Chapter # 35
1.
Sol. 2.
Sol.
Magnetic Field due to A current
Objective - I
[1]
A vertical wire carries a current in upward direction. An electron beam sent horizontally towards the wire will be deflected (A) towards right (B) towards left (C*) upwards (D) downwards ,d m/okZ/kj rkj esa Åij dh vksj /kkjk izokfgr gks jgh gSA rkj dh vksj {kSfrt fn'kk esa Hkstk tk jgk bysDVªkWu iaqt fo{ksfir gksxk(A) nka;h vksj (B) cka;h vksj (C*) Åij dh vksj (D) uhps dh vksj C A current-carrying, straight wire is kept along the axis of a circular loop carrying a current. The straight wire (A) will exert an inward force on the circular loop (B) will exert an outward force on the circular loop (C*) will not exert any force on the circular loop (D) will exert a force on the circular loop parallel to itself. ,d /kkjkokgh oy; dh v{k ds vuqfn'k ,d lh/kk /kkjkokgh rkj j[kk gqvk gSA lh/kk rkj (A) o`Ùkkdkj oy; ij vUnj dh vksj cy yxk;sxkA (B) o`Ùkkdkj oy; ij ckgj dh vksj cy yxk;sxkA (C*) o`Ùkkdkj oy; ij dksbZ cy ugha yxk;sxkA (D) o`Ùkkdkj oy; ij vius lekukUrj cy yxk;sxkA C
i
i
The straight wire will not exert any force on the circular loop. F i L B
3.
Sol.
4.
A proton beam is going from north to south and an electron beam is going from south to north. Neglection the earth’s magnetic field, the electron beam will be deflected (A*) towards the proton beam (B) away from the proton beam (C) upwards (D) downwards
,d izksVkWu iqat mÙkj ls nf{k.k dh vksj xfr'khy gS rFkk ,d bysDVªkWu iaqt nf{k.k ls mÙkj dh vksj xfr'khy gSA i`Foh ds pqEcdh; {ks=k dks ux.; ekuus ij] bysDVªkWu iaqt fo{ksfir gksxk (A*) izksVkWu iaqt dh vksj (B) izksVkWu iqat ls ijs (C) Åij dh vksj (D) uhps dh vksj A Proton contain the +ve charge & electron contain the -ve charge. So the electron beam will be deflected towards the proton beam.
A circular loop is kept in that vertical plane which contains the north-south direction. It carries a current that is towards north at the topmost point. Let A be a point on axis of the circle to the east of it and B a point on this axis to the west of it. The magnetic field due to the loop (A) is towards east at A and towards west at B (B) is towards west at A and towards east at B (C) is towards east at both A and B (D*) is towards west at both A and B
,d o`Ùkkdkj ywi mÙkj&nf{k.k fn'kk esa fLFkr m/okZ/kj ry esa j[kk gqvk gSA blds mPpre fcUnq ij /kkjk dh fn'kk mÙkj dh vksj gSA ekuk fd bldh v{k ij fcUnq A iwoZ dh vksj rFkk fcUnq B if'pe dh vksj gSA ywi ds dkj.k pqEcdh; {ks=k (A) A ij iwoZ dh vksj rFkk B ij if'pe dh vksj gSA (B) A ij if'pe vksj rFkk B ij if'pe dh vksj gSA (C) A rFkk B nksuksa ij iwoZ dh vksj gSA (D*) A rFkk B nksuksa ij if'pe dh vksj gSA
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Chapter # 35 Magnetic Field due to A current Sol. D Current goes in the clockwise direction. Curled the fingers along the direction of current then the streached thumb will points towards the magnetic field. 5.
Sol.
