21-Electromagnetic Induction

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 21.1 In the figure shown a square loop PQRS of side 'a' and resistance 'r' is placed near an infinitely long wire carrying a constant current . The sides PQ and RS are parallel to the wire. The wire and the loop are in the same plane. The loop is rotated by 180º about an axis parallel to the long wire and passing through the mid points of the side QR and PS. The total amount of charge which passes through any point of the loop during rotation is : fp=kesaPQRSoxkZdkjywiçnf'kZrgS]bldhHkqtk'a'rFk çfrjks/k'r'gSA;gvuUryEckbZdsrkjdsfudVj[k gSAbl rkjesa /k jkçokfgrgksjghgSAHkqtkPQrFk RSrkjdslekUrjgSAywirFk rkj,dghryesafLFkrgSAywidksyEcs rkjdslekUrjrFk QR,PSdse/;fcUnqlsxqtjusokyhv{kdslkis{k180ºls?kqek;ktkrkgSrks?kw.kZudsnkSjkuywi dsfdlhHkhfcUnqlsikfjrdqyvkos'kdhek=k gS& R a S (A) 0a 2r 𝑙n2 (B*) 0a r 𝑙n2 (C) 0a2 2r (D) cannot be found because time of rotation not give. Sol. (B) q = Idt   – 1 d dt  –   µ0Ia 𝑙n2 . 21.2 A wooden stick of length 3𝑙 is rotated about an end with constant angular velocity  in a uniform magnetic field B perpendicular to the plane of motion. If the upper one third of its length is coated with copper, the potential difference across the whole length of the stick is ,d ydM+h dh NM+ dh yEckbZ 3𝑙 gS tks fd blds ,d fljs ds lkis{k fu;r pqEcdh; {ks=k B esa fu;r dks.kh; osx  ls ?kw.kZu xfr dj jgh gS pqEcdh; {ks=k xfr ds ry ds yEcor~ gSA vxj bl NM+ dk Åijh ,d frgkbZ fgLlk rk¡cs ls qdko (dip) dk dks.k  , NM+ esa izsfjr fo-ok-cy- gksxk - (A*) B 𝑙 v cot  (B) B 𝑙 v sin  (C) B 𝑙 v tan  (D) B 𝑙 v cos  Sol. BH = Bv cot  = B cot  Hence the induced e.m.f. in the rod is B 𝑙 v cot  vr%NM+ijizsfjrfo-ok-cyB 𝑙 v cotgksxkA 21.15 Two identical cycle wheels (geometrically) have different number of spokes connected from centre to rim. One is having 20 spokes and other having only 10 (the rim and the spokes are resistanceless). One resistance of value R is connected between centre and rim. The current in R will be : nks ,d leku lkbfdy ds ifg;s ftuesa fHkUu la[;k esa Liksd dsUnz ls ifjf/k ij tqMs+ gSaA ,d esa 20 Liksd vkSj nwljh esa 10 Liksd yxs gaSA ¼ifjf/k vkSj Liksd izfrjks/k foghu gS½ ,d izfrjks/kd R tks Hkh dsUnz v k S j i f j f / k d s c h p t k s M + k t k r k g S A R e s a /kjkgksxh: (A) double in first wheel than in the second wheel nwljsdhrqyukesaigysesanqxuhgksxh (B) fourtimesinfirstwheel thaninthesecondwheelnwljs ifg;s dh rqyuk esa igys esa pkj xquh gksxh (C) will bedoubleinsecondwheel thanthatof thefirst wheeligys dh rqyuk esa nwljs ls nqxuh gksxh (D*) will be equal in both these wheels.nksuks ifg;ksa esa cjkcj gksxhA Sol. Since all the wires are connected between rim and axle so they will generate induced emf in parallel, hence it is same for any number of