Solutions Manual for
ELECTROMAGNETISM: PRINCIPLES AND APPLICATIONS Paul Lorrain/Dale R.
Corson
by Paul Lorrain University
of Montreal
W. H. Freeman and Company San Francisco
Copy right
©
19 79 by Paul Lorrain
No part of this book may be reproduced by any mechanical, photographic,
or electronic
phonographic recording, sy stem,
transmitted ,
p rocess,
nor may
or in the form of a
it be stored in a retrieval
or otherwise copies for public or
private use without written permission from the publisher. Printed in the United States of America ISBN
9
8
0 7167-1105-2 7
6
5
4
3
2
1
Contents
iv
Notes Chapter
1
1
Chapter
2
7
Chapter
3
11
Chapter
4
15
Chapter
5
20
Chapter
6
28
Chapter
7
34
Chapter
8
37
Chapter
9
40
Chapter 10
44
Chapter 11
49
Chapter 12
52
Chapter 13
56
Chapter 14
62
Chapter 15
66
Chapter 16
67
Chapter 17
71
Chapter 18
80
Chapter 19
85
Chapter 20
87
NOTES 1.
S o a s t o s ave s p ac e , s imp le mathematical exp r e s s ions are typ ed
on a s ingle line .
Whenever the order of the operat ions is not indi
cated explicitly by means o f p arenthes es , they are performed in the following order : multipli cations , divis ions , add i tions and sub trac t ions . Examples : l / ab
1 l/a + b ab '
l/(a + b)
l / ( a + b / c + d e / fg)
1 + alb
2.
l + b , a
1 a+b '
1 b de + a + c fg
Pro grams for drawing t h e curves in thi s Manual wi th a HP 9 8 2 0 cal
culator and a 9 8 6 2A p l o t t er are availab le free o f charge from the unders igned. 3.
Ref er enc e is made , occas ionally , to a tab l e of integrals by
Dwight.
The full reference i s "Tables o f Integrals and Other Mathe
matical Data" by Herbert Bri s t o l Dwight (Macmillan) .
P aul Lorrain Departement de phy sique Univer s i te de Montreal Montreal , C anada.
CHAP TER 1 1- 1 ( 1 . 2 )
->--+
A . B = AB cos 8
9
x
4 - 6 - 30
0
1-2 ( 1 . 2 )
->-->-
A . B = AB cos 8
2 - 18 + 1 = - 1 5 ; -0 . 650 , 8 =
cos 8
� ( 1 . 2)
-+ -+
-7-
=
A On ,
A. B + A. C
=
A Om + AO'L
-
1-4 ( 1 . 2 ) C2
=
=
D2
=
C 2 +D 2
=
+ -+
-+ +
=
A
x
l
-+ +
-+--+
C.C A . A + B . B + 2A. B 2 A + B 2 + 2AB cos 8 A2 + B 2 - 2AB cos 8 2 ( A2 +B 2 ) , C 2 - D 2 4AB cos 8 =
-+
-
B
-+
B is normal to the p lane of A
->-
and B .
2 3.1
A On
1-6 ( 1 . 3 )
->-
�
( 4+9 + 1) 2 ( 1+3 6 +1) 2
-
A. (B + C)
-+ + -+ +
130 . 5 0
AB
I ts magni tude is the area
shown hatched .
Then
I CA ->-
x
B) . c l
->-
->-
is the b a s e of the p aralle lep iped , multiplied by i t s height , or i t s
-
A
1
->-
->->Similarly , A . ( B x C) i s also the volume o f the parallelep ipe d .
volume .
1- 7 ( 1 . 3 ) The x-component is Ay (B z + C z ) - A z (B y + Cy ) ( AyB z -Az B y ) + (Ay C z -A z C y ) or ' the x-component of A x B -t A x C. The same app lies to the y- and z- com ponents . =
l - S (1 . 3) For the x-component , ay ( bx cy-b y cx ) - a z ( b z cx-b x c z )
b x ( ax cx+ ay cy +a z c z ) - cx ( axbx +ayb y +a z b z ) The corresponding equations for the y- and z- components can be found by rotating the sub s crip t s . =
1-9 (1. 4) d; /dt is perpendicular to; . Then r is a cons tant . 0 2 ; . (d; /dt ) (d/dt ) (; . ;) 2 dt) r 2 r (dr/dt ) (d/ Then dr/ dt 0 and r constant .
Also ,
=
=
=
1-10 ( 1 . 4) x = 5 00 ( cos 30 0 ) t , y 500 (sin 3 00 ) t-4 . 90t 2 2 5 0 t - 4. 90t 2 433t 2-r -r ->r = 433 t l + (250t - 4 . 90t )J , ->->--> v = 433 i + ( 250 - 9 . SOt) j , a -9 . S0j =
=
=
1-11 (1 . 5 )
V ( A . ; ) ='I ( Axx+Ayy +AZ z ) 1-12 ( 1 . 5 )
=
('O/'ox) ( ) i+ ('O/'Oy) ( ) j+ ('O/'Oz) ( )k=A
(A. V ) ; = Ax (d/'Ox) +Ay (d/'Oy) +A z ('O/'OZ) (xi+ yi+3:k ) = A
]
[
1-13 ( 1 . 5 ) a ) 'I' (l /r) = i ('O/'OX') ( l /r) + j ( 'O / 'Oyl) (1/r} + k (d /'Oz ' ) ( l /r) where r (xl. =
[
2
Now (d/dX') (l /r) = - (1/r 2 ) (dr/dx ' ) = - (1/r 2 ) (x ' -x) /r= (x-x ' ) /r 3 By symmetry , (d/dY ' ) (1/r) = (y-y ' ) /r 3 , (d/dz ' ) ( 1/r)
=
(z-z ' ) /r 3
Since x-x ' is the x-component of i, and (x-x ' ) /r is the x-component of i1 , etc , V' ( l /r) i1 /r 2 =
b) In this case , V (l/r) =t (d/dX) ( l/r) + j (d/dY) (l /r) + k (d/dZ) ( l/r) (d/dX) ( l /r) = - ( 1 /r 2 ) (dr/dX) = - (1/r 2 ) (x-x ' ) /r =- (x-x ' ) /r 3 and similarly for the other derivatives. Then V (l/r) = -i1/r 2
1-14 ( 1.5 )
-+i
-+j
k -+A x -+r x Vf Z Y df/dY df/dz df /dx -+ -+ -+ -+ + b ) A.r = ( r x Vf) ·r is zero , s ince r x Vf is perpendicular to r. -+ c) A' Vf is zero for the same reason. a)
=
=
-+-
1-15 ( 1. 8) a) V.i = (d/dX) X + (d/dy ) y + (d/dZ) z=3 2 b) The flux of r is r . r 1 4'1Tr = 4'1Tr 3 or , using the divergence theorem, for a sphere of radius r ,
-+ -+ -+
-+ 3 fs-+r.r-+1 da Tf V.rdT=4'1Tr
1-16 ( 1. 8) v. ( fA) = (d/dx) ( fA ) + (d/dy) ( fA ) + (d/dz ) ( fA ) x Z Y =cf x Z x Y Y = fV.A + A. Vf
-+ -+
1-1 7 ( 1. 8) a) fA = r 2"K , v.
(fA)
=
=
Z
2 2 2 Cd/dX) (x +y +z ) = 2x, etc 2 2 2 (d/dX) (3xr ) + (d/dY ) (yr ) + (d/dz) ( 2 zr ) 3r 2 + 3x 2x + r 2 + Y 2y + 2r 2 + 2 z 2 z 2 2 2 2 6r + 6x + 2y + 4 z 12x2 + 8y 2 + 10z 2 120
3
b) Vf = V r 2
=
Cl/3x)r 2+i +(3/3y) r 2+j +(3/3z) r 2+k
2 +r
=
V.A = 3 + 1 + 2 = 6 + . (2xi+ 2yj+ 2 zk) + + Vf 6r 2 3xi+ yj+ 2 zk) + fV. A+ + A. + + + + ( 6r 2 + 6x 2 + 2y 2 + 4z 2 = 12 0
c)
[v . (fA)] [l/L] [LL2] = [L 2] =
So V . (fA) is exp re s se d in meters squared .
(H/2 ) (l-x/R) dydx
-R
J
J
J+R(H/2 ) ( 1-x/R) 2(R2 -x2 ) 2 dx �
=
+R 2 2 � +R 2 2 � = H (R -x ) 2 dx - (Hf,R) (R -x ) 2xdx H (1I/2 ) R2 - 0 = H1IR2/2 -R -R 1- 19 ( 1 . 7) Calculate the volume in the octant where x, y, z are all posit ive . 2 2 2 2 R (R2 _ Z2 ) � (R2 _y _ z ) � R (R _Z ) 2 2 1 V/8 = dX dY dZ= (R -Y -Z ) 2 dy dz o 0 0 0 0
J J
-
=
J
JR(11/4) (R2 -z 2 ) dz
o
J J
�
(1I/4) R3 - (11/4) (R3 /3) = (1I/6) R3
1-2 0 ( 1 . 9)
J v . A dT JA. cia =
T
s
Now V . A = df(x) /dx , and A. cia is zero on the cylindrical surface . the cross-section of the cylinder is B ,
t Cdf(X) /dXJ B dx = [feb) - f(a) ] B
a
Thus b
j Cdf(X) /dXJ dX = feb ) - f(a)
a
4
If
1-2 1 ( 1 . 10) --
Set F = K/r 2 .
1- 2 2 ( 1 . The work curve is Then the
Then PE
-K/r
10) done b y zero , even taking into account the curvature of the Earth . gravitational field is conservative .
1-23 ( 1 . 12) Since the field is conservative ,
tit . dt + rit.di
a over P
b over
Q
a over P
a over
Q
1-2 4 ( 1 . 12 ) Since the value of the integral is independent of the p ath , the field defined b y the 1-25 ( 1 . 12 ) S ince the force is azimuthal ,
f 1.di = 2rrrF
".
°
and , from Stokes ' s theorem, -i>\7x F '" 0 , s o the force is nonconservative . y The curl is calculated as follows: 2 2 0. 5 Fx =- Fsin e=- F y F y = Fcose=Fx/r = Fx/ (x 2 +y where F K (x 2+ i) o. 2 . So -+ -t i k J ..... VxF = a / ax a/a y a/ az Kx Ky ( x 2 + y2 ) 0 . 3 ( x 2 + y2 ) 0 . 3 ° 0. 3 Kx = 2 2 x + 2 K2 0 . 3 2 2 1 . 3 3 O . (x + y x + (x + y ) =
tc
»
[
2 y +2 = 2 K2 0 . 3 2 _ 0 . 3 ( 2X (x + y ) ( x 2 + y2 )
. 4K k k -_ 10.6 r 5
]
0. 3 K y 2 k Y ( x 2 + y2 ) 1 . 3
1-26 (loll) ->-
i IIx A = a/3x xf(r) ->-
it
7
J 3/3y yf(r)
3/3z z f(r)
[z(3f/3y) - y(3f/3z)] i +
Nowof/3y
=
(3f/3r) (3r/3y)
.
• •
(3f/3r) (y/r) ,3f/3z
=
So IIx A = [(zy/r) (3f/3r) - (yz/r) (df/3r) ] +
1-27 (loll) ->-
i ->IIx fA= 3/3x fAx =
->-
. •
.
=
=0
it
j 3/3y fA y
3/3z fA z
[(O/3y) (fAz ) - (3/32) (fAy ) ] ! +
{
[
(3f/3r) (z/r)
[
]
= f (3A / 3Y) - (3A/3Z) + A z (Of/3y) - Ay(3f/3Z) i + 1-28 (loll)
II ·
(A x]))
(3/3x) (A D z - A z D ) Y Y + (3f3y) (Az Dx - Ax D z )
=
+ (3/3z)(Ax Dy - AYDx )
j). (lIxA ) =D (3A /3y - 3A /3z) + D (dA /3z - 3A /3x) x z y y x z
+ D z (dAy/3x - dAx /3y) -to (lIxD ) = -Ax(3D z /3y - 3DY /3z) - AY(3Dx /3z - 3D z 13x) -A z(3Dy /3x - 3Dx 13y) 1-29 (loll) -+
1I . lIxA
3/3x 3/3x A x
3/3y 3/3y A Y
a/3z 3/3z A z
6
:=
0
•
•
.
1-3 0 ( 1 . 1 2 )
Jr
7 7 E . dt =
f(
f::r
7 -3 -2 -5 ( 3tl /3 t ) .da = 2 x 10 x 10 = 2 x 1 0 V = 2 0 j1V
7 7 V x E ) .da
s
c
s
1 - 3 1 ( 1.1 3 ) 2 7 2 2 V ( V f ) = i ( 3 / 3x
2 2 3 / 3y
+
7 2 2 = i ( 3 / 3x) ( 3 / 3x
+
+
2 2 3 / 3 z ) ( a f / 3x)
2
3 / 3y
2
+
2
2
3 / 3z ) f
2
+ +
7
j ...
7
j ...
= V (V f)
CHAP TER 2 2 - 1 ( 2 . 1) COULOMB ' S LAW -31 -19 a) E e = mg , E = mg / e 9 . 1 x 10 x 9 . 8 / 1 . 6 x 10 2 2 e / 4rrE E , r = 5 . 1 m e / 4 rrE r , r b) E o O
=
5 . 6 x 10
-11
V im
2 - 2 ( 2 . 1) S EP ARATION OF PHOSPHATE FROM QUARTZ Let x b e the hor i z ontal coordinate and y the vertical coordinate , downward x
2 = ( !) ( QE /m) t ,
y = ( !) g t
x / y = QE /mg = ( Q /m) ( E / g) = 10 y
�
2x
�
100
2 -5
5 ( 5 x 10 / 9 . 8 )
RJ
0.5
mm
Referen c e : A . D . Moore , E l e c tros tatics and its App li cations , Wiley , 19 7 3 . 2 - 3 ( 2 . 3) ELECTRIC FIELD INTENS ITY 7 7 ::r By symmetry , the E = ]<; + E 2• 1 ver tical comp onen ts cancel and .-�--�--��� .... 7 2 2 0 7 .... -2 cos 4 5 Q i/4rr E ( a + a ) E
[[
= - Q/ 2
5 /2
rrE a o
J] 1
2
C1
-Q
7
C1.
.... .... C1
........
..... ....
+Q
2- 4 ( 2 . 4 ) ELECTRI C FIELD INTENSITY The charge in the r ing i s 2nrdrcr.
[
-
)2
E ach p o int in the ring is at a dE 2 2 .1 dis tance a + r from P . By sym-
p
metry , E is along the axis .
crardr / 2 E
dE
o
(a2
+
r
)
2 3 /2
R
E
J
cra 2E o
o
rdr / ( a 2
+
( cra / 2 E o ) [1 / ( a 2
r2 ) 3 / 2
+
R2 ) � - l / aJ
-7-
cr / 2E o when a
«
R
2-5 ( 2 . 5 ) CATHODE-RAY TUBE Of cour s e no t .
In app roaching
one p late an electron gains kine t i c
energy by lo s ing p o tential energy , like a b o dy falling in the gravi tational f ield of the earth . 2 - 6 ( 2 . 5 ) CATHODE-RAY TUBE Let V be the accelerating vo l tage and e the ab s o lute value of the elec tronic charg e . 2 ( 1 / 2 ) mv eV , v =
Then =
.1 ( 2 eV /m) 2 .
1
If the dis tance traveled is D , the time of f l ight is D (m / 2 eV) 2 . During that t ime the elec tron f alls by a dis tan c e 2 -19 3 2 -31 2 4 . 9 D m / 2 eV 4 . 9 ( 0 . 2 ) 9 . 1 x 10 ( 1 /2 ) g t / 2 x 1 . 6:X10 x 5 x 10 -16 m. 1 . 1 x 10 - 10 An atom has a diameter o f the order o f 10 m. =
=
2 - 7 ( 2 . 5 ) MACRO S COP I C P ARTI CLE GUN 4 -2 -16 -12 - 12 x ( 10 / 4 ) ( 1 . 5 x 10 / 1 0 ) =1 . 3 8 x 10 C Q 1 . 6 5 x 4 n x 2 x 8 . 8 5 x 10 m
( 4 / 3 ) n (10
( 1 / 2 ) mv
2
=
-18
/ 8 ) x 1000
=
5 . 2 4 x 10
4 -16 1 . 5 x 10 x 1 . 38 x 10 ,
v
Re ference: A . D . Moore , p 5 9 .
8
-16 =
kg
89 m/s
2-8 ( 2 . 5) QE /mg
ELE CTRO S TATIC SPRAYING
( Q /m) ( E / g )
=
4 3 E / g:2: 10 / 9 . 8 '" 10
=
Referen c e : A . D . Moore , pp 7 1 , 2 5 0 , 2 5 9 , 2 6 2 . 2-9 ( 2 . 5 )
THE RUTHERFORD EXPERIMENT
a) At the d i s t ance of c l o s e s t app roach , Q Q / 4TIE r is equal to the I 2 o kinetic energy : -19 6 Q Q / 4 TIE r 7 . 6 8 x lO x 1 . 6 x lO o l 2 =
r = 2 x 7 9 x ( 1 . 6 x lO b ) Q Q / 4 TI E r I 2 o c) a
2
-12 -19 2 - 13 x 7 . 6 8 x 1 . 6 x lO ) /4TI x 8 . 8 5 x 10 = 2 9 . 6 fm
= 7 . 6 8 x 1 . 6 x lO
( 4 1. 5 / 4 x 1 . 7 x lO
=
2-10 ( 2 . 5 ) a) 2 x 10
-2
-27
-13
/ 2 9 . 6 x lO
) / 9 . 8 = 6 . 2 3 x lO
-15 26
= 4l . 5 N g's
ELECTROS TAT I C SEED-S ORTING ( 1 / 2 ) g [Ct + 0 . 01 )
2
- t
J '"
2
200
mm
5 ( 0 . 02 t + 10
-4
) , t "" 0.2 s 2 The upper p e a mus t have fallen through a dis tan c e of g t / 2 , or ab out =
.
-4 b ) The average mas s of one pea is 2 0 0 0 / l00 x 3 6 0 0 x 2 4 2 . 3 2 x 10 kg. .2 -2 ( 1 /2 ) ( QE /m) t , where t ' is the t ime interval during Thus 4 x 10 =
=
whi ch a pea i s deflec ted: -2 -4 -9 5 ! t ' = ( 2 x 4 x lO x 2 . 3 2 x lO / 1 . 5 x lO x 5 x lO ) = 0 . 15 7 s . , '2 2 The p lates have a leng th L = v t + gt / 2 , with v 0 = 2g x 0 . 2 , v0 o 2 L = 2 x O . 15 7 + 4 . 9 x O . 15 7 '" 4 5 0 mm
=
2m/ s
.
Refer ence : Fluo re s c en ce-Ac tivated Cell-S o rt ing , S c ien t i f i c Amer ican ,
March 19 7 6 , p 10 8 . 2-11 ( 2 . 5 ) 2
CYLINDRI CAL ELE CTROS TATI C ANALYSER
mv /R = QE = QV/ a ,
v
=
�
CQVR/ma) 2
Reference :
Jour . Phy s . E , S c i . Ins tr . �, 403 ( 1 9 7 7 ) .
2 - 12 ( 2 . 5 )
PARALLEL-PLATE ANALYSER
Referenc e : Rev . S ci . Ins tr . �, 142 3 ( 19 7 1) , Rev . S ci. Ins tr. 4 8 , 45 4 ( 1 9 7 7 ) .
9
2-13 ( 2 . 5 )
CYLINDRI CAL AND PARALLEL-PLATE ANALYSERS COMPARED .1
In the cylindri cal analy s er , v ( QVR/ma) 2 . 2 ( R/ 2 a ) V . For a g iven ins trumen t , R and a are f ixed So ( 1 / 2 ) mv /Q =
=
and V i s a measure o f the g iven ratio . In the p arallel-p late analy s e r , f rom Prob . 2-12 , 2 QV /Q = ( 1 / 2 ) mv /Q = ( a / 2b ) V . o S o V i s a measure o f the same ratio . 2-14 ( 2 . 5 )
ION THRUS TER
Cons ider a satellite of mas s M and velo city V in a region where gravitat ional forces are negligib le .
The s a tellite ej e c t s m ' kg / s e c
b ackwards at a ve lo c i ty v wi th r es p e c t t o the satellite .
The
momentum of the sys tem ( M + total ej ec ted mas s ) is con s t ant .
Then ,
with r e s p e c t to a f ix �d referen c e frame , calling p the total momentum of the ej e c ted fuel , ( d / d t ) ( MV) + ( dp / d t )
=
0 , M ( dV / d t )
=
m'v.
