:ilJ: i'T':J;"li?,i:'j":T"1'igT.::-":1",Fil;r;ffi N:,: ,,^='7, o: a.zs ib;s'i-l It;"1::i""t (A) greater.than 0.64 pm " (B) less than 0.64 pm (C) exactly equal to 0.64 pm (D) none
ICQ 7.{.20
of these rn a nonmagnetic materiar.of conductivity o : 2 x 102 s7m, electric fier. of a propagating plane wave is given by .E : Jcos(10? t _ O.Zy)a, * 2sin(102 t _ 0.2y)a, y /m What will be the value of (A) (2 - r36eo)F/m "omple* o"#ifiu,r, of the medium ? (B) (36e6 - 12)F/m
(c) (D) HCQ 7.1,21
- p)F/m (36e0 + j2)F /m (36
Assertion (A) : All the metals are opaque. Reason (R) : skin depth of metars
(A) Both A and R are true and "t"l"lrr" range of nanometers. R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
page 447 Chap Z Electromagnetic Waves
Page 44E
nfisa ?,t.tn
Qhap 7 Slectromagnetic lVaves
In the plane z:0, electric field of a wave propagating tL *az directio f."" ,pu,"" is .Eo which is varying with time f as shown in the figure' tro(\'/rn)
0.5
1.5 2
t(p-sec)
: If the magnetic field intensity of the wave at f of 11r versus z will be
1F,sec be
IIr
then the
(B)
o
75
7b 0 75
z(m)
z(m)
I{,(A/rn)
Hr(Alrtt)
1
(D)
-150
!|4ca ?"'1.23
150
z(m)
medirrm in Electric field of an electromagnetic wave propagatirrg in a is wave jo')dt3'' The : direction is given by E" Eo(an(A) left hand circuiarly polarized (B) Right hand circularly polarized
*
(C) elliptically Polarized (D) iinearl;, Polarizerl iilca 7.t.24
the e.,lectromagnetic wave has the electric field intensitY in a Per on incident is wave EM form given by .8, : 4(o,- ie')e-ia' . The reflected the of polarization What will be the at
An
corrductor located wave
Y:0.
?
(A) Ieft hand circular (B) Right hand circula,r (C) elliptical (D) linear
An electromagnetic wave propagating in free space is incident on a perfectly at r> 0. The electric field of the incident wave in the phasor form is given by Ex:l0a"e-i(6v+8c)V/m. The net electric field of the total wave (incident and reflected both) in free space afber reflection conducting slab placed
will be (A) L0a,e-\6u-8r)V /m (C) - j20a"e-6vsin8r V/m
(B) *10a,e-{6Y-8')V/m (D) paa"e-rosin8rV/m
Electric field intensity of an EM wave propagating in free space is given by Et, :25a,e-16"+0v)V /m If the wave is incident on a perfectly conducting plane at y:0 then the magnetic field intensity of the reflected wave will be (A) (B) (# +ffi)e-tu"-8u) Alm + ffi)e-t6' 8v)A/m
-(#
(c)
(#+#),^'O
(D)
-(#
+ ffi)ea6"-8,) Alm
The complex electric field vector of a uniform plane wave propagating in free space is given by E" : (Ji o, - e,y - 2"/-B a"1e-n'o1r(4x+Jiv-22) V/-. tlru unit vector in the direction of propagation of the wave will be
@::"$@
r+t-3tut4@ (C)
-
(D)
-/Ea.*2a,)
(3o"
-
(3a,+"/i%-za,)
Cornmon Data For Q. 28 and 29 : In free space complex electric field vector of a uniform plane wave is given by
E, : (,/i a, *
o,")
e-t{{.6x-zv-sz) Y 1^
rcQ ?,1'28 The apparent wavelengths along the
(A) (B) (C) (D) ilcQ
7'1.2s
\)v^"
m m 16.7 m 28.87 m 16.7
28.87
m m 28.87 m 25m
r, y
25
28.87 m
16.7
25m 25m
and z axes are
L6.7 m
The apparent phase velocities along the
r, y and z axes are
uPt
(A) (B) (C) (D) Meq
1.73
UP"
x 1010m/s
6.93 x 108 m/s 2.77 x 107 m/s 1.2 x 10em/s
7.1"3o Which of the following uniform plane wave
1.5
x 1010m/s
6 x 108 m/s 2.4 X I07 mf s 1.2
x 10em/s
x 1010m/s x 108 m/s 1.6 x 107 m/s 1.2 x 10em/s 1
4
complex vector field represents the electric field of a
?
(A) (- i", - 2au - l{i a")e-n'u(/iv+") (e) (", -iZ", - {{ a,)e-no'r(x+'/iz)
tcl
o"f"-po2r(ria+sv+zz) [(;"* i+)^+ (r + 4)"- ilE
rol
o,le-iloz^("6,+ru+2") [(-6 - i+)**(t. t$)o,+ iJE
Page 449
Cbap 7 Electromap€tic Waves
Page 450
ilcQ
7.t.3t
Chap 7
which of the following.gairs of vector .8" and rr, field represents the electric and magnetic fiblh"vectors of a uniform plane wave
Electromagnetic Waves
E"
?
H"
(A) (i""-t2ao+ it/51)t rn"*")vlm (a,- pao- /i a")"-,(/i,+4 A/m @) (i""+ l/3 a")e-t"('+fr')Y/m (-la,+ i/i a,)e-t'('+'tr") 61* (C) (i"" - j,/i a")e-t"(,+/'") y f 11 (-/iir."- ia")e-r.("+r{4 A/n (D) (-r", - 2ao+ jlT a")e-n't"('tr"+4y 1^ (a, {B'a,)e-n:"1n"+4 g1_-},
l
-
llitcQ
?.1.12
12a,
The following fields exist in charge fre€ regions
p:
60sin(ot_t I0r) a,
wt - 2p) a6 Q: f;co"1
: Bp2cotqar+ |cosga6 lsinds ,S : in(ut _ 6r) aa R
\
The possible electromagnetic fields are
(A)
(c)
P,Q
(B) R,S (D) g,s
P,R *{.*****{<***
-
l
EXERCI$E 7.2
Page 4b1
Chap 7 Electromagnetic Waves
quF$
7,2.,1
In air, magnetic field intensity is given by H :10cos(6 x 107 t _ ky)a, Alm . Wave number /c for the EM wave will be rad/m.
QUE$ ?.2"2
An.elec\magnetic wave is propagating in certain non magnetic materiar su.ch t(at)he magnetic field intensity at a,ny point is given by .E[ : 1.5cos(10et_ bz)a, Alm The phase velocity of the wave in the medium will be x 108 m/s.
QUES 7,?,3
Magnetic field intensity in a certain non-magnetic medium is given by
II :
I/ecos(wt_ pg)a" Alm If the wavelength of the EM wave in the medium be 12.6 m then what will be the phase constant p (in rad/m) in that medium ? quss
r'z'4
rn a nonmagnetic material electric fierd intensity is given by
E :8cos(4 x
108,
-2r)arylm
What is the relative permittivity of the medium qu€$
?
?.2.s
Electric field intensity in free space is .o :l2cos(b The time period of the wave will be ns.
7''''6
In air, a propagating wave has erectric fierd intensity given by
E :9cos(4 x
108,
x
108i _
pz)a,vlm.
_ Br)a"ylm
What will be the time taken (in ns) by the wave to travel one-fourth of it,s total wave ? QUHS
?'2*? What will be the intrinsic
QuEs
?.2'8 A radio wave is propagating at a frequency of 0.5 MHz in a medium ( o:3 x 107S/m, LL,:€, = l).The wave length of the radio wave in that medium will be mm.
QuEs
?"?'3
impedance (in f,)) of a lossless, nonmagnetic dielectric material having relative permittivity e,:2.25 ?
Phasor form of magnetic field intensity of a uniform plane wave in free H" : (2 + j5)(aau+2ja")e-jp" A/m. what is the maximum elestric field (in kv/m) of the plane wave ? space is given as
Qurs 7'2"10 An electromagnetic wave is propagating from free space to a certain medium having relative permittivity 6" : 9. If wavelength of the wave in the medium be 20cm then what would be it's wavelength (in cm) in free space ?
It qu€s
Page 462
?.2.{{ If
some free charge is being embedded in a piece of glass, t}ren the "ft-pl second. rvill flow out to the surface nearly after ----( For glass relative permittivity e,:2.25, conductivity o : 10-12 S/m)
Chaf 7 Electromagnetid Wdve
eue$
:
I I
:
4). A plane wave is propagating such that I the electric field intensity of the wave is E : Eoe-"/itsin(108t - gr)onv/-. I The loss tangent of the medium will be |
?.?.t2 In a certain medium
(e"
4,
p,,
----
Common Data For Q. 13 and
o
In
14:
8, 11,,: 0.5, o : 0.01 S/m a plane wave is travelling in direction that has the electric field intensity E:0.5cos(10ezrt* rf 3)a,
a lossy
medium (e"
:
*o, at z:0.
I I
]
l
eun$ ?.a.13 What will be the distance traveled by the wave to have a phase shift of 10' eug$
So, the value
ouEs
a distance z, the arnplitude of the wave is reduced by mm. of z equals to
2.a14 After traveling
?
40%-
?.t.{i A plane wave is propagating with frequency f :50kHz in a medium ( o:2s/m, €":80, p,,:4). What will be the skin depth (in meter) of the medium
?
eur$ 7.2.{e Three different dielectrics of permittivities 4eo, 9eo and 3e6 are defined in the space as shown in figure. If the leading edge of a uniform plane wave then how propagating in a' direction is incident on the plane r:-6m much time (in ps) it will take to strike the interface defined by the dielectric 2 and dielectric 3
?
it
Dielectric I or",*r"r"
112
-i n"" se"."
:lrl
T
:-6m
I
Fu,4€o
tr: 0
I
rn, 0..
I
r:3 rn
r,
. Common Data For Q. 17 and 18 : An electromagnetic wave of 50 N{Hz frequency is incident on a dielectric medium such that it's skin depth is 0.32 mm. (Permittivity of dielectric :6.28 x l0-?)
QUE$
?.2.1? The conductivitv of the dielectric will
r.z.tt
lij*
be
x 105 s/m.
If an electromagnetic wave of 8 GHz frequency travels a distance of 0.175 mm in the dielectric medium then it's field intensity will be reduced by _*_ dB.
YIF
?"x"'!s An electromagnetic wave propagating in free space has magnetic field intensity -E[ :0.lcos(wt- l3y)a,A/m. What will be the total power (in mw) passing through a square plate of side 20 cm located in the prane
r+tl:2?
QUES ?.g.CS
\) QuEs
An electromagnetic wave propagating in a lossless medium (p, : 4po, 6r : €o , or:0) defined in the region g ) 0 is incident on a lossy med.ium (pu: to , €2:4€(), o2:0.1S/m) defined in the region y30. The electric field intensity of the incident wave in lossless medium is given by 8," :5e'Fo a,Y fm What will be the standing wave ratio ?
?.2"x'! Phasor form of electric field intensity of a uniform plane wave is given by n
"
:
(oE o,
-h
"r)
un
o4r(-2o- ,u+'/T
The wavelength along the direction of propagation is
") V I
* meter.
Qurs ?"*"2? In free space the complex magnetic field vector of a uniform plane wave is given by H" :-(fr a,+ a")e'fr'0t'(Jzx-zv-tz) A/m. Flequency of the plane wave
QuEs
will be
NIHz.
?'4.x3 A uniform plane wave in region 1 is normally incident on the planner boundary separating regions 1 and 2. Both region are lossless and e,1: pf, , €,2: ia. ltthe20% of the energy in the incident wave is reflected at the boundary, the ratio
Quss ?"2.24
e,,zf
en is
An electromagnetic wave propagating in medium I (po, er) is incident on mediurn 2 (pa, ez) as shown in figure such that the electric field of reflected wave is 1/5 times of the electric field of incident wave. The value of e1/e2 equals to
Medium
I
For €r:
*xxxt<***
t<*t<
Page 453
ehP
z
Electro4ag4etic Wavee
EXERGISE 7.3
Page 454
Chap 7 Electromngnetic Waves
o
MCO 7.3.1
What will be the direction of wave propagation in a non magnetic in which magnetic field intensity at any point is given by .EI : 3cos(a.rt- kz)a, Afm
(A) + a" direction (B) - o" direction (C) + a, direction (D) + au direction nrcQ 7,3,2
The skin depth in a poor conductor is independent of
(A) Permittivity (B) Permeability (C) Requency (D) None of these MCQ 7,3.3
Poynting vector is given by (A) ,E x .H
(C) rYrcc 7.3.4
E. H (D)H.E (B)
HxE
Poynting vector gives (A) rate of energy flow
(B) direction of polarisation (C) electric field (D) magnetic field ntcQ ?"3.5
E . H of a uniform (A) EH
plane wave is
(B) 0 (D) qH'
(c) qE' McQ
?"3's
For a uniform plane wave in the r-direction
(A) (C) MCQ 7.3.7
and
H,:0
0
Complex Poynting vector,
(A)P:ExH* (B)P:ExH* (C) P:lExH* (D)ExII
t{-
(B) r/,: 0 (D) E,: O
O
Depth of penetration in free space is (A) infinity
(c) tlltcQ 7.3,8
E,: E,:0
P
is
(B) 1/a (D) small
rca
?.3.s
Uniform plane wave is (A) Iongitudinal in nature
Page 455
ChtP ? Electromagaetic Warea
(B) transverse in nature (C) neither transverse nor longitudinal (D) r-directed
rcq
7"3.to
The dir;rytion of propagation of EM wave to obtained from
(A)'/ />4 I
H
(B) .E x (D) H
(c)lE/ rc&
?,3,{1
n
The velocity of an EM wave is (A) inversely proportional to B
(B) inversely proportional to o (C) directly proportional fo B (D) directly proportional to o
rco
?"3"{?
Velocity of the wave in an ideal conductor is
(A) (C) HCA ?"3.13
zero moderate
(B) very large (D) small
Velocity of an EM wave in free space is
(A) independent of / (B) increases with increase in / (C) decreases with increase in / (D) zero MC& ?,3.{jl
The direction of propagation of an EM wave is given by (A) the direction of .D
(B) the direction of .El (C) the direction of E x H (D) the direction of E ' H
$ce
?"3"t5
For uniform plane wave propagating in z-direction
(A) .8,:0 (B) rr,: o (C) Er:0, (D) E":0,
i,!eQ 7.3,16
H":0
Velocity of propagation of an EM wave is
(il,[ff (c)
I
MCQ 7,3,17
Hv:0
+ lto€
o
(B)
#
(D)
f;
Consider the following statements regarding the complex Poynting vector P for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(P) denotes the real part of f), ^9 denotes a spherical surface whose centre is at the point source, and o* denotes the unit surface normal on ,9. Which of the following statements is TRUE? (A) Re(P) remains constant at any radial distance from the source
(B) Re(F) increases with increasing radial distance from the source (C) #Re(P) ' (dso") remains constant at any radial distance from
Page 456
Chap 7 Electromagnetic Waves
th
s6urce
Q)
ffpre(P1
'
Q"Sa,) decreases
with increasing radial disiance from th
source
a
ItcQ 7"3.t1
uco
?.3.{9
The electric field of a,n electromagnetic wave propagation in the positir direction is given by E: a,sin(wt- 0z) + orsin(cut - 0z* f). The wave b (A) Linearly polarized in the z -direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized L^ equation ^^-.-+:^-. a28"._ t4}r=E^' . tu If a plane electromagnetic wave satisfies the ffi: df wave triropagates in the
(A) r-direction (B) z-direction (C) g-direction plane at an angle of 45' between the (D)
"-y
urco 7.3,20
r
and z direction
The intrinsic impedance of copper at high frequencies is (A) purely resistive
(B) purely inductive (C) complex with a capacitive component (D) complex with an inductive component iltcQ 7,3"21
The depth of penetration of wave in a lossy dielectric increases with increasing
(B) permeability (D) permittivity
(A) conductivity (C) wavelength uEQ 7,3.22
The polarization (A) linear (C) left hand
8XCA 7,3.23
of wave with electric field vector E
circular
0")
(B) elliPtical (D) right hand circular
7a'
+
ao)
b
It increases as frequency increases. It is inversely proportional to square root of p and o. It is inversely proportional to square root of / It is directly proportional to square root of p and o.
Which of the above statements are correct
(A)land2only (C)2and3only ?.3.24
Eoe\'t+
Consider the following statements regarding depth of penetration or skin depth in a conductor :
1. 2. 3. 4.
nco
:
?
(B)3and4only (D) 1, 2,3 and 4
Consider the following statements : 1. (Electric or magnetic) field must have two orthogonal linear components. 2. The two components must have the same tnagnitude.
3.
.
The two components must ha.ve a time-phase'difference of odd multiple of go". Which of these are the necessaxy a.nd sufficient conditions for a timeharmonic wave to be circula"rly polarized at a given point in space ? (A) 1 and 2 qnly (B) 2 and 3 only (C) 1, z ar1d s\ (D) 1 and 3 only
(,
lcQ 7.3.25 Assertion fAf :The velocity of light in any medium is slower than that of vacuum. Reason (R)
: The dielectric constant of the vacuum is unity and is lesser than that of a.ny other medium. (A) Both A and R are individually true and R is the correct explanation 9f (B) Both A and R are individually true but R is not the correct explanation
ofA
(C) A is true but R is false (D) A is false but R is true
lcQ 7.3'26 According to Poynting
theorem, the vector product E which one of the following? (A) Stored energy density of the electric field (B) Stored energy density of the magnetic (C) Power dissipated per unit volume
xH
is a measure of
(D) Rate of energy flow per unit area
lca 7.3.2?
Poynting vector is a measure of which one of the following ? (A) Maximum power flow through a surface surround.ing the source (B) Average power flow through the surface
(C) Instantaneous power flow through the surface (D) Power dissipated by the surface
rcQ 7'3.28 The electric field component of a wave in free space is given by
I Fe 73'2e
I L
E : 50sin(L07 t+ kz) any lm Which one of the following is the correct inference that can be drawn from this expression ? (A) The waye propagates along y-axis (B) The wavelength is.188.b m (C) The wave number k: 0.83 ra.d,lm (D) The wave attenuates as it travels
For an electromagnetic wave incident on a conducting medium, the depth of penetration tA) is directly proportional to the attenuation constant (B) is inversely proportional to the attenuation constant
(c)
has a Iogarithmic relationship with the attenuation constant
(D) is independent of the attenuation constant
rcq 7'3.30 What is the effect of the ea,rth's
magnetic field
frequencies in the vicinity of gyro-frequency
?
in the reflected wave at
page 4b?
chap
?
Electromigndtic waves
I lI
H
(A) No attenuation in the reflected wave (B) Decreased attenuation in the reflected wave (C) Increased attenuation in the reflected wave (D) Nominal attenuation in the reflected wave
Page 45E
Chap 7 Electromagnetic Waves
ri
0
e8*.$ ?"3.31
Consider the following statements : For electromagnetic waves propagating in free space : 1. electrical field is perpendicular to direction of propagation
2. 3. 4.
electrical field is along the direction of propagation magnetic field is perpendicular to direction of propagation magnetic field is along the direction of propagation
Which of these statements are correct (A) 1 and 3 (C) 2 and 3 iq1*{* ?,:},$e
?
(B) 1 and 4 .(D) 2 and
Skin depth is the distance from the conductor surface where the field strengtl has fallen to
(A) z' of its strength at the surface (B) e of its strength at the surfac (C) (f/e) of its strength at the surface (O) (tlre) of its strength at th surface
.4c{t ?"3.33
The depth of penetration of a wave in a lossy dielectric increases with increasing
(B) permeability (D) permittivity
(A) conductivity (C) wavelength _&{rt 7,*,34
When a plane wave propagates in a dielectric medium (A) the average electric energy and the average magnetic energy densitier are not equal. (B) the average electric energy and the average magnetic energy densities are equal
(C) the net average energy density is finite (D) the average electric energy density is not dependent on the avera83 magnetic energy density *{ i:i ?-3"3*
ntll
*
7.3"36
What causes electromagnetic wave polarization ? (A) Refraction (B) Reflection (C) Longitudinal nature of electromagnetic wave (D) Tlansverse nature of electromagnetic wave Fields are said to be circularly polarized if their magnitudes are (A) Equal and they are in phase
(B) Equal and they differ in phase by -F 96" (C) Unequal and they differ in phase by + 99' (D) Unequal and they are in phase 'i******,t
EXERCTSE 7.4
Page 4b9
_
Chap
Z
Electromagnetic Waves
A plane wave propagating in air with E : (8a, *
6ao
*
Ea") si@t+t,*ru) y f yn
is incident on a perfectly conducting srab positioned at e of the reflected wave is
(A) (B) (C) (D) rcQ
7"4.?
(- 8a,,- 6au (- 8a,* 6av (- 8a" - 6ay (- 8o,, * 6a,y -
< 0. The .E field
y
5o,)
1^ 1^ "t@t+lx+aly 5o") "t@t-lx-ai 1^
5o,)
"t@t+sx+ad
y
5a") ei@t-3r-ail V /m
The electric field of a uniform plane erectromagnetic wave in free space, along the positive r direction is given by E:n(q+ ja")e-jrr,. The frequency and polarization of the wave, respectively, are
(A)
1.2 GHz and left circular
(B) a Hz and left circular (C) 1.2 GHz and right circular (D) a Hz and right circular The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 v/m. If the
relative permittivity
oJ
the medium is 4, the magnitude of the time-average
power density vector (in
(A)
(c) 3a
7.4.4
#
# @#
#r
:
A planewave havingtheelectric field components 24 cos (3 X I08 _ py) a, -8, v/m and traveling in free space is incident normally on a rossless medium with p pn and e 9eo which occupies the region y Z 0. The reflected magnetic field component is grven by
(A) (B)
711.5
is
(B)
:
-
FQ
W/mr)
frcos(B x
108r+
y)a,A/m
x
108r+
y)a,Afm
2!6cos(3
(C)
-frcos(3 x 108r+ y)a"Alm
(O)
-frcos(3 x 10sf + y)a,A/m
A uniform plane wave in the free space is normaily incident on an infinitely thick dielectric slab (dielectric constant u, : 9). The magnitude
reflection coefficient is
(A) 0 (B) 0.3
(c)
0.5
(D) 0.8
of the
a
Page 460
MCQ 7.4.6
Chap 7 Electromagnetic Waves
A plane wave of wavelength ) is traveling in a direction making an angh with positive r-axis and g0' with positive g-ilds. The .E field of the wave can be represented as (Eo is constant)
(A) E: aoEoej("**"-i") (B) E
: avEoe4"t-n^'-*')
: arEoe4"**'*I") (D) E : suEoe\"-*'**") (C) E
rvtcQ 7"4.?
The .EI field (in A/m) of a plane wave propagating in free space is
n: o,ffcos(a.'t - az) + w(wt- 0z+$)*l is
The time average power flow density in Watts i
h
I
iico
?.4.4
I
(A)
#
(C)
50?d
Y\o
(B) 1qq rh
(D)
"
!q rlo
A right circularly polarized (RCP) plane wave is incident at an angle to the normal, on an air-dielectric interface. If the reflected wave is polarized, the relative dielectric constant
e"2 is
Linea.rly polarized
I
MgQ ?.4,S
@)
lt
(c)
2
/5 3
When a plane wave traveling in free.space is incident normally on a mediun having €,: 4.0 then the fraction of power transmitted into the medium b given by
(A)
(c) &rsQ ?.rt,to
(B) (D)
3 +
(B)
; (D) *
A medium of relative permittivity €,2:2 forms an interface with free A point source of electromagnetic energy is located in the medium al a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by space.
(A) 2r mz
(c) f{ca 7.4,{{
i ^,
I
mz
(D) z'pz
A medium is divided into regions I and II about z: 0 plane, as shown in the figure below.
Lr-1.
(B)
Page il61
Chap 7 Electromagnetic Waves
A1e-lesqromagnetic wave with electric field Er
ndgall/
on the interface from region the\rrti/rface i.s
(A) Er: & (C) 3a, *3o,0*5a, rcQ
7,4,{2
/.
:4a,,*Jaulba,"
The electric file
b
is incident
in region II at
(B) 4a,*0.75ou- I.25a" (D) -3a" *3an*5a,
The magnetic field intensity vector of a plane wave is given by H(r,y,z,l) 10sin(50000r+ 0.0042* 30) a, where ar, denotes the unit vector in y direction. The wave is propagating
:
with a phase velocity.
(A)5x10am/s (C) -1.25
tQ
7.a.{3
x
107
(B)
-3 x 108 m/s
(D)3x108m/s
m/s
Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency of 101a Hz in glass. Assume velocity of light is 3 x 108 m/s in vacuum
(A) 3 pm (C) 2 p,m
Gt 7.4.14 lf E: (a"+ ja)
Poynting vector is"ikz-htt (A) null vector
(B) 3 mm (D) 1mm and
.E[: (k/rp)(ao+ ja,)
(C) (2klwp,) a"
etkz-r't , the time-averaged
(B) (klup,) a" (D) (k/2wp,) a"
GGI 7.4.{5
A plane electromagnetic-vrave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with e ) eo . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be (A) 1202r A (B) 60zr O (C) 6002r O (D) 2atrA
EQ
7.4,16
The depth of penetration of electromagnetic wave in a medium having conductivity o at a frequency of 1 MHz is 2b cm. The depth of penetration at a frequency of 4 MHz will be (A) 6.25 dm (B) 12.50 cm (C) 50.00 cm (D) 100.00 cm
'3a
7.4.a7
A uniform plane wave traveling in air is incident on the prane boundary between air and another dielectric medium with e":4. The reflection
L T
coefficient for the normal incidence, is
(A) zero (B) 0.333l0.
(B) 0.5/180" (D) 0.333/180'
Page 462
Chap 7 Electromagnetic Waves
ffcq 7.4.18 If the electric field intensity associated with a uniform plane e wave traveling in a perfect dielectric medium is given by : 10 cos (2zil07 t - 0.1zrz) V/m, then the velocity of the traveling wave (A) 3.00 x 108 m/sec (B) 2.00 x 10E m/sec (C) 6.28 x 107 m/sec (D) 2.00 x 107 m/sec $fi*g
?.4.{* A plane
wave is characterized
(A) linearly polarized (C) elliptically polarized rn*Q 7"4.30
by.E:
k
I o,rerP)"iat-tkz. This ware (B) circularly polarized (D) unpolarized
(0.5a,
Distilled water at 25'C is characterized by a: 1.7 x 10-a mho/m e:78eo at a frequency of 3 GHz. Its loss tangent tan6 is n/m) 1e : (A) 1.3 x 10-5 (B) 1.3 x 10-3 (C) 1.3 x r0-4178 (D) 1.3 x I0-5 f78eo
$
MeQ 7.4.e1
A material has conductivity of 10-2 mho/m and a relative permittivity The frequency at which the conduction current in the medium is equal the displacement current is (A) a5 MHz (B) e0 MHz (C) 450 MHz (D) 900 MHz
rvtgo ?.4"2?
A uniform plane electromagnetic wave incident on a plane surface dielectric material is reflected with a VSWR of 3. What is the incident power that is reflected ? (A) 10% (c) 50%
r4cQ
(B) 25% (D) 75To
?'4.23 A uniform plane wave in air impinges at 45'
angle on a lossless material with dielectric constant e.. The transmitted wave propagates 30' direction with respect to the normal. The value of e" is (A) 1.5 (B) /15
(c) r,'fa
7"4,24
Two coaxial cable 1 and 2 a,re filled with different dielectric constants and ez respectively. The ratio of the wavelength in the cables (^/)t i,
7.4.25 Identify
(B) {;l€,r (D) e,zlea
which one of the following will NOT satisfy the wave equation.
(A) 5oeit''-sa
(C) cos(l+ st) f*lcQ
(B) sin [a.' (10 z + 5t)] (D) sin(r)cos(f)
?"4.26 A plane wave propagating through a medium
[e,
has its electric field given by E:0.5Xe-t'ls)rin11gsl impedance, in ohms is
.t
i*:r
b
@)'I'
2
(N lf e,Ll€t (C) etf e,z
McQ
d
(A)
377
(c)
182.s
/14'
:
-
(B) 198.5/180" (D) 1#q
8,
u,:2,ando
Bz)y lm. The
\ ,,)
*,e
r.a"z7
The time
average poynting vector,
E:24ei(at+P") auY f m in free space is
(A\
7f
(D)
7t
F
Fe
?.4.28
7.4.2S
W
/^r, for a wave
with
Electromagnetic Waves
$,
-&o"
The wavelength of a wave with propagation constant
@) '
+n /0.05
(B) lom
(C)
20 m
(D)
30
(0.1.tr
I fi.2r) m-r is
m
The skin depth at 10 MHz for a conductor is 1cm. The phase velocity of electromagnetic wave in the conductor at 1,000MH2 is about (A) 6 x 106m/sec (B) 6 x 107 m/ sec
(C) 3
x
108rn/sec
a'
(D) 6 x 108m/sec
A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incide't power that is reflected from the air-glass interface is (A) o% (B) 4%
(c)
20%
(D)
some unknown material has a conductivity of of.
4tr X
I0 ? Hlm.
(A) 15.e pm (C) 25.e pm
h
z.e.ez
mho/m and a permeabilitl l GHz is
The skin depth for the material at
(B) (D)
20.e pm 30.9 pm
The plane wave travelling in a medium of e" I, - F,: 1 (free space) has an electric field intensity of 100^/i V/m. Determine the total energy densitl' of this field. (A) 13.9 nJfms (B) 27.8nJf m3
(C) Ee I
100% 106
139 nJlm3
(D)
278 nJ f m3
7.4.33
For a plane wave propagating in an unbounded medium (say, free space), the minimum angle between electric field and magnetic field vectors is (A) o' (B) 60. (c) 90" (D) 180"
BQ 7.4"34
For no reflection condition, a vertically polarized wave should be incident at the interface bet*een two dielectrics having 6r : 4 and €z : g, with an incident angle of
t""-'(?)
(B)
t""-'(;)
(c) ta"-'(3)
(D)
t*-'(#)
(A)
cQ 7.4"35
The eiectric field comporrent of a wave in free space is given by
E:10cos(107t+ keauylm Following is a list of possible inferences 1. Wave propagates along au 2. Wavelength ,\ : 188.5 m
3.
Wave amplitude is 10 V/m
:
Page 468
Chap 7
B\ 4a, 1f
-4a, (C) 4a. tQ
in
:
4. 5.
:
iad/'m Wave attbmiat6s as'it travels Which of these infererrces can be'drawn frorn'.8 ? (B)2and3only (A) 1, 2,3, 4 and 5 (D)4and5only (C) 3and4only
Page 464 1
Chap 7 Dlectromagnetic Waves
rd*c
?"4,36
Wave number
0:33
A plarre wave is generated under'water (e : 81e6 and, pt, - p)' The wave parallel polarized. At the interface between water and air, the angle c I which there is no reflection is
Ms& ?.4.3?
(A)
83.88"
(c)
84.86"
(B) 83.66' (D) 84.08"
An elliptically polarized wave travelling in the positive z-direction in air
r ardy components
E, : 3sin(ot- Bz)Y lm E, : 6sin(c.rt- Bz* 75') V/m If the cha"racteristic impedance of air is 360 f), the
average power per
area conveyed by the wdve'is
(B) a w/m'? (D) 125mW/m2
(A) 8 w/m'? (C) 62.5mW/m2 h{$$
7.4,38
The intrinsic impeda,nce of copper at 3 GHz (with para'meters F: 4tr x 10-7 Hlm;e : !0-7e l36n; and o : 5'8 x 10? mho/m) wiII be (A) (C)
(B) 0.02erl2 ohm (D) 0.2eftla ohm
0.02erla ohm 0.2eht2 otun
s$cq 7.4.39
In which direction is the plane wave .E: 50sin(108 t*22)anV/m, (where is the unit vector in y-direction), travelling ? (A) along y direction (B) along y direction (C) along z direction (D) along z direction
t.lcQ 7.4.40
If.
E
: (a,+ ja)
e-jPz
, thetL the wave is said to be which one of the
?
(A) Right circularly polarized (B) Right elliptically polarized (C) Left circularly polarized (D) Left elliptically polarized
7.4.41
Page 465
Chap 7 Electromagnetic Waves
What must be angle 0 of acorner reflector, ,rr"n tn"J an incident wave reflected in the same direction ? (A) 30' (B) 45" (c) 60' (D) 90' 7.4.42
is
Whieh one of the following statements is correct ? A right circularly polarised.wave is incident from air onto a polystyrene (e,:2.7). The reflected wave is
(A) right circularly polarised (B) left circularly pola,rised (C) right elliptically polarised (D) left elliptically polarised 7.4.43
The electric field of a wave pro.pa,gating through a lossless medium (ph,8160)
is .E: 10cos(6r x 108t - br)a, What is the phase constant B of. the wave ? (A) 2r rad/m (B) 9r radfm (C) 18n radf m (D) 81rad/m
rcQ
7.{,44
If the
phase velocity of a plane wave in a perfect dielectric is 0.4 times its value in free space, then what is the relative permittivity of the dielectric ?
(A) 6.25
(c)
v
2.5
@) 4.25 (D) 1.25
7.4.ds In free space.E(r,t):60(ut-2r)ooV/m. What is the crossing a circular area of radius 4 m
w (c) 120 w (A) 480
average power
in the plane r: consta.rrt (B) 340 w (D) 60 W
7'4.46 A
?
plane electromagnetic wave travelling in a perfect dielectric rn-_edium of intrinsic impedance 4r is incident normally on its boundary with another perfect dielectric medium of characteristic impedance rl2, The electric and magnetic field strengths of the incident wave are denoted by E1 and H1 respectively whereas E and 1/, denote these quantities for the reflected wave, and E1 and, Ht far the tra"nsmitted wave. Which of the following relations are correct ?
1. E;: 2. E,: 3. E1 :
r11H;
TtH, r12H1
Select the correct arlswer using the codes given below (A) 1, 2 and 3 (B) 1 and 2
(C) 1 and 3
(D) 2 and 3
l) Ll
Page 466
?"4.4?
, A
Chap 7 Electromagnetic Waves
plane electromagnetic wave travelling in a perfect dielectric -"0,,rof dielectric constant er is incident on its boundary with another perfect
| I dielectric medium of dielectric constant ez. The incident ray makes an andc I of dr with the normal to the boundary ,,rifu,"". The ray transmitted into th I other medium makes an angle of dz with the normal. I If e1 : 2e2 and 4:60', which one of the following is cqrrect ? | (A) 02:45" (B) d,.: sin-10.433 (C) qr: sin-10'612 (D) There will be no transmitted
i I
?.4.4s
I I
I
wave
Match List I (Nature of Polarization) with List II (Relationship Betwee" .f and Y Cornponents) for a propagating wave having cross-section in the Xlplane and propagating along z-direction and select the correct answer ,
List-I a. Linear
List-II
1. X and Y components are in same phase components have arbitrary oh"* b. -Left circular 2.
#r"llj"""
:
abcd
(A) r423 (B)4123 (c) r432 (D)4t32 7"4"4s Match List I with List II and select the correct answer
:
List-II
List-I a.
Propagation con3tant
1.
{w;E
b.
Radiation intensity
2.
*@')
c.
Wave impedance
3.
E,f Hl
4.
ExH
Codes
:
abc
(A)r23 (B)432 (c)132 (D)423 7"4.5Q Assertion (A) : For an EM wave normally incident on a conductor surface the magnetic field ,E[ undergoes a 180' phase reversal and the phase of electric field ,E remains same. Reason (R) : The direction of propagation of incident wave will reverse after striking a conductor surface. (A) Both A and R are true and R is the correct explanation of A
b.;1,
I I |
c. Right circular 3. X component leads I/ by 90" d. Elliptical 4. X component lags behind Y by 90' Codes
I I | I
I
(B) Both A and R are true but R is Nor the.coriect,explanation of A (C) A is true but R is false (D) A is false but R is true
If the -o field of a plane polarized EM E: e,,E,t auEu then its If field is:
(t)
wave travelling in the z-direction is
",h- ",ft
(B)
aft
aft+ ",ft
(D)
-""ft- "'ft
e) ",*-
In a uniform plane wave, the value of EIH is (A)
(c)
liE
@ '\trn (D),/G
1
The phenomenon of microwave signals following the curvature of earth is
known
as
(A) Faraday effect (C) wave tilt
(B) ducting (D) troposcatter
which one of the following statements is NoT correct for H : o.Ee-o t,cos (1061 _ 2r) a, Af m (A) The wave frequency is 106 r.p.s (B) The wavelength is 3.14 m (C) The wave travels along * r-direction (D) The wave is polarized in the z-direction
a plane wave
with
The vector magnetic potential of a particular vrave traveling in free space is A: o',A,sin(wt- pz) wherc A, is aconstant. Tlie expression for the electric field will be given by
(A) - a"pA"sin(ut- Bz) (B) - arliA,sin(ut- pz) (C) - aruA"cos(wt- gz) (D) - a,uA,cos(ut -,pz)
L'-" t
l
f t I f
i
'*'
In free space.EI field is given as H(z,t):-Ufcoslr,r (A) 20cos(ot-t Bz)a, (B) 20cos(c.,'tt- Bz)a" (C) 20sin(r..'t* Bz)au (D) 20 sin(c..'t -f Bz) a,
If
electric field intensity phasor
fr.: 10e-Au a"V/m.The (A)4x3x108 (B) ay x B x 108 (C)rx3x1o8 (D)10x3x108
tI
pz)at, E(z,t) is
of an EM wave in free space is
angular frequency
r"u,
in rad/s, is
Page 462
Chap Z Electromagnetic Waves
:
Paeg
fa
I'IGS ?"4.$$
Chap 7
Electpomagnetic Waves
Assertion (A) : Electromagnetic t\:aves propagate being guided by par plate perfect conductor surface. Reason (R) : Tangential component of erlectric field intensity and normal component of magnetic field intensitv are zero on a perfect conductor surface(A) Both Assertion (A) and Reason (R) are irrdividually true and Reason (R) is the correct explanation of Assertion (A) (B) Bottr Assertion (A) and Reason (R) arc individuallv tnre but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true
?-d"ss A uniform
plane wave is prclpagating in a material for which e
:
4eo, p
:
7pn
o: 0. The skin depth for the material is (A) zero (B) infinity (C) 28 m (D) 14 rn
and
?.4.60
Consider the following statcments : 1. In conducting medium the field attenuates exponentially with increasing depth.
2.
Conducting medium behaves like an open circuit to the electromagnetic field.
3. 4.
In lossless dielectric rclaxation tirrre is infinite.
In charge-free region, the Poisson's equation becornes Laplace's equation. (A) 1,2 and 3 only (B) 1,3 arrd 4 only
(C) 2, 3 and 4 only ll3cl
7.4"61
In free
(D) 1, 2, 3 arrd
space
:
E (2, t) 60r'cos(c,,t Az) a, \r f rn. The average power crossing a circular area z:constant is
-
(A) 16zr wattfm2 (C) 14rwattfm2 MCQ ?,rt.s2
In free
4
\
of
?r square rnetres
in the plane
(B) l5; rvatt/rrr2
{l)) l:tru'alt/rrr''
space
E (2, t) : 72Ur cos(at - PZ) a, Vn What is the avera,ge power in Wm 2 'J
(A) (C)
7'4.63
(B) 60na, (D) 720ra.
30zlo, 90tra"
The electric field of a uniforrn plarre wave is given by : .E : 10sin(3zr x 10Ei - rZ)a,* 10cos(3r x What is the corresponding magnetic {relcl Il
(A)
$sin(32'x (n) $sin(3zr x
10if -nZ)a,,+Scos(3zr 108,
-nz)(-
Ssinl3a' x
108t
- rZ)a,*
(o) Ssin(3zr x
108i
- rZ)(-
1C)
x
10rl
o,S+dS"os(3zr
* rZ)a,,ym-l
-.nz)(-o,)Am
1
x 10sr- ,Z)-(a,)Am-,
l$cos(3zr x 10if
a,) + l*"!sin(32'
1081
x
- rZ)(a,) 10sr
Atn
- nZ)(*
1
c,,)
Am
1
Consider the following sta,tements in connection with electromagnetic waves
Page 469
Chap 7
Conducting mediurl bchaves like an open circuit to the electromagnetic fir'ld.
2. At radio and rnicrowave
frequencies the relaxation time is much less
than the period 3. In loss-less clielectric the relaxation time isEnite. 4. Intrinsic impedance of a perfect dielectric medium is a pure resistance. Which is these staternents is/are correct ? (B) land2only (A) 1 only (D) 2, 3 and 4 (C)2and3only
7.4.6s Assertion (A) : The velocity o{ electrornagnetic
waves is same is same as
velocity of light. Reason (R) : Electrons also travel with the same velocity as photons. (A) Both A and R are trtre and R is the correct explanation of A (B) Both A arrd R are true but R is not the correct explanation of A
(C) A is true but R is false (D) A is faise but R is true Which of the following is zero as tr,pplied to electromagnetic field (A) grad tliv ,4
?
(B) div grari l'' (C) div curl A (D) curl curl A tQ
7"4"6?
What is the Poynting's vector on the surface of a long straight conductor of m,dius b and concltrctivity o which carries current / in the z-direction ?
(A)-2!,t)\i,
@)#i.
t.oi. (C) ' or0-
(o)
12
r2
mc&
?,s"s{3
*il
Corrsidel the fbllt-rrving statements regarding EM wave 1. An IIIVI wave irrcident on a perfect dielectric is partially transmitted and ptlrtiall;' rellected
2. 3.
ittcidertt on a perfect conductor is fully reflected When an ENI wave is incident frorn a more dense medium to less dense rnedium at an angle eqr,ral to or exceeding the critical angle, the wave suffers total internal reflection Arr EN{
waver
Which of the statements given above are correct (A) Orrly I and 2 (B) Only 2 anr.l 3
?
(C) Only 1 and ll (D) 1, 2 and 3 M*S
7"4.*.$
A unifbrm plane wave has a wavelength of 2 cm itr free space and 1cm in a perfect dielectric. What is the relative permittivity of the dielectric ? (B) 0.5 (A) 2.0 (D) 0.25 (c) 4.0
Eleciromagnetic Sfaves
Page 470
ilca
7.4.?0
Chap 7 Electromagretic Waves
rucd ?.4.7{
meolum
the medium *tne uv"Duq'Iru of dielectric constant
15 MCq 7.4,72
it-
["o]
In the wave
equati
il**"t'r='rc Q) *rca
7.4.73
?
: p#dtp'# r *"* *g
r,
p"#
(D) All of the above
for
| I
three
I I
:
Poisson's equation finds application discharge
2'
which term is responsible
+
on Y2E
Consider the following statements
1.
:1
in
vaccum tube and gaseous
Problems
and potential
Gauss,s iaw is useful for determining field about bodies having unsymmetrical geometry'
I I distribution l
nerds
I
3'Forthepropagationofelectro-magneticwaves'thetimevaryingelectricl
4 lil1H'j,Tffi::fi":?]'l'#ls"'c correct Which of the statements given above are (A) 1, 2 and 3
I
?
(B) 1, 3 and 4 (C) 2, 3 and 4 (D) 1, 2 ard 4 ilcq 7.{.74
What is the phase velocity of plane wave in a (B) (A) ,tfTtfry
uctort?
Q)\m) tuco 7,4,75
propagating in z-direction is The instantaneous electric field of a plane wave
E(t) :la,Etcos
r,.ri
-
ouE2sinut]e-ik"
This wave is (A) Linearly Polarised (B) EllipticallY Polarised (C) Right hand circularly polarised (D) Left hand circularly polarised tt
co
7"4.76
electromagnetic wave has Assertion (A) : Skin depth is the depth by which been increased,to 37% of its original value'
Reason(R):Thedepthofpenetrationofwaveinalossydielectricincreases
i.l.
\ $
t
with increasing wavelength. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true
I
r F"
7A.77 Which one of the following is the correct terms of vector potential A ? (A)
r
L 7A-7& t
l
t
rB) v2,4 -PtA
v'?,4.
€
(D) v"4
-
r)V:-lt"
pr# :-
!
I I I
Which one of the following statements is correct ? The wavelength of a wave propagating in a wave guide is (A) smaller than the free space wavelength
7A'7* Wrich one of the following
statements is correct ? For a lossless dielectric medium, the phase constant for a travelling wave, B is proportional to (A) r"
l'r
@)
I
J;
(c) | le,
i
(D U ^[;
It
L- 7.4.80 In a lossless medium the intrinsic impedance rl :60tr and p. : the value of the dielectric constant e" (A) 2
t I
(c) Ir
xFO
pJ
(B) greater than the free space wavelength (C) directly proportional to the group velocity (D) inversely proportional to the phase velocity
F
I
-d:+--J dF:-r (c) v2A #:- pJ
electromagnetic wave equation in
7.4"81
HCQ ?,4"En
4r'
S'li"*z
4
1. What is
?
(B) (D)
1
8
An electromagnetic field is said to be conservative when (A) V'E: p€(A'q Af) (B) v'?fl: p€(A'zq Af) (C) Curl on the field is zero (D) Divergence of the field is zero Given that .E[:0.5exp[-0.1r]sin(1061 -2r)a"(Alm), which one of the following statements is not correct ? (A) Wave is linearly polarized along a, (B) The velocity of the wave is 5 x 105 m/s (C) The complex propagation constant is (0.1+72) (D) Thgm,ve is traveling along a,
For a conducting medium with conductivity o, permeability p, and permittivity e , the skin depth for an electromagnetic signal at an angular frequency o is proportional to (A) (B) tlu " (o) tlp Q) tl^[;
Page 471
Chap 7 Electromagnetic Waves
Page 472
lleea ?.4"84
Chap 7 Electromagnetic Waves ,t
The electric field of a uniform plane wgve is given by a, * 10cos(ot- rz) an(Y lm) , ( : ., is The pola,rization of the wave
(A)
Circular
i:l
:lH:''
I I
i
i
4= 10s .'. j'
'
(D) Undefined mca
7.4"s5
In
free space
E(z,t)
,Er(2,
t)
:
0.10 cos (4
x
10]
4
t
pz)
a'
A'f
m;
T.he
\s
(A) .E(z,J) : 37.7cos(4 x 10?t -'pz) a;' i ' (B) E(z,t):2.65 x 10cos(4 x 10?' *ifJ'z)a" (C) E(z,t) : 37'7cos(4 x 10?t -'P')o' (D) E(z,t): - 37'7 cos (4 x t0? t .* 04 dt
'
A plane wavewhose electric field is given bi- 'E: l00cos(r'ot : ( noimally fiom a mateiial 'A"having 'eo = 4, p,:1 and'a ,B' having €, 9i p, 4 and 'o : 0. Match iterns in List I w = = select the correct answer : .,ti
illCA 7r4.8S
,
List
',i'
a b c d
tiat rI
I
Intrinsic impedance of medium
'B'
coefficient TYansmission coefficient
Reflection
Phase'shift constaht of medium
Codes
1' 6t 2' 80zr' 3' ll7 4' 8 '7.' '
'
iA'
'
:
abcd
(A)4123 (B)2341 (c)432t (D)2143 MCQ 7,4.87
In free space E(z,t) : 5}cos(at - 0z) a, V f m and H(z,t) :5/72rcos(ut- pz)au Afm. The average power crossing a circular in plane z: coristant is area of radius ^[Zl ^ (A) 2oo w (B) 250 W
w (D) 350 w (c)
MCO 7.4,88
3oo
Consider a plane electromagnetic wave incident normally on the surface of a good conductor. The wave has an electric field of amplitude I Y lrn and the skin depth for the conductor is 10 cm. Assertion (A) : The amplitude of electric field is (l I e'z) (v lm) after the wave has travelled a distance of 20cm in the conductor. Reason (R) : Skin depth is the distance in which the wave amplitude decays to (1/e) of its value at the surface. (A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false (D) A is false but R is true
. t.4.ae Three media
i'1'
-
.i
are characterised by
1. €,:8,h:2ro :0 2' €": 1rF,: 9,o : 0 3. €,:4,F,:4ro:0
e' is relative permittivity, 11, is relative permeability and o is conductivity. The value of the intrinsic impedances of the media l, 2 and B respectiveiy
aIe
(A) 18S Q, JTT O and 1131 e (B) 377 CI, 1131 O and 188 e (C) 188 O, 1131 Q, and gTT e (D) i131 Q, 188 e, and BZT 9
7'4.eo A plane EM
(8,,H,) travelling in a perfect dielectric medium of surge strikes norrnally oR an.infinite perfect dielectric medium of surge impedance 22.lf the ref[acted EM wave i, (E",rI,), the ratios of Eif E, and H,lH, are respectivelv (A) 3 and -3 impedance
wave
'Z'
(B) 312 and r/z (C) 314 and 3/2 (D) 314 and 2/B
t'4's1 For a perfect
conductor, the field strength at a distance equal to the skin depth is xYo of the field strength at its surface. The value ,wo, is (A) Zerc
(B)
50%
(c)
36%
(D) 26% *<*******,k**
Page 473
Chqp z Electromagnetic Waves
sol-urloils
Page 474
Chap 7
7,1
Electromagnetic lilaves
!i{:L ?"1"'l
Option (C) is correct. FYom the property of phasor, we know that the instantaneous electric field is the real part of {8,"i"}.
E(r,t): R"{8,"''}
where .8" is the phasor form of electric field. Given the electric field intensity in time domain, E (*,
q
::,:":.ir,l:: ; :;l*:,, ll
:-ff"
",
ar"i(d-a46r+ C'C'
where C.C. i; complex conjugate of the 1" part' So, using the property of complex conjugates we get
E(r,t)
:2rt"{-+"-aa
"i@t-oxt
^}
: P,;e{- jEoe-ax 4u} "-i1x "iut get (1), we equation with it Comparing
E, s$L
?.'1.2
:-
jEoe-P+iF)'onY
l^
Option (A) is correct. Wave equation for a plane wave propagating in -f a" direction is given
#[-,"#:o
a^s
where o, is the velocity of wave propagation
Now from Assertion (A) the electric field is
B It
:
Eosin( z)cos(ct)a, it satisfies the wave equation
represents the electric field of a plane wave if
i.e. #- o#:o FYom
where c is velocity of wave in free space
the given expression of field intensity we have AE
dt
oft Thus, we get,
cEosin(z)sin(cl)
a2E :-ar
ort and
:-
AE
lfr. .'0,r,
a
oz
:
c'Eosin(z)cos(ci) Escos(z)cos(ct)
A2E : -{ dt
- "'ol^4 -dt
Eosin(z)cos(ct)
:o
fies the wave equation so it represents tbe Since, the electric field wovE. of 6a plane field uI .tlul(l .. Pl@[g wave. $yt-. .! Therefore, A and R both are true and R is the correct explanation of A.
[-.,,,
Option (B) is correct. Given the magnetic field intensity in free ,p*" i, .EI : o.1cos(1oet _ By)a"Alm (1) The general equation of magnetic field intensity of the EM wave propagating in a, direction is given as
II :
flscos(wt_ Bg)a"Alm
Page 475
Chaq 7 Electromagnetic Waves
(2)
Comparing equations (1) and (2) we get, direction of wave propagation, &*:ay
l
and angular frequency, So, the phase constant of the wave is
o:
a ,r:q- 10n c 3x108 : 3.33 rad/m
10e
rad/sec
(c is velocity of wave in free
Now, electric field intensity in free space is defined
space)
as
fi :_qsa6 X H where 16 is intrinsic impedance
in free
space and a6 is d.irection of wave
propagation.
So,
E --ZT7(ar) x 0.1cos(10nt- gy)a"
:-
(rn:
377
a)
37.7cos(1oet - B.lJy) a, Therefore, electric field intensity of the wave at U:lcm at t:0.1ns is E - - 37.7cos[(10,)(ro-'o) _ (s.33)(ro-,)]o,
:-
i
tol- 7,{.4
37.6a,Y
lm
Option (D) is correct. Given the instantaneous electric field in the free space is B : (bau - 6a,)cos(c,.,t- b}z)y lm So, the phasor form of electric field intensity is E" : (Ea, _ 6a,)e-iso" V f m The phasor form of magnetic field is given in the terms of electric field intensity as
n":fi@)x(E) where oa is the unit vector in the direction of wave propagation and intrinsic impedance in free space. So,
n"
Lfie)x (5o, -6a,ye-Fo'Yfm : t a,, - 6ar) v /^ #(-
176
is the
(at,:
a")
e-no"
:-fi{uo,+6ou)e-no"yfm sol
7,1.5
Option (C) is correct. For any electromagnetic wave propagating in a medium electric field leads magnetic field by an angle d,, where d, is the phase angle of intrinsic
impedance given as
ta420*
:'La€
Now, for a perfect conductor
o-1=* p
I
I I I
i
Page 476
I I
Chap 7 Electromagnetic Waves
so, electric o"ro r*r:-llretic field Iags electric field bY
45'.
'I
$sL 7"{,6 i
i
w
"l
4b'or in other words masnetic
Option (A) is correct. in phasor Given, the electric field intensity of the wave
I
field
I form
I
: (5a,* 104')e'7ax-22)V lm as So we get the directio.t of *u"" propag,ation
i
E,
i
ar:ffiffi:*#?:w iJr*
Therefore, the phasor
of magnetic field intensity of the Rlane wave
given as
H,
where
:^r(tnt) sol.
7.1.7
29.66e-il4b22)
rlo
is
I I I
:-t-:-^:^ l.--^ronno i | is intrinsic impedance in free snace
x po,+
1oa,1e-\a"-2")vlm
I I
mLlm
I
Option (B) is correct.
is given The time average power density of the EM wave
g*"
I
-
1 : Laax E,
:-
I
I
as
I
: {^oo
I I
whereEisthemagnitudeoftheelectricfieldintensityofthewaverottsI and the unit vector in the direction of wave propagation impedance in the free space' So, we get
JgTry (4ao-J9"\ : (D ' ave - 2(t2oir) \ ./zo ) $0r-
?"1"8
To
is the intrinsic
I l
L48.9a'- 74'L5a"Watt/m2
Option (B) is correct. ,q., tne given electric {ield vector has the amplitude Eo e2J6 a'+ JE ar- a") So in the same direction the wave will be polarized'
:
sol- 7.{.9
Option (A) is correct. Flom Maxwells' equation we have
V x.l9 Given
oft So,
:-#
E :100cos(ut- Bz)a'
E: 100Bsin(a,'t- Bz)au :-#- v x,E : loopsi".(wt- Pz)a, Yx
Therefore the magnetic flux density vector rs
g
: -199€."or1 I no1sin(ot - Bz)a,d't :
ut ,/
-
pz)oo
(u:ffi\' P'reo
:3xlOlocos(art-Ar)ou
sol-
?.'!.{0
Option (A) is correct. -.: Poynting vector in an EM field is defined
.:.
r
.i,
phge 4?Z
as
Chap Z
IP:EXH .
ElectromagneticTfaves
where .E is electric field intensity and .EI is the magnetic field intensity in the region. Now, the electric field intensity in the region is given
as
E :100cos(wt- Bz)a" and as calculated in previous question the magnetic field intensity in the region is
B: Jx
1010cos(art
pz)ao
-
So, the poynting vector in the field is
P:Ex# :
[100coi(c.,
[3
t- Bz)a"lt
x
10'0cos.(al- Bz)a'] P4
IAz\a, - pt f4"os2(ut- 3 xP41012" os2(wt- \-' \ V lln
Bz\a. t--t--'
sol ?"1"1{ Option (B) is correct. Time average stored energy density in electric field is defined
w":f,eofl,.
as
E-,
where -8, is the electric field intensity in phasor form and
.OJ
is its conjugate.
Therefore, the average stored energy derisity in the region is
tr, =
f
(s sin nr"-t"12
"-t'4""
q) . (Ssinnre*r/'dfr""
au)
t
:ff-si^zTr lol. ?.i.'ta Option (B) is correct. Given the electric field
'
,E: In phasor form, E" :
l0sinzrysin(6zr 10 sin
x
zr y2-h/2 s-i/
108,
p"?
-JEm)a,Vfm
a"
So, from Maxwell's equation, the magnetic flux densi[y in the phasor form is given as
6,:!1v " Ja\
x
E")
:6tr x 108 as determined from the given expression of .8. i'/ira So, B" : @in n y) e-,t e*+ i6r{id (costr y) e- 6 e-fr n, ou #*
where a
Therefore, the time average energy density stored in the magnetic field will be
' of'
'tj)n
:
fi{a, ' nI)
where .Bl is trre conjugate of
B,
,^:ffi125*5osin2zrr)
sol z-1.,t3 Option (B) is correct. The reflection coefficient of the wave propagating from medium 2 is defined as'
1
to medium
i
Chap ? I
I i
I
f:m
Page 478
where ?h and ?12 are the intrinsic impedance of the two mediums lespectively. So, the reflection coefficient for the wave propagating from free space to a dielectric medium is given as
Electromagnetlc Waves
i
l
I I ]
I
n
' --T-rlo q+rh where 4 is intrinsic impedance of the dielectric medium and ry is intrinsic impedance in free space. since the intrinsic impedance of the dielectric medium is given as
l
- [7'"'r-\/ ttre-t/ 4€o-'n2 rnl2 - rn 'F
So, we have
no/2
*
rn
_rl2-7 __1
- Ll2+r-
3
Therefore, the magnitude of electric field of reflected wave is
E, : fEo:-* $oL
?.t.'ra
(Eo is the magnitude of incident field)
Option (D) is correct. Time period of wave propagating in a medium is given as :
f :T
whefe or is the angular frequency of the wave.
Given the magnetic field intensity in the free space is 0.3cos (wt - By)a" Alm .E[
:
So,
at t :Tl8 the magnetic field intensity is .Er
: 0.3cos("$ -
0u)",:
0.3cos(+
- gu)",
(T:2rlu)
Il
:0.3cos(0y-"1+) or, get the plot of fI versus y as shown below we Therefore
$0t ?.1"15 Option (D) is correct. Phase velocity of the medium,
up
:7.5 x
107
m/s
F, :4'8 o :0
Relative permeability,
(lossless medium) Conductivity SincephasevelocityofanEMwaveinamediumisdefinedas vp ",_ -
c
I
F"€,
where c is the velocity of wave ir1 air, p. is the relative permeability of medium and e, is the relative permittivity of the medium. So, we have
,le'il
thl
7.5
x
I07
:
(c:3x108m/s)
oIt s,'= 3.33 Now the intrinsic impedance of the medium is given
_ n: '' \/t_@ o *.iue -
:377
. as
:
lm:4b2.4e
(o:
0)
g77
0)
E":gsfl.z,r,Vl^
(1)
and the general equation of electric field phasor of an EM wave propagating is
E": pos-it3ia"yfm
e)
So, comparing the equations (f
) and (2) we get direction of wave propagation, ak:- a, and phase constant, 0 :Q.Jrad,/m and from the Maxwell's equation, the magnetic field phasor of the
given as
n,:rrloix
war,-e is
n"
where 4 is the intrinsic impedance of the medium and o; is the unit vector in the direction of wave propagation.
so,
u" : -4#4(- o,) x (ben., a")
: ffi"nt,
au
:
(at,: - a,)
r|.oiefli, aumA/m
and the angular frequency of the wave is given as a : pup : (O.BXZ.5 x i0?) :2.25
x
I0z
so, the magnetic fierd intensity of the EM wave in time domain is H(r,t) :Re{H"eto'} : 11.05cos(u.,f * 0.,8r)a, ' : 11.0bcos(Z.ZS x 107t+ 0.Jr)anmAlm
7.1.16 Option (B) is correct. Given the field intensities of the plane wave
E(r,t)
:
H(r,t): So,,we get
lEl
:
goocos(5
x
t06rt
-
as
l3r)arV
lm
- pr)a"ylm 900, l,Hl: 1.9, cu: 5 x 106n 1.9cos(5
x
106zrf
Now, the intrinsic impedance in the medium is
,t
:g:
lIIl-
gog
1'9
:473.2 -:ru'r
and phase constant of the wave in the medium is
o
:9-up
5=x 1o6zr : o.224m-r
7x107
Therefore,
o: m ,l:,/E:rh
and
0
Since, for a perfect dielectric
0
=fi:Z/t"r,
Comparing the equation (1) and (2) we get,
Chap 7
Electromagnetic Waves
Given the electric field intensity in the phasor form is
in a, direction
Page 479
(1) (2)
Page 4E0
"
Chap 7 Electromagnetic Waves
,*w):[*61p6 +.ffi, + L37
Again ftom equa.tioq (1)
,, ssl
7,1"t?
,,- -= (.H)\ X 5.37 : :_ (DY \4Tt.7 ) \t/ r," .
3.4
Option (A) is correct. plane wave propagating General'equation of electric field intensity of a Eo an4 '{requency o is gi free space l.t - o" direction having amplitude as: F : E.cqs(art* fln)a" is thq unit vector in th where B is phase con$tant oi tt'u wave and a' in i*"ir"* or pohrization of wave and Qince itt" Bnl wave is polarized *4" direction. So, !