[2]
I B
A
Consider the situation shown in fig. The straight wire is fixed but the loop can move under magnetic force. The loop will fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA lh/kk rkj dlk gqvk gS] fdUrq ywi pqEcdh; cy ds izHkko esa xfr dj ldrk gSA ywi -
(A) remain stationery (B*) move towards the wire (C) move away from the wire (D) rotate about the wire (A) fLFkj jgsxk (B*) rkj dh vksj xfr djsxk (C) rkj ls ijs xfr djsxk (D) rkj ds ifjr% ?kw.kZu djsxkA B Fforce per unit length LLenght of the segment of loop BMagnetic field. Force on the side AB & DC is equal & opposite in direction to i1. Force on AD
M0 i1 i2 a towards the wire (west direction) 2d Force on BC FAD
0 i1 i2 a
FBC
2 d a
towards the East direction
Net force on the loop due to wire Fnet = FAD + FBC + FAB + FDC
0 i1 i2 a 0 i1 i2 a 2d
Fnet
2 d a
FAB FAB
0 i1 i2 a2
2d d a
0 i1 i2 a 1 1 2 d d a
Fnet
0 i1 i2 a
2d d a
towards the wire
So the Loop will move towards the wire. F q V B
F qvB sin angle between v & B is zero So F = 0
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D i1
A a
i1
i2 a
C B
a
Chapter # 35 Magnetic Field due to A current 6. A charge particle is moved along a magnetic field line. The magnetic force on the particle is (A) along its velocity (B) magnetic field only (C) both of them (D*) none of them ,d vkosf'kr d.k dks pqEcdh; cy js[kk ds vuqfn'k xfr djok;h tkrh gSA d.k ij pqEcdh; cy gksxk (A) blds osx ds vuqfn'k (B) dsoy pqEcdh; {ks=k (C) ;g nksuksa gh (D*) buesa ls dksbZ ugha 7.
Sol.
8.
Sol.
A moving charge produces (A) electric field only (B) magnetic filed only (C*) both of them (D) none of these ,d xfr'khy vkos'k mRiUu djrk gS (A) dsoy fo|qr {ks=k (B) dsoy pqEcdh; {ks=k (C*) ;g nksuksa gh (D) buesa ls dksbZ ugha C A moving charge produces electric field as well as Magnetic field. A particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by a the path described by the particle is proportional to (A) the velocity (B) the momentum (C*) the kinetic energy (D) none of these ,d le:i pqEcdh; {ks=k ds ry ds yEcor~ ,d d.k iz{ksfir fd;k tkrk gSA d.k ds }kjk r; fd;s x;s iFk dk {ks=kQy lekuqikrh gksxk(A) osx ds (B) laosx ds (C*) xfrt ÅtkZ ds (D) buesa ls dksbZ ugha C
r
Sol.
2
mv qB
9.
[3]
,
mv m2 2 area = r2 = 2 2 v qB qB
area v2 & K.E. v2
Two particles X and Y having equla charge, after being acceleration through the same potential difference circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is ,d leku vkos'k okys nks d.k x rFkk y ,d leku foHko ls Rofjr gksus ds i'pkr~ le:i pqEcdh; {ks=k esa izfo"V gksrs gSa rFkk Øe'k% R1 ,oa R2 f=kT;k ds o`Ùkkdkj iFkksa ij xfr djrs gSaA x rFkk y ds nzO;ekuksa dk vuqikr gS (A) (R1/R2)1/2 (B) R1/R2 (C*) (R1/R2)2 (D) R1R2 C
r
mv qrB v qB m
vx
qR1B velocity of ' x 'particle m1
vy
qR 2B velocity of ' y 'particle m2
They accelerated through same potential difference mean their K.E. is same.
1 1 m v2 m v2 2 1 x 2 2 y
q2R12B2 1 q2R22B2 1 m1 m 2 2 2 m12 m22 R12 R 22 m1 m2 ma ss of x m1 R1 ratio of mass of y m2 R2 10.
2
Two parallel wires carry currents of 20 A and 40 A in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be (A) towards 20 A (B*) towares 40 A (C) zero (D) perpendicular to the plane of the current nks lekukUrj rkjksa esa 20 A rFkk 40 A /kkjk,¡ ijLij foijhr fn'kkvksa esa izokfgr gks jgh gSA bu nksuksa rkjksa ds Bhd e/; esa ,d vU; rkj fLFkr gSA ftlls /kkjk dh fn'kk 20 A okys rkj dh /kkjk ls foifjr gSA bl ij yxus okyk pqEcdh; cy gksxk (A) 20 A dh vksj (B*) 40 A dh vksj (C) 'kwU; (D) /kkjkvksa ds ry ds yEcor~
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Chapter # 35 Magnetic Field due to A current [4] Sol. B Magnetic field direction due to (20A & 40A) wire is downward direction according to midway wire. F I L B
Magnetic force on midway wire due to Both (20A & 40A) wire is towards 40 A wire. 11.