I t i s thi s quan t i ty that i s called the thrus t . is n o t the force ( d / d t ) ( MV) . a) F v
2
=
m ' v , ( 1/ 2 ) m ' v
= 2 I V/ ( I /ne ) m,
b ) ( 2 x 1 . 7 x lO c) F F d) V t
-27
�
( 2m ' p ) 2
Q / 4'1TE R , o Q /I
=
= 10Q
2
=
( I /n e ) m .1
( 2mV /ne ) 2 1
4 - 19 ! x 5 x lO / 1 . 6 x lO ) 0.1 2
=
3 . 2 6 x lO
-2 N
P
=
2 p /v Q
Note that the thrus t
In the las t equation
IV , m '
F
m ' v , (1/2) m ' v
0,
0 , or MdV / dt + VdM/ d t + m ' (V-v)
=
MdV / d t - m ' V + m ' V - m ' v
=
�
2P / ( 2 IV/m , ) 2
4'1TE RV O
5 . 6 x 10
(2m/neV)2 p
-'12 4 x l x 5 x l0 47r x 8 . 85 x lO
=
-5
1
�
(2m ' / I V)2 P
=
-6 5 . 5 6 x lO V
s
Reference : R . G . Jahn , Phy s i c s of E l e c t r i c Propu l s i on . 2-15 ( 2.5 )
COLLOID THRU S TER
Referen c e : The E l e c t r ical P r opuls ion of S p ace Vehi cles , A . W . B right and B . Makin , Contemporary Phys i c s 14 (19 7 3 ) p 2 5 .
See also S ta t i c
Electrification 1 9 7 5 , The Ins t i tute o f Phy s i c s , London , 197 5 , p 4 4 .
10
CHAP TER 3 3-1 ( 3 . 1) e
=2
ANGLE SUB TENDED BY A LINE AT A P OINT
arc tan ( a / 2b ) ......
3-2 ( 3 . 1)
......
SOLID ANGLE SUB TENDED BY A DISK AT A P OINT
The r ing of radius
r
and width
dr s ub tends at P a s olid ang l e [ 211rdr / (b 2 + r 2 ) J co s e 2 21Trdrb / (b 2 + r 2 ) 3 /
If b
2 2 " 2 11 [ 1- ( 1+ a /b ) -2J »
a, n
3-3 ( 3 . 2 )
�
O.
-c
LL
0, n
If b
a
2 11 .
If n
11 , a
2
GAUS S ' S LAW
�
Gaus s ' s law can only tell us tha t the net f lux of E emi tted
No .
b y a d ip o l e i s zero , s in c e the net char g e i s zero .
For examp l e , the
average radial E is zero. 3-4 ( 3 . 2)
SURFACE DENS I TY OF ELECTRONS ON A CHARGED B ODY
a)
=
cr
=
E E o
8 . 85 x lO
-1 2
x 3 x l0
6
=
2 . 7 x lO
-5
C /m
2
-10 2 2 Thus the b ) E ach a t om o c cupies an area o f ab out (3 x 10 ) me ter 19 . numb er o f a toms per square meter is ab out 10 -5 - 19 c ) T h e numb er o f e l e c trons p er s quare m e t er i s 2 . 7 x 10 / 1 . 6 x 10 , 11
or 1 . 7 x 10 o r 1 . 7 x lO
14 -5
The numb er of free e lectrons p e r atom is 1 . 7 x 10
. •
THE ELECTRI C FIELD IN A NUCLEUS
3�5 ( 3 . 2 )
= 1 . 2 5 x 10 - 1 5 ( 12 7 ) 1 / 3 = 6 . 2 8 x lO -15m . A t the center , 2 r 2 2 3 ( p / 2 E ) R = Q /f/;ITfR / 3 ) J (R / 2 E ) = 3 Z e / 8TfE R v o L
R
=
3 x 5 3 x 1 . 6 x lO
- 19
0
0
/ 87r x 8 . 8 5 x lO
-12
x 6 . 2 8 x 10
- 15
At the s urf ace , 7 2 2 2 V / 1 . 5 = 1 . 2 x 10 V , V = ( p / E ) ( R / 2 - R / 6 ) = pR / 3 E center o o 2 -19 -12 -15 2 Q / 4TfE R = 5 3 x 1 . 6 x 10 E / 4Tf x 8 . 85 x lO ( 6 . 2 8 x lO ) o 21 V /m . = 1 . 9 x 10 THE SPACE DERIVATIVES OF E , E , E x Y z From Gaus s ' s law , aE / ax + aE / ay + aE / a z p/E z x 0 y 3- 6 ( 3 . 2 )
S ince V x
3- 7 ( 3 . 2 ) ->-
We s e t E If p
=
E
= 0,
aE l ay z
=
aE / a z , aE / a z y x
PHYSICALLY IMPOSS IBLE FIELDS
= Ek
->-
0 , V ·E
Als o , V x E
=0
0 and aE / a z
=
and aE / ax
-+
=
So, if p = 0 , E i s uniform.
If p
�
aE / ax , aE / ax z y
0, V · ->-
Als o , V x E
E
=0
=
p/E
0
=
O.
aE / ay
and aE / a z
and aE / ax = aE / ay
O.
= p/E = 0,
0
. as b ef o r e .
Then E is a function of z , but ind ependent of x and y.
12
14
/ 10
19
� (3.4)
I ON BEAM
2 2 From Laplace ' s equation , 3 V / 3x - ( p / 2 E ) X +Ax+B . O S ince V = a at x 0, B
V
=
=
=
2
o
O . Als o ,
V / a+pa / 2 E 0 0 2 Finally , V = ( V / a+pa / 2 E ) x - px / 2 E ' o o o
V
=
o
- ( p / 2 E ) a +Aa and A 0
-(P / E )x+A ,
-p / E ' 3V / 3x O
2
= - dV / dx
E
3-10 ( 3 . 4)
=
=
- (V / a+pa /2 E ) + px / E . o 0 0 A UNIFORM AND A NON UNIFORM FIELD
r 100 1/
(b)
0 .--____--:-____..,;;. 0,1
£
-s�o -/OCOL--____________
a) V 1000 x , E -1000 . 2 2 4 b ) d V / 3x = -10 , 3V/ 3 x =
=
a at x 0, B 4 2 V = - 10 x /2 +1 , 5 00x .
S in ce V
=
=
-
S ee Figs . a and c . 4 4 2 -10 x+A , V - 10 x / 2 + Ax+B
= O.
=
S ince V
=
100 at x
S e e Figs. b and d .
13
=
0.1 , then A
=
500 .
VACUUM DIODE (4Vo /9S 4/3 ) X-2/3 , 'dV/ 'dx (4V0 /9 S 4 / 3 ) 3x l/ 3 + A 4/3 4 /3 Ax = V (x/s) 4 /3 Ax V ( 12 Vo /9 s ) ( 3 / 4 ) x + + 0 S ince V Vo at x s , then A = 0 , V V o Cx/s ) 4 / 3 1 b) J pv , ( 1 /2) mv2 eVo ' v (2eVo /m) 2 1 J - ( 4E 0 V 0 /9s 2 ) (2eV0 /m) 2 _ (2 5 /2E o/ 9 ) (e/m) ! (V0 3 /2/s 2 ) 3 -2.335 x 10 -6Vo /2// 3 - 11 (3 . 4)
---:;-;2V/ 'dx2
=
=
=
=
=
=
=
=
=
=
F�
G----------¥ ..TF2 . .. . 1 ... ... T
b
_____ __ _
C
,.
�
....
....
__ _ _
I I I
0
Q
3-l3 ( 3 . 7 )
O
IMAGES
=
�t
=
J
=
3-14 ( 3 . 7 ) IMAGES a) At some point P on the con ducting plane , E 2 Q / 41fE o (D2 + r2 ) cos 8 2 2 3 /2 = 2Q D/41fE (D + r ) O 2 2 3/2 a - E E = -QD /21f (D + r ) o =
[
]
Q
14
�-
---"A
0T a
_________
B
G)
..... .
-� -
=
t-- a
C
J
.. I
I I I I I
e
2 -7 E A - (Q/41fE o a )J ->2 -T E - (Q / lt1fE o a )J B ->2 2 3 /2 (2a-7l + a-7J ) E Q /41fE o ( a + 4a ) C ->2 2 3 / 2 ( -2ai->- + aj->-) E Q /41fE o ( a + 4a ) D ->2 E tot (Q /41fE 0 a ) (-2 + 2/5/S) j ->- ( 1 . 82ll Q / 41fE o a2 ) j ->-
:
- - __ _
"
e
,," e
- -
t
-
,."" -
o
"
- -
...
P�-
-
,� I
:r I
-
...
............
E
-Q
I ...... � - - - - - - - - �. -
: I
0
f
2 2 b ) - 21TrdrQD/21T (D 2 + r ) 3/ o
-QD
f
2 2 2 rdr/ (D + r ) 3/
-Q
o
CHAPTE R 4 4-1 ( 4 . 1 ) THE PERMITT IVI TY OF FREE SPACE From Cou l omb's l aw , [ E o ] [Q 2 /FL2 ] = [ Q 2 / (FL) LJ = [Q 2 /(Q 2 / C ) L ] = [ C /L ] , where the brackets indicate that we are concerned on ly with the dimens ions , and where F , L , C , s tand for Force , Length, and 2 Capacitance . Note that FL is an energy , l ike Q /C . =
4-2 (4 . 1) THE EARTH'S ELECTRI C FIELD a) C 41TE 0 R 7 . 1 x 10 -4F '" 700jlF 2 b ) Q 41TR2CJ 41TR2E 0 E ( 6 . 4 x l 0 6 ) x l OO/9 x l 0 9 c) V 41TR2E E /41TE R = E R 6 . 4 x l 0 8 V =
0
0
4 . 5 x l 05 C
=
Reference : Richard Feynman , Lec tures on Physics , 2 , Ch 9 , Addis on Wes ley. 4-3 ( 4 . 2 ) PARALLEL-PLATE CAPAC I TOR I f there are 3 p l ates , 2 C 8 . 85 x 10 -1 ( 2A / t) � V =
l
L---_
With four p l ates , C 8.85 x 10 -12 (3A /t) , etc. =
For N p l ates , l2 C 8 . 85 x 10 - ( N- l ) A!t F 8 . 85 (N- l ) A / t) pF 4-4 The A�
----�O
o�---� ------I�
( 4.2) PARALLEL-PLATE CAPAC ITOR 2 plate separation might be 1 mm . Then 10 -1 10 -4 , or one square centimeter .
15
8 . 85 x 10 -12A/I0 -3 ,
4-5 ( 4 . Z ) PARALLEL -PLATE CAPAC I TOR C ' = C aCb / ( C a+ Cb ) = E o S / (a+b) = E S / (S-s ') O //J//L/////// // //at-/ The capacitance is l arger , but it T is independent of the pos ition of 'h, coudu" 'ug p1u', .
I
�1 ' 1 � 1
4-8 (4. 3)
.--+ ---I
///T///?T/77Td7
ELECTROSTAT I C ENE RGY
V
11 1------1
a) w l /W Z b) W /W Z
[ ]
( QV l /Z) / ( QVz /Z)
=
( Q IV /Z) / (Q Z V/Z)
=
Vl /VZ
Q /QZ
C l V"/C V Z
C l /C
Z I
4-9 (4 . 3) ELECT ROSTAT I C ENE RGY a) The energy that is dis s ipated z Z Z Ql + QZ . Ql QZ Z C l + Z CZ - Z ( C I + C Z )
r
=
1
b ) The energy is dissipated by Jou le heating in the resis tance R of the wires . Let Q l O and Q ZO be the charges at t = 0 , Q l and Q Z the charges at t . C l dis charges in to C Z . Then Q l /C l - Q Z /C Z = IR , Q l + Q Z Q l O + Q ZO Q , 1 = dQ Z /dt ,
=
=
16
S ince Q 2 d Q2 /dt
=
=
[
Q 20 at t
Q 20 - 1/
=
0, A
��:i /C2J
Q 20 - 1 /C + 1 / C 2 1
=
[ - ( 1 /C l+ l /C2 ) /RJ exp [ - ( 1 /C l+l /C 2 ) t /RJ
[ -Q20 ( 1 /C l+ l /C2 ) /R + Q / C lRJ exp [ - ( 1/ C l+1/ C2 ) t /RJ
W
[ -Q 20 /C2R + Q 10 /C l RJ exp [ - ( 1/ C l+l /C2 ) t /RJ 2 [ Q lO /C lR-Q20 /C2R] 2 ( dQ 2 /dt) R = - 2 ( 1/C +1 /C ) /R R( O-l) 2 l
f
o
( Q l O /CrQ2 0 /C2 ) 2 2 ( 1 /C l + l /C2 )
2 (Q lO C2-Q 2 0 C l ) 2C l C2 ( C + C2) l
This is the result found under a , excep t that the initial charges are now called Q lO and Q20' ins tead of Q l and Q2' P ROTON B OBB R ( 1 /2 ) pVd, ( 1 /2)
4-10 ( 4 . 3) a) W
=
J
o
=
J
o
R 2 2 2 (p2 /E o ) (R / 2 - r /6 ) 4 rrr dr
Use the above result , rep lacing 1/4rrE o by G 2 = 3GB /5R G 22 2 c) 3 x 6 . 6 7 x lO - ll x (7 . 33 x l0 ) 2 /5 x 1. 7 4 x l0 6 = 1 . 24 x l0 9J . d) p ( 100 0 /1 . 7 x lO - 2 7 ) 1.6 x lO -19 9 . 6 x lO lO C/m3 2 2 5 I f R is the radius of the sphere of protons, 4rrp R /15 E 0 1 . 24 x 10 � 2 2 10 1/ 5 9 0 . 17 m [l5E o x 1 . 2 4 x lO /4 11" ( 9.6 x 10 ) ] R
b)
W
=
=
=
=
=
4-11 (4 . 5 ) ELE CTROSTATI C MOTOR Reference: A. D . Moore , E lectros tatics and its Appli cations .
17
4-12 ( 4 . 5 ) ELECTROSTATI C PRESSURE a) V = Q /41l"o R , E = Q /41l"o R2 , V = E R = 3 x 10 6 /0 . 05 = 1 . 5 x 10 5 V b) The pressure is 0 2 /2"o (Q / 41lR2 h N0 = (Q/ 41l"0 R) 2 ( "0 2 /R2 ) /2 e 0 ( 1 . 5 x 10 5 ) 2 " /2R2 ::::< 40Pa ::::< 4 x 10 - 4 atmosphere. 0 =
4-13 ( 4 . 5 ) PARALLEL-PLATE CAPACITOR Let each p l ate have an area S . Then the capacitance changes by dC = d ( " o S / s) = - (" o S / s 2 ) ds . Let ds be positive . The capacitance decreases and a charge dQ V l de l = (" o�S /s 2 ) Vds returns to the battery. Thus dWB = - (" o S /s 2 ) V�ds = - " 0 E 2 Sds . The energy s tored in the fie l d increases by dWe d ( " oE 2 sS /2) = d ( " o V 2 S / 2s ) = - (" o V 2 S /2) ds /s 2 = - " o E 2 Sds /2 . The mechanica l work done on the system is Fds S (" oE 2 / 2) ds " o E 2S /2 . =
=
4-14 ( 4.5) PARALLEL-PLATE CAPACITOR See the preceding prob l em . Here, dWB dWe d (" E 2 s S /2 ) = (" 0E 2 S / 2) ds O2 2 Fds ( " oE S / 2) ds , F "o E S / 2
o
and E is cons tant .
=
4-15 (4 . 5 ) OSCILLATIN G PARALLEL -PLATE CAPACITO R a) The energy stored in the capacitor is 2 2 Q V/ 2 = CV / 2 = ( E o A /x) V /2 . Let x > x o . Then the tota l potentia l energy is W mg (x-x o ) + k (x-x o ) 2 / 2 2 + E A ( 1 /x-1/x ) V /2 + VL'lQ , o o where L'lQ is the charge fed into the battery because of the decrease in capacitance : 2 2 VL'lQ = -V L'lC = -V "oA ( 1 /x-1/x o ) . The battery gains energy if �C is negative . Thus W mg (x-x o ) + k ( x-x o ) 2 /2 - E oA ( 1 /x-1 /x o ) V 2 /2 2 ( x-x o ) [mg + ( x -x 0 ) k/2 + " 0AV /2xx ] 0 =
18
=
2 [ (x-x ) / x ] [kx /2 o
+
(mg-kx / 2 ) x 0
.
+
2 E AV / 2x 0
b ) There are thr ee d ownward forces and , at equilib r ium , 2 2 + k ( x-x ) + E AV /2x o o Also , at equilib r ium ,
mg
dW / dx
=
mg
k ( x-x ) o
+
+
O.
0
] 10-2.
10-'
vv
2 2 E AV / 2x =0 o
c) The relation F - ( dW/dx) eq comes from the cons ervation of =
(b)
energy for a small displacement near equilib rium . dW/ dx
In s e t ting
) , we as sume that the W (x) curve app roximates a eq 2 (K/ 2 ) (x-x ) in the reg ion near the p o in t of s tab le eq equil ib rium . Thus 2 3 2 2 K ( d W / dx ) k - E Av /x 0 eq ' eq 2 3 k - E AV /x 0 eq 6 . 16 Hz . w ' f m =
K (x-x
p arab ola W =
=
[
=
J
4- 16 ( 4. 5 )
!
=
HIGH-VOLTAGE GENERATOR
a) The charg e den s i ty on the plates , and hence E , remain cons tan t when t h e p lates a r e sep arated . Then the increase in energy i s 2 2 The mechani cal work done is the force , t E S / 2 , € E S s (n-l) / 2 . o o multiplied by s (n-l) . b ) Ref eren c e: A.D. Moore, E lectrostatics and i t s Appli cations , Chap ter 8 . 4-1 7 ( 4. 5 )
INK-JET PRINTER
rr
2'ITE V / tn ( R /R ) 2 l o 3 2 b ) ( 4/ 3 ) 'IT ( 2R ) / R = 32 R / 3 l l l a) A
=
C'V
=
c) Q
32 R A / 3 I
d) m
1000T , Q /m
Q /m e) v
=
6 4'ITE VR / 3 tn (R /R ) o l 2 l =
Q / lOOOT
6 4'IT e VRI / 30 0 0 tn (R /R ) (4/ 3 ) 'IT (2 Rl ) 2 l o
2 E V/ 5 0 0 R tn ( R / R ) l 2 l O -5 -5 2 -3 -12 x 100 / 5 0 0 ( 2 x 10 ) tn (5 x 10 / 2 x 10 ) 8 . 8 5 x 10 5 -4 10 x 10
=
10 m / s
19
=
8 . 0xlO
3
-4 C/kg
f) A drop let remains in the deflec ting field during 4 xlO- \ . During
that time it is s ub j ected to a transverse force QE and i t s accelera2 5 -4 The transverse d e f l e c t ion tion is QE /m , or 8 x 10 x 10 , o r 8 0 m / s is
-3 2 2 at /2 = 8 0 ( 4 x lO ) /2 = 0 . 64 mm
The transverse velo city at the far end of the deflec ting p lates is -3 at = 80 x 4 x lO = 0 . 32 m/s Re ference: S p ecial is sue o f the IBM Journal o f Res e arch and Develop men t , January 19 7 7 .
CHAPTER 5 CONDUCTION IN A UNIFORM MEDIUM
5-1 ( 5 . 1) cr
=
cr -t-ax / s , E o
5-2 ( 5 . 2 )
=
J / cr = J /(cr -t-ax / s ) o
RE S IS TIVE FILM
Let the film have an area a R = a /crat
=
5-3 ( 5 . 2)
2
and a thickness t .
Then
l /cr t . RE S IS TOJET
The thrus t is m ' v , wher e m ' is the mas s ej ected p er s e cond , and v is the exhaus t velo city . S e e the s o lution of P r ob . 2 - 14. 1 1 2 � ( 6 0 0 0 m ' )� 1.9 N. m ' v /2 3 0 0 0 , v =(6 0 0 0 /m ' ) , m ' v =
=
Then
=
Reference : Rob e r t J . Jahn , Phy sics of Elec tric P ropuls ion , p 103 . -
5-4 ( 5 . 2 ) JOULE LOSSES - 2 5 2 RP 10 x O . 2 5 , V /R = P , V
B
=
V = 158 V
20
..