(trn
and we get,
=
_:_
(tr,
(in free space
P = Eocos(ut+lr)a,
P: i)
wave is given as Therefore, the ryagnet.ic field intensity of the
p :fr(o,) x (E)
wave prop4gation and rp is tbc where or is the unit vector in the directiorr of intrinsic irnpedanie of the wave in the medium'
so,
Ir -frt-"")
x
= ffcos(,,t $oL
?"1.{s
[Eocos(r"'t
+ir)a"l
(o*: - a')
+lr)a,
Option (A) is correct. a plane wave propagating in General equation of blectric field inteqsity of free space is giveor as
:
E = Eocos(ut- k' r)a"
lc is the wave where o" is unit vector in direction of polarization'
: t' propagation with amplitude k : A^u r: aa,+ 11ro't+ zo" is the pmition veetor' origin to point (1,1;1)since, the warreis prqprgiing in tlre direction from in the direction of wave
So,
and since the fleld is polatized parallel
So,
*"
to r-z plane
=ffi
where
m
and 'n are const
Now,theelectricfie}clofwaveisalwaysperpendiculartothedirection propagatior,r of EM wave" So, we have
k'a-=0
lu
Igi &i
3211
, lrna' + nn,l =
g
[7\---;6-ll l{ara rn*n =0 ]
'ln :" n polarization of the wave is Therefore, the unit vector in the direction of
1 q, -
*o,*(-m\d,"'
az__sz
Paqe 4E1
puttingarthe"","",l"ffiffl,*]ftl""oricnerd,rrJ:;? o'\ rro" +uo^.+ za-\11a,-:a"\ 4 : Eo"orlrt-',(o'* 9=* ') .'('o'+a%*'")]1-m-) c\ lz I
: Eocos[r,.'i -
sol
7.{.19
ft;(*
+r
. 4l(W)
Option (A) is correct. For the microwave experiment the angular frequency is w :2rf :2n X 10 x 10e : 2zr X 10lo o. 6.25 x 707 So, (t€
(/:10GHz)
2zrXr.l0loxixasSx1oT : I.I2 x 108 >> 1
Therefore, the skin depth of the material is
6-r-A-\/E-
(olue >>
"w
1)
: \/ 2rx 1010x1x4ztx L0-7x6.2bx107 : 6.36 X 10-7 m : 0.636 F,m Thus, for the successful exppnirr\ent, wfdth of coatin$ must be greater thhn skin depth i.e.
sol. 7.{,20
r > 0.636 t > 0.64'pm
Option (p) is correct. Given, the electric field intensity of the plane wave .E = 3cos(107 t - 0.2y) a"+ ?sin(107t - 0.2y)a"V /n comparing it with the genera.I equation of electric field of a plane wave, we get Angular frequency, Phase constant,
a =107 0 =A.z
So, the phase veloCity qf the prqpagating wave
io-q
_u_10t_. u':7:ffi:5x1o7m/s ^, ort
-+:=5x10?
where c is velocity of wave in air and e" is the relatiyg permittivity of the medium. So,
^ /3 x 108\'
":tE*toE/:"o
Therefore, permittivity of thg ryediur.r,r is
€=its=36e0 Now, the complex perraittivity of the medium is given ac-e
where a4d Thus,
,,r, -JE
: €:36g0 -r, o 2XlO7 a 10' : e" (36e6 - P)F l^ €'
4s
Chap ? Electroriragnetic Waves
Page 4E2
sc'. 7.{"2{
option (A) is
correct.
1
Conductivity of all the metals are in the range of mega siemens per meter I and frequency of the visible waves are in the range of 1015 Hz. So, we can I
Chap 7 Electromagnetic Waves
assume
I
Conductivity of a metal = 106 S/m I Frequency of a visible. wave - 1015 Hz I Now, the attenuation constant of a wave in a certain medium is given as : I
":"uTfl@-1
Since for a metal, o
So,
>>
I
a€
I
":r.r[E^/E --T - \/@
Therefore, the skin depth of a metal is
r_1_t2__E - e- \/ ,tto - t/ 70ttx4rx
v
10-7x
106
= lnm Thus, the skin depth is in the range of nanometers for a metal and that's why the wave (visible wave) can't penetrate inside the metal and the metals are opaque.
i.e. (A) and (R) both are true and (R) is the correct explanation of (A).
sol
7"1.22
Option (D) is correct. since, the wave is propagating in free space so, the velocity of the wave is 3 x 108 m/s and the amplitude of magnetic field intensity in z:0 plane is given as
rro
: p! \o
Therefore, the plot of magnetic field intensity IIe versus time plane is as shown in the figure below :
t in z:
0
t(psec)
Since, the wave is propagating in * a" direction so' an amplitude which exists in the plane z: O at any time f must exist in the plane
z:(7 x So, the amplitude of IIo
10
6- t) x 3 x
108m at
will be equal to the Ht at t:
l:lpsec. 1 Jr,sec
for the plane
z:(106-t) x3x108m Thus, the plot of 1/r versus z will be as shown in figure below
Hr(Alm)
Page 4E3
Chap 7 Electromagnetic Waves
-300 -150 u 30L 7.1.23
r50 z(m)
Option (B) is correct. Given the electric field intensity in phasor form
E"
:
Es(ar- io,,)s-io'
So, the instantaneous expression of electric field intensity
"
will
_',i;!::::] .,,in (wt)t|
e-
be, i a.
:=Y",I,17 Eo (o, cos(cuf ) + o,sin( ,,t)) e-i1' Therefore, the magnitude of the field is
lEl:@ :
or,
l&f +lE,l'
Eo
which is a circular equation i.e. the wave is circularly polarized. Now, the insta,ntaneous angle 0 that the field E makes with y-axis is given
tanl
:
as
Eosinat
bocosat
0:ut
ort
Therefore as the time increases,
E
rotates from y
to z as shown in
figure
below:
and since the direction of wave propagation is in *a, direction so, the rotation from y to z obeys the right hand rule. Thus, we conclude that the field is Right hand circularly polarized. sol-
7,,t"24
Option (A) is correct. Given the phasor form of electric field intensity,
E,
:4(a"-
ia"1s-ioo So, thc electric field intensity of the reflected wave
E* : ffL(a,- jo,))"tun where
l'
will
be
is the reflectian coefficient at the interface. Therefore,
:4(-
o,"+ ja,)eiBY (for perfect conductor f :- 1) and the instantaneous expression of the electric field of reflected wave will
E*
be
,B
: Re{4(- a"* ja,)(cosat+ jsin,lrt)}eiB!
- 4(-
cos(at) a" - sin(wt) a,) Therefore, the magnitude of the reflected field is
Page 484
Ctap,7.
lEl:@
Dlectromagnetic Waves
or,
ejP'
IE
l'+l0rl':+
which is a circular equation i.e. the wave is circularly polarized' Now, the instantaneous angle 0 that E makes with z-axis is given
as
tanl:ffi 0:ut
So, as time increases, electric field -E rotates from z
figure below
to r
the
as shown
:
since the direction of wave propagation is a,long - aal sQl the rotation from z to r follows left hand mle. Thus, we conclude that the EM wave is LHC (left hand circularly) polarized.
sol. 7"t.25
Option (C) is correct. Given the electric field intensity of incident wave, 8," : !Oa"e (6Y+ 8x) So, the direction of wave propagation is
K :6ay*8a, Since the wave is incident on the perfect conductor so, the magnitude of the reflected wave is given as'
(f:-
E,o:-Ett:-Ija,
1 for perfect conductor)
The direction of wave propagation of reflccted wave wiil be along (6a, as shown in figure below :
-
8a,)
n:o Therefore, the field intensity of the reflected wave is
E" :-
10@'
''-i(6u-8t) T\us, the net electric field intensity of the total u'ave in free space after reflgction will be E" : 8,, * Eu : r1a"s {9u+a') 1 [- 10a" e-(6e-82)] : I}a"e fru (e-r8" - t') - - 720a"e rosin8r V/m
"
sor
7,,*.2s
Option (A) is correct. Given, the electric field intensity of the incident wave, Et, :25a"e-{62+8?) V/m So, the direction of the wave propagation is
k:
Pege 485
"Chep ? Electromegnetic lVaves
Ga,"-f 8au
Since the wave is incident on a perfect conductor so, the magnitude of the electric field of the reflected wave is
Ero
: -
(reflection coefficient,
E;o
** The reflected wave will
25a, propagate
in 6o. - 8a, direction
as shown
l-:-
1)
in figure
below:
Y:l) So, we get the electric field intensity of reflected wave as
E" :
-- 25A, e i(6'-8u) Y f n Since, the rnagnetic field intensity of a plane wave in terms of electric field intensity is defined as
H
:lbnx rlo' "
El
where o1 is unit vector in the direction of wave propagation and 4s is the intrinsic impedance of free space. So, the rnagnetic field intensity of the reflected wave is given as
H,":f,,@^yE,,) where, So, we get
on
H,"
6a, 8a,, : lc i: ffi,-- : (0.0a, - 0.8o,)
-'ffi1t0.e""- 0.8ar) x (- 2sa,e-'to'-8n))] : *ror, [(- 15 au - 2o a,) e- t(6"
: - (# + sst- ?"i"37
-8!) e- x6 "
67T.9')
au)1
A/m
Optiorr (B) is correct. The general expression fbr phasor form of electric field vector is E,: Eoe 4t*B'!-a:) Comparing the given field with this expression we get,
0,r* 0,a* l3"z: o.o1z'(-3r* /Ey-2z) So, the propagation vector is
k
:
V (13,r+
/ry + 0,2):
0.01ur(-3
a,* JE au-
Therefore, the direction of the propagation of the wave is
Jo")
k
-3a"+'/3ou-2a" -to-L!'-LA 9+3+4
- - k -r'
Page 486
Chap 7 Electromagnetic Waves
:|1-n,+/3ou-2o,) sol- 7"{,28
Option (D) is correct. FYom the given expression of the field vector, we have the propagation vector,
k
: *(JE a,-2ayr, y
So the phase constants along
o,:*,
an.d z-axes are
on:-r#,
D
3o,)
g,:-*
Therefore, the apparent wave lengths along the three axes are
:T: #^: #: ^,
28'87 m
\-rE-l
:ffi:pg1:125m \2r2r \": sol.
7.1,29
:
2r
21501
w n:.+rt:+? ll-rr/l
:
16'7 m
Option (B) is correct. As determined in previous question, the propagation vector of the plane wave is
k:#(J3a,-2au-3o") Therefore, the direction of wave propagation is
o,:I So
-
&-?'--e
the phase constant along the direction of wave propagation is A
: k' at:0'I6r
Therefore, the angula,r frequency of the propagating wave is
u
:
apg
:(3 x 108) x (o.tOr)(In free space up: 3 x 108m/s)
:1.51 x
108rad/sec
So, for the determined values of apparent phase constants question, the apparent phase velocities are given as .. (t 1.I1 X 108 : 6.93 x 108 m/s
up":
in
previous
p, - lJJr\
\E/
u"-6, ^, -u -1.51 L108:6x10sm/s , ano sol.
7.1,30
l-'*-l
-. r.., _ 1.bL X.108 :+^. :4x 108m/s up,:8,:l+l
Option (D) is correct. The necessary condition for the vector field E - Eoe-iP to represent the electric field intensity of a uniform plane wave is
k'Eo:g where lc is the propagation vector of the wave and
.Eo
is the amplitude of
the electric field intensity of the plane wave. Now, we check all the options for this condition. (A) From given data we have
k :,/5 an* Eo
So,
:- ja,-
given
a"
2o,u*
j/E
a"
k. Eo:-2^/3+jJg+0
(B) Flom given data we have
: a, - J2au k:e,l/Ea" k. Eo:1-3*0 Eo
So,
,/E a"
(C) Flom given data we have
E,
k
so.
: (/E + t|)".+(r + 4)"-
ili
o"
: ^/i a,*3ar*2a,
k.Eo:t*++r++-plE+o
(D) Flom given data we have
E,
: (-Ji - i+)".* (t - j+)"*
jJE
a,
k:"/ia"*3ar*2a"
So,
k.
Eo
q /; /o --3- jt**B-j+L+pJE:0
So the vector represents electric field vector of a uniform plane wave.
sol ?.d"s1 Option (D) is correct. For the field vectors -8" and
.EI" defined as
E, : Eoe ja and H, : l{us-i? The condition that it represents the field vectors of a uniform plane
Eo'Hs:Q,
Eo'k:0and Hs. k:0
wave is
k is the propagation vector of the plane wave. Now, we check the all given pairs for this condition where
In Option
:- ja, - 2ar* j,/5 a" Ho - a, - J2aa - ,/E a" k:,/Ta,,*a" Eo. Hn:- j+il- jg:0
(D)
and So
Eo
k:- j/i+ j,/E:g Ho'k'^/E-/5:o
Eo. Therefore,
it
represents the field vectors of a uniform plane wave.
sct ?.'t"3* Option (A) is correct. For a propagating electromagnetic wave, the field satisfies the following Maxweell's equation.
V..E:0 Y xE:-0-P +o dt Now, we check the condition for the given fields as below.
page 48? Chap Z Electromagpetic Waves
*''
Page 4E8
Chap-7 Electrpmagnptic. Waves
So, and i.e.
P
{'} 6osin(ut* ratr)a, V.P:0 V x P--600cos(a.,f*10r)au*0
-*',n.
is a possible EM field.
agarn,
So,
and
Q
: Q"or( ut-
2p)aa
V. Q:0 v x Q :|ftlroo.s(r.,r-
2p)!a"*
o
i.e. Q is a possible EM field
R :Sp2cotSao+lcosSa*
So, i.e.
v
v.R:f,Strt"otOyff+o
R is not a possible EM field. ,S
Sp,
v
:
lsindsin(cuf
.s=J
r" stn
-
6r)aa
0^sirtut-arf
i.e. S is not an EM field. Thus, the possible EM fields are P and Q. ******:k****
\Y'6) +o Or
sol.uTloNs 7,2
Page 489
ehap 7 Elecirbriraineiii'Waies
Correct answer is 0.2 . Given the magnetic field intensity,
Ir :
l0cos(6 x 107r _ ky)a" Alm Comparing it with the general equation of magnetic field. g : /locos( ut - ky)a, A/m
:6
get,
ar x So, the wave no is,
We
k sol.
7.2.2
1'07
: t: H#
:0.2
(c is the velocity of wave in free
space)
Correct answer is 2. Given magnetic. field intensity in the non magnetic medium is .Er
Comparing
We get,
and
:
1.5cos(roet- Sz)a,Afm
it with the general equation of magnetic field 1Y : f/scos(wt - Bz)a" Alm
intensity
: 10e rad/ sec 0 :5.
ru
So, the phase velocity of the wave
in the medium is given
as
,r:ft:#:2x1oqm/s sol.
7.2.3
Correct answer is 0.5 . wavelength of an electromagnetic wave with phase constant is defined as
p in a medium
A:2I -p so, the phase constant of the wave in terms of wavelength can be given
p sol-
7.2.4
:+ : &:
o.b
(,\:
rad/m
Correct answer is 2.25 Given the electric field intensity in the nonmagnetic material E : gcos(4 x 108, _2r)arylm Comparing it with the general equation of electric field We get,
g : flgcos(wt- pr\ou A]m a:4 x 108rad/s
and
0 :2rad/m
as
12.6 m)
as
So, the phase velocity of the wave in the medium is given by
'up:fr:2xL08mfs since the medium is non magnetic so, of the medium is given as
ltr:
pa and the relative
permittivity
Page 490
€,
Chap 7 Electromagnetic Waves
$fiL
?.2.5
:2.25 :(eY: /l tx lqY \uol \2 1oE/
Correct answer is 12.57 The general equation of electric field intensity of an EM wave propagating in a, direction in a medium is given as gr : Bocos( wt pr)au Alm .
Comparing
-
it with the given
expression of electric field intensity, we get
w:b x 108rad/s So, the time period of the EM wave is
, $sL
?"4,$
T
:+:a
, ,Zn.^u : 5x10'
12.57 ns
Correct answer is 3.93 . The general equation of electric field intensity of an EM wave propagating in o, direction in a medium is given as Comparing
B : Eocos(wt- pr)arA,lm it with the given expression of electric
field intensity, we get
u:4 x 108rad/s So, the time period of the wave in air is given
21 2tr u -4x108
' : Since
15.71 ns
in one time period the wave travels its one wavelength
taken by the wave to travel
t s{}L
?"2"?
as
)/4
:T:+
so, time
3.93 ns
Correct answer is 251.33 . Intrinsic impedance of any material is given
Tl:@ - t/ o* where
())
distance is
p is permeability, o
as
jae is conductivity and e is permittivity of the
medium. Since the given material is lossless, nonmagnetic and dielectric so, we have
O:0 p:po and
(lossless)
(non magnetic)
(ur:2'25)
€:€7€s:(2.2b)eo
Therefore the intrinsic impedance of the material is
s{il.
?.2.s
Correct answer is 0.8
llo
1.5
-
377 1.5
:
(^: ,[* : srr a)
251.3 O
.
Given, Flequency of the wave propagation,
: 0.5 MHz : 0.5 x 106 Hz o:B x 1o7S/m "f
Conductivity of medium, Relative permeability of medium, Pr:er=1 So, the angular frequency of the wave propagation is w :2rf :2n X 0.5 x 106 : zr x 106
and ---- we-
get o--
orx 107 a€ - zr X 106.3X 8.85 X 10-12 :0.1 x 1013 >> 1
Page 491
n-t""t--**rr"";"T;"1
Therefore, the phase constant of the propagating wave is given as
_T tra _\/ - fd4a
(olwe >>
1)
iffi
v-
:7695.29ndf m So, the wavelength of the radio wave in the medium is
,2n
.oL
2.2.s
^:?:o.8mm Correct answer is 9.08 . Given the magnetic field intensity of the plane wave in free space is H" : (2 + jl)( an + 2ja,) e-jB, Af m Flom the Maxwell's equation, the maximum electric field intensity of the plane wave is given as
:nolHl ,Ei I tmu I lmd where ri6 is intrinsic impedance in air and field intensity of the plane wave.
lf/ [*
is the maximum magnetic
I
Now, the maximum magnetic field intensity of the plane wave is given
III l^* : /ii; Ef
as
where .EIf is the complex conjugate of the magnetic field phasor.
So,
lIrL*: :
r/2g x 2o :24.1Alm Therefore, the maximum electric field intensity of the plane wave is lE l-* ,hlIJ l** 377 x 24.I 9.08 kV/m
:
tot.
z"2.to
:
:
Correct answer is 60. For an EM wave propagating in two mediums, the waverengths of the wave in two mediums are related as
!- tq E-r/ €,
where \ and )2 are the wavelengths of EM wave in two mediums with permittivity rr and €2 respectively. So, the wavelength of plane wave in free space is given as
&_ T: \/tI,, \:
\,/e,
where ) is the wavelength of the wave permittivity e".
So, SoL
?.2.{1
k :20"/g :
in the
medium with relative
(,\:
60 cm
Correct answer is 20. Given,
Conductivity of the glass, and relative permittivity of the
o
glass,
:
ro-12
€, :2.25
S/rn
20cm,
6":
g)
pel1itti$ty
Page 492
So, the
Chap 7
€ : €o€r:2,258o Therefore, the time taken by the charge to flow out to the surface is € (2.25) x (8'85 x 1o-L)
Electromagnetic Waves
of" gJass
is
-'-o = : 19.9 = 20 sec
soL
7"2-t2
,
)
I
Correct answer is 0.52 . Given the electric field intensity of the propagating wave,
E
:
"
: *NP/m and. o :
pr)arV lm (11 propagating plane il wave The general equation of electric field intensity of a" direction is given by (2\ E - Ese-o'sin(wt- Pr)aoYlm Comparing equation (1) and (2) we get, Eoe-"/3sin(108t
-
108rad/sec
So, the attenuation constant of a propagating 'ryave is given as
:,J +l/'*(#l *'] ^:,F+(#J "
Let
ffu,e,(n*
Therefore,
(rn-1):
ort
Now, we put o
1)
u2Q2
uf pn€oU,€,,
= 113,'p,= €;:4) u:
108 so, we get
^ 1_zx(rftf 4 '- (ro8f x (af
cB
JA-L--
('o
:
3 x 108\2 ro-7:2(YZ" " rtrx+
\
1
G
:3x108m/s\
/
a:|+t _9_g
14
fiw:& Thus, so|.
7.2,13
loss tangent
: #:
O.SZ
Correct answer is 8.33 . Flom the field intensity we get,
u:!}er
is given that, p,: 0.5, So, the phase constant, and
it
o:
0.01
S/m,
e,: 8.
0:, :
10ezr
poeo(exos)[Fm.,l
I
:
20.95
Page 493
*"*'i6 i'6 frrit a phase shift of L0' Pz:I0":ffirad
Let the distance travelled by the So,
Cbap 7 Electromagnelic tYaves
7r z-: 18t-@os4:6'rrmm -813 Correct answer is 542. The attenuation constant of a propagating wave in a medium is defined
as
":,J+WIW_\ Now, from the given data {rre have 1f,,:0.5, o : 0.01 S/m, e, : 3.
'.
So,
:1gnn
.u
,f@
:'0.9425 Initially the amplitude of the electric.field : 0.5 So, after travelling distance z amplitude of wave :0.5e-"". Therefore, the distance travelled by the wave for which the amplitude of the wave reducedby 4O% is evaluated as 60 0'5e-"' '= U.5 ^t" x ffi
,'
(0.e+2s)z
(amplitude reduces
9.6
-
z:6&rstt(#)
oft
to 60%)
:s42mm
Correct answer is 0.796
skin depth (d) of any medium is defined as the reciprocal of attenuation constant (a) of a plane wave in the medium
6:1
Le.
i
d
:"ffi
I
The attenuation constant of.the plane wave in the medium is given
Now,
022 u€ 2trfe,e6 2r x b0 X 103 X g0 x g.gb x 10-12
i.e.
6
:8991.8
So,
ae d
))
1
>>1
-\/
IAnr
(o/ue>>
/t
2
:0.4n
v-
Therefore,
sol. 7.2.{6
as
6:1:-j-:o.Zg6m (\. t)-47f
Correct answer is 0.06 . Velocity of the wave in free space is
' c:
x
^/*:3
1o8m/s
So, the velocity of the wave in dielectric 1 is
uPl: - \/tET€'
c
7)
in dielectric 2 is c tltn oo:/guo:g
The velocity of wave
Page 494
Chap 7 Electromagtetic Waves
The velocity of wave in dielectric 3 is c un:t/ tun B^:n
Therefore, the time
f taken by the wave to strike
the interface at
r:5m
t:hltz-lh _ 6 -3 -2 3 x tbr-;/'-;F
I
l i l
: (0.02 + o.o2 + o.o2) x 10-6 :
$oL
?"4.{r
0.06
Jr,sec { 1
Correct answer is 0.99
.
Given
wave, , f :50MHz:50 x 106 Hz Skin depth of the dielectric medium, 6 :0'32mm : 0'32 x 10-3 m Permittivity of dielectric, F :6.28 X 10-7
Flequency of the propagating
So, the conductivity of the dielectric medium is given as
: $0L ?.2,{8
0.99
Correct answer is
x
105
S/m
-60.
Flequency of the wave,
"f:8GHz:8x10eHz z :0.775mm : 0.175 x
Distance travelled by the wave,
10-3 m
Permittivity of dielectric, Lr:6.28 X 10-7 and as calculated in previous question the conductivity of the dielectric medium is
o
:
x
S/m So, t\e attenuation constant of the wave in the dielectric medium is
o
:
': :
0.99
105
Jnfw 3.95
x
104
NP/m
Therefore, the reducing'fH"to, of the field intensity distance z is 201ogne-"'
sol. ?"2.{9
:
20logro
e-(3'e5
x 10r)x
(0 175
x 10 3)
in dB after travelling
- -
60 dB
Correct answer is 53.31 . Given, the magnetic field intensity of the EM wave propagating in free space.
:
,EI 0.1cos(r,.,t By)a, Alm So, the time average power density of the EM wave is given as
po,"
-
: |rorr o,
is the intrinsic impedance in free space and 1/ is the magnitude of magnetic field intensity in free space. where
so,
r7s
poo":|$zo"){o.ty", :0.6tra,
(no: L20r,I1:0.1)
l
Therefore, the total power passing through the square plate of side 20 cm is given as Putot
: P"," ' dS I - Pou" . San
S : (0.2tr : 0.04 m2 (Side of square and an is the unit vector normal to the plate given as
:
0.2 m)
:%ilu
a,
so,
ptotot
J2
:(o.atra,).
: :
[r
*(#)]
0.05331Watt 53.31mW
sol z.z.zo Correct
answer is 10.025 Given, the electric field intensity of the incident wave,
Ei"
-
ge,-rtu
6"y
l^
So, we get the phase constant of the wave
or,
Atr lJl -
as
t)
:r i"/@Yi : s
f/
(A:9\ ''
w,u"
uP
,:) Now, the intrinsic impedance of the lossless medium is given
n,: {T,:rr/t:2qo:754 and the intrinsic impedance of lossy medium is
,n:lrnl/!, where, the magnitude of the intrinsic impedance is given as
:
60tr,r :95.48
' (15.18I/.
and the phase angle of the intrinsic impedance is
tan2lr,
or
0,n
: +:3.77 u€z :
37.57"
So, the reflection coefficient of the wave is given as
n
n2q2+
:
rh ry
95.48
/37.57"
uo-1*lrl -1+0.8186 -1-ll-t -1--o-31s6tl 10.025
- 754
95.488.57" +754
0.1886/171.08" Therefore, the standing wave ratio is
:
chap
Z
Elictromagnetic Waves
where ,9 is the area of the square plate given as
i.e.
page 49b
as
?"2"r{
Page 496
94up z
Co;fec '; ,ftgfrg1..ilv-.:{4$ . As calcrllqfrp&,.lriffiidous'question we ft ave tle pro.pagatiqn -vectqr frorn given data as
" Eldtromagnrctic,Waves
:
0.04n(- 2a, - 3a, i JE a,) direction'of wave prcipagation is 'the O:a4tr(-2a, - 3au * : lc
.and
a,
Ga") 4_ (-za,- zq+,/E a")
-ffi
_ *2a,_B.au+,/5
a"
+
Therefore, t'he phase constant along the direction of propagation is
0:k'ax . -2a, = [o.oarr(-z a,--1'3au* 6r,)] (
:
\au *,fi - -T-
a"
0.16n
So, the wavelengttr along the direction of wave propagation is
A:4:12.5m lJ $oi-
?"2"22
Cerrect answer is 24.
Flom the given expression of magnetic field vector we get, B,r* fty* A"z :0.\Atr(,/i r- 2y - 32) So, the propAgation vector of :the plane wave is k: y (0,r* lty* 0"2)
:
0,04r(/E a, -
2au
-
3o")
and the directisn of wave propagation is
^ lc _ 0.0{*(./Q e, - 2a, -r-E-@
3o")
_(J3a,-2au-3a,) 4
Therefore, the plrase constau! alE4g the direptlo4 of wave propgga{ion is
13:k'qn-0'L6zr Since'the wavd is prop4gating in'ftee qpace so it's phase velocity will be
up:l x 1oqq/s 9:3x108
or,
11
So,
tle
freqpe5rcy' .of t4e plq.ng wpve is
(3 x 108X0.167r) f" :ffi:2'4xro7Hz
(+o
:2rfl
:24MHz s$L
7,2,23
Correct an$wpr ip 17.9 Since, 20% of t$9 anelgy in fihe incident w.pvg is reflected at tfrg bgundary. So, we have,
t nt2:100 lr I 20 of,
lf I :
^1A.2
:+
0.447
t
wl;rcre 1- is have
the rcflect;ion coefficieit^:atttil'rft"dh;hh interface. Therefore. rlz
Tt rh+, Tt
:*
0.447
(€,r: dt,€,2:
!a+
:+
rt2)
0.447
1t'o
ltr '-l+0.447 -TTTW IL :2.62 or 0.3g ttn l'Lt2
9o
So,
g-
7.2.24
:
€r7
(UeY: \Wt
0.056 or LT.9
1
Correct answer is 2.25 . Intrinsic impedance of 1"t medium is
Tt: !qfN and the intrinsic impedance of 2"d medium is
!€z 't': ^E So, the reflection coefficient at the interface of the two medium is given as
'r --Tz-rlt n2+ry /W
/
tT
tT
tt'o
'-" : {--pz--l,q fp^_ M \/er'\/e, 1
ort
5
!e, !e, lr 11 \/er'\/e,
1
5
5+1 5
--I
t a-
(given
": +)
t-/Ez
Var
r+'!€tF 2
o /ez !€t
6
4
Page AQf
. Chap.7 Eleetiondgieitc Wav6s
0.442
^{@z-^ffr :* n'{t+q'{@
lt,r . ltQ
we
!ez
o €, -r 4 -'lqE ct
*xxxx*{<*x<*x
(py rationalization)
soLUTloNs 7,3
Page 498
Chap 7
1
Electromagnetic Waves
soL
7,3.,t
Option (A) is correct. Given magnetic field intensity in the non magnetic medium is .E[ : 3 cos(ot - kz)a, Af m The negative coefficient of. z is (wt - kz) shows that the wave is propagating in * a, direction.
sol.
7"3.2
Option (C) is correct. Attenuation constant for a plane wave with angular frequency ar in a certain medium is given as
":,,/+l[lw-']
(1)
Since for a poor conductor, conductivity is very low i.e.
o(t
O
<
o aa1
ue
So, in equation (1) using binomial expansion we get,
":"^/+[r+]1;y-r]
tudl o o tu 2 Jrue - 2! e
Thefefore, the skin depth of the poor conductor is
6:!:Z E a oy'LL which is independent of frequency
scl
7.3.3
Option (A) is correct.
$oL ?.3.4
Option (A) is correct.
sol.
7.3.5
Option (B) is correct.
$oL ?.3,6
Option (C) is correct.
soL
?.3.?
Option (A) is correct.
sol-
7,3,8
Option (C) is correct.
sol
7.3.9
Option (B) is correct.
so|.
7.3.{O
Option (A) is correct.
sol
7.3.{l
Option (A) is correct.
sol. 7.3.{2
Option (A) is correct.
30L ?.3.{3
Option (A) is correct.
sol.
Option (C) is correct.
7.3.14
(c..').
(olue << I)
sol.
7.3.1$
Option (D) is correct.
.sol
7,3,t6
Option (C) is correct.
' '". '.; .l .,',; ,"
'
Page 499
Chap 7 Electro-a gnetii Waves "
sol. 7.3.{7
Option (C) is correct. Power radiated from any source is constant.
sol
Option (C) is correct. Given, the electric field intensity of the propagating wave ,E : a,sin(wt- pz) * arsin(wt_ 0z+r12) so, we conclude that the wave is propagating along a, direction and the field components along o, and a, are equal.
7.3.{8
I.e.
Er:Ey
Therefore, the wave is circularly polarized. Now we will determine the field is either right circular or lefb circular. The angle between the electric field .E and r-axis is given as Q
:tan'(rffi) - +^--rTcost''f 1 :|-ut zr
so, with increase in time the tip of the field intensity moves from y to r in a, direction therefore, the wave is
-axis and as the wave is propagating left hand circularly polarized. sol- 7.3.{9
Option (B) is correct. We have
d'8"
-aFdv --'^zo2E,
As the field component ,8" changes with z so, we conclude that the EM wave
is propagating in z- direction.
sol
7,3.20
Option (D) is correct. Intrinsic impedance of a medium is given
as
qSince, copper is good conductor i.e. a
))
&16
q:r/ry:/yrc:
Thus, the impedance will be complex with an inductive component.
i
:
so, we get
SOt :.S,zf
Option (C) is correct. The depth of penetration or skin depth is defined
as
6:-L i.e. or,
/"fw
6. -L /7 6oJi
(\:
"/f) So, the depth of penetration (skin depth) increases with increase in
wavelength.
sol
7.3,22
Option (A) is correct. Given, the electric field intensity of the wave (zrt) : flo si(wt+ 9z) ar + eo ei@t+ Bz) o,! ...(1) Generalizing E(z) : a,Er(z) * auE2(z) ...(2) comparing (1) and (2) we can see that E1(z) and E2(z) are in space quadrature but in time phase so, their sum .E will be linearly polarized along a line that E
Page 500
chp
makes an angle
i, wiih r-axis as shown below.
7
Electlo&gnotic.Wavec
sol
Option (C) is correct. The_Skin depth of a conductor is defined
?.s,23
as
"1 "-Jm So, statement 2 and 3 are correct while are incorrect.
sol.
$oL
Option (C) is correct. For circular polarization the two orthogonal field componen ts must have sarlre magnitude and has a phase difference of 90'. So, all the three statements are necessary conditions.
7,3"24
Option (A) is correct. ' Velocity of light in any dielectric medium is defined
7,3.25
tb
j i
as
1_ 1 _c "-rG-7G; -7where c is velocity of light
in vacuum and e" is dielectric
constant of the
medium. Since e"
)
1, so u
< c. Therefore, both A and R are true and
R,
is correct
explanation of A. $oL
?.3.26
Option (D) is correct. The poynting vector is the instantaneous power flow per unit area in an EM wave and defined as
P:ExH So,
sol.
EX
.E[ is rate of energy flow (power flow) per
unit area.
7"3.2? Option (C) is correct. Poynting vector represents the instantaneous power density vector associated with the EM field at a given point.
P:ExH
I.e.
$oL
7.3.28
Option (B) is correct. Given, the electric field intensity of the wave irr free space,
E: Comparing
5osin(107
t+
kz)a,,y f rt
it with the general expressiorr
of electric field defined
as
B: Essin(at- l3z)arYlm We get,
(1) The wave propagates in - a" direction along z-axis. (2) The wavelength is given as
^:i :+:
(3) Wave number. k ' (4) The $oL
*#i":188.5m ,*"at:0.033 it travels.
wave doesn't attenuate as
?.3"2$ Option (B) is correct. An electromagnetic wave incident on a conductirrg medium has the depth of penetration (skin depth) defined as
i.e. inversely proportion to attenuation cqnFtant., ?"3-34
Pa!e.5O1
Option (C) is correct. The gyro frequency is the frequency whose period is equal to the period of revolution of an electron in its circular orbit under the influence of earth's magnetic field. So, the radio wave at frequency near f is attenuated by the earth's magnetic field. (Since, there is a resonance phenomena and oscillating electron receive more and more ene.rgy from incident wave.) Option (A) is correct. An EM wave propagating in free space consists of electric and magnetic field intensity both perpendicular to direction of propagation.
taa,
Option (C) is correct. Skin depth (6) is the distance through which the wave amplitude decreases
toafactor et or7fe.
Option (C) is correct. The depth of penetration of wave (skin depth) in a lossy dielectric (conductor) is given as
-a-laG
6--1-
1
So, the skin depth increases when
(1) permeability decreases (2) conductivity decreases (3) frequency decreases
i
Since. the wavelength of the wave is given as ,,
A:!+JT i.e. .l*l
;
So, as
30L
?,3"34
,\ increases, / decreases and therefore, skin depth increases.
Option (B) is correct. For a good conductor,
o
: 0: /"fW
Since, the skin depth is defined as
6:* o, d :tr
@:0)
Now, the phase constant of the wave is given
as
. :t 2tr
11
So, we have
s{fi-
?,3,3S
.1^ u: p:2n
It
is defined for a good conductor.
Option (D) is correct. The polarization of a unifcrrm plane wave described the time varying behaviour of the electric field intensity vector so for polarization the field vector must be transverse to the propagation of wave. i.e. Transverse nature of electromagnetic wave causes polarization.
sol.
7.3,3S
Optiorr (B) is correct. Fields are said to be circularly polarized if their components have same magnitudes but they differ in phase bv +90'. **********>k
Chap 7 Electronagne0ic"Wavi*
solurloN$
Page 502
Qhnp 7
7.4
Electromagnetic Waves
60L
?.4.1
Option (C) is correct. Electric field of the propagating wave in free space is given as B : (8a,* 6a, + 5a") e!@t+3"-n') v l^ that wave is propagating in the direction (- 3o'*4a)' So, it is clea,r Since, the wave is incident on a perfectly conducting slab at r:0. So, tbe reflection coefficient will be equal to -1. 8," : (- l) E^ :- 8a. - 6au - 5a" r.e. Again, the reflected wave will be as shown in figure below :
-3arl4a,
Flee space
r:0 i.e. the reflected wave will be along the direction field of the reflected wave will be
E, : (-
8a,
-
6au
-
3a,*
4au. Thus, the electric
5o") ei@t-\x-4i V
l
lm
I
sol.
?.4.2
Option (A) is correct. Given, the electric field intensity of the EM wave as E : lO(au r ja,) e-r25' So, we conclude that the wave is propagating in a, direction and the y and z-components of the field are same. Therefore, the wave is circularly pola.rized.
Now, the angle formed by the electric field with the z-axis is given
as
0:at with increase in time the tip of the field magnitude rotates from z Lo g -axis and as the wave is propagating in o, direction so) we conclude that the wave is left circular (i,e., left circular polarization). The phase constant of the field is given as n _wt-c So,
o< 2rf
(0:25)
c
25x3x108 r" - 25xc 21 2x3.14 : l.2GHz $oL 7,4,3
Option (C) is correct. Intrinsic impeda,nce of EM wave
,t
:,/T:
r[E^
:ry:60tr
I
Time average power density of the EM wave is"giv""
Page 503
*
Chap 7
po,":L"r:t#
Electromagnetlc Waves
11 -ztTni-w sol-
7.4"4
(E:IY
lm)
Option (A) is correct. In the given problem
(Fleespaceg<0)
n': n/H : l20tr Reflection coefficient at the medium interface is given
_400tr-120n _ '. _q2-rlr rh*nt - 40n-t720tr -
as
1
2
As, given the electric field component of the incident wave is
Er :24cos(3 x 108 - 7u)a" that the incident wave is propagating along a, direction and the angular frequency of the wave is u:B x 108rad/s phase So, the constant of the wave is given as So, we conclude
,, _va _tt-lbB_' 3x108
u
Therefore, the reflected wave will be propagating electric field component is given as
E, where
E16
fEncos(3
x
(0:
108+ g)
r rad/m)
is the maximum value of the field component of incident wave.
i.e. So, we
:
in - a, direction and its
:24a, n* :-|lZ+cos(3 x
E6
have
108
+ g)0,]
:_
12cos(B x 108+ gr)o, Therefore, the magnetic field component of the reflected wave is given as
H"
:!(ar q'- x E,\
where 11 is the intrinsic impedance of medium 1, and ar is the unit vector in the direction of wave propagation. So, we get
n,:
nonLfo,
x (-12cos(3 x
108+
:,ficos(3 xt08+u)a" sol 7"4.5 Option (C) is correct. The intrinsic impedance of the wave is defined ,t
'
rc
-
I \/e -
as
y)"")l
where p is permeability and e is permittivity of the ,rnedium. Now, the reflection coefficient at the medium interface is given r _ rlz_ rlt
Page 504
c@P7, Electr6dagnetic Waves
'
rl2+
Substituting values for
4
rh
and rp we have
,:r/r=-/E-:r-{E:t-{g ^/++J+" r+J;- 1+Js
as
I (t': r)
:- 0.5 ll-l:0.5
or,'tt-
I I
$oL ?,4,6 Since, the wave is propagating in a direction making an angle 90" with positive y-axis. So, the y-component of propagation constant will be zero. As the direction of propagation mak6s dn angle 30' with positive r-axis so. we have the propagation constant of the wave as "' '
7:
p cos
30'r
-F psin 30'
gr
,
where B is the phase constant of the wave..So, we get
'
^r
-
=++.tTto =$'+\u
Now, in all the given options the direction of electric field of the wave is Sd; considering that direction we'get the field intensity of
giverr along the wave as
",
E sol-.?.d7
-
auEsei@i-t)
-,p,f"'-(*"fr)l
Option (D) is cor,t'ect. Since, the given field intensity,have components in o".and c, d.irection so, the magnitude of the field,intensity of the plane wave is
'
I Hl'
: HZ+'H'. = (fl.(*l: (#l
So, the time average power density of the EM wave is given as Po,"
soL
7"4"8
:rkr! r' :ryffif:
Option (D) is correct The Brewster angle is given
tanln
@ watts
as
: ' f@ Y€t
tan60': -\/ tE 1
Of $oL 7.4.9
€rZ:B
Option (A) is correct. The reflection coefficient a,t the medium interface is given
'r --rlz-\r rh+ry-So, the transmitted'power is
lE-J*
_L4E _r-{4 _ /..+{E - r+G-TTIT--s
P,:(7_lruPt OI'
as
4:(t -t)o:$e n8 Pr-g
1
rol ?.4''o option (D) is
,;'|
+: + Je, - Jz g :45" =t
or
r rr:
rr
":',T;l=
t.
The configuration is shown below. Here .4 is point source.
A
,zl\ /t\ rr:i:__;:;) ozitll-J-=-\
'
Now
AO:1m Flom geometry BO :1 m Thus, area: nf :rx tor.
7.4.'ti
OB:n
i-
ri2
Option (C) is correct. Given, the electric field of the EM wave in medium 1 as Fq :4a,*3.or*5a" "As the medium interface lies in the plane components of the ehctric field are
i.
F11
r:
0 sg, the tangential and normal
:Saul5a"
Ero :4e, Now, from the boundary condition we know that the tangential component of electric field is uniform. So, we get
and
Ezt : Ett:3au*5a, Again from the boundary condition the norma-l component of displacement vector are equal.
i.e. Or or ot
Dzn: Dn e2fi, : e1fin 4eo4^ :3€o4a",
&,,
:3a,
Thus, the net electric field intensity in medium 2 is
&:4r* ta-12
Ezn:3a,*}au*5a"
Option (C) is correct. Flom the expression of the magnetic field intensrty of the EM wave, we have Angular frequency,c.r : 50,000 Phase
constant, 0 :0.004
So, the phase constant of the wave is given as
_!4_ 5x10a _rr< xru ,, v1^7^tm/s "r-F -4X10-:L.zo ]oL
7.4.t3 Option (C) is correct. Refractive index of glass
nn
:
{11,e,
:
l.s
FYequency f :t1taHz
c:3x108m/sec The wavelength of the l}La Hz beam of light is
Page 505
. QhBp z, Electromagnetic Waves
!I ^:1:%.g:3x10-6 as
Page 506
Chap ? Electromagnetic lVaves
So, wavelength of the
light beam in glass is given
+: "f#{ :2 x 1o-6 m Option (A) is correct. :
sgl-
^,
7.4.14
I
The time average poynting vector of the EM wave is defined
po,":{netz x r/,']
I
as
I
I where, .8, is the phasor form of the electric field intensity and IfJ is tne I complex conjugate of the phsor form of magnetic field intensity. So, we have I
E"
x H"' : (a"+ jau) sjk"- j't
ia,t " he
(-, u) jrl: "."1hPoo":fn"1r, x .Er,-]: o
:
Thus,
o
an)
"-!kz+iat
I
I j
j
$*r-
7,4,15
Option (D) is correct.
we I i
have
j
: "ffi:5: i;]+J lfl:Z3
vswR
or
Ij
;
l
As the wave is normally incident on the interface so, the reflection coefficient
I
will be real (either positive or negative). Now, for a wave propagating from medium 1 to medium 2 having permittivities sr and e2 respectively. (i) If e2 > e1, the reflection coefficient is negative
!. :
(ii) If
e
then, the reflection coefficient is positive. Since, the given EM wave is propagating from free space to the dielectric material with e ) es, therefore €2
1-
f :-?3
or.
or'
rlz,\ rh+ rlt --2J rn- 72Ur 2
;;+m;:-g
So, sol
?.4,ts
rh
:24tr
Option (B) is correct. The skin depth (6) of a material is related to the operating frequency (/)
n b- n fcr
Therefore,
*
1
l1
n lL- \/4 2561
tT
6D: t/ i"25:12.5cm $oL
?.4"17
Option (D) is correct. The intrinsic impedance of a medium with permittivity e and permeability
p
is defined as
\ TE n:\/7 So, the reflection coefficient
is given
Page b07
at the boundary interface of the two mediums
as
-r:ry-l!-/#-/4 rh*rh IE+JE :-l-'fe, tiG: -1-/4 m : +: 0.333/180"
since
€':
4
d
r.4.1&
Option (B) is correct. We have E(z,t) :
10 cos (2zr
x
107
t
u :2tr X 107, 0 :0.In
So, we get
-
0.Itrz)
Therefore, the phase velocity of the wave is given
Dt 2.4.{s
:
'o
fr:u#Y:2
as
x 108 m/s
Option (C) is correct.
g : (0.5a,*
We have So,
and i.e.
aneG)"i(ut-kz)
its components along r and y-axis are
lE,l lnrl
:
9.5"i@t-*z)
:
s6 sj{"-t"")
ln,l + la,l
Since' the components are not equal and have the phase difference of so7 we conclude that the EM wave is elliptically polarized.
rf2
z.A.za Option (A) is correct. Loss tangent of a medium is defined as
tan6
o
:
ae
where o is the conductivity e is permittivity of the medium and c.r is operating angular frequency. So, we get
(w:2trf)
J-
7.4,21
LT x10-4 x g
3t1otx
x
10e
39
:
1.3
x
10_5
Option (A) is correct. The required condition is lr"1 :;rr1 i'e. the conduction current equals to the dispracement current. so, we get
lJ"l:lJdl
oft oft
laEl : lue t1 o :2nf€oe,
r_
'
_
o
(u:2rf, 2o
2T X €o€r 4tf€oe. 9
x
10e
X2 x
10-2
:45 x 106:45
MHz
e
:
e"ee)
,
Qhln
?
Electromagnetic Waves
Page 508
sol.
7,4,22
Chap 7 t1
Eleciromagnetic Waves
Option (B) is correct. VSWR (voltage,standing.wave latio) of the.trans4ission,line is defined
a.s
where -l- is the reflection coefficient of the transmi3sion line. So, we get od -
or,
1+li-1 I
(VSWR:
l
1-TlT tt
3)
lrl:0.5
Therefore, the ratio of the re{lected power strength to the incident power is given as
,P,
n:lr|:o'25'
Thus, 25% of incident power is reflected. sol.
?.4.23
Option (C) is correct. The fig is as ghown below
:
As per snell law
sin4 sin4 sin30"
1
-:---:-;-6
t_
srn4c
1
1
1
J2
--,rr
/^
!Er
-o
sol.
7.4.24
Option (B) is correct. Since, the phase constant is defined
as
'n2tr
P:T:a\/Pe
So, the wavelength in terms of
,2r A:----
a\/
1"-!\/e
oft So, we get
sol.
7.4.25
l' \z-l
permittivity of the medium can be given
as
P€
tE; e'
Option (C) is correct. A scalar wave equation must satisfy following relation
tE-,*olE^ "o
af
where
a*
,o
:o :
fr
...(1) (Phase velocity of the wave)
Basically ar is the multiply factor of frequency,
/
and p is multiply factor
of zor
roty.
Page 50Q
So, we can conclud.e that expresSidn equation (1) (i.e. the wave equation). sol- 7.4.t6
Option (D) is correct. In a lossless dielectric
bption (C) does not satisfy
(o:0) *edio*, impedance is given by
\:r/T where
p is permeability and e is perrnittivity of the medium.
So, we get
n:@ ' \ €o€, :120r.x
: 3(}L 7.4,27
I20n
x Utr, :'188.4
O
Option (A) is correct. Given, the electric field intensity of the EM wave as ' E - 24ui@t+P,) auy f m Now, the time average poynting vector for the EM wave is defined
as
(r":
*)
where q is the intrinsic impedance of the medium and ar is the direction of wave propagation. Since, from the given expression of the field intensity we conclude that the wave is propagating along
, : H? so|.
7"4,?8
o")
* o,
:-To"
So, we have
(o*:- o",lE l:24Y
lm)
Option (B) is correct. Given the propagation constant of the wave So, we get
of'
'l:Q*i0:O.1rr*.fi.2n 0 :0.hr 2f : o.zn
Therefore, wavelength of the propagating wave is
A:&:1om sol.
?.4.29
Option (A) is correct. Skin depth.of the conducting medium at frequencS
.fr
:
10
MHz is given
as
-ffi
6_ or OIt
70-2
1
/" i iot
Il,O:'7t
10-3
lou
x
(,fi
t
"
Now, phase velocity at another frequency
", _ Putting p,o: l0-3 f r
,,
ehspz
Electromagnetic Waves
("fr:
1000MHz) is
@
^,/ Lto in the above expression, we get
,,:ffi=ox1o6m/sec
:
10
MHz)
PaS"e 510
soL
7,4.30
Chap 7 Electromagnetic Waves
Option (C) is correct. Reflected power P, of. a plane wq,ve in terms of incident power
P;
AS
where,
f
is the reflection coefficient at the medium interface given as
rh_ rlt -tr - -ETz*qt
(2!
where 41 and q2 are the intrinsic impedance of the two mediums (air and glass) respectively. Since, the refractive index of the glass is 1.5 i.e.
flz : Ct/ lla€z: I.5
where
Pa
:
e2
-
(3)
Ltn
(Permeability of glassi
crao
(Permittivity of glass|
So, putting these values in equation (3) we get
and
Jd :1.5 tu" ,h: ,/;:
rh
rh 1.5
T^
!Er
Therefore, from equation (2) we have
&- tru : 1 , 1'9 :-'!5 r :-#+rh-l+1'5Thus, from equation (1) the reflected power is given
(for \^"^ free space 11: ryf as
e:(ff"e or' *0L
7.4.31
D
E:n%
Option (A) is correct. Skin depth of a material is defined
"{_-
as
1
,/;lF
Putting the given values in the expression, we get
(_ ' _;ER 1x10ex4z-x10-7x106
I
*aL
7,4.32
:
15.9 u,m
Option (C) is correct. The energy density in a medium having electric field intensity ,E is defined a^s
:
where e is permittivity of the medium|el E l2 So, due to the field E:700\/; V/m in free space, the energy density is
we
," : : $0L ?,4,33
]1a.ss 1.3g
x
x 1o-"Xloo./;f 10-7 Jf m3
:13gnJ/m3
Option (C) is correct. For a uniform plane wave propagating in free space, the fields E and E are every where normal to the direction of wave propagation or and thei direction are related as o,6XAg:q', i.e. the angel between electric field (ar) and magnetic field vector (o") i always 90'.
sol
7,11.34
Option (B) is correct. The incidence angle of an EM wave for wliich there is no reflection is calied Brewster's angle. For the vertically polarized wave (parallel polarized wal.e) the Brewster angle is defined as tan0s11:
{?,
So, for the given dielectric medium we get
tanlBu: ,Fn
or, EOL 7.4.35
: tan-t/i\
d"tt
\2)
Option (B) is correct. Given, the electric field component of the EM wave propagating in free space,
.D : locos(107 t + kz)ouy lm The general equation of electric field component of an EM wave propagating in a, direction is given as So, we conclude
or
.E : Eocos(ut* kz)ouylm that the EI\I wave is propagating in o. direction. : 107 rad/s
2rf : 107 107 I __E
So,
:
.\
?:%g
X 2tr :188.5m
i.e. wavelength of the wave is wave amplitude, Eo :70Y lm wave
number'
:+: #: #
k
:0.033 ,ua1^ The wave doesn't attenuate as it travels. so, statement (2) and (3) are correct.
?.rt.36 Option (B) is correct. The incidence angle of a plane wave for which there is no reflection is called Brewster's angle. For the parallel polarized wave, Brewster's angle is given as
tanfll1,:
,/?
where €r and 62 are the permittivity of two mediums respectively. so, for the given parallel polarized plane wave the incidence angle (4) ro, ,ro reflection is given as
or,
tanli: {# 4
: tan-t(+)
a for no reflection a :90o - 0t
Therefore, the angle
:
83.66'
is
Page
5ff
Cbrp 7 Electromagnetic Wewr
,
-P.age
sor- ?.4.32'optio,{'@JifS,in-1dfut.
ffZ
7' "OhaP nr"co.ori"g"*id
'cio"r', tii"Yffiil{#ibtic iinpedance of air, ? :360O g, :lidin(ot-,pz)v lm
Waves
'So,
Ev =- p sin(ari
:-"
pz
*
75")Y I m
the time average piriler per'unit area is 1, -LW'ove-2 Tl -2^
(D
(a'f +ln'l) 360
(3r+6r) : ,1z .,n --T60L'6.25 x ro-zW/*? :62.5'mW/-' soL
7,4.38
bpiion (A) is'correct. 'Op'rirhing hequ'ehcy 'Midiirm"pafimeiers,
.f :'3'GHz : 3. x 10e Hz p = 47T i<'ro-'H/m e
:
Lo-e 136r
o, :5.8 x
10?S/r4
So, we have inirinsiC impedance dedned as
trl=ffi:
5.8
r.( :2.02 x
Ztr
o,:|tan-'(#) =fxtun-' / =2ng:4 So, $oL
7.4.39
L0?
t;lo'tE il" \tO?T
10_20
The phase angle of intrinsic impedance is given
| .. r
x
x
5.8
lr".
1t
as
t
x
107
r\ w " 10-e
n:ln1{")o.\2et'/aQ
bption (D) is correct. Given, the elecilic ffeH of a plane wave,
F:
5osin(108r+ 2z)quY lm Compaiing it with tirp genqral exprps'siop electric field of a plpne wave traveiling in a, directipn given aq
F : Erdb(at,* 1z)q We Sp,t the directiou pf propagatip4 qf
sol.
?.4.40
t\e
givep plane S'ave is = s".
Option (C) is eorrept. Civen thg Bleetric field,
F:
(g,"+ !qr)e-jP'
Sq, it is clgar lhat y-compor.reni of field lea{s tfe r-component by 90" and the wavg p,fpPagaies'ip z-$irgption. The components are same. So, the tip of electric fipld traVefge in crypulal pa,th in the plpckwipe directiqil and wave prop4gates in z-directipn as ohpwn in figule.
Page bt3
Chap 7 Electromagnetic \ilaves
Therefore,
it is negative circularly
polarized wave
or (left hand polarized
wave). ?"4"dt
Option (D) is correct. Consider the reflector is of angle 0 : 90' for which the incident and reflected wave is shown in figure.
So' it is clear that the incident and reflected wave both makes same angle o with the z-axis i.e. reflected w,aye in sarne direction.
?.,e.4? Option (A) is correct. Since, after reflection the phase of both r aruJ y components so the reflected wave wilr be also right circularty poi*ir"a.
will be reversed
?"ri"43 Option (C) is correct. Given,
E : 10cos(6zr x 11: ltl
Electric field intensity of the wave Permeability of mediurn,
108t
-
Permittivity of medium, e :81e0 trlom the expression of the electric field, we get the angular frequency
:6r x
u
as
I08
The phase velocity of the wave is given
up-
br)ou
as
I
r:
,/ Pe
-ffi
1
:--3x108 9--: +, k:]-:Bxlosm/s) p,oeo
So, the phase constant of the EM wave is
r.t_u_62rx108 P - -
:
107% lgzr radfm
"/
Page
Cbap
5r4 7
Electromagnetic
sor-
r.A.u
Option (A) is,correct. . ^ ,
phase
Given, the
Waves
.
veloaityoi the
plane wave in dielectric is 0.4 times ,r"
in free SpaCe ap :0.4c Sinee, the phase velocity of a medium having
"tj
i.e.
p is defined
(1)
permittivity 6 and permeability
as
,r-- .{ I t',
So,
putting it in equation (1) we get
:0'4c
1
J
$oL
7.4.45
(lt:
lr^lt,,e
:
eo)
Poe,eo
/1\' :6'25 - : (oz)
@:
-):'1 { Poeo
Option (B) is correct. iven, the electric field in free space, E(t,t) : 60cos(of - 2r)anY lm So, we get thd.magnitude of the electric field as Eo
:60
The time average power density in the electric field is given as (6oY o -L4d-1'ave-2rlo-2..72Ur
Therefore, the average power'through the circular area of tadius 4 m is
Poo":(P*")x(trf)
:! sol.
?.4.46
x{#
"
n(+l :240 w
Option (C) is correct. The relation between electric and magnetic field of the reflected, transmitted and incident wave is given below E; : \tHt
E,:-qrH, E1 : r12H1 So, (1) and (3) are correct while (2) is incorrect.
sol.
?.4.47 Option (D) is correct. From snell's law,
nrsin4
,/ taqsinfa ,/ 1aQ$ sin6o'
-
nasinlz
: ,lffisin22 : ,/ 1tae2sinZ,
sind2-(r"+):./t-s>t which is not possible so there will be no transmitted wave. soL
7.4.46
Option (C) is correct. (1) Consider Er is r-component and E2 is y-component so, when .Er and Ez wlll be in same phase. The wave will be linearly polarized.
(a-+
1)
E1 and Ez will have any arbitrary phase difference then it will be elliptically polarized.
(2) When
(drz)
I
l
(3) when -fi
leads
E, by 90" then a.rf increases,eounter
the wave is right circularly polarized.- ,
(c-3)
clockwise and so
7 Eldctromagnetic Waves
(4) when E1 lags Ez by 90" then the tip of field vector E will
sol.
7.rt,49
traverse
circularly in clockwise direction and left circularly polarized.
(b-a)
Option (A) is correct. (a) Propagation constant for a perfect conductor is
'y:a+iB where
a: g: ,/qy
a--+1
(b) Radiation intensity of an antenna U(0,0)
:
is defined as
fPo,"
:,'w:(fi)tnr (c)
b'+
:
qH6
q:#
so|.
7.4,50
2
Wave impedance of an EM wave is defined as E6
c+3
Option (D) is correct. An incident wave normal to a perfect conductor is completely reflected in the reverse direction. The magnetic field intensity of reflected wave is same as the incident wave whereas the electric field intensity of reflected wave has the 180" phase difference in comparison to the incident field. (.r-:- 1 for conducting surface).
i I
sol
7.4.5{
Option (A) is correct. Given,
Electric field intensity, E : E,a,I Euau The direction of wave propagation, (trh : Az So, the magnetic field intensity of the EM wave is given as
H:%xE where, 4 is the intrinsic impedance of the medium. putting the expression for electric field in equation, we get
H : Q x (E,a,t sol-
7.4.52
Er%)
:
|(n,oo
-
Option (A) is correct. In a uniform plane wave the field intensities are related
Eua,)
as
E:TH where 4 is intrinsic impedance given as
,l:.@ \/ o * j,ue Assume the medium is perfectly dielectric
of,
,l: /! E_ ro
E: \/;
Page 515
'Chap
-
(o:
0). So, we get
Page 516
sol
?"4"s3
Chap ? Eloctromagnetic lVaveq
Option (B) is correct. The higher frequency (microwave) signal is continuously refracted on tbe ground as shown in figure.
This phenomenon is called ducting. soL
7"4"54
Option (D) is correct. Given, the magnetic field intensity of a plane wave, H : O.be o t, cos(106f - 2r)a. (r) The general expression for magrrctic field intensity of a plane wave travelling in positive r-direction is
H
-* H11e "'cos(urf - 0r)o, Comparing the equation (1) and (2) we get, Wave
frequency,
cr,'
Wavelength,
^
:
106
rad/
:+:T
(2)
sec
:3.14
m
and the wave travels in *r-direction. Since, the magnetic field inteusity points toward o, direction and the wave propagates in * a, direction. So, directiou of electric field intensitv will be AE :-
A* X A,a:_(O, X A"):
Therefore, the wave is polarized irr intensity). sot"
A,
o, direction (direction of electric field
?.4"5s Option (D)
is correct. Flom Maxwell's equation, For a varying magnetic field B, the electric field intensity .E is defined as
YxE
--0P dt
Since, the magnetic flux density
B in terms of rnagnetic
vector potential is
given as
B:Y xA So, fiom the two equations we have
haA
E :-'.:-
A : a"A,sin(ut-
Given,
7"4,$6
V:0)
pz)
n :-S[o,A,sin(at- 0r)] :* a,uA,cos(wt- !32)
So,
stll.
(For V
dt
Option (A) is correct. Given, the magnetic field inte'sity of the wave propagating in free space,
H(z.t): - Uf cos(, t.t So, we conclut-le as
rlz)a,
direction of propagation,
:-
(trk
Q,
direction of magnetic field. aH:au So, the direction of electric field intensity is given
Page 517
.
a6: as X a7 :aux(-a"):-o, and the electric field arnplitude is given as,
t
:[rirat- {cos(ut +
oz))
:-
20cos(c,'t+ l3z') So. the eiectric field vector of EM wave is E(z' FST
!
i
TT
t)
:
20 cos(d
* Az) a,,
Option (A) is correct. Given, the electric field intensitv of EM wave in phase form E" : l\e-AY a,
as
So, we get
the phase constant. A :4radlm Since, the wave is propagatirrg irr free space, therefore, the angular frequency a,' of the wave is given as
" 1r{ *108X4) :4 SOL
?",$.,$E
stx-
?"d,"$t,
6
:
as
'--.1-:. "/
"ftto
Since, conductivity of the material is o
6
So, we get
ssL
?."$"6s
x 3 x lo8rad/s
Option (A) is correct. A and R both true and R is correct explanation of A. Option (B) is corrcct. Skin depth of a material is defined
:
0.
' infinity
Option (B) is correct. (1) In a conductirrg medium as the wave travels its amplitude is attenuated by the factor e-o' (i.e. attemrated exporrentially). (2) Conducting medium rloesn't behave as on open circuit to the EM \eld. (3) In lossless dielectric (o: 0) relaxation time is defined as
l,:9 ( ) In charge
free region
v,
v
+
cn
(p":0).
Poissions equatiorr is generalised as
:*?
V2V:0 which is Laplace equation. Therefore only statement 2 is incorrect. g${" ?.*"sd
Option (B) is correct. For a given electric field in liee space the average power density is defined as pou,
: 2L ltrto :f(99:l-2I2Ur : tbrwattlm,
Chap 7
Electrouiagneiic Waves
as
Page 51E
$oL
?.4.62
Chap 7
Option (B) is correct.
E
Given,
Electromagrretic Waves
: l2}rcos(ut-
Since, the wave is propagating
pz)a,
in o, directions so, the magnetic flux
densit-v
of the propagating wave is
tt -e*x E -a,xfl20Tcos(c..'f - 1t)o"] \o \o : cos(r,,lt - 0r)o,
(ax:
Therefore, the average power density of an EM wave is defined
o"\
as
poo":]n"1r x rr.)