spokes. pw¡fdlHkhrkjifjf/kvkSj,Dlsy¼/kqjk½dschptksM+sx;sgSaAblfy;soksizsfjrfo-ok-c-lekukUrjesamRiUudjsaxsAblfy, ;gfdruhHkhla[;kesaLiksdgksalcdsfy,cjkcjgSA 21.16 A constant force F is being applied on a rod of length '𝑙' kept at rest on two parallel conducting rails connected at ends by resistance R in uniform magnetic field B as shown. ,d'𝑙'yEckbZdhNM+dksnkslekUrjpkfyrjsy(rails)dschpizfrjks/kR}kjktksM+kx;kgSA;gfudk;pqEcdh;{ks=kB esa fp=kkuqlkj j[kk x;k gSA fuEu esa ls lgh dFku pqfu, : (A) the power delivered by force will be constant with time cy}kjkiznÙk'kfDrle;dslkFkfLFkjgS¼vpj½ (B) the power delivered by force will be increasing first and then will decrease cy}kjkiznÙk'kfDrigysc<+rhgScknesa?kVrhgSA (C) the rate of power delivered by the external force will be increasing continuously ckácy}kjkiznÙk'kfDrdhnjyxkrkjc<+jghgSA (D*) the rate of power delivered by external force will be decreasing continuously. ckácy}kjkiznÙk'kfDrdhnjyxkrkj?kVjgh gSaA dP Sol. dt dF.v = dt Fdv = dt = Fa as 'a' is decreasing continuously hence the rate of power delivered by external force will be decreasing continuously. D;ksfd'a' yxkrkj?kVjgkgSvr%ckg~;cy}kjkiznr'kfDrnjijyxkrkj?kVsxhA 21.17 A uniform magnetic field exists in region given by B  3 ˆi  4 ˆj  5kˆ. A rod of length 5 m is placed along y axis is moved along x axis with constant speed 1 m/sec. Then induced e.m.f. in the rod will be: ,d {ks=k esa ,d leku pqEcdh; {ks=k B  3 ˆi  4 ˆj  5kˆ fo|eku gSS ,d NM+ ftldh yEckbZ 5m gS vkSj y v{k ds vuqfn'k gS] dks 1 m/sec. dh vpj pky ls x-v{k ds vuqfn'k xfr djkrs gaSA rc NM+ esa izsfjr fo-ok-c- gksxk - (A) zero (B*) 25 volt (C) 20 volt (D) 15 volt → Sol. e = (v  B).𝑙 e = [ ˆi  (3ˆi  4ˆj  5kˆ)] .5ˆj  e = 25 volt Ans. 21.18 In a L-R growth circuit, inductance and resistance used are 1 H and 20  respectively. If at t = 50 millisecond, current in the circuit is 3.165 A then applied direct current emf is : L-R o`f) ifjiFk esa izsjdRo vkSj izfrjks/k Øe'k% 1 H vkSj 20  gSA ;fn t = 50 millisecond ij ifjiFk esa /kkjk 3.165 A, rc yxkbZ xbZ fn"V /kkjk dk fo-ok- c- gksxk : (A) 200 V (B*) 100 V (C) 50 V (D) Datais insufficient to findout the value. eku fudkyus ds fy, vk¡dM+s vi;kZIr gSA 1 Sol. Time constant le; fu;rkad = 20 so i = 0.633 i = 0.633 E max R = 50 msec  E = 3.165  20 0.633 = 100 V 21.19 Figure shows a square loop of side 1 m and resistance 1 . The magnetic field on left side of line PQ has a magnitude B = 1.0T. The work done in pulling the loop out of the field uniformly in 1 s is fp=kesa1mHkqtko1çfrjks/kdkoxkZdkjywiçnf'kZrgSAjs[k PQdscka;hvksjpqEcdh;{ks=kdkifjek.kB=1.0T gSA1sesaywidks{ks=kls,dlekupkylsckgjfudkyusesafd;kx;kdk;ZgS& (A*) 1 J (B) 10 J (C) 0.1 J (D) 100 J Sol. W = (L)F = L × LB L2B2V = L × R = 1 J 21.20 A and B are two metallic rings placed at opposite sides of an infinitely long straight conducting wire as shown. If current in the