5-5 ( 5 . 4)
VOLTAGE DIVIDER
The curren t flowing through R V
i
I(R + R ) , V o I 2
=
5-6 ( 5 . 4 )
=
l IR , V /V o i 2
and R is I . 2 R / (R + R ) l 2 2 =
P OTENTIOMETER
See P rob . 5-5 . 5-7 V'
S IMPLE CIRCUIT V - ZV [ R / ( R + R ) J I I Z
=
5 - 9 AMPLIFIER a) R
and R carry the s ame curren t l 2 I = ( V - V ) /R = ( V - V ) /R , iA l i iA o 2 ( V + V /A) /R o i l
-R / [ R + (R + R ) / A J l 2 2 l
V /V o i V /V o i
V ( - 1 / A - l ) /R , o 2
=
�
-R /R if A 2 l
»
=
- (R /R ) / [ l + 1 /A + ( R /R ) /A J . 2 l 2 l
1 and if R /R « 2 l
A.
The gain R /R mus t l l
the refore b e much less than A .
12
resistors - -
expx
=
e
2
--
4 3 1 + x + x /2 ! + x / 3 ! + x / 4! + 1 + 1 + l/Z + 1/6 + 1/24 +
5-11 ( 5 . 5 )
...
...
TETRAHEDRON
a) By symmetry , the currents through ACB and ADB ar e equal.
The
p otential at C and D is half-way b e tween the p o t en tials at A and B . b ) ZR/2
=
R , in p arallel wi th th e R b e tween A and B .
The resis tance
b e tween nodes A and B is R. 5- 12 ( 5 . 5 )
CUBE
a) By symmetry p oints BED are at the s ame p o tential . P o ints FCR are 21
at ano ther p o tential .
F
b ) The r es is tance f rom A to BED is R/3 . R/6.
That from BED to FCR is
BIE-__ p'-;(",
That f rom FCR to G is R / 3 .
The resis tance is 5R/6 .
.,;Ji==---t----;J H D
5-13 ( 5.5 )
CUBE
a) P oin ts FBDR .
Branches FD and DR can be either removed or short
circuited . b ) Remove tho s e b ranches. tance 2R/2 = R.
Around the inner s quare , we h ave a resis
Around the outer s quare , the resis tance i s 3R.
Thus
we have R and 3R in p arallel and the resis tance is 3R/ 4 . 5-14 ( 5 . 5 )
CUBE
C�______,G
Dis tort the cub e as in Fig. a . Then , by symmetry , B and E are sho r ted .
F
B
at the s ame p o tentia l and c an be Similarly , C and R can
be shorted .
Now r edraw the
f igure as in b and c.
The resis
tance to the righ t o f th e dotted line is 0 . 4 R and
(0)
R = R ( 1. 4R) / ( R+ 1. 4R) = 1 . 4R/2. 4 AD = 0 . S83R G
R
D
22
5 - 15 (5 . 7 )
LINE FAULT LO CATION
Let th e len g th of th e line be � and let a b e the r e s i s tan c e of one meter of wire . 2 ax + R
Then
= 5 5 0 / 3 . 7 8 = 145 . 5 , R
s
s
=
l45.5 - 2 ax ,
2 ax + R 2 a(�-x) / [ R + 2 a(�-x) J = 5 5 0 / 7 . 2 s s
=
76 . 39 ,
ax[ R + 2 a(�-x)J+ R a(�-x) = 3 8 . l9 [ R + 2 a(�-x) J , s s s 2 2 2 aXR + 2 a x� - 2 a x + R a(�-x) = 3 8 . l9R + 7 6 . 3 9 a� - 7 6 . 3 9 ax . s s s C anceling the axR
t erms and sub s ti tuting th e value o f R in the s s firs t equation , 2 2 2 2 a x � - 2 a x + (145 . 5 - 2 ax) a� 3 8 . l9 (145 . 5 - 2 ax) + 7 6 . 3 9 a� - 7 6 . 3 9 ax , 2 2 - 2 a x + l52 . 8 ax + 6 9 . lla� - 5.5 5 7 O. -3 2 7 -3 Now a 1 /5.8 x lO '1T(1 . 5 x lO ) = 2 . 4 39 x lO . =
=
S olving , x
5 - 1 6 (5 . 7 )
=
7 . 8 1 8 kilometers .
UNI FORM RE S I S TIVE NET
a) From Kirchoff ' s voltage law , the sum o f the currents flowing into o i s zero .
Thus
(V -V ) /R + (V -V ) /R + (V -V ) /R + (V -V ) /R B O A O D O C O
0,
V o = (V +V +V +V ) / 4. A B C D
b) For a three-dimensional circuit we have 6 r e s i s tors conne cted to 0 and
5 - 1 7 (5 . 9 )
v· .<.
POTENTIAL DIVIDER
For the left hand mesh , V -I R-(I -I ) R = O . i l l 2 For the middle mesh, (I -I ) R + I R + I R = O . 2 2 2 l Then 1
2
=
V / 5 R , V = (V /5R) R ::;: ViS. i o
23
S IMPLE CIR CUIT WITH TWO SOUR CES (Rl + r ) I , - rI2 = V , - rI l + (R2+ r ) I2 V I = 1 1- 12 = (R2 -Rl ) V/ [ Rl R2+ r (Rl +R2 ) ] 5-18 ( 5.9 )
DELTA-S TAR TRANSFORMATI ONS Equating the voltages , VA- VB Re ( ID- I C ) RA ( IB -I C ) + � ( IA- I C ) , VB - V C Ra ( ID - IA) � ( I C - IA) + Rc ( IB -IA) ' =
( 1)
=
(2)
RC ( IA- IB ) + RA ( Ic- IB ) ·
(3)
5 - 19 (5.10 )
V C - VA = � ( I D- IB )
Rewriting , - IARB - IB RA + I C(RA+ �-Re ) + IDRe - IA ( �+R C-R) - IB Rc - I C� + IDRa - IARC + IB (RC +RA-\ ) - I CRA + ID\
0,
( 4)
0,
(5)
O.
(6)
Eliminating ID from Eqs . 4 and 5, - IA�- IB RA + I C (RA+ �-R C ) + (Re / Ra ) [ - IA ( �+R C-R) + IB RC + I C �]
=
0
( 7)
IA [ - � - (Re /R) ( �+RC-R)] + IB[ -RA + ( R/R)RC ]
Thus � + CRe / R) (�+RC ) - Re =
� + (R/R)� + RA - Re
= °
0,
RA /RC = R/Ra
( 10 )
( ll)
Only two of these equations are independent . Combining the firs t two , ( 12 ) � + (RA /RC ) � + RA = Re' Re = ( �RC + R CRA + RA� ) /RC Or , settin g
R
=
l/ G , Ge
=
GAGB / ( GA+GB +GC )
24
( l3)
5-2 0 (5.10)
(
a
DELTA-STAR TRANS FORMATI ON S
4K
5K
)
Redraw the circuit as in Fig. a and transform the le f t hand delta in t o a s tar , as in Fig . b , with 4000 x 100 0 / 7 0 0 0 2. 8 9 krl.
R
=
(8 / 7 ) 1000rl, R
(4 / 7 ) 1000rl, R 2
=
3
(2 / 7 ) 1000rl,
OUTPUT RE S I S TANCE OF A B RIDGE CIRCUIT
5-2 1 (5 . 12 )
The output resis t ance is the res is tance one would measure at the terminals o f the voltme ter if the source were repla c ed by a short circuit.
This is R/2 + R / 2
5- 2 2 (5 . 12 )
=
R.
INTERNAL RE S I S TANCE OF AN AUTOMOB ILE BATTERY
The headligh t s , tail ligh t s , e t c draw ab out 15 A. resis tan c e of the b a t te ry is ab out (1 / 15) � . b e caus e a cranking mo tor draws , s ay 2 0 0 A.
H ence the internal
This is much too large , A normal automobile bat -2 rl .
tery has an in ternal resis tance of the order o f 10 Referenc e :
S tandard Handb ook for Electrical Engineers , S e c tions
2 1 and 2 4 . 5-2 3 (5. 1 4 )
DI S CHARGING A CAPACITOR THROUGH A RE S I S TOR
From Kirchof f ' s voltage law , Q / C RdQ /dt - Q / C
5-2 4 (5 . 14 ) V
o
=
Q/C
=
=
0, Q
=
=
RdQ / d t.
Q exp(- t!RC) , V o
RAMP GENERATOR
It/C
25
=
Thus
V exp (- t / RC) o
100 exp (- t )
5-Z 5 ( 5 . 14) CHARGIN G A CAPACITOR TH ROUGH A RES I S TOR The energy supp l ied by the source is 00 00 CV Z Vldt V ( dQ /dt) dt V dQ CV . Ws =
J
J
=
=
0
0
J
0
The energy s tored in the capacitor for t The energy dis s ipated in the resis tor is
J
z R [ (V/R) eXp (- t/RC) ] dt
=
o
5- Z 6 ( 5 . 14) RC TRANS IENT a) I V/RZ + CdV/dt , Vs IRl + V =
=
R1 C dV/dt + ( l +Rl /RZ ) V
=
=
Z is CV / Z .
+ 00
Z CV / Z
(V/RZ + CdV /dt) Rl + V
Vs
S ince V 0 at t 0 , V [ Vs / (1+Rl /RZ ) ] { 1 - exp [ - ( 1 +R l /RZ ) t / R1 C] } [RZ / (Rl +RZ ) ] { 1-exp [ - (Rl +RZ ) t /R1RZ C ] }V s =
=
=
The time cons tant is R1 RZ C/ (Rl+ RZ ) , or C / ( l /Rl+ l /RZ ) . For t + 00 , V = RZV s / ( Rl +RZ ) . b ) Now , at t 0 , V = RZVs / (Rl +RZ ) . The capacitor dis charges through RZ and =
5- Z 7 (5.14) RC DIFFERENTIATIN G C IRCUIT V i = Q / C + RI�Q /C V o RI RdQ /d tRiRCdV / dt o I.. =
=
v·A.
Z6
QI t-- -,----.f C
5-28
DIFFERENTIATING A SQUARE "lAVE
5-30 ( 5 . 14)
RC INTE GRATING CIRCUIT
The curren t flowing in to the cap acitor is I . V V
i o
=
RI + Q / c
=
Q/c
( l /RC)
!
o
(V -V ) /R i iA v +V /A i o
=
V dt. i
INTE GRATING CIRCUIT
5 - 3 2 ( 5 . 14)
i
RdQ / d t + Q / c '" RdQ / d t , t
=
5-31 ( 5 . 14 )
V
=
INTEGRATING CIRCUIT =
C (d / d t ) ( V
-V ) iA o
-RC ( l+l /A) dV / d t , o
-RC ( l+l/A) dV / d t - V lA o o -RCdV / d t if A o
»
1 and if V lA « o
27
RCdV / d t. o
5 - 3 3 ( 5 . 14 )
PULSE-COUNTING CIRCUIT
a) Dur ing a puls e , the vol tage
V
acr o s s C
i s ap proximately equal l to V and Q � C V ' Af ter th e p l p fir s t puls e , the voltage acro s s
is C V / C ' The pro c e s s l p 2 2 repeats its elf . T h e voltage
C
acro s s C
in creas es by C V / C 2 l p 2 at e ach puls e .
t
CHAPTER 6 6-1 ( 6 . 1 ) a) p
P iN
b) s
p /Q
THE DIPOLE MOMENT p 23 6 -7 10 / [ 6 . 0 2 x 10 x (3 . 5 /l2 ) 10 ] =
5 . 7 x 10
-37
16 x 1 . 6 x 1 0
- 19
=
=
5 . 7 x lO
5 . 9 x 10
- 19
-37
Cm
m
Th e diameter of an atom is of the' order of 10 - 10 m . 6-2 ( 6 . 2 )
THE VOLUME AND SURFACE B OUND CHARGE 'DENSITIES
J-17
J
T
. PdT + CJ da b S
6-3 ( 6 . 2 )
o
BOUND CHARGE DENS ITY AT AN INTERFACE
In the figur e , we have shown the
CD
two media s ep arated, for clari ty. On the face of 1 , CJ On 2 , CJ
b2
6-4 ( 6 . 4)
-+ -+ = P ' ( -n)
2
bl
n
=
COAXIAL LINE
Cons ider a volume
T
of diele c tr i c h aving the shape shown in the
figure,
28
fV.EdT fE.ta T =
S
where S is the surface b ound ing
T.
The surfaces A and B are the only ->-
->-
ones where E·da is no t zero . if their radii are r
f->- ->-
E·da
S
T
=
- ( A / 2TrE r ) r 6L o A A
V·EdT ->
6-5 ( 6 . 7 )
Then,
and r , A B +
( A / 2TrE r ) r 6L 0 B B
o and , s in c e r , r
B
A
o
and 6 are arb i trary ,
V·
->-
E
=
O.
COAXIAL LINE
a) Near the inner conductor , E
A / 2TrE E R r o l
l
If A is the charge p er meter and C' the capaci t ance per meter , A
C' V
Thus E 5 x 10
6
l
=
[ 2TrE r E o / in ( R /Rl ) J V 2
= V/Rl in (R2 / Rl )
5 0 0 /R in ( 5 x 10 l
=
-3
1 . 7 7 2 x 10
/R ) , Rl l
-5
m
b ) One should use No 34 wire . c) C '
=
2Tr x 2 . l x S . S5 x lO
6 - 7 ( 6 . S)
-12
/ in ( 5 / 0 . l 6 )
CHARGED WIRE EMBEDDED IN DIELE CTRI C : THE FREE AND B OUND CHARGES
a) Ins ide the diel ectric , D
=
3 3 . 9 3 pF /m
A / 2Tr r , E
=
A / 2TrE E r r o
On the inner s urface of the diel ectric ,
On the outer surface ,
29
6-8 ( 6 . 8)
PARALLEL-PLATE CAPACITOR
5
a) We can treat this prob lem as if we had two c apacitors in series C
E A/ ( s- t) , C z = E E A / t l = o r o
= C l C2 / ( C l+ C z )
C
=
[:�:
(
E E 0A r
o
A
[,:" ;lJ '
E E 0A r t + (s-t) E
Er 1 --+ t s-t
t ( s- t)
=
l!G
" ' oA ,
J
E A o r
1 l- ( t / s ) ( 2 / 3 )
6-9 ( 6 . 7 ) a) S ince the only free charge is Q , Eq . 6 - 1 7 g ives us tha t Z -9 Z -11 2 Z 10 / 4 rrr = 7 . 9 6 x lO / r C /m D = Q / 4Tfr =
b o th in s ide and outside the die l ectri c . Ins ide the die le c tric ,
Z Q / 4TfE E r r o -11 2 -11 2 7 . 9 5 x 10 /3r 2 . 6 5 x 10 /r
D/E E r o
E.
1=
=
=
Out s ide th e diele c t ri c , 2 2 D/E Q / 4TfE r 8 . 94/r o o o
E
=
To f ind V , we s e t V
o
=
=
Outs ide the sphere , V
0.1
0 at infinity .
=
o
=
Q / 4TfE r o
=
9 . 00 / r V
At the surface of the s phere , V = 4 5 0 V. R 2 Ins ide the s phe r e , V = 4 5 0 + ( Q /4TfE E 0 r ) dr i r r
f
3 00 + 3 . 00 / r
c) L e t us apply Gaus s 's law to a smal l element of volume at the s urface.
The bound charge on the element of area da is
30
'\
2 ( Q / 4 rrR ) ( l- l/E ) , r
;;; E (E -E . ) o 0 l
as previous ly .
da
The dis con t inui ty in E is due t o the b ound surface charge .
6-10 ( 6 . 7 )
CHARGED DIELE CTRI C SPHERE 3 2 ( 4 / 3 ) rrR P / 4 rrE r f o 3 At r R, V ( 4 / 3 ) rrR P / 4 rrE R f o 2 3 Ins ide the sphere , D ( 4 / 3 ) rrr P / 4 rrr f Out s ide the s phe r e , E =
=
=
=
V
At the center ,
2 R p /3E + f o
J
o
2 3 R P / 3E r f o 3 R P /3E R f o rp / 3 , f
E
=
=
2 R P /3E f O rp / 3 E E o f r
R ( r P / 3E E ) dr f o
6-11 ( 6. 7 )
MEASURING S URFACE CHARGE DENS ITIES ON DIELE CTRI CS
6-12 ( 6 . 7 )
VARIABLE CAPACITOR UTILIZING A PRINTED CIRCUIT B OARD
Reference : Journa l of Phy s i c s E 2, 412 ( 19 6 9 ) .
Y
C /
� z
(3) In al l c a s e s , dC 10
y
a)
y
=
E E yd z / t , d C / d z r o
-3
3 x 8 . 8 5 x 10 =
l1»
dC
-12 dz
9 -9 10 10 = 37 . 7 26 . 55
=
=
9 10 dC 2 6 . 55 dz
rom
31
E E y/t r o
y =
b)
9 10 -8 10 x 2z 2 6.5 5
6-13 ( 6.8)
0.75 3 z m
EQUIPOTENTIAL SURFACES
a) No . b) o
= D . The char g e den s i ty is po s i t ive on the s i d e wh ere the f + ve ctor D poin ts away from the shee t . 6-14 ( 6.8)
v·n
=
S ince P b
=
NON-HOMOGENEOUS DIELE CTRI CS
V(E rE o"E)
VEr"
E 0 v·l
E (E v·l l· VE
+ r o r + 0 , V·E" 0 and P " 0 b
=
=
) = 0
-(liE r )l' VE r
FIELD O F A SHEET OF ELE CTRONS TRAPPED I N LUCITE -7 a) Th e t o t al free charge i s -10 C, .t> -3 -4 -7 x 2 x lO -10 125 x lO P f -2 . 000 x 10 - 2 e/m 3 6-15 ( 6. 9 )
e
b ) From Gaus s ' s law , D on each s ide, in the neutra l region , is one hal f the free charg e per square met er : -4 -7 D = - ( 1 / 2 ) 10 / 2 5 x 10 n 2 -5 Clm - 2 . 00 x lO P E
n n
(I-liE r ) Dn (D -P ) lE n n O
- 7 . 0 6 2 x 10
- 1 . 3 7 5 x lO
5
-5
C /m
2
D lE E n r o V im
S ince E = -dV Idx , and s ince n n -3 0 at x = 6 x 10 n 5 7 . 0 6 2 x lO x - 4 , 2 3 7 V V n
V
.D
+ + + c) At b o th sur f ac e s , P points inward , like E and D , and -5 2 -1 . 3 7 5 x 10 P 0 C /m b n =
32
d) Ins ide the charged region ,
v.nc
=
dD / dx c
P
f
=
- 2 . 000 x lO
-2
3 C /m , D
=
-2 . 0 0 0 x lO
-2
x C /m
2 O.
The c ons tant of in tegrat ion is zero b e caus e D changes s i gn at x
So D ->-
V ·E
c
o at x
O.
dE /dx c
( P + P ) / C. = - 7 . 0 6 2 x lO f b o
Als o , 8
2 V /m , E
c
8 = - 7 . 0 6 2 x lO x V /m
The con s t ant of in tegration is again z e r o , f or the s ame reason . From P oi s s on ' s equa tion , 2 2 8 2 2 V V d V /dx = - ( P + P ) h = 7 . 0 6 2 x lO V/m b o c f c 8 2 dV / dx = 7 . 0 62 x lO x V /m c The con s t ant of integration is zero , s ince dV / dx c 8 2 V = 3 . 53l x l0 x - 3 88 4 V c
-E
The cons tant of in tegration is now chos en to make V e) S e e curves. f) The s tored energy is 2 x ( 1 / 2 ) g) N o . 6-16 ( 6 . 10 )
-��
E
6-1 7 ( 6 . 7 )
f
o
3 lO-4 P V 2 5 x lO dx f c
Thus
c
10
V at x n
c
1 . 88 3 x 10
-4
-3
J
SHEET ELECTRET
+
-
+
E=O
+
+
+
tP
T E��:o
E=O
+
+
+
+
+
+
! E = CTb/ Eo:: P/Eo
RELATION BETWEEN R AND C FOR ANY PAIR OF ELE CTRODES
Let th e area of one plate b e S and the spacing s . c.
r
c.
0
/a
33
Then
0=0 +
Q =0
0=0
.