: ]11tzo""os(r.rr- 0r)o,) x (cos(c.rf - p4q)l :60na" sol.
?.4.6s
Option (A) is correct. Given, the electric field intensity is
E:
10sin(3zr
x
108t
-rz)a,*
So, the magnetic field intensity is given
E H -&*X rlo
: 7.4.64
ffisin(3zr x
10cos(Szr
x
108,
-nz)an
as
(Direction of propagation is 108,
-
rz)ao+
$cos(3zr x
108,
-
a1
:
4,)
nz)(- a,)
Option (D) is correct. (1) For a perfect conducting medium the transmission coefficient is zero but a medium having finite conductivity transmission coefficient has some finite value. So it doesn't behave like an open circuit to the electromagnetic field.
(2) Relaxation time in a medium is defined
as
t, -Ge Which in turn given the values in the range of 10-20 sec. While the radio frequency wave has the time period '?' in the range of nsec to psec. (10-e to 10-12) So the relation time at radio frequency/microwave frequency is much less than the period. (3) For a lossless dielectric (o:0) and so,
T,- 9tcn (4) Intrinsic
impedance of a perfect dielectric (o
:
0) is
q : ,/T which is a pure resistance. So, the statement (2), (3) and (4) are correct.
?.4.65
Option (A) is correct. In free space electrons and photon both have the same velocity 3 x 108 m/s. So, the velocity of electromagnetic waves is same as velocity of light. So A and R both are true and R is the correct explanation of A.
?.4.66
Option (C) is correct. Flom Maxwell's equation for an EM field, the divergence of the magnetic flux density is zero.
V.B
l.e.
-.0
,
Page 519
v.(vx.4) -0 div curl A
sol
7,4.67
Chap 7 Electromagetic Tgaves
:0
Option (A) is correct. Electric field intensity due to the current element is defined
E : !o
as
-!^-o" nb"o
The magnetic flux density due to the current element is given
HI:
as
2n6o,
So, the poynting vector of the field is
P:ExIf
r_r :- *sus,'
,PF;uo sol
?.4.68
Option (D) is correct. All the three statements are correct.
sol.
7.4,69
Option (C) is correct. Wavelength of a plane wave in any medium is defined as
),:2 -f where Since,
So,
: phase velocity / : frequency of the wave a, : * !Er uo
+
,\ .,
!cr
2- t€; T-\/A €r sol.
7.4.70
r
:4
Option (D) is correct. The velocity of an EM wave in free space is given as 'u": Q:3 X 108m/s and the characteristic impedance (intrinsic impedance) is given
as
Z":^E:l2;tr Veo so both the terms are independent of frequency of the wave i.e. remain unchanged.
sol
7.4.7'
Option (D) is correct. Given, electric field intensity .E
:5cos(loet+Boz)a,
So, we conclude that,
and
w:70e, and g :9 since
up
p:
39
I- : --9:1 (clJe,)
Page,5!0
(For non magnetic nredium
'' . : I Pl\' -/30 x 31 10E\' : sr "'-\r/-\ 1oe )-
Chap ? Eleeiromagnetic Waves
sor-
ur: -Ly n='
?.4.?2
Option (C) is correct. For attenuation of the wave the medium must have some finite conductir-it-s dP i o so this terrn b o. In the given wave equation the term p"ff irr-rolrro, responsible for the attenuation of the wave.
?.4.?3
Option (B) is correct. The statement 1,3 and 4 are correct while statement 2 is incorrect Gauss's law is applicable only for symmetrical geometry'
7"4.74 Option (D)
is correct.
InaGoodconductor
B:{lrfry
So,
u,
Phase veloci IY
: fr:,rl
4 po
?"4.?5 Option (B) is correct. Given, the electric field intensity of the plane wave is E(t) : l4lcos wta, - E2sin wtar]e- ik" Since the components of the field are and i.e.
sot-
lE-l:E, t'l lnol: n,
So, the wave is elliptically polarized
lE"l + lE,l
?.4"?6 Option (D)
is correct. skin depth is defined dielectric, For a lossy
as
r- '\ u-2t so, as the wavelength increases the depth of penetration of wave al-so increases.
i.e. Reason (R) is correct. The Skin depth is the depth by which electric field strength reduces to I : 37% of its original value i.e. Assertion (A) is false.
sol
7"4"77
Option (D) is correct. The electromagnetic equation in terms of vector potential Y2
$oL
?.,4,78
A
-
pr# :-
A is given
as
p,J
Option (A) is correct. The wavelength of an EM wave propagating in a waveguide is defined
as
where ,\' is the wavelength of the wave in unbounded medium(free space), I is the cutoff frequency of the waveguide and / is the operating frequency. Now, for a propagating wave in the waveguide, the operating frequency is higher than the cutoff frequency. i.e.
. f f"+ f"lf <1
Putting it in equation (1) we get
I
l < r'
Page 521
i.e. W'ar,'elength of a propagating wave in'a wa,ve guide is smaller than free
ssi*
SpaCr:
the
wavelength.
?"4.?s Option (B) is correct. For a lossless dielectric medium
o:0 and propagation constarrt,
".t:a* jH:{J"p\"+tr] a+ jJ - j'{t;; i.o. g o G, 0 :,rfii sel.-
T.d"s*
Option (C) is correct. l-or a lossless rnediurn
(o: 0) intrinsic irrrpedance 't, /lt" tl: I /ile :_ V/Uu i,.,\/
:
60r
is defined
as
[T 12ur(e,
€r':4 s{3L
?"i$.s'! Option (C) is corrcct. A field is saicl to be conseivative if the curl of the field is zero.
selL
?.{"r}a
Option (A) is correct. Given. the rlagnertic field iritensity, -E[ : 0.5e-0,'sin(lOot - 2r)a" Af m Cornparing it u,'ith general expression of magnetic field intensity of wavt: propagating in o, dirr:ction given as
H
.= Hoe-n'' sin(c"t
-
p,t:)o,,
\&b get
(i) the
(it)
alt
1: (trk' : a,
: &,,
!!I'' :5 x 10;m/s
So, dircction of polarization, &E
ss{.
?"4.$3
:-(e1,
or) :-(o, X
a,)a,a i.e. wave is polarized along ar.
Optiorr (C) is correct. Skin depth of any conducting rnedium is defined
6 So,
as
-t-- J nfpo
at a given frequency u:2trf
A*_'1.-anddo,-!: ro /t, $sr-
?"*"{t{i
Optiorr (A) is correct. Given, the clectric field interrsity of the plane wave,
(
_ E:
ChapT Eleotromagnetic T[aves
l0sin(10c^rt
- rz)a,f
10cos(u,rt
- rz)a,
Page 522
So, the field components are
Chap 7 Electromagnetic Waves
and
sol
?"4.rs
since,
So
the polarization is circular-
Option (D) is correct. In free space electric field intensity is defined as E :_qo(ao X H) where a* is unit vector in the direction of propagation. Given, II : o.10cos(4 x 107t - pz)a, Alm So, the direction of propagation, a1,: a, and we have,
${}t- ?,4.ss
E,: 10sin(10a;t- rz) E, : l0cos(wt- rz) lA,l:lArl
E
-:*
x (0.10cos(4 x 107 t- gr)",)l (rro: 37.7cos(+ x to?t- gr)a, 3771a"
377 O)
Option (B) is correct. Given the electric field in medium .4 is -l? : 100 cos (wt - 6trr) z In medium ,4., €,:4, F,: 7, In medium B, €,:9t Fr: 4, So,
(a) intrinsic
impedance of medium
'B' is
,":#:rre:!rt2or:8or (b) Intrinsic
(a-2)
impedance of medium ',4,' is 'to
:,/! : r[F^: ] 72on : "
6otr
So, reflection coefficient,
lA : 802'- 6011 : 1 r :- !!,1"+ rt^ - EGloz + 60r 7 (c)
(b-3)
Tlansmission coefficient,
2rr, 2 x Sotr 8 -i:nB+ry:Eotrro;:7 (d)
Phase shift constant of medium ,4 is given from the field equation as
0 sor
7.4.87
(c-a)
:6r
Option (B) is correct. Average power density in an poo"
(d -+ 1)
EM
:1o"(*
wave is defined as
x rd)
:!*5ox#:3.316 So, the average power crossing a circular area of ra.dius Por"
: (s.sto)(zr( /%f): sol
7"4.88
J24 m is
: P*"(nf)
250
watt
Option (A) is correct. Electric field amplitude, Eo:lYllri' Skin depth, 6 : 10cm:0.1m So, the attenuation constant of the wave in the conductor is
o:i:10 1
Page 523
Now, the electric field intensity after travelling a distance z inside a conductor is
E
:
Eoe-""
where, Eo is the field intensity at the surface of the conductor. so, the distance travelled by the wave for which amplitude of electric field changes
to (.Ile')
(v/*) -;"1ff " e
Bnr,-r}z "ez
-
I
E!
:2 z:20cm
702
Alternatively, since the skin depth is the distance in which the wave amplitude decays to (l I e) of its value at surface. So, for the amplitude to be 7 f e2 of the
field at its surface the wave penetrates a length of. 26 :20 cm. So A and R both are true and R is correct explanation of A.
7.4"ae Option (C) is correct. For any media having conductivity,
o:0.
the intrinsic impedance is given
AS
,:r/T:,[Tro
sol-
?.rt-g{}
For media
1,
For media
2,
for media
3,
: fiGz7): 188 Cl n, : ,pr@72) : 11319 n, : ,[f@TZ): s77 g n,
Option (A) is correct. For an EM wave a medium incident on another medium, reflection coefficient is defined as
. _E,_ H, '-E--E r
and So. vv'
Z 7 -Tz-ry rlr+ Tl, --22,Z+Z :-I3
4:-E E --iI- -L3
#: 3 and ft:sol-
?"4"s1
s
Option (A) is correct. For aperfect conductor conductivity 1 o So, the skin depth of the perfect conductor is
6:_L:o ,/ nfPo *>t***r<:kt<**
:
cp
Chap 7 Eiectromagnetic Waves
-l
I
I
l
l
TRANSMISSION LINES
IIITRODUCTION Thansmission lines are the guided conducting stmctures which are used in power distribution at low frequencies in communications. The main aim of
this chapter is to provide the overall concepts of rlansmission Line theory.
r Tlansmission lines parameters: primary and secondary constants. o tansmission line equations: input impedance, reflection coefficient. o Expression for characteristic impedance, propagation constant, and velocity of wave propagation for various transmission lines: lossless line. distortionless line. a a
a
Standing waves and standing wave ratio. Smith chart: determination of line characteristics using Smith chart. Transients on transmission line: bounce diagram.
TRANSMISSION LINE PARAMETERS The equivalent circuit of a transmission line is a distributed network. This consists of cascaded sections and each section consists of a series resistance -R, series inductance tr, shunt capacitance C, and shunt conductance G. One section of the equivalent circuit is shown in Figure 8.1. Let us study the primarv and secondary constants for transmissipn line.
lr'l:rirp !.1 : Equivalent Circuit of a tansmission Line
8.2.1 Primary
Constants
L, c, and G are called the primary constants of the transmission line. The prirnary constants of a transmission line are The distribufed parameters R,
defined as follows: 1. r? is defined as loop resistance per unit length of line. 2. tr is defined as loop inductance per unit line length.
)
i.e.
Z-ll,
Now, the input irnpedance at a distance
)14
from the load is defined
as rhf
:4 -ZL
zr,
And sirrcc the transrnission line is open (Zr: *) So' Zn :0 which is purely resistive i.e. R is correct statement. In a lossless line voltage or current along the line are not constant. i.e. A is not a correct statement. soL
e"d.s? Option (A) is correct. The characteristic impedance of a transmission line can be defined as below.
7 - lR+iaL "u_fG+W Zu
-
,/
Z^Z*
oV* ztt_ f So, all the three statements are correct.
sst s"d.** Option (C) is correct. Given, load impedance of the transmission line is
Zr :0 So. the
(Short circuit)
input impedance of the lossless line is given '7 ? I ZtI .iZohn Al\ "'n -- "u\zof rttt""ljl ) o 1 jZotanpll ) : jZstanBI
as
: ",\__2,
$sN.
s.4.*4
Option (A) is correct. Given,
Characteristic impedance, Zo :600Q Load impedance, Zr. :900Q So, the reflection coefficient of the transmission line is given
, -Zt-2, ,r_TrlZ, _900-600 - 9Q[J6ffi -
as
u'z ^o
sst #.d.*$ Option (B) is correct. The pr<-,pagation constant of a transmission lirre is defined
as
1=u\R+1rL)(G+1,,) and the characteristic impe
7 _ fR+jeL to:le+j;z 7_ I _R+jaL -dTJwe : j
"o s$l-
&.s.s$
Option (D) is correct. Characteristic impedance of a lossless transnrission line is defined as zn ^Fz-7., where 2., is open circuit impedance and 2,, is short circuit impedance.
ftttr g1rt lb
Following are some important points about modes of wave propagation:
Page 624
Chap
I
Wavt{,aides
lil'arisvcti,e electric (TE) and tansverse magnetic (TM) modes are cornmonly referred to as wav€guide modes since' they are the only modes *fri"n can exist in .n enilosed guiding structure. 2: I}.s[rlsferse electromagnetic
(TEef) modes cannot exist on
single
conductor guiding structures. 3. TE and TM modes are eharacterized by a cut-off frequency below which they do not proPagate. 4, TE and Th{ modes can exist on transrnission lines but are generaly undesirable (higher order modes). 'TEM mades are sometime.s qalled transmission line modes since they {). are the d.ominant modes on transmission lines. 6. Tlansmission lines are typically operated'at frequencies below the cut-off frequencies of TE and TM modes so that only the TEM mode exists. 7.
R,
9.3
'Quari-fEM 'm6{ss:are modes which appronimate tiue TEM rnodes r,*hen the frequency is sufficiently small. The mode having lowest cut-off frequency is called dominant mode.
PARALLEL PLATE WAVEGUIDE
The paraliel plate waveguide is probably the sirnplest type of guide that can support TM and TE modes; it carr also support a TEI\I rnode, since it is forrned frorn two flat plates or strips, as shown in Figure 9.1. The plates are assumed to be infinite in extent in the r-direction. It is assumed that the wave propagates in a, direction, and the fields do not vary in the o" direction. The characteristics of the various modes of propagation are exa,mined in following sections.
1:
i1,r.r;r l).1:
Parallel Plate Waveguide
9.3.1 TE Mode For TE morle irr a parallel plate waveguide, the non-zero field components are expressed in phasor form as
H,":
H2(u)e-1"
: B,cos(T)u"
H, : Il:(y)"-r', : |B,sin(!\L)e "
i{
3. C is defined as shunt capacitance between two wires per unit lengtL 4. G is defined as the conductance per unit length due to the dielect
Page 526 Chap
E
Tlansmission Lines
medium separating the conductors. The values of R , L , C and G depend on the geometry of transmission Ii characteristics of the dielectric material and in some cases on the frequer
For coaxial, two-wire, and planar lines, the formulae for calculating t values of R, L. G, and C are provided in Table below. Table 8.1: Primary Constants of TYansmission Lines at High FYequencies
CitA ti
,,
:,,
;l$6i,t,{l{l
L,ine
,,|,,,:,,1...:.,l
-': :.rt$.,t,,tit,t'll.l,lllt'
',','j&',' :ra*Il
- '
rrrr'rrr"
"
::.W.:,";:t:l,ti',:',:.,.:,::::,,,,::,,t:.;,,,,:::i);,,,:,:,,:.:
t ,-
lnlr,
\a;l:ataa., 8;:.:.:::::1.
-.t
I
::::::::::,,:::.,,. .:a::,::::.4:t::,:;.:
€,$lm}',111,
..ff',i.,tu''< <
-d
*-!W3!e
-ffi
reW,+n[!J Lyff'try-re'* ffi*' "'Wffi';As*d;ffiiffi 6W f#% 'ryffi,hr; ffi
"€.w"
j:*s.*{wiiffif
(")
";""* -.;tSM c}}ffi
ffi
(b)
.<*_._-,-,t
(c)
Figrn'r: 8.2: Ttansmission Lines: (a) Coaxial Line, (b) Two-wire Line, and (c) Planar
8.2.2 i,
F
Li !
Secondar rdary Constants The secondary constants of a. transmission line are 1. Propagation constant, 7
2.
Characteristic impedance,
zi1
L. Propagation ragation Constant t t
Propagation constant of a transmission line can be defined as follows: Propaga DEFIt{lTlOtrl ilolt Ir
I I I
t
I
I
l*^
For a transmission line with primary constants R, L, C and G, propagation constant is given by
7:
.".G
Page 527
@+ 1"t')(G + ItC)
CbaP
DEFINITION II
Intermsofattenuationconstant'oandphaseconstant'B'thepropagation constant is defined
as
7:a+
jp
DEFINITION III
Foratransmissionline,ifthevoltageandcurrentatsourceendbeV",l,'and thevoltageu,'d"""""tatloadendbeV1'16t'henthepropagationconstant for the transmission line is defined as
.y:20r"*,.(+)
: zoros',(f)ae 2. Characteristic ImPedance
Characteristicimpedanceofatransmissionlinecanbedefinedasfollows: DEFINITIOil I
R, For a transmission line with primary constants characteristic impedance is defined
za
L, C
and G, the
as
: \,tGeries@ ;E[nt-admittance
: I 1ETFT G+ L'te ll Foratransmissionline,iftheforwardvoltageandforwardcurrentatany pointbeVon,Io*;thenthecharacteristicimpedanceofthetransmissionline
DEFlNlTlol{
is given bY
o Vn* ao:T DEFINITIOX III
Foratransmissionline,ifthereflectedvoltageandreflectedcurrentatany pointbeVo,Io;thenthecha^racteristicimpedanceof,thetransmissionline is given bY
I.3
-v; zo:_
h_
TRANSMISSION LINE EOUATIONS as shown in Figure 8.3. If the voltage Consider a transmission Iine of length I locuiion on the line be v and 1, respectivelv,
and current at any arbitrary then the transmission line equation is defined
d'Y dz"
'
d2I^
as
: r,v _
-rz1
dz"
wherelisthepropagationconstant.Thesolutionofthelineequationscan be obtained in Phasor form as
V,(z):Vs*e"+Voe'" rr+ v2 I,(r) : Is+ s-1'* Io e1' : h"-'" - oo "" where
E
Ilansmission Lines
Vo*
and
ff
travelling are the voltage and current for forward wave
Page 52E
Chap
E
+4,
d.irection, pnd 7n and di are the voltage and current reflected wave travellilg along -* o," direction.
along
-j
Ttansmission Lines
+1, +
Source
L,
I/o
I
Load
I l
I
i"'lgr:r.*r
!.ll: Equivalent Circuit for defiling Transmission Line Equation
8.3.1 Input Impedance At
of Tbansmission Line
source end of the transmission line. we have
: V"(z - 0) : Vo'* Vi to:1,(z-0):W;%Ls
Vr
So, ttre input impedance of transrnission line is given bv
o -_V"(r:0) _ Zu(Vf + Izt) ""' I"(z - 0) (V,f - Vo)
The input impedance can also be expressed
2,.
as
: z,ltr:,I i :ixiil,,l
where Zo is the characteristic impedance. I is the length of transmission
7
lir
is the propagation constant, and 21 is the load inrpedance given by
z':Yl- Ir. where
77,
is the load voltage, and -
V1
: V"(2:
is the load current defined as
17.
l)
:Vo*ett ny-"tl 11
: I,(2: :Ya*-"
[1
""
!
Vn
2,,
""
Following are some impcirtant points related
to input impedance
transmissiorr line.
FOIltr$ Ts nEilFlilBxn 1. , 3'ar ,* shsrted tran*rnissi,ort
line,
(Zio)
Zr.:
Ze: A, the input impedance is : Z.ranhll
"@,
2.
For open circrdted line.
3.
(Zn)o' : Zocoth-Yl From the abole two results, we have
(Zi^)*(26)*
4.
For matched line,
Zt-
the inprrt impedance is given by
:
Zfi 26,Lhc input impedance is Zrn
-
Za
r
Reflection Coefficient
Page 52e , rt
At the load termina.l,,the voltbge reflection coeffftient is defined of voltage re{lection Wave to the incident wave, i.e. n V;e''
as the
ratb,
'L - vte:i
which ean be further expressed
as
n -Zr-Zo 'r-Zr+Zo The current reflection coefficient at any point on the line is the. neg?tive the voltage.ieflection coefficient at that.point, so we have Io'e't _rn L
rb'Ssr-ess
oJ.
-_fitr
rRnilsiltsglolt qfi
E
A transmission line is sd,id'to be lossless if
the' conductors of, the line,ar.e. perfect, or oc: oo and the d,i'electric medium between the, lines 'is lgsslbss, et , o 4 :0. Some important characteristics of a lossless tranbrnissiort line ate given below
Primary Constantb of a Lossless Line For a lossless transmissircn line,
1. 2.
lr
Series'riasistance is zero, i.e.
-B:O
Shunt conductance is zero, i,e.
G:
0
Setondary Constants of a [ossless Line Using the above condition for lossless line, ve obtain a generalisd expression for secondary constants as follows:
Itopagation Cotrst
1
ory
: ,t@ + vuL){G + jwe : \/$ +MQT6q : julEO = jP:iaJLC
Thus, the attenuaiion and phase aonstants sf transmission line is giryen by
0:0,9:uJLC Characteristic ImPedance The characteristic irnpedance of a lossless tra,nsmissiotr line is obtdined
Zoa
-\/ 4.S
/-T C
Velocity of Wave Propagation in a Lossless Llne The velocity of propagation in a lossless transmissiron line is given by
,r: u1 p:m ,4.4
Input Impedance of a Lossless Line We have just derived the propagation constaart for a lossless line as
as
chp
8
lbwiseimfiD€8
r "Y:jp
Page 530
Chap 8
,
flansmissisl Linsg
So, we have
tanhTl
: tanhj|l: jtanpl
Thus, the input impedance for a lossless line is obtained as
ry : zoLZ;+ ry[Zt* Zotanhill
z,i"
Z;iunh.yl l I Zt -l jzotan gll
, - "olzJ jzLtanml 8.5
DISTORTIONLESSTRANSMISSIONLINE A transmission line is said to be distortionless when the attenuation a is frequency-independent and the phase constant, p is linearly on the frequency.
8.5.1 Primary
Constants of a Distortionless Line
For a distortionless line, the required condition is satisfied when the constants are related as
RG
T:E 8.5.2
Secondary Constants of a Distortionless Line Using the above condition for distortionless line, we obtain expression for secondary constants as follows:
Propagation Constant The propagation constant of a distortionless transmission line is obtained
1
: r/@+ jufi@T j,Q : f nc(.+)(,++
:/nc(t.+):a*io Thus, we obtain the attenuation and phase constants
as
:,/ RG g:r/LC o
Characteristic f mpedance The characteristic impedance of a distortionless line is obtained
as
- tR-+ iaT 4u-\/ G+pC
- | FGrlil e(+ rdfq tE tT :\/G:t/e 8.5.3
Velocity of Wave Propagation in a distortionless Line The velocity of propagation in a distortionless transmission line is given by 1 .. _a_ _B_JE
",
The comparison between the propagation parameters of transmission lines are given in table below.
various
Table 8.2: Propagation Parameters for Different Types of Lines
Ftr t
Page 531
Chap 8 I!'ensnission Lines
iittr,6l,lffi: :
a
',s.
g
ls#f6,.
zo
,ffi',*..ffi :.($.
r&
.9,
E.6
STANDING WAVES IN TRANSMISSION LINE
If the receiving end of a transmission line is not perfectly matched, there will be reflection of the voltage and current. As a consequence of reflection, a stand,i,ng waue r;rray be visualised as an interference between the incident signal I/6+ at a given frequency, travelling in the forward direction, and the reflected signal tr/6 , at the same frequency, travelling in the reverse direction. Standi,ng waue rat'io is defined as the ratio of the maximum voltage (or
current) to the minimum voltage (or current) of a line having standing waves.
V^^" 1-"" 1+ltLl
" -E
-4;-irll-,l
Q_vmu_rmu_'tl'L
Following are some important points about standing waves: POIIITS TO REMEMAER
,.1,,,,,,1,St
.*a@,,,,a,r*.so;nl1
,:l$X**l;i*,wi*.'l X
but never moves literally.
along the line is constant, .lMe.
Bo
,fur,q*plitude
h[e'Sl
8.7
bubre
SMITH GT{ART The Smith chart is basically a graphical indication of the t*o"dr""" transmission line and of the corresponcling reflection coefficient as one "r.l -oo'o I along the line. The Smith chart is constructed within a circle of unit radils I (l i. I 5 1) as shown in Figure 8.4. f,. and f, are the real and imaginary purrc I the reflectio"
tr*hndtdrtiht,tffi$
-ltt:Tl',il:
r,+
jri
@l
(8
uI
i trt:t,s-lll:l,s:m
I
lrl:0, s:1
tr;igurr: I.,1:
Unit Circle on which Smith Chart is Constructed
Let the norlatized_^y ,,,,
-
for the transmission line
;:^"nce ,t : Z,: r* .lir Since, the reflection coefficient is defined
I
be
I ...(8.2)
a^s
:ffi:ffi1:t; r,:fo--$:e-t:zt'-r
"'(8'3)
So, by solving equations (8.1), (8.2), and (8.3) we get
2,.:r*i,:f*,Pfi
8,7.1
..(84)
Constant Resistance Circles Equatingthe real parts in equation (8.4), we get the expression for r-circles on Smith chart
as
_*[, _ [r ,=rl' + r? i
These circles are also called constant'1, resistance circles. Figure g.5(a) shows typical r-circles for a transmission line. Some important properties of r-circles are given below. PROPERTIES OF T.GIRGLES t 1
ir
i
"t; 2.
tfw.ee$tfe$:of:rit1''i1"i1*1es'1ie on the
The
r:
C
,,', , . ,r ':
'
0 circle having a unit radius and centred at the origin, is the
largest.
3. The r-circles become progressively
smaller as r increases from 0 r;fl * 0) point'fiai apon iircuit. 4. All r-circles pass througtr ttre (J'" : L,l;:0) point.
.ir',,r;,,r l lra
at the
(fl.,==
,
hn h*rb (r, t12)
(u) ,
liiguri:
r.i:
(b)
Typical Circles on Srnith Chart: (a) 'Iypical r-circles, (b) Typical tr-circles
i
J,7.2
Constant Reactance Circles Again, equating the imaginary parts in equation (8.4), we get the expression
for r-circles
as
lf *[.,- ]1":[]l' lr.-l These circles are also called constant resistance circles. Figure 8.5(b) shows typical r-circles for a transrnission line. Some important properties of r-circles are given below. i i
I t
l.
,,,
**,prrrE*ell*-.,,,
',,;
i;;ti;::;:;i!;;;i'i.,,;iti rii'
L',The,ce , - 'of an'r*eides"lie,on the J-.= I line; tfuqry.fsr:s,>.9iL ,r..,,.,,.{induiiive',ieaet gl ,,.abo1g.,1t}&.. +elti*;t*.hd.,.!}i!f$',SoI.:-,ri{,0 (capacitive reactance) lie below the f'-a:ris. .
h
I I
t
: z:::':'::
me: :A * 0',cirole b'eiomiig''the &a&tr*']l'],-,',].
.;::,
r', r r
-':
r'"r':'
,e,r'tt" ,iiecome progree*i.vely, sniailgf asllcl,,ix*!ryi4'hsq,O" "-ctrcles oo, ending'xrili'g{ *r fl.r; :,0i,,poii1t .. : ; tnward ,9ry1_ :itcuit, ::r4, 911, p-ass tbroq :thd,(1.:; *:,,1; fi,*-.0):.p4&ii. :,:,: : , :: , . ,..:: , . l :':t : :t, . ' . .
_,r,,, .:
"*;rcles
t
.
.'.::aa
.'.'
Application of Smith Chart Smith chart can be used for the following purposes: 1. To find the normalised admittance from normalised impedance
a,nd
vice-versa.
F
2. 3. 4. 5. 6. 7.
To find the parameters of mismatched transmission lines To find the standing wave ratio for a given load impedance' To find the reflection coefficient. To find the input impedance of a transmission line. To locate a voltage maximum on a transmission line'
To design stubs for impedance matching. The methodoiogy to determine some Iine characteristics using smith chart is given below.
hf
Page 534
Chap
tullFf$ry!$rF ras, trrftE €Hamcr*Rlsrrcs
E
Ilansmission Lines
% is xSt
':i$:rffu9,,,{6$
.sti :€I
step
t*:.
:
ry::'llYagU.
:.Seith.chert at,point P where:'r
ee*i.
:
d:'circle
5a,,
.'..,1.i,;:r.::::rl.illi
Z; + &oa, 8.8
TRANSIENTS 01{ TRANSTIISSION LINE
When a pulse generator or battery connected to a transmission line is it takes some time for the current and voltage on the line to reach steady values. This transitional period is called the transient. Tyansit time is defined as the time for the waves traveling in the positive direction to reach the load and interact with it, i.e. switched on,
T:l where
I
ap
is the length of the transmission line, and 0 wave along transmission line.
,r.,0
is the velocity of the
Page 535
Chap 8 flnnspis5i6a [,ia6s
f&l
Instantaneous Voltage and Current on I!'ansmission Line
After ?
secondsT the wave reach the load. The voltage (or current) at the load is the sum of the incident and reflected voltages (or currents). Thus,
V(l,T)
and
I(I,f):
the generator,
so
:
V+
* V-: V* lr.V:(t+
1+* I-
: Io- lr.Io:(I *
tr)V" fL)l"
where l-z is the load reflection coefficient. The reflected waves V- : lr,V and 1- :- lt lo travel back toward the generator in addition to the waves % and 1" already on the line. At time t - 27, the reflected waves have reached
: V+ * V- : lclt V+(t + ft)V : (1+ f r.t fcf t)V" I(0,2r) : 1+ * I- : - lc(- ILI") + (7 * rL)l" :(1_ ft-ttr,fc)l"
V(0,27)
and
where J-6 is the generator reflection coefficient given by
Zo- Zo rc:Zo+Z Again the reflected waves (from the generator end) trz+ : lcltV, and I+: le lr,Io propagate toward the load and the process continues until the energy of the pulse is actually absorbed by the resistors Zn and 27. 8.8.2
Bounce Diagram Instead of tracing the voltage and current waves back and forth, it is easier to keep track ofthe reflections using a bounce diagram. The bounce diagram
Page 6fG
chrp
E
Xtagrnitdoa
fin6
consists of a zigzag line indicating the position of the voltage (or curr wave with [email protected],the gbnerator end, as shown in Figure 8.7. On bounce diagram, the voltage (or current) at a,ny time may be determined by adding those values that appear on the diagram above that time.
f:fr, z: I
l: lc z:0 f
=
0
f:-lz z: I
f:-lc
z:0 ,:0
2tr
-tctlt"
rcr.lu" 4tt
rir?.u
.
i i
rZrT
6tr
I
(")
(b)
I
l
i:'igurt ii-7: Bounce Diagrarn for (a) Voltage Wave, (b) Currerrt Wave
I
I
l *
f**********
iI
:i
I
'I ",2'ili '' ,t,' ' "',,
:l#
"
I
PrggSt?
''€'lbp
E
llennmisgisa Lin€s
HCg 8,{,1 I
i
I
-t
[co
a.1.2
Assertion (A) : A sinusoidal voltage u;: Vscos(4 x 10an.t) is applied to the input terminal of a. transmission line of length 20 cm such that the wave propagates with the'Velocity c : 3 x 108 m/s on the line. It's outprrt voltage will be in the same phase to the inpirt voltage. Reason (R) : Tlansmission line effects can be"ignored if.{ < 0.01. where" wave.
I is the length of transmission
(A) (B) (C) (D)
arid R both are true and R is correct explaruif,ionrof A. and R both are true but R is not correct explanation of A. is true but R is false
A A A A
is false
HCd 8.r.4
The space between the strips of a parallel plate transmission line is filled of a dielectric of permittivity, e,:1.3 and conductivity, o = 0.If the wi{th of the strips is 9.6 cm and the separation between them is 0.6 cm then the line parameters G' and C' will be respectively
(B) 0.02mS/m, 0.14nF/m (D) 1.8 mS/m, 0
"
Inductance and capacitance per unit length of a lossless transmlssion line are 250nHlm and 0.1nF/m respectively. The velocity of the wave propagation and characteristic impedance of the transmission line are respectivelv.
(A) 2 x 108m/s, 100O (C)2x108m/s,50f,)
A
(B) 3
:0.2k0/m, L' :
propagation constant, 7 will be (A) (0.5 * j1.2) m-'
(C)
+p.5)m-1
(1.2
t08m/s, 50fl 100f)
at a
frequency 6 x 108 ra
(B) (0.10 * p,.4)m-l (D) (2.4+p.10)m-t
The parameters of a transmission line axe given as R' : Lo?lrn', L' :0.IpH/m, C' : I0 pF/m, G' : 40pS/m. Ii ttre transmission line is operating at a frequency, a:1r2 x 10erad/s then the characteristic impedance of the line
MCQ 8.t.6
x
(D)3x108m/s,
transmission line operating
parameters R'
illcQ Lt.s
ind ,\ is the wavelength of the
but R is true
(A) 0, 0.02 nF/m (C) 0, 0.18nF/m
htt t''' J fr
line
(A)
50
(c)
100
-
wiil
be
J2Q
- j|o
(B) 4 - j100f) (D) 100 + rto
A parallel plate lossless transmission line consists of brass strips of width u and separatbd by a distance d. If both ,u and d a,re doubled then it's
characteristic impedance will
(A) halved (C) not change
(B) doubled (D) none of these
Page 538
r**{l
8"1.7
Phase velocity of voltage,wave
$c&
8.1"8
A
Qhap 6 T!'onsmission Lfurcs
in a distortion less line having characteristic impedance, Zo : 0.1kO , and attenuation constant' o : 10 mNP/m is ap:0.5 x 108m/s. The line parameters ,t' and L' will be respectively (B) 10kQ/m, 2p"Hlm (A) 1A/m,0.5nH/m (C) 2Olm, 1pH/m (D) 1A/m, 2p"Hlm distortionless line has parameters .R' :4o,lm and G' :4 x 10-4S/m. The attenuation constant a,nd characteristic impedance of the transmission line will be respectively
(B) 100 NP/m1 4 x 1o-2 f) (D) 0.01NP/m, 25C)
(A) 25NP/m, 0.01O
(C)4x10-2NP/m, loo0 n*c0 s.t.E
A
150
f) transmission line is connected to a 300 C) resistance
and to a 60 V
DC source with zero internal resistance. The voltage reflection coefficients at the load end and at the source and of the transmission line are respectively (B) -1, -1 (A) - L, 713
(c) rl3, t13
Mcg $.1"{$
A purely
(D) 1/3,
-1
resistance load, Zr, is connected to a 150 0 lossless transmission it has a voltage standing wave ratio of 3. The possible value
Iine. Such that
of Zr will be
(B) 450 O (D) none of these
(A) 50 CI (C) (A) and (B) both
voltage generator with or(t) - 3cos(er x 10ef) volt is applied to a 100 O lossless air spaced transmission line. If the line length is 10 cm .and it is terminated in a load impedance ZL:(200-1200)Q then the input impedance of the transmission line will be (B) (i2.5 - j12.7)a (A) (50 - jso.s)c, (D) (25 - j25.\a (c) (25.4 - 125) o
8.1'{1 A
8.{.{2
Phase velocity of a voltage wave in a transmission line of length I is oo. If the transmission line is open circuited at one end and short circuited at the other end then the natural frequency of the oscillation of the wave will be
(t) Tf ; n:0,1.......oo e) Q!#1"1 n: r,2,8,....a ereQ 8"t.{3
(D)
P!#!"
;n
: 0,1,.......*
fff; n: 1,2,3.....oo
At an operating frequency of 500H2, length of a transmission lihe is given' by l: .\/4. For the same transmission line the length at l kHz will be given by
(A)
(c) MCO A-1.{4
@)
': *
l: +
'
(B)
I:+
(D) none of these
A transmission line is operating at wavelength '^'. If the distance between successive voltage minima is 10 cm a,nd distance between load and first voltage minimum is 7.5 cm then the distance between load and first voltage maxima is
(A) .\/8
(c) 5ll8
(B) 3^/8 (D) ^14
i
'GQ
8,'tr,tS
A z-polarized transverse erectromagnetid wave (TEM) propagating along a parallel'plate transmission line filled,of perfect diereciric t; ;; direction. Let the eiectric ahd nragnetic fierd of the wave be E and. -Er respectively. Which of the following is correct relation for the fields.
(rt#:o
@+:;
(C) Both (A) and (B) IGQ
S,{,16
lcQ 8"{,'t?
(D) none of these
Distance of the first voltage maximum and first current maximum from the load on a 50 o lossless transmission rine are respectively 4.b cm and 1.5 cm ' If the standing wave ratio on the transmission rine is ^9: 3 then the load impedance connected to the transmission line will be
(A)
(e0
(c)
-
(30
- to)o
jr2o)Q
(B) 10 f, (D) (40 - 730)c)
Total length of 50 Q lossress transmission rine terminated in a load. impedance zL : (30 + J15) f) is / : 7 as shown in figure. The total input impedance across the terminal AB will ^120be
,7\ 20
zL:$0+fl')
rcQ 8,1.18
(A)
(38.3
(c)
(64.8
- 164.8) 0 - 738.3)fl
a
(B) (1e.2 - j32.4)a (D) (32.4 - jrs.2)a
Assertion (A) : The input impedance of a quarter wavelength long lossless line terminated in a short-circuit is infinity. Reason (R) : The input impedance at the position where the magnitude of the voltage on a distortionless line is maximum is purely real. (A) A and R both are true and R is correct explanation of A. (B) A and R both are true but R is not the correct explanation of A. (C) A is true but R is false.
(D) A is false but R is true. Common Data For e. lg and 20 : A voltage generator with oo(l):10cos(gzr xr07t-30) and an internal impedance zs: 30 o is appried to a 30 o iossless transmission rine that has a relative permittivity e,:2.2b and length, /: 6 m. :Q &{.{9
If the line is terminated in a road impedance, 21:(g0- j10)o, then what will be the input impedance of the transmission line ? (A) (0.05 - p.01) CI (B) (50.62 p3.a8)e
(c)
!Q 8.{.20
(e2.06
-
21.80)
CI
+
(D) (23.14 +75.a8) A
The input voltage of the transmission line will be (A) a-acos(8?T x 107r+ 22.56") (B) a.acos(8rr (C) a.acos(8n x 707 t- 22.56")V (D) a.acos(8n
i lorr x
102,
-g7.44")v _ g0.) V
Page 539
Chap 8 Tlansnissi6n I/i[q6
Common Data For Q. 21 and 22 : Two equal load impedances of 150 o are connected in.parallel through a of transmission line, and the combination is connected to a feed transmissil lien a^s shown in figure. All the lines are lossless and have characteristh impedance Zo: 700Q.
Page 540
Chap 8 Tlansmiesion Lhes
150
M{:& A,l.2t
0
The effective load impedance of feedline (Zi,') equals to
(A) (7.04 - it7.2qa (B) (35.20 + 7s.62)f) (c) (35.20 - i8.64a (D) (s.62 + 135.20) {-l MGQ 8.1.22
The total input impedance of the feedline (line 3) will be
o (c) (215.14+j113.4)CI
(A)
lrlcQ s"t"23
(2.15
- jr.rr;
(B) (215.14 - ji13.4)o (D) (107.57-756.7)O
GHz voltage generator with 7,, : 150 volt and an internal resistance zg:100o is connected to a 100o lossless transmission line of length l: 0.375,\. If the line is terminated. in a load impedance zt: (100 - J100) o then what will be the current flowing iu the load ?
A
0.3
(A) 0.67cos(e x rOst- 108.4") (B) 0.67cos(6zr x 108t- 108.4") (C) 75cos(3 x 1o8r- 108.4') (D) 0.67cos(6zr x t08t- 135") MCO 8.{.24
i,tcQ 8.{"25
The input impedance of an infinitely long transmission line is equal to it's characteristic impedance. The transmission line will be (B) lossless (A) slightly Iossy (D) (B) and (C) both (C) Distortion less
At time f : 0 unit step voltage generator 7, with an internal R, is applied to permittivity € : Rs
\;,
resistance
a 100 Q shorted transmission line filled with dielectric of 4€o as shown in figure
Short
circuit
The voltage 'raveform for any'iime figure below
lEoat
the sending end is sown in
Page 541
Chap
E
Ilansmission Lines
6V 1.5
V 0
Vn and,B, will be respectively equal to (A) 30volt, 19.2Q (B) 3S.4volt,60f) (C) 60volt, 38.4f) (D) 19.2volt, 30Q
Cornmon Data For e. 26 and 27 : A 1'5 m section o an airspacd lossless transmission line is fed by a unit step voltage generator % : 30 volt with internal resistance Rg:200O. The transmission line is terminated in a resistive load Zr,: b}e and characterized bY
Zs:100f).
rcE 8:t.26 The
bounce diagram of the transmission line
l": l/3
z:0
Tr:-r/3 z: Lbm
will
be
Il :1/3
\:-1/3
z:
L:
5ns
I.JM
10 ns
10 ns
(A)
(B) lll
ns
30 ns
20 ns
ft:-I/3 z: Lbm
I
tr:
= 1/3
113
Z:L5m
5ns
5ns 10 ns
(c) 15 ns 0.74 Volt
(D)
-i.rr
vori
15 ns
Page 542
rR,Chal '
s
a.'i"c? The
instantaqeous. voltage waveform
u(t)
at the, sending end
of
the
transmission line.will be
TYansmission Lines
o(t), Volt
t 10
(B)
(A)
i.
6.67 5.6
10
10
t(nsec)
t(nsec)
r4.4 13.3 10
'(D)
(c)
1o
s..r.zs
t(nsec)
1o t(trt"c)
The SWR circle' LrLz' is shown on the smith chart for a lossless transmission line.
If line'is
terminated in a load.,Zr.= 50e.then the possible value of the characteristic impedance of the line will be
(A)
0 (c) 20 0 125
(B) 250 0 (D) (A) and (C) Both
Page 543
Chap
E
IhansmiEsion Lines
Common Data For Q.29 to 32 : A lossless transmission line characterized by Zo:100O is terminated in a toad Z7: (100 +yb0) O
8.1.2e
The reflection coefficient of the line will be
(A) 4.+e-'''u' (C) +.+et'a'
8.{.3o
8'{'31
The input impedance at a distance of 0.35,\ from the load will be
(A)
(0.61 _
(c)
(61
fr.22)o
- r2.2)o
(B) (61 +p.2)a (D) (0.61 + J0.022)a
The shortest length of the transmission line for which the input impedance
appears to be purely resistive will be (A) 0.25^
(c) McQ 8,1.32
(B) o.24en6' (D) 0.24en6'
0.106^
(B) 0.456^ (D) 0.544)
The first voltage maximum will occur at a distance of
(A) 0.106.\ from toad (B) 0.1441 from load (C) 0.106.\ from Generator (D) 0.144) from generator 'o"o
u'@
jlaismission line of characteristic |inductor as shown in the figure.
impedance 50 e is terminated by an
A positive wave with constant voltage Vo: l volt is incident on the load terminal at t:0. At any time f the resulting negative wave voltage at the load terminal will be (A) (1 - 2e-2,t)Yott (B) 12"-ru, - 1)Volt (C) ze-25'Volt (D) 1"-2" - 1)Volt
8't.34 A
transmission line has the characteristic impedance zn and the voltage standing wave ratio is The line impedance on the transmission line at ^9. voltage maximum and minimum are respectively.
(^) z,s, (C) zrS,
?
zoS
zos
@)
?,
@)
?,?
I
Page
i4a
HCA Ll"3$
Chan,Q Ttansmissior Lines
Consider th; iir."e;d$iiiii{s'&iirtri'sic impedanc€s ?r r \2 and,4, respettlvd as $hown in tit" figl?e1 Wtiai'will be the thickness 't' and intrinsic impedanc ,rlz,,ofthe, medium 2 for which the refleitbd wave having wavelength'')'ir
eliuriuated in medium 1 are
(A) (B) (c) (D) s.t.36
intrinsic impedance
thickness'f'
42
frt'Tt
,6Jn' {nJ,t' {nnt
^14 ^12 ^14 ^12
A transmission line has characteristicp impedance 100'Q and standin$'warc ratiq 3. The distance between the first voltige maximum and load is 0.1251 . Load impedar,rce of the transmission line is
(A)
(Bo
+to)f,
(c)
(3'o
- j4o)o
8"r.3? A
(B) (60 + i8o) o (D) (60 - j80)o
10OO lossless transmission line with
it's parametet L' :0.25pH1rn
and
A
c' :100PF/m is terminated by it's characteristic impedance' 15V voltage source with internal resistance 50 O is connected to the transmissirr line at t:0. Plot of the voltage on the line at a distance 5m frdm th source against time
will
be
I
I j
10v (D)
mco B,{,3S
lossless transmission line terminated by a load impedance Zt, * Zo is connected to a D.C. voltage source. Ttre height of the first forward voltage pulse is yi. If the voltage reflection coefficients at the load and source a.re respectively Ir, and "ln then the steady state voltage alross the load is
A
'r[ffi] (c) yr(f#)
tB)
1aj
(D) *x*xxr******
vr(#)
yr(1+)
EXERCI$E 9.2
Page 545
Ghap
t
Ilalrgdidlon I{Bet
QUES 8"2,1
A
transmission line is formed of coaxial line with an inner conductor diameter of 1 cm and an outer conductor diamet er of 2 cm. If the conductor has permeability p":2p' and, conductivity o":l!.6 X 10?S/m then it,s resistance per unit length for the operating frequency of. 4GHz will be
----
CI/*.
QUES 8.2"2
A transmission line formed of co-axial line with inner and outer diameters l.b cm and 3 cm respectively is filled with a dielectric of permeabirity :2p.. 1r It's line parameter .L' will be equal to ____ nH/m.
ou€s
A co-axial transmission line is filled with a dielectric having conductivity, o :2 x 10-3 s/m. If the inner and outer rad.ius of the co-axial line a,re
9"2.3
1/8 cm and Ll2cm respectively then the conductance per unit length of the
transmission line will I I
qu€s
be
mS/m
I'n'4 If permittivity of the dielectric filted inside the coaxial
transmission line having inner and outer diameter 1 cm and 4 cm respectively is €: ges then the capacitance per unit length of the line will be pF/m.
____
QUES 8.2"5
A parallel plate transmission line consist s of 2.4 cm wide conducting strips having conductivity, o:1.16 x 108S/m and permeability p: pa is operating at 4 GHz frequency. what will be the line parameter ft''(in him )?
quxs
8.2"S
,4. parallel plate transmission line is formed by copper strips width w: 4.8 cm separated by a diotance d: 0.Bcm. If the dielectric fiiledofbetween the plates has perrneability, pr :2po then what will be the ind'ctance per
unit length (nH/m) of the transmission line QUr$ s,2.7
?
A 1 GHz parallel plate transmission rine consists of brass strips of conductivity o : 6-4 x 107 s/m separated by a .dielectric of perryrittivity e : 660. If the axial component ,and transverse component of the electric field in the transmission line is .8, and E, respectively then E" I Eu equals to x
10-5.
QUES
Lt.s
Qucs
8'2'e Amplitude of a voltage wave after travelling a certain
After traveling a distance'of 20 m along a transmissfon line, the voltage wave remains 13 % of it's source amplitude. what is the attenuation constant (NP/m) of the transmission line ? distance down a
transmission line is reduced by 87 %. If the propagation constant of transmission line is (0.5 + p. )tben the phase shift in the voltage
Page 546
Chap E Trmmission Lines
-u'*l
degrees.
ours $,2.{o A transmission line operating at 5 GHz frequency has characteristic impedance Zo:800 and the phase constant 0:l.\rad/m. The inductance per unit nH/m. Iength of the transmission line will be
I
I I
----
I I of 6volt and minimum magnitude of.2.4volt. The reflection coefficient of I the transmission line is I I euEs 8.2"{2 An insulating material of permittivity e : 9es is used in a 25 O lossless co I axial line . If the inner radius of the coaxial line is 0.6 mm then what will be I
eu;s
a,n.ri
The voltage wave in a lossless transmission line has the maximum
it's outer radius (in mm)
ouss
magnitude
t
I line of cha,racteristic impedance Zo:25{l is I connected to a loa.d impedance ZL: (15 - J25)C). What will be the standing I
s.z.rt A
lossless transmission
. wave ratio on the line
?
I I
euE$
qucs
s.t.t4
lossless transmission line is operating at a frequen cy of.2MHz. When the line is short circuited at it's output end, the input impedance appears to be equivalent to an inductor with inductance of 32 nH but when the line is open circuited at it's output end, the input impedance appears to be equivalent to a capacitor with capacitance of 20 pF. What is the characteristic impedance (itt O) of the transmission line ?
A
section of a 100O lossless transmission line terminated in a 150O ^14 load is preceded by another .\/4 section of a 200 0 lossless line as resistive shown in figure. What is the input impedance, 26 (in Q)?
s.2"is L
150 Q
Cornmon Data For Q. 16 and 17: A load impedance Zt:(0.3 *"ro.s)k(-) is being connected to a lossless transmission line of characteristic impedance zo: 0'5 ko operating at
wavelength,\:4cm.
s.z.i6 What will be the distance (in cm) of the first voltage maximum load
?
from the
I I
l I i i
,
I QUss
6.r'{?
The distance of the first current maximum from the load will be
cm.
QUE$
a'2.r8 A
QUES
s,2.1s A voltage generator Vs:b}}volt with an internal resistance Zc:100A is applied to a configuration of lossless transmission lines as shown in figure. The power delivered to the load Zu will be watt.
QuEs
voltage generator Vg:Ib}V with an internal resistance Zs:I}}e is connected to a load zt : LS} e through a 0.15.\ section of a 100 o lossless transmission line. What is the average power (in watt) delivered to the transmission line ?
Zn:I50
Q
Zn:I50
Q
8"2'20 An infinitely long lossy transmission line with characteristic impedance Zot:200 O is feeded by a ),12 section of 100f) lossless transmission line as shown in figure. If a voltage generator V"c: 4y with an internal resistance zc: 700 Q is applied to the whole configuration then the average power transmitted to the infinite transmission rine will be mw.
Zoz:I0O
Qi
Zot:200
Q
Commsn Data For Q. 2l and,22 A unit step voltage generator is applied to a 100 f,) airspaced rossless transmission line at time, f : 0. At any time, r > 0 the voltage waveform at the sending end of the transmission line is shown in the figur" berow , z
Page 547
Chap 8 Tbangmission Lines
rage
Ftap
rje
&uns
8.x"2{ The length of the transmission line will be
meter.
E
Tlarsmlsdon Irines
eur* e.s-ss The unit step generator voltage connected to the line has an internal resistance Rs:100f). What will be the load impedance (in O) connected to the transmission line
?
Common Data For Q. 23 and 24 : A quarter wave dielectric of thickness 'l' and permittivity 'e ' eliminates reflections of uniform plane waves of frequency 1.5 GHz incident normally from free space onto a dielectric of permittivitv 16e0. (Assume all media to have F: F4) qu€$
8"2"23 The relative permittivity of the dielectric coating equals to
QuEs
8.*.24 What is the thickness'f ' (in cm) of the dielectric coating
?
60 CI transmission line, terminated by a load of 180 O is connected to a 0. The internal resistance of the source is 120 0. The 100 V DC source at f
ours s.?.ar A
:
steady state voltage across the load
eux$
s.z"ts At t :
0a
50
Volt D.C.
will
be
volt.
30 f,) is connected characteristic impedance having a load of 45 Q
source
with an internal resistance
to a transmission line of 15 fl . The stea.dy state load current for the transmission line Common Data For
is
Ampere'
q.27 and 28 :
of an unknown length terminated in a resistance is with zero internal resistance. The plot of input a battery to 6 V connected
A
transmission line
current to the line is shown in the figure below
-5 mA
QuE$
8.2.2? The characteristic impedance of the transrnission line will
be
Qtlxs E'2"2s The load resistance terminated to the transmission line will be *xt<********
CI
c)
xxHRGtSX 9.3 trts&
Page bZ{i
Chnp 8 tlangmie3$ii'f,isog
E"3"'t
which one of the following statement is not correct for a transmission line ? (A) Attenuation constant of a lossless line is always zero. 1l (B) Characteristic impedance of both lossless and distortitnless ]ine is real (c) Attenuation constant of a distortionless line is always zero. (D) Both (A) and (C).
iil*Q E"*.?
The wavelength on a lossless transrnission line terminated in a short circuit is ). what is the minimum possible length of the transmission line for which it appears as an open circuit at it's input terminals ?
(B) (D) ^12 M$*
^14 A"3"3
A transmission line of length / is short circuited at one encl and open circuited at the other end. The voltage standing wave pattern in the transmission line
will be
(A) -l
0pcn circuit
'-
(B)
*Short
circuit end
end
m
open circuit end
Short
circuit end
(c) Open
'
circuit end
n4c& &"3.d
A
Short
circuit
end
end
lossless transmission line is terminated
possible length of the line for which
tennirrals is
(A)
(c) MeQ 9,3.5
Open
circuit
^12 ^
it
Short
circuit end
in a short circuit. The minimum
appears as a short circuit at its input
@) (D) o^14
If the load impedance in a transmission line is 100 * j2o0 e and characteristic irnpedance is 100 O, the normalised load impedance is (A) 1+.720 (B) 10000 * 720000 Q
(c) 1+ pooa (D)
100 +
j2a
Page 550
Chap 8 ftensmiss,iqn I/ins5
MCQ 8.3.6
If the load impedance in a transmission line is zz, and impedance, reflection coefficient
(N\" ,", (rr+ ^), a) rc\
ZL .n
is
zo is
@9j4 ' \ztzo)
***xx**x***
(D) "Zt
&
the characterkli
i
EXERCISE 8.4
Page 551
Chap
E
Ilrnsmision Line
rcQ
9.4",1
A
coaxial-cable with an inner diameter of 1 mm and outer diameter of 2.4nrn is filled rvith a dielectric of relative permittivity 10.89. Given lh:4tr x 10-7 Hfm, e0:#F/m, the characteristic impedance of the
cable is
f'
(A)
330
(c)
143.3
(B) 100 o (D) 43.4 o
c'
transmission line with a characteristic impedance of 100 O is used to match a 50 Q section to a 200 0 section. If the matching is to be done both at 429MHz and 1GHz, the Iength of the transmission line can be approximately (b) 1.05 m (A) 82.5 cm (D) 1.75 m (C) 1.s8 cm
rca
a.d"t
A
rcc
8,4.3
A
transmission line of characteristic impeda,nce 50 f) is terminated by a When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be rf 4 radians. The phase velocity of the wave along the line is (B) 1.2 x 108 m/s (A) 0.8 x 108 m/s 50 Q load.
(C) 1.6 5GA
A"4"4
x
108
m/s
(D) 3 x 108m/s
A transmission line of characteristic impedance 50 O is terminated in a load impedance Zr,.The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of ),14 from the load. The value of Zr, is (B) 250 f' (A) 10 0 (c) (1e.23 + j46.15) O (D) (1e.23 - j46.15) o
:
HCo 4.4"5
If the scattering matrix [S] of a two port network is [,9] , then the network is (A) lossless and reciprocal (B) lossless but not reciprocal (C) not lossless but reciprocal (D) neither lossless nor reciprocal
HCQ 8.4.6
A transmission line has a characteristic impedance of 50 Q and a resistance of 0.1O/m . If the line is distortion less, the attenuation constant(in Np/m)
l::ffiS.?ff]
is
ffcq
8.4.7
(A)
5oo
(c)
0.014
(B) 5 (D) o.oo2
In the circuit shown, all the transmission line
sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 O line is
j
?r*#2 Chap E fbalonfoshF LittGt
mca s.jt"8
t-:
Z.=50 Q
z,l
MCA a"4.9
MCQ 8.4,lll
Z.:50
Z.:50 A
:;[: ntgo 8"4.{{
;1
Q
(")
,",
[? l]
-i]
[_+
The parallel branches of a 2-wire transmission line are terminated in 100 O and 200 o resistors as shown in the figure. The characteristic impedance of the line is Zo:500 and each section has a lengtn of )' The voltage reflection coefficient f at the input is
\ P3ge 553
Cbp t Ilorsmission LiDcs
(A)
.7
(B)
-Jg
.5 (c) J7
xcQ
8.4.12
(D)
+ +
A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is (A) 10 Watts (B) l Watts (C) 0.1 Watts (D) 0.01 Watt
utcQ 8.4.{3
Characteristic impedance of a transmission line is 50 Q. Input impedance of the open circuited line is Z*:100 + j1500. When the transmission line is short circuited, then value of the input impedance will be
(A) 50 c' (B) 100 +j150 o
(c)
7.6e
+ il1.54 0
(D) 7.6e - j7r.540
Comnon Data For Q. 14 to l5 Voltage starrding wave pattern in
a
:
lossless transmission line with characteristic
impedance 50 and a resistive load is shown in the figure.
\ a.4.14
The value of the load resistance is (A) 50 0
(c)
8.4"t$
\/2
12.5
0
(B) 2oo 0 (D) 0
The reflection coefficient is given by
(A)-0.6 (c) 0.6
(B) (D)
-1 0
Page 5!4
Chap
MCO 4.4"t6
E
Tlaosmission Lines
MGA A"{-{?
Meq 8.4.18
(A) (B) (C) (D) MCA 8.4.19
adding an inductance in series with Z adding'a capacitance in series with Z adding an inducta^r/ce in shunt a"$oss Z adding a capacitance in shunt anross Z
A
lossless transmission line is terminated in a load. which reflects a part of the incident power. The measured vswR is 2. The percentage of the power
that is reflected back is (A) 57.73
(c)
0.11
(B) 33.33 (D) 11.11
- --t
rGGl
s.4'2o A short - circuited stub is shunt connected to a transmission line as showrr in fig. If Zo:50O, the admittance Y seen at the junction of the stub and the transmission line is
Page 555
Chap
E
Tlnnsmission Lines
Y
(A)
(B) (0.02 - p.01) mho (D) (0.02 *fl) mho
(0.01
- fl.02) mho (C) (0.04 - p.02) mho HGQ $,4,2t
The VSWR can have any value between (A) 0 and 1 (B) -1and +1 (C) 0 and m (D) 1 and oo
s.4.22 In an
impedance Smith chart, a clockwise movement along resistance circle gives rise to (A) a decrease in the value of reactance
a
constant
(B) an increase in the value of reactance (C) no change in the reactance value (D) no change in the impedance
a.4.23 A transmission line (A)
RL:h
is distortionless
if (B)
(c) LG: RC
ftr:
GC
(D) .RG: LC
s'4.24 The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are impedance of the line is, (A) 25 CI
(c) $cQ
8.4.25
75
100
Q and
o
25
O respectively. The characteristic (B) 50 o (D) 1oo 0
In a twin-wire transmission line in air, the adjacent voltage maxima are at 12.5cm and27.Scm. The operating frequency is (A) 300 MHz (B) l GHz (C) 2GHz (D) 6.28 GHz
s.4.2s In air, a
line of length 50cm with L:L}VHlm, at 25MHz.Its electrical path length is (B) ) meters (D) 180 degrees
lossless transmission
C: 40 pF/m is operated (A) 0.5 meters (C) rlZradians
J
Page 556
A transmission line of 50 o characteristic ir4redarrce is t<:rminated q'ith 100 0 resistance. Ihe minimum impedance measured on the line is equal (A) o0 (B) 25 {^t (c) 50 o (D) 1oo 0
adsf; s.4"t?
Chap 8 Ilansmission Lines
MEQ
8.4"18 A very
lossy, \f 4 long, 50C) transmission line is open circuited at load end. The input impedance measured at the other end of the approximately (A) 0 (B) 50 f) (C) (D) None of the above '>o
rxsa 8.4,?s
A lossless transmission line having 50 0 characteristic impedance and lengtl )/4 is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is (A) 0 (B) 0.02 A
tc) xca
-
(D) none of these
The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Ze respectively. The velocity of a travelling wa\-e
8,4"30
on the transmission line is
'' i,tcE
(A) ZoC
@#
ZU ,n, 1.-i c
@fr
8'4.3{ A
line, shorted at one end, presents impedance at the other end equal to
(q
(A)^14 Zo
(c)
*
lt
zo
(D) 0 where Zo is characteristic impedance of the line. rrrlcQ 8"4.33
Ill*Q
A
75 f) transmission line is first short-terminated and the minima locations are noted. When the short is replaced by a resistive load -R7,, the minima locations are not altered and the vswR is measured to be 3. The value of -Bi is
(A)
25
Q)
225A
o
8'4.33 If maximum
(B) (D)
50
f)
250 C'
and minimum voltage on a transmission line are 4 V and
2
V
respectively, VSWR is
rv!s& 8.4,34
(A)
0.5
(c)
1
(B) 2 (D) 8
An ideal lossless transmission line of Zo: 60 f) is connected to unknown . If. SWR: 4, find Zt (A) 240 CI (B) 480 0 (c) 120 (D) 100 c) .
o
nticc s.rt"35
Loading of a cable is done to
1. 2. 3. 4.
Increase its inductance Increase its leakage resistance Decrease
its leakage resistance
Achieve distortionless condition
Z1
-'t
I
(A) 1, 2.3
and 4
Page 557
(B)land3only
Chap
E
Thansmission Lines
(C) 2and3only
(D)land4only rco
I'd'36
Given a range of frequencies, which of the following systems is best for transmission line load matching ? (A) Single stub (B) Double stub (C) Single stub with adjustable position (D) Quarter wave transformer
lcQ s'4.3? A line of characteristic impedance . The VSWR on the hne
50 f,) is terminated
at one end by +j50
f,)
rs
(A)
1
(B)
(c)
0
(n)
-r
lca s.4"3ll At UHF
short-circuited lossless transmission lines can be used to provide appropriate values of impedance. Match List I with List II and select the correct answer using the code given below the lists :
List
I
a. l<^14 b. Al4 < I <
c. I: d. l: Codes
\14
List
II
1. Capacitive 2. Inductive
^12
3.0 4.