wire is slowly decreased, the direction of induced current will be : AvkSjBnks/kfRodNYys,dvuUr:ilsyEcsrkjdsnksukasrjQfp=k uqlkjj[ksx;sgSaAvxjrkjesa/kjk/khjs&/khjsls de gks tkrh gSA rc izsfjr /kkjk dh fn'kk gksxh : (A) clockwise in Aand anticlockwise in B Aesa nf{k.kkorZ B esa okekorZ (B*) anticlockwiseinAandclockwise in B AesaokekorZ vkSj B esa nf{k.kkorZ (C) clockwise in both A and B AvkSjBnksuksaesanf{k.korZ (D) anticlockwiseinboth A& B AvkSj B nksuksa esa okekorZ Sol. The field at A and B are out of the paper and inside the paper respectively. {ks=kAvkSjB Øe'k%dkxtdsckgjvkSjvUnjgSA As the current in the straight wire decreases the field also decreases. tSlsghjs[kh;rkjesa/kjkdegksrhgS]pqEcdh;+{ks=kHkhdegksrkgS For B : B dsfy,  B The change in the magnetic field which causes induced current (B) is along (+)z direction. izsfjr/kjkmRiUudjusokyspqEcdh;{ks=kesaifjorZu(B),(+)zfn'k esagksxk Hence, induced emf and hence current should be such as to oppose this change B. vr%izsfjr/kjkvkSjizsfjrfo-ok-c-blizdkjgksaxsfd BesaifjorZudkfojks/kdjukpkfg,A Hence, induced emf should be along – z direction which results in a clockwise current in 'B'. Similarly, there will be anticlockwise current in 'A'. Hence (B). vr%]izsfjrfo-ok-c-–zfn'k esagksukpkfg,AtksfdBesanf{k.korZ/kjkmRiUudjrkgSA blhizdkj'A'esaokekorZ/k jkmRiUugksxhblfy,(B). 21.21 A vertical conducting ring of radius R falls vertically with a speed V in a horizontal uniform magnetic field B which is perpendicular to the plane of the ring : ,dÅ/okZ/kjpkydNYykftldhf=kT;kRgS]Å/okZ/kjVpkyls,d{kSfrtlekupqEcdh;{ks=kBtksfdNYysdsry ds vfHkyEcor~ gS] esa fxj jgk gS rks : (A) A and B are at same potential AvkSjBlefoHkogS (B*) C and D are at same potential CvkSjDlefoHkogS (C) current flows in clockwise direction /kjknf{k.korZcgrhgS (D) current flows in anticlockwise direction /kjkokekorZcgrhgS Sol. (B) When the ring falls vertically, there will be an induced emf across A & B (e = Bv (2r)). Note that there will be a potential difference across any two points on the ring of line joining them has a projected length in the horizontal plane. tcNYykÅ/okZ/kjfxjrkgSrcAvkSjB dschp,dizsfjrfo-ok-cy(e= Bv (2r)) mRiUugkstkrkgSA ;gk¡ijfjaxdsfdlhnksfcUnqvksadschpftudks,djs[k lstksM+usij{kSfrtryesa,diz{ksiyEckbZvkrhgS]fo-ok-c-gksrk gSA For example, between points 'P' & 'Q' there is a projected length 'x' in the horizontal plane. mnkgj.kdsfy,'P' &'Q'dse/;{kSfrtryesaiz{ksiyEckbZ'x'gSA  P.d. across P & Q is : foHkokUrjP vkSjQ ds chp gS & V = B v x. But for points C and D : x = 0. Therefore; P.d. = 0. Hence (B). ysfdufcUnqvksaCvkSjDdsfy,: x= 0. blfy, foHkokUrj ; = 0. blfy, (B). 