CHAP TER 7 CONTINUITY CONDITIONS AT AN INTERFACE
7-1 ( 7 . 1)
D = A /2�r , b o th ins ide and out s ide the dielectric , E = A /2�E E r ins ide, i r o E = A /2�E r out s i de . o
0
V is continuous at the s ur f ac e , but i ts slope dV /dr is smaller ins ide than outs ide . CONTINUITY CONDITIONS AT AN INTERFACE 2 Q / 4�r , b o th ins ide and outside the dielectric, D 2 Q / 4 �E E r in s ide , E. 1 r o E Q / 4�E r2 outs ide . o o 2 ( Q / 4 � E R ) Cl- l / E ) · Thus , at the surfac e , E - E o i r O V i s continuous at the surface , b u t i t s s lope is smaller ins ide 7-2 ( 7 . 1) =
=
the dielectric . 7-3 ( 7 . 2)
ENERGY S TORAGE IN CAPACITORS 2 -6 6 CV /2 = 10 x 10 /2 = 0 . 5 J
W
QV/2
mgh
0.5 J, h
7-4 (7.2)
=
0. 5 / l x 9 . 8 = 51
ENERGY S TORAGE IN CAPACITORS
For Mylar , W I
=
=
mm
3 . 2 x 8 . 8 5 x lO 5 3 3 . 2 x l0 J /m
- 12
8 2 ( 1 . 5 x lO ) /2
One would us e the geometry shown in the figur e .
We need an ab s o lute
mlnlmum of one kilowatt-hour . Then 6 we need 3,600 x 1 0 0 0 , or 3 . 6 x 10 J . As suming 10 0 % e f f i c i ency , whi ch i s
unrealis tic ( the ac tual ove rall ef-
f i c iency might b e, s ay 25 %), the capaci tor would have a volume o f 3 11 m . The d ens i ty of Mylar b eing app roxima tely equal to that o f
water , th e cap a c i tor would have a mas s o f 11 tons , whi ch is ab s urd !
34
7 - 6 ( 7.1)
B OUND SU RFACE
CHARGE DENSITY
7 - 7 ( 7 . 3 ) EXAMPLE OF A LARGE ELECTRI C FORCE 7 2 2 -12 The force per s quare me ter is E E E /2 = 3 5 x 8 . 85 x lO ( 4 x lO ) /2 r o 5 =2 . 5 x l0 p a . The force i s 2 . 5 atmos ph eres . 7-8 ( 7 . 3)
P ERPETUAL-MOTION MACHINE
We have four shee t s o f charge as in the figur e .
Sheets a and b are
(J -+
- O"b
+
+
+
+
+
+ a
b
c o in c ident and are s i tuated in th e Cho o s ing the
fields of c and d .
+
righ t-hand dire c ti on as p o s i tive ,
+
+
the field at the pos i t i on of a and
- (J--
b is E
a,b
[(J -
=
(J /2E - (J /2E 0 0 b
( l- l / E ) (J ] r
=
=
( 1 /2E ) 0
(J /2E E r 0
Then the forces p e r uni t area on a and b are 2 2 - (l- l / E ) (J /2E E (J /2E E ' Fb = -(J(Jb /2E r E 0 r r 0 a r o
F
=
S imilarly , E F
c c
(J /2E
o
- (J /2E + (J /2E = ( 1 /2E ) �2(J - ( l- l / E )(J ] = ( 1+ 1 / E r ) (J /2E o r o o O b
( 1+ 1 / E ) (J(J /2E b r o =
=
2 _(J /2E
2 ( l+l / E ) ( l- l / E ) (J /2E r o r o
F inally , F +F +F +F b a d c 7-9 ( 7 . 3)
0
SELF- CLAMPING CAPACITOR
2 2 2 2 2 ((J /2E ) S ( S /2E ) ( VC / S ) ( S /2E ) ( V /s ) ( E E s i d ( E E /2) S o 0 0 0 r o 8 2 2 2 2 -12 -2 x 4 . 3 8 x lO x 3 6 x lO / = E E S V /2t = 3 x 8 . 85 x lO r O 4 -4 2 2 x ( 7 . 62 x lO ) 1 . 1 x lO N This is a very larg e forc e . I t is approximately the we ight of a mas s
F
=
=
=
=
of one t on . 35
C
7-10 ( 7 . 3 )
ELECTROS TATI C CLAMP S Z 5 a) ( l / Z ) E E ( V / d ) Z x lO P a , d r o
--
Z
=
=
15 )lm �
b) The E in the Mylar is 3000 / 1 . 5 x 10 8 5 6 . 4 x 10 V /m 3 . Z x 3000 x 10 / 1 . 5
5
V /m .
Then the E in the air is
Z 5 Z x 10 P a , d 8 . 4 )lm (l/Z ) E E (V/d) r o Reference : S ta t i c Electrifi cation 19 7 5 , p . Z 1 5 . c)
=
CALCULATING AN ELE CTRI C FORCE BY THE ME THOD OF VIRTUAL
7-11 ( 7 . 3 )
WORK Let the f orce be F .
Assume a vir tual d isplacement dx .
Then the
work d one by the battery is equal to the mechanical work done , p lus th e increase in th e stored en ergy , these two quan t i t i es b e ing equal . Fdx
Thus Z E E 2V /Zs r o
F
7-lZ ( 7 . 4 ) IlE
Z
=
=
Z ( V / Z ) E E 2dX /S r o 6 -I Z -3 3 x 8 . 85 x 1 0 x 0 . 1 x 10 /Z x 10
ELECTRI C FORCE
( d / ax) E
7-13 ( 7 . 4)
Z V dC/Z
Z d (V C / Z )
d (QV/ Z )
Z
i+
( a / ay) E
Z
j+
ELECTRI C FORCE
=
c a / a z ) E�
ZE aE / axi +
1 . 3 3 x 10
. . .
-3
N
ZEllE
S ee P r ob . 7-10 . The mechani cal work done is equal to the increase in ele c tric B o th energies are supplied by the b a t tery . Z Z Z d (V C / Z ) V dC/Z (V / Z ) ( E - l) E 2dx / s r o -3 6 -lZ Z (V / Z ) ( E -l) E 2 /S ( 10 / Z ) ( 3- 1) 8 . 85 x 10 x O . l / lO o r -4 8 . 8 5 x 10 N
energy . Fdx F
=
7-15 ( 7.4)
ELE C TRI C FORCE ON A DIELE CTRI C
( A / Z � E E ) /r r o From P r ob . 4-6 , V ( A / Z � E E ) 2n ( R /R ) Z l r o Z Z 3 Thus E V / r 2n ( R /R ) , dE /dr [ V / 2n (R /R ) ] ( - Z / r ) Z l Z l
From Gauss's law , E
=
=
Th e force is directed inwards . F'
Z
Z
( E -l) E V / 2n ( R /R ) r r o Z l 3 3 3 3.2 x 10 /r N /m
3
Disregarding the sign , 8 . 85 x lO
36
- 12
2 6 3 x 1 . 5 x 6 2 5 x 10 / 2 n 5 x r
-3
6 3 3 . 2 x 10 N/m 3 The gravi tational force p er cub i c meter is 9 . 8 x 10 N/m . S o 3 6 (Electric force)/ ( Gravitational force) 3 . 2 x 10 /9 . 8 x 10 330
b) Near the inner c onductor, F '
=
3 . 2 x 10
/10
-9
=
3
=
7-16 ( 7 . 6)
=
DISPLACEMENT AND P OLARIZATION CURRENTS
From S e c . 5 . 14 the voltage on the cap a c i t or is V
V [ l-exp ( -t/RC) ] , E = V /s C
c
D
= E E E r o
=
E E ( V/s)[ l- exp ( - t/RC) ] r o
dD/d t = E E ( V/S) ( l/RC)exp ( - t/RC) r o P = ( E - 1)E E = ( E -1)E ( V/s) [ 1- exp (- t/RC) ] o o r r dP/dt
=
[ ( E -1)E V/RC s ] exp (- t/RC) o r
7-17 (7 . 7)
DIRECT ENERGY CONVERSION
E E A/s =8000 x8 . 5 x10 r o -6 Q = C V = 340 x 10 x700 C l B 1 C
1
-12
/2 xlO
-4
Q V /2 - Q V /2 = 0 . 2 48 ( 3 500-700)/2 l 2 1 1 4 -4 6 W = 2 . 9 x10 x 2 x10 x30 = 1.74 xl0 J th 4 W /W 0 . 02 = 3 47/1 . 74 xlO e th W
e
3 40flF
347 J
The e f f i ciency is only 2 percen t . References: S . L. Soo , Dire c t En ergy Conver s i on , p 18 4 ; P ro c . IEEE , 51 , 8 3 8 ( 19 6 3 ) .
CHAPTER 8 8-1 ( 8 . 1)
-;-
MAGNETIC INDUCTION ON THE AXIS OF A CIRCULAR LOOP 2 3/2 IJ
fl NI } /2 ( } + z )
o
41T x1 0
2 1T X 10
-7 -7
7xm-?
2
2 3/2 x 100 x 1 x 0 . 1 /2 ( O . 01x z )
/ ( 0 . 0 1+ l )
3/2
-0.3
37
o
+0·3
SQUARE CURRENT LOOP a 2 2 B = 8 ( � o I / 4rr) d�cos 8 / (a +� ) 8 - 2 ( 8 . 1)
f a 2 2 S ( � o I /4rr) f a d� / (a +� ) 3/2
d.£
-
o
1
2 2 � o I /rra
20
o
20
8 - 3 ( S . l) FIELD OF A CHARGED ROTATING DISK a) E O/E ' b ) a vO = wrO O c) A ring of radius r and width dr acts as a current loop . S o , from S ec . S . 1 . 2 , R B = � 0 (wrodr) /2r �wRo o /2 =
=
f
=
o
10 - 6 / S . 85 x 10 - 12 1 . 13 x l0 5 V/m B = 0 . 5 x 4rr x lO - 7 x l0 3 x O . l x 10 - 6 6 . 2 S x lO - ll T d) E
=
=
=
S- 4 ( S . l) SUNSPOTS a) The current loop between rand r + dr carries a curren t 2rrrdro (w/ 2rr) = wrodr . At the center , R B ( � 0 /2) wrodr/r = � o wRo / 2 =
o
f
o
=
2B / � o wR
=
2 x O . 4 /4rr x lO - 7 x 3 x lO - 2 x l0 7 = 20 /3rr
The elec tron density is ( 2 0 / 3rr) / 1 . 6 x 10 - 19 r:;t 10 19 m- 2
b ) The current is the to tal charge divided by the period : I [ rr x 10 14 x (20 /3rr) ] / ( 2 rr /w) ::::; 3 x 10 12 A =
c) The negative charge o f the electrons is neutrali zed by quasi s tationary positive ions .
S- 5 ( S . l) HELMHOLTZ COILS B = 2� oNI} /2 (}+} /4) 3 /2 (0 . 8) 3 /2 � 0NI/a = S . 9 92 x 10 - 7NI /a References : Durand , Magnetos tatique , pp 4 4 , 2 70 ; O ' Dell , The Electro dynamics of Magneto -Elec tric Phenomena , Appendix 4 ; Rubens , Rev . S ci =
38
Ins tr. 1.§., 243 (19 45) 8 - 6 ( 8 . 1) HELMHOLTZ COILS a) In the northern hemisphere the magnetic field points downward . In a N- S plane , looking W , the coils are oriented as in the figure .
\
(5) Q \
c) 8 . 9 92 x lO - 7 NI /a s x lO - 5 , NI s x lO - s / 8 . 9 9 2 x lO - 7
\
\
=
5 5 . 6 At .
d) Try a current o f 2 amperes so as to make the numb er of turns as small as possib le . Then we need at leas t 2 8 turns in each coi l . Then R 2 8 x 211 x l x 21 . 7 x lO - 3 3 . 82$1 =
=
V 7 . 6 4V , P 4 x 3 . 82 15 . 3 W for each coil . No cooling is required for this size of coil . =
=
8 - 7 ( 8 . 1)
B
=
=
LINEAR DISPLACEMENT TRANSDUCER 2 ( )l o la /2) l/[ a 2 + (z - a) 2 J 3 / 2 2 2 3 /2 _ 1 / [ a + ( z+a) J
{
}
()l o la 2 /2) [ l/ ( z 2 - 2az+2a 2 ) 3 /2 2 2 3 /2 J _ 1/ ( z +2az+2a )
()l o I /2a) [ 1 / ( z ' 2 - 2 z ' +2) 3 / 2 3/2J 2 (z ' z / a) - 1 / ( z, +2 z ' +2) =
8 - 8 ( S . 2 ) THE SPACE DERIVATIVES OF B \l.t = d B /dX+ d B /d Y + 3 B /3 z 0 z y x
IN
A S TATIC FIELD
=
d B y /3 Y is positive . By symmetry , 3 B x /3 x is also positive . d B z /3Z is necessarily negative .
S- 9 ( S .3) MAGNETI C MONOPOLES Q *H£ S . 2 7 x 10 - 15 x ( 10 /411 x 10 - 7 ) x 0 . 16 1 . 05 x 10 -S / 1.6 x 10 - 19 6 6 Gev =
=
39
1 . 05 x 10 - 8
J
Then
MAGNETIC FIELD OF A CHARGED ROTATING SPHERE Q / 4�r 2 , V Q / 4�E OR, a E o V/R av ( E o V/R) wR s in 8 E OWV sin 8 2 3 ( 2 /3 ) E 11 wV o ( E 0 wV s in 8 Rd 8 ) (R s in 8 ) /2R o 0
8 - 10 ( 8 . 4) a) a
b)
ex
=
=
c) B
=
J\
=
=
=
o
d) B ( 2 /3) 8 . 85 x lO - 12 x 4� x lO - 7 x 2� x ( 10 4 /60) x lO 4 The field is parallel to the axis of rotation . =
e) m
=
7 . 7 5 x lO - 11 T
J [ ( E oV/R) (2�R Sin 8) Rd8 J ( wR sin 8 ) R sin 8 � ( 4 / 3 ) �R3 E 0 wV �R3 E o wV f s in 3 8d8 ( 1 /2)
�
=
( 4 /3) rr 10 - 3 x 8 . 8 5 x 10 - 12 x h ( lO 4 /60) 10 4 g) (�/4) 10 - 2 I 3 . 882 x 10 - 7 , I 4 . 943 x 10 - 5A f) m
=
=
9 - 1 ( 9 . 1)
2 3 . 882 x 10 - 7 Am
=
CHAPTER 9
DEFINITION OF 110
9 - 2 ( 9 . 1) MAGNETIC FIELD OF A CURRENT - CARRYING TUBE a) B 110 I /2rrr =
b ) A is parallel to the tube and in the same direction as the current c) B inside is zero
�+ + gives the d) +A is as above . I t is not zero . For any curve C , �A·d� flux linkage . If the curve is entirely situated ins ide the tube where B 0, the integral is zero and A mus t be uniform . I ts value + is of no interes t , s ince B O . =
=
B 9 - 3 ( 9 . 1) MAGNETIC FIELD CLOSE TO A CURRENT SHEET .. 1- - - - - Cons ider the dashed curve ,
- - - -
I I
010000 oro � I '-
40
____ ____
I .1
9- 4 (9. 1) a)
0
=
I
b)
B
V AN DE GRAAFF HIGH - VOLTAGE GENERATOR 3 2 E o E 2 x 8 . 8S x 10 - 12 x 2 x 10 /10 - 3 3 . S 4 x 10 - S C /m 2 3 . S4 x 10 - S x O . S x 1T x O . 1 x 60 3 . 33 6 x 10 - 4A . =
=
4 1T x 10 - 7 x ( 3 . 336 x 10 - 4 /0 . S) /2
9 - S ( 9 . 1) B
=
=
=
=
y
=
4 . 19 2 x 10 -19r
SHORT SOLENOID a +L 2 2 3 / 2 2 ()lo la /2) (N/L) d z / [ a + ( z - Z) ] --�-�-----+�Z�--�-r.��L � L -L +L ()loNla 2 / 2L) d ( z - Z) /[ a 2 + ( z - Z) 2 ] 3 / 2 -L +L z2 ()l oNla /2L) z 2 3 a 2{a 2 + ( z - Z) } /2 -L
()l oNI /2L)
J
) I
=
J [
J
L- Z L+Z + 1 [ a 2 + (L - Z) 2 ] 3/2 [ a 2 + (L+Z) 2 ] 3 /2 I
FIELD AT THE CENTER OF A COIL R2 +L /2 ()lo nI /2) x 2 dxd z / (x2+ z 2 ) 3/2 R1 -L / 2
9 - 6 ( 9 . 1) a) B
4 1T x 3 . 3 36 x 10 - 11
()l onI /2) ()l onI /2)
J J
JR2 [x2 (xx�+z z 2 ) ;J +L /2 dX
R1
J
R2
R1
[
L
-L/2
2
Ldx 2 ; 2 (L /4 + x )
1 R2 ()l onIL / 2) 2n{x + (L 2 / 4+x 2 ) 2 } R1 1 2 2 2 ()lonIL /2) 2 na+ (a + { l+ ( l+ S ) 2
�
L
L
+-
2
J
Note : Integrating firs t with respect to x would be much more dif ficult . 41
b ) The numb er of turns is L (RZ -R1 ) n , and the average length of one turn is Z rr ( Rl +RZ ) /Z . Thus the length of the wire is Z !I, = L ( R - R ) nrr (R +R ) = rr R - R Ln = Vn , z Z l l Z where V is the volume of the winding . Also 3 Z !I, = Zrr (a - 1 ) (L/ZR ) R n Zrrn (aZ - l) SRl 3 l l
[
/)
CURRENT DIS TRIBUTION GIVING A UNIFORM B The field inside the hole is the same as if one had two superposed current distributions : a uniform current dens ity throughout the cross - section , p lus a current in the oppos ite direction in the hole . The current in the full cylinder is I a IR Z / (RZ _a Z ) . From Ampere ' s law , B ax - � o I ay / Z rrRZ , B ay � 0 I ax/2rrRZ 9 - 7 ( 9 . 1)
=
The current in the small cylinder is Ib Bbx � o \ y / Zrra Z � o l aY /ZrrRZ , Bby = - � o I a (x-b) /ZrrRZ B ax + B bx
Bx
By
=
B ay + Bby
0,
Z � o I ab /ZrrR
=
� o Ib /Zrr (RZ - a Z )
The field is therefore uniform ins ide the hole . Note that B is proportional to b . Thus B o when b changes s ign with b . Also , when a + R, b + 0 and B O .