\12
oo
:
abcd (A) 21.43
(B) 3142 (c) 2413 (D)34r2 llcQ 8..4'3$ Consider the following statements regarding a transmission line : 1. Its attenuation is constant and is independent of frequencv 2. Its attenuation varies linearly with frequency 3. Its phase shift varies linearly with frequency 4. Its phase shift is constant and is independent of frequency Which of the above statements are correct for distortion less line (A) 1, 2, 3, and 4 (B) 2and3only
?
(C) land3only (D) 3and4only meq
8.4"40
The reflection coefficient on a 500 m long transmission line has a phase angle of -150'. If the operating wavelength is 150m, what will be the number of
voltage maxima on the line
(A)
0
(c)
6
?
(B) 3 (D) 7
_J
I
I .
Page 558
tucQ 8"4.41
Chap 8
with
regard
correct?
Ilansmission Lines
to a transmission line, which of the following statements b jI --l "
(A) Any impedance repeats itserf every ),/4 onthe smith chart. I (B) The swR : 2 circle and the magnitude of reflection coefficient :0-El -l circle coincide on the Smith chart. (c) At any point on a transmission line, the current reflection coefficient -- hl -I the reciprocal of the voltage reflection coefficient. (D) Matching eliminates the reflected wave between the source and the
]
matching device location.
MCQ S.4"42
It is required to match a 200 o road to a 4b0 e transmission line. To reduce the swR along the line to 1, what must be the characteristic imped.ance of the quarter-wave transformei used for this purpose, if it is connected directly to the load ? (A) e0 ko (B) 3oo CI
(q ?o r!,!cQ 8,4.43
(D);o
The load end of a quarter wave transformer gets disconnected thereby causing an open-circuited load. what will be the input impedance of the transformer ? (A) Zero (B) Infinite (C) Finite and positive (D) Finite and negative
MCq 8,4"44
A lossless transmission line of characteristic impedan ce zs and. length l< is terminated at the load end by an open circuit. what is its input ^14 impedance Zin?
(A) Zn": jZstanpl (B) Zo": jZncot pl (Q) (D) rr{C& 3.4.45
Zn":- jZstanBI Zo":- jZstanpl
Which one of the following statements for a short circuited loss free line is not correct ? (A) The line appears as a pure reactance when viewed from the sending end (B) It can be either inductive or capacitive (C) There are no reflections in the line (D) standing waves of voltage and current are set up along rength of the lines
l$c{t
8.4"46
Match List I (Load impedance) with List II (value of Reflection coefficient) and select the correct answer using the code given below the lists
List-I
a. b. c. d.
H
Short Circuit Open Circuit
Line characteristics impedance 2 x line characteristic impedance
List-II
1.0 2. -1 3' +1 4. +t/3
:
: a 2
Codes
(A)
b t
(B)43r2 (c)23r4 (D)4732
c 3
page db9
d 4
Chap8 TransmiEEionlineB
rcQ 8.4.4? When the reflection coefficient equals (A) Zero (C) 3 f,cQ
I/0"
whati, tn" VSWR? (B) 1 (D) rnfinite
8.4"48 If the reflection coefficient is 1/b, what is the corresponding vswR
(^) 312 (c) 512
@)
215
rcQ 8'4'4e Which one of the following is the characteristic impedance of transmission line ? (^) (B)
{RlG
lossless
{qc
g lrTrc-
rco 8'4'50
?
(B) 2/3
@ /R/c
I (Quantity) with List II (Range of Values) and select the correct answer using the code given below the lists : Match List
List-I a. Input Impedance l. b Reflection coefficient Z. c. VSWR 3. Codes
List-II _ 1to
*1
1to
oo
to
oo
0
:
abc (A)231 (B)321
(c)312 (D) 2r3 tlcQ
s'4'5{ A quarter
ucQ
0'4'52
wave impedance transformer is terminated by a short circuit. What would its input impedance be equal to ? (A) The line characteristic impedance (B) Zero (C) Infinity (D) Square root of the line characteristic impedance
Scattering parameters are more suited than impedance parameters to
describe a waveguide junction because (A) the scattering parameters are frequency invariant whereas the impedance parameters are not so
(B) scattering matrix is always unitary (C) impedance parameters vary over unacceptably wide ranges (D) scattering parameters are directly measurable but impedance parameters are not so
Page 560
Chap
iltcQ 8"4.53
E
Tlangnbeion Lines
ln a transmission line the reflection coefficient at the load end ,, ,rt"" J 0.3e-F0'. What is the reflection coefficient at a distance of 0.1 wavelengttl
*;;,,
towaros source
ill MCQ 3"4.54
i
,
l3l
3,:',::.,,""
To couple a coaxial line to a parallel wire,
lll
it
li*:
is best to use a
I
I
:
I
I
(C) Directional coupler (D) Quarter wave transformer
!\t!c{t 8"4,5s
sfi**
4.4"56
I I A plane wave having r-directed electric field propagating in free space u,toog I the z-direction is incident on an infinite electrically conducting (perfea I conductor) sheet at z: 0 plane. Which one of the following is correct ? | (A) The sheet will absorb the wave I (B) There will be r-directed surface electric current on the sheet I (C) There will be g-directed surface electric current on the sheet I (D) There will be magnetic current in the sheet. For sea water with o : 5 mho/m and e, : 80, what is the distance for radio signal can be transmitted with 90% attenuation at 25kHz ?
(A) 0.322 m (C) 32.2 m
M*0 4.4"57
which
(B) 3.22 m (D) 322 m
Consider the following statements regarding Srnith charts : 1. A normalized Smith chart applies to a line of any characteristic resistance and serves a^s well for normalized admittance
2. A polar coordinate Smith chart contains circles of constant lzl and II circles of constant /z 3. In Smith chart, the distance towards the load is always measured in clockwise direction.
Which of the'statements given above are correct (A) 1, 2 and 3
?
(B) 2 and 3 (C) 1 and 3 (D) 1 and 2 MCe 8"4.58
A
(100
-
J75)O load
is
connected
to a
co-axial cable
of
characteristic
aI I2GHz.In order to obtain the best matching, which one of the following will have to be connected ? (A) A short-circuited sub at load (B) Inductance at load (C) A capacitance at a specific distance at load (D) A short-circuited stub at some specific distance from load impedance 75 ohrns
n$cQ 8"4,59
In a line VSWR of a load is 6dB. The reflection coefficient will be (A) 0.033 (B) 0.33 (c) 0.66 (D) 3.3
I i l
xcq a"d.6o Zr,:200C) and it is desired that Zi,:50O. The quarter wave transformer should have a characteristic impedance
C)
(A)
100
(c)
10,000
of
Ch.p
(B) 40 O (D) 4 o
Cl
2. Z6n:* ja for a shorted line with I: U4 3. Znn: Zo for a matched line of any length (C) 1 and 3 HcQ
8.4'62 The input
:
(D) 2 and 4
impedance of a short circuited quarter wave long transmission
line is (A) purely reactive
(B) purely resistive (C) dependent on the characteristic impedance of the line (D) none of the above mcQ
8.4.s3 A transmission
line of output impedance 400 Q is to be matched to a load of through a quarter wavelength line. The quarter wave line characteristic impedance must be 25 C)
1a) ao
(c) MGo
o
400
(B) 1oo o (D) 425 o
o
s"4.64 The input
impedance transmission line is
(A) (C)
of ,\/8 long short-circuited
zero capacitive
section of
a lossless
(B) inductive (D) infinite
tlcQ 8.rt.6s Match List I (Parameters) with List II (Values) for a transmission line with a series impedance Z: R' 1- jal'Qlm and a shunt admittance Y: G' + jec'mho/m, and select the correct answer :
List-I a. b. c.
Characteristic impedance
Z. {ZF input impedance Z^ when the line B. {W
Propagation constant
The sending-end
7
is terminated in its characteristic impedance
Codes
:
abc
(A) 311 (B) 233 (c)21.2 (D)t22
Lbt-II 1. .{Zy
26
Zs
t
ftansmiBEion Lines
Hce 8.rr"6,r Consider the following : For a lossless transmission line we can write : 1. Zon:- jZs for a shorted line with l: \18
Select the correct answer using the codes given below (A) 1 and 2 (B) 2 and 3
papEof
Page 562
Chap
ilrGQ 8.4.66
Which of the following conditions will not guarantee a transmission line
E
Ilansmission Lines
(A) (B)
d
?
,t: G: 0 RC: GL
(C) Very low frequency range (ft >> uL,G >> uC) (D) Very high frequency range (R << uL,G << uC) ruGQ 8.4.67
In an air line,
adjacent maxima are found operating frequency is
at
12.5
cm and
37.5 cm.
(A) 1.5 GHz (B) 600 MHz (C) 300 MHz (D) 1.2 GHz iltcQ 8,4,68
Fig. I shows an open circuited transmission line. The switch is closed at l: 0 and afber a time t the voltage distribution on the line reaches shown in Fig. II. If c is the velocity in the line, then
zo "+ llii E+lii +-c I
z:O z:l Figure I
",rl Ii I z:0
i-
z: I
Figure
II
(B) t: tlc (D) t < 2Ilc
A
75 f) transmission line is first short-terminated and the minima locatiom are noted. When the short is replaced by a resistive load -R1, the minimr locations are not altered and the VSWR is measured to be 3. The value of -Es,
(A)
is 25
o
(c) 225a MCO 8.4.70
i
:
(L) t < tlc (C) LIlc > Il c 8-4-69
th*
+
l-l-
rtco
tirrr
(B) 50 rl (D) 250 o
For a lossy transmission line, the characteristic impedance does not depend on
(A) (B) (C) (D) tutcQ 8.4.7r
the operating frequency of the line
theconductivity of the conductors conductivity of the dielectric separating the conductors length of the line
If the maximum and minimum voltages on a transmission line are 4 V 2V, respectively for a typical load, VSWR is (A) 1.0 (B) 0.5
(c)
2.0
(D) 8.0
and
Mce
s.4"?2 A transmission line is distortionless if (A)
Q) MCQ 8.4.73
If
RG: LC
-
8:g
RC: GL
(D)
ft:
Chap
(c)
G
(B) 4 (D) 2
1.5
A signal of 10 V is applied to a 50 ohm coaxial transmission line, terminated in a 100 ohm load. The voltage reflected coefficient is (B) 1/3 (^) 114
(c) rl2
(D)
1
MCO 8.4"?5
A transmission line of characteristic impedance of 50 ohm is terminated by a load impedance of (15 - j20) ohm. What is the normalized load impedance ? (A) 0.6 - jo.8 (B) 0.3 - jo.6 (c) 0.3 - jo.4 (D) 0.3 + jo.4
tllco
Two lossless resistive transmission lines each of characteristic impedance Z are connected as shown in the circuit below. If the maximum voltage on the two lines is the same and the power transmitted by line A is Wr, then what is the power transmitted by the line B ?
8.4.?6
22
(A) 4W
(c) 2w MCQ 8.4,?7
Z/2 LineB
Line A
(B) 3 r44 (D) 1141
transmission line section shows an input impedance of 36 0 and when short circuited and open circuited. What is the characteristic impedance of the transmission line ? (B) 50 o (A) 100 o (D) 48 o (c) 45 c,
A
64 Cl respectively,
MGQ 8,4.?8
Consider the following statements for transmission lines : 1. When a transmission line is terminated by its characteristic impedance the line will not have any reflected wave.
2. 3.
E
Tlansmission Lines
reflection coefficient for voltage be 0.6, the voltage standing wave ratio
(VSWR) is (A) 0.66
MCQ 8.4.74
Page 563
(B)
For a finite line terminated by its cha,racteristic impedance the velocity and current at all points on the line are exactly same. For a lossless half wave transmission line the input impedance is not equal to load impedance.
Which of the statements given above are correct ? (B) 2 and 3 (A) 1 and 2 (D) 1, 2 and 3 (C) 1 and 3
-4'!-..'14-
iltco
Page 564
Chap
8"4.7s
E
lhdnsrision
Lines
What does the standing wave ratio (SWR) of unity impiy ? (A) Tlansmission line is open circuitud (B) Tlansmission line is short circuited (C) tansmission line's characteristic impedance is equal to ioad impedance (D) Tlansmission line's characteristic impedance is not equal to load impedance
rurcQ 8.4.80
:
half centre to centre spacing, r: conductor radius and e : permittiritr of the medium. Which one of the following is equal to the capacitance pe unit length of a two-wire transmission line ? h
(A)
(c)
@
1f€
8.4.8r
For a line of characteristic impedance Zs terminated that Zn: Zo/3, what is the reflection coefficient ]-z ?
(A) 1/3
8.4.82
.$ff=Y
.(,8=)} in a load of Zp sttch
(B) 2/3 (D) -r12
(c) -113 scQ
.-"{(*)
3re
.*"{{*) ilcQ
,-C
(B)
A transmission line has ,R, L,G,C distributed parameters per unit length of line. If 7 is the propagation constant of the line, which one of the following expressions represents the characteristics impedance of the line
w E+Fr rc\
G
+ tuc_ 'v
r3n
4lt'L
?
I
rotfffir,
MCO 8.4,83
What is the value of standing wave ratio (SWR) in free space for reflection for reflection coefficient f :- Il3 (A) 213 (B) 0.5 (c) 4.0 (D) 2.0
MCd 8.4.84
What is the attenuation constant a for distortionless transnrission line
(A) o:0 (C) ilGQ 8.4.85
o:
(B)
t,e
ft \/
?
o: Rl[g
(D)cv:
fry
A 50 f) distortionless transmission line has a capacitance of 10-10 f/m. What is the inductance per meter ?
(A) 0.25 pH (B) 500 pH (C) 5000 pH (D) 50 pH
itco
8.4.46
The open circuit and short circuit impedances of a line are 100 What is the characteristic impedance of the line ? (A) loovec, (B) 100o (c) rcol/t9 (D) 50 CI
0
each.
{$ca s.d"s?
atcQ E,4,s$
!r,tc& 8.4,*$$
A
load impedance of (75 - j50) is connected to a transmission line of characteristic impedance Zo:75 0. The best method of matching comprises (A) A short circuit stub at load (B) A short circuit stub at some specific distance from load (C) An open stub at load (D) Two short circuited stubs at specific distances from load When a lossless transmission line is terminated by a resistance equal to surge impedance, then what is value of the reflection coefficient ? (B) -1 (A) 1 (c) 0 (D) 0.5 A lossless transmission line of length 50 cm with L : 70 pHlm, C is operated at 30 MHz. What is its electric length (pl) ?
MCQ *.4"9{J
40
pF/m
(B) 0.2) (D) aOr
(A) 20)
(c)
:
108"
Which one of the following is the correct expression for the propagation constant in a transmission line ? (B)
(A)
(C)
(R
-
j,'L)(G
- rrc)
(D)
(c
- pc)
(R+ juL)(G+ jac)
nllca *r,4,9'!
Assertion (A) : In a lossless transmission line the voltage and current distributions along the line are always constant. Reason (R) : The voltage and current distributions in an open line are such that at a distance ),14 from the load end, the line looks like a series resonant circuit. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true
tltc& *.4.$?
Consider the following statements : Characteristic impedance of a transmission line is given by
1. ^E+4, V G+.!dL 2 ^[Z*2",
(R, L, G and C are line constants)
(2""
'
and,
2""
are
the open and short circuit impedances of the
Iine)
3.
V' I I' ,
( I/'
and
I'
are the voltage and current of the wave travelling
the positive g direction)
Which of these are correct (A) 1,2 and 3 (C) 2 and 3
?
(B) 1 and 2 (D) 1 and 3
terminated at the load end by a short circuit. Its input impedance Z, is (A) Z,:- jZstanl3I
(B) Z": jZllcot PI (C) Z, : jZ1'tan l3l (D) Z,:- jZ1'cotpl
in
Pagp 565
Chrp
E
I}enrmhrftn Lirec
Page 566
ilrcQ 8.4.94
Chap E flensmirsign l,irs6
A
with characteristic impedance of 600 ohms is terminated in a purely resistive load of 900 ohms. The reflection coefficient loss-less transmission line
is
MCQ 8,4.95
(A)
0.2
(c)
0.667
A transmission line has R, L, G and C distributed parameters per unit length of the line, 7 is the propagation constant of the lines. Which expression gives the characteristic impedance of the line ?
(^)
EltuL
Qst# MCQ 8.4.96 I
McQ
(B) 0.5 (D) 1.5
@)u# (D)
The open circuit impedance of a certain length of a loss-less line is 100 O. The short circuit impedance of the same line is also 100 O. The characteristic impedance of the line is (A) loo/t CI (B) 50 o (c) toolJt o (D) 100 c)
s'4.e2 In the relations s: ffi; the values of ^9 and l- (where ,9 stands for wave ratio a.nd f is reflection coefficient), respectively, vary as (A) 0 to 1 and -1 to 0 (B) 1to oo and -1to *1 (C) -1to *1and 1to oo (D) -1to 0 and 0 to 1
mcQ 8.4.98
Consider the following statements : The characteristic impedance of a transmission line can increase with the increase in 1. resistance per unit length 2. conductance per unit length 3. capacitance per unit length 4. inductance per unit length Which of these statements are correct ? (A) 1 and 2 (B) 2 and 3 (C) 1 and 4 (D) 3 and 4 ********r<*x
$oLUTIONS g,{
Page56?
tllt7htyIfYlltlJllrI
Ch,pE Ttnnrllissiea lrirres
sot.
s"'t.t
Option (A) is correct. Given
the input voltage, o, : V6cos(4 x l}art) and length of transmission line, I :20 cm : 20 x 10-2 m So, the angular frequency of the applied voltage is
u:4 x 10azr and the wavelength of the voltage wave is
^ Therefore,
:7:r#
(f
: #)
*:9##a :ffi
(in free space ?rp:3
x
1o8m/s)
:1.33 X 10-5
Since, * 'o'ot So the effect of transmission line on the voltage wavg is negligible i.e. the output voltage will be in the same phase to the input voltage. Thus, A and R both are true and R is correct explanation of A.
sol.
4.1.2
Option (C) is correct. The width of strips, w :9.6cm Separation between the strips, d, :0.6cm permittivity Relative of dielectric, 6, : 1.3 Conductivity of dielectric, o=0 So, the conductance per unit length of line is given as
: :
9.6
n' v :-ewd :- ror,#:(g.g5 x 10-12) x : 1.84 x 10-10F/m :0.18nF/m aoL
8.1.s
10-2 m
0.6 X 10-2 m
G':T:o and the capacitance per unit length of the line is given
x
o=o as
1.3
x
Sfi-i$
Option (C) is correct. Inductance per unit length, L' :250 nH/m : 250 X 10-e H/m Capacitance per unit length C' :0.7nF/m : 0.1 x 10-e F/m So, the velocity of wave propagation along the lossless transmission line is given as
1_1 "o-E-r-m :2 x 108 m/s
The characteristic impedance of the lossless transmission line is given
as
I
Chap
E
llaormirsion
(for lossless line,
x10_T
Lis
:"':1
I
. Tr ffir1oT oo:nl e:10.1
Page 56E
.R'
:50O sol. 8,{.4
Option (B) is correct. Given the operating angular frequency of the transmission line is
u:6 x 108rad/s and the parameters of transmission line are
R' :0.2kA/m:200Q/m L' :4pH/m:4 X 10 6H/m
G':8pS/m:8 x 10-6S/m C':4pF/m:4x 10-12F/m So, the propagation constant of the transmission line is given as
(R'+ juL')(G'+ j,")c')
t-
:
: : sol. 4.t.5
(0.10 + j2.4) per meter
Option (C) is correct. Given the Operating angular frequency of the transmission line, u : I.2 x 10e rad/s and the parameters of transmission line are
R'
:70elm
L' :O.ApHlm:
x 10-6 H/m x 10-12 F/m G':AopS/m:40 x 10-oS/m C'
:
l0 pF/m
:
0.1
10
So, the characteristic impedance of the line is given as 0.1
Zo:
:100_lo 30L 8.1.6
x
10-6)
10
x
10-
Option (C) is correct. The width of strips : tl Separation between strips : d So, the characteristic impedance of lossless transmission line is given 7 ,tF
-A\/ 'o -d
as
€
when d and w is doubled, the characteristic impedance of the transmission line will be given as
7-'-2d tF-z "o -tW\/ E : Therefore, the characteristic impedance
sol. 8.t.7
Option (D) is correct. Attenuation constant, Characteristic impedance, Phase velocity,
oo
will remain
: Zo : ap : o
10
same.
mNP/m
0.1kC)
0.5
x
: 108
:
100
10 2 NP/m
fl
m/s
Since the transmission line is distortion less so, the resistance per unit length
of the transmission line is given
R, :
aZo :
as
page ico
(tO-r)(tO0) : 1O/m
chap
and the inductance per unit length of the lossless transmission line is
given
AS
L, sal.
8.{.8
::h:---1Q0 d: o, -
:2v'Hlm -2u,H/m
Option (C) is correct. Given the parameters of distortionless transmission line are
:4{llm G':4 x 10-4S/m R'
and
So, the attenuation constant of the distortion less transmission line is given AS
o
:,fH G' : J4 x 4 x I0: :4 x l0 zNp/m
and the characteristic impedance of the distortionless transrnission line is given as
r--4--' - :,/ ttr 3 : rl A" r0= : 1000
distortionless line
Zo
sot.
4.1"s
Option (D) is correct. Load impedance, Zr, :300Q Characteristic impedance, Za : I50Q So, the reflection coefficient at the load terminal is given
a"s
r,:zt-$:199,i99:1 2,11 Zs - 300 + 150 - 3 and the reflection coefficient at generator end is given as
n -Zn-Zo ts-ffifi where Z, is internal impedance of the generator. Since, it is given that the internal resistance of the generator is zero (i.e., Zn:0) so, we get
0-150 rn:o+150: sol 8.1"{0 Option (C)
I
is correct.
: S:
impedance, Voltage standing wave ratio, Characteristic
Zo
SO{L
3
Since, the load connected to the lossless transmission line is purely resistive so, phase angle of the reflection coefficient of the line will be
0r:0 or r Now, the magnitude of the reflection coefficient is given
rnt_,s-1_ i''
as
3-1
nr i:s+1:3+1:u'a
So, reflection coefficient of the transmission line is
f :l llI le,fr' : 0.5e! or 0.5er : 0.5 or -0.5 For
l-:
0.5 the load impedance of the transmission line is given as
z, and for
l-:-
:
z,f+#1
: rro[ffi] : aso cr
0.5 the load impedance of the transmission line is given as
Ti"n"mi$ion
E
Lim
Page 570
Chap
: z,lf$1 :
z,
tso[,
*L**]:
E
so
o
Therefore, the possiblevalues of load impedance connected to the transmission
Tlansmission Lines
line are
:50 O or 450 O
Zt' $oL
8.l.tt
Option (D) is correct. Load impedance,
z,
Characteristic impedance,
Zo
: (zo0 - 72oo) o
:700Q
I :10cm:10 X l0-2:0.1m
Length of transmission line, Generator voltage, So, we get the angular frequency
x
uoft) :3cos(a'
10et)volt
c..,:ZfX109 and the phase constant of the wave on the transmission line is
(t fi x 10e ^ : ,r:5t-loT: 't x 0.1:5 PI
10n' -F
(in air
up:3 x
108
m/s)
:+
Therefore, the input impedance of the lossless transmission line is given as
,^:^(Hitm) .,nnl 200 _ :t'r\@/
:(25 $0L
8.'1.12
i200
+ i100tan(zr/3)
\
125.a)0
Option (C) is correct. The natural frequency of oscillation of a wave in a transmission line of length I which is open circuited at one end and short circuited at other end is given as
f,
:Q#9,
n
:
1.,2,t,....a
where oo is phase velocity of the wave.
sot. 8,t.{3
Option (C) is correct. The dimension of the tra,nsmission line will remain same at all frequencies i.e. I will be constant but as it is defined in terms of wavelength which changes with the frequency so , the expression for length wiII vary in terms of wavelength ,\. The wavelength of a wave is defined in terms of frequency
/as
):; where, c is the velocity of wave in free space so' at
J:
500
Hz we have
):560 Therefore, the length of transmission line is c
'-r - ^4- 2000 Now, the wavelength at frequency, f
\-
^
c
- 10TO
: lkHz:
1000H2 is given as
Since, the length of the transmission line
will be same as determined in
equation (1). So, we get
Chap
r c ':2ooo 30L $"1,{{
(c/1000) )
:---2-:z
(from eq. (2))
Option (A) is correct. Given, the distance between successive maxima and minima is 10cm.
\f2 : I0cm
l.e.
) :20cm Now, the distance between first minima and load is
/-i' :
7'5 cm
l*r>) So, the distance between
first maxima and load will be
/,"*:4oin 30L S.{"15
-i:T.r-/.:ffi*
: ^-+ t
Option (C) is correct. Since, the TEM wave is z-polarized i.e. the electric field of the wave is directed along *a,. l.e. dE:a" and the direction of wave propagation is along a, l.e. ak: a, So, the direction of magnetic field intensity will be AU - A*X AB - o,rY. Az -- Oa As E is in * a" direction and .EI is in - a, direction so, we can consider the two vectors as
: 8"e," H :- Ha&u E
and
(1) (2)
Now, from the maxwell's equation in phasor form we have yxH:jueE (for perfect dielectric
ar a! a ld dy )4,
a
l0 -4, It
o:
0)
az dz
:
jueE"a.,
Using equation (1) and (2)
o
gives the result as
0Hu
-aT
-0
Again from Maxwell's equation we have
VxE:_jupH la, o,, a"l
a al.:* I'tPHrou la I ar du a" I lo o E,l dE.' aE, : _ __;- 4.- dr 4". PFfluAu da
__
So,
it
gives the result as
0!, 0y
Page 571
:o
Thus, Both (A) and (B) are correct.
Using equation (1) and (2)
E
Tlansmission Lines
Page 572
$0r- &"{"{$
Option (C) is correct.
Chap 8
Given,
Tlansmission Lines
l-* The position of first voltage maximum, Position of first current maximum(voltage minim&),1-i. Standing wave
: :
4.5 cm
1.5cm
: 3 ^9 Zs :50Q
ratio, impedance,
Characteristic since, the distance between a maximum and an adjacent minimum is discussed
in previous question.
l-*-
t.e.
l-t"
:
)/4
as
\14
4.5-1.5:\14 ,\ :
So,
12 cm
Again the distance of first voltage maximum from the load is given
as
, - 0r\, n\ qn*47tT 2
4.5:or!12) +o 417
(For
n:
0)
tr:T Now, the magnitude of reflection coefficient is given
as
lrl:fi+:3#:?:os So, the reflection coefficient of the transmission line is
r :1r /0r
0.5 ( 3rf2 : 0.5eF"12 :--f.5 Therefore, the load impedance of the transmission line is given as
:
":::;Ji;-:50[ffi] $oL s,{,'l?
Option (D) is correct. Characteristic impedance, Load impedance, Length of transmission line,
Zo
:50(l
Z,
:
(zo + j1b)
C)
I :7)J20
Since, the transmission line is lossless so, the attenuation constant is zero
a:0 'f : (t+ i0: i0
i.e.
oft
Therefore, the input impedance of the lossless transmission line is given as
0:
(u:Tl
"I
I I t
:
"I (32.4
j7t
-
jr9.2:)o,
3()L 8.1.18
Option (B) is correct. In the assertion (A) given, Length of the transmission Load impedance, so, we
Page 573
Zr,
get
t3,
Chap E llansmission Lines
t : \14
line,
:0
: (+)(i):
(0:2trl\)
+
The input impedance of the lossless transmission line'is given ,7 tZtljZrt"p!\ v _ -"\Zo+ Lm:Lol=
as
jZrtan0l )
. ljZstan|\ : zol:t'\:
J@
\zol"
Now, we consider the reason part, Distance of the maxima from load is given l^u*
:
(07
as
+ znn) l2p
d; is the phase angle of reflection coefficient p is the phase constant of the voltage wave and n :0rI,2,.... Therefore, the input impedance at the point of maxima is given where,
2,. -tt
as
: zo(#4#\ (, -"\r*/ : r,(:*ll'':n::^'.'^*'\ t_;1a;w;=n"') (f:lfle/')
::" /'j-]l'\ : ""\1-lrli Znl .
\
so, zin is real if. zo is real and since, zs is always real for a distortionless line' Thus, zi^ will be purely real at the position of voltage maxima in a
distortionless line. i.e. A and R both are true but R is not the explanation of A.
8.l.le
Option (D) is correct. Length of transmission line, Characteristic impedance, Relative permittivity, Load impedance, So, we get the angular frequency,
l:6m Zo cr
:30Q a.aQ --toR
zL:(30 - j10)o
a :8r X
I07
and the phase constant of the volatge wave along the transmission line is
p_ti_82rx107 up cfJe, _ 8tr x 707 x ,/2.25 _
P,:h:i;)
'
3x10o
gzr
x
107
x
1.5
3x108
2tr
-5 fl :+
or,
x 6:2.4r
rad,
Therefore, the input impedance of the lossless transmission line is given as
,,.:^(Hffipj) :
(2J.r4 + 7b.a8)
:rr(rffi)
C) ,{
,lt
!
l'f, I I
Page 574
$ot- 8.1.20
Chap 8 Tlansmission Lines
Option (C) is correct. Given the generator voltage to the transmission line, vn(t) : Locos(8n x 107f - 3o') So, in phasor form the generator voltage is %e
:
lQs-r3o'
and as determined in previous question, the input impedance of
Iine is
:
(23.74+ i5.a8) fl voltage, we draw the equivalent circuit for thc input the determining for so, transmission line as shown in figure below :
zi,
zs
Usingvoltagedivision,wegettheinputvoltagetothetransmissionlineas
u,: vs"(*A)
, or,
vs,,^:
v",(*z)
(in Phasor form)
:.u,uo'(##ffi:r.) :
(10 e-Fo)(o .44 e''nn'
)
- {.{s-P2'56' line is Thus, the instantaneous input voltage of the transmission un^(t) :RelV,,i,e,tf : 4.4 cos(Sn x 107 t - 22.56") volt sol- s.{.2'l
Option (C) is correct. Given,
:
Zn: '; Zn: 150 Load impedances to the line I and 2, ft:: lu: \f $5 ft Length of the transmission lines 1 and 2, be Zln and No*, *e consider the input impedance of line 1 and line 2 be the input so' identical are line Znz rgspectively. Since, the transmission given as and impedances of the transmission lines 1 and 2 will be equal (lossless transmission linel zrnr: znz: f)
rr(?+mT)
+aoot.n(T+)]
Irso : tttlr..r,r*4441
(0:2rl\l
: :'"'[to;-o"6;1 'oo[tto*'100""(3)l
:(70.4- iL7.2qa
be equal to tbc Therefore, the effective load impedance of the feedline will
equivalent input impedance of the parallel combination of the line 1 and
i.e.
Zt' : Zr,rll Zn* _ zbL _ 00.4 - j17.24)
5(,L a.l.22
(35.20
-
78.62)
Zm:
Znz
o
feedline is
: (95.20 _ 79.62) A ot :(\)t0.3.\):6.61
ZL'
Therefore, input impeda.nce of the feedline (lossless transmission line) given as
ry_
,
/ Zr't jZotan0I\ )
"0\6T iZltanql
'*(#.ff## 0
(21,5.14
-
00tan(0.6n') 62)tan(0.62r]
jl13.4)
8.t.23 Option (B) is correct. .f : 0.3 GHz :0.3 Zt : (t00 - J100) O Zo : l00Q
Operating frequency Load impedance,
Characteristic impedance Generator voltage in phasor form, I/,g : 150 volt Internal resistance of generator Zc : I00Q Length of the transmission line, I : 0.375.\ So, the input impedance of the lossless transmission line is given
,,":rr(ftffi) I roo - /oo * :1001
:
,^^.[too-i 1(1oo it00) A
x
10e
Hz
as
jrootan(40.875^\ ]
\^
- rroo)tan(f
/
I
0.375^),|
(200 +
Now, for determining the load current, we draw the equivalent circuit for the transmission line as shown in the figure below :
Therefore, the voltage across the input terminal of the transmission line is
given as
V,,n
: V""(-Z--\ '"e\Zs+ Zi")
E
Ila$mission Lin€s
Option (B) is correct. Given the length of the feed line, l: 0.3) and as calculated in above question, the effective load impedance of the
so,
Page 575
Chap
--2-------r-
:
2.
: rso(m#ffiifirolo) :
Page 5?6
Chap
"'
1oo' t
" Since, at anv point, on the transmission line voltage is given as
E
Tlansmission Lines
V"(z) : V{ 1"-ia" * feitu) where 7o+ is the voltage due to incident wave, I is the reflection of the transmission line at load terminal and z is the distance of the poid from load as shown in {igure. So, for z:- I (1) Vs,.n : Vt (etBt + fe-i1')
zs
,
:o.szs\
ar ll
Z.+l
Io'
I |
zo:loo
zL:G00-
Q
j100)
a
I I I
z:-l Now, the reflection coefficient of the transmission line at load terminal is
r:/j{:tffi
:o'45e-n343'
l- and V",i, in equation (1), we get 13. : frra3" e-.{Tlo.rtt^y; V { ("4Tr10.375r) + 0.4b e 106. 1 eF
Putting the value of
VT
=
106.1eP
13"
;twl O^;:pt.n-;4s
ar -i135" :loe"
The current at any point on the transmission line is given
I"(z)
as
:%k-*" - t"i,")
So, the current flowing in the load
(at
z:
0) is
t,,:fig _ r) : tffi(t - 0.45e Fr.nr') af
:
-11,71
Q.$l
"-lrozt" Therefore, the instantaneous current at the load terminal
il(t) :
will
be
P1e{I't'evt}
x 0.3 x 10er- 108.4") :0.67cos(6zr x 108t- 108.4")
:0.67cos(27T
$$t-
8,1"24
Option (A) is correct. Given, transmission line is of infinite length i.e' l: oo. and infut impedance of the transmission line is equal to its characteristic impedance l e.
Zon:
Zo
Since, the input impedance of a transmission line is defined as
-lZt*Zntanh"Yl\ Z,n: Lo\z;+zrtffi1t
So,
)
tr:tr(ffiffi)
Solving the equation, we get
tanhTl
:
1
\t'1
erl
_ e-rl _. t
Pa6c 67?
;IT;4 e-11
:
Chap E Trrscafoeion Llder
O
Since, l: @. So for satisfying the above condition propagation constant 7 must have a real part.
i.e.
real part of 7
*
0
a*0
ort
Q: a+ i0)
As the attenuation constant of the voltage wave along the transmission line is not equal to zero therefore, it is a lossy transmission line. sol-
4"1.25
Option (B) is correct. Observing the waveform we conclude that at the sending end voltage changes at t: tr. The changed voltage at the sending is given as
a(t)
:
v{ + ft v{ + lnfr,vt at sending end at t = 0, lt and d
(1)
are the reflection where V6+ is voltage coefficients at the load terminal and the source terminal respectively. So, we get
(Zt:
-1 'n
and
-
RO_ Z,
(Zo:
Rn* zo Putting these values in equation (1), we get
u(t,): Vr*-V{-lnVr* u(t') :- 7nY1
a) Ro)
(2)
From the shown wave form of the voltage at sending end, we have t'(t1)
:
6 no1tr
Vt :24volt Putting these values in equation (2), we get 6 : - fn(24)
ls :- 4 Ro-Zo _ E;+z; -- -^
of,
.
- '-'At I :
Rg:60f)
(zo: tooQ)
0 as the voltage just applied to transmission line, the input impedance
is independent of 21 and equals to Zs (i.e. Znn: Zo at t= 0). Therefore, using voltage division the voltage at the sending end is given as
v{
: v,(Erta)
24: %(60+%0) n,
8.1.26
:'1$1-!4:
Option (A) is correct. Length of the transmission line, Internal resistance of generator, Characteristic impedance, Generator voltage, Load impedance,
l:
(Vt
38.4 volt
1.5
m
Rs:200O % :1A0Q
: 30 volt Zr :50{l %
-
24vok)
Page 578
I r,:m:333i133:*
So, the reflection coefficient at the load terminal is
r,:ffi:f3;+33:_+
Chap 8 llansmission Lines
and the reflection coefficient at the source terminal
is
I
Again as discussed in previous question at time, f : 0 as the voltage is just applied to the transmission line, the input impedance is independent of zs and equals to Zs (i.e. Znn: Zo at f : 0). Therefore, using voltage division the input voltage at the sending end is given as
v{
: v,(qrt%) : ro
" (zoo*oo):
I
1o,,ort
Now, the time taken by the wave to travel from source terminal to the terminal (or load terminal to source terminal) is given as
load
r:I
I I I I I I I I I
I is length of transmission line and c is the velocity of the voltage wave along the transmission line. So, we get where,
t :,*%u: 5ns
I
]
l
Therefore, for the interval 0 < t< 5ns, the incident wave from source to load and will have the voltage
Vi :
10
will be travelling
volt
For the interval 5 ns < t <
10 ns an additional reflected wave travelling from load to source and will have the voltage
Vl : ltV{: -lrq : For
10
source
ns
15 ns
3.33
will
be
volt
the wave reflected by source resistance travelling frorn
to load will be added to that has the voltage
V{
: lnfz; :-
g'g33
:-
1.11
volt
For 15ns < f < 20ns again the wave reflected by load travelling from load to source will be added that has the voltage
V;
: ft v; :+:
0.37 volt
This will be continuous and the bounce diagram obtained between source (atz:0) and load (at z:!.5m) will be as shown in figure below: Z:0.75m
=Ll3
I
i
Ir,: -l/3 Z:1.5m 5ns
15 ns
t
\ sol.
a.1.27
Option (C) is correct. From the bounce diagram that obtained between source terminal (z:0) and load terminal (z:7.5 m) in previous question, we can determine the voltage u(t) at any instant by just summing all the voltage waves existing at any time f . Since, for interval 0 < t < 10 ns only a single voltage wave with I/l* : 10 volt exists at sending end so, the voltage at the sending end (z: 0) for the interval is
u(t)
:
: 10 volt for 0 < t< 10ns < t < 20 ns, three voltage waves with y1+ : 10 volt
Vr+
again for the interval 10 ns
, Vl:- 3.33volt and V{:- 1.11volt exists voltage at the sending end for the interval is u(t)
:
V{ + Vl -t V}
10ns
:10 -
3.33
-
1.11
:
at the sending end so, the 5.6
volt
Thus, the obtained voltage wave form is plotted in the figure below
10
5.6
10 sol. a.{.2a
Option (D) is correct.
t(nsec)
for
Page 579
Chap 8 Tlnngmission Lin€s
I I
i
Baep
As shown in the smith chart, SWR circle meets the 4 axis (real part of reflection coefficient) at ,Lr and .L2 respectively. So, We have the two possible values of normalised impedance (real values of z1). at L1 zn :2.5 al Lz zn :0.4 is defined as normalised impedance the Since,
[80
Ctnp.i ftansmisclo. Llnot
Load impedance
pL-
Characteristic impedance
?:2.5
So, we have
Lo'L
v_ 201
oIt
Zr,
- ntr Z ..)
:fr:
zoo
: fu:
Similarly,
,r,
or,
Zor:fu
g,4
Loz
_ 50 -_ .,.r". ,oo ro Q.4
Therefore, the two possible values of the characteristic impedance of the lossless transmission line are 20 f,) and 125 O. sol- 4,t.29
Option (B) is correct. We can determine the reflection coefficient of the transmission line using smith chart as explained below : (1) First we determine the normalized load impedance of the transmission line a^s
- - Zt ,r:"d
100+J5o
:1*fl.5
(2) Comparing the normalized impedance to its
general form
zt:r*jr where r is the normalized resistance (real component) and normalized reactance (imaginary component). we get
i:
r is the
r:1andr:0.5
(3).Now, we determine the intersection point of r:1 circle and r:0.5 circle on the smith charge and denote it by point P as shown in the smith chart. It gives the position of normalized load impedance. ( ) We j-oin the point P and the centre O to form the line OP (5) Extend the line OP to meet the r: 0 circle at Q. The magnitude of the reflection coefficient of the transmission line is given as
lr t:
OP
oQ
_ n, g'Acrr. - u'"" --2.icm (6) Angle of the reflection coefficient
at point Q
in
degrees is read out from the scale
as
0r :76'0" (7) Thus, we get the reflection coefficient of the transmission line
f :lfll- t/0r:0.22e'16
as
Ir
tf
rlrl|l+?r
ALTERNATIVE IUIETHOD:
Reflection coefficient of the transmission line is defined n _ Zt- Zo _ 100+750- 100
as
' -z;Tz; -1To+F0Fr00 :
0.24
/lS::
0.24en6'
which is same as calculated from smith chart.
sol
8.{.30
Option (C) is correct. As shown in the smith chart in previous question normalized load impedance is located at point P. So, for determining the input impedance at a distance of 0.35) from the load we follow the steps as explained below : (1) First we draw a SWR circle (circle centered at origin with radius OP)
(2) For finding input
impedance at a distance of 0.35) from load we move a distance of 0.35,\ on WTG scale (wave length toward generator) along the SWR circle.
(3) Since, the line OP corresponds to the reading of 0.144,\ on WTG scale so, after moving a distance of 0.35) on WTG scale we reach pt 0.144^* 0.35): 0.494^ on WTG scale. The reading corresponds to the point A on the SWR circle. (4) Taking the values of r and r-circle at point ,4 we find out normali?ed input impedance as zrn
(5)
: r* jr :0.6I+ j(-0.022):
0.61
-
J0.022
Therefore, the input impedance at a distance of 0.35) from load is given .7 21n _^ -
:
arn
17
Ll)
100(0.61
_
j0.022)
:
(61 _
p.2)A
ALTERNATIVE TETHOD: We can conclude the input impedance at
Page 582
Chap
l: v - ,7 /Zt* jZotan7l\ ";" "o\joa iZjan6l)
E
l}ansmission Lines
I
roo + Jbo +
:lool ---[t,
I
I
0.35) directly by using formula lossless transmission line
jloota"t/?o.ss.\\
'l
\1= /l
+lloo 1:ro;ta"(2f
0.35)),|
: (61_ 12.2)0
l,
as calculated above using
$oL
8"{.31
with chart.
Option (C) is correct. For determining the shortest length of the transmission line for which the input impedance appears to be purely resistive, we follow the steps as explained below : (1) First we determine the WTG reading of the point denoting the normalized load impedance on the smith chart. Flom the above question, we have the reading of point P as 0.144,\ on WTG circle. (2) Since, the resistive load lies on the real axis of reflection coefficient ( l-,-axis). So, we move along the SWR circle to reach the ,[-axis and denote the points as A and B. (4) Since, point B is nearer to the point P so, it will give the shortest length of the transmission line for which the input impedance appears to be purely resistive. (5) Now, we have the reading of point B on WTG scale as 0.2b^. So, the shortest length for the input impedance to be purely resistive is given as the difference between the readings at point B and P. i.e.,
l:0.25),-0.144^
:
0.106)
80L 8,'r.32
Option (A) correct. The voltage maximum occurs at the point where the SwR circle intersects the positive f axis on smith chart. The SWR circle of the load impedance intersects the positive 4 axis at point B as shown in the Smith chart. So, the point B gives the position of first voltage maxima. As calculated in previous question the distance between point B and point ,4 on the WTG scale is 0.106,\. Therefore, the l"t voltage maximum occurs at a distance of 0.106) from load.
sol- 8,{,33
Option (B) is correct. At any time f , the currents of positive and negative waves are respectively 1+ and 1- and the voltages of positive and negative waves are respectively I/+ and I/- as shown in the figure.
r+V+1 ' - Zo-
Zn
(I/+: l Volt)
Page 5E3
V
and
Chap E flensniqsign Lines
Zn
Now, the voltage and current across an inductor are related as
,:L# v++v :2*e.+I-)
7+v':r*l+l
(V+
:
1,
Zo:
50)
I dvIt v :_%_dT -25dt
:
=dV=, I+ V
Taking integration both sides we get ln(1 + V-) :- 25t+
(t+
Cr
whereGisaconstant
(1) - 1"-25t wave is incident at t:0 so, at f:0+ the current through inductor is zero and therefore, from the property of an inductor at f: 0+ the current through inductor will be also zero.
Since, the voltage
tz-)
(I/*)
(1++1-)",r_n*:o
l.e.
t%- uLL*:'.:o
:0
lh-81*_o*
(Iz+: l Volt)
: l volt
So at l: 0+, Zo Putting it in equation (1), we get
(t+r):1 A:2
Thus, the voltage of the reflected wave is
V- : (2s-n, _ sol
s,1,34
1)
Volt
Option (A) is correct. The voltage of positive wave in transmission line is Vs+. So, at the voltage maxima, magnitude of the voltage is given as l%
l-*
: llzo* l[t + r]
and at the point of voltage maxima the current will be minimum bnd given AS
l/,1*,":\V_ rl So, the line impedance
at the point of voltage maxima will be
z^*:ffk:z'(+#) : zos Now, at the voltage minimum the voltage magnitude
:
is
(t:, \
_t#)
l% l-" IYJ l[1 - i-] and at the point of voltage minimum current will be maximum and given as,
11,,_*:sv*rl
Pare 5E{
and the line impedance at the point
gbP a
z^io:#k :
Ilmrnbeicn r.noe
sol. 8.{"35
will
be
z,(#+):?
(r:ij#)
Option (A) is correct To determine the required quantityT we note that for a particular line of characteristic impedance zs, the product of the line impedances at two positions (two values of d) separated by an odd multiple of \14 is given by
{zta}{z[a+(r" -
')+]]
:{"(ffi)}i^r,{HFB} :^[+ffi\=Hffii] :^[=ffiilffir;+t :*l*ffiltl;#l :23
As the intrinsic impedance of medium 1 is 41 and that of medium 3 is 43 so,
for required match, thickness the medium 2 is given as rl7lz
8.{.36
:
rl2z or
t is ),14 and the intrinsic Tl2
impedance
t _ 0r) -r, nA Inu - 47( 2 so, for l"t voltage maxima we have n voltage maxima as
:
I
f l/!1is
0 and so) we get the position of first
-G --0"A 0.125)
:o{\ :T "'-2 4r + g,
The magnitude of reflection coefficient is defined in terms of
S-1 3-1 lrnrl:3+l:B+T:z
swR
1
So, the reflection coefficient of the transmission line is
i-
: lf l/!t:t"^,, : *
Therefore, the load impedance of the transmission line is given as
"":t;:n \1-+/ :
of
: {rlrrt,
Option (B) is correct. Distance between load and first voltage maxima, /-* :0.125) Characteristics impedance, Zo : I00Q Sta^nding wave ratio, ^9:3 Position of voltage maxima (k*) in terms of reflection coefficient
, ,rn*
(a)
(60 + 780)0
as
sol. 8.t,3?
Option (D) is correct. Given, the transmission line is terminated by its characteristic impedance i.e.' Zr' : Zo so, there will be no reflected wave and therefore, the height of the voltage pulse
will be given
as
W : #h
(Zn
: ffi+# : As the wave travels in the
*Z
-
internal resistance of generator)
lovort
direction along tra.nsmission line at velocitv
'r:j=-:+ L'C' [email protected] x ,/
10-6)
x
:2 x 108m/s So, the voltage pulse will reach at l: 5 m at time, 5 : zc ^- ns t'(t: tt-loT So, at l: 5m for 0 < f < 25ns, V:0 and for t ) 25 ns V:V{:lOVolt
(100
x
10-12)
Therefore the plot of voltage against time at a distance b m from the source is as shown in graph below.
soL
8.1.39
Option (D) is correct. As the first forward voltage pulse is
[+
so, the
first reflected pulse voltage is
Vl:ftVf The 2"d forward pulse voltage is given
as
V{ : lnVl: fsLV{ The 2'd reflected pulse voltage is given
as
V; : lr.V;: fnf?V{ So, summing up all the pulses at load end for steady state the load voltage as
,t=
(l +
oo) we get
v{+vl+V{+v;+... rzJ[r + n t ryrt + rnfi -y ....1 tz,*[(r + tntr* 4r] +....)+rr(r + rsrL+.
'r[(T=+/]).(,+E)]
v{(#h) X)t<**X******
)]
Page [E5
Chap
E
Ilangmission Lines
soLUTloNs 8.2
Page 5E6
Chap
E
Dansmission LiD€g
sol.
8,2.1
Correct answer is 0.788
.
Given,
Inner diameter of coaxial line, 2a : lcm+ o: 0.5 x 10-2 m and outer diameter of coaxial line, 2b :2cm+ b- 10-2 m Permeability of conductor, F" :2Fo Conductivity of conductor, o" :17.6 x 1.07 S/m Operating frequency, f :4GHz:4 X I}eHz So, the resistance per unit length of transmission line is given as : 1\
:Zil Ffitl no,-r -i\a*t1
1 T"x(+xto\x(zx+"xtotir 1 , 1 r l.oi;lo--TiJ )
-- fr\l : s$L
8.2.2
0/m
0.788
Correct answer is 277.
Inner diameter of coaxial line, 2a :1,.5 cmJo: 0.75 x 10-2 m 2b :3cm9 |- 1.5 x 10-2 m Outer diameter of coaxial line, Permeability of the filled dielectric, p :2lrn So, it's inductance per unit length is given as
r,'
:ftn(*):ry%fg'(#i#) :2.77 x
sol.
8.2.3
Correct answer is 9.1
10-7
H/m
:
277
nHlm
.
Given,
Inner radius of the coaxial line, a :718 cm: 1.25 x 10-3 m Outer radius of the coaxial line, b :7l2cm: 5 x 10-3 m Conductivity of dielectric, o :2 X 10-3 S/m So, the conductance per unit length of the transmission line is given x (? \!0-s):9.1mS/m n,
r :-27ro -2tr ,.4H9; :, hrT:
$oL
8"2.4
Correct answer is 361. Given,
Inner diameter of coaxial line, 2a : Icm=)o:0.5 x 10-2m Outer diameter of coaxial line, 2b : Acm+b- 2 x 10-2 m : Permittivity of the dielectric, € 9€o So, the capacitance per unit length of the line is given as ZTe Z1f 2tre 2rx9x8.85x10-12 xY x ^r
u:lrrT:---l"m :
3.61
x
10-10
F/m
:
361pF/m
as
sol
s.2.S
Correct answer is 0.9722
.
Page 58?
Given,
Width of strips, Conductivity of strips,
u :2.4 x
Permeability of strips, Operating frequency, So, the parameter -R' is given
o
:
l.r
-Po
f:
1.16
Chap 8
10-2 m
x
108
Tbansmissien IJins5
S/m
4GHz: 4 x l}e Hz
as
R':2.@ uv o
__2 M 2.4 x lj-'tt/
1.16
x
108
:0.9722Qlm sot- 8,?.6
Correct answer is 157. Strips width,
w d
Separation between the plates,
:4.8 :0.3
cm
:4.8 x 10-2 m 0.3 x 10-2 m
cm :
Permittivity of dielectric, l.I :2p4 So, the inductance per unit length is given as L,
sok
a.2.7
_2x4zrx10-7x0.3x 10-2 -pd w 4.8 x 10-2 : 7.57 x 10-7 H/m : I57 nHlm
Correct answer is 7.22
.
Operating frequency,/
:
l GHz :
10s
Hz
Conductivityto :6.4 x 107S/m Permittivity,r :6€e Axial component of electric field. : E, tansverse component of electric field : 4 so, the ratio of the two components for the transmission line is
+: En-{ fud o :-\/ tr"rfrrfi4x10:
sot
*_g.s
Correct answer is 0.10
7.22
x
(u
10-5
.
The amplitude of voltage wave after travelling a distance transmission line is given. as
V:
:2rf)
/
along a
Voe-o'
where % is the amplitude of the source voltage wave
Now,
in the given problem, after travelling 20 m
transmission line the voltage wave remains 13% of
distance along the
it's
source amplitude.
So, we get
: e-"Po) V
Voe-"t
:
ISVo of.
Vo
o.1g
a :0.10Np/m
sol.
8"2,9
Correct answer is 561. Given the propagation constant of the voltage wave
'y:a-ti0:0.5+P,A
(/:20m)
So, we get the attenuation constant of the wave
Page 5E8
a:0.5
Chap 6 Tlansmiesisn Linss
and phase constant of the wave along the transmission line is
0
:2'4
Since, the amplitude of voltage wave after travelling a distance transmission line is given as
V:
I
along a i
Voe-"I
where Vs is the amplitude of the source voltage wave. Since the amplitude of a voltage wave after travelling a certain distance down a transmission line is reduced by 87% so, for the given transmission line we have
L\ e-"1
:
l/se-or:(t
- ffi)
*
:0.13
r:]r"(#)
:a.oa-
Therefore, the shift in phase angle for the travelled distance is given as o'l 360" t ,_ p,\T)
A
: (2 4X4 08)(#) sol.
8.2,10
Correct answer is 3.81
:561'
.
f :SGHz:5
Operating frequency, Characteristic impedance,
Zo
X 10eHz
:80Q
0 :7.5 rad/m
Phase constant, So, the inductance per
unit length of the transmission line is given
r,-FZo - 1.5x80 a 2rx5x10e : 3.81nH/m sol. 8"2.tt
(w:2rf)
Correct answer is 0.43 . The maximum magnitude of voltage wave,
: /*io :
V**
volt 2.4volt
6
The minimum magnitude of voltage wave, So, the standing wave ratio on the transmission line is given o -:b*- V*,^ -
as
as
J.-:r.s - "'"
2.4
Therefore, the reflection coefficient of the transmission line is evaluated as
f -,S-1-2.5-1:n.43 -S+1-25+f-' $oL 8,2.12
Correct answer is 2.1
.
Zo
Characteristic impedance,
:25Q
o : 0.6 mm : 0.6 x 10-3 Inner radius of the coaxial line, e:9eo)e,-9 Permittivity of insulated material, Now, the characteristic impedance of a lossless coaxial line is given as
t, : rym(l\ Je, \al
where b is the outer radius of the coaxial line. So, we get
25-60'r-l -
b
6"'\uo;
1
tb=7
b:(0.6
OI,
Y1g-t)"zs{stao
Page 589
:0.0021 m:2.1mm
sol
Correct answer is 3.65 Load impedance,
8.2"13
Chap
.
zL:(r5-125)A
Characteristic impedance Zo :25{l so, the reflection coefficient of the transmission line is given
as
zo _05-i25)-25 -r _zr zt* Zo (rb * ps)+25 :0.57e-J7s'8"
Therefore, the standing wave ratio of the transmission line is determined as 'I rl
s:ff:i+*#:365 8.2.'14
I-l
Correct answer is 40. Given,
Operating
frequency, f :2MHz:2 X 106Hz
So, the angular frequency of voltage wave is
a :2rf:4tr
x IO6 rad/sec When the line is short circuited, input impedance is Z:; : jurL (Equivalent to 32 nH inductance)
:
j(4tr
x
106)(82
x
10*,)
:
fr.4e
When the line is open circuited, input impedance is doc
-," -
1
(Equivalent to 20 pH capacitance)
3UC
:@:-i3979'9Q Therefore, the characteristic impedance of the transmission line is given as zo J zi;-zi;
:
: rrre{Fqsl8g
:40e sot.
8.2.ti
Correct answer is 600. Given, the length of the transmission lines 1 and
-
2
ft:
12: \f { So, the input impedance for line
1 is given as
:
2..:23,-(100f zialzL -T50-: -2O0., 3 ,' F}om the shown arrangement of the transmission line it is clear that the effective load for line 2 will be equal to the input impedance of line 1. l.e,
Zt'
:
Ztnt:
Sn
Therefore, the input impedance for the whole combination is
z -z&z- QooY z*:il:ffii:600o sol-
4.2.{6
E
Thansmission f,ines
Correct answer is 1.56 . The voltage maximum exists at the point where the incident and the reflected voltage wave both are in same phase and the distance of voltage maximum from the load is given as
,
Page 590
0"), . 4lr'
ho-
Chap 8
n).
(1)
2
where dp is phase angle of reflection coefficient, ,\ is the wavelength of the voltage wave and n:0,7,2,.... Now, the reflection coefficient of a transmission line is given as ' p zr'- zo (o'3 -P'5)- o'5 2f,12; 1o: - p5;1 s'b
Tlansmission Lines
:-
-:
_ -0.2 - fl.5 _ n qr7.-7e.8"
-i-3-75 0r
:-
So, from equation (1)
for
l
e.
79.8"
n:
0 we have
,
x
4x
47r---79.8" 4tr ^u--fu^ :- 0.44 x 10-2 m
70-2
T u "180"
=
which is negative (i.e. the point doesn't exist). Therefore' the l"t maximum voltage will exist for n:l and the distance of the 1't maximum from the load is
i.e.
,^*:T*i
(n:
:- 0.44 x 10-2 +2 x : 1.56 x 10-2 m : 1.56 cm sol
a.2.{7
1)
10-2
Correct answer is 0.56 . In a lossless transmission line, the current maximum lies at the same point where the voltage minima lies and similarly, the current minima lies at the same point where the voltage maxima lies as shown in the figure below :
3^14
(Dista.nce
from load)
^/2
^14
Now, it is clear from the figure that the distance between two adjacent maxima and minima is ),14
i.e.
I**
- l-r :4 -4
Since the maximum voltage wave lies
l*o :
at a distance
1'56 cm
So, the distance of 1"t voltage
minimum (the distance of 1't current maxima)
from the load will be
h-
:
Lo
- I :1.56
-t : o.so"ro
Thus, the distance of 1't current maximum from the load is 0.56cm.
sal
8"2.18
Correct answer is 27. Generator voltage in phasor form, Internal impedance of generator, Load impedance
%o
Page 591
:150V
: Zr, : I: Zo :
Chap
l00Q I50Q
Zg
Length of transmission line, 0.15,\ Characteristic impedance I00Q so, the input impedance of the lossless transmission line is given
''":"(ffiffi) _,nn[ 150*/00tan(+15) -'*[@] : too(ffiffi)
as
I
: {ro.r -
1',2.7)a
Now, for determining the power delivered, *u dru* the equivalent circuit for the tra,nsmission line as shown in figure below : xin
zs + uin
Znn
Using voltage division, we get the input voltage as
z^ rZ** Zo=\) 82'5 - i32'7 : 150/ sz-fuftioo
V.nn:
vn(
r
So, the current at the I".rn
input current is
\
)
:
7
tt
1's
t ta"
"-
77'8e-itl+s" ::Lti; -:1i,ffiffi7 - o.81eto16'
Therefore, the average input power delivered given as
P,,:|n
to the
transmission line
fu,n^Ifnl
: f n" 11Zf .S"-rrr'ao')(0.81 e-io.16')] :27 Watt sol-
8,2.{S
Correct answer is 153.1 . Since, the lengths of line 1 and line 2 are So, the
h: la: \12 input impedance of the line 1 is given
as
,,"r:rr(*+Em) jzotan(+ll
z^lz',* - "lz'+E;taneil]
@
: z,(2#):'u Similarly the input
;t"::3"
of line 2 is given
as
E
Ilansmission Lines
:
2zrlA)
Ctap
lt -
26,r:
Page 502
15gg
The effective load for line 3 will be equal to the equivalent impedance of tbe parallel combination of input impedances of line 1 and line 2.
E
Ilansmission Lines
Zr,'
1.e.
:
Zn
:ry
tll
Zo*
:75Q
input impedance for line 3 is given Zon: Zr' :75d1 Therefore, the input voltage of line 3 is So, the
v,i,
as
(Length of line
3,1: )/2)
: v,,(ZkZ) :5oo(zs+%o)
I
:2I4.2Bvolt
l
and so the current at the input terminal of line 3 is ,
1",,n
:V:Zn
2.86
A
Thus, the average powel delivered to the lossless transmission line 3 is given as
Pn:P',efV",n ltn^l
:|* Q14.28) x (2.86) : 306.11Watt Since, the transmission line is lossless so, the power delivered will be same and given as
Pt i$t"
:
8.!.20 Correct answer is 17.8
Pz:
+:+
x
306.11
:
to
each load
tr53.1Watt
.
As discussed in previous question the input impedance of infinitely long lossy transmission line is equal to it's characteristic impedance. So, the input impedance to line 1 will be Z,a : Zot: 200O From tbe shown arrangement of the transmission line it is clear that the effective load impedance for line 2 will be equal to the input impedance of
line i.e.
1.
Zr"
-
Zoa: 200 O
Since the letrgth of the line 2 is be equal to its load t.e.
il ll
I
Z;,2
)|2 w, the input impedance
: Zn:200o,
of line 2 will
(I:
^12) as Therefore, the reflection coefficient at the load termirral of line 2 is given
an-Zrr-Zoz-200-100 -1 -Z;2+Z;-200TT00 -3
I
Now, the input voltage of line 2 is determined by using voltage division rule AS
I z*'z \ V"jn_ t',*\j*ae)
-*\200Tf00)*3 [;
i
t
Again, the voltage at any point on line 2 is given as (lossless line) V"(z) : V{ (ejt3' + fe-iB") where I/e+ is voltage of incident wave B is phase constant df the voltage wave
and z is distance from load. So, for
v"(z) 8
S
:
z:-A/2
Page
vt("+(+)+ re Px(i))
: v{(e-'* feF)
(v,(r):
:;x1_ri5
I/o+-8"
:-
V,a at
z:-\12)
1
(.:$)
2volt
Therefore, the incident average power to the line 2 is given as
: ZZ;: ztloo : 2o mwatt P:,,:W So, the reflected average power at the input terminal of line 1 (load terminal
of line 2) is
P:,
:lr
f
pk: (+i
X 2o
: 2.2 mwatt
Thus, we get the transmitted power to the line 1 as
Pk sol
8.2.2'l
: Pl,- Pk:20 -
2.2:'17.8 mWatt
Correct answer is 600. consider the length of the transmission line
/
as shown
in figure below.