21.22 Two identical conducting rings A & B of radius R are in pure rolling over a horizontal conducting plane with same speed (of center of mass) but in opposite direction. A constant magnetic field B is present pointing inside the plane of paper. Then the potential difference between the highest points of the two rings, is : nks le:ipkydoy;AvkSjB ftudhf=kT;k,saRgSpkyls{kSfrt pkyd/kjkryij 'kq)yksVuh(rolling) xfrls ,dnwljsdsfoijhrfn'k esaxfrdjjgsgSaA,dvpjpqEcdh;{ks=kBtksisijdsrydsvUnjgSArcnksuksaNYyksads mPpre fcUnqvksa ds e/; foHkokUrj gksxk : (A) zero'kwU; (B) 2 Bvr (C*) 4Bvr (D)noneof these buesalsdksbZugha Sol. Considering a projected length 2R on the ring in vertical plane. This length will move at a speed v perpendicular to the field. This results in an induced emf : e = Bv(2R) in the ring. ,diz{ksfiryEckbZ2RÅ/okZ/kjryesaNYysijysusijA;gyEckbZv pkyls{ks=kdsvfHkyEcor~pyrhgSA;g,dizsfjr fo-ok-cymRiUudjrkgSA e = Bv(2R) in the ring. NYys ds vUnj In Ring "A" : eA =B(-v)(2R) In Ring "B" : eB = B(v)(2R) fjax "A" esa : eA =B(-v)(2R) fjax "B" esa: eB = B(v)(2R) Therefore, potential differnce between A & B = B(v)(2R) - B(-v)(2R) = 4 BvR. Note : there wil be no p.d. across a diameter due to rotation. blfy, A vkSj B ds chp foHkokUrj= B(v)(2R) - B(-v)(2R) = 4 BvR. uksV:O;kldschp?kw.kZudsdkj.kdksbZfoHkokUrjughagksrkgSA Alternate – Considering rotation of diameter about lowest point : B(2r)2 e = = 2Bvr in A (since pure rotation). 2 and e = – 2Bvr in B. Hence (C) fodYi – O;kl ;k ?kw.kZu fupys fcUnq ds lkis{k ysus ij :- B(2r)2 e = 2 =2Bvr A esa(pwafd'kq)yksVuh) vkSj e=– 2Bvr B esa blfy, (C) 21.23 An inductor L and a resistor R are connected in series with a direct current source of emf E. The maximum rate at which energy is stored in the magnetic field is : ,dizsjdRoLvkSjizfrjks/kR,,dEfo|qrokgdcydsfn"V/kjkL=ksrlstksM+sx;sgSa¼Js.khØeesa½rks]pqEcdh;{ks=kesa vf/kdre ÅtkZ lap; dh nj gS (A*) E 4R (B) E R (C) 4E2 R (D) 2E2 R Sol. The graph of current is given by : /k jkdkxzkQfuEulsfn;ktkrkgS: i = i0 (1 – e–t/)  di dt = i0 –t/  Energy stored in the form of magnetic field energy is : pqEcdh;ÅtkZds:iesalafprÅtkZgS: UB = 1 Li2 2  Rate of increase of magnetic field energy is :  pqEcdh;{ks=kÅtkZdsc<+usdhnjgS: R = dUB dt = Li di = 0 dt  (1 – e–t/) e–t/ dR This will be maximum when dt = 0 dR ;g vf/kdregksxh dt = 0  e–t/ = 1/2 Substituting : izfrLFkfirdjusij% Li2  1   1   Li2 L(E / R)2  E2 R = 0 1     = 0 =   = max   2   2  4   4(L / R)  4R 21.24 In the circuit shown in figure, the switch S was initially at position 1. After sufficiently long time, the switch S was thrown from position 1 to position 2. The voltage drop across the resistor at that instant is : fp=kesafo|qrifjiFkesa]izkjEHkesafLopS '1'fLFkfresagSA,di;kZIryEcsle;dscknfLopSdksfLFkfr'1'ls'2'esa cnyk x;k gSA ml {k.k izfrjks/k ij foHkokUrj gksxk 2 1 (A) zero (B*) E (C) E LC (D) none of these R Sol. When the switch is at position 1 : tc fLop fLFkfr '1' esa gS % 1 Li 2 = B 2 LE2 2R2 Just after the switch is shifted to position 2, current  = that instant P.d. across reisistance will be : E is flowing across the resistance. Hence, at R fLopdks'2'fLFkfresacnystkusdsrqjUrckn]/kjk= E izfrjks/kesacgjghgS]vFkZr~izfrjks/kijfoHkokUrjgksxk- V = R = E .R = E R Hence (B). blfy, (B) 21.25 A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field dB is increasing at a constant rate of dt (tesla/second). An electron of charge q, placed at the point P on the periphery of the field experiences an acceleration : ,d le:i pqEcdh; {ks=k B csyukdkj {ks=k ¼ftldh f=kT;k R ½ esa fo|eku gSA dB pqEcdh; {ks=k vpj nj dt (Vslyk/lsd.M). ls c<+ jgk gSA ,d vkos'k tks {ks=k dh ifjf/k ij fcUnq 'P' ij gS] dk Roj.k gksxk : 1 eR dB 1 eR dB (A*) 2 m dt towards left cka;h rjQ (B) 2 m dt towards right nka;h rjQ eR dB (C) m dt towards left cka;h rjQ (D)zero'kwU; Sol. If we consider the cylindrical surface to be a ring of radius R, there will be an induced emf due to changing field. ;fngecsyukdkjlrgij'R'f=kT;kdhoy;ds:iesafopkjdjsarcblesaifjofrZr{ks=kdsdkj.k,dizsfjrfo-ok-cy gksxk- → d dB E.d𝑙 = dt = – A dt dB dB R dB  E(2R) = – A dt = – R2 dt  E = 2 dt  Force on the electron. bysDVªkWu ij cy F = – Ee =  eR dB  acceleration Roj.k = 1 eR dB 2 dt 2 m dt As the field is increasing being directed inside the paper, hence there will be anticlockwise induced current (in order to oppose the cause) in the ring (assumed). Hence there will be a force towards left on the electron. isijdsvUnjdhfn'k esa{ks=kc<+jgkgS]blfy,¼dkj.kdkfojks/kdjusdsfy,½NYysesaizsfjr/kjkokekorZgksxhAblfy, ,dbysDVªkWuijck;harjQcygksxkA 21.26 AB is a resistanceless conducting rod which forms a diameter of a conducting ring of radius r rotating in a uniform magnetic field B as shown. The resistors R and R do not rotate. Then current through the resistor R is : AB ,dizfrjks/kghupkydNM+gStksrf=kT;kdhpkydoy;dkO;klcukrsgq,fp=k uqlkj,dle:ipqEcdh;{ks=kB esa ?kwe jgh gSA izfrjks/k R1 vkSj R2 ?kw.kZu ughs djrs gSaA izfrjks/k R1 ls /kkjk gS : (A*) Br 2 2R1 (B) Br 2 2R2 (C) Br 2 2R1R2 (R1 + R2) (D) Br 2 2(R1  R2 ) Sol. The equivalent diagram is : rqY;fp=kgS& The induced emf across the centre and any ponit on the circumference is : dsUnzvkSjifjf/kijfdlhfcUnqdschpizsfjrfo-ok-c-]gS% → | e | 1 = 2 B  𝑙2 = Br 2 Br 2 2  Current through R = 2R1 Ans. R1 lsizokfgr/k jk= Br 2 2R1 mÙkj 21.27 AB and CD are fixed conducting smooth rails placed in a vertical plane and joined by a constant current source at its upper end. PQ is a conducting rod which is free to slide on the rails. A horizontal uniform magnetic field exists in space as shown. If the rod PQ is released from rest then, ABvkSjCDfLFkj?k"kZ.kfoghujsy,dÅ/oZryesagStksfdbldsÅijhfljsij,dvpj/k jkdsL=k srlstqM+hgaS ¼fp=k esa½PQ ,dpkydNM+jsy(rails) ijpyusdksLora=kgSA,d{kSfrt,dlekupqEcdh;{ks=kbltxgfo|ekugSAfp=k esa vxj NM+ PQ dks fojke ls NksM+k tkrk gSA rc (A) The rod