0,
and B
=
9 - 8 ( 9 . 1) SADDLE COILS This current dis tribution is ob tained by superposing two full cylinders of current flowing in opposite direc tions . x Let I be the current flowing through the coil . Then the cur rent I ' that would flow through one complete circle is related to I as follows : ( I = l - ( Z /rr) cos - \ a/R) - ( a/R) ( l - } /RZ ) � I ' \ Ins ide the lef t -hand circle , at the radius r , the B due to that s ide is B � o ( I ' /Zrrr) ( r Z /R Z ) ( � o I ' / Zrr) r/RZ . S o
�
J
[
=
42
+ y 2 ] ! IR2 , B 2 = ( )lo I 12'!TH ( a-x) 2 + y 2 ] ! IR2 = -B ly I [ (a+x) 2 + y 2 ] ! + B 2 y I [ ( a-x) 2 + y 2 Fb 1 1 B l cos 8 l + B 2 cos 8 2 B l ( a+x) / [ ] 2 + B 2 ( a-x) / [ ] 2 )lo I ' a/'!TR2 I
Bx
By
o
=
S o B is uniform and parallel to the y-axis . 9-9 ( 9 . 1)
a)
TOROIDAL COIL
J J B . da = r t '!T 2'!T)l(R+pNI cos 8 ) pdpd8 0
0
( )l0 NI /2 '!T)
0
rt
0
0
'!T
d8 R+ p cos 8 pdp
The integration with resp ect to 8 mus t be done with care , taking into account the two branches of the curve . We integrate from -'!T /2 to +'!T/2 , where cos 8 is pos itive , and then from '!T/2 to 3'!T /2 , where cos 8 is negative . Then '!T/2 d 8 3'!T /2 d8 '!T/2 d8 '!T /2 d 8 2'!T _d_8--:: + + -=:':: '::: -= R-p cos 8 8 cos R+p -:::8 cos p R s 8 c R+ P o + R+p c o s 8 - '!T /2 -'!T/2 '!T /2 o - '!T /2
J
J
J
=
-p I (R�- p 2 ) !
tan )
43
J
J
tan ( 8 / 2 ) 1
\
] '!T/2
-'!T/2
Tr/2 f R-pd8cos 8 -Tr/2 J
o
=
2 Tr _-=d-=-8_ R+ p cos 8
4 1 ) arc tan �-p 1. + arc tan �+p ! 2 2 _ (R _ p 2 ) ! \ (R _ p 2 ) 2 (R P ) 2 1
Since arc tan a + arc tan b = arc tan [( a+b)/( l-ab )] , Tr d 8 2Tr 4 ( Tr / 2 ) 4 2R/ (R2 _ p 2 ) ! tan arc 2 -p 2 ) = (R2 _ p 2 ) ! - (R2 _ p l ! 2 2 2 2 R+p cos 8 (R _ p ) ! l- ( R -p ) / (R 0 r Thus 1> ( � oNI /2Tr) f 2 2Tr 2 1 p dp � oNI [ R- ( R2 -r 2 ) 1.2 ] (R _ p ) 2 o
}
1
t
The integration is more difficult wi th Cartes ian coordinates . 1':\ . b) B = � oNI [R- (R2 -r 2 ) ! ] /Trr 2 at the radius V\ S et a{ = R + p cos e . Then � oNI /2 Tr I/{ � 0 NI [ R- (R2 -r 2 ) 1.2 ] /Trr 2 � � r 2 / 2 [ R- ( R2 -r 2 ) ! ] r 2 / 2R[ 1- ( 1-r 2 / R2 ) ! ] =
=
=
For r 2 « R2 ,
� II<
'"
r 2 /2R(r 2 /2R2 )
=
R
CHAPTER 10 There is an interes ting ar ticle on the crossed-field mas s spec trometer in The Journal of Phys ics E , S cientific Ins truments , Volume 10 ( 19 7 7 ) page 458 . 10-1 ( 10 . 1) THE CYCLOTRON FREQUENCY mv 2 /R. a) The cen tripe tal force being BQv , BQv b) Then W viR = BQ/m . c) BQ/2Trm = l x 1 . 6 x lO -19 /27f x 2 x 1 . 7 x lO -2 7 = 7 . 5 megahertz . 10-2 ( 10 . 1) MOTION OF A CHARGED PARTI CLE UN A UNIFORM B -+ The velocity component p arallel to B is unaffected . The component -+ normal to B gives a circular motion as in the preceding problem . 44
10 - 3 ( 10 . 1) MAGNETIC MIRRORS The figure shows part of a helical orbit for a positive particle . The particle drif ts to the right . The magnetic force points to the lef t . After a while , the drif t velocity will also p oint to the left . Reference . There is a good article on the magnetosphere in Contemporary Physics , �, 165 ( 19 7 7 ) . 10 - 4 ( 10 . 1) HIGH ENERGY ELECTRONS IN THE CRAB NEBULA a) W 2 x l0 14 x 1 . 6 x lO - 19 '" 3 . 2 x lO - 5 J . 3 . 2 x lO - 5 /9 x l0 16 3 . 6 x lO - 22 kg , mc 2 /m 0 c 2 3 . 6 x lO - 22 / 9 . l x lO - 3l 4 x lO S
b) m
=
m/mo
=
c) R
=
d) ( 2 1T x 3 . 4 x l0 13 /3 x lO S ) / ( 2 4 x 3 , 6 00)
=
S . 2 days
10 - 5 ( 10 . 1) MAGNETIC FOCUSING a) An electron goes through one full circle in a time T 2 1T/W . During that time it travels a dis tance L vx T . S o 1. 1. L ( 2eV/m) 2 2 1T (m/Be) 2 3/2 1T (mV/e) 2 /B , B 2 3 /2 1T (mv/e) ! /L =
b) B IN '
=
=
2 3 / 2 1T ( 9 . l x lO - 3l x 10 4 /1 . 6 x 10 - 19 ) ! /0 . 5 = 4 . 2 4 x lO - 3 T
B / �o
=
3 37 3 A
10 - 6 ( 10 . 1) DEMPSTER MAS S SPE CTROMETER a) mv 2 /R = BQv , mv BQR , (2m ( 1 / 2 ) mv 2 ) 1.2 1. QR2 B 2 /2V ( 2mQV) 2 BQR, m
BQR
1
b) B ( 2mV/Q) 2 /R For HI + B l ( 2 x 1 . 7 x lO - 2 7 x lOOO/1 . 6 x lO - 19 ) ! /0 . 06 0 . 11 T . 0 . 09 4 T . '
45
=
7 . 7 x lO - 2 T .
Note that 6m/m 26B /B . Thus , for large m ' s , where 6m/m b ecomes small from one iso tope to the nex t , 6B /B becomes even smaller . =
c) B
1,.
( 2mV/Q) 2 /R ( 2xl . 7xlO -2 7 AxlOOO / l . 6xlO -19 ) ! /0 . 06
( 3 . 4xlO - 5 / 1 . 6) !A! /0 . 06 = 7 . 6 8xlO - 2A! where A is the atomic weight . This value of m is approximate . =
10- 7 ( 10 . 1) MAS S SPE CTROMETER mv 2 /R BQv , R mv/BQ x = 2R ( 2 /B ) (m/Q ) v The time of flight from A to the target is TIR/v TI /B (m/Q) . During that time the ion drifts through a distance y = ( 1 / 2 ) (QE /m) ( TIID / BQ) 2 ( TI 2 E /2B 2 ) (m/Q) . Reference : Rev . S ci . Ins tr . �, 819 ( 19 7 4) . =
=
z�o
=
=
=
10-8 ( 10 . 1) HIGH-TEMPERATURE PLASMAS 1,. a) mv2 /R BQv, R mV/BQ ( 2mmv 2 /2) 2 /BQ . : 0 . 225 m =
=
=
b ) A D+ ion has the same velocity , but half the mas s , so R 113 mm . Reference : Glas s tone and Loveb erg , Controlled Thermonuclear Reactions pages 156 and 395 . =
10-9 ( 10 . 1) HIGH TEMPERATURE PLASMAS a) By symme try , B can only be azimuthal . But the line integral -+ -+ of B · d � over a circle perpendicular to the paper and with its center on the axis of symm e try mus t be zero , s ince the net cur rent is zero . Then B O . S imilarly , B = 0 ins ide the inner cylinder . =
b ) See figure . c) I t bends downwards .
46
0
0
0
0
B
=
0
--
0
0
0
0
0
0
r/ ///////_//// /////
®
(f)
®
(£) (t) B ":. 0 0
(±) (±) B "= 0
(t)
(t)
(±)
®
0
d) I t b ends upwards
e) They als o return to the discharge .
Reference : G1as s tone and Loveb.erg , Controlled Thermonuclear Reactions, p 2 78 . 10- 10 ( 10 . 1) a) I
ION BEAM DIVERGENCE
vA , A = I /v
Q ( A/21TE o R) = QI/21TE 0Rv
b ) QE
QV ( ll o I /2 1TR) = QIll o V/21TR d) QE - QvB (QI /21TR) ( 1 ! E o v - ll oV) c) QvB
=
10-11 ( 10 . 1) ION THRUSTER S ee the s olution to Prob . 2-14 . Here , the force exerted on the ej ected fuel , in the reference frame of the vehicle, is BIs , or m ' v . I 2 R I 2 ( s /oA) , where A is the area of one of the electrodes , =
n
C or D .
J
/[
_2_ l / ( l+P D /P G) = 1 1 + D oA BIsv 1 / ( 1 + 2m ' /oB 2 T)
Also , 2 I / oABv
=
2oE /oBv
2J /oBv
=
1 / ( 1 + 2BIs /oB 2 w)
2E /Bv ,
n 1 / ( 1 + 2E /Bv) . As v increas es , n ->- l , and n ", l for v » 2E /B . 10-12 ( 10 . 3 ) GAMMA ( 1- S 2 ) ! 1 / 1.01 , S 2 =
1 - 1 / lt0 1 , S =, 0;{9 9504 .
10-13 ( 10 . 5 ) REFERENCE FRAMES 1 1 / ( 1-1/4) 2 1 . 155 1 . 155 ( 1-1 . 5x10 8x1) -1 . 7 32 x 10 8 m
Y
=
=
Y1 = 1 m , z2 = zl = 1 m 1 . 155 s . 1 . 155 ( 1-1/2�3x10 8 ) =
47
10-14 ( 10 . 5 ) REFERENCE FRAMES 1 . 15 5 as above y \ 1 . 7 32 x 10 8m , 1 . 155 ( 1+l . 5xlO 8xl)
1m , t l
zl
Yl
1 . 15 5 s
..;- ..;- ..;- ..;..;- ..;- ..;..;- ..;-
10-15 ( 10 . 8) HALL EFFECT
a) v
vxB
=
-M(E +vxB) , where i
vx 0
=
vxi + v j + v z k Y vx
j
k
v Y 0
vz
=
B
[
-M(E -vxB ) ' v z O . Y M(-E x +MBE ) -ME x +M2 BE "';l "';l -M[E x-M(E -vxB ) B ] Y 1+M2 B 2 l+M2 B 2
-M(E x +v B ) , v Y Y
b) vx
� [f
=
MB ( -E x MBE y ) l+M2 B 2
-M y -
=
Jy
=
)
]
]
2 2 MBE M _ _ (E +MBEx ) - M 2 Ey + l+M2 B 2 y 1+M B 1+M2 2 neM(E x-ME B) / ( l+M2 B 2 ) Y neM(E y +ME xB ) / ( l+M2 B 2 ) (b / a) MVxB ( 10 -3 /5xlO - 3 ) 7 x 1 x 10 - 4 1 . 4 x 10 -4 V . -M
c) Vy
=
=
v Y
Jx
]
-M E xi+E j+v B t-vxB r , Y Y
�
:
__
=
d ) When E
0, Y Jx = neME x / ( l+M2 B 2 ) , I x b cneM(Vx /a) / (l+M2 B 2 ) , R Vx II x a ( 1+ � B 2 ) /b cneM, LIR/Ro M2 B 2 Let us calculate the mob ility in copper . J p v aE , v (a /p ) E . · . a /p a /ne 5 . 8 x lO 7 / 8 . 5 x 10 28 x 1 . 6 x 10 -19 Thus the mob ilitY :Ls -3 . = 4 . 3 x lO References : H . H . Wieder , Hall Generators and Magnetores is tors ; =
=
=
=
•
=
=
=
48
=
=
H . Weis s , S tructure and Applications of Ga1vanomagnetic Devices . 10-1 7 ( lO . 8) V vBa
ELECTROMAGNETIC FLOWMETERS
=
CHAPTER 11 11-1 ( 11 . 1)
a) Bv 9, b) Zero .
=
BOAT TESTING TANK 2 x 10 -5 x 20 x 3 1 . 2 mV =
11-2 ( 11 . 1) EXPANDING LOOP a) I = Bvs /R b ) (B ls ) v (Bvs) 2 /R c) I 2 R = (Bvs ) 2 /R. The power exp ended to move the bar appears as heat in the resis tance R . =
11-3 ( 11 . 2)
INDUCED CURRENTS
t
t
Reference : Rev . S ci . Ins trum . , � , 1581 ( 19 7 7 ) . 49
11-4 ( 11 . 1) INDUCED CURRENTS a) Counterclockwise b ) Counterclockwise c) S ince the flux linkage is con -+ -+ s tant , and s ince v x B = 0 , the induced electromotance is zero .
11-5 ( 11 . 2)
'1J
=
dfl/dt
INDUCED ELECTROMOTANCE NA( dB /dt) = 100 x 10 - 2 x 10 -2 x
sin ( 2Tf x 60 t)V We have disregarded the s ign .
2Tr
x 60 s in ( 2 Tf x 60 t)
3 . 77
11-6 ( 11 . 2 )
ELECTROMAGNETIC PROSPECTION
B
J
a) The induced elec tromotance is azimuthal . Over a circle of radius r , 2TfrE Tfr 2B o wsinwt , E = (rB 0 w/2) s inwt . The induced current density is a E and is also azimuthal J (arB o w/2) s inwt . =
=
b) With our s ign convention , a positive J gives a posi tive B . At t 0 , dB /dt 0 and J = O . Then , as B decreas es , J increases as per Lenz ' s law , etc . =
=
11- 7 ( 11 . 2 )
INDUCTION HEATING a) -d� /dt - Tfr 2 (d /dt) ( V o N ' I 0 coswt) b ) The length of the conductor is 2 Tfr and its cros s-section is Ldr . =
50
Hence R c) dP av
=
21Tr /crLdr 2 2 ( 11 U WN ' I 0 ) o 2x2 rrr /crLdr
=
[Ilo2rrcrw2N ' 2
1
0
)
2 / 4 Lr 3 dr ,
Th e average va 1ue 0 f S ln equa 1 to 0 . 5 . , 2 wt b elng ' R d) P av ( ) L r 3 dr ( ) LR4 / 4 = ( 4rr x lO -7 x 21T x 60 x 5000 x 2 ) 2
J
=
o
( 1T x lO 5 /16) ( 6 x lO -2 ) 4 x l
=
5 . 71
W
Note The power dis s ipated in the winding is I 2 R , where R is its resis tance . The conduc tivity of copper being 5 . 8 x 10 7 siemens per meter , if there are n layers , the cross-section of the wire is rr (n /5000) 2 /4 and R 2 1T x 6 x lO -2 x 5000 / [ 5 . 8 x l0 7 1T (n/5000) 2 / 4 J , I 2 R '" 10 3 I 2 /n 2 . =
I f n 10 , I 2 R � 40 W . Reference : S t andard Handb ook for Elec trical Engineers , p 22-28 and following . =
z
11-8 ( 11 . 2 ) INDUCED ELECTROMOTANCE y vt ( w A /21T) t , wt = 21TY / A Y + A /4 � B o s in ( 2 1TY / A) s inwt ( A / 4) dy y =
=
J
[
B o sinwt ( A/4) ( A /2 1T) -cos ( 2 1TY /A)
] y+
A/4
y U 2 /81T) B o s inwtC -cos ( 2 rrY / A+1T/2) + cos ( 2 rrY /A) J
CU"=
( A 2 /8rr) B o ( s in 2 wt + s inwtcoswt) (B VA /4) ( s in2wt + cos2wt) O
11-10 ( 11 . 4) THE TOLMAN AND BARNETT EFFECTS In the reference frame of the conductor , the force on a particle of charge -e and mas s m is Reference : Landau and Lifshi tz , Electrodynamics of Continuous Media , p 210 . 51
11-11 ( 11 . 5) ELE CTRIC CONDUITS From the definition of A (Eq . 8-18) , A is p arallel to the wire . Then aA/ a t is also parallel to the wire and , if there is a single wire , there is a longitudinal induced electromotance in the wire . Reference : S t andard Handb ook for Electrical Engineers , Sec . 17-11 . 11- 12 ( 11 . 5) THE POTENTIALS V and A -+
-+
-+
S ince Vx (A - VC) =: V x A , B is not affected . Also -V(V + ac/ at) - ( a / a t) (A - VC) =: -vv - aA / a t and E is not affected either .
CHAPTER 12 12-1 ( 12 . 1) [H ]
=
[ Jl o ] [L ] , [ Jl o ]
[H] / [ L ]
12-2 ( 12 . 1) MUTUAL INDUCTANCE Assume a current I in the wire . toroidal coil is
The flux linkage through the
( Jl o IbN /2TI) �n C l+b /a) ,
M
=
( Jlo bN /2TI) �n ( 1+b /a) .
12-3 ( 12 . 1) MUTUAL INDUCTANCE a) From S ec . 8 . 1 . 2 , coil a produces at b a magnetic induction B = JloN a I a a 2 /2 C } + z 2 ) 3 /2 . S o M = "' ab /I a = Jl oN a a 2 /2 ( a2 + z 2 ) 3 /2 Nb TIb 2 TIJl oN aNb a 2b 2 /2 ( a 2 + z 2) 3 / 2 b ) M varies as the cosine of the angular displacement . c) No .
{
}
=
12-4 ( 12 . 2 ) A OUTSIDE A SOLENOID The magnetic flux inside the solenoid is TI R2Jl oN ' I . Then the electromotance induced in a loop of radius r > R coaxi al with the solenoid is TIR2 Jl oN ' dI /dt = 2TIrdA/dt , A = (Jl o /2r) N ' R2 I . 52
12-5 (12 . 2) A INS IDE A SOLENOID The magnetic flux inside a loop of radius r < R , coaxial with the solenoid , is ( d/dt) ( rrr 2 )JoN ' I) 2 rrrdA/d t , A ( )J o /2)N ' I. r . 12-6 ( 12 . 3) MAGNETIC MONOPOLES The flux due to the current mus t exactly cancel that due to the monopoles . Then -15 LI nNe* , I nNe* /L 100 x 1 , 200 x 8 . 26 x 10 / 75 x 10 -3 1 . 322 x 10 -8 A =
_
12- 7 (12 . 3 ) ROGOWSKI COIL Reference : Glass tone and Loveberg , Controlled Thermonuclear Reactions , p 16 4 ; Rev . S ci . Ins tr . �, 6 6 7 ( 19 71) . 12-8 (12 . 3) INDUCED CURRENTS a) In the azimuthal direction , R 2 rra/crb£, B )J o I / £ , � L � /I )J o rra 2 / £ . b ) L 4rr x lO -7 x rr x 2 5 x lO -6 /1 = 9 . 87 x lO -11 H R 2 rr x 5 x 10 -3 /5 . 8 x 10 7 x 10 -3 x 1 5 . 42 x 10 -7 " =
=
L /R
=
1 . 82 x lO -4 s .
12-9 (12 . 3) COAXIAL LINES In the annular region , B is due only to the current in the inner conductor , from Ampere ' s circuital law . Thus R2 dr/r B )J o I /2 rrr , � ' = ()J o I /2rr) Rl =
J
12-10 ( 12 . 3) COAXIAL LINES From Ampere ' s circuital law , the B in the annular region between the conductors is the same at all frequencies . Inside the conduc tors there is more field at lower frequencies . Hence Wm is larger at low frequencies and L £f > Lhf · 53
12- 1 1 ( 12 . 5) LONG SOLENOID WITH CENTER TAP 2 2 2 2 LAC ( � oN /�) rr R , LAB LBC ( � oN /2 �) rrR , M O . Our formula for a long solenoid is based on the assumption that B is � o (N /�) I ins ide , and zero outs ide . Wi th this assumption , the coup ling coefficient k is zero , and M is zero . =
12-1 3 ( 12 . 7)
=
=
VOLTAGE SURGES ON INDUCTORS
I +
R L
( R + Rs ) I + LdI /dt "" Rs I + LdI /dt , 100 "" Rs I - 10 4 4 Th e v o1 tages acros s Rs an d 1 are b oth ab out 10 V . a) V
=
b ) Connect the diode as in figure b . current is I - (V/R) e -Rt /1 10 e-t .
Upon opening the switch , the
=
12-14 ( 12 . 7) TRANSIENT IN R1C CIRCUIT In circui t a, V/V s 1-exp (-t /RC) 3 In circuit b , 1 10 4 x 10 -6 / 4 2 . 5 x 10 =
=
=
1d 2 Q / dt 2 + RdQ /dt + Q / C
=
Vs .
H,
As in P rob . 12-1 5 , the particular solution is Q .1 1n 2 + Rn + l / C 0 , n = - (R/2L) ± (R2 / 41 2 - 1/1C) 2 , 3 R/2L 100 /5 x 10 2 x 10 4 1/1C . =
=
=
=
Thus the two values o f n are equal and Q (A + Bt) exp (-Rt /21) + V s C . =
At t
=
0, Q
=
0 and A
=
-V s C . Also ,
I dQ /dt exp (-Rt /21) [B - (R/21) (A + B t) J . At t 0 , I = 0 and B - (R/21) V s C . =
=
=
54
Vs C
-Vs C ( 1+Rt /2L) exp (-Rt / 2L) + V s C 2 /RC , Q Vs C[ 1 - ( 1+2t /RC) exp ( - 2 t /RC ) J ,
Finally , Q
S ince R/2L
v
=
Q / C , V/Vs 1- ( 1+2 t /RC) exp ( - 2 t /RC) . Summarizing , V/V s l-exp ( - t /RC) , for circuit a . =
V/Vs
=
1- ( 1+2 t /RC) exp (-2t /RC) , for circui t
S etting t /RC = t ' , V/V s l - exp ( - t ' ) , for a , V/Vs 1 - ( 1+2 t ' ) exp ( - 2t ' ) , for
b.
/. 0 b.
The figure shows Q /V s C as a function of t for the two circuits . The charges are the same at t 1 . 2 6 RC . The circuit with the inductor is slower at firs t , and then faster . The inductor is not us eful . =
0
12 - 15 (12 . 7 ) TRANS IENT IN RLC CIRCUIT a) LdI /dt + RI = V, I Aexp ( - Rt /L) + V/R S ince I 0 at t 0 , I (V/R) [ l - exp ( -Rt /L) J 2 b ) LdI /dt + RI + Q / C V , Ld Q /dt 2 + RdQ /dt + Q / C The particular solution is Q vc = 10 -4 .
t'
3
=
=
=
=
=
lO [ (l - exp ( -lOt ) J .