The generator voltage is applied .to dlre transgrission line at time f : 0 for which the voftage at the send!4g etd is o(O) : 10 volt (at f: 0) After time at:Aps the volgage u(d) at the sendiqg end changes to 6v. This change in the voltage w.ill be calised onry if the reflected voltage wave from the load cornes to the sendipg end. so, the time duration fqr the change in voltage 4t sending end can be.given as
41 :(time taken by inoident
*
to reach the lqad) (time talgp by refleeted wave to leach sendlng.end from wavo
the load)
oI' where
A,t
=L+L:21 up I)p
(1)
Up
I is the length of the traasmisgio4 line
sending terminal) and
(distance letween load and
is p[ase velocity of the arave along the transrBissjon line. Since, the line is air spaeed so, uo
UP:c:3X108m/s Putting it in equation (1) we get
2I 4Ps: 3xL08 " Thus, length of the transmission line is
r_3x108x4x10-6
':ffi:6(x)m
bOS
Chap 8 flrnsmisgiqn Li1sg
({t:
a ps)
Page 594
sol. a.:"22
Correct answer is 42.86
.
Let the load impedance connected to the transmission line is Zr, so the equivalent circuit for the transmission line will be as shown in figure belor' :
Chap 8
Tlmmission Lines
Es:100
O
Sending end
Zo:lOO
Q
zL
Since, the internal resistance of the generator is equal to the characteristic impedance of the line Rg
].e.
: Zo: 100 O
So, the reflection coefficient due to source resistance will be zero and therefore, the change in voltage at sending will be caused only due to the reflection coefficient at load terminal given as Au(t1 : 7Yo' where, 7s+ is amplitude of the incident voltage wave and f is the reflection coefficient at the load terminal. Since, the change in voltage at, t:4 ps is
:6 - 10:-4 : \0f
Au(t) So, we get
r
:-#:-
(Zl- z:o\ :\Zr'+
Zo)
4
(7t*
:
10
V)
0.4
0.4
(Zo:
effi):-04
100o.)
ZL-700:-0.42t-40 Zt :42'86Q
sol.
4.2.23
Correct answer is 4. As determined in previous question, for a wave travelling through the three mediums of intrinsic impedances \t, qz and 43, the condition for matching dielectric (the intrinsic impedance of medium 2 that eliminates the reflected wave in medium 1) is
,t, : ,/ rlrrl, all the media have ;l : Since, equation can be rewritten
t@_
t/ er-
,t,l0
so, for the dielectrics
as
u#)G/#)
where e2 is the permittivity of the medium 2.
€ So, the relative
sot- 8.2.24
(o: 0) the above
(,: '/T)
:4€o
permittivity of the medium 2 is €, :4
Correct answer is 2.5
.
The thickness 'f ' of the dielectric coating for the perfect matching (the condition for eliminating reflection) is given as
t: j
(quarter wave)
where
,\ is the wavelength of plane wave. The wavelength in
frequency is
terms of
Chap
\:lt-f
flanimissisl
where oo is the phase velocity of the wave in the propagation medium which is given as
,o so, at frequency
/:
3 x-108 : l: 1 s ru 1o8 r'v rv +JPe Jpoheo 2 1.5 GHz the thickness of
the dielectric coating is given
as
,_up : _: 1.5x108
So,
sol.
8.2.25
4 zft;itft5 : o'25m : 2'5cm
'
Correct answer is 60. Characteristic impedance,
Zo
Load impedance,
Zr.
:60Q
:780Q % :100V Zs :120Q
Voltage generator,
Internal resistance, So, the first forward voltage pulse will be
6o--^\too:190vott Vf: / Zn \" (60-_r \z;T4)v,: ' ,zu7 ,r
The reflection coefficient at load terminal is given
tr-4-A-180-60-1 - z;Tz,: r8o-TTo :
as
11
2
The reflection coefficient at source terminal is given
Zo-Zo 120-60-l l' - 2,;6: T2o +60:3
Therefore, the voltage across
as
the load at steady state is given by the
expression as determined in previous question
vt(#h)
v,:
t.e.
100/ 1 ++ \ : -r\1-Gxt/
:l$q rt"9:60vort $EL
8,2_26
Page 595
Correct answer is 0.67
.
Voltage generator,
: 50 Volt Zc :30{l Zo : 15Q %
Internal impeda.nce, Characteristic impedance, Load impedance, So, first forward voltage pulse is
v{
Zt :45Q
:(#z)v,:(mrfm;, :ry
Now, the reflection coefficient at source terminal is
n -Zo-Zo 30-15 ro-/ot-fi:m+lb:5
1
and the reflection coefficient at load terminal is
, _zr_zs_45_15
'L -
1
Z;T4:6115:2
E
Linsg
Page.E96
Chap
So,
at steady state
(r:
oo) voltage across load is given as
v,: v,.(|:ifi.) sol 1-+ \ =
S
Trirnsdilbblon trities
a\r:6157 :$" E"t:3ovolt
Therefore, the current through load at steady state is given as
r, sol.
a.2,27
:g:
gg: ?:0.67 td
LJ.
d
A
Correct answer is 80. Since, the internal resistance of the battery is zero so, the l"t forward voltage pulse is
V{
: Vn:6Volt
and from the plot we get the first forward pulse current
as
1i :75mA Therefore, the characteristic impedance of the transmission line is given
as
.:goo zo:yi: -jl-tE 6LF-'
sol.
8.2.28
Correct answer is 262.85 . Reflection coefficient at source and load end are given
as
r -Zn-Zn--' to -Zo+7o--' and
Now, from the plot of input current (current at generator errd) we get,
Vl :75mA
(1)
V'*-Vl'fV{:-5mA
and
(2)
where, Vr+ is the first forward voltage pulse, V1 is the first reflected voltage pulse and I/2+ is the second forward voltage pulse. So, putting the values of these voltages in terms of reflection coefficients we get Vt*
- ftV{ * ft fsllr* :yi(l- lt- ft):-
r-2n
or,
5rnA 5mA
:-+ f
/i)
(r,:-1) (vr*
:75 mA)
n:f5
For determining load resistance of the line the reflection coefficient is written
in the terms of impedances
as
zL-zo_g Zr* Zo - 15
z,-80 Lffi:fr
(Zo:80Q
ZL(15-8):80x8+15x80 Thus,
Zr.
:262.85Q
*
******
r<* r<*
as calculated
in previous question)
SOtUTlOlrl$ 8,3
Page 597
Chap
E
*aosnirgi{rn Lin€s
$sr,
$.3".t
Option (C) is correct. Characteristic impedance of a transmission line is defined
as
and the propagation constant of the transmission line is defined as
^t:a+jp:w where,
a is attenuation constant p is phase constant unit length of the line G' is conductance per unit length of the line .L' is inductance per unit length of the line C' is capacitance nce per unit length of the line -R' is resistance per
R, R':G':0
Now, for lossless line,
So, the characteristic impedancer of lossless transmission line is
-tT
oo: tld and the propagation constant of lossless transmission line is
1:q* j0: jaJTC OI
O:0
Therefore, the attenuation constant of lossless line is always zero (real).
i.e, statement (A) is correct. Again for distortionless line,
R, i::c
G,
So, the characteristic impedance of distortionless line is
E'u-ld-lG, ry
til
and the propagation constant of the distortionless iine is
7
Of'
:
rv
* i0: ^/Ed
a:^/Hd+o
+
I,t{Td
Therefore, the attenuation constant of distortion less line is not zero but it is real. Thus, (A) and (B) is correct statement while (c) is not a correct statement. $sL
s,3"?
Option (D) is correct. Load impedance, 21 - 0 Input impedance,Zin : @
and,
wave length
(Short circuit) (Open circuit)
: )
Now, the input impedance of lossless transmission line is defined as
iZoraiQl\ :Z,\ffi)
z, - z^lhi
Page 598
Chap
E
Ilansmission Lines
where,
I is the length of
the transmission line and
p is the phase constant
of the voltage wave along the transmission line. So, we get :- Zo@ + jZotan0l) (Zo+ fltan0l) cn : jZotanBl or, Since, for a practical transmission line, Zs + oo so, we have tanPl : a (for minimum length) or, PI : rl2 Therefore, the minimum required length of the transmission line is
t-T'l:4r. t:ixb:Znn : \14 sor.
8.3.3
Open circuit end 8.3.4
\
Short
circuit end
Option (A) is correct. Given, the transmission line is terminated in short circuit i.e., 21:S and line should be short circuited at its input terminal i.e. Z;,- 0. The input impedance of a lossless transmission line is defined as
So,
z^: 4(?li4j44\ -"\zo* jZ;tanPI ) o: z,(wmp!)
(zr: o, zn:
:0 pl : 0, T, 27r,..........
jtanBl
Since, Iength of transmission line ca,nt be zero i.e', I
r:b_ t:@*!N. t:i
sol 4.3.5 Option (A) is correct. 3oL
8.3.6
)-/
Option (C) is correct. Since the transmission line has one short circuited and one open circuited end so at the short circuit end voltage must be zero while at open circuit end voltage must be maximum. So the voltage standing wave pattern will be half sinusoids with zeros at short circuited end and maxima at the open circuited end.
.W
eol-
(B:Zn\
I
Option (A) is correct.
xxx********
*
0 so, we get
o)
soluTloNs 9.4
'
Page 599
Chap
E
llansmiesi6n Lines
ssl
S"4".1
Option (B) is correct. Characteristic impedance of a coaxial cable is defined
: !" [EmlL\ e"'\a) 6 - outer cross sectional
as
Ztt
where,
--+
a
So,
inner cross sectional diameter
Zo: t/F6-hlL\ e6e, "'\a
)
4trxl0-7x36zr
-105x : s$L
$$.c,.?
100
10s9
f)
'(+)
Option (C) is correct.
Since.
I
diameter
Zo:J422 100:r/50x200
i
As this is quarter wave matching so, the length of the transmission line would be odd multiple of ),/a. Now,
For For
/t: $:
7
:
(2rn+ 1)+
:
429M11z,
" 1GHz,
T+z: +zg?ffi7
o'174m
'':i+4:di*%:o'075m
Now, only the length of the line given in option
both lr and
:
(c) is the odd multiple of
/2 as
(2rn* r;
:
(2rn-r 1)
: LF -
1',58 h
:9 21
lD
Therefore, the length of the line can be approximately 1.bg cm. s{}L
n,4.:3
Option (C) is correct. Length on the transmission line, Operating frequency, Phase difference,
d :2mm
f :I0GHz 0:nl4
since the phase difference between the two points on the line is defined
t:To
where .\ is operating wavelength and points. So, we get
d is the
,* 4:-Xa A:8d:8X2mm:16mm
as
distance between the two
Page
Chap
600 E
rlansmission
Therefore, the phase velocity of the wave is given
a,
Lines sot 8.4.4
as
: f),: 10 X 10e x 16 X 10-3 : 1.6 x 108 mf sec
Option (A) is cottect. Since, voltage maxima is observed at a distance of )/4 from the load and we know that the separation between one maxima and minima equals to )f 4 so
voltage minima will be observed at the load. Now, the input impedance at the point of voltage minima on the line is defined as fzn
]"*: ?
where, Zs is charanteristic impedance and S is the standing wave ratio on the line. Therefore, the load impedance of the transmission line (equal to the input impedance at load) is given as
Zt :fZnl^io sol-
4.4.5
: ?: $ :
Option (C) is correct. For a lossless network,
fO
ff
(Zo:
5052,
^9:
5)
:1
l&'l'+l&'1'
Since, from the given scattering matrix we have
/0" , Sn: 0.9 /90" Sn : 0.9 /90", Sr, : 0'1/90' (0.2)'+ (0'9)' + 1 ,9rr
So, we
get
=
0.2
Therefore, the two port is not lossless. Now, for a reciProcal
network,
Sn:
Szr
As for the given scatbering matrix we'have stz
:
szr:0'9 /90"
Therefore , the two port is reciprobal.
sol.
4.4.6
Optiol (D) is correct. For a distortion less transmission line cha,racteristics impedance
-m
zo: t/ G
(1)
Attenuation constant for distortipnless line is
o
:,fRG
(2)
So, using equation (1) and (2) we get
o:*:?#:o.ooz sol.
8.4.?
Option (B) is correct. For a lossless transmission line, the input impedance is defined
v -- uolzo vlzr'+ jz"tan0ll * jzt tan Pl ]
ain
Now, for the guarter wave (,\/4) Iine we have Load impedance, Characteristic
impedance,
Length of the line,
:30Q Z":30{l
Zr.
l:I
as
:t*(f f): *
tanpl
So,
(u:'f)
Therefore, the input impedance of the quarter wave line is
,-r':lm"l:d:6oQ :0dl
Load impedance,
Zt
Characteristic impedance,
Zo :30o,
(Short Circuit)
,,-) -g
Length of the line, So, we get
:t*,(f *):
tanpt
Therefore, the input impedance of the
Z;*
-
jZ"tanBI:
r
,\/8 transmission line is given
as
frO
The equivalent circuit is shown below
:
60+130 O
The effective load impeda"nce of the 60 O transmission line is
Zt :60 *730 So, the reflection coefficient
'
at the load terminal
is
- 160+F +To l: m ', _lZr-z"l - lZ;'r2; | _160+:3-6ot 1
Therefore, the voltage standing wave ratio of the line is given as
: r-7.i :-t+ln s:1+l{l
-ffi:
sol.
a.4.a
L'64
Option (D) is correct. The transmission line are as shown below. Length of all line is
R
Zr:50
Q
Zr:50
Q
The input impedance of a quarter wave defined as
,z
o2
"i" -Lo -Z
()/4)
Chap 8 Ttansmission Lines
I Zt t;zl Tltan7l'r""1-o' znn Now, ror A/8 transm
Page 601
)
lossless transmission line is
Page 602
Chap
where, z() is the characteristic irnpedance of the line and 21 is the load impedance of the line. so, for line 1 we have the input impedance as
E
Ilansmission Lines
:4: # : f) Similarly, for line the input z,:*:#:2ooo zo,
2oo
2,
impedance is
So, the effective load impedance of line 3 is given as
:
Zn
Zrll
Zn
:2000
||
2000
:
100f)
Therefore, the input impedance of line 3 is
,o:*:#:2be soL
8.d.,$
Option (D) is correct. The input impedance of the lossless transmission line is defined
as
:'"\2.+ jzia@D) zn-7lzt+iz"tun(p')\ since, the given transmission line of characteristic impedance zs:75o is short circuited (21: 0) at its one end. Therefore, the input impedance of the line is
26": jZ"tan(pl) Now, the operating wavelength of the line is
so,
^:f:3i#:0.1 morlocm fl:+t:?o"r:F
Therefore,
Zi,
: jZ,tan[
(/:3GHz)
(l:
l cm)
,
Since, Zotan(zr/s) is positive so, Zin is inductive. $gL 8.4.{O
Option (C) is correct. The 2-port scattering parameter matrix is
,:
[i;l f;;]
zo) - z" (50 | 50) - 50 :- (5oll :-'o -:(Z'll soF 50 6p;;4 I
i 1
sn:sz'::dft#!r":#Hr1% (Ztll z")- z" _ (b0 ll50) - b0 ,r^^ _ (zrll z") + z" (50 lTo;1_ 56 : - 3 1
soL s.4"*t
|
Option (D) is correct. The input impedance of a quarter wave (r: )/4)rossress transmission line is defined as Zo^
:4 -Zt
where, zs is characteristic impedance and, z7 is the load impedance of the line. So, we have the input impedance of line 1 as
znt
: *:
#:
ru
Similarly, the input impedance of line 2 is
: h: #: r2.s
zta
Page 6oJ
Chap 8
The effective load impedance of the line 3 is given 21 : Zi,1ll Z;,2
as
Tlansmission Lines
: 25 ll r2.5: + So, the input impedance of the 50 O transmission line is
,, :@x)":300 T
Therefore, the reflection coefficient at the input terminar is given
r :4-A- z;+%:
3oo
-
50
300+io-
as
:- 75
ssl s"4"'ta Option (A) is correct. We
have
l0logGo
:
10 dB
Gp:lo The power gain of the antenna is defined
as
__t "p n
Proa
where P,"a is the radiated power of the antenna and. p", is the input power feed
to the antenna. So, putting all the values we get
9:&e -1W or sol-
8.rt"{3
P,od
:10
Watts
Option (D) is correct. The characteristic impedance of a transmission rine is defined as Zfr Z* 2"" where Zo" and 2"" are input impedance of the open circuited and is short circuited line. So, we get
:
5ox5o bo " -z&- z*- 100+750: zTSj:49# :7.6e - tr.skj
"""
sot.
8.d"t4
Option (C) is correct. From the diagram, VSWR is given
4 oc :ffi:1:4 -V^u-
as A
Since, voltage minima is located at the load terminal so, the load. impedance of the transmission line is given as
sol
s"*.ir
zt :lzn l^,^: ?
:5f :
tz.s
o
Option (A) is correct. The reflection coefficient at the load terminal is given
r :ftjfu:1;*i*3:-06
sol.
8.4.ts
Option (C) is correct. The given circles represent constant reactance circle.
sol s.4.{7 Option (C) is correct.
(zo: as
sofl,
,9:
4)
The ratio of the load impedance to the input impedance of the transmission line is given as
Page 604
Chap
E
Va-Zo
Ttensmission Lines
Vn -
: +^u, :463@:
v,
sol.
8.4.t4
Zno
6o
v
Option (A) is correct. Suppose at point P imPedance is
:
Z
r+
i(-I)
we move in constant resistance circle from point P in clockwise direction by an a,ngle 45", the reactance magnitude increase. Let us consider a point
If
Q
at
P in clockwise direction. It's impedance
45" from point
is
h: r-0'5j 4: Z*0.5j
or
Thus movement on constant
r - circle by an / 45" in CW direction is the
addition of inductance in series with Z. sol.
8.4.te
Option (D) is correct. The VSWR of a tr'ansmission line is defined
as
"o-1-lrl -1+ll-l where
l- is the reflectibn coefficient of the transmission line. So, we get
2:1-!!l '-t+1ri
(s:2)
tr,l: lr l:5 1
or
Thus, the ratio of the reflected and incident wave is given
P,
as
_tTp_l
z:t, t_g P,:+
or
i.e. lJ.Il% of incident power is reflected. soL
8.4.20
Option (A) is correct. The input impedance of a lossless transmission line is defined
o z'": Now, for
)/2
as
o lzr'+ jz"tan0l] toQo+
iz'tanfr
transmission line we have
l:Al2
and
Zn :100(7 So, the input impeda,nce of the )'12 transmission line
z": z"W+im:
is
zn:roo{,
(u:T\
For .\/8 transmission line, we have
I:\18 and
Zn
:0
(short circuit)
So, the input impedance of
.\/8 line
is
Page 605
zlo+iz'tanil- jZ":il}{l zinz=2"ffi:
2rt -r)
Thus, the net admittance at the junction of the stub is given
:4-+]-
v
Lint
as
Zin2
:#+i*o:o'01-P'02
100
sol- 8.4.2{
0
Option (D) is correct. VSWR (voltage standing wave.ratio) of a transmission line is defined
as
1+]"* _
-r_f
where
f
is the reflection coefficient of the transmission line that varies from
0 to 1. Therefore, so|.
sol. sol.
8,4,22
8.4.23
8,4.24
,S varies
Option (B) is correct. Reactance increases, if we move along clockwise direction resistance circle. Option (C) is correct. A transmission line is distortion
8,4.25
leSs
if
LG:
in the constant
RC
Option (B) is correct. Z" =^/
sol-
o.
from 1 to
ZocZi: ft00 x 2b :
10
x 5 : 50e
Option (B) is correct. we know that distance between two adjacent voltage ma:
|:zz.s-t2.s ort
),:2x15:30cm
Therefore, the operating frequency of the transmission line is
f : x: Lt#n1: $oL
8.4"25 Option (C) is correct. Electrical path length
where
A
:
pl
: T, l:
50 cm
l GHz
(c:3X1010cm/s)
Chap
E
flansnfusiel l,insg
tr
Page 606
Now, the operating wavelength ,\ of the transmission line is given
Chap 8
\:9:1" T T
Tlansmission Lines
as 1
1
lJ--
JLC
_ 1 " _____1_ 25 x 100 Jtox 10-6 x 40 x 1o-12
JLC
-5x107-:2m 25 x 10' So, the electric path length is
fl:+ sol-
8,4.27
x
50
x 19-':frradian
Option (B) is correct. The input impeda.nce at the voltage minima on the transmission line is defined as
fzo*]^^:
?
where S is standing wave ratio along the transmission line. Since, the reflection coefficient lz, of the transmission line is given as
!9:g:l ,,:4:4:199, - 150 - zr.* zi - 100+50
3
So, the standing wave ratio of the line is
-r-+ _r -r-lnl _1+l-
uq_1+lrrl
Therefore, the minimum input impedance measured on the line is equal to
Izn $oL
8.4,28
l*":9
:zsa
Option (A) is correct. For a lossy transmission line the input impedance is given :'^l Z t + iZotunk=til=l
'' -
Load impedance, 21 = Length of line, I-
Zo,:
So,
"olz;w;iffiill
(open circuited at load end) ^14
^*4m]
Zo : jtanhfu-n" sol
8,{.29
as
(ta"rtf
- "";
Option (A) is correct. Input impedance of a lossless transmission line is given by
2n
:- "olz'+-izLtan0|] z^l4ii4t""-41
where
Zo
-
and
Zt 'LoadimPedance I - Length of transmission 0 :2nl\
So, we
Charateristic impedance of line line
have PI:++:+ Zt
:0
(Short circuited at load)
I and
Zo
:50{l
Page 607
Therefore, the input impedance of the transmission line is
Chap 8
a.:,'[ffiffi]:*
Tbansmission LineE
i.e. infinite input impedance and thus, the current drawn from the voltage source
sol
4"4.30
will be zero.
Option (B) is correct. For lossless transmission line, the phase velocity is defined
:v:ffi
as
1
'r*Q -
...(1)
characteristics impedance for a lossless transmission line is given t7 on: -\/[!c
as
...(2)
So, from equation (1) and (2) we get
', sor-
8"4,31
1
_1
frThJ4:
z;e
Option (C) is correct. Input impedance of a (\l\ transmission line is defined
as
"
o," --ZB Z, where zs is characteristic impedance of the line and z7 is load, impedance of the line. Since, the ),14 lineis shorted at one end (i.e. Zr. : 0) So, we get,
Zi' : zrliYl$: t u2L sol-
a.d.32
a
Option (A) is correct. voltage minima of a short circuited transmission line is located at it's load. As the location of minima is same for the load -Rr (i.e. the minima located at rB1,) so, the first voltage maxima will be located at \14 distance from load. 0"A Now,
,
...(1) 4T where l^* is the distance of point of maxima from the load,, 07 is phase angle ofreflection coefficient and ) is operating wavelength ofline. So, putting the value of l-o is equation (1), we get
^*-
\ _er^ 4- 4tr
0r:r Now, the standing wave ratio of the line is given as
1+lt-l
or, i.e.
"e _- 7-lr
3:1+l?l -1-lr'l
(s:
3)
lr'l:112
fr:lfrlrer:LA:-*
The reflection coefficient at the load terminal is given
-!2 - tu-75 Rr,+75
as
(Zt:
Rt), (Zo: 75Q)
Page
608
Chap 8 rlansmission
-
Rr,
Lines t*{}1-
s. :3
: 2Rr, - 150 : 3Bt 75 + Rr' :
-
75
25 Q
Optiorr (B) is correct. The VSWR (voltage standing wave ratio) in terms of maxima and minima voltage is defined as
o - l7-*l -+ -" "-lY*;l-2-4 $t}t s"4.34 Option (A) is correct. Characteristic
impedance,
Zo
SWR So, we
S
have
:60Q
:4
++ : S: 4 L.D:9:
0.6
The reflection coefficient at load is defined t-
as
n _Z;T% ,L -Zr-Zo
So,
O.a:ffi tr:ffix
60
:240O
sol &"4.3s Option (D) is correct. Loading of a cable is done to increase the inductance as well as to achieve the distortionless condition. i.e. statement (1) and (4) are correct.
$:L
s.4.3s
Option (C) is correct. Single stub with adjustable.position is the best method for transmission line load matching for a given frbquency iunge. :
$*L
8.4.3?
Option (B) is correct. The reflection coefficient at load terminal is defined n Zr'- Zo +j50 - 50
as
--.F0T-50'L-z;TA -j
Therefore, the standing wave, ratio.is
* vsw*_1+lrrl_1+l_* vuYvrr --[l/;1-1=Tsot-
fi.4.3f
Option (A) is conect. Given, the load impedaace is short circuit l.e. So,
Zr =0
input impedance for lossless line is given
,,^: Now,
for
I
<
r,(ffiffi)
+ pt<
as
= izotangt
t
^14 So, tanpl is positive and therefore, Z;n is inductive
For *.,. i-t
a.+2
tanpl is -ve and therefore, \ oo aud therefore
Zm:
- Pt:r ':i tanBI:0 and therefore, Zin-
@
cn4
0
d-3
For
s"*.s$
Option (C) is correct. For distortionless transmission line'
o:JRG, 0:aJLC and for lossless transmission line,
o
:0,
0: o,/rc
So, for both the type of transmission line attenuation is constant and is independent of frequency. Where as the phase shifb p varies linearly with frequency
rr,,.
i.e. statement 1 and 3 are correct.
s.4.48
Option (D) is correct. Given,
I :500m
Length of transrnission line,
/0, :- 150' ,\:150m
Phase angle,
Operating wavelength,
Consider the reflected voltage wave for the lossless transmission line terminated in resistive Ioad as shown in figure.
Since, the reflection coefficient has a phase angle
-
15O' So, the wave lags
by 150'angle.
The voltage wave has the successive maxima at each So, the
total no. of
)/2
distance,
maxima:ffi -
500 (150i2)
-
i.e. 6 maxima and remaining phase angle
02
"3
:4
Pa€E Cet
Chp t I}ansrirsimLlE
t:i- 0t:5
For
tanpl:
b-1
fi, is capacitive
x
360"
:240"
F}om the wave pattern shown above we conclude that the remaining phase (240") will include one more maxima and therefore the total no. of maxima
Page 610
Chap
E
Thansmission Lines
is 8(}L 8.4.41
7.
Option (D) is correct. Reflection coefficient at load terminal is defined
as
n -Zt-Zo 'L-Zrj^ For a matched transmission line we have
Zt --
Zo
lr, :0
So,
i.e. matching eliminated the reflected wave between the source and the matching device location.
sol
8,4,42
Option (B) is correct. Consider the quarter wave transformer connected to load has the characteristic
impedance Z'o as shown in the figure.
So, we have the input impedir,nce,
,7
1tm
jz'stan(\X))l
(z'oY o,lzr+ -z; - 2'l z,+Eianey3;1-
This will be the load to 450 O transmission line' ryt (Z'of (Z'of
r.e.
oL:
and for matching
Zo :
zL:-20il
Z'r,
a5s:(?:^t - 200 Z,O: /G@pool sol
8.4"43
: 3oo r)
Option (A) is correct. Given and
Zv-a
t:+ ,,,:
-rq+
8.4,44
(quarter wave)
t,(Hffi):-
__ jz,cot(+)(*) 5('L
(open circuit)
(27
izocotgr
' a1
:o
Option (D) is correct. Length of transmission line r<^14 Load impedance, Z1 :a So, the input impedance of the transmission line is given
(open circuit) as
v '7 lZr'+ jZotan7l\ "* - "o\^T izianql ) 1 \ -ol - "'\it""@) :- jZscotpl SOL 8.,{.it5
Option (C) is correct. Given,
Page 611
(zt: *1
l}ansmission Lines
Chap
0
impedance, Lineparameters, Attenuation constant,
:0 R:G:Q o :0
Load
So, the
(lossless line)
Zn
input impedance of the line is given Zi" : jZstanBl
(short circuit) (loss free line) (loss free line)
as
o
i.e. pure reactance
Statement (A) is correct. since tanBl can be either positive or negative So zi, can be either capacitive or inductive. Statement (B) is correct. The reflection coefficient at load is
rr:4=4:-1*o Ltt Lo
So, the reflection exists.
Statement (C) is incorrect. and since the standing waves of voltage and current are set up along length of the lines so, statement (D) is also correct.
sol.
8.4.46
Option (C) is correct. (a) short circuit (Zt: 0)
So
7r-Zr-Zo--1 - z;+z; --
(b) Open circuit
So,
(Zt: *1 zt'- Zo l, - z;+z;-
(c) Line characteristic impedance
(a-2)
L
(b-3)
1 L
(Zr:
7, -- z:-4: o
7o; (c. -+ 1)
Zo*
(d) 2 x line characteristic impedance (Zr:2Zo) 2Zn_ Zn 1
fi
sot- s,4.47
Tffi2;: t
Option (D) is correct. Given, reflection coeffi cient,
It :7/o'
sol
8.4.48
so,
1+l vswR: 1+ll-r i=5: r
Option (A) is
correct.
-p1:
Given, reflection coefficient,
so, sol
8.4,49
l-:
'
1
+
vswR:1*lli:9:l 1-lfl-4-2
Option (C) is correct.
oo
(d'+
a)
I
Characteristic impedance of transmission line is defined
Page 612
Ohap
t
[T+.loL on: t,-:C f
TlairsmlssiDn Linos
line (,R So, for lossless transmission -G+
v_trC
:
G
:
as
0)
"o_\/ 4.4.$t) Option (C) is correct.
(a-3)
Input impedance has the range from 0 to oo. VSWR has the ran$e from 1 to oo Reflection coefficierrt (f) ranges from -1to *1.
s.4"5{
(c -+ 2)
(b-+
Option (C) is correct. Input impedance of a quarter wave transfonrrer is defirierl
1)
as
-za : T,
zt"
where Zo is the characteristic impedance of the line and Z; is fhe lua
s.4.sz
Option (A) is correct. The scattering parameters linearly relate the reflected wave to incident wave and it is frequency invariant so the scattering parameters are nlore suited than impedance Parameters.
8.4.$3
Option (C) is correct. Given, the reflection coefficient a,s 11 : g'y"-tto" At any point on the transmission line the reflection coefficient is defined
l(r) : ft.e
as
2"'
where .4is the distance of point frorn load.
z
So,
:0.1\
(Given) t'\;
: (0.3r: r3(]')(('-?'.t(0 f(z) : lrs : g.3s- ito"1s tt't"1 : 0.3e j1o2' - o.3e'2t'3 2r(tt
1)))
(Assurne o
:
0)
5(}L 8"4,54
Option (A) is correct Balun is used to couple a coaxial linc to a parallcl '*'irc.
$oL
8,4"55
Option (B) is correct. The reflection coefficient of the conducting sheet is I : - ! where as the transmission coefficient is l- : 0. So. there will be r-directed surf'ace uurrclll on the sheet.
sol.
e.4.s6
Option (B) is correct. Given,
Operating frequency, Conductivity,
f :25kHz o : Smholm e" :80
Relative perrnittivitv, The attenuation constant for the medium is defined
as
n: r,FF
(o >>
c"'e)
Chap
_
0t:
The attenuated voltage at any point is given
:
po)
as
Voe,"L
(1)
where I/6 is source voltage and I is the distance travelled by wave Since. the radio signal is to be transmittecl with 90% attenuation so, the voltage of the signal after 90% attenuation is
V : Vo- 90% of Vo : g.1yn Comparing
it with equation
(1) we get (0.11
.${}t- $.4-r{7
: r "'
_ It :-
or,
In(0.1)
_""J.27 m
ffi1zi:
Option (D) is correct.
In smith chart, the distance towards the load is
always measured in anticlockwise direction. So, statement 3 is incorrect while statement 1 and 2 are correct. ${3t
$.i!-sa{
Option (D) is correct. Giverr,
h :75(l
Characteristic impedance, Load impedance,
zL:(7oo-jrs){)
The condition for matching is
Z'r' : Zu load impedance of the transmission line after conrrecting an additional circuit. So, the best rnatching will be obtained by a short circuited stub at some specific distance frorn load. where
S$L &,!i-It
z't is the equivalent
Option (B) is correct. Given, the voltage standing wave ratio in decibels is VSWR, in decibels : 6 dB
or,
: : 'S
201o916.9
So. the reflection coefficient
6
(t0)6/20:
2
at the load terminal is given
as
:311:;i:o'33 Ir-S-1 '-1 ${:rL tt"4"rii}
Option (A) is cc-,rrect. Input impedance of a quarter wave transformer (lossless transmission line) is defined as
Z*:4 Zr.
wlrere zo is rhe characteristic impedance of the line and Z1 is the load impedance of the line. So. we get
zo: J7;"2,.: /(50(zoo) :1ooo s{}e- $,4.4$'t
Option (D) is correct. (1) Given,
E
Ilanemissiqn Lines
:0.7025 V
Page 613
Page 6L4
Length of line,
Chap 8 l}ansnission Lines
Load impedance,
l:)/8 Zt :0
So'
p,
: (+)(+):
+
Therefore the input impedance of the lossless transmission line is
t,^:t,(Hffi):r,(tu#) :
jzo
(i.e., incorrect statement)
(2) Given, Length of line, Load impedance, Zr. -- 0
t:\14
n:(+)(+):+
So,
Therefore the input impedance of the lossless transmission line is
z - v IZL+izoLanPl\ "o\/oa jzrtan6l)
"n"
:,,(!ry):r."
(i.e., correct statement)
(3) Given, Length of line, Load impedance, 21 :a
l:\12
pr
So,
: (+)(+): "
Therefore, the input impeda,nce of the lossless transmission line is
,,":rr(Zr*Um) : Zr(#)
:- i."
(i.e., incorrect statement)
(4) Matched line have the load impedance equal to its characteristic impedance
i.e.
Zr, _t7
-LO
So, for the matched line the input impedance is Znn
sol.
8.4.62
o (Zr.* jZstanPll- ,? -- "o\z;+izian0l )- "o
(i.e., correct statement)
Option (C) is correct. Given,
line, I : \18 Load impedance, 21 - 0 7t r2n'tl lt So, Pl - \T/\8/ -4 Length of
Therefore, the input impedance of the transmissi'on line is
z,^: If the line is distortion
z,(Hffiffi): less (i.e.
o:
z'*',ntt
0) then, the input impedance of the line
is
So,
Zio : jZslanpl: jzo it will depend on characteristic impedarlce as the line is resistive
reactive.
or
s9L
&.4.63
Option (B) is correct. Since, a transmission line of output impedance 400 o is to be matched to a load of 25 c) through a quarter wavelength line. so, for the quarter wave line we have Input impeda"nce,zin:400o (same as the o/p impedance of the matched Iine)
Zt :25Q
Load impedance,
l:\14
Length of line, The characteristic impedance of quarter wave transmission line is zs that connected between the load and the transmission line of output impedance 400
0
as shown
in figure.
25Q
So,, the
input impedance at AB is given
,r,:tr(Hm):q
Therefore,
sol
8.4.Sd
as
:
Zo
JZn
h :
,/ 4oo
x 2s:
100 Cl
Option (B) is correct. Given,
I:\/8 :0
Length of transmission line, Load impedance,
Zr,
p,:(+)(+):+
So, we get
(Short circuited line)
Therefore, the input impedance of lossless transmission line is given as
t'' :
t'(*ffi):
a(fi): in
which is inductive
so, the input impedance of ,\/8 long short-circuited section of a
lossless
transmission line is inductive. $(}L 8.4.6$
Option (C) is correct. In list I (a) Characteristic impedance of a transmission line is defined
zo:
r.:L tz C+Fe : \/ R+
(a-+ 2)
Y
(b) Propagation constant of the line is given
a^s
as
1 : ^/(N1rfi@+q"Q:,/Zy
(b-+ 1)
(c) Sending end input impedance is
,,"_r,(ft1ffi) Given, So, we get the
Zr,
: Zo
(terminated in characteristic impedance, Zs)
input impedance
as
Page 615
Chap E llansmisslgn Lingg
Pgge
6rf
Chap
E
Ilanemis3ion Linej
/t7 Zn: Zo: \/Y sol.
4.4.66
("'2)
Option (C) is correct. For a distortionless transmission line, the attenuation constant (a) must be independent of frequency (r^.,) and the phase constaut (B) should be linear function of (u)
cu.
R:G:0
For this condition propagation constant is given I.e.
a-s
.y:a-t j0:KR+Jrire+t o :0 and fi: rJ LC
q
As the attenuation constant is independent of frequency and the constant is linear function of (b)
cu
so,
it
phase
is a distortionless transmission line.
RC:GL R _G
L_C
This is the general condition for distortionless line for which o : r/EC and, p: r,/ LC (") R >> uL,G>>wC
.f
:a+
j{3:^/RC i.e. o :,,/EG, and p: g Since, B is not function of r.r so, it is not the distortionless line.
$oL 8.4.67
(d)
R<
i.e.
o
:0
and 1J:
,'frc
Option (B) is correct. Distance between adjacent -maxima of an EM wave propagating along transmission line is ),f 2. So, we get
a
^f2:(37.5-12.5) ),f2 :25cm
i.e.
l:5Ocnr
Therefore, the operating frequency of the line is
c 3x108 rt -^-50x10-2 : 600 MHz 3('L
S,/t.68
Option (C) is correct. Forward voltage wave along the transmission line is given
vi::a:F:4 - Zo-l Z,t" -
as
2
As the transmission line is open circuited at its load terminal
(21:
co)
so.
the reflection coefficient at the load terminal is
Page 61?
Chap
E
flansmissietr Linsg
Therefore. the voltage travelling in reverse direction is
_ r.t/'-E _Z v,, _tLvo The tirne taken by the wave to travel the distance between source alcl loacl terminal is giveri as
, ,.t
-l -E
I is the length of transmission line and c is velocity of propagatirrg wave. Now, from the plot rve observe that at z: 0, voltage of the line is tr',ll rvhcre as at z: /, voltage is E therefore, it is clear that the voltage wal'e has where
been reflected from the load
i.e.
but not reached yet to the genera,tor.
('L . t.4.(:
$$*
8"d.,{{t)
Option (C) is correct.
s*i-
*.4"?&
Option (D) is correct. The characteristic impedarrce fbr a lossy transrnission line does ncit depentl on the length of the line.
${3L *"41,?t
Option (C) is correct.
s*L
Option (B) ig correct. A clistortionlbss transmission line has it's parameters related
$.4,??
RG
T:E
or ssl.
&."{"?3
RC:GL
Option (B) is correct. Given the reflection coefficient
<-,f
the line is
f :0.6 Srr. the voltage standing wavc
ratio is defined
as
swR:!tf,+:J+99:+ 1-li-i -1-0.6$$L
S".4"?S
Option (B) is correct. Characteristic irnpedance, Load impedance, Forward voltage
Zo
:50Q
Zr :700Q 7+ : l0V
So, the reflection coefficient of the line is given as
100_50:3 ', -_zr,_za_ TlrZ,, - Too++ 50
1
sol.
*_4"?$
\ Option (C) is correct. Characteristic irnpedance, Zo :50Q Load impedance, Zr-15-.i20A norma\ized So, the \oad impedance is given as
',:2;:H
-rH
:0.3-io.4
as
the reflection coefficient at the load terrninal is
Page 617
lL
Chap
E
Tlnnsnission Lines
Therefore. the voltage travelling in reverse direction is
-.r/+_ZE _tLvo V/o-_ The tirne taken by the wave to travel the distance between source and load terminal is giveri as
, -l ,,t-c where
/ is the length of transmission line and c is velocity of propagatirrg
wave. Now, frorn the plot we observe that at z:0, voltage of the line is r!',t! rvhere as at z: l, voltage is E therefore, it is clear that the voltage wal,e tras been reflected from the load but not reached yet to the genera,tor. I.e.
L. t.4.c
(
*$&L S,d"{i*
Option (C) is correct.
ssl-
Option (D) is correct. The characteristic impedance f'or a lossy transmission line does rxit
e"4,"?8
$$N- S,4"?1
Option (C) is correct.
s#,- ti"4"7*
Option (B) ig corrcct. A distortionlbss transmissiorr line has it's parameters related
RG D:E
or $st-
&,4-y3
RC:GL
Option (B) is currect. Given the reflection coefficient of the line is
f :0.6 So. the voltage standing wave ratio is defined as
swR:ftlfl -1-0.6r- fl :1+99:+ s$t-
&"4-?d
Option (B) is correct. Characteristic irnpedance, Zo :50Q Load inpedance, Zr. : I00Q Forward voltage 7+ : l0V So, the reflection coefficient of the line is given as 1Q,Q-1[Q -Zu 4+zo --= ro'o + 50 :-
t, :- LssL
&.d."?ti
1
3
Option (C) is correct. Characteristic irrrpedance, Zo :50{l Load impedance, Zr,:!5-j20A So, the normalized load impedance is given as
,,:fi:#- i#:o.z_
io.4
as
Page 61E
$0:* 4"4.?6
Chap 8 Ilansmission Lines
Option (A) is correct. Since both the transmission lines are identical except that the loads connected to them are 22 and Zl2 respectively. Let the maximum voltage across the loads be 7- So, the power transmitted to the loads are
Po:l& -22
1UL
6"4.{
/
:- YL zl2
and
P,
Given,
Pa:
So,
vA
and
'e - (Zl4--ltr -+vvl
W,
: (22)W
n _ V:" _2ZWr _Attl
Option (D) is correct. Given, the short circuited and open circuited input impedance 2".".
--
36{L,
a,s
20.".:64{l
So, the characteristic impedance of the transmission line is defined as Zo
: J Z,,Z*. : /36 x 64 :
48o
Option (A) is correct. (1) Zr, : Zo (line terminated by So, reflection coefficient
its characteristic impedance)
r:ffi:o
i.e. no any reflected wave.
: Zo l':0
Zr'
(2)
and so, there will be no reflected wave and the wave will have only forward voltage and current wave which will be equal at all the points on the line. (3) For a lossless half wave transmission line
Zn:Zr So, statement 3 is incorrect while statements 1 and 2 are correct. ;*ill*
S""1.7S
Option (C) is correct. Since, the standing wave
ratio of the wave is 1.
SWR :7 So, expressing it {n terms of reflection coefficient, we get l.e.
1+li-l
1-ij-l -1 lr-l *0 Zr,
-
Zr,*
Zo Zo
-0
Zt'
:
Zo
i.e. characteristic impedance is equal to load impedance. ${}!-
4"4"&8
Option (A) is correct. Given, the two wire transmission line has Half center to center spacing Conductor radius
: n: t
: r
{u
Page 619
Chap 8 Tbansmission Lines
d
So, the capacitance per
.,
unit length of the line is defined
C: :
ltc
.,/W=]
b*"[* sol.
a.4"a{
Option (D) is correct.
r-
Reflection coefficient,
ZnZa*
: 4-
4+
sol.
8.rt.82
as
Zo Zo
zo zo
:_
tl2
Option (B) is correct. Propagation constant, ^Y- (n+ at)(G + The characteristic impedance of the transmission line is given
i'c)
v-F+W I G+fue
as
oo:
o R+ iuL "o_ .y sol.
8"4,83
Option (D) is correct. Given the reflection coefficient,
f:-l
3
So, the standing wave ratio
1+l.f _. "a _-1-lrl _4 -2-z I
sol-
8"4,84
Option (B) is correct. For distortionless transmission line
and so,
sol-
8.4.85
RL e:e the attenuation constant, o:,/RG-Fg, :R![g
Option (A) is correct. Capacitance per unit length, C : 10-10 F/m Characteristic impedance, Zo :50Q Now, for distortionless line the characteristic impedance is given
.7-tr oo_t/e
5o: - Vt4 1o-to So, the inductance per
unit length is
as
Page 620
:
7,
(50f x (10 r'r) :
pil/rn
0.?5
Chap 8 t"ensmission Lines
$$x- s"6.&*
Option (B) is corrcct. The characteristic impedance Ze in terms of open circuit impeclance 2,," and short circuit impedance .Z"" is defined as
2,,: [z^Z*
:
:100Q ${:*i* s"4"s?
(Given
v(loo)(loo)
Zo": 2,,,:
100
I
Optron (B) is correct. Given.
:
I,oad irnpedance.
zr.
Characteristic impedance.
Zo:75Q
(75
-
j50)
0
Since. for matching the load impedance is equal to the characteristk impedance (i.e., 21: Zs) so, vre have to prodqcc a1 additional impedalce u{ *:50 at load to rnatch it with transmission line. Therefore" for matchirrg the t,rarrsruission line a sirolt circuit stub is connected at sorne specific distance from load. S*i;
&,4.{iS
Optit-,rr (C) is correct. Given,
: Zt:
The load impedance l.e.
Surge impedance Zu
So, reflection coefficient of the line is given as
'r-Zt-Zo - Zr*Zu
-0 $$L
{!"4.{i$ Option (C) is correct. Given,
/:50cm:0.5m
Length of transmission line, Operating frequency,
.f:30MHz:30x106H2
Line parameters,
L:I}pH/rn:10x106H/m
and
C :40 pF/m : 40 X
10-'12
So, the phase constant of the wave a,long the transrnission line is
u
:{!^'rox
1oo,/.(10
_9"
xlrT4o-x lo ]
5
Therefcire, r$&r- fi"4.1is
fl :+
X 0.5:0.6a
:
108"
Option (D) is correct. Piopagation constant in a transmission line is defined
1 : {@+i,fi@T1,Q
$lrtlL a"4"sd
as.
Option (D) is correct. For a series resonant circuit the required conditions are (1) The angular frequency is
**
1
lrc
(2) The total equivalent impedance is pure resistive
F/m
Page 622
Chap
%
So,
:.reooxloO: looo
E
Tlansmission Lines
sol.
8"4,9?
Option (B) is correct. The range of sta,nding wave ratio 5 and reflection coefficient l- is defined
or $OL B;t,S8
lrl<
as
1
-1 < .f < 1 and 1 < ^9 < co
Option (C) is correct. Cha,racteristic impeda,nce of a transmission li4e is defined as
_ rE+Fr |
,7 "o G+fue So, Zo can increase with increase in resistance or inductance per unit length.
xx***r.{.*r<**
t\
CTIAPTER 9 WAVEGUIDES
9.,1
INTRODUGTION
The propagation characteristics of guided waves are different from wave propagation characteristics in free space. The main objective of this chapter is to provide the conceptual behaviour of EM waves in guided structures. The topics include:
o o
Fields and propagation characteristics between parallel plate wavegrddes. Fields and propagation characteristics in hollow rectqrgular and circular waveguides.
o TE, TM, and TEM waves o Cavity resonators st.2
MODES OF WAVE PROPAGATION
"
For time-harmonic fields, assuming wave propagation along the z-axis, the electric and magnetic fields can be written as
E(r,y,z) :lE,(r,a)+
E"a")eip" H(r,y,z):lH,(r,a)* H"a,fe-i,"
.
.
where the first terms Er(r,y) and - H1(r,y) represent the transverse components and the second tetms E" and H" represant the longitudinal components of the electric and magnetic fields, respectively. Colsidering the expression of field components, we define the foliowing modes of wave propagation:
Thansverse Electromagnetic
(TEM) Modes. In TEM mode, the electric and magnetic fields are transv€rse to the direction of wave propagation with no longitudinal components, i.e.
E,: H,-0
Tl'ansverse Electric (TE) Modes
In TE mode, the electric field is transverse to.the direction of propagation
(no longitudinal erectric field component), while the magnetic fierd has both transverse and, longitudinal components, i.e.
E":0, Tbansverse Magnetic
H, +
0
(TM) Modes
TM mode, the magnetic field is transverse to the direction of propagation (no longitudinal magnetic fierd componenJ), while the electric- field -;- has both tiansverse and longitudinal components, i.e. In
H":0,
E"
+
0
; n*
:
: Y B,sin(ff)e'
E2(g)
Page 625
"-r" where the amplitu de Bn depends on the strength of excitation of the particular TE wave. Propagation Constant Propagation constant for TE" mode in pa,rallel plate waveguide is grven by
1: ![W_",tu
Cut-off fhequency
The cut-off frequency for
TE,
mode in a pa,rallel plate waveguide is given by
rt"-rm n
9.3.2 TM
Mode For TM mode in a parallel plate waveguide, the nonrzero field mmponents are
t", : E2(a)e-j": Ansin(T)ur" Ho Eo,
:
Iq(U)"-r'
: ffA,cos(T)"r"
: El (y) "- r" : - I A,"o"(W) "
Propagation Constant Propagation constant for .'/
Cut-off flequency
TM"
mode is gil/en by
: \m-r'tu
The cut-off frequency for TM, mode in a paralhl plate waveguide is given by
rt" -
n
2bG.
9.3.3 TEM Mode For TM, mode in parallel plate waveguide, if n = 0 then we get
En:'A
i.e. only the transverse componenls I{o, apd
is the TEM mode.
4" exist. Thus, the- TMo ry,gde
Propagation Constant
Propagation constant for TEM mode (TMe mode)
ip a parallel pfate
waveguide is obtained as'
:/a-w2p.e Cut-off Fleguency
The cut-off frequency for TEM mode (TMo mode) in a parallel plate waveguide is
/.t
:0
f" Various propagation parameters summarized in table below.
of parallel plate
waveguide are
Chep 0 Wawggidc
Table 9.1: Propagation Parameters of Parallel Plate Waveguide
Page 626
Chap 9 Waveguides
RECTANGULAR WAVEGUIDE Rectangular waveguide is one of the earliest type of transmission lines used to transport microwave signals and are still used today for many applications. The hollow rectangular waveguide can propagate TM and TE modes, but no TEM waves, since only one conductor is present. Figure 9.2 shows a rectangular waveguide with its cross-section of sides o and b.
Figure 9.2: Cross-sectional View of a Rectangular Waveguide
'Let us analyse the characteristics of the TE and TM waves in rectangular waveguide.
9.4.1 TM
Modes For the TM case in rectangular waveguide, the non-zero field components are
: E* : 8," : E^
H""
k
Eo"(r,a)
_ "-,"
n,ri"(:H!)rr"(ry)"
,
-_a(T)a""'1l.-),t"(ry)"El(r,y) e-t, :-a(T)*,t"1ff. E\(r,v) e-1'
: r1(r, y)
)*,(ry)"","
e- 1"
- +(+)
",r"
(ff
)* "(ry)
"","
H*
where
o,
: 4(r,y1e-',, :-ff (T)a""'1+-),"(ry)"-
:[ry]'.[+]'
Page 627
Chap
I
\ilaveguides
Propagation Constant
For TM-, mode in a rectangular waveguide, the propagation constant is defined as
k, m,
ar,.d
n:
Case 1: Cut-off
ti
If
:
a2tte:[ry]'
.[+]'
"Y:0 ot q,:0:0
Then,
The value of
r,,'
that
this called the cutoff angular frequency c..r"
causes
; that is,
'":h llry]'.[+[
No propagation takes place at this frequency. Case 2: Evanescent
: u2tre .[ry]' .lTl'
If
ti
Then,
'y:et 0:o
In this case' we have no wave propagation at all. These nonpropagating modes are said to be evanescent. Case 3: Propagation
rr
t*
Then,
'Y:
:
a2
tre
,[ry]' .[T]'
j0, a:o
So, the phase constant B becomes
components will have the factor
Cut-off Flequency The cut-off frequency for
TM-,
"-tz
:
i?z.
"
mode in rectangular waveguide is given by
...(e.2)
Cut-off Wavelength The cut-off frequency for
TM*,
^":!f;:
mode in rectangular waveguide is given bv
, *g
-5-
i-:-
Page 62E
Phase Constant
Cfcrp e
Using equations (9.1) and (9.2), the phase constant terms of f, as
$hafrriidc
B
:r{G^Fffi
p
: p'rE-W
p
can be written in
Intrinsic Impedance The intrinsic wave impedance of the
^E,Er'Y ?rru : Eu:-fi:
TM-,
mode is obtained as
:#:#[-w
or where q' medium.
-7u,
Jru : n'lE-W : / lt l€
is the intrinsic impedance of a uniform plane wave in the
TE Modes For the TE case in rectangular waveguide,,the non-zero field components are
Ho
:
rr2(r,a)
",,"
_
n,*"(try!)*,(ty)"""
: (r, y) "-'" - #(ry) r,"" (ff u,' )'r "(ff) En, : E!(r,y) e-,' - -tVW)n,,i,(!TL)*"(tt;)"," H," : H2 (r, a) r,,-(#)""' (ry) "-'" - #(T) ""' H,, : rf;(r,y)e-,"#(T)r,*,(ry),t"(T)" ,, E*
E9
Propagation Constant
'
For TE-, mode in
the propagation constant
is
defined as
Cut-off flequency
-\. . .The cut-off frequency
for TE-, mode in rectangular waveguide is given.by
r": ,h /l+l';f",f Cut-ofr Wavelength The cut-off wavelength for TE-n mode in rectangular waveguide is given by 2
Phase Constant
The phase constant
p for TE,n, mode is also given
bv
where
g'
: r^/G
trr is the phase
medium.
"onrtuot
for the uniform plane wave in the
Intrinsic Impedance The intrinsic wave impedance for TE,,n mode is given by
7tE: :Jwhere
rt'
:,/t4€ r, ,r'/1;1{'J{),,r'o"our'"e of a uniform
plane wave in the medium. Following are some important points about wave propagation rnode in rectangular waveguide:
Ihble 9.2: Propagation parameters of Rectangular Waveguide
P'*anptei Plqsq Corstad
r:+Mm:i:;;
.*j
ur=$:15 'lJ
rt:Tt'fry I
9.4.3
Wave Propagation
in Rectangular Wavegurde
The propagation parameters of TE and TM waves inside a rectangular
waveguide is defined below.
Guide Wavelength
The wave components inside a rectangular waveguide has a different
wavelength from that of the plane waves. This wavelength along the uris of
qrt
Et
Page 630
the guide is called the waveguide wavelength and is given by
Chap 9 Waveguides
where
)'
is the wavelength in the medium given by
'U' 1 n\/ :7:7G
where u' is the medium velocity. Phase Velocity
The phase velocity is the velocity at which loci of constant phase
are
propagated down the guide and is given by O,
"e--9 p -.u
,
..(e.3)
Group Velocity The group velocity is the velocity with which the resultant repeated reflected waves are traveling down the guide and is given by "n
-
1
T4FA
,
:U'ffi
...(e.4)
Following are some important points about wave propagation in rectangular waveguide:
9.5
CIRGULAR WAVEGUIDE
A hollow metal tube of circular
cross section also supports TE .and TM g.3 waveguide modes. Figure shows the cross-section geometry of such a circular waveguide of inner radius a.
Page
Gll
ChT
t
Waveguidet
l:igule
1.,1.3;
Cross-sectional view of Circula.r Waveguide
9.5.1 TM Modes For the TM case in circular waveguide, the non-zero field components are
: fif;s-t": C,J,(hp)cos(nb)e-1" Er" : Ef;e-1" :-tOO C,Ji(hp)cos(nd)e-1" En
Eo"
:
E!e' ,,
:
Hn"
:
If;e-l,
:-
H6"
: If
e-1,
g:C^J^(hp)sin(n')"-,"
:-
h"p j?uun
\ '/ h'p C,J*(hp)sin(ng\e-,"
jgn,
C,J,i(hp)cos(ny)e-1,
where C, is an arbitrary constant that depends on the field strength of excitation, and J"(hp) is the Bessel function of the first kind defined as
r^(hp):ft,#flffi* Phase Constant
For TM,- mode in a circular waveguide, the phase constant is defined
p:JE:R:/*ffi
where
k: u"/G, and. P, J,(P, ) L
is the
rn16
as
root of J,(r),i.e.
0
Cut-off Flequency The cut-off frequency for
TM,-
mode in a circular waveguide is given by
f. : -Jc:: r( 2tr./ 1te -!t" 2ra"[1,te Intrinsic Inpedance The intrinsic impedance for TM,* mode is given by
^ rl.ru where ,7' medium.
Ea-T'0 :-8,jI;:_Tr: i
: ,[ tt I € is the intrinsic impedance of a uniform plane wave in the
Page 0ll2
9.5.2
Chap 9 Wrvo1uilet
TD Modes For the TE case in circula.r waveguide, the non-zero field components are Ho : f!f;s-'t, : CJ J"(hp)cos(nfi)e-l"
: -{
Ho"
:
Hr"
= I|e-r' :
Eo"
:
E*"
= E\e-,'
1,
Hf;e
Ef;e "'"
:
C,'
Ji(hp)
cos(n4y s-',,
J,(hp)sin(nd)"-,"
#ci
ffic,i J,(hp)sin(r,Q)"-,"
: #r;r,'(hp)cos(ng)e-l'
Phase Constant
For
TE"-
where
mode in a circular waveguide, the phase constant is defined as
k: w^fG,
ar'rd.
Jr'(P^-'1
-
Pn'
is the rntu root of
J,'(r),
i.e.
g
Cut-off fbequency The cut-off frequency for
TM"lc"
mode in a circular waveguide is given by
Pn' t _ t"-2nG-r";G Intrinsic TrnFedance The intrinsic impedance for TM"^ mode is given by
' where ,' medium.
E.6
Trt
E" _ rl'k =fr:-fi:T Eo
: .liJe is the intrinsic imped.ance of a uniform
plane wave in the
WAVEOUIDE.RESONATOR \dig.'yeguide resonatols
can be considered as a rectangular waveguide with both bnds closed by a conducting wall. The interior dimensions of J resonator ate.a,b, and c as shown in Figure 9.4. I
I I
t I
i i I
i I I ;
i
I'igure 9.,4: Dimensions of a Wayggude Resonator
I
As there is no unique longitudinal direction in a resonator, we choose the as the reference direction ofpropagation. rn actuality, the existence of
I
z-axis
I
conducting end walls al, z = o a"rid z: c give rise to multiple reflections sets up standing waves; no wave propagates in an enclosed cavity.
I
I I
i
*\ il\ I
t
a^nd
9.6.1 TM Mode Page
For TM-'o mode in a waveguide resonator, the non-zero field components are given as
: Eo sin(:r)"^(W) *, (ry)
- : - #(TX T) *' (w),^(rc)* (ry) " :*"'" (ry)*(ry) (#) '," #(TXT) "" : ,., #(T)n,"i"(T)*,(ry)" *(ry) u
Hu"
where
:#(T) *""" ( w)""(T)
* :lryf .[ry|
""-
(ry)
Resonant trYequency
The resonant ftequency for by
TM*,,
mode
r:#ffi
in a waveguide resonator is given
:#'m.W;W
Resonant Wavelength
The resonant wavelength for TM-no mode in a waveguide resonator'is given
by
f* :4: r ^,iEJGY+1xY+Gf 2
',
:
'.
r'*:'"
r.f ir
TE Mode TDnnp mod.e
in a waveguide
are given as
"
resonatox,' the non-zero fierd components
: r/s (ry)*(ry)*" (ry) - : #(,T) u, *"(T) -' (ry),
11*
cos
^(ry)
:#(T)*"'" W)*.(t#) "" (ry) , - : - #(TXf * (ry) *(ry) *' (T) "'" ) ,," : (ry)l^(W) *' (ry) E o"
1
#(TXT)
Resonant trbequency
Chap
I
lilaveguides
Eo
For
GIl
i
"
"''
The resonant frequency for TE-no mode in a waveguide resonator is given bv
r:#,m*W;W
Page 634
Resonant Wavelength
Chap 9
The resonant wavelength for TE-oo mode in a waveguide resonator is given by
Waveguides
I,: 9.6.3
,r@ie,yi ?f
Qualitv Factor The quality factor of a resonant cavity is a means of determining the loss. It is also a measure of the ba,ndwidth of the cavity resonator. The quality factor may be defined as
'" Pr.T - * Pr. - ^WW : 7ll
is the period of oscillation , P1 is the time-average power loss in the cavity, and W is the total time-average energy stored in electric a,nd magnetic field in the cavity. Following are some important points about waveguide resonator:
where T
,F** *
*rt<
rF***,t
EXERCISE g.{
Page 635
Chap 9 Waveguides
M{}& $"{",i
An electromagnetic wave propagating in an airfilled 10 has it's electric field in phasor form given as
x 4 cm waveguide
E* : b sin(2\tr r)sin(2ltra) e-jp" v I m What is the mode of propagation of the EM wave ? (A) TM,1 (B) TM1' (c) TE,1 (D) TE1' MCQ 9.1.3
The electric field component of an electromagnetic wave propagating in a rectangular waveguide is given in phasor form as E : Eosin(b0zrr)sin(4\ry) e-,, y m f "" The ratio of field components 8,"/E* will be equal to (A) 1.25 cot (50tr r)tan( Otr y) (B) 0.8 cot (5Urr)tan(4}tr y) (C) 1.25 tan (40n r) cot (50r y) (D) 0.S tan ( 40rr) cot(5}ny)
MCQ S.1.3
An EM wave is propagating in TEM mode in a parallel plate waveguide filled : 2.25, p,: 1). If the waveguide operating at 10 GHz then the phase constant and the group velocity of the wave will be respectively (A) 4.5 x 108rad/s, 139.6m/s of a dielectric (e"
(B) 139.6rad/s, 4.5 x 108m/s (C) 2 x 108rad/m, 3t4.2m/s (D) 314.2rad/m, 2 x 108m/s
;
ntsa g"{-d
A parallel plate wave guide has the plate separation b : 20-mm is made with s-lass (e" :2.r) between it's plates. If the guide is operating at a frequency f : 16 GHz then which of the following modes will propagaie ? (A) TMI (B) TM3 ; (c) TE, (D) all the three
me*
In a symmetrical slab waveguide, the phase velocity of TEr mode at cutoff is ur1. So, the phase velocity of TM2 mode at cutoff will be (A) ,o' (B) 2u,r
9,,t.5
s+ ua{& s.{"$
(D) ^/i u,,
An a x b airfilled rectangular waveguide is operating at a frequency, f :5GHz. what will be it's dimensions if the design frequency is 10% Iarger than the cutoff frequency of dominant mode while being 10% lower than the cutoff frequency for the next higher order mode ?