PQ will move downward with constant acceleration PQ NM+ vpj Roj.k ls uhps tk;sxhA (B) The rod PQ will move upward with constant acceleration PQ NM+ vpj Roj.k ls Åij tk;sxhA (C) The rod will move downward with decreasing acceleration and finally acquire a constant velocity NM+?kVrsgq,Roj.klsuhpstk;sxh]fQjvUresa,dfu;rosxizkIrdjysxhA (D*) either A or B. ;k rks A;kB → → → Sol. There is a force FM  (d𝑙 B) acting on the rod carrying a current . By the rule of cross product, this force is vertically upward. F.B.D. of the rod : → → → /k jkizokghNM+ijcy FM  (d𝑙 B) yxjgkgSA otzxq.kuQydsfu;els]cyÅ/okZ/kjÅijgksxkA NM+ dk F.B.D. (Freebody diagram eqDr oLrq js[kkfp=k) : F – W = ma a = The magnitude of acceleration will be constant, but the direction will depend on the mass of the rod. Hence (D) is correct option. Roj.kdkifjek.kfu;rgSijfn'k NM+dsnzO;ekuijfuHkZjdjsxhA blfy, (D). 21.28 A conducting ring of radius r with a conducting spoke is in pure rolling on a horizontal surface in a region having a uniform magnetic field B as shown, v being the velocity of the centre of the ring. Then the potential difference V – V is - ,drf=kT;kdkpkydNYykftldhLiksdHkhpkydgS],dlekupqEcdh;{ks=kBesafp=k uqlkj{ksfrtryijyq<+dk;k tkrk gSA v NYys ds dsUnz dk osx gSA rc foHkokUrj V0 – VA gS - (A) Bvr 2 (B) 3Bvr 2 Bvr (C*) 2 (D) 3Bvr 2 Sol. Considering pure rolling of OA about A: the induced emf across OA will be : → | e |  B(r)2 . 2 From Lenz law, O will be the negative end, while A will be the positive end. Br 2 Hence v – v = – 2 Hence (C) is correct option. OA dh A ds ifjr% 'kq) ykSVuh xfr ij fopkj djrs gSaA OA ds e/; izsfjr fo- ok-c- gksxk - → | e |  B(r)2 . 2 ysat+dsfu;elsO_.k RedfljkvkSjA/kukRedfljkgS blfy, vr% (C). – v = – Br 2 2 21.29 A metallic ring of mass m and radius r with a uniform metallic spoke of same mass m and length r is rotated about its axis with angular velocity  in a perpendicular uniform magnetic field B as shown. If the central end of the spoke is connected to the rim of the wheel through a resistor R as shown. The resistor does not rotate, its one end is always at the center of the ring and other end is always in contact with the ring. A force F as shown is needed to maintain constant angular velocity of the wheel. F is equal to (The ring and the spoke has zero resistance) ,d /kkfRod NYyk ftldk nzO;eku m vkSj f=T;k r gSA ftlds lkFk ,d m nzO;eku vkSj r yEckbZ okyk LikWd Hkh yxk gS fp=kkuqlkj viuh v{k ds lkis{k dks.kh;  ls ,d leku yEcor~ pqEcdh; {ks=k B esa ?kw.kZu djrk gSA fp=kkuqlkj bldk dsUnzh; fljk NYys dh ifjf/k ij ,d fcUnq ls izfrjks/k R ds lkFk tksM+k tkrk gSA izfrjks/k ugha ?kwerk gSA bldk ,d fljk ges'kk ifjf/k ij rFkk ,d dsUnz ij gSA ,d cy F ifg;s dks vpj dks.kh; pky ls ?kqekus ds fy, t:jh gSA F cjkcj gksxk ¼tcfd NYyk o rkj dk izfrjks/k 'kwU; gS½ & (A) B2r 2 8R (B) B2r 2 2R (C) B2r 