V
=
The complementary function is Q Aent , n [ - R± (R 2 - 4L / C) ! J / 2L - 5 ±10 3 J. , Q exp (-5t) (BcoslO 3 t + Ds inlO 3 t) + 10 - 4 . . S lnce Q 0 at t = 0, B - 10 -4 , Q exp (-5t) ( - 10 -4 cosIO 3 t + DsinlO 3 t ) + 10 -4 , dQ /dt exp ( - 5 t ) ( 5 x 10 - 4 cosIO 3 tHO -1 s inlO 3 t - 5Ds inlO 3 t +10 3 Dcos lO 3t) I Als o , 1 = 0 at t O . Then D -5 x lO - 7 Q = exp (-5t) ( - 10 - 4 cosIO 3 t -5 x 10 - 7 s inlO 3 t ) + 10 - 4 ;::; -10 - 4 exp (-5t) cos IO 3t + 10 -4 . I -10 -4 exp ( - 5t) ( - 10 3 s inlO 3 t) + 5 x 10 -4 exp ( -5 t ) coslO 3 t -4 = 0 . lexp ( - 5t) s inlO \ + 5 x 10 exp ( - 5 t ) coslO \ ;::; 0 . lexp ( - 5t) s inlO \ . =
=
=
=
=
55
CHAPTER 13 13-1 ( 13 . 1) MAGNETIC FORCE BIL = 5 x 10 -2 x 400 x 5 x 10 - 2
--
13-2 (13 . 1) MAGNETIC FORCE a) Let the wire have a cross-section a , and let the current density be J . For a length o f one meter , BJa p ag , J pg/B = 8 . 9 x l0 3 x 9 . 8 x lO - 4 8 . 7 x l0 8A/m2 8 . 7 x l0 8 a A I 1 / 5 . 8 x 10 7 a . "/m b) R' p' ( 8 . 7 x l0 8 a) 2 /5 . 8 x l0 7 a 1 . 3 x lO lO a W/m 10 - 8 , then p ' 130 W/m . The wire will become very hot . If a =
=
=
Convection will spoil the measurement .
c) In the Northern hemisphere there is a South magnetic pole . lines of B point S outh . The current mus t p oint Wes t .
The
d ) At the p oles the lines o f B are vertical and the magnetic force on a horizontal wire is horizontal and perpendicular to the wire . 13-3 (13 . 1) MAGNETIC FORCE -F 50 x 100 x 0 . 5 x 10 -4 x s in 70 a
0 . 235 N
=
13-4 ( 13 . 1) +
F
=
+
MAGNETIC FORCE +
I a y),h d,Q, a x B
=
+
+
-I aB x ':I',h d,Q, a
O.
13 - 5 ( 13 . 1) ELECTROMAGNETIC PUMPS Cons ider an element of volume dxdydz , as in the figure , with the current flowing along the z-axis . + The current is Jdxdy . Both J and + B are uniform ins ide the infini tesimal element of volume . The force per unit volume is
56
-
-
8
J dx
dx
dy
HOMOPOLAR GENERATOR AND HOMOPOLAR MOTOR
13-6 ( 1 3 . 1)
v
2 B wR /2
=
=
1 3 - 7 (13 . 1 )
=
I x ( 3 000 X 2 1T / 6 0 ) 0 . 2 5 / 2
=
39 . 2 7 V .
HOMOPOLAR MOTOR
The curr ent has a radial comp onent p o in t ing towards the axi s .
The
a z imuthal c omp onent o f the current g ives a B p o in t ing to the righ t . The whe e l turns counterclo ckwis e . 13-8 ( 1 3 . 2 )
MAGNETI C P RE S SURE -
-
- - -
-
- -
.
-----c:)--::=::----
---{Br--.__ --------.. - - -
- _.
�
---{B,---
------c:)r----
------c:)1----
a) We h ave p r e c i s e ly the s i tuation des crib ed in S e c . 13 . 2 .
b ) Inside the inner s o lenoid , there i s zero magnetic f i eld . B e tween 2 the two s o l enoids th e f ield is B . The magnetic p r e s s ur e B / 2 � pushes o inward on the inner s o lenoid . 13-9 ( 13 . 2 ) 2
a) B / 2� o
MAGNETI C P RE S S URE 2
=
B / 2 X 41T x lO
�
4 B
2
-7
2 -7 -5 P a '" ( B / 2 x 41T x lO ) 10 a tmo spheres
atmospheres .
b)
57
c) ( i) The pres s ure is always equal to the energy den s i ty ( ii ) The e l ectric "pres sur e " we have cons idered is as s o ciated with th e fact tha t lines o f force are under tens ion .
This "pres s ur e " is
(We have no t cons idered the r epuls ion b e tweeen 2 electric lines of force , whi ch gives a p o s i t ive pres s ure of E E / 2 . o For examp l e , if we have two electric charges of the s ame s ign , one 2 can f ind th e corre c t force o f r epul s i on by in tegrating E E / 2 over o the p lane half-way b e tween the charges , where the lines of force clash) . always attractive .
( iii) The magnetic p r e s s ur e we are concerned with here is as s o ciated with the lateral repul s i on b e tween l ines of force .
This p r e s s ure is
repuls ive .
I n P rob . 15-6 we are concerned wi th the tens ion in the 2 lines of force , which gives an at trac tive "pressure" of B / 2 )1 . o ( iv) In pratice , the electric " p res sure" is nearly always negligib le
while magnet i c pres sure is o f t en larg e . For examp le , a large E of 6 10 V /m g ives an electric " p re s s ure" o f 5P , wh ile a large B o f 1 T 5 g ives a magnetic p r e s s ure of 4 x 10 P . 13-10 ( 1 3 . 2 )
MAGNETIC PRE S S URE
0 ins ide , from Ampere ' s circuital law . a) B = )1 I /2 �R out s ide , B o 2 , 2 2 ( 1 /2 )1 ) ( )1 I / 2 �R) = ( )1 / 8 � ) ( I /R) . o o o -7 2 8 -8 7. b) P ( 4 rr x lO / 8 � ) ( 9 x 10 / 2 5 x lO ) = 5 . 7 3 x 10 Fa :::; 5 . 7 3 a tmom spheres . =
Thus P m
Re ference : J . Phys . D . App 1d Phy s . �, 2 1 8 7 ( 19 7 3 ) . 13-11 ( 1 3 . 2 )
--
MAGNETI C P RE S S URE 5 -7 "" 4 x 10 P a "" 4 atmospheres .
a) P = 1 / 8 � x 10
b ) The p r e s s ure would b e unchanged , s ince B is uniform in side a long s ol eno i d . 13-12 ( 1 3 . 3 ) ENERGY S TORAGE 2 3 12 -12 E /2 8 . 85 x lO x 10 / 2 4 . 43 J /m o 2 5 3 -7 1 / 8 � x 10 B / 2 )1 = 3 . 9 8 x l0 J /m o E
13-13 ( 1 3 . 4 )
=
MAGNETI C PRE S S URE
a) The magneti c force i s 2 �R2p ' It a c t s through a d i s tance dR . m Then the work done by the magnetic force is 2 �R2p dR . m 58
b ) The mechanical work done i s Z Z Z 'TTR f(, (B / Z )l ) d R = Z 'TTR j(, [<)l NI / f(, ) / Z )l J dR o 0 0
Z 2 'TT )l I N RdR / ,Q" 0
=
Z
c) I (Nd 1»
= INd ( 'TTR )l NI / f(,) o
13-14 ( 1 3 . 4)
FLUX COMPRE S S I ON
a) As the tub e shrinks in d i ameter ,
E x p l o . i ve
an a z imuthal current is indu c ed that main t ains the enclosed mag netic f lux app roximately con s t ant . Z Hence B B (R / R) . o 0
Tu b e
�
b) B
c) B
�
)l a , a o 9 10 A/m
�
10 ( 10 / 1)
=
3 -7 10 / ( 4 'TT X 10 )
=
(1) Z
10
=
3
(Z)
T.
d) The s o lenoid maint ains a con s t an t B
i n i ts interior . The curren t o in the tub e in creas e s the induction ins i de the tub e to B . Thus the increas e in magne t i c energy i s lM m
(
Z Z Z 'TTR L B -B o 'TT x ( 10
-4
Jj
Z )l
Z = 'TTR L
o
(
R
o
4
/ R4 _ 1J B o Z / Z )lo '
4 Z -7 /4 ) x O . Z ( 10 _ l) 10 / 8 'TT x lO
�
6 x l0
6
(3) ( 4)
J.
The s ource s upp lies an extra en ergy oo Z 6W N I ( d 1> / d t ) dt = NI 6 1> NI 'TTR ( B -B ) , s 0 0 0 0 0 Z Z Z Z Z Z 'TT R (B L / )l ) R NI IlR R , R - 1 B R -1 B , o 0 / 0 0 0 0 1 0
J
(
J
1
Z Z 2 ( 'TT / )l ) LB (R -R ) o o o �
1 . 3 x l0
5
=
=
=
( 'TT / 4'TT x lO
-7
[ ' )
f
R
R o
J.
(7)
Note that
J
[
Z Z Z 1TrL ( B -B ) / Z )l dr o o
( 'TTL / )l ) B o 0 6W
m
(6)
2 -Z -4 ) O . Z x lO ( 10 -10 ) / 4
Th e exp l o s ive s up p l i e s an energy -
(5)
Z
[
R
0
4
- ( 'TTL / )l ) o
Z Z /ZR + R / Z _ R
= 6W + W exp 1 s
0
z)
�
J[ / R
R
R
o
6 x l0
0
6
4
J.
4 r _l
h
) 0
Z
rdr ,
(8)
(9)
Note also tha t , although the magnet i c f ield j us t ins ide the s o l enoid
59
is unaffected by the current I in the tub e , the current I produces a a A/a t in the s o leno id that makes W
'" O . The exp los ive s upp lies mos t s We have neglected the mechanical en ergy required t o
of the energy .
crush the tub e , acous tic energy , e t c . 13- 15 --
( 13 . 4 ) PULSED MAGNETI C FIELDS 6 6 2 a) W = ( B / 2 ]1 ) V = 4 x 10 J . C o s t ", $ 8 x 10 o m
6 6 6 b) W 4 x 10 J 4 x 10 / 3 . 6 x 10 '" 1 kwh . m depending on p r evai l ing rates . =:
c) p
=:
=:
2 B / 2 ]1
13-16 ( 1 3 . 5 ) a) W m
=:
'" 4 x 10
o
9
P a '" 4 x 10
MAGNETI C ENERGY
I ' / 2 + I fl /2 b b / a
I ( fl + a aa
=:
b) W m 13-17 ( 13 . 5 ) a) W m
=
LI / 2
=
13-18 ( 13 . 5 )
1 (11) / 2 =:
2 INTIR ]1 NI / 2 £
=
( V /L) t 2 2 (V / 2L) t
2 L ( Vt /L) / 2
=
Q/C
=
2 i i i ) CV / 2
a)
2
E E /2 r 0 2 B / 2 ]1 o E
=
2 2 (V /2L) t
=
( I/C) t
i i ) ( 1 /2 ) [ ( I t /C ) I J t
13-19 ( 1 3 . 4 )
2 2 2 ]1 TIR N I / 2 £ o
0
V, I
=
2 11 / 2
i) V
I fl / 2
=
i i ) ( 1 / 2 ) (Vt /L) Vt
b)
\ a ) /2 + I b ( flab + \ b ) / 2
ENERGY S TORAGE
i) LdI / d t
} ii)
atmospheres .
ENERGY S TORAGE
2
b) W = I (N ¢ ) / 2 m
a)
4
Co s t '" 2 to 10 cents ,
=
2 2 ( 1 /2C) t
2 C (It/C) /2
=
2 2 (I /2C) t
ENERGY S TORAGE 3 . 2 x 8 . 85 x 10 6 4 / 2 x 4TI x 10
2 7 b ) ( TID / 4 ) L x 2 . 5 x 10
=
-12
-7
x 10
16
/ 2 = 1 . 4 x 10
= 2 . 5 x 10
7
J /m
5
J /m
3
3
7 10 13 2 ( TID / 4 ) 2 0 D x 2 . 5 x 10 = 10 x 3 , 6 00 = 3 . 6 x 10 , 60
D = 45 m , L
900 �.
=
2 5 0 atmospheres A = 1TDL = 1T x 45 x 9 00 = 1 . 2 7 x 10
5
m
2
Referen c e : Foner and S chwar t z , Superconduc ting Machines and Devi c e s , p 41 . SUPERCONDUCTING P OWER TRANSMI S S ION LINE
13-20 ( 13 . 6 )
10
o 6
N /m
2 Il I / 2 1T ( D / 2 ) o
b) B
2 B /2 Il c) W
2 -7 11 5 2 -2 ll o I / 2 1TD = 4 1T x 10 x ( 10 / 2 x 10 ) / 2 1T x 5 x 10
( 11 I /2 1TD) I
a) F
o
= 6 4 / 4 1T x 10
( 1 / 2 ) 11
=
=
2
C o s t = 5 x 10
2 Il I / 1TD = 8 x 10 o -7
= 5 x 10
= ( 1 / 2 ) ( 6 . 6 x 10 -3
( 8 . 2 5 x 10
10
7
J /m
-7
-7
5 -2 x 5 x 10 / 5 x 10
8 T.
3
5 2 6 x 10 ) ( 5 x 10 )
8 . 2 5 x 10
10
J
6
/ 3 . 6 x 10 ) = $ 115 .
Referen c e : P r o c . I . E . E . E . , Ap ril 19 6 7 p age 5 7 . 13-21 ( 13 . 8 )
ELE CTRI C MOTORS AND MOVING-COIL METERS
a) F
=
NIBb , p e rpendi cular to
F
=
NIB a , in the ver t i cal
F
=
F
=
1 b o th 1 and B
o
2 direc tion
1
NIBb , p erpendi cular to
3 b o th 3 and B
b
NIB a , in the vertical 4 dire c t i on = 2NIBb ( a s in 8 / 2 ) = NIB ab s in e
b) T
13-22 ( 1 3 . 8 ) I d
a) T ->-
b) T
MAGNETI C TORQUE
ISB s in e
=
->-
I d (B S cos e ) / 3 8
->-
mxB
61
� 2 I
3
-
B
13- 2 3 ( 13 . 8 ) a) iJl
ATTITUDE CONTROL FOR SATELLITES
NBA cos 8 , T
=
=
I ( a / a 8 ) (NBA cos 8 )
=
NIBA s in 8
S ee figure for P r ob . 13-22 . b ) IN
=
T /BA s in 8
13-24 ( 13 . 9 ) nw
f
s
10
-3
/ 4 x 10
-5
2 x ( 11 / 4 ) 1 . 14 x O . 0 8 7 3
2 8 0 At
ME CHANI CAL FORCE S ON AN I SOLATED CIRCUIT
I ( d iJl / d t ) d t
2 n (LI / 2 )
nW m
=
=
=
I n iJl
2 ( 1 /2 ) I nL
=
2 ( 1 / 2 ) I n ( iJl jI )
Thus the mechani cal work , nw s
=
( 1 / 2 ) I n iJl
nW , is equal to nW . m m
CHAPTER 14 14-1 ( 14 . 2 ) C!
e
MAGNETI C FIELD OF THE EARTH
N
M s in 8
=
If the sphere carried a surface
+
charge den s i ty cr and rotated at an angular velocity
+
w , we would
have Cl
=
crv
crwR s in 8 , crwR
14-2 ( 1 4 . 3)
=
+
M
+
EQUIVALENT CURRENTS
s
+
The surf ac e current density is the s ame as if a toroidal coil were wound on the torus , and C!
14-3 ( 1 4 . 4)
e
=
M.
EQUIVALENT CURRENTS
The equivalen t curren t s are equal and in opp o s ite dir e c t ions .
-+-
B
=
Thus
0 ins ide the tub e . --
M
62
14-4 ( 1 4 . 4)
+
+
+
DIELE CTRI C S AND MAGNETIC MATERIALS COMPARED
+
+
+
+
+
+
+
( oj
+
+
+
+
+
+
+
( b)
a) b ) D i s unaffec ted , E i s reduced by c)
+
The energy is minimum .
E:
r
'
S e e Figs . a and b .
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
(C)
(d )
d) e) H is unaff e c ted , B is increased by ll ' r f) The energy is minimum . S e e Fig . e .
S ee Figs . c , d
Th e loop i s in s tab le equilib r ium .
(e) 14-5 ( 14 . 4 )
MAGNE T I C TORQUE
The magnet acts like a s o leno id . 14-6 ( 1 4 . 4 ) Boo s in 8 cos 8
S ee P rob s 13-22 and 13- 2 3 .
MEASUREMENT OF M 00
s in 2 8
The f i e ld i s larg es t a t 8
45 degrees .
63
14- 7 ( 14 . 4 ) a) V
MI CROMETEORITE DETE CTOR
( d / d t) (Ml ) b
2 2 11)1 N a b O a ( d/dt) 2 2 3 /2 2(a + z )
I
z
b
b ) S ee the figure Reference : Rev . S c i . Ins tr . 42 6 6 3 ( 1 9 7 1 ) 14-8 ( 14 . 4 )
ME CHANI CAL DISPLACEMENT TRANSDUCER
a) S ee curve . 2 2 5/2 b ) xz / (x + z )
4 0 . 9 5 x l0 z ,
3 2 5/2 = 9 . 5 x 10 , O . l / ( O . Ol + z ) z = 14 . 4 mm
Reference : H . Wieder , Hall Genera tors and Magnetores i s tors , p 9 5 . 14-9
( 14 . 5 )
MAGNETIZED DISK
Out s ide , we can us e th e field of a curr ent loop : B
ex
=
2 3 /2 . Mt } / 2 ( } + z )
2 3 /2 , H )1 Mt } /2 ( } + z ) ex o
Ins ide , we have the s ame value of B , wi th z
14-10 ( 14 . 6 )
2
2 « a :
T OROIDAL COIL WITH MAGNETI C CORE
Let N ' be the numb er of turns p e r meter in b o th cases . 14-15 , H
=
N i l in b o th cases .
Wi th the air core , B
From Eq . )1 N ' L a
th e magnetic core , B is larger by a factor )1 . r Th e equivalen t curren t s f l ow in the s ame direction as I . 14-11 ( 14 . 6 )
EQUIVALENT CURRENTS
64
With
b ) On the inner surfac e ,
a
M
=
e
=
B / Il - H o
=
X H m
( Il - l) H r
=
X I / 2 rrb , m
in the s ame direc tion as the curren t On the outer surface , c) B
=
+ V ·H
;<
e
=
X I / 2 rr c , in the opp o s i te direction . m
ll o I /2 rrr , as if the i ron were ab s ent . THE DIVERGENCE OF H
14-12 ( 14 . 7 ) V .B
a
';7 · ( ll ll B) = 1l V Il . li + ll ll v . li 0 r o r o r 0 if VIl
r
14-13 ( 14 . 8 )
;<
0 and if VIl
r
=
0 +
is n o t p erp endi cular to H .
THE MAGNETI ZATION CURVE
Interp olating logar i thmically b e tween the p oin ts marked 2 x 10 4 2 x 10 , 4 3 6 . l x 2 x l0 1 . 2 2 x l0 . Il r =
and
=
14-14 ( 14 . 8 ) a) H
b) V
3
=
ROWLAND RING
5 00 x 2 . 4 / 2 rr x O . 2 ,::; 1000 A/m , B ,::; 0 . 5 T -4 x O . 5 ( 10 / 2 . 4 ) N S dB / d t 10rr x lO
Nd CJi / dt
14-15 ( 1 4 . 9 )
=
=
6 . 6 mV
THE WEBER AMPERE-TURN
A web er is a uni t of magnetic f lux , and d CJi / d t is a voltage . Thus a web er is a vo l t s econd .
The numb er of turns is a pure numb er .
S o [web er ] [ ampere ] = [ volt s e c ond ] [ amp e r e ] 14-17 ( 14 . 9 )
=
[watt ] [ s e cond ]
=
[ j oule]
TRANS FORMER HUM
The hum is due to magne tos tric tion Reference : S tandard Handb ook for . Elec trical Engineers , S e c 11- 9 6 and f o llowing . 14-18 ( 14 . 9 )
P OWER LOS S DUE TO HY S TERE S I S
The area of the loop is app r oximately 2 . 8 x 16 , or 45 W /m
65
3
cycle .
THE FLUXGATE MAGNETOMETER AND THE PEAKING S TRIP
14-19 (14 . 9 )
Referen ce : H . Zij ls tra , Experimental Methods in Magne tism , Vo l 2 , p 3 7 , B r an d t , Introduc tion to the S o lar Wind p 145 ; M . S tanley L ivings ton and John P . B lewe t t , P ar ticle Accelerators , p 2 7 6 .