(A) o : 3.3 cm, b :2.7 cm (C) a:0.37cm, b:0.3cm
(B) a: 1.1cm, b : 0.9 cm (D) o: 0.5 cm, b : 0.4 cm
Page 636
Chap
ingcl c"t.7
I
Waveguides
electromagnetic wave is propagating in a pa^rallel plate waveguide operating at TMl mode. The magnetic field lines in the gz-plane will be (Assume the positive r-axis directs into the paper)
An
"12 r
,
3trl2
2tr
?t'
*x '.6)
,o
mcQ s.t.8
An EM wave is propagating at a frequency 'f in an air filled rectangular waveguide having the cutoff frequency '/.'. Consider the phase velocity of the EM wave in the waveguide is uo. The plot of. (clu) versus (i//) *itt Ue(c is the velocity of wave in air)
(A)
(B)
("/
r)
(f./
MCA 9.1.9
r)
An electromagnetic wave propagating at a frequency '/' in free space has the wavelength ')'. At the same frequency it's wavelength in an airfiIled waveguide is \. If the cutoff frequency of the waveguide is $ then the plot
of
(\/I)
versus
(///"; *itt
P{e
ue
Ctap
()ri ))
i\ _ _ _ _i_ _ _ _ _ _ _ _ _
I
_
i
Ulf")
Ull.)
(!l f")
(fl f")
(c)
Common Data For Q. 10 and 11 : A parallel plate waveguide is separated by a dielectric medium of thickness b with the constitutive parameters e and p
".o
*.r.ro If the cut-off
MCQ 9"t.t1
frequencies
of the waveguide for the modes TE1, TE2 and
TEs are respectively u)",, Q"z, r.r". then Which of the following represents the correct relation between the cutoff frequencies ? (B) ," 1u", 1u"" (A) ,", : 0)"2: oJ"t (D) ,/r"1a,: ,", (C) ,", ) u",) u",
curve (graph of frequency f versus phase constant wayeguide for the modes TM2, TMa and TMa will be The
f-B
(B)
(A) ,,
'fz
/'
h
I,
/' (c) 't,
(D) j;
h
h
I
Waveguides
,}..
(B)
437
p) of the
Page 638
Chap
&acq
s,{.{:
I
Waveguides
A rectangular waveguide operating in TEro mode has the phase constant B. If the average power density of the guide in this mode is P,, then what will be the relation between P",, asd, 0 ? (A.) P"" o 0 (B) P^, a P2
(c)
P,"
o
1
'B
(D) P,, is independent of B Common Data For Q. 13 and 14: A asymmetric slab waveguide has the different mediums above and below the slab as shown in figure. The regions above and below the slab have refoactive indices nz arrd ns respectively while the slab has refractive index nr
mc& g"'!."r3
2.8, n2 - 7.7 , nt : wave propagation will be
2.1 ,
(A) 48.6"
,
If
n1
(c) rKs(l g-{"{{
:
then the ininimum possible wave angle for the
4t.4"
(B) (D)
37.4" 54.1"
If the refractive
indices of the mediums are related as maximum phase velocity of a guided mode will be
n1
) nz> nz the
(A) n3c (B) clnl (C) clnt (D) cln, (c is velocity of wave in free space) n6g& $.{.{5
The first four propagating modes of a circular waveguide are respectively
(A) TMo1, TE21, TEo1, TM11 (B) TM1l, TMzr, TM62, TMp (c) TE11, TEzr, TMrr, TEo, (D) TE11, TMor, TE21, TEol Common Data For Q. 16 and L7 : A microstrip line has the substrate thickness MC& 9,{"{S
If the characteristic width of microstrip. (A) 2.83 cm (B) 0.28 cm (C) 0.36 cm (D) 0.14 cm
d:
0.316
cm with
impedance of the guide is 100 O then what
e
,:2.2
will be the
trcQ
s,1.{7 If the transmission line is operating at a frequency, :gGHz then the f effective permittivity e" and guide wavelength ,\, will be
(A) (B) (C) (D)
ee
t.TO 2.8s 0.158 18.87
^s cm 2.83 1.26 cm 18.87 cm 0.158 cm
page63g Chap 9 Waveguides
**{<+*****x*
.\.:
.r1
Page
6z10
Chap
I
EXERCI$E 9.2
Waveguidee
QUES 9.2,1
An airfilled rectangula,r waveguide is operating in TM mode at a frequency twice the cutoff frequency. What will be the intrinsic wave impedance (in O
)? qur$
e.?'2 An EM wave is propagating in
quc$
e.2.3
TMzr mode in an air filled 10 x 4 cm waveguide at a frequency of 7.5 GHz. What will be the phase constant (in rad/m) of the EM wave ? An airfilled 5 x 3 cm rectangular waveguide is operating at TEro mode at frequency of 3.75GH2. The group velocity of tbe propagating wave in the waveguide will be __-__--_ x 108 m/s. rectangular wavegurde with the dimensions a:2.5cm, b:5cm is operating at a frequengy f :15 GHz. If the wave guide is filled with a lossless dielectric with p" = 1., €,: 2 then the wave irnpeda,nce of propagating TE26 0. mode in the waveguide will be
rxrF$ s.2,4
A
{}irE$ s,2.5
Cutoff wavelength of a parallel plate waveguide for TMz mode is 2 mm. If the guide is oferated a,t a wavelength ,\ = 0.1cm then the no. of possible modes that can propagate in the waveguide is _______
&Lr€s 9.2,6
A lossless parallel plate waveguide is operating in TMs mode at frequencies as low as 15 GHz. What will be the dielectric constart of the medium between plates if the plaie s€paration is 10 mm ?
otles
e.2.7
The cutoff frequency of TMr mode in qn air filled parallel plate wave guide is 2.5 GHz..If the guide fs operating at wavelength .\ - 3 cm then.what will be the group velobity gt.fE, mode will be x 108 m/s.
-------
auE$ s,2.8
A symmetric slab waveguide flas a slab thickness d: 5lm with refractive indices pr:3, rh:2.5 ps qho-wr-r in figure. The phase velocity of the TEr mode at cutoff will be x 108 m/s.
\ Common Data For Q. 9 and 10i A strip line transmission line has the ground plane separation, and filled of a material with e": 8.8. guEs 9.2.9
:
0.632 cm
Qurs
9.2"1t An airfilled,
line is operating at a frequency f be it's guide wavelength (cm) ?
and
c:
:
lossless cavity resonator has dimensions
20 cm.
B
GHz then what will
o:30cm, b:25cm
What is the resonant frequency (in MHz) of
TE161 mode ?
s":.tz An airfiIled cubic cavity
resonator (a:b: c) has dominarrt resonant frequency of 15 GHz. The dimension of the cavity resonator is cm.
quc$ e.2"13
A parallel plate waveguide
,oper,ating at a frequency of 5 GHz is formed of two perfectly conducting infinite plates spaced 6 cm apart in air. The maximum time average power that can be propagated per unit width of the guide for TM1 mode without any voltage breakdown will be MW/m. : (Dieigclric strength of air 3 x 106 V/-)
e.2.14 An air filled parallel plate wave guide has the separation of 12 cm between it's plates. The guide is oper'ating at a frequency of 2.5 GHz. What' is the maximum average power per unit'width (GW/m) of the guide that can be propagated without a voltage brea.kdown for TSM mode,?
quc$ e.a",rs A parallel plate waveguide filled of a dielectri. (r": 8.4) is constructed for operation in TEM mode only6ver the fggqueiicy range p < .f < 1.8 GHz. The maximum allowable separation betweed.the flates will.be cm,
Co"'mon Data For q. 16 apd"L7 : A parallel plate waveguide having plate separation b: 14.1mm is partially filled with two lossless dielectric with permittivity e,1: 2 and €z: 1.05. QuEs
s.2.'t$ The frequency 'fo' at
Guss
e.2.17 How many TM
Qurs
e.2.t8 A symmetric
whiCh the TMr mode propagates through the guide without suffering any reflective loss is GHz.
€n:2.I d:20
modes
that can propagate in the guide at the frequency $
?
dielectric slab waveguide.with it's permittivities €,^:2.2 and is operating at wavelength, .\:2.6pm. If the slab thickness is
pm then how many mbdes can propagate in the slab
?
Ml
Chap 0 Waveguller
impedance of the transmission line is 35 then what will be the width (cm) of conducting strip ?
s.2.{o If the transmission
QuE$
b
If characteristic
Quss
quss
Page
e.z,le A symmetric
dielectric slab waveguide with it's refractive indices n1 a\d ft2 is operating at wavelength, .\:3.1pm. If the slab thickness is 10pm and nz:3.3 then what will be the maximum value of n1 for which it supports only a single pair of TE and TM mode ?
Page 642
Cbap 9 Waveguides
QuEs
3.2.20 An air filled waveguide is of square cross-section of 4.5 cm on each side. The waveguide propagates energy in the TErz mode at 6 GHz. The wavelength of the TE22 mode wave in the guide is
QuEs
e.2.2t
cm.
An attenuator is formed by using a section of waveguide of length I operating below cutoff frequency. The operating frequency is 6 GHz and the dimension of the guide is a:4.572cm as shown in figure. What will be the required length I (in cm) to achieve attenuation of 100 dB between IIP and OIP guides
?
u-u-
Fropagating
Propagating 'vfave
QUE$
e'2.22 A rectangular waveguide with the dimension, a:1.07cm is operating in TEro mode at a frequency, f - 10GHz. If the waveguide is filled with a dielectric material having 6": 8.8 and tan6:0.002 then the attenuation constant due to dielectric loss will be dB/m.
'
QuEs
Common Data For Q. 23 and 24 : A rectangular cavity resonator with dimensions o:2.5cm, c : 5 cm is filled with a lossless material (p: ltr, e : 3eo).
9.2.23 The resonant frequency of the cavity resonator for
b:2cm
TE1s1 mode
and
will
be
GHz. QUE8
e.2.24 If the resonator is made of copper then the quality factor for TE1s1 mode (Conductivity of copper r: 5.8 X 10? S/m)
QU:a
e.2.25 An air filled circular waveguide for TE11 mode will be
ouEs
is
has it's inner radius 1 cm. The cutoff frequency
GHz.
(P'rr:1'841)
s.2.26 A cylindrical cavity shown in the figure below is operating at a wavelength of 2 cm in the dominant mode. The radius a of the cylindrical cavity will be cm. (Por: 2.40b)
PtEs al3
Cary
t
Wercgrider
**x****<**x*
EXERCI$E 9.3
P{p M4
(lrp9
t
Wrrgddi
r{tcs
e.3.t
Assertion (A) : In a waveguide operating below cutoff frequency there is no net average power flow down the waveguide' Reason (R) : Propagation of energy requires a propagating mode' (A) A and R both are true and R is correct explanation of A' (B) A a.nd R both are true but R is not the correct explanation of A.
(C) A is true but R is false. (D) A is false but R is true' tvtce
9,3,2 An a X b rectangular waveguide is operating in four different
modes
as
TMrr,TMrz,TEroandTE20'Ifa:2bthentheascendingorderofthe
operating modes for their cut-off frequencies will be (A) TEm < TE:o ( TMtt ( TMrz (B) TEm ( TMtt ( TEzo ( TMrz
(C) TMrz ( TMtr ( TEzo ( TEro (D) TEro ( TEro ( TM'z ( TM" Mca
e.3.3
Consider the following statements 1. TEM mode can not exist within a hollow waveguide'
2. Any of the TM mode can't be the dominant mode of propagation
in
rectangular waveguide. The correct statement is
(A) only 1 (C) 1 and 2 both mce
s.s.4
(B) onlY 2 (D) None of these
The lowest order TM mode that can exist in a cavity resonator is
(A) TMru
(B) TN{i1o (D) TM1ol
(c) TMol1 Ince
rurce
e.3.5 If the dimensions of cavity resonator are equal (i.e. o: b) then the lowest . order TE mode will be (B) TEloo (A) TEoll (D) (A) and (C) both (C) TE1ol resonator has the dimensions a> b> c. Which of the amanged in ascending order with respect to their are modes following
e.3.6 An air filled cavity
resonant frequencies
?
(A) TMrro, TEou, TEror (B) TEolr, TMrio, TEror (C) TMloo, TMror' TMttt (D) TMiro' TEror, TEorr
llcQ
e'3.7
When a wave is propagating in r-direction, TE wave has E" equal to
h!"Gg . Gbr
4
(A) zero (C) E"
(B) (D) Ey+ E,
muqddce
ItdsQ
s.3.s
For a wave propagating in z-direction in a hollow rectangular waveguide, TEM wave has (A) E,:0 (B) H,: o (C) all components of .E and H are zero (D) E, : 0, H,: 0
Msa
e.3"e
The cut-off frequency of TEM wave is
(A) (C)
zero infinity
(B) / (D) that of TEro
McQ
e.g.tt)
MCQ
9,3"Xd If 'f *n: Ct-,, i6r a WavegUide
The dominant mode of propagation in an air-filled waveguide of 2 dimensions operating at 10 GHz is (B) TE,l (A) TEol (D) TE11 (c) TElo
(A) propagation takes place (B) no propagation takes place (C) energy is increased (D) energy is decreased XX**<*t {<*t
t
x
Lcm
EXERCT$E 9.4
Page 646
Chap 9 .Waveguides
ltco
9.4-{
The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is
:3cos(2.094
x
102r)cos(2.618
The phase velocity
uo
of the wave inside the waveguide satisfies
11,
x
102y)cos(6.288
x I0lot-
pz)
1.2 cm
(A) uo> c (C) 0< ircQ
9.4-2
(B) ?,: (D) oo:
uplc
s
g
The modes in a rectangular waveguide are denoted by TE* or TM-. where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. which one of the following statements is TRUE? (A) The TM16 mode of the waveguide does not exist
(B) The TE16 mode of the waveguide does not exist (c) The TMls and the TEls modes both exist and have the same cut-off frequencies
(D) The TM16 and the TMsl modes both exist and have the same cut-off frequencies
lltco 9-{-3
The electric and magnetic fields for a TEM wave of frequency 14 GHz in a permittivity 6" and relative permeability are given by E: Ers!('t-2eu"d a"Y f m and ,E[ - g"i@t-28oru) a, A.f m. F,:7 Assuming the speed of light in free space to be 3 x 108m/s, the intrinsic impedance of free space to be 1202r , the relative permittivity s, of the medium and the electric field amplitude ,8, are (A) r, 3, Er: I20r (B) r" : 3, Ep: 3602r homogeneous medium of relative
:
(C) r" iltco
9.4.4
:9, Ep:
(D) r,
3602r
:9, Ep: I20n
Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ?
P: Coaxial
(A) (B) (C) (D)
Q: Cylindrical
Only P has no cutoff-frequency Only I has no cutoff-frequency Only .B has no cutoff-frequency All three have cutoff-frequencies
ffi
R: Rectangular
Mcq
ItrlcQ
4 cm a.nd b : 3 cm) to be operated in TE11 mode. The minimum operating frequency is (A) 6.25 GHz (B) 6.0 GHz (C) 5.0 GHz (D) 3.75 GHz
s,rl.s A rectangular waveguide of internal dimensions (a:
$"4"$
is
The .E field in a rectangular waveguide of inner dimension o X b is given by
, : #(+)n,,i"(T)sin(r.,t - g") a, Where Ilo is a constant, and o and b are the dimensions along the r-a>cis and the y-axis respectively. The mode of propagation in the waveguide is
(A) TEro
(B) TM11 (D) TElo
(c) TM,o McQ
s.4.7 An air-filled
rectangular waveguide has inner dimensions of 3 cm x 2 cm. The wave impedance of the TE26 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance T0 : 377 Q)
c2 (c) 4oo 0
(B) 355 o (D) 461 o
(A) 308
l{lco
e.4-a A rectangular
wave guide having TEro mode as dominant mode is having
a cut off frequency 18 GHz for the mode TE36. The inner broad - wall dimension of the rectangular wave guide is
(A) f cm (C) f cm IttcQ
(B) 5 cm (D) 10 cm
e"4"e Which one of the following does represent the electric field lines for the mode in the cross-section of a hollow rectangular metallic waveguide
(A)lltlllllllllll. l_
(c)lllllllllllill"
(B)
l:l l:l
?
,
(")l-l a I
!-l
McQ
9.rl"lo
The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TEro mode is (A) equal to its group velocity (B) less than the velocity of light in free space (C) equal to the velocity of light in free space (D) greater than the velocity of light in free space
McQ
9*4"{{
In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz (A) To increase the sensitivity of measurement (B) To transmit the signal to a far-off place (C) To study amplitude modulations (D) Because crystal detector fails at microwave frequencies
page 64? Chap 9 waveguide
a
Page 64E
McQ
Chap 9
e"4'12 A
rectarrgular .lnetal wave guide fiIIed with a dierectric material of relative €,: 4 has the inside dimensions ti;- x r.2 cm. The cut_off frequency for the dominant mode is (A) 2.5 cHLz (B)
permittivity
Wavquidea
5.0 GHz
(C) 10"0 GHz Iucc
e"4.13
$ca
e'4't4
(D) 12.5 GHz
The phase velocity for the TE,q _mode in an air_filled rectangular waveguide is (c is the velocity of plane waves in free space) (A) less,than c (B) equal to c (C) greater than c (D) none of these
The phase verocity of wave propagating in a holrow metal waveguide is (A) grater than the velocity of hght irr-fr""
,pr."
(B) less than the velocity of light in free space (C) equal to the velocity of light free space (D) equal to the velocity of light in free ilcQ
e'4'15
The dominant mode in a rectangular waveguide is TE16, because this mode
has
(A) the highest cut_off wavelength (B) no cut_off (C) no magnetic field component (D) no attenuation. MGQ
MGq
e'4'tc
A TEM wave is incident normally upon a perfect conductor. The .o and ---- - *^.r field at the boundary will beiespectiv"ty, (A) minimum and minimum (B) maximum and maximum (C) minimum and maximum (D) maximum and minimum
e'rl't7
rectansular waveguide has dimensions i.'|
(A) 5 GIIz
(c) tucQ
cH;
cm x 0.5 cm. Its cut-off frequency
(B) 10 GHz (D) 12 GHz
e'4't8 Assuming ,
ileQ
15
1
r/
s'4'to
perfect conductors of a transmission line, pure TEM propagation is NOT possible in (A) coaxial cable
(B) air-filled clindrical waveguide (C) parallel twin-wire line in air (D) semi_infinite parallel plate wave guide Inilicate which one of the following will NoT exist in a rectangular resonant
cavity.
(A) iE,io (C) TMrio
(B) TEml (D) TMll1
MGQ g.{.20
The ratio of the transverse electric field to the transverse magnetic field is
called as
Chap
(A) wave guide impedance (B) wave guide wavelength (C) phase velocity (D) Poynting vector MCQ 9.4.2r
Consider
a
ItcQ
s.4.22
rectangular waveguide of internal dimensions 8cm X 4cm. h would be (B) 16 cm (D) 32 cm
'Y: ./(+y +(W
-et
€
represents the
rectangular waveguide for (A) TE waves only
(C) TEM waves mcq 9,4.23
constant in
with the symbols
having their standard. meaning, cut-off frequency (frequency below which wave propagation will not (occur) for a rectangular waveguide is
(D)
A
^h{w.w h,[wiw
standard air filled waveguide w4-rg7 has inside wall dimensions of 4.755 cm and b:2.2l5cm. At 12 GHz, it will support (A) TEro mode only (B) TElo a^nd TE2s modes only ' , (C) TEro,TE2sandTEsl modes only
a:
(D) TEro,TE2e,TEsl iitcQ 9.4.25
a,nd TE11 modes
consider the following statements relating to the resonator : 1' The cavity resonator does not posses ais *ury *od." "urriiy u, th" waveguides does. "or.erponding 2. The resonant frequencies oi cavities are very closely
3.
spaced.
The resonant frequency of a cavity ,"ronuto,
Uu
"*r, its dimensions. Which of the above statements is/are correct ? (A) 2 and 3 only (B) 2 only (C) 3 only (D) 1, 2 and l,tcQ 9.4,26
a
(B) TM waves only (D) TE and TM waves
(B)
rucQ 9.4.24
.hurrg"d by altering
3
The correct statement is
(A) Microstrip lines can support pure TEM mode of propagation but shielded coaxial lines cannot
(B) Microstrip lines cannot support pure TEM mode of propagation but shielded coaxial lines can
(c) Both microstrip lines mode of propagation
I
Waveguides
Assuming an Hro mode of propagation,
(A) 8cm (C) acm
Page 649
and shierded coaxial lines can support
p're TEM
(D) Neither microstrip lines nor shierded coaxiar lines can support pure TEM mode of propagation.
Page 650
Chap
MCQ 0.4.27
I
Waveguides
An air-filled rectangular waveguide has dimensions of a : 6 cm and 6 : 4 cm . The signal frequency is 3GHz. Match List I with tist II and select the correct arlswer using the code given below the lists :
I
List
;\
TEto
b.
TEb
List
1. 2. 3. 4.
c.
d.
II
2.5 GHz 3.75 GHz
4.506 GHz 4.506 GHz
abcd
(A) r234 (B)4231 (c)1324 (D)4321 9.4"28 For plane *ui" proprgating in free
space or two conductor transmission line, what must be the relationship between the phase velocity uo, the group velocity us and speed of light c ?
(A) ,o ) (C) ,o : $ca
9,4,29
c) c:
(B)ro1c4ug (D),olus
us 'us
Consider the following statements : In a microstrip line
1.
Wavelength ,\: *, where e" is the effective dielectric constant and the free space wavelength.
\
is
Electromagnetic fields exist partly in the air above the dielectric substrate and partly within the substrate itself. 3. The effective dielectric constant is greater than the dielectric constant of the air. 4. Conductor losses increase with decreasing characteristic impedance. Which of the above statements is/are correct ? (B) 1 and 2 (A) 1, 2 and 3 (D) 4 only (C) 2,3 and 4
2.
rilcQ 9.4.30
Match List I with List
II
and select the correct answer using the codes given
below the lists.
List II Modes of Propagation
List I Type of lSsnsYnission Structure
a. b. c. d.
Strip line Hollow rectangular waveguide Microguide Corrugatedwaveguide
Codes
(A) (B)
(c) '(D)
a
b
c
d
2
1
3
4
4
1
3
2
2
3
1
4
4
3
1
2
1. 2. 3. 4.
Quassi TEM
Pure TEM
TE/TEM Hybrid
).",",,
Assertion (A) : TEM (Tlansverse Electromagnetic) waves cannot propagate within a hollow waveguide of any shape. Reason (R) : For a TEM wave to exist within the waveguide, lines of 11 fierd must be closed loops which requires an axiar of E which is not present "o-porrJrrt
in a TEM wave. (A) Both A and R are individuaily true and R is the correct explanation of A (B) Both A and R are individually true but R is not the correct explanation
ofA
(C) A is true but R is.false (D) A is false but R is tiue !i.?$& S,;t.S3
A standard waveguide wRg0
has inside wail dimensions of a: 2.2g6 cm cm. What is the cut_off waveguide for TEor mode ? (A) (ej z.zaacm (C) 2.032 cm (D) 1.857 cm .,,
and
rscQ 9,4.33
b
:
1.016
a.\T2cm
when a pa'rticular mode is exited in a waveguide, there appears an extra electric component, in the direction of propagation. In what mode is the wave
propagating
?
(A) Tlansverse electric (B) Tlansverse magnetic (C) tansverse electromagnetic (D) Longitudinal MC& S,4"34
Consider the following statements : For a square waveguide of cross-section 3 m X 1. at 6 GHz dominant mode will propagate.
3
m it has been found
2. at 4 GHz all the mode are evanescent. 3' at 11 GHz only dominant modes and no higher order mode will propagate. 4. at 7 GHz degenerate modes will propagate. Which of the above statements are correct ? (A) 1 and 2 (B) 1, 2 and 4 (C) 2 and 3 (D) 2, 3 and 4 Mca
e'd'35
Match List with List below the lists.
List
a. b. c.
I
(Mode)
List
Evanescent mode
Dominant mode TMro and TMor
Codes a
b
c
1
2
3
1
3
2
(c)
2
3
1
(D)
2
1
.)
(A) (B)
II and select the correct answer using the codes given 1. 2. 3.
II
(Characteristic)
Rectangular waveguide does not support No wave propagation Lowest cut-off frequency
Page 651
Chap 9 Waveguides
e.4.s6 Assertion (A) : A z-directed rectangular waveguide with dimensions 3cm X lcm will support propagation at 4GHz.
Page 652
Chap 9 lVaveguide
Reason (R)
:
,t?
*(ryY
*(Tf : (+),where
)
is the wavelength.
(A) Both A and R are individually true and R is the correct explanaric A.
(B) Both A and R are individually true but R is not the correct of A.
(C) A is true but R is false (D) A is false but R is true IrrcQ 9,4.37
ilcQ
Which one of the following is the correct statement A rectangular courial line can support (A) only TEM mode of propagation (B) both TEM and TE modes of propagation (C) either TE or TM mode of propagation (D) TEM, TE or TM mode of propagation
?
9.4.38
A rectangular waveguide (A) is gradually deformed first into a circular guide (B) and lack again into a rectangular waveguide (C) which is ori through 90" with respect to (A) If the input mode is TE16, which mo&i excited in the output waveguide (C) ? (A) TElo (B) TEol (c) TE1l (D) TMl, [dcQ 9,4.39
The dominant mode in a circular waveguide is a
:
(A) TEM mode (B) TM61 mode (C) TE21 mode (D) TErr mode rRcQ 9.4.40
The cut-off frequency of the dominant mode of a rectangular wave guib having aspect ratio more than 2 is 10 GHz. The inner broad wall dimensirr is given by :
(A) 3 cm (C) 1.5 cm mcq
9.4.41
(B) 2 cm (D) 2.5 cm
In a waveguide, the evanescent modes are said to occur if (A) The propagation constant is real (B) The propagatioir constant is imaginary (C) Only the TEM waves propagate (D) The signal has a constant frequency
:
Assertion (A)
: A microstrip line cannot
support pure TEM mode of
propagation. Reason (R) : A microstrip line suffers from various forms of losses. (A) Both A a.nd R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false
(D) A is false but R is true r$GQ 9.4.43
Consider the following statements relating to the cavity resonators
1.
:
For over-coupling the cavity terminals are at voltage maximum in the input line at resonance 2. For over-coupling the cavity terminals are at the voltage minimum in the input line at resonance 3. For under-coupling the normalized impedance at the voltage maximum is the standing wave ratio 4. For over-coupling the input terminal impedance is equal to the reciprocal of the standing wave ratio Which of the statements given above are correct ? (A) 1 and 2 (B)3a,nd 1 (C) 1 and 3 (D) 2 and 4 ffiGA 9,4.44
Consider the following statements relating to the microstrip lines 1. Modes on microstrip lines are purely TEM
:
2. 3.
Microstrip line is also called open strip line Radiation loss in microstrip line can be reduced by using thin high dielectric materials 4- conformal transformation technique is quite suitable for solving microstrip problems Which of the statements given above are correct ?
MCQ 9.4.45
(A) 1, 2 and (C) 1, 3 and
4
Match List
I
(B) 2, 3 and 4 (D) 1, 2 and 4
3
(Dominant Mode of Propagation) with Libt
transmission Structure) and select the colrect answer
List-I
a. b. c. d.
Coaxial line Recta.ngularwa,veguide
Microstrip line Coplanar waveguide
Codes
(A) (B)
(c)
(D)
List-II 1.. TE
:
a
b
c
d
1
4
2
3
4
1
3
2
1
4
3
,
4
1
2
3
2. 3. 4.
Quasi TEM
Hybrid
TEM
:
II
(Type of
Page 653
Chap
I
IVaveguides
Pa.ge 654
&frSe Lit"46
Chap 9 Waveguides
For TE or TM modes of propagation in bounded media, the phase velocity (A) is independent of frequency (B) is a linear function of frequency (C) is a non-linear function of frequency (D) can be frequency-dependent or frequency-independent depending on the source
MA{!
S"4"4?
n{s& s.4.68
A waveguide operated below cut-off frequency can be used as (B) An attenuator (A) A phase shifter (D) None of the above (C) An isolator Assertion (A) : The quality factor Q of. a waveguide is closely related to its attenuation factor a. Reason (R) : Normally attenuation factors obtainable in waveguides are much higher than those obtainable in transmission lines. (A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is MCS S.4,{9
Assertion (A) : The greater the 'Q" the smaller the bandwidth of a lesonant circuit. Reason (R) : At high frequencies the Q of a coil falls due to skin effect. (A) Both A and R are true R is the correct explanation of A (B) Both A and R are true R is NOT the correct explanation of A
(C) A is true but R is false (D) A is false but R is tt\e n.l*q 9,4,$s
For a wave propagation in an air filled rectangular waveguid.e' (A) guided wavelength is never less than free space wavelength
(B) wave impedance is never less than the free space impedance (C) TEM mode is possible if the dimensions of the waveguide are properll-
.
chosen
(D) Propagation constant is aiways a real quantitv ts{{} s.4"s{
When a particular mode is excited in a wave-guide, there appears an extra electric component in the direction of propagation. The resulting mode is
(A) transverse-electric (B) transverse-magnetic (C) longitudinal (D) transverse-electromagnetic ru{:{t 9.4,52
For a hollow waveguide, the axial current must necessarily be (A) a combination of conduction and displacement currents (B) time-varying conduction current and displacement current (C) time-varying conduction current and displacement current
(D) displacement current only
i[cQ
9.4.$3
As a result of reflections from a plane conducting wall, electromagnetic the velocity of light in
waves acquire an apparent velocity greater than space. This is called
(A) velocity propagation (C) group velocity
lltco
(B) normal velocity (D) phase velocity
Assertion (A) : A thin sheet of conducting material can act as a low-pass filter for electromagnetic waves. Reason (R) : The penetration depth is inversely proportional to the square root of the frequency. (A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is the correct explanation of Assertion (A) (B) Both Assertion (A) a.nd Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A) (C) Assertion (A) is true but Reason (R) is false (D) Assertion (A) is false but Reason (R) is true
tscQ 9,4.55
Consider the following statements in connection with cylindrical waveguides
constant is real and wave does not
propagate.
2. 3'
At intermediate frequency the propagation constant is zero and wave cut off. At high frequency the propagation constant is imaginary and wave
propagates.
4.
At transition condition the cut-off flequency is inversely proportional to the eigen values of the Bessel function for the respective TEn, mode. Which of these statements is/are correct ? (A) 1, 2 and 3 (B) 2 only (C)2and3only (D) 2, 3 and 4 luco
9.4.56
How is the attenuation factor in parallel plate guides represented (A) o : Power lost/power transmitted
?
(B) o : 2 X Powerlost/powertransmitted. (C) o : Power lost per unit length/ (2 x power transmitted) (D) a : Power lost/ (power lost * power transmitted) MCQ 9.4"57
which one of the following statements is correct? A wave guide can considered to be analogous to a (A) low pass filter (B) high pass filter (C) band pass
filter
(D) band stop filter
*+*******x*
Chap 9 Waveguideo
9.4.54
1. At low frequency the propagation
Page 655
be
)
$oLurloNs 9.{
Page 656
Chap
I
Whveguides
ssl-
e.1.1
Option (A) is correct. Given, electric field intensity of the propagating wave is (1) E- : 5 sin(20zrr)sin(25zr 11 e-i?" Y f m So, we conclude that the wave is propagating in o, direction. Since, the wave has it's component of electric field in the direction of propagation so, the waveguide is operating in TM*o (Tla,nsverse magnetic) mode. Now for determining the value of nz and n, vre compare the pha.sor form of electric field to its general equation given as.
8"":Eosin(ff)si"(ff)"tu
(2)
where a and b a,te the dimensions of waveguide and since, the waveguide has the dimension 10 x 4 cm so, we get
o : 10 cm and b: 4cm *o.1,"o*Ouring equation (1) and (2) we get
rtnr -2onr + m:2 a
ry:25ry
) n:!
Thus, the mode of propagation of wave is TMr,'
s*L
e"1.2
Option (A) is correct. Since, the electric field Somponent exists in the direction of propagation so it will be operating in TM (Transverse magnetic) mode. So, for the TM mode the electric field components in phasor form are given as
r ano
Er --r 08," uBh2 or r1E" r, Ey":_FZy
Since, the given electric field component is E
*-
Eosin(5Qnr)sin(4ozrgr)
So,
8,,
:-
fi(50a'),Eo
and
E:,
:-
#(4ozr
) E',
cos (50 n
n)
e--
VI
m
sin( }n y) e- "
sin (so n n).cos (40r y) e-'"
Therefore, the ratio of the components is
E* : ^ ffi cot ffi -50n
:
sol.
s.'r.3
1.25 cot
(5Oz'
r)tan( }ry)
(5Ozrr)tan(40ny)
Option (D) is qorrect. Relative perrnittivity of dielectric Rel"ative permeability of the dielectric,
e, :2.25
p,:
1
Operating frequency, f : 10 GHz : 1010 Hz so, the phase constant is in TEM mode is operating the waveguide Since,
o
:,^/G :
:Tlia:Ti#
Chap 9 Waveguides
1_
"g-up-
will be equal to its
phase
c
I
\/ --I,t€
:2x108m/s rx$L $,d,d
Page 657
(1.b)
JI4.2 rad/m
The group velocity of the wave in TEM mode velocity in the unbounded dielectric medium l.e.
x
lJ,€,
Option (D) is correct. Plate separation, b:20mm:20x10-3m Relative permittivity of medium, c -O1 Operating frequency, : f l6GHz For propagation of wave the operating frequency must be greater than the cutoff frequency of (TE), or (TM), mode of paraller plate waveguide 1.e. > (f"),
f 2 f 'r ' ' zu/weor,
" < Zfbrfttreoe" ^-2x16x10ex20x10-3 3x10
X J2.T
n < 3.09 So, the maximum allowed mode is
n:3 since, all the modes given in the option are in the range, therefore, all the three modes will propagate. !t{11 s { {
Option (A) is correct. The phase velocity at cutoff is independent of the mode and equal to the phase velocity of a plane wave in unbounded media. Since, in the given problem the phase velocity of TM2 mode is to be determined for same waveguide so, the phase velocity of TM2 mode will be equal to that of TMr mode. l'€.' Up2 : Upl
sst
Option (A) is correct. consider the dominant mode of the waveguide is TEro. since, the cut-off frequencv for TE-" mode is defined as
s",r"s
n:dn{W;W
So, the cutoff frequency for the TEro mode is
(.[]o :
3 " *: *
(for airfilled waveguide
c: tl ,/G)
\T w, the next higher order mode of the waveguide will be TEor so, it's
cutoff frequency is given
as
(f")rr: * For the given condition design frequency will be
f:
1.1("tlo: 0.9(.fL
Since, the operating frequency of the waveguide is
f :5GHz:5 x
70eHz
Page 658
,.\ : ---TA5 x 70e c : 5x70s
So, we get
\J" ho
Chap 9 Waveguides
-TT-
2A
:
2x(5x 5x10e : --T35x10e c t6 : -T.g-
(/"L
and
(3 x , _:7t
9""t.?
x
ro8)
' sol.
3.3 cm
(b
x
q.g :2.7 cm 103) -
Option (C) is correct' Since, the waveguide is operating at TM' mode so, the phasor form of magnetic field of the EM wave will be given as
H*
: ffHocos(ff1e-'a"
Since, the waveguide is operating
So,
H*
: ff
at TMr mode (i'e'
Ho"o"(ff)'-
ri':
1)
jP'
Therefore, the instantaneous magnetic field intensity of the wave is given as
n,
: n"{# n, cos(ff) "- *" ",,1 : - ff n,""r(ff)sin(ut - 0z)
At
f
:o
H,:ff"or(ff)sin(Bz)
As the EM wave is propagatingin y-z plane so, in TM mode the -components of the magnetic field intensity will be zero'
Hu: lI' :
l.e.
(1)
y
and
z
g
Thus, the field will have the component only in z-direction for which we sketch the field lines in y-z plane. Flom equation (1), we conclude that the field intensity ll,depends on the values cosines and sines of the trvo variables defined as
t,.ut :l-u" [f ve 0 < y.< 0.5 cos(t') 0.5
[+ne ' s\nPz:t_r"
021r r
O
<
using these values we get the sketch of the field lines in the Ez-plane shown in the figure below where r-axis directs into the paper'
rl2
r
3tr/2
2.n a
as
sol. g".f.i
Option ($) is correct. The pha# velocity of the EM wave in the guide is defined
I
Page 659
as
Chap 9
ar-9 -p
Waveguides
where ar is the operating angular frequency and, is the phase constant B inside the airfilled waveguide given as
p:tF-(ffi
^. U4:-
So, we get
or
waveguide
0J
7/t (f,lff _
(tl.(+i
:'
The above equation is the equation of a circle. so, the graph betwe en (cf an'd
(f"lf) will be as plotted below:
ao)
(f"/ t)
sol
9.{.9
Option (A) is correct. wavelength for a propagating wave inside the waveguide is defined
where
p is the
where
I
as
-tr ^o:4
phase consta,nt of the wave in the waveguid.e given as
is the cutoff frequency of the waveguide and .f is the operating frequency ofthe waveguide. So, we get \So:---L
"/G ytt-(fl
,2tr1 Ao:_-=-:
ulto€0re
:
(for airfilled guide p
':(f)(ffi)
r:16; or
\' :
^
(f/f:f ulf"r
-
Thus, the plot between
L
(flil
and
(\/,\)
is as sketched below
:
pa,
e:
es)
I'
Page 660
Chap
I
Wbveguides
(fl f")
sol. s"l.t0
Option (B) is correct. The propagation constant
(f) i" the parallel ^f +,j pe : (Tl
plate waveguide is defined
as
(1)
Since, for lossless medium propagation constant is given as
^l
: jp
(attenuation constant, o
:0)
Putting it in equation (1), we get
-0'+Srr:(Tl At the cutoff
frequency,
u):
u)c phase
constant is zero (i.e.,
p:0).
So, we
get
err:(Tl w. -- WT b"/ pt So,
for TE1 mode
w^
'-
7f
hr/--1tr
u..: ' bJ22 ttt ,""-' - -3!: bJ pe
So TE2 mode
For TE3 mode
Comparing the three expressions we get, u)c]
sol
9,{.11
.Option (A) is correct. As calculated in previous question, the expression for the operating frequency of the wave in the waveguide is given as
-0'+J"t:(Tf ,ip'e: f *(Tl f'(i : 0' * (Tl
or So,
Ps1
for TMz mode
(n:
2)
For TMr mode (n: 3) and for TMa mode
(n:
4)
f : #Gl':'+ t1;Yl f': nJrrl'r'+e(f I] f :#Gl6'+ t6(flJ
Thus, for the above obtained expressions for the frequencies at clifferent modes, we sketch the f -p curve as shown beiow :
-!
I
Page 661
Chap 9 Waveguides
ft f, f,
.ss*- s,'*"1?
Option (A) is correct. For a rectangular waveguide operating in TE16 mode the phasor form of electric field is given as E,. : Eosin(tm\"-'r'
u* :-fOnosin(kr)e-ia, H,":iK "Qp
Eocos(kt:\ert,
Since, the wave is propagating
exists in the
in TE mode so, no any other field component
waveguide.
Now, the average power in an EIVI wave is defined
a,
\ as
: $n"i E" x H!)
Since, 11* hns a factor j. So it would lead to an imaginary part of the total , power when cross product with E, is taken. Therefore, the real power in the ' case is found through the cross product with complex conjugate of ,F1* as
below
:
p",
Po,g 0
Thus.
**!-
s"'1.1i:i
: jR"{ Eo, x Hl,} : i#6uA"tn2 (kr) a"
Option (A) is correct. The wave angle must be equal to or greater than the critical angle of total reflection at both interfaces. So, the minimum wave angle in the slab is determined for the greater of the two critical angles determined at two interfaces. Since. ns ) TL2 means the critical angle
It
9.r
:
will be greater for
sin
$-.1"-*d
media and given as
'/&\:48.6' \.nt I
Therefore, the minimum possible wave angle s&!-
q
Option (D) is correct. Phase velocity of a guided mode is defined
will be 48.6'.
as
ur:# So, rnaximurn phase velocity for the guided mode is "'n'u^
where
-
s)
(1)
ar,rn
0*i, is the rninirnum
pha^se
constant given
as
(2) 0^o : nikosind-io in free number is the wave where d,"io is the minimum possible wave angle, fto space and n1 is the refractive index of propagating media (slab)' Now, from the given relation for refractive index, we have
Page 662
Chap 9 Waveguides
'na ) ns in previous question the minimum wave
angle will be n2) which and of nr interface (i.e. the at determined by larger critical a.ngle
so, as described is given as
sind-io
-
sir
l.rr:fr
Putting it in equation (2), we get
0*: uldff:
nzlt
Again putting the value of p,,,i' in equation (1), we get Upmu
soL
9.1.15
_
(velocity of wave in air,
nDleo rk
Option (D) is correct. In a circula,r waveguide, cutoff frequency for
TE-'
,: fr)
mode is given as
' P'^n
t Jcmn:t;;G
and the cutoff frequency for TM-n mode of the waveguide is given as
'
t
Jcmn:
D lmn
,";G
where a is the cross sectional radius of waveguide, P'*n and P^n are the roots of the Bessel's equation. Their values are related as listed below in increasing order
P'n.,-Pu1P'r, I P'rr.: Prr (
P':.nz- Pzt/- P'u and so on' p so, for the corresponding values bf ^'rnd P^n, we get the increasing order of the modes with respect to their cutoff frequencies as shown below on the frequency axis : TEnt
f" TMrt
Thus, the first four propagating modes are respectively TErr, TMor, TErr, TEor or TMrr sol. 9.t.16
Option (B) is correct. Given
d
Thickness of substrate,
:
0.316 cm
c
Relative permittivity of substrate, -)9 Zo :100Q Characteristic impedance of line, The width to thickness ratio (wld) is defined as
w a:7r1 8eA
where o
: 30,[+ * r[#r(o.x + +)
for
ff <2
Now, we assume # < 2. So, we get
Page 663
EJt , t2.: ro-v ---z-"1#;f;10
: 100 A_
23+
W):"'
and therefore, the width to thickness ratio is
#
:a%:
As the obtained value of and we have
s.1.1?
2
(wlQ is less than 2 so, our assumption
# : o.sgo u:(0.896)xd : (0.896) x (0.3i6) :
Oft
$sL
(
o'8e6
was correct
0.288 cm
Option (A) is correct. As calculated in previous question the width to thickness ratio is
7: w
o'gge
So, the effective value of
'P
permittivity is given
_r.*1 e"-1 1 --2----2--REy L-f _(2.2+I),z.z_ t
as
-u
-_T----2--7=T :1.7b8
1
v^'0'896
Therefore the guided wavelength of the EM wave is
where
/
/ e.f ^s:-+-
is operating frequency and c is velocity of wave in free space. so,
we get
\ ^'-@ffi;1sxro) 3x108
:
2.83 cm
t(*********rt
,f
:8GHz
Chap 9 Iilaveguides
sol-urlol{s
Page 664
Chap
I
9,2
Waveguides
sol.
9.2.1
Correct answer is 327. The intrinsic impedance of an airfiIled waveguide for TM mode is defined
as
.,. FJ-t")?rM-":_,tu|,_\_ri Since, the operating frequency is twice the cutoff frequency l.e.
f
:2f".*,
So, we get the intrinsic wave impedance as rne;^,
sol-
9.2,2
:377
:32.6.49Q =
Correct answer is 120.7 . The dimensions of wave guide,
327 Q
o:10cm:0.1m b:4cm
and,
The mode of propagation,
TfL
:2, n:7
f :7.5GHz:7.5 x up: c:3 X 108m/s
Operating frequency,
10e
Hz,
(air filled)
Unbounded phase velocity, So, the cut-off frequency of the waveguide is given as
: il(,+l *(+l
:tn.--r@;g
Therefore, the phase
;*.t;1? tf:
wave inside the waveguide is defined as
E-W:''t!cl re 0:9 -i,l'\//f : #iou x :
sot- 9.2.3
Correct answer is 1.8
120.7
ro'g
,t(if - 1a.sY
rad/m
.
Dimensions of wave guide and
o:5cm:5x102m b :3cm:3 x 10-2m f :3.75 GHz : 3.75 x 10e Hz
Operating frequency, Since operating mode of the waveguide is TEls (i.e., rn: 1 and the cutoff frequency of the airfilled waveguide is given as-
f":ir/WJW 3x108
10e
The group velocity of the EM wave in the waveguide is given
,r:"ffi
as
n: 0) so'
:3 x 108 : 1.8 x 108 m/s
Page 665
'-(*#ihl
s$t*
$.9"d
Chap 9 Waveguides
Correct answer is 323.
a:2.5cm:2.5 x
Dimensions of waveguide,
10-2m
b :5cm:5 X 10-2rn f:l5GHz:15x10eHz o:0 (lossless dielectric)
and
Oferating frequency, Conductivity of medium, c Relative permittivity, -c Relative permeability, F, :1 The operating mode of the waveguide is TEzo mode (i.e., tn:2 and n: 0) So, the cutoff frequency of the waveguide in the TE2s mode is given as
f,:rl;rtrfTflT 3x10b
- 2rT;
:
8.5
x
D
z
V \z.s x
,, ro-'?/
10e
The wave impedance for the TE2s mode is given
rrEx:
qnt
0?q
as
FS:r'{*(Tfr)
Correct answer is 9. Given, the cutoff frequency for TM2 mode is
()"L:2mm:2x103m for tM, or TE,
Since, the cutoff wavelength waveguide is defined as
(1"),
r: :_2b XJ r,
mode for a parallel plate t
(1)
where b is the'separation between parallel plates of the waveguide and €,. is relative permittivity of the medium. So, putting the known values in the expression, we get'
2
x ro'
'
:+/;
(n:
2)
b:Lug_-* ! e,
Now, for a',y n mode to propagate the operating wavelength must be than or equal to the cutoff frequency.
less
l.e.
So, from equation (1) for the propagation of wavelength
the relation
):0.1crn
as
x
10-2
o.l x
1o-2
0.1
.4J; er h v
-
u, /=?r2Lj-'-J+ I cr
, -
,*,e have
Page 666
Chap
"<#i# n<4
I
Waveguides
Therefore, the possible modes that can propagate in the waveguide are TEM, TEi, TE2, TE3, TEa, TM1, TM2, TM3 and TMa Thus, there are nine possible modes that can propagate in the waveguide.
snl
9,2,6
Correct answer is
9.
b:10mm:102rrr
Plate separation,
Minimum operating frequency, ./-,' : f": lSGHz: 15 x 10eHz Since, for TM, mode of parallel plate waveguide, cutoff frequency is defined AS
(f")_ So,
: ,G n,
for (TM), mode (n : 3) we have the cutoff frequency
(t),
: ,;rffi
as
3
15x1oe-sx(sItO 2 I0-'/;
/+:3 €, :9
Of $oL 5.2.?
(.[),: /.'
x
Correct answer is
2.
Cutoff frequency of (TM) mode, (r)r : 2.5GHz:2.5 x 10e Hz Operating wavelength, I :3cm:3 X 10-2m The cutoff frequency of (TE), mode of the parallel plate waveguide is given as
(/").
:3(i) 3
x
2.5
x
10eHz
:7.5 x
10eHz
Since, the operating frequency of the waveguide is defined as
f :* where ,\ is the operating wavelength. So, the operating frequency of the parallel plate waveguide is
/:##:
loroHz
Therefore, the group velocity of TEs mode is given
as
(r,),:hrf_(H :3 x 108 :2 x sol.
9,2.S
Correct anslrer is 1.2
108
'-(%#ql
m/s
.
At cutoff the mode propagates in the slab at the critical angle which means that the phase velocity will be equal to that of a plane wave in upper or lower media of refractive index n2. so, the phase velocity at cutoff will be ,o
: h: gjjq :
r.2
x
108
m/s
scL
$"2.e
Correct answer is 0.295 . Relative permittivity of material,
line, Characteristic impedance, So, Jd zo: /es (gs) : Separation between strip
Since,
^/1zo <
Chap 9 waveguides
103.8
120
to separation ratio of strip line
Therefore, the width
page 662
6" : 8.8 b : 0.632 cm Zo : BS
transmission line is
given as
w : 30r -o {e,Zo -0.44I
w - 3o+- - 0.44r os3z:@XBs)-u.rar w :0.295 $sL
e.z.{s
Correct answer is 3.37 . Guide wavelength of a stripline is defined as,
^r:-=\/ €,J where, c is velocity of wave in free space, / is the operating frequency and e, is the relative permittivity of the medium. So, we get
' : -ix--lE(/8.8X3 x
: 3.32 cm
,\,
sol $.r.{,
Correct answer is 901.4
10'g)
.
In an airfilled cavity resonator, resonant frequency is defined
r* So, for TE161 mode
:
. (Tl * (Ifl''' h\+'t' (^: p:1, n:0)
p
/,0,
the resonant frequency is
: rXrlo'[(rr+,-f +1"j--f]',' : :
$oL
9.?.r?
as
9.01388
x
108
Hz
901.4 MHz
Correct answer is 1.41 . The resonant frequency of a cavity resonator is defined
f*-rmn,
as
:- --Ll(ry\,n -- tnY -\"/ tlP\211/2 2/ pr€rL\ ) \T ) .|
Since, o : b: c so, the dominant modes are TEror or TEs11 or TM116 . Therefore, taking any of rn)n or p equal to zero, we get the resonant frequency as f*np
:
hu+r
. (+r * (trl'''
p _ 3 x lo8lzl't' rmnp2 larl
(a: b:
15x1oe:a+t{"+ i.e. sol s.2.13 Correct
a:L.41 X10-2m:1.41 cm a: b: c : l.41Cm answer is 414.
Operating frequency,
f :lGHz:5 x
70eHz
c)
Separation between the plates b :6cm:6 So, the cut off frequency for TM1 mode is given as
Page 668
Chap 9 Waveguides
X 10 2m
rl
'' -
:
zutJaeo
3x108 -:2.sxloeHz 2x6xIO-2-''"
For a parallel plate waveguide, phasor form of components of electric field and magnetic field intensity of a propagation wave are given as Eu"
and
:
Escos(T)u""
H,":-- -L
co.(Ty)"-r
^'l'-(+l
\
So, the average power is given as
Puu":[]n"1n"xHl)as
:
I' t{e ",,)(H'")} (w
d,v)
:tI' - j=+ir.cos'(ff)au
'-\j) Ea r,1+ cos(ry) ,^. _ ,u
^tl
_u -2
;F-1q-r'--z-"': r;m Et
The maximum power propagation will be due to the maximum electric field in the medium (the dielectric strength of the medium). so, we have tbe maximum average power as
(P*")^*
:*
x touf
;Fg (a
((Eo)**: 3 x
Putting all the values, we get the average power per unit width (P',")-* 6 x lo-2 r. - (3 x 1o6f
w
-
: $0t
Le"',d
Correct answer is 1.432
n
4
4.135 X 108
106
v7m
'
as
l:::;:fi r2orlt-(f) :
aL4MW
lm
.
plates, b :72cm : 0.12 m Operating frequency, f : 2.bGHz : 2.5 X 10e Hz For the TEM mode, phasor form of electric and magnetic field components Separation between waveguide
are given as
Er,
:
H,,
- -In
So, the average power propagated
,
P","
Eoe-1"
"-
''
in the waveguide is given
- Io"{tn"x n{}as
_ f r, - J ZR"{-(A"Xn*)ia"
as
-T :- I'i@'tF"A'* : ffi'u
Page 669
The maximum electric field, without any voltage breakdown is defined as the dielectric strength of the medium as given and as the dielectric strength of air is
Chap$. .. .:r_. Waveguides .
.:
'
(Eo)-*:3x1o6v/m So, the maximum average power propagated in the waveguide is
(P",")-*
: z=(tx1-t^o='L (tzor) wft.r2\
Therefore, the maximum time average power propagated per unit width in
the waveguide is (Po,"),"*
w:
-
x
1.432
10e
t.432GW s&L $,p"1ii
lm
Correct answer is 3.45 . Maximum operating frequencv Relative permittivity of medium, The cutoff frequency in TEM mode is
(TE), or (TM)" mode is given
e,
:
1.5
x
10e
Hz
:8.4
"fl: 0 and the cutoff frequency in
as
(f"),: So,
"f*o
2b7G
for TE1 or TM1 mode (n: 1) we get
/f\ \rrh _ --l
tilfiG
to be operated only in TEM mode. So, the operating frequency must be less than (l) while it must be greater than 0 (cutoff Since, the guide is
frequency in TEM mode). i.e.
0<
Oft
"f
/< (r),
<--]: 2br/ paese,
b<_L 2f,/ pae6e, As the frequency inside the waveguide ranges in 0 < f < 7.bGHz, therefore, the maximum allowable separation between the plates is
c A 2.f^*Je,-
:
sst
s,2"ts
0.345
m
:
3x108 2
x
1.5
x
10e
oorr
:
1.5 GHz)
3.45 cm
Correct answer is 12.8 . The Brewster's angle for parallel polarized wave is given tan0611
(/**
x /&4
as
: ,/?,
: tu'-'( {ry):35.e'
The cutoff frequency for TMimode in 1't medium (permittivity
:
€r1) is
given as
3x108 (rI: %tra= 2xt4.7x1O-3,/2
:
7.52
x
10e
Hz
d
./\
670 I Waveguides Page
So, the frequency for which there is no any reflective loss is given as / f\
Chap
fr: .o,9
where
I
d: 90" - ds11. So, we get ,^ _ 7.52x l}e _ 7.52 x 10e _1ra/rrr: rz'o\'rtL Jo -;orcoo - dBr) - sin3tg-
is ray angle that has the value,
ilOTE: Brcwslcr'$ a.rigL: is ilx: in<;irltnt nrrgic ol a pl;ln,: rrirlo ivhith thr.:r'e is rio a,nv rcficrl,itin in ihc rnerlirrnl.
$EL
9"3.{7
.
i:rl,
llu' irrtr:r'lilrr'r;l'trvo lur:dirlrrs
i,.r:
Correct answer is 1. As we have determined in the previous question, the value of /6 is lo : l2'8GHz and the cutoff frequency for TM1 mode is
(t)
:7'52GHz
So, the cutoff frequency
for TM2 mode will be
(f"\r:2(f"\ : lLofGHz
Since, the operating frequency /e is below the cutoff frequency for TMz mode so, TM2 mode or the higher modes can't propagate at the frequency
.
S. Therefore, only one mode through the waveguide. sOL
9.2.18
(TMr) can propagate at the frequency $
Correct answer is 10. For a propagating mode TM" or TE" the cutoff wavelength of the symmetric dielectric slab is defined a-s
, _2d,G,r_ €, .\c_ .rc---1 n_l where e"1 and e,2 are the permittivities of dielectrics and
d is the
slab
thickness. So, we get
\ -2 x20x:roar/y2-yt n._T _
1.26
x
n-l
10-5
Since the operating wave length must be lower than or equal to the cutoff wavelength
) < I,
i.e.
Therefore, for the propagation of wavelength
):
2.6
pm in the dielectric
slab waveguide, we have the condition as
2.6x10-6.1.26x_10-5 n- |
n_I<1.26x10-"5 2.6x106 < n- 1 4.85 n < 5,85 So, the possible values of z for which the propagate in the waveguide a,!en:1,213,4, modes as follows
wavelength ) : 2.6 p,m can b. Thus, we get the possible
:
TE1, TE2, TE3, TEa, TE5 TM1, TM2, TM3, TM4, TMs and as TEM doesn't exist in the dielectric slab waveguide so, total 10 modes
,}
can propagate for the operating wavelength.
sol- 9.2"{9
Page 671
Chap
Correct ansv/er is 3.304 Cutoff wavelength for symmetric slab waveguide is defined as,
()"),
: 2dG"=u
(1)
where d is the thickness of slab n is the propagating mode, ra and €,2 dre the relative permittivites of the mediums. Now, the refractive indices of the two mediums can be given as
: n, : n1
and
"/
e,1
nfe*
So, the equation can be rewritten as
()"t:2d'm4 Since, the waveguide supports only a single pair of TE and TM modes. i.e. supports n: 1 mode and denies all the higher modes. Therefore, the operating wavelength ) must be with in the range.
it
()")r>)=()"),
l.e.
(\)r
(2)
and (,\)z are the wavelengths for mode respectively. Putting n : 1 in equation (1) we get where
n:1
and
n:2
(.\") : o" Therefore, the condition obtained in equation (2) reduces to ()"1
^
3.1
x
10-6
r
:
_!.1 X_Uj_ rrd_@e < -2x10x10-6 zr < 3.304 Thus, the maximum value of n1 is 3.304. ,
sol
9.2,20
Correct answer is 5.3 . Given, the cross-section dimension of the waveguide is
a: b:4.5cm The cut off frequency of the rectangular waveguide is defined
(I)*,
:rGl\;) I
as
lrmv+(t)l /n\21't/2
So, the cutoff frequency for TE22 mode of waveguide of square cross section is
r _ 1 [12\r,t2\r]r/, n"-rG[\;i -\r/] - .,a .,^ 108 x /-L :2GHz "/0.045
The phase constant of the wave inside the waveguide is given
0
: r,/ raeol, -(4Yl'''
t \/ / 2rx6x10e[' /2\2f1/2 3x1o_[r_(6/J J
_
:
L1847
x
102
m-l
Therefore, the wavelength of the TE22 mode wave is
as
I
Waveguides
Page 672
^
Chap 9
:T : #*:
b.3og
x 1o-2: 5.3cm
Waveguides
Correct answer is 20.67 . Given, the operating frequency of the waveguide is
f :6GHz: 6 x 10e Hz So, the wave number in the waveguide of dimension
'a' is given
as
2a"x 6 x-10e :40r 3x108 Now, thr. l.r,-nuation constant of section of waveguide (attenuator) ritth dimension o/2 is given as
k
:2nf c -
(a:
C\:
:
- @otrf :
(#*l
0.04572m!
55'63 t'{p/m
since, the total required attenuation is 100 dB along the attenuator so. se have
dB :20loge-ot where I is length of the attenuator. (length travelled by wave in the smail
-
100
section of waveguide). Therefore, solving the equation we get, 1g-s
, $&L S.*.*p
-
"-at
: #*
:o.2o17 :20.6T cm
Correct answer is 6.12 . Dimension of waveguide
: 1.07 cm : : : Operating frequency, "f 10 GHz Permittivity of dielectric, 6" : 8'8 tand :0.002 and a
0.0107m 10
x
10e
Hz
:
1010
Hz
The phase constant of the EM wave inside the waveguide is defined
:
as
1"6
P where k is the wave number in the unbounded medium given as (ko is wave number in free sparei tc : ^/lleo
''F-
:1[a.af!
(o:?#l
:(/8.8)Ti# :
621.3
m
(c:3 x 108mfst
1
So, the phase constant of the wave along the waveguide is
P:trcnsYGffi :547.5m-1 Therefore, the attenuation constant due to dielectric loss is given as _ t*ta''6 _ (621.3f (o.oo2) Q4 2(547.5)
2p
:iHiliit $.*r* $"2,n3
\
Correct answer is 3.87 . The resonant frequency for
TE-',
mode is defined as
n: So
for
+(!:121"
TE161 mode the resonant frequency of the
r.
Page 673
x
3.87
10e
(t,
=
Waveg;lftis
to,e =
3eo)
Hz
3.87 GHz
Correct answer is 7733. The qualitv factor of TEror mode is defined
Q""':
dhap'9
cavity resonator,is
:W[(#*l*(uhil"'
: : $QL 9"A.A4
,hlffY.fff
(a2
6l2b(a"
+
as
c')abc
+ c')+
ac(ct
+
c')l
where d is skin depth g iven as
6where
f, -
1
taG
resonant frequency for the defined mode. 4n x 70-7 P4: c": Conductivity of copper So, we get the skin dept has
5_: 1.06 X 10-6
1
Therefore, the quality factor of the resonator is
,, sol.
9.2"?5
[(z.rf +(sfl(2.5x2)1b) x 1o-2 (1.06 x 10-6)[2 x z{(z.sf+ (5I} + (2.5x5x(2.sf + (s)r}l :7732.7 = 7733
Correct answer is 8.79 . Given, the inher radius of the guide is
a:1-cm:0.01 m TE-, mode of a circular
The cutoff frequency for J
rr,nn - P'^n - 2ra"/ 1te
where p'^n is the rnth root of Bessel's function Now, from the given data we have
P'n
waveguide is defined as
(J'^:
0).
:
I'841 So, the cutoff frequency of TErr mode in the circular waveguide
r trll :-
is
1.841
2*(lo-\7G
_3x108x1.841 2r x 10-2
: sol
9.2.26
8.79-
X
10s
Hz
:
8.Zg GHz
Correct answer is 0.765 . The resonant frequency of TM-,r mode in cylindrical cavity is defined
: -l - n4t-Wl;e f,,,,,t: 2"7G1\ r /+(z/
as
where a is radius of cylindrical cavity, d is height of the cylindrical cavity and p^, is the root of Bessel's equation.