Two parallel, long wires carry currents i1 and i2 with i1 > i2. When the current are in the same direction, the magnetic field at a point midway between the wire is 10T. If the direction of i2 is reversed, the field becomes
i1 30T. The ratio i is 2
nks lekUrj ,oa yEcs rkjksa esa i1 rFkk i2 /kkjk,¡ izokfgr gks jgh gS rFkk i1 > i2 gSA tc /kkjk,¡ ,d gh fn'kk esa gS] rkjksa ds e/; fLFkr fcUnq ij i1
pqEcdh; {ks=k dh rhozrk 10T gSA ;fn i2 dh fn'kk ifjofrZr dj nh tk;s rks {ks=k 30T gks tkrk gSA vuqikr i dk eku gksxk 2 Sol.
(A) 4 C
(B) 3
(C*) 2
(D) 1
Magnetic field on the mid way wire due to i1& i2 is
0i1 0i2 10T 2d 2d
....(i)
Magnetic field on the mid way wire due to i1& i2 is
0i1 0i2 30T 2d 2d
....(ii)
From equation (i) & (ii) we get :-
20i1 20i2 40 & 20 2d 2d i1
40 d 0
ratio of 12.
20 d 0
i1 2 i2
Consider a long, straight wire of cross-section area A carrying a current i.Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the current with a speed = (i/nAe) and separated from the wore bu a distance r. The magnetic field seen by the observer is very nearly A vuqizLFk dkV {ks=kQy ds ,d yEcs rFkk lh/ks rkj ls i /kkjk izokfgr gks jgh gSA ekuk fd blds ,dkad vk;ru esa n eqDr bysDVªkWu gSA VªkWyh ij cSBk gqvk ,d izs{kd rkj ls r nwjh ij /kkjk ds foifjr fn'kk esa = (i/nAe) pky ls xfr'khy gSA izs{kd }kjk izsf{kr pqEcdh; {ks=k yxHkx gksxk (A*)
Sol.
& i2
0i 2 r
(B) zero
(C)
0i r
(D)
20i r
A Magnetic field due to long wire carrying a current i
0i 2r
Objective - II
1.
The magnetic field at the origin due to a current element i d placed at a position r is ewy fcUnq ij fLFkr ,d /kkjk vo;o i d ds dkj.k fLFkfr r ij pqEcdh; {ks=k -
0i dl xr (A) 4 r 3
(B)
0i dl xr 4 r 3
(C*)
0i rxdl 4 r 3
(D*)
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0i dl xr 4 r 3
Chapter # 35 Sol. CD i d l r 0 B 4 r3 A B B A 0i d l r B u r 3 2.
Sol.
3.
Sol. 4.
Sol.
Magnetic Field due to A current
[5]
When id l placed at a position r then magnetic field. (By property of the cross product)
Consider three quantities x = E/B, y = 1/oo and z = / CR. Here, l is the length of a wire, C is a capacitance and R is a resistance. All other symbols have standard meanings (A*) x, y have the same dimensions. (B*) y, z have the same dimensions (C*) z, x have the same dimensions (D) None of the three pairs have the same dimensions rhu jkf'k;ksa x = E/B, y = 1/oo rFkk z = / CR ij fopkj dhft;sA buesa rkj dh yEckbZ] C /kkfjrk rFkk R izfrjks/k gS] vU; ladsrksa ds lkekU; vFkZ gS (A*) x, y dh foek,¡ ,d leku gSA (B*) y, z dh foek,¡ ,d leku gSA (C*) z, x dh foek,¡ ,d leku gSA (D) rhuksa esa ls fdlh Hkh ;qXe dh foek,¡ ,d leku ugha gSA ABC Force1 = qE Force2 = qvB Dimension of Force1 = Dimension of Force2 [qE] = [qvB] E x v (i){v velocity} B
1 c (iii){c velocity light} 0 0
y
RC gives the lime constant of the RC circuit
z
l l velocity RC t
....(iv)