CHAPTER 15 15-1 ( 1 5 . 2 )
RELUCTANCE
( 1 / 2 ) LI
W m
2
15- 2 ( 15 . 2 ) L
=
N
/ I
=
=
( 1 / 2 ) (LI ) I
( 1 / 2 ) (N
2 ( 1 / 2 ) /NI
RELUCTANCE 2
N /NI
15-3 ( 15 . 3 ) a) = B A g g
=
2 N /
G(
CLIP-ON AMME T ER I (2 rrR-L ) / V V A + L /V A g r o g o
�
V IA/L ' B 0 g
�
V I /L 0
g
Without the iron core , B i s V I / 2 1fR and is much smal ler . o b ) The p o s i tion of the wire is unimp ortan t . 15-4 (15 . 3 ) For V B = =
r
=
MAGNET I C CIRCUIT
500 ,
NI 2 1fR / V V +t / V r 0 g 0
fl NI o 21fR/V H r g
-7 x 5 00 x 2 . 4 -3 2 1f x 0 . 2 / 5 0 0+ 10 41f x 10
0 . 43 T .
66
1 . 508 x 10
-3
1 . 2 5 7 / 5 00+10
-3
Thi s B is too large ; for � B
=
1 . 50 8 x lO
-3
1 . 2 5 7 / 5 2 5+ 1 0
-3
r
=
500 , B
=
=
r
525 , B
0 . 45 9 T , ins tead o f 0 . 5 o n the graph .
15-5 ( 1 5 . 3 )
MAGNE TORE S I S TANCE MULTIPLIER
1 5 - 7 ( 1 5 . 4)
RELAY
F
=
2 (B / 2 � ) A o ( 4 7T x lO
15-8 ( 15 . 3)
-7
=
(A/ 2 � ) ( � NI /t )
x lO
0
-4
Try �
525
r
0 . 44 T .
This B is again too large ; for � B
0 . 32 T .
0
2
=
=
0 . 38 T .
Try �
r
= 550
Thi s i s s at i s f ac tory .
( � A / 2 ) (NI / t ) 0
2
4 -2 -3 2 / 2 ) (10 x lO / 2 x lO ) = 0 . 16 N .
MAGNET I C FLUIDS
a) Th e magnetic f lux is concen trated in the f lui d , where B i s large enough to g ive an appreciab le magnetic pres s ure . b ) S ee the l i terature p ub lished by Ferr o f luid ics Corp .
CHAPTER 16 16-1 (16 . 1)
RE CTIFIER CIRCUITS COMPARED
b
67
1 6 - 2 ( 16 . 2 ) V
!mS
16-4 ( 16 . 2 )
i
V
0
o
a) V ' O c)
f
( l /T)
RMS VALUE O F A S INE WAVE
T
[
r[
0
2
[
2 co S Wtdt
RMS VALUE
b ) 4V
( l /T)
2
/2
}
�
=
V o ( 1- 2 t /T)
� 2 V
V
0
2
/WT
Jf : 2
0
O
] 2 dt } �
=
2 os ada
� ( Vo /T )
[r
y
2 ( 1- 2 t /T) c:lt
,
V / 2 2 , so that the me an s quare d) For the comp l ete s ine wave , V O rill S 2 value i s V / 2 . For the half s ine wave , th e mean s quar e value i s 0 2 = V /2 . V o / 4 , and V O rms 16-5 ( 1 6 . 3) THREE-WIRE S IN GLE-PHASE CURRENT 339 V 2 2 ( 2 40)
----r-
=
16-6 (16 . 4)
THREE-PHASE CURRENT
c o s wt + coswtCO S ( 2 1T / 3 ) - s inwt s in ( 2 1T / 3 ) + c o s w t COS ( 4 1T / 3 ) - s inwt s in ( 41T / 3 ) cos wt [ 1+ cos ( 2 1T / 3 ) + co s ( 41T / 3 ) ] - s inwt [ s in ( 2 1T / 3 ) + s in ( 41T / 3)] 0 0 cos wt [ l+ c o s 120 + c o s 2 40 ° ] - s inwt C s in 120 + s in 2 4 0 ° ] 0 0 0 0 coswt [ l+ cos 120 +c os ( - 120 )] - s inwt [ sin 120 + s in ( -120 )] 0 0 c o s w t ( 1-0 . 5-0 . 5 ) - s inwt ( s in 120 - s in 120 ) = 0 1 6 - 7 ( 1 6 . 4)
ROTATING MAGNETIC FIELD
a) Us ing trigonome tric fun c t ions , B
x
= B [ co s w t - 0 . 5 co s (wt+ 2 1T / 3 ) - 0 . 5 c o s (wt + 4 1T / 3 ) ] o
B [ co s w t - 0 . 5 (cos wt c o s 2 1T / 3 - s inwt s in2 1T / 3 ) o -0 . 5 (cos wt cO S 4 1T / 3 - s inwt s in41T / 3 ) ] = =
B [ co s wt - 0 . 5 (-0 . 5 cos wt - O . 5 /3 s inwt) -0 . 5 ( - 0 . 5 c o s w t+ 0 . 5 !:3 s inwt) ] ° 1 . 5 B cos wt °
68
B
B 0[ 0 . s l3c O S ( Wt+21T/3) - 0 . s l3 coS (Wt+41T/3) ]
y
O . S !3B o (cosWtc O S21T/3 - sinwtsin21T / 3 - coswtcos41T/3+ s inwts in41T/3)
O . S I3B o (-0 . Scoswt - O . S /3 s inwt + O . Scoswt - O . S l3 s inwt) -2 ( 0 . S I3) 2 B o s inwt -l . SB 0 s inwt 7 7 7 S o B 1 . SB coswti - 1 . SB s inwtj , B 1 . S B O ' O O =
=
=
Using exponential functions , B x B o [ expj wt - O . S expj (Wt+2 1T 13) - O . S expj ( Wt+41T 13) J , B o expj wt ( 1-0 . S expj 2 1T/ 3-0 . Sexpj 41T/3) ,
b)
=
=
=
=
By
B o exp j wt [ 1 - O . S (-O . S+O . S l3j ) - 0 . S (-0 . S-0 . S /3j ) J , 1 . SB o exp j w t ,
B 0 [ 0 . S l3expj (wt+21T/3) - 0 . S !3expj (wt+41T/3) J , O . S I3B o expj wt (expj 2 1T 1 3-expj 41T 13) ,
O . S I3B o expj wt (-0 . S+0 . s l3j +0 . S+0 . S /3j ) , 2 ( 0 . S I3 ) 2 B o j expj wt , 1 . SB 0 expj (Wt+1T 1 2 )
Thus B
•
=
1 . SB o (coswt! - s inwtj) as previous ly .
16-8 (16 . 4)
a)
1
DIRE CT - CURRE NT HIGH VOLTAGE TRANSMI S S ION LINE S
2 (V / 2 2 ) I Sp o .1 2 b) 3 (V /2 ) I o TP
2V I ' I o DC Sp =
2V I ' I TP o DC
.1
2 2 1 DC
.1
( 2 / 3) 22I
69
DC
0 . 9 4I
DC
ELECTROMAGNET OPERATING ON ALTERNATING CURRENT
16-9 ( 16 . 5 ) q,
rms
=
NI
rms
jQ=
1 6 - 10 ( 16 . 6 ) a) 1 + 2 j
=
-1 - 2j
1 / ( 1+ 2 j )
j s inx
=
2
=
=
5;
=
- 3 + 4j
=
1 / C- 3+ 4 j )
=
-0 . 12 - 0 . 16 j
i Nw , from P r ob 15- 2 .
=
( - 3-4j ) / (- 3+ 4j ) ( - 3-4j ) = ( - 3-4j ) / ( 9+ 1 6 )
( 1+ 2 j ) ( 1+2 j ) / (1+ 4)
=
-0 . 6 + O . 8j
COMPLEX NUMBERS
COMPLEX NUMBERS
4 5 3 2 1 + jx - x /2 ! _ j x /3 ! + x / 4 ! + j x /5 ! 1 3 - J. X / 3 '
+ jx
16-13 ( 16 . 6 ) expj 1l
1 +4
=
1 - 4 + 4j
=
16-12 ( 16 . 6 )
=
I
2 . 2 3 6 £-1 . 10 7
2 . 2 3 6 L 1 . 10 7
cosx
rms
2 . 2 3 6 L 4 . 2 4 9 ; 1 - 2j
1 6 - 11 ( 1 6 . 6 )
=
V
COMPLEX NUMBERS
( 1+ 2 j ) / ( 1- 2 j )
exp j x
=
2 . 236L 2 . 034 ;
=
2
./ wL ) /Q.
rms ,
2 . 2 3 6 L 1 . 10 7 ; - 1 + 2 j
b ) ( 1+ 2 j ) ( 1- 2 j ) ( 1+ 2 j )
(NV
•
COMPLEX NUMBERS
C O S 1l + j s in1l
16-14 (16 . 6)
=
-1
COMPLEX NUMBERS
a) Multip lication by j in creases the argumen t e by 1l / 2 radians .
b ) e increases by 1l .
c ) e decreases b y 1l / 2 .
16-15 ( 16 . 7 ) THREE-PHASE ALTERNAT ING CURRENT o o 0 cos wt - c o s w t c o s 1 2 0 + s inwts in120 cos wt - c o s ( wt+120 ) 1 o 0 coswt ( l + c o s 6 0 ) + s inwtsin 6 0 0 / 2 ) cos wt + 0 2 / 2 ) s inwt o 0 3 /2) c o w t + C 2 ) s inwt ( co s 3 0 coswt + s in30 s inwt) 3 2 c o s ( wt-1l / 6 ) 3 2 cos ( wt-30 ) =
: [o!
�
F
=
J
=
=
o!)
70
CALCULATING AN AVERAGE POWER WITH PHASORS T ( l /T) V0 coswtI 0 cos (wt+�) dt
16 - 16 ( 16 . 7 ) a) P av
=
(v I 0 /T) o
I
o
I coswt ( coswtcos � - s inwtsin�) dt T
[
o
!
" (V 1 0 /T) C O ", c " e 2 wc" - eln " (V o I 0 /2) cos l.!(
{
coewMnw" C
!
(v I 0 /T) cos�(T /2) o
The integral of sinwt C O E Wt over one period is zero . b)
P av
=
(1/2) R c fv0 exp ( j wt) I 0 exp - j ( wt+�) L
J
=
( 1 / 2 ) V0 I 0 cos
CHAPTER 17 17 - 1 ( 17 . 1) a) Z
=
wC R + J. wL + RR/j + 1/ j wC
- Rj wC) R + J' wL + R(l R2 w 2 C 2 + 1 2 + J WL - 2 R2 wC2 R + 2 2 R2 R w C +1 R w C +1 =
1
.[
2R at f = 0 and Z + j wL at f + 00 R + R/ (R2 w 2 C 2 +1) 10 + 10 / ( 100x41T 2 x10 6 x25x10 - 18 +1) = 20(1 wL - R2 wC / ( R2 w 2 C 2 +1) 2 1Tx10 3x5x10 -3 - 100x21Tx10 3 x5x10 -9 / 1 31 . 4(1 0 3 7 . 2 (1 � = arc tan ( 3 1 . 4 / 20) 5 7 . 5 ( 2 0 2 +3 1 . 4 2 ) ! 1Z1 Z
b)
IMPEDANCE
=
=
=
c) I y l
d) e)
f) g)
P
No
l/ I Z I 201 2
2 . 69 x 10 - 2 S , 0 . 20 W
is induc tive (posi tive) for L true
X
X
is zero at f
-5 7 . 5
� =
O.
71
>
0
R2 C / CR2 w 2 C 2 +1) , which is always
1 7 - 2 ( 17 . 1) Z
=
REAL INDUCTORS
( R+j wL) / j wC R+ j wL+1 /j wC
2 ( R+j WL) [ ( 1-W LC) - Rj wC J 2 2 2 2 2 ( l-w LC) + R w C
R+j wL 2 . RJ wC-w L C+1
2 2 . R+J w [ L ( l-w LC) - R c J 2 2 2 2 2 ( l-w LC) + R w C
2 2 2 2 R ( l- w LC) + Rw LC + j w[ L ( l-w LC) - R c J 2 2 2 2 2 ( l- w L C ) + R w C 17-3 ( 1 7 . 3)
COMPENSATE D VOLTAGE DIVIDER
R /j wC 2 2 R + 1 / j wC 2 2 R / j wC R / j wC 1 1 2 2 + R +1 / j wC R + 1 / j wC 1 2 1 2
V /V a i
17-4 ( 17 . 3)
R R 1 2 + R j wC +1 R j wC + 1 2 2 1 1
RC FILTER
0. "1
10
0,/
a) Call V ' the po tential at the conne c t i on b e tween A and C v /V ' o
=
D / ( C +D) ,
B ( C+D) , B ( C+D) + A (B+ C+D) BD V /V = ( V /V ' ) ( V ' / V ) = o i i B ( C +D) + A ( B + C+D ) o b) A V /V a i
R, B
=
=
=
l / j wC , C
l/jwC , D
=
R/j wC R/j wC + (R+ 1 / j wC)
2
=
2
=
BD AB + (A+B ) ( C+D)
R R
R + R j WC + l / j wC + 2R
72
=
1 3 +j (RIDC - l / RwC)
I vo /V1.. I
When
is maximum at RwC
I v o /v1.. 1 =
b road .
1 7 -5 ( 1 7 . 3) r
=
Z
RT = Z
R
=
R i' =
=
Then V I v . = 1 / 3 . 0 1.
1.
( 1 / 3 ) / 2 2 , RwC = 0 . 30 3 or 3 . 30 .
The p a s s -b and is very
MEASURING AN IMPEDANCE WITH A PHAS E - S EN S ITIVE VOLTMETER
V / V = Z / (R ' +Z ) , 2 1
R ' r / ( l-r) = R ' (a+b j ) / ( l-a-bj ) ( a+b j ) ( l-a+b j ) 2 2 ( I-a) + b
S e t t ing Z
RT
1
2 a (1-a) - b + j b ) 2 2 ( I-a) + b
=
R + jX, 2 a_a _b 2 2 1 + a + b _ 2a 2
1 + Ir
2 r co s e _ r X 0 o 2 1+ r - 2r c o s e , i' o 0
I2 =
x
-
, "'jiV =
2a
b
1+ IrI
2
-
r s in e 0
1+ r
o
2
_ 2r cos e 0
Reference :
E lec troni c s , July 2 5 , 1 9 7 4 , p 1 1 7 .
1 7- 6 ( 1 7 . 3 )
IMP EDANCE BRIDGE S .
THE WIEN B RIDGE
73
2a
1 R /2 , R C ' the firs t equation is s atis f i ed For R R , C 4 4 1 3 2 3 and the s e cond one yields R wC 1. 3 3 =
=
=
=
17-7 (17 . 3)
LOW-PAS S RC FILTER
_0 I1z � 0r-___________________15
a ) V /V ' 0
b)
l
=
I V /V . I o l
c) db
l / j wC R + l / J' " ,C =
� 2 2 2 l/ (R w C +1) 2
2 0 log l V Iv . o l
=
1 Rj wC + l '
w
17-8 (17 . 3)
=
O
1 50 H z O ------------------___ �
I V0 / V l. I
::; 1 / RwC l. f Rw C
8 2 2 -8 -' 1 / ( 10 X41T xf x10 + 1 ) 2
I.
PHASE SHIFTER
o
_ I ?O°L-0, J
-:-___�
____
V
I. �
Rw C
ItJ
)
R-1 / j wC [ R l / j WC v V R+1 / j wC i = R+ 1 / j wC i o = [R+1 / j wC
V /V . o l
=
exp
2 j arc tg ( l /Rw C)
17-9 ( 1 7 . 3) a) V
=
cr
=
d E a a
=
MEASURING SURFACE POTENTIALS =
d E ' d d
E E + E E E r o d o a
E E + E E ( d E /d ) o a r o a a d 74
R+ j /wC R-j /wC
=
»
1
•
2 2 -' 1 / ( 41T f +1) 2
E
( a / E ) / ( l+ E d / d ) o r a d
a
b) a .
E E o a
a / ( l+ E d /d ) r a d
c) V
d E a a
( ad / E ) / (l+ E d / d ) r a d a 0
J.
d) IR
( dQ / d t ) R
=
C
( 10
IR
( 10
=
-13 -13
V ( dC / dt) R
=
dC/dt
/ 2 ) ( l+ exp j wt ) ,
( 10
- 13
/ 2 ) j w exp j wt
/ 2 ) j w ( exp j wt ) RV
References : S tatic E l e c tr i f ication 19 7 5 , p ages 1 7 3 , 1 8 2 ; Catalog o f Monroe E l e c troni cs . 1 7- 1 0 ( 1 7 . 3 )
RE FERENCE TEMPERATURE S NEAR AB S OLUTE ZERO
Let I
be th e currents in the primar ies and in th e s e condaries ,
and I
p s respec tively .
Reference :
=
O.
R I ' j wM I = R I ' M /M 2 p l 2 2 s 1 S
j wM I 1 p
Then
At balance , V
=
Rl /R2
Rev . S c i . Ins tr . 4 4 , 1 5 3 7 ( 1 9 7 3 ) .
1 7 - 11 ( 1 7 . 3 )
REMOTE-READING MERCURY THERMOMETER
a) S in c e C » C , the c ap a c i t ance of C and C in s e r i es may b e s e t l l 2 2 equal to C and 2 V ' /V
s
R R + l / j wC
=
where C
2
=
2
Rj wC / (Rj WC +l) , 2 2
varies l inearly wi th the temperature .
b ) S in c e V ' i s to vary linearly wi th C , w e mus t have that Rw C « 2 2 and then V ' « V S
1,
c) C is a cylindrical c ap a c i t or wi th an out s i d e radius o f , s ay , 2 mm . 2 The mercury column has a radius o f , s ay , 0 . 05 mm . Then C
2
::::; 2 1fE E L / 9,n (2 / 0 . 0 5 ) o r "" (2 1fx8 . 85xlO
-12
=
2 1fE E L / 9,n 4 0 , o r
x3 / 9,n 40 ) L "" 4 5 L pF .
We have s e t E
"" 3 . Here L is the leng th of the mercury c o lumn ins ide r the e l e c t rode C . S et ting L "" 1 0 0 mm , if R » j wL ' R » j wL ' l ls 2 2s
75
d) C
'" 5 pF , R « 1 /w C
2
. 7 -1 2 1 / 2 -rr x 10 x 5 x 10
2
3000 11 .
S et t ing R '" 3011 , RwC '" 1 / 100 , V ' /V '" 1 / 100 , V' '" 0 . 1 volt . s 2 Reference : Review o f S ci en t i f i c Ins trument s � , 1 9 5 ( 1 9 7 6 ) . 1 7 - 12 ( 1 7 . 3 ) S ince wL «
WATTMETER
Z , we may s e t th e voltage acro s s the load equal to that
a t th e s ourc e .
Als o , s ince R » Z , we may set the curren t through 1 the load equal to the curren t supplied by the s ource .
The coil produces a B that is proportional to , and in phas e with , the curren t through Z . The vo l tage acr o s s R i s R / ( R +R ) t imes the voltage acro s s the 2 l 2 2 s ource . Then
v ' av
=
0 0
KV I (co s tp/ 2 )
1 7- 1 3 ( 1 7 . 3 )
=
KV
I costp rms rms
TRANS IENT SUPPRE S S OR FOR
AN
INDUCTOR
We can calculate the voltage acros s the induc tor , after the swi tch is opened , in ano ther way .
We consider a clo ckwi s e mesh curren t in
L and C . a) Firs t , we f ind Q and I as functions of t , with th e swi t ch open . From Kirchoff ' s vo ltage law , 2 2 C , Ld Q / d t + 2 RdQ / d t + Q / C
LdI / d t + 2 RI + Q / C
O. 2 Then Ln + 2Rn + l / C
=
Try a s o lution of the form Q Q exp n t . 0 2 2 2 -R/L ± (R /L - l /LC ) -R/L , s in ce R =
n
=
=
=
=
L/C .
0
The two roots are
equal . Then Q
( A+B t ) exp ( -Rt/L) .