Page 674
Chap
I
-
Waveguides
Since, the.domina,nt. mpde
in
cylindrical cavity
is
TM616 so,
the cutoff
frequency fot domiharit mode is
t
rcoro
Por --r;;Jffi-z?ta -Potc
Therefore, the cutoff wavelength for dominant mode is given as )",0'o
:
./c010 =L 2 x l0-2 _2tra
Por
u--
: :
(2.40b\(2
x
a lt
7.65 x'10-3 0.765 cm
****i<*****x
1o-2)
('\",s16
:
2
cm)
\
$ol.urtoN$ 9.3
Page 675
Chap
I
Waveguides
so|-
9.3.1
Option (A) is correct. A wave mode propagates in a waveguide only if it's frequency is greater than cutoff frequency. If there is no any propagating mode inside the waveguide then energy in the propagating mode is zero. So, average power flow down the waveguide below cutoff frequency is zero. i.e. Both the statements are correct and R is correct explanation of A.
sol
9.3,2
Option (A) is correct. In an a X b rectangular waveguide, cutoff fiequency for (TE)-, or (TM)-, mode is defined as
(f")*,
- 2/G 1
Now, for TM11 mode
l,:#ffi*W:#^tH;W a:2b
:/'(#)
Similarly, for TM12 mode
For TE16 mode
For TE2e
f""
:
f",
:
mode f"n:
r)t2
* (')y:n(r",r*L)
1
,7G 1
2J pe
,G 1
(*l
1
2aJ
p,e
:21 1 \2a/
\
pe )
So comparing cutoff frequencies of all the modes we get the modes in ascending order of cutoff frequencies as TEro(TEzo(TMll
9"3"3
Option (C) is correct. Statement L Suppose on the contrary the TEM mode existed. In this case the magnetic field must lie solely in the transverse rgr-plane. The magnetic field lines must form closed paths in this transverse plane, since V . H - 0. FYom Ampere's law, the integral of this transverse magnetic field around these closed paths must yield the axial conduction or displacement current. B:ut E": 0 for the TEM mode so, no axial displacement current can exist. Also, since there is no center conductor so, no axial conduction current can exist. Therefore statement 1 is correct. Statement 2 The dominant mode is the mode that has lowest cutoff frequency. Now, f,,-, is clearly minimized when either Tn or n is zero. Since, TMsl or TM16 mode doesn't exist so, TM mode can't be the dominant mode of propagation in rectangular waveguide.
It
Ps-"9?,9
is also correct stat'brnent'.
Chap 0
qe""t"tdrq
sol-
9"3.11
Option (B) is correct. For a TM-,' mode, neither rn rrot n can be zero otherwise all field components vanish, however p can be zero. So, the lowest order TM mode is TM1o.
$0L
9,3,S
Option (D) is correct. Since, TEn,nn mode of cavity resonator can have either rn : 0 or n : 0 (but not both at a time) where as p can't be zero for TE mode so, the Iowest order of TE mode is TEott if a b As the dimensions of the cavity resonator a,re equal TEror and TEorr are lowest order mode.
s6L
9-3"6
Option (D) is correct. Given, the dimensions of cavity resonator arc related
(a: b) so, both the
as
a>b>c
(1)
The condition for propagating TE and TM modes in a cavity resonator are as follows
:
(1) for fM*,r mode, neither m nor n can be zero however p can be zero. (2) For \E*nn mode p can't be zero but either rn or n can be zero (but not both at a time) The resonant frequency of TM-,, or TE,n,o mode in a cavity resonator is defined as
I ltm\2' lnP' 1Pf|/z f'"u -t/p;l\,r/-\1,/ -\c/l So, comparing the resonant frequency for the different values of
m,nandp
using the relation defined in equation (1), we ge.t the lowest order rnode will
be
TMlo
and the ascending order can be written as below
TMrroi TErori TEoli TErrr sol.
9"3.7
Option (A) is correct.
col
9"3.8
Option (C) is correct.
sok
9.3.9
Option (A) is correct.
sol- s.3.{o
Option (C) is correct.
sol-
Option (B) is correct.
9"3,1{
-
TMrtr
:
ssil,rvtsXs 9.4 s$L
$.,4"$
r 1!'i:.! ': i ,
vqge,a!7, Chap g tltcrecddeg
Option (D) is correct. Given, the magnetic field component along the z-direction as H, :3cos(2.094 x 102:r)cos(2.618 x 102g)cos(6.283 x 10rot- Pz)
So,
:2.094 X l3u :2.618 x a :6.283 x
13,
102
lo2 1010
rad/s
For the wave propagation inside the rectangular waveguide,
a
: ^lVc'o.] - 1i+ ni
Substituting the values, we get
p ' 'Since, l.e.
p is irnaginary up
/
^;;. - i^io.z _ : y.l (6.:2P+:g:I \ 3x10" I Q.os42 + 2.618) x 10,
= j26l
so. mode of operation is non-propagating
:0
sol e.4.2 Option (A) is correct. TMrr is the lowest order rnode of all the TM-, modes. $01
s.4.3
Option (D) is correct. From the given expressions of E and fif , we can write,
d :280r
or
'f, :rtnn
- ^: rlo
So. the wave impedance is given as
't,:lEl__E, - Hl- 3
-_tzotr G,
(1)
Since, the operating frequency of the wave is
f:
l4GHz
So, the operating wavelength of the wave can also be given as
c 3x108 3 "r ,trJ-G14xloe-l4o,/; 1_ 3 140 - Tn/; /i
or OI,
---
--
--
6":9
From equation (1) we have
rT ) E,: 72otr
-1 : J9
u+ sot"
e"4.4
Option (A) is correct. Rectangular and cylindrical waveguide doesn't support TEM modes and have cut off frequency. Coaxial cable support TEM wave and doesn't have cut off frequency.
Page 678
sol-
Chap 9
s.di.s .
Option (A) is correct. Cut-off Fbequency for TE-" mode of a rectangular waveguide is defined
r : r,/
Waveguides
So,
as
(+I.(fl
for TE11 mode (m:l,n:1)
the cutoff frequency is
(c:3x108cmsl aol.
g.rt.s
Option (A) is correct. Given, the electric field intensity of the wave inside rectangular waveguide
, : #(+)n"t"(+)sin
(c..'t
-
94
r
ou
This is TE mode and we know that
Buosin(T)*'(#) it with the given expression the propagating mode is TEzo. So, comparing
sol
9,4,7
we get
m:
2 and n
:
0. Therefce-
Option (C) is correct. The cut-off frequency for the TE-, mode of the waveguide is defined
t"
as
: iJ (+l. (ff
So, the cutoff frequency of
the TE2s (m:2, n:0)mode
is
2 t :-c/tn\-3x108i\;): elt'L " of,g:- lo GHz
(o: 3clfl
I"
Therefore, the wave impedance of the TEzo mode is given as rlt
30L 9.6.8
:
rh
TT_ '-(r#*i
Option (C) is correct. The cut-off frequency of TE-n mode is defined
:400f,)
(/:30Glrrtt
as
f":irrcfTw So, the cutoff frequency
of TE3q (m: 3, n : 0) mode
: i(+) 18 x 10e : t#nq*
is
f"
or
or sol-
9.4.9
3(}1- 9.4-.to
(l:
1s
o:h*:f"-
Option (D) is correct. Option (D) is correct. For any propagating mode inside a rectangular waveguide the velocities related as
ap>c>us i.e. the phase velocity of the wave inside the waveguide is greater than velocity of light in the free space. $oL 9.4.{t
Option (D) is correct. In a microwave test bench, the microwave signal is modulated at because crystal detector fails at microwave frequencies.
sol
9.4"r2
Option (A) is correct. The cutoff frequency of
rE-"
mode in a rectangular waveguide is defined as
7 flinn,/-nS I : rTE^/(;/ +(t)
Page 679 Chap g Wavegrldes
since in the given rectangular waveguide o > D so, the dominant mode is Tffs and the cutoff frequency for the dominant mod.e is given as
r:#r[WiW
(#:")
:riE/HFH
: t#g SOI- 9.,t.13
:2.bGHz
Option (C) is correct. Phase velocity of an EM wave inside an air-filled rectangular waveguid.e
,rry
where c is velocity of EM wave in free space I is the cutoff frequency of the propagating mode and / is the operating frequency. Since, for a wave propagation the operating frequency must be greater than the cutoff frequencv. l.e.
f>l
Therefore, the phase velocity of the wave will be always greater than the velocity of wave in free space. l.e. sQL s,4.14
up)
c
Option (A) is correct. In a hollow metal wave guide
where
-
?p
uo
)
c
)
ag
Phase velocity
c -+ Velocity of light in free space. tre -+ Group velocity so, the phase velocity of a wave propagating in a hollow metal waveguide is greater than the velocity of light in free space. sol.
9.4,15
Option (A) is correct.
In a wave guide dominant highest cut-off wavelength.
gives lowest cut-off frequency and hence the
sol
9.4,16
Option (C) is correct. As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary.
sol
9.4,{?
Option (C) is correct. cutoff frequency for TE^n mod.e in a rectangular waveguide is defined
.I
up nm^z.
- z\/
as
(;/ *(t/
/n^2
since, for the given rectangular waveguide a > 6 so, the dominant mode frequency of the dominant mode of rectangular
is TEro and the cutoff waveguide is
t:#:##:15
X
1oe
:15
GHz
(For air
up:3 x
108)
/
PageffiQ,t
$sL
9.4"tS
Option (D) ls conect.
ln TE mode E":0, at all points within the wave guide. It
chep,9'
implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide.
Wav,ggi{ee,
sol
s,4.t$
$oL 9.4.ttl
Option (A) is correct. In a rectangular resonant cavity TE-*r, mode must have its mode TErro doesn't exist in the rectangular resonant cavity.
p:
1. So, the
Option (A) is correct. The transverse electric field and transverse magnetic field inside a waveguide are related as
Or
lEl:
'tlH
where 4 is intrinsic impedance
I
, :EJ - lHl
i.e. the ratio of transverse electric field to the transverse magnetic field is called waveguide impedance.
sol- 9.d.2t
Option (B) is correct. Cutoff wavelength for H-, mode of a rectangular waveguide is defined
r_
as
2
where a and b are the dimensions of waveguide. So, for the Hro mode (rn : I, n: 0), the cutoff wavelength is
,2 - -T=-i:: ,u -
/(i) :16cm
$oL
9"4.2e
(o:8cm)
*o
Option (D) is correct. A rectangular war,-eguide supports TE and TM waves where as it doesn't support TEM waves. The propagation constant for TE or TM waves inside a rectangular waveguide is defined as
.,:W
$oL 9.4.23
Option (B) is correct. Cut-off frequency for TE-n or TM-, mode inside a rectangular waveguide is defined as
1- /(ux* +1rm'> r ::r;TG'/ \ )*(-a-)
'
Where o and b are the dimensions of rectangular waveguide. $oL 9.4.24
Option (D) is correct. Cut-off frequency for TEro mode is
f",o:_L:frWiW 1 3x108..r r : ---T " (4.755t-10=z/ :
3.16 GHz
Cut-off frequency for TEor mode is
(m:l,n:0)
/,,:(s+u)" (_"f,;.)
Page 681
I
Chap WavftiHee
:6.77 GHz cut-off frequencv for TE11 mode is
,,:L1d" :7.47
GHz
and the cut-off frequency for TE26 mode is
: i+1c
f*,
Since the operating frequency
, Therefore, all the modes
x
1nr5* loa
f :L2GHz
will propagate.
: 6.3 GHz
so, we have, fao, far,
f"n, f"ro)
f
}*OTX:
riir:\(:iril]it
$oL
$.4.2S
r.r1;1.i1,1 r1"i,
rl]tc,,'ilt
ltai.l! sx\. r;l>r_l*l; i
l_ll :l rr],1r.1,"
Option (D) is correct. Consider a rectangular waveguide has dimensions a : b and the corresponding resonator has the dimensions a: b: d. now take and operating point that has frequency / iust greater than the cut-offfrequency for m:n: 1. So we have the propagating modes in waveguide. TEol, TE1o, TEu and TMrr. Where as the propagating modes in resonator are TEorr, TEror, TMrro Therefore the cavity resonator does not possess as many modes as corresponding waveguides. As the resonating frequency of a TE-^o or TM-,, mode is defined as
,*:iG{ef.WW
so for the different modes (different values of rn, n andp) the resonant frequency are very closely spaced and also the resonant frequencies of cavity can be changed by altering its dimensions. sal. 9.4.ts
Option (B) is correct. N{icrostrip lines cannot support pure TEM mode but shielded coaxial lines can support pure TEM mode.
s{tl
9"4"2?
Option (A) is correct. Given,
,f :3GHz:3 X
Operating foequency, and
f"^,
:hffirw
f^o
_3x1010 --T-
(u-
1)
f*, -3x1010 2
(b-
2)
The cut-off frequency for
So,
for
?E1s,
for TEu,
10eHz
a :6cm b :4cm
Dimensions of waveguide
TE^,fTM-,
--
mode is defined as
Page 6E2
For
Chap 9
ZE11
or TMt11,er,
: q!ry/ffi
Waveguides
(" r* i,.
13.4,3&
Option (A) is correct. Phase velocity of a wave propagating in a waveguide is defined
-
3,d
-
4)
as
The group velocity of thb wave' pfopagating in waveguide is defined
as
'-$l
where c is the velocity of wave in free space, I is the cutoff frequency and / is the operating frequency. As the operating frequency / is always grater
than cutoff frequency ,up
s{:}t- s"4"*$
l. so, comparing ) c) us
the above two expressions we get
Option (C) ii correct. In a microstrip line operating wavelength is defined where,
\
as
),: \r/1
is free space wave length and e" is the effective dielectric constant. So, Statement 1 is correct.
The electromagnetic fields exist partly in air above the dielectric substrate and partly within the substrate. Statement 2 is correct. The effective dielectric constant of microstrip line is e" and given as
l
s"4"so
Option (C) is correct. Stripline carries two conductors and a homogenous dielectric. So, it supports a TEM mode (Pure TEM). a--+ 2 Hollow rectangular waveguide can propagate TEM and rE modes but not TEMmode. b-3 Microstripline has some of its field lines in the dielectric region and some fraction in the air region. so it cannot support a pure TEM wave instead the fields are quasi-TEM. c-+1
${}L S"4"${
Option (A) is correct. wave doesn't have an electric component in its direction of propagation consequently there is no longitudinal d.isplacement current. The total absence of a longitudinal current inside a waveguide leads to the conclusion that there can be no closed loops of magnetic field lines in any transverse plane. Therefore, TEM waves cannot exist in a hollow waveguide of any shape. i.e. Both A and R are true and R is correct explanation of A.
A TEM
ssL
9"{,?E
Option (C) is correct. Given, the dimension of waveguide is a:2.2g6cm, b : 1.016 cm. The cut off wavelength of the guide, for TE_" mode is defined as
L_ So,
m 2
Page 683
Chap
I
Waveguides
for TEor Mode the cut off wavelength of the guide is
ffi 2
:2b:2.032cm sol.
$.4"33
Option (B) is correct. Since, the electric component is existed in the direction of propagation. So the electric field is not transverse to the propagating wave and therefore the mode is transverse magnetic (TM mode).
sot
s,4.34
Option (B) is correct. Given, dimension of waveguide a: b: 3 cm and so the dominant mode is either TE61 or TE16 mode. So, the cutoff frequency for dominant mode is given as
r r'rnv'-.nP f",-rffi\/ \o/ -(t/ 3x108.: ---2^ Stl0_t 1
(for TE61 or TEls mode)
:5GHz at 6 GHz dominant mode will propagate. Statement 1 is correct. At 4 GHz no modes will propagate so the modes are evanescent at 4GHz. Statement 2 is correct. At 11GHz along with the dominant mode TErr mode (f, : b/r) will also So,
propagate.
Statement 3 is incorrect. Degenerate modes are the different modes that have the same cut off frequency and at 7 GHz frequency TE61 and TEls propagates that has the same cut off frequency i.e. Degenerate modes propagate at 7 GHz. Statement 4 is correct.
sol
9,4"3$
Option (B) is correct. Evanescent mode - No wave propagation dominant mode is the mode that has lowest cutoff frequency. Rectangular waveguide does not support TM61 and TMls mode.
A-2rB-3,C-L
sol
9.4.3s
Option (D) is correct. Assertion (A) : Given the dimension of waveguide, a: 3 cm, So, the dominant mode (TEro) has the cutoff frequency.
b
:
l cm
.,e:L11q.(r*r") :lGHz
f
:4GHz < f"
So, at 4 GHz there is no propagating mode. i.e Assertion (A) is false. Reason (R) : The wave equation for the rectangular waveguide is defined as
:(+)
J
"Page 6E4
Chap
for
I
a:3"
b:'1'wehave
k?-(lnirr-ffr
Waveguides
:(f)'
So, Reason (R) is also false.
s$t" 9.4,37
Option (D) is correct. A rectangular coaxial line can support all the three modes (TE, TM or TEM).
stlt
s.4"3&
Option (B) is correct. Consider the rectangular waveguide (A) has the dimension o X b after deforming into waveguide (C) the dimension is changed to b x o and so the input mode TE16 is charged to TE01. (Since the frequency of mode must rernain same for both the wavpguide dimensions).
$sL
$"4"3S
Option (D) is correct. The dominant mode in a circular waveguide is TErr.
$sL
9"4,4$
Option (C) is correct. Consider the dimension of inner broad wall of waveguide is a (i.e. o > b). So, the dominant mode will be TE16. Since, the cutoff frequency of the TE-, mode is defined as
f,
:
I
ltmP,tftP'tt/2
,J:t',^l\;) + (t) l
So, for dominant mode (TEro) we have
1ox los
:ffi;l(*f.(BIl"
(l :
10
GHz)
1ox10e:L50"*
,:##:r5cm snl*
e.4.rll
Option (A) is correct. Propagation constant in a waveguide is defined
as
Since, for the evanescent mode of waveguide the operating frequency is less
than the cutoff frequency. l.e.
f
o,f.t
So, for this condition the propagation constant
7 is purely real.
$0L
9.4,S2
Option (B) is correct. Microstrip lines consist no ground plate and so the electric field lines remain partially in air and partially in the lower dielectric substrate. This makes the mode of propagation quasi TEM (not pure TEM) Due to the open structure and presence of discontinuity in microstrip line, it radiates electromagnetic energy and therefore radiation losses take place.
s$L
9.4.43
Option (C) is correct. Statements 1 and 3 are correct.
$s!.
9,4.44
Option (B) is correct. Modes on microstrip lines are quasi
TEM (not purely TEM).
So
the l't
\ statement is incorrect rvhile rest of the stat,ements,are correct. StlL
$.4".1$
Option (D) is cc.'rrcct. (a-+ a) Coaxial line + The dominant mode of propagation is TEM. (b-1) Rectangular waveguide -+ propagating mode is TE or TM. (c-2) I\{icrostrip line + The mode of propagation is Quasi TEM. + propagation The propagation mode is hybrid of Coplanar waveguide (d-3) (TE',,,+ TM",,)
sol
s.4.4s
Option (C) is correct. The phase velocity of TE or TM mode is defined
as
c
where
c
--+
Velocity of wave in free space
,f -
cutoff frequency operating frequency So, 'uo is a nonlinear function of frequency. ,f"
$sL
s.4.47
Option (B) is correct. A waveguide operated below cut off frequency can be used as an attenuator'
sol.
9-4.48
Option (B) is correct. Quality factor of a waveguide is defined
:
as
i.e. Q closely related to a. Also the attenuation factori obtained in waveguides are much higher than that in transmission lines. So, both statements are true but R is not the correct explanation of A.
A
sol.
!t.4"49
ffi
Option (B) is correct. Quality factor (Q) of a resonator is defined
Q: or,
Bandwidth
as
w
Resonant freqeuncy (/")
:*.
+
Therefore, the greater the 'Q', the smaller the bandwidth of resonator Q is also defined for a resonator as
a:*
where p is phase constant and o is attenuation constant of a resonator given AS
so
u:./'a^d -\/ 2 o,*
a.I
So, at higher frequency the Q of coil falls due to skin effect. sQL 9.4"50
Option (A) is correct. Guided wavelength of a propagating wave in rectangular waveguide is
\-
)
kC!aE
ct?t Wawguilcs
where ) is fr*e space wavelength and \ is cutoff frequency. Since, for piopagation the operating wavelength must be less than cut off frequency
Page 686
Chap 9 Waveguides
,\
t.e.
)r2)
So, we get
So, for a wave propagation in an air filled rectangular waveguide, guided wavelength is never less than free spa,ce wavelength.
sot
9,4.5t
Option (A) is correct.
(TM mode) consists of magnetic field intensity perpendicular to the direction of propagation where as the electric field intensity may be in the direction of propagation. Tbansverse magnetic mode
soL
9.4"52
Option (D) is correct. Since the conduction current requires conductor along the axis and a hollow waveguide doesn't have a conductor along its axis. So, the axial current is due to displacement current only.
SOL g"rt"Sg
Option (D) is correct. Electromagnetic waves propagating in a medium (bounded that has the velocity greater than the velocity in free space (velocity of light in space) is given as
up:
or sSL
S.4"54
u,
) C
The velocity u, is called phase velocity of the wave.
Option (A) is correct. A and R both true and R is correct explanation of A.
s$L Lrt,55
Option (A) is correct. Statement 1, 2 and 3 are correct.
sol
Option (C) is correct. Attenuation factor in a parallel plate waveguide is defined (Power lost per unit length) ^, _ n _
!t"4"56
as
"-tP,-@
sol
9.4,57
Option (B) is correct. Since the waveguide has a cutoff frequency f, below which no wave propagates while above /" all the waves propagates so it can be considered as high pass
filter.
*******xx**
-t
_. ,IO.{
_
G,HAPTHR tO ffi
ANTENNA AND RADIATING SYSTEMS
INTRODUCTION Antenna is a radiator and sensor of EM waves. The main aim of this chapter is to provide the fundamentals of antennas. They include: o Antenna definition, functions, properties and parameters
o o
Basic antenna elements
Radiation fundamentals, radiated power and radiation resistancc of current elements, dipoles and monopoles Directional characteristics of basic elements liirtenna arrays: array factor of two-element array and uniform lV -element array FYiis equation for received power. .
. o o
1O.2 ANTENNA BASICS An antenna is the device which receives or transmits the electromagnetic It may be a piece of conducting material in the form of a wire, rocl or
waves.
any other shape of excitation. Following are some functions and propertie-s of an antenna:
L0.2.L Types of Antenna According to characteristics, there may be two types of antenna: transmitting and receiving antenna. In the following texts, some typical antennas are described.
1. Wire Antenna: Three different (i) dipole antenna, (ii) loop antenna, and
types of wire antennas are
(iii) helical antenna, as shown in Figure
__J ;-]
10.1.
& re
f;age 688
€rqp t0
;MHdr andfiatfattug
.$ystins
I''iguri-, I {1.1 :
wire Antennas (a) Dipole Antenna, (b) square Loop Antenna and (c) circuiar
Loop Antenna, (d) Helical Antenna
2.
Aperture Antenna: A horn antenna is an example of aperture antenna. may be regarded as flared out or opened out waveguide. It may,be of different types like (i) Sectional ,O plane horn, (ii) Sectional 11 plane horn, (iii) Pyramidal horn, and (iv) Conical horn as shown in Figure 10.2.
It
(a)
(b)
(.)
(d)
l:iqrrr'l lri"'J; HornAntennas (a) Sectional .E planeHorn, (b) Sectional Pyramidal Hcn:n and (d) Conical Horn
3.
ff
planeHorn, (c)
Microstrip or Patch Autenna: It consists of a rectangular or square metal patch on a thin layer of didlectric (called substrate) on a ground plane. The radiating patch may be square, circular, elliptical or rectangular in shape as shown in Figure 10.3.
Ligrrrr.. 10.1}: Microstrip or Patch
Antenna: (a) Square patch, (b) Circular patch
4. Array Antenrras: Yagi-Uda
Antenna is an example of array antenna. Yagi-Uda antennas are the most high-gain antennas and are knovrn after the name of Professor S. Uda and H. Yagi. Figure 10.4 shows the YagiUda antenna.
Reflect
or (Parasitic element)
Page 689
ehep ven element
pl Di.""to,
IV '-il------
Systems
(Parasitic element)
td
ll
U
l:t r1r.r
iir i'
Yagi-L]da Antenna
Parabolic Reflector Antenna:
5.
A structure of a parabola is two
dimensional, but in practice, a parabolic reflector is a three-dimensional curved surface obtained by rotating a parabola about its axis as shown
in Fieure 10.5.
f,:r.rrri: ril i;,'Parabolic Reflector Antenna
10.2.2 Basic Antenna Elementb Following are some basic antenna elements:
1. Alternating Cunent
Element or Hertzian Dipole : It is a very short linear antenna in which the current along its length is approximately constant.
2. 3. 4. 5.
Short Dipole: It is a linear antenna whose length is less than ,\/4 and the approxirnate current distribution is triangular. Short Monopole: It is a linear antenna whose length is less than .\/8 and the approximate current distribution is triangular. Half wave Dipole: It is a linear antenna whose iength is ),12 and the current distribution is sinusoidal. Quarter Wave Monopole: It is a linear antenna whose length is )/4 and the current distribution is sinusoidal.
L0.2.3 Antenna Parameters Following are some important antenna parameters: 1. Antenna Impedance
It
is tiefirred as the ratio of input roltage to input current, and given by Zo
:
10
Antenna and Radiati.g
Ro+ jX ,to-:V
L
Page
Qhap
690 10
Here, the reactive part X" results from fields sulrounding the antenna. The resistive part, ft, is given by
Antenna and Radiating
Systems
Ro
where
.R1
: Rt*
represents losses
ft.ua
in the antenna, and -B,aa is called
radiation
resistance.
2. Radiation Resistance
Radiation resistance is defined as the fictitious or hypothetical resistance that would dissipate an amount of power equal to the radiated power. ft,"a
:
Power-{adiated J
rms
: # l r-"
3. Directional Chaqnsteristics
Directional characteristics are also called radiation characteristics or radiation patteln. These are of two types: (a) Field Streiglh'{attern: It is the variation of the absolute value of field strength asjb function of d. (b) Power Pattern: It is defined as the variation of radiated power with d. More geneially, an antenna ra.diation pattern is a three dimensional variation of the radiation field. 4. Effective Length of Antelna
It
is used to indicatethe efftctiveness
of EM
energy. .
ofthe antenna,
as a radiator or receiver
.r
(u)3fJ,;:i"tfiigl"_9,#trffl-il*f
ffi;':J"li-T,#1*,:lJ;
and which radiates the same field-streng"th as the actual antenna. Mathematically, effective length of transmitting antenna is defined m
tr r : 1 (rom)r*
.
rH
= i'J-lVY'
(b) Effective Length of Receiving Antenna: It is defined
as the ratio of the open circuit voltage developed at the terminals of the antenna under the received field strength, f , that is,
e*t(.,:+ I1.
Effective
.
lergtt'of an antenna is always
less
than the actual length.
.(-
' 5. Radiation Intensity i tt is defined as the power radiated in a given direction
per unit solid angle,
i.e.
U(0,0): t
..t
,
\o : intrinsic impedance of the medium,
wherei
-
f P'"a:#
r:
(C))
radius of the sphere, (m)
' P :power radiated-instantaneous E - electric field strength, (V/m)
.6; Directive Gain
Directive gain is defined as the ratio of radiation intensity in that direction .to the average radiation intensity, that is,
a;-rJM
--P,ua
\ Ptge 691
'f,/here P,"a is the radiated Power'
CbaP 10 Artenna and Radiating
7. DirectivitY
It
to the average is defined as the ratio of the rtrdrimum radlation intensity
radiation intensitY, i.e
max{t/(d,d)} a, .. ,-., u:---U*-- udru 8. Power Gain The power gain of an antenna is defined as
G'
: +u@'o)
where Pr is the total input power given by P1
.
: P,n:* P1
where P,"a is the ra'diated posnr' and
4
is ohmic losses in the antenna'
9. Antenna EfliciencY
'Itisdefinedasthatratioofradiatedpowertothetotalinputpower,i.e'
^ Tt
P,^ : :9%tL:- -P; - f+ PtGo
L0. Effective Area
Theeffectiveareaofareceivingantennaistheratioofthetime-average the time average povler power received r" (or delivered to t6e load) to
! ,'
density P'u" of the incident wave of antenna' i'e'
o":# / sYe
Thi, expression can be further generalised as \2
A" =hGo(o,0) 11. Antenna Equivalent Circuit
with resistance Ro, inductance Figure 10.6 shows an anterina equivalent circuit .Lo,andcapacitanceCoconnectedinseries.Theantennaconductancepeak occurs slightly away from resonant frequency' co
F'igrrrc 1{J.6: Antenna Equivalent Circuit
{0.3
RADIATION FUHDATENTALE
Beforeexaminingtheradiationpropertiesofanantenna,weshouldfirst the radiation of electromagnetic understand the physical process that causes waves.
10.3.1 ConcePt of Radiation
in an electromagnetic The Poynting vector (instantaneous power density) wave is given bY P
: E"x H,*
SYstens
f
i Page 092
So, the
total power passing out of the spherical surface of radius r -) co
Chap 10 Antema and'Radiating Systems
Ptota.r
: ir*t'|o"lt.
x
-Er.
is
*)rJS
where S : 4rf is the spherical surface area. This is the r nergv per unit time that is radiated into infinity. Thus, E\{ wave radiations constitute both electric and magnetic fields. Keeping this relatiorr in tnind. we analyse the following three results. POII{TS TO REMEMBER :,'l
2 J.
A stationary charge will not radiate A charge moving with constant velocity will not radiate A time-varying current or acceleration (or deceleration; of chargc will radiate.
10.3.2 Retarded Potentials The potential are usuallv esta,blished due to tinre varying fielcl orily after some amount of propagation time. This propagatioll tirne depends on the distance between the point of the potential frorrr their sources and velocity of propagation of electromagrretic fieltls. As a re,"ttlt. the potr:ntials are retarded by a time In general for any medium. the retarded potentials are defined as
and
v:hl!]u, .t:fr|,fra,,
.ftO
1)
where I/ is the electrostatic scalar potential. A is tire uragttetic vector potential, p, is the static charge derrsity, J is thc' static current density. and R is the point of observation from p,, and J rospectivelv. IIr the above expression, [R,] and [J] indicates that every I appr:a,ring in the expression for p, and ..I respectively ha^s been rerplacrxl hv a retardetl time
t'
: t- & 1)
where u is the velocity of wave propagation jrr thc tuerlirrrrr.
'
ISETHODOLOGY: TO OETERilINE FTELO vECToRs
POTEHTIALS Step 1: Obtain the time varying
-
vcctor poterrtial
Uslltc
RETARDED
{C) using equation
(10.1).
,$bpp,tX,.:fhenr,obtain magrretic.field vector ( .EI) using the relation
H =lr(v p' x A) Step
3:
Obtain electric field vector (.8) from Maxwell's cquatiou
#:|1v dt €' x rry Step
4:
as
\
Finally, determine.the timc-averale pc,u'"r clensit.y in the field It
po,: jne(ax It.*)
as
14.4
RADIATION FROM A HERTZIAN DIPOLE
Page 693
An infinitesirnally smail current element is called a Hertz'ian di,pole. Hertzian dipole is not of much practical use, but it is the basic building block of any kind of antennas. Consider the Hertzian dipole shovin in Figure 10.7, located at origin. The infinitesimal time-varying current flowing in o,-direction in the Hertzian dipole is 1 : locosc,.'t where c,,r is the angular frequency of the current. The field components at point P due to the Hertzian dipole is
n,,
: lrde;t (# * $)u,,,
D tndcos|l d ",- ' 2nr \rr.0,, Lo'
i,)" oT t
...(10.2a) ,,,
--. tud'si\g(4*4=.L,\, ,u, 4nt \rr - -rl." Uf)'
...(10.2b)
...O6':"1
Thus, the electric field is in th-e (r; d) plane,-whereas the magnetic field has @ coniponent.frfy."Tne fields can be classified-into three categories: 1.. Raciiation fields (spa,tial va.r.iation 1/r), 2. lnduction.iieids (spatial variation Ilf ), and
3.
Electrostatic fields (spatial variation
l/t')
In the foilowing texts. the field cornponents are generalised for near and far zone.
Il:rdiation frorn llertzial Dipole
10.4.1 Field Components at Near Zone
folr << \f 2tr. In near zone, electrostatic fields (spatial variation 1/r3) dominate. Hence, neglecting thelf r and 1/l terms in equatiorrs (10.2a), (10.2b), and (10.2c) we get \Ve define the near zone
, aTl(l
Ho":o fi,"=-'I'tdcosQ"it' lrt"sf E,. -- - ,l'1lsitt? , ,r Ane.rro
10.4.2 Field Components at Far Zone \&'e define the far: zone
for r' >>
),f
2r. In far
zone, radiation field (spatial
Chap 10 Antenna and Radiating System.s
l
variation llr) is the dominant term. Hence, neglecting the terms in equations (10.2a), (10.2b), and (10.2c) we get
Page 694
QhnF 10 Antenna and Radiating
llf
and
7lf
: iIoA-FlLo " 4Tr "-ia' E, :0 Er, : iloq'=q^Yo €ar "
ur"
Systems
and
47f
",u' .::.ll..-],ll]:]j]]:]:]].]'']]]:.'.:i.].
,,r,44$,,:,
be
eitt lt: ugti |i$l
i**.i.fr i,,a*ffWd:
&.@!b
is.,s,i'..$st,l
10.4.3 Power Flow from Hertzian Dipole The time-average radiated power from a Hertzian dipole is defined
as
P,,a:*t*l
...
(10.3)
where I: Iocosat is the current through the Hertzian dipole, d is the la,rgest dimension of antenna, q is the intrinsic impedance of the medium, ,\ is the wavelength in the medium. In free space, intrinsic impedance is r1o:129n , so the radiated power in free space is given by P."a
:
\r/'o"
4gnz1dt2
10.4.4 Radiation Resistance of Hertzian Dipole The radiated power in a Hertzian dipole is equivalent to the power dissipated in an imaginary resistance -R,.a by current 1: locoscuf . i.e. P,ua
:
13*rR"*
: $Io'R,^a
...(10.4)
So, from equations (10.3) and (10.4), we obtain the intrinsic impedance of Hertzian dipole as
ri."a
('l\' : 2+ - T-\r/ |s" -2'ln
In free space, intrinsic impedance is 46 : 720tr , so the radiation
resistance
in free space is given by
R,,a:ggnzld'f
\f/
{0.5
DIFFERENT GURRENT DISTRIBUTIONS IN LINEAR ANTENNAS Following are some possible current distributions in linear antennas:
10.5.1 Constant Current along its Length Constant current distribution is possible only in Hertzian dipole (i.e. when length of antenna, d < A). In Figure 10.8, the constant current distribution
of the linear antenna has been shown bv dotted line.
Page 695
Chap 10 Anteuna and Radiating SYstems
j1*3rffii1"" d<<)
Figrrnr l.{i.li: Constant Current Distribution in Hertzian Dipole
10.5.2 Thiangular Current Distribution Tliangular current distribution is possible in either a short dipole or a short monopole.
1.
2.
Short dipole: The antenna with length d< is called short dipole. ^14 current distribution. Figure 10.9(a) shows the short dipole with triangular Short monopole: The antenna with length d < Al8 is called short monopole. Figure 10.9(b) shows a short monopole with triangular current distribution. Current
distribution
d<^14
ffi" Current
,\
1..
I ,' / IV'-.' /
l{J.:.}:
plane
Image
(") Figur{,
Perfect
conducting
(b)
Tliangular Current Distribution in (a) Short Dipole (b) Short Monopole
10.5.3 Sinusoidal Cunent Distribution Sinusoidal current distributions is possible either in a short dipole or a short monopole. The sinusoidal current distributions for the short dipole and short monopole has been shown in Figure 10.10.
f-t
I| i current
[-,
**l -l /l-.
/ distribution ffi'wffiG;ffi
I
I
currenr
12"'ibution
t" \. Perfect / conducting r plane
I
Image
(")
t
(b)
Figruc l0.l{}: Sinusoidal Current Distribution in (a) Short Dipole and (b) Short Monopole
{0.6
Page 696
RADTATTON FROl{t SHORT DTPOLE (d
< AI4r* -
_
Qhnp 10
Practical elementary dipole is a centre fed antenna having ttre length that is-very small in wavelength. The current amplitude in such antennas decreases uniformly from maximum at the centre to zero at the ends. For the same current, the short dipole of length dL will radiate onlv quarter a^s much power as the current element of the same length which has current 1 throughout its length. This is because field strengths are rerluced to Li2 arrtl hence power density is reduced to ll4. Therefore, the ra
Lntenna and Radiating Systens
:
(ft,"a)rr'ortatpo* t .'-
_
't
",^ \f/ = r6b(*l
-tP"
1O.7
(R.ud)"u,,"'t"lo*"'t ] X
RADTATTON FROM SHORT MONOPOLE (d
< I/8)
Figure 10.11 shows a short monopole antenna with triangular crrrrent distribution. The monopole of height d or short vertical antenna placed on a reflecting plane produces same field strengths above thc plane as the dipole of length I:2d,, when both are fed with same current. However, the antenna radiates power in hemispherical plane above the reflecting plane. radiated power is half of that of dipole. Therefore. the radiation resistance of the short monopole antenna is given by ( R.ua).ro,, *o'opole
: ]
tft
*n) -,,-, o,*,,^
:10"r(*)
Itr -_-]
1.. ,"
I
/l
Perfect
conducting
/"
plane
Image
figrlrtl
{O.8
l{.1.1.1:
A Short Monopole Antenna with tiangular Current Distribution
RADIATION FROM HALF WAVE DIPOLE ANTENNA The half wave dipole derives its name from the fact that its length is half a wavelength (d: As shown in Figure 10.12, it consists of a thin wire ^12). fed or excited at the mid-point by a voltage source connected to the anrtenna via a transmission line.
10.8.L Power Flow from Half lMave Dipole Antenna
"
The time-average power density in the electromagnetic wave radiated from half wave dipole antenna is given bv
rr*
qfrcos'($ cos?\
: _ Bpisfu:%
Therefore, total power radiated from half wave dipole is given l,! trrkr:.!, the surface integral of power density over the spherical surface of radius r. i.e. , P"ua
: f n",'
In free
o.(1'218)
intrinsic impedance is
sp&ce)
by the half wave rlipole in free
Pluo:
r7s
:
720rr Q. So, ttre radiaterl po!,{::l-
sprace is
WxI.2I8:36.54ff
t'\.
lr Il{
:
lr
-
/
lr
t'It
Current
dirt.ibrrtiort
I
I
:
I"cos
,32
d:Al2
l/ lt
Lrr'
(u)
(1,)
r:rrll tij.J j: Haif Wave C-'alculating the Fields .i;
Dipole Anteuna (a) Current Distributiorr. (b) Getrrnotn'
10.8.2 Radiation Resistance of Half Wave Dipole Antenna The racliation resistance for the half u'ave dipole antcnira cau directly
l.,,.,gir.
li.,,
l:r
a,s
. llrad D -21^t r 1,i _ 2 *lqqafr') = Io'
IO.9
73
()
RADIATION FROM QUARTER WAVE MONOPOLE ANTENNA The quarter wa\re rnonopole antenna corrsists of a half-wavc tiiitole altr:l1rii located on a conducting ground plane, as shown in Figure 10.111.
\
V .,
I:
I"cos 0z
I
i
i Figrur-
ll1.
Perfect conducting plane
lll: A Quarter-wave
Chap 10 Antenna antl Radiating Systems
aS
Solvirrg the integral by Simpson's or the 'Ilapezoidal rule we get,
P'"a:(it#)
Page 697
N,Ionopole Antenna
l
i
Page 698
10.9.1" Power Flow from Quarter'Wave Monopole Antenna
Chap 10
The time-average power density P* for a quarter wave monopole antenna will be exactly same a,s described for half wave dipole, i.e.
Autenna and Radiating Systems
\
@ - ffifcos'?(fcosd)l "'- 877L--s."To ]
,
As the monopole is fed by a perfectly conducting plane at one end, it radiates only through a hemispherical surface. Therefore, the time-average radiated power from the quarter wave monopole is P,ua
: f f."'
aS
:\!; * - 4n I Yrzcos2(zLcos?) sin0 Solving the integral by Simpson's or the Ttapezoidal rule we get,
D - lrtk'\ . P'^d: (#-/
(o'6oe)
In free space, intrinsic impedance is 46 : l20r Q. So, the radiated power by the quarter wave monopole in free space is P,"a :78'27102
10.9.2 Radiation Resistance of Quarter Wave Monopole Antenna The radiation resistance for the quarter wave monopole antenna can given directly as
n,*:?
be
lD
I
-ry#e=36.b4f) IO.IO ANTENNA I i
ARRAY
An antenna array is a group of radiating elements arranged to
produce
particular radiation characteristics. Here, we first consider the simplest case of two-element array and then generalize it to uniform linear arrays made up of many identical elements. ,
\\
10.10.1 Twoelements Arrays The simplest array is one consisting of two identical radiating elements spaced a distance apa,rt, as shown in Figure 10.14. Let us assume that the far-zone electric field of the individual antennas be in the d-direction a^nd that the antennas are lined along the r-axis. The antennas are excited with a current of the same magnitude, but the phase in antenna 1 leads that in antenna 0 by an angle d.
t*_d_*l liigrrr<'
1{}..1,1:
Two Elements Array
Array Factor
Page 699
The array factor of the two element array is defined
lr' :
2cos[
[{0a
rine
"o"
6
Chap 10 Anterna and Radiating
as
+ il!"ilP
Systems
So, the normalized array factor of the two element array is given by
rAF |
: I
In the ,Fl-plane,
0:
:
-'(@5@)
I
|
".'(+)
I
r12, and the normalized array factor becomes
rAF |
:
Total Field of an Array
1""'(@-#{)l
: -'(5) |
I
The total field of an array is equal to the field of single element located at the origin multiplied by an array factor. Thus, in general, the far field due to a two-element array is given by E (total) : (E due to single element at origin) X (array factor) Resultant Fattern of an Array
At point P (far-field zone) in Figure 10.14, the total electric field is given by
,, : 4#"or6"-iar In
"iat,
Zcos(E!$
t!)
above expression, lcosdl is the radiation pattern due to a single element, whereas the normalized aryay factor, lcos[](Bacosl*o)]1, is tfre radiation pattern the array wopld have if the elements were isotropic. These may be regarded as "unit pattern" and ttgroup pattern", respectively. Thus
the "resultant pattern" is the. product of the unit pattern and the group pattern, i.e. Resultant pattern : unit pattern X group pattern This is known as pattern multiplication, and it can be used to sketch, almost by inspection, the pattern bf an array,
L0.10.2 Uniform Linear Arrays Consider an array of identical antennas equally spaced along a straight line. The antennas are fed with currents of equal magnitude and have a uniform progressive phase shift along the line. such an array is called a un'iform l'inear array.
Array Factor Array factor of an ly'-element array is defined
as
. NiIt sm-zIAFI: .-Esmi
where
iP
:
pdcos$* a
Following are some important points about an N-element array:
.:tAF.i;i;
',
r
v,.
Page 700
Chap 10
'':r_l
:rr'
:
2''"fhC biiilsipdturiaximiixi6icui! riirhen {'
.:
Antenna and Radiating Systems
,0,* 0d cosd
'{:os:$ 3,
-| *
1.*.
a
- :ffi
Wh;n.leFj.;0i'lgf1,[*inulls.(or,,zeros);i,e,
Y:t*,
:'
.
k:!,2,J,...
where fr is not a multiple of y'f .
A broadside array has its maximum radiation directed normal to the axis of the array, i.e. $:0, @: 90" so that a : 0. 5; ,Ao.end-fire pfl,,qy,lla 'its maximum radiation directed along the axis
.''..'ofiire,.'arrry':i'e....41.*.:'6;d:[1*thata:[ur..
... lO.l1FRllS EQUATION
_
.._..:-=',._l
.-.--.
_,_.-
Friis trarrsmission formula relates the power received by*orre antenna to the power transmitted by the other. Let the transmitting antenna has eII'ective area A"l ancl directive gain Qo,, and transnrits a total power Pr(: P.^a). Also, assume that the receiving antenrra has effective area of A"' and directive gain G7,, and receives a total power of P". So, at the transmitter, we have
_ 4rf Po," -pP^," : J-uco, 41(f t-t L'dr
...
(10.5)
is the time-average power density of the incldeilt" wave at the receiving antenna. Therefore, at the receiver, we get the power where
Puu"
p,:
Puu.A",:#G0,.r,,"
...(10.6)
Substituting equation (10.5) into equation (10.6), we get
P,:
Ga,C.flnn;fr,
This is the FYiis equation antennas are separated by r > either antenna.
for the received power) provided the twcr 2&l^, vrhere d is the largest dimension of
***********
ffiKffiffisxstr'$o-{
Page 701
Chap 10 An'tenna and Radiating
tffidi]iffi
$tcs
$:'5terrrs
"!*.r,.1 A Hertzian dipole of length )/50 is located at the origin' If a point P of located at a distance r from the origin then for what value will be in radiation zone'
(A)
,:!
(C) Both (A) and
(B)
(B)
":
'r'
is
the Jroint
+
(D) none of these
to a satellite in space' e$** 1$."r"2 A certain antenna is used to radiate a 0.2 GHz signal
Giventheradiationresistanceoftheantennais3l.6fl.Theantenrrais (A) half wave diPole (B) quarter wave diPole (C) one-fifth wave diPole (D) none of these
Q6mmor Data For Q. 3 and 4 : :0'2GHz.' A Hertzian dipole is operating at a frequencY' f ,sfi# ,1*."r.3 what will be the rnaximum effective area of the dipole
?
(A) 0.54 rn'? (B) 1.07 m' (C) 0.18 rn' (D) 0.27 m'? r*{:& .!itr..r.4
power density of antenna receives 1.5p,W of power then what is the the incident wave ? l (A) 8.331,W/*'
If the
(B) 5'56 PW/*' (C) 1.40PW/*' (D) 2.793 PW/ur2 Ms{&
If power of acertain antennawith an efficiency of 95%is 0'8Watt' directivit'y it's the antenna has maximum radiation intensity of 1w/sr then
.!*".t"s The input
q'ill be (A) 5.26
(c) tv!s*
,*.,{"s
0.76
(B) 16'53 (D) e'55
the rtirectivitv An antenna has maximum radiation intensity of 1'5 W/Sr. If
oftheantenhaisD:20.g4thenradiatedpowerofantennawillbe (A) 1.11w (B) o.3o w
w (D) o.eo w
(c)
0.26
to"t.? Three element airay that
Page 702
Chap 10
2lol0o Iolo" +\12-o-\12-t
Antenna and Radiating Systems
has the current ratios
1
:2:1
as shown in figure
loloo
The resultant group pattern of this array will be same as the two element antenna array with (B) o: 180", d : \12 (A) a: 0, d,: \f 4 (D) o : 180', d: 2.\ (C) a:0, d,: \f2
$cq {*.{.s When the two three-element a.rrays with current ratio 1:2:7 by \12 then it forms (A) Four element array with current ratio 1 : 3 : 3 : 1 (B) Three element array with current ratio 2 : 4 : 2 (C) Four element array with current ratio 3 : 1 : 1 : 3 (D) Three element array with current ratio 1 : 3 : 1
are displaced
{8"1.s A Hertzian dipole of length
,\/100 is located at the origin and fed with a current of z(t) : 0.5sin108, A. A point P is located at a distance r from the dipole as shown in figure. What will be the magnetic field at P ?
I
ll . I
(B) 1.15cos(ro8t+ oo') (D) 2.30sin(ro8t+ oo")
(A) 1.15sin(108t+ 90') (C) 1.15 sin(108t - 90')
ll
tsc&
{o.{.1o Directive gain of Hertzian dipole (A) 1.5sin'?0
(c) nfiso
to.l.{t
antenna is
+
(B) 3sin'z0 (D)
lsin'?d
Two Hertzian dipole antennas are placed at a separation of a's shown in figure below :
d: \12 on
-axis to form an antenna array
i
d:\,/2
I
i
,l t
I
t; ,i'
t +
{
L": Io /0" and the 2"d antenna carries field pattern of the antenna array the resultant then a current 12,: Io /180"
If the 1"t antenna
(' '' t^--
'i)
il
*.y ,
will
be
carries a current
(A)
(B)
(D)
\\
1o't'12 An antenna array is formed by two Hertzian dipoles placed at a separa,tion of )14 as shown in figure. The current fed to the two antennas are d, and
\.
12, respectively.
T
\14
t If
Iz" is lagging 1r, by an angle will be
antenna array
rf2
then the resultant field pattern
o1'
\
\
t'
6
\:'
(A)
"i
(B) 1:3*
rtr I
1-(c)
(D)
't -'a.
. --i_ _-.1
.-t-
-
...-
a'
:
,i
{ Page 704
${$& 1qr.*"'r3
The group pattern function of a Iinear binomial array of /f-elements
as
shown irr figure is
Chap 10 Antenna and Radiating Systems
i.,12
I"la I
10"
1"1 d
(A)
(c)
tsr[-'(
[cos(P@e!#s)]'-'
oonry
*")1"
(D) [cos(Bdcoso+o)]''
[cos(44541CI)l'.'
Common Data For Q. 14 and 15 : Maximum electric field strength radiated by an antenna is 6 mV/m measured at 40 krn from the antenna. ik
ir{,r 1'! 1
l's If t}re antenna
rarliates a total power of 100 kW 100 kW therr the directivity
of antenna is (A) -2.02 dB
(c) ;,:ri tn
(B) e.6 dB (D) -20.18 dB
o.ooe6 dB
r.t* If the efficiency of the radiation is 95 Ya tlten it's maximum (B) s.4 x 10-3 (A) 9.12 x 10-3 (D) 9.6 x 10-3 (C) 0.11 x 10-3
powcr gain is
Common Data For Q. 16 and 17 : A radar with an arrtenna of 1.8 m in radius transrnits 30 kW at a frequency 3 GHz. The effective area of the antenna is 70 % of it's actual area. :'i;:.* -::j
l 1{; If thc minimum detectable power is 0.13 mW for a target of cross sectiorr 1.25m2 then the maximum range of the radar is
(A) (C) rfr
ti
1{::.i
1?
584.3
(B) 1270 n (D) e77.8 rn
m
2e2.1m
The average signal power density at half of the range of rarlar will be (B) 69.80 Wfrn? (A) 350.25 \Y lm2
(c)
80.6e
w/m'z
(D)
250.35 W
l*'
Common Data For Q. 18 and 19 : A rnetallic wire of crr-rss sectional radius 6 mm is wouid to form a small circular k-'op of radius 1qr with 10 turns. Conductivity of metallic wire is o:2.9 x 10r S/m.
\
1
.
srcQ
{o"{"{8 If a 0.5 MHz will be (A) 2.37 x (C) 2.37 x
McQ
t{lcQ
lo.t.te
uniforrn current flows in thB loop then it's radiation resista.nce
Cbali ro
(B) L.42 x 1o-3 O (D) 4.53 x 10-4 O
10-6 O 10-4 Q
Page ?0b Antenna and Radietisg Systems
Radiation efficiency of the antenna will be
(A)
18.36%
(c)
10.8e%
(B) 0.101% (D) 0.055%
1o.'t.2o The polar radiation pattern of a .\/B thin dipole antenna is
(D)
Common patE Fo1 Q. 21 qp"d,Q2..: Twp qhort antennas at the origln in free Epace carry idg&ticq,I gqrents cos.,tA. one iq the p, direefior.r and pther in itre a, Jirection.
ffcq
to.t.zt If both the antennas
arp of length 0.1m qqd way,sleugfih is .\: 2rm theq the electric field .s, at the disrant goiuts f(0;0,,1000) and e(1000,0.,Q) .iii be
at point F
at poipt
(A) - j(1.2 y lQ-2)6-nooo o,Y lp (B) j(\.2 x 1g-z)r-noooa,.Y/fq (C) * j(I.z.x 10-?)e1lmoa" Vfm (D) j(|.2 x 16-z)r-rroooo"Vfm ilrce
{o.r.22 .E at point (0,1000,0) at (A) 9.92(a,-t a,)mylvL (C) 1.2(a,* a,)r'rrYfm
t:0
will
- i(1,2
x
I
10-2) e-'looo
a,Y
In
o.\ f m *i(l.2 x 10-2)e-rrooo a"V/^ x 10-2)e+oooqYlm i(1.2 X 1g-?)6
be
tlmo
^1.2 (B) -9.92(q, * o")mYfm (D) *12(a,* a"),mYfm
d
Page
706
Chap
10
Cornmon Data For e. 28 and 24 ! In a free space short circuit vertical cuqent.el€ment is located at the origin in free space. The radiation field due to the element at any point is given as ot":\ l?sinde-rto"i ;_
Radiating svstems Antenna and
v
ilcq {o.,t.2s
!
.86"
at poinl
p (r:100, g : T/2, dL
(A) o.zs"ooo" Y l(C) o.te-"ooo"Vl^ ilrsQ
\
116) is
(B) g.zs-rlooo" V lrn (D) 0.1siro00"yl^
to"1.24 If the vertical element is shifted to a point
P(100,4,ff)
"n"r,g.,
,o
(A) O.te-aooV V l*t (C) O.te-p'" V lF l|tco
lm
(B) (D)
(0.r,+,f) ,t"", \ g.1e-rtooo" en';" Y g.1s-rtooor e*n.sr
,86,
at point
/m
r,^
{o'{'25 An antenna
is made of straight copper wire of length 1cm carrying current of frequency 0.3GHz. If the wire has a cylindrical cross section of radius 1mm then the ratio of the radiation resistance to the ohmic resistance of wire will 5" ft"oa 'v tLt (A) 11 (B) 6
(c)
17
(D)
****t ****xx
5
EXERGISE
Page 707
Chap t0
'9.2
Antenria and Radiating Systems
qu€s '1o"2.{ A qua"rter wave monopole antenna is operating at a frequency, f The length of antenna will be meter. Quss
:2bMHz.
{o.2.2 A half wave dipole antenna is located at origin as shown in figure below. The antenna is fed by a current i(f):33.3"osof mA. What will be the electric field strength (in pV/m) at point p ?
to'2'3 The transmitting antenna of a radio navigation
system is a vertical metal mast 50 m in height inducted from the earth. A source current is supplied to it's base such that the current amplitude in antenna decreases linearly toward zero at the top of the mast. The effective length of antenna will be meter.
ouEs 1o.2.,t1 A vertical antenna of length 7.bm is operating at a frequency, f The radiation resistance of the antenna is f).
:2MHz.
ours to'2.5 The current in a short circuit element of length t:0.03) is given by
tt'
for
Ir,
b, f;
IQ):]2
0
What will be the ra.diation resistance (ia O) of the element
?
to'2'6 A dipole antenna radiating at 100 MHz is fed from a 60 O transmission line matched to the source. What will be the length (in meter) of the dipole that matches the line impedance at the signal frequency ?
ques
to'2'7 A time harmonic uniform current Iscos(2T X 10?t)
--' ---.--_-----_
anteqggf radius 30 cm. Radiation !q1p
rn0- -,-.'---
s:l
--
flows in a small circular
resistance of the antenna is
// 7
Common Data For Q. E and I : An antenna is a center fed rod having cross sectional radius 4cm and conductivity o :2.9 X 107 S/m. The length of the antenna is 30 m.
{ase Jop Chap lQ Antenna and Radiating Systelhs
If a 0.5 MHz current flows in the antenna then the loss resistance of the antenna is 0.
QuEs
10.2.8
ouEs
ls'2.9 What is the percentage radiation efficiency
QUE$
{o,2'{o A
of the antenna
?
100 MHz uniform current flows in a small circular loop of radius 20 cm If the loop is made of copper wire of radius 5 mm then it's loss resistance will be o.
.
(conductivity of copper, o : 5.8 x
10.2.11
QUEa
10?
S/m)
A quarter
wave monopole antenna is connected to a transmission line of characteristic impedance Zo:75O. What will be the starrding wave ratio ? (Input impedance of quarter wave monopole is Zn: (36.5 + J2I.25)0)
{o'2.{2 Radiated power of a vertical antenna is 0.4 kW. What will be t}re maximum electric field intensity (in mV/m) at a distance of 10 krn from the antenna ?
Qurs to-2.t3 Aquarter.u.r"r.torropoleantennaisfedbyacurrent i.(t):41.7.osrr.,f mA. The average power radiated by antenna is mW. QuEs'
to.2.14 A dipole antenna in free space has a linear current distribution. If the length ' of the dipole is 0.01,\ then the value of current 16 required to radiate a total power 250 mW is Ampere.
QUE$
{0'2.15 A monopo}e antenna in free space has the length of the antenna 0.02). The antenna is extending vertically over, a perfectly conducting plane and has a linear current distribution. What value of 16 (in Ampere) is required to radiate a total power of 4 W ?
OuEs
to.z.tc What is the directivity of quarter wave monopole
?
oux$ lo.2.l? An antenna has a uniform radiation intensity in all directions. What is the directivitv of the antenna ?
ours {o.2.{8 Normalized radiation intensity of an antenna is giverr by [sind 0 < 0 < 7T12. 0 < 6 < 2tr u(e.o) :
lo
otherwise
What will be the directivity of antenna
?
quEs 1o.2.'ls An antenna has the uniform field pattern given by
u(a)
f4 :to
0<0
rfl
Page?O9 1 ChaP 10
whereU(0)[email protected]? L*"nna.nd
n;dhtiDg SYetems
eircs ."!s"*.x$ Directivity of Hertizian monopole antenna is
Common Data For Q' 21 to 23
----
:
AtransmittingantennaisbeingfedbyacurrentSourceofamplitude 1o:50Aand"frequencyf:l80kHz.Theeffectivelengthofantennais
20m. r
at a distance 80km *rr*s {s"x.xr what will be the maximum field intensity (in mv/m) from the antenna
?
is &uss 'i*.3.4* The time average radiated power of the antenna
----
kW'
antenna &i!rs ,rE.?.*3 What will be the radiation resistance (in O) of the
e{Jr$ .,*.e.;4
A short circuit current element of length l: 0.06) distributed
]
The radiation resistance of the antenna will tre
----
the current
ca,rries
as
I(z).:"['- ?l' &{rss .i$.e"x$
?
^' ohm'
-L"'
L
at a frequency A 2 cm long Hertziarr dipole antenna radiates 2 w of power of 0.6 GHz. The rms current in the antenna is t<**t<* ***(*
*X<
----
Ampere'
I
I
EXERGISE {O.3
Page 710
ChsF 10 Antenna and Radiating Systems
rucQ to.3.{
McQ
{o.3.t
Input resistance of an antenna is (A) R" (C) .8" * ,?r
Horizontal dipole has the directional cha,racteristics of (A) circle (B) figure of eight
(C) four lobes lllcQ
{0.3.3
(D) ellipse
Retarded vector magnetic potential has the unit of
(A) wb/m
(c)
(B) v/m (D) wb-sec/m
v/m'?
ilcq to"3,rt Rediation (A) 73 Cl
resistance of half wave dipole is
(B)
, (c) 80l(*l ilcq ro.3.5 Effective length (c) 10.3.6
of half wave dipole is
(B) < ^12 (D) 0.6)
^12 0.55A
Far-field consists of
(A) (C)
f term 'r" 1 term
McQ
mcQ
10.3.7 Induction field
to.3.8
36 c'
(D) 2e2A
@) > Mca
(B) ft, (D) .R, - ftr
(B) (D) consists of
term
fi r
term
1,term
(A) 'r
1t"r-
(B)
(C)
4r term
-T
(D)
12
(B)
+
Radiated power is proportional at
(^)
R?
(C)
I,
(D) ******r<***(x
/
term
EXERCISE 10,4 Antenna and Radiating Systems
McQ
{o"4.t
The radiation pattern of an antenna in spherical co.ordinates is given by
u@):cosndl o
ts.4.?
(B)
12.6 dB
(c)
11.5 dB
(D)
18
dB
For a Hertz dipole antenna, the half power beam width (HPBW) in the E -plane is (A) 360'
(B) 180' (c) 90' (D) 45"
ro"4.3 At 20 GHz, the gain of a parabolic dish antenna of L meter and7lTo
efficiency
is
(A) 15 dB (B) 25 dB (c) 35 dB (D) 45 dB
10.4.4 A
dipole is kept horizontally at a height of f above a perfectly ^12 infinite ground plane. The radiation pattern in the lane of the conducting dipole (,E plane) looks approximately as
(A) (c)
6.
d2_
.b a
(B)
A
(D)
1o.$.5 A mast antenna consisting of a 50 meter long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of600 kHz. The radiation resistance of the antenna in Ohms is t
t
@)+
(B)
Q)+
(D) 20#
/ Page 712
Chap 10
lucQ
to'4'6
AnteDDe and Radiating Systems
Two identic4 and parallel dipole antennas.are kept apart by a distance of in the H - prane. They are'fed with equal currents but the right most ^/4 antenna has a phase shift of *90". The radiation pattern is given as.
(A)
lrtcQ to"4.?