A long, straight wire carries a current along the Z-axis. One can find two points in the X-Y plane such that (A) the magnetic fields are equal (B*) the directions of the magnetic fields are the same (C*) the magnitudes of the magnetic fields are equal (D*) the field at one point is opposite to that at the other point ,d yEcs ,oa lh/ks rkj esa Z-v{k ds vuqfn'k /kkjk izokfgr gks jgh gSA x - y ry esa fLFkr nks fcUnqvksa ds fy;s dksbZ O;fDr izsf{kr dj ldrk gS] fd (A) pqEcdh; {ks=k ,d leku gSA (B*) pqEcdh; {ks=kksa dh fn'kk,¡ ,d leku gSA (C*) pqEcdh; {ks=kksa ds ifjek.k ,d leku gSA (D*) ,d fcUnq ij pqEcdh; {ks=k dh fn'kk nwljs fcUnq ij {ks=k dh fn'kk ls foifjr gSA BCD A long, straight wire of radius R carries a current distrobuted uniformly over its cross-section. The magnitude of the magnetic field is (A) maximum at the axis of the wire (B*) minimum at the axis of the wire (C*) maximum at the surface of the wire (D) minimum at the surface of the wire. R f=kT;k ds ,d yEcs ,oa lh/ks rkj ls gksus okyk /kkjk izokg lEiw.kZ vuqizLFk dkV esa ,d leku forfjr gSA pqEcdh; {ks=k dk ifjek.k gS (A) rkj dh v{k ij vf/kdre (B*) rkj dh v{k ij U;wure (C*) rkj dh lrg ij vf/kdre (D) rkj dh lrg ij U;wure AC The magnitude of the Magnetic field is maximum at the surface of the wire.
0i 2d The magnitude of the Magnetic field is nimumum at axis of the wire. B0 = 0 BA
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Chapter # 35 Magnetic Field due to A current [6] 5. A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field (A) increases linearly from the axis to the surface (B*) is constant inside the tube (C*) is zero at the axis (D) is zero just outside the tube ,d [kks[kyh ufydk dh yEckbZ ds vuqfn'k gksus okyk /kkjk izokg] bldh lrg ij ,d leku :i ls forfjr gSA pqEcdh; {ks=k (A) v{k ls lrg dh vksj jsf[kd :i ls c<+rk gSA (B*) ufydk ds vUnj fu;r jgrk gSA (C*) v{k ij 'kwU; jgrk gSA (D) ufydk ds rqjUr ckgj 'kwU; gksrk gSA Sol. BC The magnetic field is constant inside the current carrying hollow tube. (due to Ampere’s Law) The magnetic field is zero at the axis of current carrying hollow tube. 6.
Sol.
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero. (A*) outside the cable (B) inside the inner conductor (C) inside the outer conductor (D) in between the two conductors.
,d lek{kh;] lh/kh dscy esa dsUnzh; pkyd rFkk cká pkyd ls leku /kkjk,¡ ijLij foijhr fn'kkvksa esa izokfgr gks jgh gSA pqEcdh; {ks=k 'kwU; gksxk (A*) dscy ds ckgj (B) vkarfjd pkyd ds vUnj (C) cká pkyd ds vUnj (D) nksuksa pkydksa ds chp esa A The magnetic field is zero outside the cable due to Ampere’s Law. B dl 0iinside
Magnetic field at distance ‘r’ is
B 2r i i 0 0
B=0 7.
Sol.
A steady electric current is flowing through a cylindrical conductor. (A) the electric field at the axis of the conductor is zero (B*) the magnetic field at the axis of the conductor is zero (C*) the electric field in the vicinity of the conductor is zero (D) the magnetic field in the vidinity of the conductor is zero ,d csyukdkj pkyd ls LFkk;h fo|qr/kkjk izokfgr gks jgh gS (A) pkyd dh v{k ij fo|qr {ks=k 'kwU; gksxkA (B*) pkyd dh v{k ij pqEcdh; {ks=k 'kwU; gksxkA (C*) pkyd ds lehi fo|qr {ks=k 'kwU; gksxkA (D) pkyd ds lehi pqEcdh; {ks=k 'kwU; gksxkA BC The magnetic field at tha axis of the cylindrical conductor is zero. The electric field in the vinicinity of the conductor is zero by the Gauss Law.
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