Set
Q
Q
N ow
I
S et
at t o dQ / d t
I
=
10 a t t O. Then B 1 + RQ /L , 0 o exp ( -Rt /L) ( Q +l t+RQ t /L ) , o o o exp (-Rt /L) [ - (R/L) ( Q +1 t+RQ t iL) + 1 + RQ . /L J , =
Q I
O. Then A Q . o exp (-Rt /L) [ - (R/L) (A+B t ) +B J
=
0
0
0
exp ( -Rt/L) [ I - (R/L) ( I +RQ /L) t J . 0 o 0
0
0
and 1 . S e t I and I the 0 o L C currents through the inductor and c apaci tor b efore t 0 in the
b ) Now let us f ind a rela tion b e tween Q
=
76
directions shown , s o as to give a clockwis e current in the right hand mesh when the switch is open . Then V 1 = � 2 12 cos [ wt-tan-1 ( w1/R) ] , 2 1 ( R +w 1 )
=
Q I
=
+
Vo -1 2 l/w 2 C 2 2i cos [w tHan ( l/RwC) ] ) (R + d Q /dt , Vo -1 2 2 2i sinew t + tan ( l / RwC ) ] , 2 W(R +l/w C ) 2 2 2 -Vo R/ (R +w 1 ) ,
o Vo
-1 2 2 � sin [ tan ( l /RwC) ] 2 W(R + l /w C ) 2 Io Qo -
2 2 2 R (R w C +1) C (R 2 + w 2 1 2 )
If the source supplied be VC , and I o / Q o would Since 1 0 -RQ o /1 , Q =
-R/1
=
�
-1 / ( 1C) 2 .
DC , ins tead of AC , I o would be - Vi R , Q o wou l d l again be -l/RC , or -1/ (1C) 2 . Q o exp (-Rt /1) , I = I o exp (-Rt /1)
c) The vol tage acros s the inductor , after the swi tch has been opened , is 0, RI + 1dI /dt = RI o exp ( ) + 1 (-R/1) exp ( ) desp i te the fact that I decreases exponential ly with time . Reference : R eference Data for Radio Engineers , p 6 - 12 . =
17-14 ( 1 7 . 3 ) SE RIES RESONANCE R + j ( w1-l /wC) Z ->- _00 j for w ->- 0 , Z R for w 2 1C = 1 , Z ->- ooj for w ->- 00
Z
=
77
j
y
R
X
PARALLEL RESONANCE
17-15 (1 7 . 3)
Q
1 4 Ie liz.
-100
a)
°
t 90
1 00
Z
(R+ j wL) (1-w 2LC-Rj wC) 2 2 2 2 (l-w 2 LC) + R w C
Phase of
Z
Z
=
R + j w [ L ( 1-w 2 LC) - R2 C J (l-w 2 LC) 2 + R2 w 2 C 2
21Tf[ 10 - \ 1-41T 2 f 2 10 -9 ) _ 10 -4 J ( ) 2 + 10 -10 41T 2 f 2
�
{ 100 + 4 1T 2 f 2 [ J 2 } � ) 2 + 10 -10 41T 2 f 2
��������- �
J /R
= arc tan w[
Z
=
R 2 ( l-w LC) 2 + (R2 w 2 C 2 )
Z
Imaginary part of Magnitude of
-IfI
R+j wL RJ. wC-w 2LC+ 1
(R+j wL) / j wC -_ R+j wL+1/j wC
Real part of
I O � j.f'Z.
b ) X = 0 when [ J 0 , or when ( l-R2 C /L) /LC L ( l-w 2 LC) R2 C , 1 - w 2 LC R2 C /L , w 2 f [( 1-R2 C /L) /LC] � /21T = [ ( 1-10 - � /10 -3 ) /10 -9] � /21T = 4 . 7 7 =
=
=
=
=
c) 1 - w 2LC
1 - 41T 2 x 6 4 x 10 6 x 10 -3 x 10 -6 1 - 41T 2 x 64 x 10 -3 = -1 . 5 27 10 + j ( 2rrx8x10 3 ) [ 10 -3 (-1 . 52 7) - 10 2 x 10 -6 ] = 10 - 81 . 7 8j 2 . 584 2 . 33 2 + 10 2 x 41T 2 x 64 x l0 6 x 10 -12
Z =
d)
kH z
Z
=
=
R ' + l/j wC ' , R '
=
=
3 . 86 9 , C'
=
0 . 629
]IF
Reference : Philips Technical Review 31 No 4 ( 19 7 1) .
78
S TAR-DELTA TRANSFORMATION
17-16 ( 17 . 4)
,o,,��� ,, -, - 0 .1 5 92
T1
J
- 8.063 X
10-7 j
- 2. 533
10- 4
X
17-17 ( 17 . 4)
STAR-DELTA TRANSFORMATION
17-19 (17 . 4) R + Z 0, R
B RID GED-T
=
1 /R
=
A
=
-Z ,
- [Zj W C - wZ C Z ( r+j wL)] -Zj wC + wZ C Z r + j W3 LC Z -Y
=
Then wZ C Z rR
=
1 , wZ LC
=
Z.
R
79
('
17-20 ( 17 . 5 ) MUTUAL INDUCTANCE Trans form the circuit into the one shown in the figure of the preceding pag e . 10 3 I 1 - I 2 -- C ! ) x1 . 9 j 2 rrx 10 3x 0 . 9 J 5+ 10+2rrx10 ' 2 2 2 1 = 5 / 1 5+5+100rrj l = 5 / 1 10+314j l = 5 / ( 10 +314 ) 2 = 1 . 5 92 x 10 -2A = 15 . 9 2 rnA
(5
Ul �
<
-;-
_
(!)
0 �
....
CHAPTE R 18 18-1 ( 18 . 1) DIRE CT - CURRENT MOTO RS a) R eplace V ' by a resis tance R ' V ' /I . Then the power supplied by the s ource is IV = I ( I R+I R ' ) The first term represents the var ious los s es and IV ' is the useful power . c) At no load , V ' -+ V , 1 = (V-V ' ) /R -+ O , B -+ O . Since V "" V ' ''' wB , w -+ oo . d ) IV ' increases . The motor slows down and V ' decreases , s o I increases faster , R ' V ' /I decreases and the efficiency decreases . =
18-2 ( 18 . 2) POWER-FACTOR CORRECTION 600 / 100 6 . 00 Q a) I z l 6 x 0 . 65 3 . 9 0 $I , R x = 6 x sin (arc cos 0 . 65 ) = 4 . 55 Q b)
I
viz
=
V(R-j x) / l z I 2
(600 /36) (3 . 9 0-j 4 . 55) The in-phase componen t is (600 /36) 3 . 90 65 . 0 A . The quadrature comp onent is 76 . 0 A , 2 lagging . Check : 6 5 . 0 2 + 76 . 0 2 = 100 •
80
�x R
c) VwC = 7 6 , C 7 6 / ( 600x21rx60) 366 ]l:F . This capacitor would cos t ab out $ 400 . 00 . R eference : S tandard Handb ook for Elec trical Engineers , 5 - 9 8 and 16-185 . =
=
18-3 ( 18 . 2 ) POWER-:FACTO R CORRECTION WITH :FLUORES CENT LAMPS The in-phas e component of the current is 80 /120 = 0 . 6 6 7 A . S in ce cos 'i' 0 . 5 , 'i' 60 degrees . The current is 2 x 0 . 66 7 1 . 33 A The reac tive current i s 1 . 33 sin 60 0 1 . 16 A . Then VwC 1 . 16 , C 1 . 16 / (2�60x120) � 20 ]l:F . References : Henderson and Marsden , Lamps and Lighting , p 32 5 S tandard Handbook for Electrical Engineers , 19 -33 . =
=
=
=
18-4 ( 18 . 3)
=
ENE RGY TRANS :FE R TO A LOAD
R
L
R(I
c
a) rI
RI ' + (Q ' / C) , rj wQ (r /R) Q , xco (r/ R C) Q
Q'
�
b)
W =
=
=
Rj wQ '
fVIdt f VdQ co fYd
0
+
Q'
X
one cycle one cycle
81
/C
�
Rj wQ '
t-
j)
V o coswt ,
c) i) For a resis tor , V Q (V 0 /wR) sinwt . See Fig . R
I
(V o /R) coswt
dQ /dt ,
=
V0 coswt , Q
ii) For a capacitor , V S ee Fig . C .
CV 0 cos wt .
iii) For an inductor , V Vo coswt , Ldl /d t Vo coswt , 2 Ld 2 Q /dt 2 Vo cos wt , Q - (V o / W L) coswt . See Fig . L . =
=
=
=
iv) For a resistor in series with an inductor , with R jwQ
=
=
I
=
Vo exp j wt / (R+j wL) l
=
( V o /2 2 R) exp j (wt-1T/4) ,
(V/ R) exp j wt / ( 1+j )
=
j wL ,
;L
Q ( V o /22 wR) exp j (wt-1T/4- 1T/2) See the fourth figure . Reference : Rev . S ci . lns tr . �, 109 ( 19 7 1) . =
lS-5 ( lS . 3) ENERGY TRANSFE R TO A LOAD , Let the voltage across G , at a V given ins tant , be V . The voltage at y is then approximately equal to V . Let the current through Z , at a given ins tant , b e l , and the pulse duration be T . Then the energy dis sipated in Z during a pulse is T Vd Q W VIdt one cycle o
J
J
ale
The voltage a t x i s Q /C . Then the spot on the os cilloscop e s creen des cribes a curve as in the figure . The area under this curve is proportional to the above integral . Reference : Rev . S ci . lns tr . �, 1004 ( 1974) .
S2
A positive X2 is equivalent to a negative X in the primary . lS- 7 ( IS . 4) MEASUREMENT OF THE COEFFICIENT OF COUPLING k With the secondary open , Z oo Z l j wL 1 . With the secondary short- circuited , 2 2 Z o j WL l + w i /j wL 2 j wL l - j wL lj WL 2k /j WL2 2 Z O /Z oo 1 - k . =
=
=
=
=
lS - S ( lS . 4) a ) Z in
=
REFLECTED IMPEDANCE 2 2 Rl + j wL l + W M / (R2 +j WLZ ) 2 ( 1+j R2 ) R2 / R2 +1
[
b)
J
10
0.1
lS-9 (IS . 4) MEASURING THE AREA UNDER A CURVE Draw a line around the periphery of the figure with conducting ink . Then measure the voltage induced when the Helmholtz coils are fed , say at 1 kilohertz . The sys tem can b e calib rated with a circle or with a rectangle of known area . Reference : Rev . S ci . Ins tr . 41 , 1663 (19 70) . lS-lO ( lS . 4) a) R
=
I sec b ) Ipri
9v / Acr
=
S OLDE RING GUNS 10 -1 /4 x lO - 6 x S . S x lO 7
(100 /4 . 3 x lO -4 ) ! 100/120 = O . S A .
=
4S0 A ,
S3
=
4 . 3 x lO -4
V se c
=
>l
4S0 x 4 . 3 x lO -4 ",, 0 . 2 volt
18-11 (18 . 4) CURRENT TRANSFORMER Disregarding the s ign , the induced electromo tance is d�/dt , with b+a (v o I /2 rrr) 2 adr ( V o a /rr) Itn[ (b+a) / (b-a) J , b-a
I
v
( v a/ rr) tn[ (b+a) / (b-a) J dI / d t .
=
o Reference : Rev . S ci . Ins tr . �, 32 4 ( 1 9 7 5) . 18-12 ( 18 . 4) INDUCED CURRENTS Rt t /aA 2rra/ iba =
=
The tube is a single- turn s olenoid .
Hence L
18-13 ( 18 . 5 ) EDDY-CURRENT LOSSES S ee the s tandard Handbook for Electri cal Engineers , Sec . 2- 7 4 . 18-14 ( 18 . 5) EDDY-CURRENT LOSSES For a s olid core , the p ower loss is P I = V2 /R f'O (d� /dt) 2 / ( 4a/abL) (abL/4a) (diJ> /dt) 2 . If the core is split into n laminations , insulated one from the othe� P n n [ d (iJ> /n) /dt J 2 / ( 4an /abL) P l /n 2 . Reference : S tandard Handbook for Electrical Engineers , Sec . 2- 7 4 , 9 1 , 9 2 , 9 3 and following . =
=
=
18-15 (18 . 5 ) HYS TERESIS LOS SES Place the laminations inside a sole( R- A ) / f noid and measure the resis tive part R of the imp edance of the solenoid as a func tion of the fr equency . Then B R A + B f + Cf 2 , where A is the DC resis tance . Then (R-A) / f B + Cf . A plot of ( R-A) /f as a function of f gives both B and C . =
=
84
18-16 ( 1 8 . 5 )
V
a)
CLIP-ON AMME T ER
NAdB / d t
=
NAwB
3
10 ( 0 . 6 4 x 10
-4
=
NAw� � I / 2TIr r 0
) ( 2 TI x 6 0 ) ( 4TI x 10
-7
4 -2 x 10 ) / 2TI x 1 . 5 x 10
=
3 . 12
b) Loop the wire c arrying the unknown current s everal t imes around the core .
CHAPTER 19 19-2 ( 19 . 3)
MAXWELL ' S EQUATIONS 6 .5
-+
-+
-+
� V x (H + M) - E � a E / a t o o o Dividing by � -+
VxM
=
-+
-+
-+
±
J + ( a / a t ) ( E E + l' ) o f
1 9 - 3 ( 19 . 3 )
-+
and c anc eling V x M on b o th s id es ,
o
-+
=
-+-
� ( J + ap / a t + V x M) , from Eq . 14-20 . o f
=
-+
-+
J + a D / a t , f r om E q . 6 - 5 f
MAXWELL ' S EQUATIONS
Us e the equat ions o f the pr evious prob lems , s e t ting
B
=
V
V
r 0
a, a / a t
19-4 ( 19 . 3 ) VxB- E
V
o 0
=
jw
n
=
E E E, r 0
MAXWELL ' S E QUATI ONS
aE / a t =
V
j
0 m
-+
T aking the divergence of b o th s i de s and r ememb ering that V ' V x B -+
for any vector B ,
85
=
0
V
19-5 ( 19 . 3) MAGNETIC MONOPOLES AND MAXWELL ' S EQUATIONS a) Taking the divergence of the equation for'V'x E and remembering that the divergence of a curl is always equal to zero , ->-
( 3 / 3t) (V .B)
b)
=
-v . !* , V .!*
=
-
From the equation for the curl o f
JV x E · da f E. J2 -J!*.�
s
3p */ 3t .
-+
=
=
-+
-+
E,
-1 * .
c
19-6 ( 19 . 3) 11
1N . m
1W
11/ s
1C
lAs
1V
lJ /C
111 1S
1V/A 1 11-1
1F
1C/V
1Wb
1Vs *
1T
1Wb /m2
1H
=
1 kg(m/s 2 ) m 1 kg m2 /s 3
=
1 kg m 2 /s 2
1 kg m2 /As 3
l (kg m2 /s 2 ) /As 1 kg m2 /A2 s 3 1A2 s 3 /kg m2
=
=
=
lAs / (kg m2 /As 3 ) 1 kg m2 /As 2
=
1 kg /As 2 1Wb /A**= 1 kg m2 /A 2 s 2
*From the fact that , in a changing magnetic field , the induced voltage is equal to the rate of change of the magnetic flux **From L
=
'1! /I
Let us check a) j wLI
=
V
. 1 kg m2 s -Z--Z A glves A s
=
kg m-2 , ---A s3
Correct
The energy s tored in a capacitor is CV2 / 2 . 2 kg -m-2 = A2 s 4 kg m2 Correct 2 2 kg m s A s3 b)
--
[ ]
86
Then
�Itored
c) The energy kg m s
2
2
2 2 kg m _ - 2 2 A A s
LI 2 /2 .
in an induc tor is
Corre c t
2 The p ower l o s s in a res is tor i s I R .
d)
Then
2 2 kg m Corre c t = A2 � 2 3 A s3 s 2 . e ) w LC lS a pure numb e r . Then 2 2 4 l � � - l . Correct 2 2 2 2 s A s kg m
Then
etc .
20
CHAPTER
20 - 1 ( 20 . 4) E
=
E exp o
where E
o
PLANE WAVE IN FREE SPACE
j ( w t- z /'te..) , and H
o
H exp o
=
j ( wt- z / J:) , x,
are ind ependen t o f
a) Then , from V · E
( a / ax)
H
= 0,
� oxexp j ( ) ]
0 , ( a / ay)
=
y, z , t and have no z-comp onent
[E oyexp j ( ) ]
=
0
Thes e equatiop s are iden t i t i e s
!I
b ) We have s imilar equation s for H . c ) From V -+
-+
0
0
i
E
x
Thus ,
x l
o
ag / at ,
k
-�
°
ag/ at ,
0
Y
-+
H
20 - 2 ( 20 . 4)
E
=
0
0 -j / x
-+
0
,...
-+
-j w�H , k x E -+
-+
=
a l/ a t , k x H =
�
E
x 0
-E
-+
j
-+
wJ:H °
k
=
o
max
= 10
-+
�J cH °
-+
max
8
=
10
30 (2 rr x 3 x 10 ) 0 . 1 /3 x 10 = 5 4 . 4 87
-+
-j w� H o
cE
o c o s 3 0 ( dB / d t ) 6
=
0
y
LOOP ANTENNA
10 ( d 1> / d t ) lO c o s
-+
E
C-j /1e") k x l = x
-+
i
a/ az
d) From V
1rmax
-�
-+
j
E
=
o c o s 3 0 ( dE / d t ) mV .
max
/c
2 0- 3 ( 2 0 . 6 ) P
P OYNTING VE C TOR
a) Jl9- = 3 . 8 x lO
iU b ) ,e9f -s c)
E
=
'&E = =
/ 47r x 49 x lO
16
= 2 . 6 5 x lO
-3 2 5 E 1 . 5 3 x lO V im E rms ' rms =
2 2 2 2 = E /E = ( l /R _ ) / ( l / R ) , E /E = RS /R _ E S s E S S E E S
q
E
26
5 11 8 ( 7 x 10 / 1 . 5 x 10 ) 1 . 5 3 x 10 = 7 0 0 Vim
2 . 6 5 x lO
-3
x ( 700)
2
= 1 . 3 x l0
3
W/m
2
3 4 2 6 0 x 1 . 3 x lO ( cal / 4 . l9 ) / ( 10 cm ) = 1 . 8 6 calorie /minute centi2 meter
Thi s quan t i ty i s called the s o lar cons tant We have neglected ab s o rp ti on in th e atmosph ere . The average daily f lux at the ground , in the United S tate s , is ab out 2 0 . 4 calorie /minute centime ter . Reference : Ameri c an Ins t i tute of Phy s i c s Han db o ok , 3rd e d , p 2-143 . 20-4 ( 2 0 . 6 )
SOLAR ENERGY
f
At the surface of the earth , -6
=
1 . 3 x 10
7 3 P = �A / 5 0 , A = 5 0 P 0 = 5 x l0 / 1 . 3 x l0
3
2 W /m , from P rob . 20-3 . �
4 i l0
4
2 m ,
or a s quare 2 0 0 me ters on the s ide .
20-5 ( 2 0 . 6 )
.Rf1
=
CE
o
E
2
P OYNTING VE CTOR 8 2 2 -12 = 3 x 10 x 8 . 8 5 x 10 x 2 0 = 1 . 0 6 W /m
In one s e c ond , the energy ab s orb e d by one s quare meter of the copp er she e t is 1 . 06 J .
This energy will increase the temperature of one kilogram of copper by 1 . 0 6 / 400 ke lvin . In one s econd the temp e ratur e of the she e t rises by 100 x L 06 / 400 = 1 . 06 / 4
=
0 . 2 6 5 kelvin .
20-6 ( 2 0 . 6 )
P OYNTING VE CTOR
a)
GJA CD oAjot Or ct - -
---
- --88
b ) Energy f lows into the field . 20-7 (20 . 6)
"*
.J.av
P OYNTING VE CTOR
=
20-8 ( 2 0 . 6 )
P OYNTING VECTOR
N e ar the s ur face o f th e wir e , E
= =
I R ' , H = I / 2 rra , 2 I R ' / 2 rra
l B" x It I
Thus the p ower loss per meter is 2 I R' 20-10 ( 2 0 . 6 ) E
=
J
2 rr€ V / �n ( R /Rl ) ' E 2 o
A H
J
COAXIAL LINE R 2 A / 2 rr€ r , ( A / 2 rr € r) dr o o R l
I
V , ( A / 2 rr € ) �n ( R /R ) 2 l 0
=
V,
v/ r�n ( R / R ) 2 l
I / 2 rrr R2
R
[ v/ r �n ( R /R ) J (I / 2 rrr) 2Tf r dr 2 l
VI
l
20-11 ( 2 0 . 7 )
220
x
10
2200 W
RE FLE CTION AND RE FRACTI ON , FRESNEL ' S EQUATION S
S e e E l e c t romagnetic Fie lds and Wave s , S e c 12 . 2 . 2 .
89