Consider a lossless antenna with a directive gain of +6dB. it the total power radiated by the antenna will
power is fed to (A) a
mW
(C) 7 mW illcQ to.4.a
MCA 10.4.9
If 1mW of be
(B) 1 mw (D) i/a mw
Two identical'antennas are praced in the 0:r/2 plane as shown in Fig. The elements'have equar ampritude excitation with 1g0" polarity difference, operating at wavelength ). The correct value of the magnituJe of the far_ zone resultant electric field strength normalized with that of a single element, both computed for d:0, is
(A)2"":(+)
(B)
(c) 2cos(T)
(D) 2sin(#)
2,i"(+)
A person with receiver is 5 km away from the transmitter. what is the distance that this person must move further to detect a 3-dB decrease in signal strength (A) ea2 m (B) 2070 m (C) ae78 m (D) 5320 m
MCQ to.4.10
A medium wave radio transmitter operating at a waverength of 492 m has a of height r24.wriatis the radiation resistanle of the antenna? i:T"j""-'*na \ ) .o nL (B) 36.b CI
(c) 50 tttcQ t{l.4.tl
o
(oj
ze
cr
In uniform linear #ray, four isotropic radiating
elements are spaced )/4 apart' The progressive phase shift beiween ifr" the main beam at 60' off the end _ fire "rJ-""rr;;"a for forming is : (A) -t' (B) -f radians (C) -+ radians (D) _+ radians
ificq
lo'4'{2 If the diameter of a )/2 dipole antenna is increased from )/100 to A/50, then its
(A) bandwidth increases (C) gain increases McQ
{0'4'13
IttcQ
Chap 10
(B) bandwidth decrease (D) gain decreases
Antenna and Radiating systems
For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz,the minimum distance required for far field measurement is closest to
(A) 7.5 cm (C) 15 m
(B) tb cm (D) 1b0 m
to'4't4 An electric field on a place is described by its potential V :20(r + r where r is the distance from the source. The field is due to 1
2)
(A) a monopole (B) a dipole (C) both a monopole and a dipole (D) a quadruple
lllcQ
to'4'tr A transmitting antenna radiates 251W isotropically. A receiving
antenna
located 100 m away from the transmitting antenna, has an effective aperture of 500 cm2. The total received by the antenna is
(A) 10 pw (C) 20pw
(B) 1pw (D) 100pW
McQ
10.4.{6 The vector H in the far field of an antenna satisfies (A) V .If :0andV x .E[:0 (B) V.H+}andV x H+0., (C) V..E[:0andV x H+0 (D) V . H+ 0andV x I[:0
lllco
to'4'{7 Theradiationresistanceof acircularloopof oneturnis resistance of five turns of such a loop
(A)
(c) McQ
c) o.o5 o 0.002
,
0.01 O.
Theradiation
will be (B) 0.010 (D) 0.25 Q
to'4'lo An antenna in free space receives 2 pW of power
when the incident electric field is 20mY/mrms. The effective aperture of the antenna is (A) 0.00b m2 (B) 0.0b m2 (c) 1.885 m, (D) 3.77 m, ,
McQ
1o'4'1e The maximum usable frequency of an ionospheric Iayer at 60" incidence with 8 MHz critical frequency is (A) 16MHz .1n) {MHz ,/3 (C) 8 (D) 6.e3 MHz
MHz
McQ 10.4.20 The far field of an antenna varies
r '+
(A)
, r,
(c) +
'
with distance (B) +
r
,r-,, 1
(D)+ ' "/r
as
page 71J
ancl
( I
I 1
I
Page 714
ffcQ
Chap 10 l I
l
t$.4"?{ The critical frequency of an ionospheric layer is 10 MHz. What is the maximum launching angle from the horizon for which 20 MHz wave will be
Antenna and Radiating
reflected by the layer
Systems
?
(A) 0'
l
(c)
j
(B) 30' (D) 90'
45"
l
l
ntcq
to.rr":*
The directivity of a ),12 long wire antenna is
(A) 1.5 (c) 2 ll}ca
10"rt.23
l l I
i I I I
I
I
MSQ {$.rt.?{
@)
1.66
lt
The characteristic impedance of TV receiving antenna cable is 300 O. If the conductors are made of copper separated by air and are 1 mm thick, what is the phase velocity and phase constant when receiving VHF channel 3 (63 MHz) and VHF 69 (803 MHz) ? (A) 1.32 rad/m and 17.82 ndfln
(B) 1.52 radfm and (C) 1.52 rad/m and (D) 1.32 rad/m and
l
(B)
rad/m 17.82 radfm 16.82 radfm 16.82
An antenna located on the surface of a flat earth transmits an average power of 200 kW. Assuming that all the power is radiated uniformly over the surface of a hemisphere with the antenna at the center, the time poynting vector at 50 km is
(A) Zero
I
I
(c)
I
McQ
fp wl^'
B\ "
La-Wlm2
(D)
*",r, wl^'
avera,ge
'17
{o.4'25 An antenna can be modeled as an electric dipole of length 5 m at 3 MHz. Find the reduction resistance of the antenna assuming uniform current over the length. (A) 20 (B) 1f, (c) 40 (D) 0.5 0
ruset {o.4.t6
A dipole with a length of 1.5m operates at 100MHz while the other has a length of 15m and operates at 10MHz. The dipoles are fed with same current. The power radiated by the two antennas will be (A) the longer antenna will radiate 10 times more power than the shorter one,
(B) both.antennas radiate same power. (C) shorter antenna will radiate 10 times more power than the longer antenna
(D) longer antenna will radiate ./10 titn"r more power than the shorter antenna MCO 10"4.27
element has length r:0.03), where ) is the wavelength. The radiation resistance for uniform current distribution is
A short current (A) o.o72io (B) 8olo (c) 72a (D) 80 c,
MCq {0.4.28
In a three element Yagi antenna (A) All the three elements are of equal length (B) The driven element and the d.irector are of equal rength but the reflector is longer than both of them (c) The reflector is longer than the driven element which in turn is longer than the director (D) The reflector is longer than the driven element which in turn is ronger than the reflector
MCQ {O"S.29
Multiple member of antennas are arranged in arrays in order to enhance what propertv ? (A) Both directivity and bandwidth (B) Only directivity (C) Only bandwidth (D) Neither directivity nor bandwidth
mce
If the total input power to an antennaisffi,the radiated power is W., and the radiation intensity is @, then match List-I with List-II and select the correct answer using the code given beloW the lists:
{0"4.3$
List-I
a. b. c. d.
Power gain
Directive gain Average power radiated
Efficiency of the antenna
Codes
List-II
1. W/W 2. W,lan 3. 4"4/W, 4. 4"d/W,
:
abcd (A)3427
(B)432r (c) 3412 (D)4312 $Ge {o.4.3{
Where does the maximum radiation for an end_fire array occur? (A) Perpendicular to the line of the array only (B) Along the line of the array (C) AT 45" to the line of the array (D) Both perpendicular to and along the line of the array
MCO 1O"r1,32
As the aperture area of an antenna increases, its gain
(A) increases (B) decreases (C) remains steady (D) behaves unpredictably MCQ 1$"4,33
which one of the following is correct ? Normal mode herical antenna has (A) low radiation efficiency and high directive gain (B) high radiation efficiency and low directive gain (C) low radiation efficiency and low directive gain (D) high radiation efficiency and high directive gain
7lb t0 XChap Page
.
Antenna andtadiating
Systems
Page 716
rdlsq
Chap 10 Antenna and Radiating Systems
MoQ
to"4"34 For taking antenna far field pattern, what must be the distance .rt, between transmitting and receiving antennas ?
(A)
,q
r+
(B) n
r+X
Q)
Rr#
(D) ,?
,Y
to"4.35 A transmitting antenna has a gain of 10. It is fed with a signal power of 1W. Assuming free-space propagation, what power would be captured by a receiving antenna of effective area 1m2 in the bore sight direction at a
distanceoflm? (A) 10 w (c) 2w mcQ
(B) 1w (D) 0.8 w
{o'4.3s The F}aunhofer region where the pattern measurement of transmitting antenna has to be taken from a distance of 4, where D is the maximum aperture dimension and ) is the free-sprce wfvelength. What is the region generally known as (A) The near field
?
(B) The far field (C) Quiet zone (D) Induction field mcq {0.4.37 Match List I (Typ" of'Antenna) with List II (Example) and select the correct answer using the code given below the lists :
a. b. c. d.
List-I Aperture
List-II
antenna Circularly polarized FYequency independent Isotropic anteirna
Codes
1. 2. 3. 4.
Helical antenna
Point source antenna Log periodic antenna
Microstrip antenna
:
abcd
(A) 324t (B)4732 -(c)3142 (D),4 2 McQ
3
1
to.4"38 A TEM wave impinges obliquely on
€r2: l).
30' (c) 60' (A)
a dielectric-dielectric boundary (er1 The angle of incidence for total reflection is (B) 45.
(D) 75"
Mca 10.4.39 In a four element Yagi-Uda antenna (A) There is one driven element, one director and two reflectors (A) There is one driven element, two directors and one reflector (C) There are two driven elements, one director and two reflectors
(D) All the four elements are driven elements
:
2,
or abore' compared to illeQ {o"4"4o Assertion (A) : For extremely high frequenc.r'rang€s useful' linear antennas, aperture antennas a're more
Reason(R):Thelargertheeffectiveareaofanantenna,thesharperisthe radiated beam.
(A)BothAandRaretrueandRisthecorrecterrplanationofA (B)BothAandRaretruebutRisNoTthecorrecterplanationofA (C) A is true but R is false (D) A is false but R is true srritten in $ca ro.4.4t The current distribution along a traveling wave antenna can be the form (A) lZ:los-i1'
(c) l(z) :
Me&
McQ
l,
(B) l(z):lsirBz (D) l(z) : Icos(ot
-
Jz)
at UHF/ {$.4.42 Following antenna is frequently used for local area transmission VHF (A) Ground monopole (B) Tirrnstile antenna (C) Slot antenna (D) LooP antenna a t{r.4.43 For frequencies up to 1650kH2, the tra,nsmitting antenna used is (A) Parabolic dish (B) vertical antenna (C) Yagi antenna (D) turnstile antenna
Mca {0.4.44 The radiation field of an antenna at a distance r varies (B) 1lr'z (A) 1/r
(C) Llrs
(D)
Ilra
as
J
McQ 1o.4.4s The wave radiated by a helical antenna is
(A) linearlY Polarized (B) right circularlY Polarized (C) left circularlY Polarized (D) elliPticallY Polarized Mcp ,!0.4"46
In a certain microstrip patch antenna' the unexcited patch is of length tr, e. width w, thickness of the substrate being ft, and its relative permittivity . Then, the capacitance of the unexcited patch is
(L) LWle,h (C) e,LWlh
(B) LWleoe,h (D) ese,LWlh
The ueq ro.4.4? A radio communication link is to be established via the ionosphere'
virtualheightatthemid-pointofthepathis300kmandthecritical frequencyi'gNaH,.Themaximumusablefrequencyforthelinkbetween the stations of distance 800 km a^ssuming flat earth is (B) 12 MHz (A) 11.25MH2 (D) 25.5 MHz (C) 15 MHz
Page ?17
ChaP l0 Anterna and Radiating SYstems
Page
Chap
718 10
r&*{t
Radiating svitems Antenna and
ro"4.4a Assertion (A) : Program broadcast by radio stations operating in the medium wave band of 550 to 1650 kHz situated at long distance in excess of 500 km cannot be heard during day-time but may be heard during night time. Reason (R) : In the night-time, radio waves reflected from the F-layer suffer negligible attenuation since D-and E-layers are absent during the night-
time. (A) (B) (C) (D)
'
rrllco
----;
-.- of A Both A and B are true and R is the correct explaf,ation Both A and R are true and but R is NOT the correct explanation of A A is true but R is false -: A is false but R is true i.
ts.4.4s Assertion (A) : For an end-fire array, the
'
"rrrr"ni
in
successive antennas
must lag in phase. Reason (R) : Radiation of successive antennas will cancel along the axis. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false (D) A is false but R is true Mee
{s.4.5o Assertion (A) : The radio horizon for
space wave is more than the optical
horizon. Reason (R) : The atmosphere has va,rying density. (A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A (C) A is true but R is false (D) A is false but R is true McQ
to"4"5{ What is the radiation equal to ? (A) 2 o
(c) ?tncQ
0.6
resistance of a dipole antenna
o
)/20 long approximately
(B) 40 ct (D) 20 o
{0,4.52 Consider the following statements about the effective length of a half wave dipole (Elevation angle 0 is measured from the dipole axis) 1. Effective length is a function of d
2. 3. 4.
Effective length is maximum for 0 :
:
rl2
Maximum effective length is larger than physical length Effective length is the same for the antenna in transmitting and receiving modes.
Which of the statements given above are correct ? (B) 2, 3 and 4 (A) 1, 2 and 4 (C) 1, 2 and 3 (D) 1, 3 and 4
sol.uTloNs 10.{
Page 719
Chap 10 Anterna and Radiating Systens
sgL
ts.t.'i
Option (C) is correct. The borrndary between near and fat zone is defined by the antenna) as
-,0-_2d2 )
r:
ro (distance from
where d is the length of dipole.
So, the near and far zones of the field are as following Near zone for r ) rs and far zone for r ) ro
:
Now, for the Hertzian dipole of length .\/b0, we have
_
'u
Since
and
_2(^150f -
,:? , t
":*
-\--:m
,o
^
So, both the positions are
10.r.2
.l
at far zone(radiation zone).
Option (C) is correct. Operating frequency, f :0.2 GHz Radiation resistance, R,oa:31.6Q So, the operating wavelength of antenna is
:
0.2
x
10e
Hz
3x108 :7: fit-ior:1'5m \:9-' Now, the radiation resistance of the antenna is defined R'od
So,
as
: """ 80'?(4\' \)/
putting all values we get
31.6: 80i(4Y """
\)/
dl
i=o'z dr=+ i.e. Antenna is one
*o.1.3
fifth wave dipole.
Option (D) is correct. Operating frequencv. f :0.2GHz So, the operating wavelength of the Hertzian dipole is
^:?:u%XS: Now, the effective area of the dipole is defied as 'q"
r 5m
: #Go
where Ga is the directive gain and since the directive gain of Hertzian dipole
v Page 720
is 1.5sin2d so, pritblng thisvalue, we get
Chap 10
o" :(ti*Y
Autenn4 and Radiating
(1.bsin2d)
:
o.2T sinzg
Systems
Therefore, the maximum effective area of the dipole is Ae,w : 0.27 m2 (maximum value of sin0 is 1)
.
sol
{o"t.4
.
Option (B) is correct. The time average power density of the incident wave is defined in terms of received power as
(D tave- _p,
Ae
where, P, is the received power and 4, is the effective aperture area and as calculated in the previous question, the ma:
A" :0'27
m2
So, we get the average power density of the incident wave as
x 10-6 KA,,\r//-2 ,@ ave _ -1.5 0.22 _-( u.vv sol
to.{.5
Option (B) is correct. Maximum radiation intensity, Efficiency ofantenpa, Input power of antenna, So, the output ra.diated power is given P,oa :
TPin:
(0.95)
[/^* : 1W/Sr
\:95% : 0.8 Watt
P," as
x (0.q) :
0.76
Watt
Therefore, the directivity of antenna is evaluated as
: $oL
16.53
10.1.6 Option (E) is correct. Maximum rp.dia-tlpp
inlen*t{,
U^* = 1.5 {/Sr D :20.94
Directivity of antenqa; Since, the ditectiv-ity of'anterlna is delined as
{
:ry**
So, radiated power o{ lhe antenna,is giv"e+
q,s
_azr(1.b) _NS{ tprad. _
:0.QWatt eot.
{0.{.r
Qption (C) is gorregt. f,he thr.ee elpment antenpa g,rralr has the cqrrent ratio 1 :2:1
'"(''
*1, 'l:(!"
^1r-!0"
We can split the piddle elgrqent to two elqrqrerrts each of them carrying current.[s/0' as showp bgloly.
3r
1 -\/2-2.-\f2-*.
Page 721
4
Chap t0 ADterna and Radiatiag
Now all the four elements are carrying current Io /0" and separation between them are d: \12. So, this array can be replaced by two array antenna with two elements as shown below :
t,2 +
3,4
la<
Since the currents are in same phase, so the phase difference between the currents will be zero.
i.e.
o:0
and separation between the antennas as obtained from the above shown figure is
:
d,
\12
stll ls-'!.s Option (A) is correct. As shown below the three element array displaced by ),12.
1:2:l +\12-o
L:2:I "r
Now we split all'tfie elements with current F^/
a-a -'-------------'-+A/4---.\-aa\
16
as showrl below
:
l aa
\ The three current elerrrents /s located at the same position can be treated as the single element carrying current 31s as shown below :
I^
r1-r
,2
3I^ ;o
31.
r io
1'o
r
I
rn...........-io
Thus, the current ratio will be 1 : 3 : 3 : 1 of the four element array. $01
'lo.'l"s '
Option (A) is correct. Current in the dipole, t(4 : 0.bsin 108t A Lerrgth of the dipole, d,I : A1100 So, the magnitude of the current flowing in dipole is 1o
:
0'5
and from the shown figure, fie get
r : 100) and d: 60' Now, the magnetic field components at any point dipole located at origin are defined as
and
(r,lr|)
due to hertizian
Hs": I{n-g n^" : il:7dl sinoe-ia' '
47fr
where /s is the magnitude of current flowing in Hertzian dipole, dl is the length of dipole and p is phase constant. So, putting all the given values, we get
svstems
rPage 722
do
Chap 10
t)(+X#)sin6o",-(,f
"^ -- azr(100,\) j uJE _ 2 - (4 x 104)^ , 2rc - 2r X 3 L10u : 6r. o: ,:__ld_
'tQs
Astenna and Radiating Systems
As,
;r,oo^y
(.u
:
108
rad/s)
Therefore, H* : 24n*fr " + :1.1486 x 10-6epo'A/m Thgs, the net magnetic field intensity at point P will be
H:
: sol.
io.{.lo
rm(Ho,e'' ao) : 1.1486 x 10-6sin(crt + 90') 1.15sin(108' + 90') pA/m
Option (A) is correct. The field intensities of Hertzian dipole antenna are defined H^"
: il:7dl sinoe-io' 41f r
E6"
:
So, average radiated
\H6"
OtT"t of the antenna is given
po,":f
RelE,
x rr,
|
:
as
:L1#l$"r"',
the radiation intensity of the antenna is defined U(e,o)
as
as
fPoo"
:(IodI! -
So, the
32/2 ^B2sin2o total radiated power of the antenna
is
, P*a:fu@,il(sinr.dfrdfi) : I" I^ g#r1oP2sin3odfrdg
:S^o'(+) Since, the directive gain of the antenna is defined as vd -
Proa
Therefore, we get the directive gain of the Hertzian dipole antenna as
Go:ffi sol-
'to.{.{l'
:fsin'a
Option (A) is correct. Ir, : Io/0" Current in 1't antenna, Iz, : Io /180" Current in the 2"d antenna d, : \12 Separation between two antennas So, the phase difference between the two currents is 0 = 180": zr rad' The unit pattern function of a Hertzian dipole antenna (i.e., the unit pattern function of both the antenna) is where d is angle with z-axis ft(e) : lcosd I
The field pattern of /r(0)
ha-s
treen plottett belos
PagE 7:13
:
ChaP 10 Adenna and Radiating Systems
Now, the group pattern function of the two antenna is defined
as
f,(o) :cos[](0dcose + o)] is the phase difference, B is phase constant and d is the separation between two antennas. So, we get
where
a
f,(o)
:*'[;(++"oso
+
'r)]
:
+ zr)] "or[]{rr"osa This field pattern is plotted as below :
Therefore, the resultant pattern
/(d) ofthe antenna array will
be drawn by
just multiplying these two patterns l.e.
f(0)
:
tn(d)lx [/,(d)]
Thus, the obtained plot for the antenna array has been shown below
sol. {0.{.{2
:
Option (C) is correct. d:\14 Separation between the two antennas, a ---rf2 Phase difference between the currents, (i.e., the unit pattern a,ntenna dipole a Hertzian of pattern function unit The is function of both the antenna)
Page 724
Chap
f'(0)
l0
:
lcosd
where d is angle
l
This field pattern has been plotted below
Antenne and Radiating
with z-axis
:
Systems
Now, the group pattern function of the two antenna is defined as
f'(0) :cos[](6dcos0 + where
o is the phase
cv)l
difference between the currents
in the dipole, B
is
phase constant and d is the separation between two antennas. So, we get
r,(o)
: "",[](+]".,, - +)] : *,[+(+",,, _ +)]
It's null (zero) will be at 0: zr and maxima will be al 0:0'. pattern $(d) is as plotted below
So, the field
Therefore, the resultant pattern f(0) of the antenna array will be drawn by just multiplying these two patterns l.e.
f(0):tn(d)l xlf,(o)l
Thus, the obtained pattern for the antenna 6rray has been shown below
aol
{o.1",t3
Option (A) is correct. The normalized array factor for the antenna is given
(Atr!,
:;it * Netu +{{],::v
as
1 ....rnr-l)v l
where
t!:(l3dcos1+a)
:
and
:
x
1
+ N+
l(d-,l)
+
N(N- il(nr-
2)
Page 725
+
Chap L0
: (1 + lf ':2"-' (AF),:#tt + "*l*-'
so,
:
ett'tz1N-tre-t{/z
#l
: #i
cosltl2lN
Antenna and Radiatiag Systems
*
eiv/2lN-1
t
Therefore, the group pattern function of the array is
:l
f(o)
oa*'_e
+'1
"",1
l"-'
s*t- '$s.{.'r* Option (D) is correct.
- 6 mV/m : 6 X 10-3 V/m Location of point of field maxima, r : 40 km : 40 X 103 m Total radiated power is P,oa :100 kW : 10s W Maximum electric
field,
Eo,*
The average radiated power of an antenna is defined
as
Poo":]ne{a x rrJ} So, the radiation intensity of the antenna is given as
:
U(e,4)
fPo,"
: |n"1n" x rr,.) I
Therefore, the maximum radiation intensity of the antenna is
rr_*:i!_n"1n.xrr.-)
:
:ti#l
: fi{u^*f $!19#
x
(o
(r:+) x
ro{f
(rn:
r2otr)
Since, the directivity of an antenna is defined as
:4r_!^* I
D
rad
so, we
ger
x lglf x (6,x to-3l , - an x (+92xl20rxl05
:
0.0096
Therefore, in decibel the directivity is given as
l0logroD:-20.18dB $$t- 't&.'$"'ts Option (A) is correct.
consider the maximum power gain is Go and directive gain is Ga so, the radiation efficiency is defined as
or,
,1,: G" d : G, \,Gd : (0.9b)Gd
(q,:
gb%)
:
G^^*)
Therefore, the maximum power gain is
Gp,**
: :
: (0.9b), 0.95 x (0.0096) : 0.00912
(o.os)G4**
:9.12 x
10-3
(D
Prge 726
Ctap
{o.l.ts
10
Adenna and Radiating
Option (A) is correct. Minimum detectable power,
tansmitted
Sy*ems
P*i' :0.13mW Praa :30 kW : 30 X
power,
f :}GHz:
Operating frequency, Target cross section,
3
x
103
10e
W
Hz
o :7.25m2
o :1.8m since, the effective area of the antenna is 70% of it's actual a,rea so. the effective area of the antenna is Radius of antenna,
,4":# x(tra2) : (0.7) x (r x (t.8f) :7.I25m2 As the maximum range is the point where the received power is equal to the minimum detectable power. So, the received power by the target located at its maximum range is P, : Porin
:
mW : 0.13 X 10-3 W Now, the operating wavelength of the antenna is
: ^
0.13
?:, "*-i$ : 0'1m
So, the directive gain of the antenna is given as Go
4tr x (7.!ZS) :4r4" - l'?- - --lo'T- :285otr
Since, the maximum detectable range of the antenna is defined as rmu
1*l'/o lX -t6y-,
:
G'oo
where P' is the received power by the target located at its maximum range. So, putting all the values in the above expression, we get
rw
(t.zs)
: [(0.lF(2850rf
30
x 10, l'/o
[email protected]=l
: ttn'''' ^
1a.1.17 Option (D) is correct. As calculated in previous question, the maximum detectable range of radar is
tioo So, half of the range
:
584.3 m
will be at the position
,:lr"u:292.2m Therefore, the time average power density at half of the range of the radar is
(D tauc-
-
GaP,,o
_
(28502r)
:
250.35 W
4Tf
x
30
x
----A"{Lrf 1o.l.l8
Option (C) is correct. Cross sectional radius of wire Radius of the circula"r loop,
103
/^, : b:
a
6
mm
1m
:
6
x
10-3 m
: 0.5 MHz : N:10
Operating frequency,
"f
No. of turns,
0.5
x
106
Hz
Page 727
Chap 10 Antenna and Radiating Systens
So, the operating wavelength of the antenna is
3x1!::6x102m 1_c_:0.5x10" n:.f Therefore, the radiation resistance of the antenna is given as R'oa
:ltl'x :
102
SZO"t(ff
x
320
x f (=-l-"\o :2.37 x \6 x 10'l
1o-4f)
{$"1.1s Option (D) is correct. As calculated in the previous question, radiation resistance of the antenna is R,od
:2.37 x
10*4 O
So, the surface resistance of the antenna is given as
R.-^FA "vo
ffi
- \l 2.9 x 107 :2.6I x 10-4 f, Therefore, the loss resistance of the antenna is Rr"
:Nx/g\n, \0/ l-\x2.6rx1o-4 :1ox/ 10-'/ \6 x
=
0.435
f)
Thus, the radiation efficiency of the antenna is
n-.,: ,lr"o =R'oo=:0.055% - Rroa+ Rt to.{"eo Option (A) is correct. Radiation function of the dipole antenna of height h is defined
as
F(o):!g!@gFi Since, the height of dipole antenna is h
lr(d)l:l
: )/8. So, we get
cos(t.25 cos d) -
cos (1.252r)
sin d
This function has been drawn as to obtain the pattern shown below
'to.l.ai
:
Option (C) is correct. Since, the point P(0,0,1000) lies along the axial direction of antenna carrying current in o, direction, so it's contribution to the field will be zero. Now for
d
Page 72E
Chap r0 A.denna and Badiating Systems
the antenna carrying current along a, direction, we have Amplitude of the current in antenna, Io : 4A Length of the antenna, d/ : 0.1m
The position of point below:
p is r:1000
So, the electric field component E6" :
and
g:90"
(i(t) :4cosr..'f A)
as shown
in free space is defined
in the figure
as
esH6,
: ,k(W#singe:,e,) )(
- ^r'z,"4r(1000) ? lfjllsi
:
i7.2 X
as
E"
.
t
) *atrooo
"-
(u:T)
10'2 e-ilooo V
Since, rest of the components of field
field
n eo
/m will be zero so) we get the net electric
: Ee,e,e: j(1.2 x 10-re-rroooX_ :- j(I.2 X 10-2)e-irooo a,y f m
o,)
similarly, at point q(1000,0,0) the contribution due to antenna carrying current along r-axis will be zero while the electric field due to arrtenn* alon! a, will be
E,
:_
j(r.2
x
l0-2)e-itooo a"Y
l^
1o.t.22 Option (B) is correct. since, the antenna are carrying current arong c, and, a, while the point is located at y-axis so, both the antenna will contribute to the field. Therefore, summing the fields obtained due to the two antennas in previous question, we get,
E" :- j(I.2 x I0-2)e rooo(o,+ o,) in the time domain E(t): Re(Z,e^*) : (I.2 x 10-2)sin(c,,,f _ 1000)(o" + a")y /m Thus, the field at f : 0 at point (0,1000,0) is So,
E :-(9.92 x
10-3)(o,
* a)y
fm
{o.{.23 Option (C) is correct. The field component due to the current element is given
E, : Qsin Qs-ironr So, at
point
p (r:100,d :
7T12,
6
: r/6)
as
6r,
rtlU qr
:
,'o"r'oor Ssin($1e IUU \ZI : o.re ,'ooor v/m
tu*emdfrft
Sy*rr
,is.'t.24 Option (B) is correct. Since, the vertical element is shifted from origin to a point I : 0.1 on the g-axis the distance of point P from the two locations of antenna is approximately same and therefore the magnitude of field courponent, I Ee" will be same in both cases but the phase angle will change due to the change in location of current element. Su, the field intensity at point P due to the new location of vertical element is given as I
Eo*:lEs,le-11o"('-t)
(1)
where I is the difference between the length of point as shown in figure below :
P from two locations
(0.1,r12,'nl2)
P(100,n/2, n16)
Now, using geometry we get the length
I
:0.1-'(5)
I
as
:0.05
Putting the value in equation (1), we get the field component
t' s6L
4$,'1.2S
" -oo:) ),'^"'^T^7Tl,|u,
Option (A) is correct.
: /: r:
wire, Operating frequency, Cross section radius,
d,l
So, radiation resistance is given
as
Length of
^
1cm 0.3
as
:
0.01 m
GHz:0.3 x
lmm :
10eHz
10-3 m
R,oa:8ol(*l
.
: 8or'f +gFY : o o78gr o \0.:^to'/
Now. the ohmic resistance of the wire is defined
Rr:J- o2raf where
o
--r
Conductivity o - Radius of the cross section 6 - Skin depth tr - Length of the wire Since, the skin depth of the wire is given as
as
(^:
i)
T Page 730
A_ "_
Qhrp 10 Antenna and Radiating Systems
: So, we
get
Rt:
:
I
/;IW Jr(0.3 x
10'g)(4zr
3.82
x
10-6
(5.8
x
70')(2r
x
10-?Xb.8
x
107
0.01
x
10*3X3.82
x
10-o)
0.0072 O
Therefore, the ratio of the radiation resistance to the ohmic resistance of wire will be
ffi:ro'977 = lt R,oa
***:f ****rf **
\
sol.urtol{s to.z
Page 731
Chap 10 Adlennr and Radiating Systems
s$L "t{}.:.{
Correct answer is 3. Given, the operating frequency of the antenna is
I :25MHz Since, the antenna is quarter wave monopole so, the length of the monopole
antenna
will be given
as
l:U4 where
)
is the operating wavelength of the antenna given
as
3 x 108- t' ):9-f-25x106-^" Thus, we get the length of antenna
as
,12 [-
4:Jm
sol
t{t,2.2
Correct answer is 50. Given, the current fed to the antenna is
i(t)
:
83.3 cos r,.rf mA
So, the magnitude of the current flowing in the antenna is 1o
:
83'3
x
10-3
A
and from the figure we get the location of point
P
as
r:100Km:105m
and
0 = nf2
Therefore, the electric field strength at point
lpr" I
P is given
as
: (1202r)(83.3
x
10-3)cos(f cos 90')
2z'(105)sinf
_
(t2ozrX83.3
x ro-3)(r)
2zr(105)(1)
: $oL 10.t"3
b
x
10-5
:
b0pV/m
Correct answei is 25. Since, the amplitude of current decreases linea"rly toward zero atr the top the current amplitude at a height z above the plane is given as
so,
I(z):r,(r-t) is amplitude of source current and h is the height of the antenna. Therefore, the effective length of the antenna is
where
16
l"
z\,1'
lt - J,fo \' E)"'
:lr-#l^": h-.f:9:zs^
(given
h:5om)
r
ry Page732 chaP 1o
$€ll",
{*,*."{
Antenna and Radiating
Systems
Correct answer is 1.97 Length of antenna,
.
: T.bm f :2MHz:2 X 106 Hz
dl
frequency,
Operating
So, the operating wavelength of the antenna is
^:i:H-t,,3:
:r5x102
Therefore, the radiation resistance of the antenna is given as
: *,? (*l : 8or(ljTlo_J :
R.oa $r?;,
lt*.?.fi
r sz er
Correct answer is 0.4 . Since, the crtrrent ha"s the step distribution and both the current levels are distributed for eqrtal intervals so, the average current will be given as
lt+
k : #:0.75{r
I^"
Since, the average current flowing in the anterrna is 0.75 times the uniform current 1o therefore, the radiated power will be (O.ZSI times of the value obtained for 10 and due to the same reason the radiation resistance will down to (0.75f tirnes the value for a uniform current.
i.e.
:(o.zs),[sol(*i]
R,oa
:
0.5625[80t(0.03I]
:
0.4
o
s*i- $q]"x.,n Correct answer is 0.827 . Since, the dipole must match the line irnpedance.
i.e.
:
R,ua
Zrt
where Zo is characteristic impedance so, we get
/i"; :
60
sor(ff : (jl--Y :
8o',?
\c/J I
: 1$.?.?
oo 60
:[#, (r-3*%i]*
dr
ljijrr-
f)
0.827 m
Correct answer is 3.076 . Current flowing in the antenna, Radius of the circular
(/: looMHz)
,(t): b:
loop,
locos(2a.. 30 cm
:
X
107t)
30
x
10-2
m
So, we get the operating frequency of the antenna as
t :Ui#:707
Hz
The operating wavelength of the antenna is
Since,
) ))
^
:1: %s:30m
b so, the radiation resistance of the antenna is given as
D _ :^F 52 : Z20na
rtrud
__-_ n where is area of the circrrlar loop. ^9 ___r
(S:
rb2)
-q@ : $*L
s*.?.&
0.003076
Correct answer is 0.031
hF nlt ff," ll
(30I
0:
l-dfrftE
3.076 mQ
.
antenna, Conductivity of the antenna, Length of antenna, Operating frequency Cross sectional radius of
:4cm:4x10-2m
a
:2.9 x :30m
o dl
f :
107
0.5 MHz
S/m
:
0.5
x
106
Hz
So, the surface resistance of the antenna is
ost mdt-4rx loj
D /i ,rr-V - Ffw o-lW
:2.61x
10-n O
Therefore, the loss resistance of the antenna is given as
*,
: o,( ,*") : (2.61 x 10-')(. , iorro ,) :
${},. t0"x.$
0.031Q
Correct answer is 98.6 . The radiation resistance of the antenna is defined
H'od_.",,
on:z
t
as
d'l\2
\)_i
where dl is the length ofthe antenna and
)
is the operating wavelength. So,
we get
Rro,r:nr?x(+l
:
Q,:
cl.f)
: t .ot (sx10") ) -" \ Therefore, the radiation efficiency of the antenna is : 98.6% n, : VfuV,: rSz + Bo#,. / 30 x-qr 4--1CY
o-U.OSr
'xqr"*.{s Correct answer is 0.104
.
Given,
Operating frequency,
/:
Radius of circular loop,
b:20gm:20 x 10-2m
Cross sectional radius of wire,
o:5mm:5x10nrn
100 MHz
Conductivity of copper, o : 5.8 The surface resistance of antenna is given as
x
107
:
108
Hz
S/m
t) - lrfpo 1-nx ro^ x 4n x lo "'-\, o -V5f x1o-:2.61x 10-3 CI
'
So, the loss resistance of the antenna is
Rt,- lb\."" \4i **"*"1{
-
2o
x
1o
' x 2.6r x 10,3 :
5t-10-5
o.1o4f)
Correct answer is 2.265 Given that the quarter wave monopole antenna is connected to transmission Iine. So, the load impedance of transmission line will be the input impedance .
Syr-
/ Page 734
of monopole antenna.
Chap 10
Zr,:
l.e.
Antenna and Radiating
Since, the
Systems
Zn
input impedance of quafter wave monopble antenna is
Zu,:(36.b+j2r.zb)a So the reflection coefficient of transmission line is given as
zt - zo _ $a5 + p.t.z5)- 75 t, -_ Z;+Z; +r2Tjtr ?b
l3ab
:0.3874 < 140.3' Therefore, the standing wave ratio along the transmission line
is
s:1+lll "'""" " - l-lr1-:2.26s sOL
ll),2.'r2
Correct answer is 19. Radiated power of an antenna is defined tP^, rtd
:-
F=3lY 72n
as
(1)
,oB,
where .I is the current in the antenna, dl is the length of the antenna and B is the phase constant. Now, the maximum electric field intensity at a distance ,? from the antenna is defined as
lE,l-*
:(#)ry
(2)
So, comparing equation (1) and (2), we get
lEr
l.*
: +/soP^ : mjlos/9o x 0.4 x 103 t
: $oL
{o.2"{3
19
(P,.a:0.4 kW)
mV/m
Correct answer is 63.5 . Given, current flowing in the a,ntenna is :41.7cosc..'f i(t) mA So, the magnitude of the current flowing
Io
in the antenna is
:47.7 mA
Now, for a quarter wave inonopole antenna, radiation resistance is R',a
so' the averase
= 73{l
T#1,1": : ;': IITT ;--*
;
:63.5mW sot.
{o,2.{4
Correct answer is 5.03
Total radiated power, Length of antenna,
.
P,oa
:250 mW:
0.25
W
dl :0.01)
Now, the radiated power of an antenna in terms of current Is flowing in the antenna is defined as
p,aa:|trrY^,"0
(1)
where ft"oa is the radiation resistance of the antenna. Since, the current is Iinearly distributed over the antenna So, we get the average current in the
.
antenna as
r
the average current flowing in the antenna is half of the uniform current 16 therefore, the radiated power will be ]ttr of the value obtained Since,
for uniform current in equation (1) l.e.
Prad
: i(ir;
^,,,1
o.2s:fx4"*l(*l Oft
{o.z.ts
:
l"'(1ol)(o.o1f I3 :25.33 1o : 5.03 A
o.2b
Correct answer is L4.2
.
:0.02\ P*a:4W dI
Length of antenna,
Total radiated power) since, the monopole antenna is extending over the conducting plane so, the power will be radiated only over the upper half space a,nd therefore, the radiation resistance of the antenna will reduces to half of its value
i.e.
R,od:1[*",(*)]:
*r(*l
As the current is distributed linearly. So, the average current in the antenna is
I*n:+ the average current flowing in the antenna is half of the uniform current 16 therefore, the radiated power will be ftir of the value obtained for 10. Since,
i
e
:i(*ll**(*ll I 2x4 2P'oa lr/2:['#Gfu] ' -[
p..d
1r/2
or,
":1,*61
{0"2.16 Correct answel is 3.28
:t42A
.
For a quarter-wave monopole antenna pattern function is
f(0)
:
cos[(zrl2)cos d]
sin 0 So, the normalized radiation intensity of the quarter wave monopole antenna
is given
as
u(0,0)
: f (0) _
cos'z[(?r/2)cosd] sin2d
Therefore, the maximum radiation intensity is
[/*:1 Now, the power radiated by the qua,rter wave monopole antenna is evaluated as P,od,
Page 735
Chry r0
Io ,aug __ 2
: f {ute,Ol}{sinodad|}
ArmrdBrtrg SFL-
'\&
V
Page 736
: I','' I^ 2{E:y-rlsiuodfrd.Q : (lzr)(o.ooo)
Chap 10 Antenna and Radiating Systems
Therefore. the directivity of quarter wave monopole antenna is
p :AnlrnI r,trl
: ts.*.'t?
@;ffioel:3'28
Correct answer is 1. As the radiation intensity in all directions are sarne so,
U(0,0)
:
r1o,"
where, U(e,d) is radiation intensity in a particular direction and [/", is the average radiation intensity. So, the directive gain in a particular direction is
G,(0.6\ -o\"\vr
:ulT'q) -r - uou" -- au u^,,.-'
Therefore the directivity of the anterrna is
D : G1,n*:! silL
**"4.{E
Correct answer is 2.546 FYom
.
the given value of radiation intensity, we get maximum radiation
intensity of the antenna
as
U^o:I So, the radiated power of the antenna is evaluated as Prod
:f
u1e,|lr"inodfrdo
fn/2 rhr
J L' : J'=' J'!3'"t)(sinodfra61
'.
' {s,*.{s .
,
:
Therefore, the directivity of antenna is
D
_4rV_*:n\(!.):2.b46 !'u=v p*,t -
,t
Correct answer is 4. Given, the field pattern of antenna,
the,",.,,,Jj:::ok", f2r
",:f";ffi
ft/3
P-a:lJ6:o Js:I : 4x :4n
.
0<0
[4 so,
lz -
is given
as
U(O)sin?dflrla
s
2trf_
cosd][/3
!:
Therefore, the directivity of the antenna is
D ${3f-
: 4T!-:4n.(4) : P*o 4tr -
o T
,$.?"9$ Correct
answer is 3. The field intensities of the Hertzian monopole are defined
and
D ne'
:,nhljdt. -frr
H,. '
," : il:frdl 4Tr sinoe
^ slnae''"'
,tr
as
So, the time
averageportrofem Po,":|n"1a x 8".) :+(#f
Now, the radiation intensity of the antenna is given
:
u(o,o)
fPo,"
as
:ffi^O2sin2l
So, the maximum radiation intensity is
u^*:Srn|t
(maximum value of
As the radiated power of an antenna is given P,oa
sind:
1)
as
: f U1e,611"i"0dfrd,4) <
where, the integral is taken in the range 0
0
i,0
<
< d <2n for Hertzian
monopole. So, we get P'oa
: t"r' 1^ U@Iqsl32sin2lsinldld'o
S
:
o'
^
"",t"', [(,f
-)(I';r)]
4tr1 :127'70'" - UratY ^ pz |l\T/ Since, the
directivity of an bntenna is defined
D
:
as
bp,^
So, putting the values obtained above we get the directivity of Hertzian monopole antenna as
n 1a,p.zl
: (+r !T(l), :z l3)
Correct answ€r is 2.83 Current arnplitude
.
Operating frequency, Effective length,
1o
:50A
.f
:
180
kHz
I:20m
-
180
x
10e
Hz
-R:80km:8x10am Location of the observation point, So, the maximum field intensity at the observation point is given as lE,l**
:ffi,^g:ffin|x2t"
As, the operating wavelength is
^:1:rr'#%:+
and so the phase constant is
A:+:ft*G:72tr
x
10 {
Therefore, the maximum field intensity at the observation point is
l&[* :
:0.002827
t*.z.zz
x
4" x&l'
:
2.83
(1202r)
x
L2r
x
10-4
x2x
20
mV/m
Correct answer is 0.43 . The time average power density of antenna is defined
as
J
Page 738
Poo":**"{t, x rr,-}
Qhap 10 Antenna and Radiating
So, the time average radiated power is given as
Systems
** :-t;
i' r'?r,A" J; ":ff',:':*
:+I"''
I*^(M#4f "
:j*,""r,(Mi&l
eonao
[li"en
:1r X l20tr x/50 x I2trx10-ax2x20 .2 4n \ )"t : 426.37 W : 0.43 kW $oL
to.2.as
Correct answer is 0.34 . As calculated in previous question the time average radiated power is
P,oa:0'43kW Amplitude of the current in the antenna is 1o
:50A
So, the radiation resistance of the antenna is given as
p _2p,"0 _ 2 x 0,43_x 103 :0.34e tLrad,__F:___lsof__:L sol. 'to.2.24 Correct answer is 0.71 Radiation resistance of a short circuit current element is determined
R,ad,:8ol(*l R,,joa
where
as
I is the length of dipole
:8ol(a*l :2.84e
But, as the current is not uniform so, we determine the average current through the element. Now, from the given expression of current in the
'\'
element, we get
, tI+22\ tr(r) -,0\--l-/ and
lr(z)
for
: ,^11--221 "\ I )
!<
for0<
z<
0
r=t
Therefore, the average current in the element is given as
.
r _-1,(z)+Iz(z) __2__
^orn
since, the average current flowing in the antenna is half of the uniform current 16 therefore, the radiated power will be ]th of the value obtained for rn and due to the same reason the radiation resistance will down to th of its value. -1,
i.e. $sL
LR^o),",
:
IR,"o:
I " 2.g4 :0.71 f)
1a.2.25 Correct answer is 1.26 . Length of antenna,
d,l
:2
cm
:
0.02 m
-T
-{power, Operating frequency, Radiated
:2W .f : 0.6 GHz :
P,oa
page ?89
0.6
x
10e
Hz
QhaF 10 Antenna and Radiating
So, operating wavelength of antenna is
^
:
?:
*is:
svstems
0.5 m
Therefore, the radiation resistance of the antenna is given as
R.ad:so",(*i
:8o",(%fl:W
As the radiated power of the antenna is defined as p,od,
:l{rrYn,",
:(1,.^.,.YR,oo
So, the rms current in the antenna is
: tl ffi: Ir.^".:E;:E | 6e76:1'26A ***********
(
r,.^.":
tol/z)
$ol.urlo].|$ {o.3
Page !4Q
Chap 10 An{enna and.Radiating Systems
sol.
18.3.{
$oL
10.3.2 Option (B) is correct.
sot-
to.3.3
Option (A) is correct.
$oL
'to.3.4
Option (A) is correct.
sor-
{0.3.5
Option (B) is correct.
sor-
"to.*.s
Option (A) is correct.
sol-
{0.3-?
Option (A) is correct.
Option (C) is correct.
sol ro.3.8 Option (C) is correct.
***
t
***
V
$otuTloN$
lg.4
Page741 Chap 10 Antenna and R^ailieting Systenis
sol to.4.t Option (A)
is correct.
The directivity of an antenna is defined
as
D:W U o.r.
where
U-* is the maximum radiation intensity of the antenna and (,,"
is
the average radiation intensity. Since, the given antenna has the radiation pattern
tJ(0):gostg
('
So, the maximum radiation intensity is
U**:1 The average radiation intensity is
rT Uo"
:- ii1 Jf F@'d dQ
: *ll'.
I'h(e'')'in',*
*l
l l f'" ["'' uur u''v@@t/] -- [ilJ' I "ororsir'/,dfrdb] 1 lr-l costdll"/' _ - 4nl'" \ il/-|. 1 t'- z"[-o+Jl -- 47r Lttl " ' 5l | ,.2tr :4trn5:m 1
Therefore, the directivity of the antenna is
D:#:10 or, scl.
{4.4.3
D(in dB)
:
:
10log 10
10
dB
Option (C) is correct. The beam-width of Hertzian dipole is 180" so, its half power beam-width is 90'.
sol"
{0.4.3
Option (D) is correct. The operating wavelength of the antenna is 3 1_c_3x108_ :fr;fto:2g-o o: f
U:2oGHz)
Therefore, the gain of parabolic antenna is given as
G'
: 'lf (K) :
or,
l0logro
0.7
x
,?
(+l: 30705.4 \lod/
(efficiency, rt
:
70%)
Gp:44.87 dB
/
-=
Page 742
QheF 10 Antenna and Radiating
sol 10.4.4 Option (B) is correct. Using the method of images, the configuration is as shown below
o
Systems
T
\/2
Perfectly conducting infinite ground
I
T
\/2
I
Here
d: ), 0 :
zl, thus,
pd: hr
So, the array factor of the antenna is given as
A.F.:
: sol {o.4.$ Option (A)
*,elorfltl
"o"l4rylal : si'(n'"or,r/)
is correct.
at conducting ground. So, the power will be radiated only on the half side of the antenna and therefore, the radiation resistance of the antenna will be half of its actual value and given as Since, the antenna is installed
R oa
sol to.4.6 Option (A)
: il*,t rw;l : 4ol(0#1orl : T"
is correct.
The array factor ofthe antenna is defined
Here,
: *'(P4ryjt) d: +
and
o :90"
as
A.F.
rhus,
A'F'
: ."(i$ry) :
"or(1,
ir.l +
+)
(u
: T)
The option (A) satisfy this equation.
scl {o.4"? Option (A)
is correct. The directive gain ofan antenna at a particular direction
Go(0,6):gW9
(d,/)
is defined as (1)
Since, for lossless antenna Proa
So, we
get
:
Pin
P,od :4,:
1 mW
Again the directive gain of the antenna is given
l0logGa(0,5) :6 dg So, Gd@,Q):3.98 Putting it in equation (1) we get the total power radiated by antenna 4trU(0,Q) : P,oaGa(0,d) : 1m X 3.98 : 3.98 mW
sol to,4.8 Option (D) is correct. Normalized array factor is given
A.F.
: zl"o.4l '1"""2
i
as
as
.
-\
-r/
: Bilsin?cos@ * d 0 :90',
where,
t[
: J2 s, 6:45",
page 74t
chap
1o
Antenna and Radiating
d,
'
svstems
6 :180"
so,
A.F.
: ,l*"tl : r".'[@ry#@] : z"orfffliscos+5' . ryl :2cos[T+eo'] :2ri"(T)
$er1
t6.4.s
Option (B) is correct. The signal strength (power) at a distance proportional to the distance r.
i.e.
from an antenna is inversely
P-+r
So, '
r
+:4 d:4
(1)
Since, 3 dB decrease -+ Strength is
Therefore, f; :
halved
(10'zto: 100'3: 2)
z
Substituting it in equation (1), we get
t- -ri s2 :|r/i rz km :
or
(n:5km) 7071 m
Thus, the required distance to move is
d
: rz- rt :7071 -
5000
:
207I m
sor- d0.4,.16 Option (B) is correct. We
), :492 m
have
and height of
antenna,
d,l
:124 - = *
So, it is a quarter wave monopole antenna and radiation resistance of a quarter wave monopole antenna is 36.5O. sot-
10.4.'l'l Option (C)
Wehave ' where
is correct.
{:pdcos?*6 a :4 4
(1)
Distance between elements
t :0 0 :60"
Because of end fire
Putting all the values in equation (1) we get
o:Txlcos6o'*d:$x!+a
or
6
:-t
$01 ,o.4.'12 Option (B) is correct. For a dipole antenna we have
BW cr --=-L-
(Diameter)
/
Prye 744 Chap lO A!0err[a aDd Rediating
So, as diameter increases Bandwidth decreases.
sol- 10.4.13 Option (D) is correct.
Systems
Fa,r field region for an antenna is defined for the distance
r
from the antenna
as
^_2d, ,-^ where d is the largest dimension of the antenna and ,\ is the operating wavelength. Now, the operating wavelength of the antenna is given as
or
:
c 3x108
T: zt-lbr:
3
40
-
So, for the closest far field we have
1s.4.'14 Option (C) is correct. We know that for a monopole its electric field varies inversely with 12 while its potential varies inversely with r. Similarly; for a dipole its electric field varies inversely as r3 and potential varies inversely as 12. In the given expression both the terms (i + +) are present, so, this potential is due to both monopole and dipole.
{0.4.15 Option (D) is correct. Power received by an antenna is defined as
P.'
:4o x A" Atrt'
where Pr is the power radiated by the transmitting antenna,
r
is the distance
between transmitter and receiver and A" is the effective apertufe area of the receiving antenna. So, we get
E, _ 25I x 500 x
,4"
10-4
: 500 Qmr,r :'i;;; : q##Fr:
100 p,w
1o.4.{c Option (C) is correct.
i
i
Magnetic field intensity in terms of vector potential is defined
H
:Lv
where ,4 is auxiliary potential
So, and $oL
as
x,4
I,L
l:
!
''
,function.
.
V . H - V . (V x A): g Vx.EI:VX(Vx,A) l0
,to.4.{z Option (D) is correct. Radiation resistance of a circular loop is given
as
R,od:$rn["#] i : Rroaq N2 where N is number of turns. Since, the radiation resistance of a circulalloop is 0.01Q.
i.e.
:
Xa
so, we get
tle
n€r
hP
(LOIO
redidin nsi*rc
fuz: Nt X 8,1 :
of tbe fite turns of such loop as
(S)t
x
0-01
(l{:5)
:0.25O
io.4.{a Option (C) is correct.
power Aperture area of a receiving antenna is d€fin€d in t€rms of received
Aperture
Area:po _Etr -
Since,
P,
:+
So,
A" ^
10-6 3.14 Lzv x 12ox :-2x - {ZO-2 x>< 1o-6".. -Z1lT-: rgq2 ^
E12
I
a"s
"fi"ffi
p
A ne
Un: t20n is intrinsic irnpedance of space)
\-6 /
2x10-6x12x3.L4:1.884m2 :----_-.8-400 x 10 sol. lo.4.t9
Option (B) is correct' defined Maximum usable frequency(/-*) in terms of incidence a,ngle (i) is
as
J*:+ t
sln
where
$ is critical
?
frequency. So, we get
"f_*:ffi:,fu,:ftrvr*, \-r-i
sol-
'10.4,20
Option (A) is correct. Far field
sot.
10,4.2t
o1
r
Option (B) is correct. The maximum usable frequency is given
as
J-:+ t
sln
where
i
20 X 106
of, ort
soL {0,4,22
?
is launching angle and
srnu
:
10+ sln
"6
is critical frequency bo, we get
106 ?
:21
i :30"
Option (B) is correct' The directive gain of half wave (.\12) dipole antenna is given
as
cos'($cosg)
6, So, the
:1.66--ffg
directivity of the arrtenna is D : Ga,*
Since, the maximum value of the function of. ),l2long wire antenna is
D
: :
1.66(1) 1.66
=##0).
7{5
Cb4 r0
is 1' So, the directivity
A*errna and Radiating SYstems
7 I I
Pase
chap
7a6 10
Aatenna and Radiating
sysrems
sGL
{o.4.r3 Option (D) is correct. Since, the EM waves are travelling in free space, So the phase velocity wave will be equal to the velocity of light in free space.
ofthe
: c So, at frequency, .f : 63 MHz (Channel 3) 3x108 :4'76m : i:c ffit-iot wavelength, ^ So, phase constant,B :T:1,.32rudf m i.e.
ap
:
and at frequency, "f
803
MHz (channel 69)
:1- ES*%:
waverength, ^
:*
So, phase constant,p
o.3z4m
:16.82 rad/m
sct '18.4.?4 Option (D) is correct. Since, the antenna is located at earth so, power radiated to the hemisphere will be half of the transmitted value. 2a%tW loo kw P, :
i.e.
:
+:
Now, the average poynting vector (power radiated per unit area) at a r from the antenna is given as
distance
P*":4u Trwhere
a, denotes the direction of Poynting vector.
So,
for
r:50km,
we
have
P..":#S:& *:*o,trw,o.' sst xo.4.25 Option (A)
is correct.
Radiation resistance of a dipole antenna is defined R,od
as
:sol(*l
...(1)
Given, The length of d,I :5m operating frequency, .f :3MHz :3 X 106H2 So, the operating wave length of the antenna is given as
dipole,
),
t 14: loo m :9: J ?3x10"
Putting these values in equation (1) we get R,od,
:80l(#l :#
:r.e7
- 2e
sal. 1o.4.zc Option (B) is correct. The radiated power of an antenna is defined
as
p,aa:ir3^,,0 i.e.
P,oa
d R,oa
Now, the radiation resistance of the antenna is given as
l.e.
R,od,
:80r?(4\' --"
nrto
o
\^/
9
...(i)
rx:i
Since So, we
get
R,oa
q
f
Page 747
Chap 10 Airtenna and Radiating
...(ii)
(d,lY
Combining eq(l) and (2) we conclude that
p,aa. (a$(yf Now, for the 1"t antenna we have
(drx/) :
(1.5X100
x 106): 1.b x
108
for 2"d antenna
(d,t\(l):
(15X10
x 106):
1.5
x
106
Since, the product of length and frequency axe same for both the antenna So, the power radiated by both the antenna.s will be same.
$01 1s.4.2? Option (A) is correct. Given the length of current element, l:0.03,\. So, the radiation resistance of the system is given
R'od:80l/l\' -- \)/ : 801/0.q3.\\2 :
\^
I
0.072)2
a.s
f)
sol 't0.4.28 Option (C) is correct. In a three element Yagi antenna there are one
'
reflector, one folded dipole (driven element) and one director. The length of reflector is greater than driven element which in turn is longer than the director.
sol. {o.4.ze Option (B) is correct. Antenna arrays are formed to produce a greater directivity i.e. more energy radiated in some particular direction and less in other directions.
sol to.4.3o Option (A)
is correct.
Input power
:W
Radiated power
-w
Radiation intensitv
-6
So,the power gain of the antenna is
G,: 4nd w
(a-3)
Ga: 4"P W
(b.'a)
Directive gain of antenna is
Average radiated power of the antenna is
t,o _w, -G
(c-2)
The efficiency of antenna is W,l:fr.
sor.
(d-+
1)
{o.4.3{ Option (B) is correct. Maximum radiation for an end fire array occurs along the line of the array.
$oL 10.4.32 Option (A) is correct. The gain of antenna is directly proportional to the aperture a.rea. So, with
Systems
t
'
increase of aperture area, received powcr increases and therefore the gain
248 ChaP l0
page
increases.
Adcnna and Radiating
Systems
BoL
{o.A,ss Option (A)
$01
10'4.34 Option (A) is correct.
is correct. In the helical antenna, normal mode of operation is very narrow in bandwidth and therefore the directivity is high. while the radiation efficiency is low.
For an antenna near and far zone are specified by a boundary defined
as
:
a,r:;tl*'*;"".,
where, R is the
a
i, tt lirgest
dimension of antenna
" and ,\ is the operating wavelength of antenna. so, any targetJocated at a distance R> + from antennu ir iti the far zone for the antenria and any target located at a distance R < $ is ln the near zone.
soL 10.4.35 Option (A) is correct. Gain of transmitting antenna,
: l0 Pr : 1W A", : !m2 r : lm
Ttansmitted power, Effective area of receiving antenna, Distance between transmitter and receiver ) so' totar
t
Gat
received';::'#:r:,,;""":'Hlffx (1)
: s.7s1ry
sol. to.4.36 Option (B) is correct.
'
that on2
ff sol-
fro
,, +
antenna so, the defined region is far zone or far field'
to.4.37 Option (B) is correct. Helical, antenna is used to provide circularly polarized wave and the log
. $oL
is called far zone for the antenna and as it is given in the Ftaunhofef'region measurement to be taken from a distance of
Since, the region
periodic antenna is frequency independent.
{0.4.38 Option (B) is correct. For a wave travelling from medium 1 to medium 2, the incidence angle 0" of the wave for which it is totally reflected by medium 2 is given as nrsind"
where
??r
- nesin9O"
(1)
and rh are the refractive index of medium 1 and medium
2
respectively. Since, refracting index of a medium having
p is defined So, putting
\
as
n:rlG it
in equation (1), we get
,/1a€rsino": {t,o€, sinl, _,{?r: g"
\,\
permittivity e and permeability
:
sin
G
'1*\: \J2 )
as'
-\ tol to'ir.s @ion (B) is correct.
Page 741
Chap 10
Anr-dlcting W.ms
Folded dipole (Driven element)
Directors
Yagi-Uda antenna must have one reflector and one driven element while it can have any number of directors. So, the four element Yagi-Uda antenna will have 2 directors, one reflector, and one driven element.
sol-
1s.4,40
Option (A) is correct. Directivity of an antenna is directly proportional to the effective a,rea and therefore larger the effective area, sharper the radiated beam. This is the reason for using an aperture antenna instead a linear antenna for extremely high frequency ranges.
sol
10,4,4,1
Option (A) is correct. Current distortion along a travelling wave antenna in general is defined as I(z,t): Iscos(cot- Pz) but when we eliminate f by taking it's phasor form, the current can be written as
4") :
[os-iaz
sol to.4.42 Option (B) is correct. T\rrnstile antenna is generally used at UHF/VHF for local area transmission.
10.4.43 Option (B) is correct. The frequencies upto 1650 kHz is in the range of medium frequency. Vertical radiators ranging from .\/6 to \15 used for broadcasting the medium frequencies as the operating conditions and economic consideration.
to.4"44 Option (A)
is correct.
The radiation field intensity of an antenna at a distance
n^ ue"
or
hIoAil _ - _Ti_r
r
is defined
as
sin1e-,o,
tr^t-rlluldtsinl I ---Ttrr oI r
uos
sol" {0"4.4$ Option (D) is correct.
f,,.r .
.r,
,111.
:r
'
The resultant field irr a helical antenna is either circularly polarized or elliptically polarized depending on the pitch angle a. *, ",flbe rad,iryt"@!il9$dlb$ a,.helical antenna is circul*tlf;.polarized only when, : ...t ._. ,*,, .:,r.
,"r;,,,#:Tftffiffih" 'i!!f
\
/ irrt,
else it is elliptically pcilarized. Irr conclusion for a general term we can say the wave radiated by a helical antenna is elliptically polarized.
/
vr
r 3e 750
nr0 ll
bna
tm
18.4.46 Option (D) is correct. The unexcited patch is shown below
and Radiating
h
The capacitance between the plates is given as e(Area of plates)
C_
(separatoin between plates)
€o€,LW -----T
sol {0.4.42 Option (C) is correct. Maximum usable freeuency betweentwo stations of distance D is defined
frur: Q+(#l
where h vertical height at the mid point of path and we put all the values to get,
I*r,:9 " to'vf{#W
f, is critical
as
frequency
:1.b x toTHz:15NIHz
10.4.48 Option (A) is correct. D-layer is the lower most region of ionosphere which is present only during {he day light hours and disappears at night because recombination rate is highest and also E-region is weekly lonised during night hour hence radiowave suffer negligible attenuation in night hour. This is the reason that the wave band which can't be heard during day time but may be heard during night time.
.r*_
to.4.4e Option (C) is correct.
-
to.4.so Option (A) is correct. The atmosphere has varying density (refractive indexj with the height from earth given
*
#.Rudiu,
o',
R:-4
"rrrr"r,rr"
of the wave
J.,n*
dlt'
Solving
it,
we get the effective earth radius (Radio horizon)
: $ actual earth radius (optical
horizon)
1o.4.st Option (A) is correct. The radiation resistance of a dipole antenna is defined
as
Since,
So,
R"ad
:sO"'(ff : z O
sol. {o.4.s2 Option (A) is correct.
i---;
t-
I
"
\ tsctive
fr*ato
Iength of e t"
:
(0)
of.
Atr:Ef
:''
0.
Lis *l 0:
-fr:*. 1.e.
maxlmum
The effective modes. So, statements, 1, 2 and 4
* bbrDdrd*.h
+ h benmitting
&e
lffie
and receiving
$atement 3 is incorrect.
I
-T Efiective length of a half
Page ?51
ehrp l0
r"(0)
futin
:?[*'!-rr;")]
of d. nh of I, is at
Antenna and Radiating SYstems
0: rl2.
tl:3: ) ' 2 A
f
b
h
than its actual
o'"I* ).
for the antenna in transmitting and receiving modes. So. statements,
1-,
2 and 4 and'ffid shile statement 3 is incorrect. ****+*xx{<**