3 2R (D*) B2r 3 4R Sol. F = BL Inducedcurrent :izsfjr/k jk (Br 2 / 2) = R  2   F = B  r B2 r 3 = b  2R  2R To maintain constant angular velocity : fu;rdks.kh;pkydkscuk;sj[kusdsfy,% F(r) = F (r/2)  F = FB B2r 3 = . Ans. B 2 4R 21.30 A closed circuit consists of a resistor R, inductor of inductance L and a source of emf E are connected in series. If the inductance of the coil is abruptly decreased to L/4 (by removing its magnetic core), the new current immediately after this moment is : (before decreasing the inductance the circuit is in steady state) ,d cUn ifjiFk ftlesa ,d izfrjks/k R vkSj ,d izsjd L dks E fo-ok-c- ds L=kksr ls Js.kh Øe esa tksM+k x;k gSaA vxj izsjdRo vpkud L/4 rd de gks tkrk gS (bldh pqEcdh; dksj dks gVkus ij), bl {k.k ds rqjUr ckn /kkjk gksxh : (izsjdRo dks ?kVkus ls igys ifjiFk LFkk;h voLFkk (steady state) esa gS) (A) zero (B) E/R (C*) 4 E R (D) E 4R Sol. Flux through a closed circuit containing an inductor does not change instantaneously. ,dcUnfo|qrifjiFkftlesa,dizsjdgS]bldk¶yDl,dnels ughacnyrkgSA  E    R  = L (i)  i = 4E Ans.   4 R 21.31 A magnetic field (B), uniform between two magnets can be determined measuring the induced voltage in the loop as it is pulled through the gap at uniform speed 20 m/sec. Size of magnet and coil is 2cm × 1cm × 2cm and 4cm × 6cm as shown in figure. The correct variation of induced emf with time is : (Assume at t = 0, the coil enters in the magnetic field) : nks pqEcdks ds e/; le:i pqEcdh; {ks=k dh x.kuk ml dq.Myh esa izsfjr foHko dks ekidj dh tk ldrh gS ftls fd nksuksa pqEcdksa ds e/; fjDr LFkku esa 20 eh-@lS- dh nj ls [khapk tkrk gSA pqEcd rFkk dq.Myh dk vkdkj Øe'k% 2cm × 1cm × 2cm rFkk 4cm × 6cm gS ¼tSlk fp=k esa iznf'kZr gS½A le; ds lkFk izsfjr fo-ok-c- esa ifjorZu dk lgh izn'kZu gS & (le; t = 0 ij] dq.Myh dks pqEcdh; {ks=k esa izosf'kr ekusa) : (A*) (B) (C ) (D) d Sol. E = dt B d(b𝑙) = dt = Bbv = B × 2 × 10–2 × 20 = 0.40 B 1 10 2 t = 20 = 5 × 10–4 sec = 500  sec 6  102 t = 20 = 3 × 10–3 sec = 3000  sec 21.32 The plane of a square loop of wire with edge length a = 0.2 m is perpendicular to the earth's magnetic field BE at a point where BE = 15 T. The total resistance of the loop and the wires connecting it to the galvanometer is 0.5 . If the loop is suddenly collapsed( such that area of the loop becomes zero) by horizontal forces as shown, the total charge passing through the galvanometer is : Hkqtk a = 0.2 eh- dk rkj dk oxkZdkj ywi dk ry ,d fcUnq ij i`Foh ds pqEcdh; {ks=k BE ds yEcor~ gS] tgk¡ BE = 15 T gSA ywi ds rkj rFkk tqM+s /kkjkekih dk dqy izfrjks/k 0.5  gSA fp=kkuqlkj ;fn ywi vpkud {kSfrt cy ls VwV tk;sa ¼bl izdkj fd ywi dk {ks=kQy 'kwU; gks tkrk gS½ rks /kkjkekih ls ikfjr dqy vkos'k gS F F (A) 0.4 C (B) 0.75 C (C) 0.9 C (D*) 1.2 C | E | B dA Sol. i = R = R dt Q =  idt = B d A = B A R R using values Q = 1.2 × 10–6 C ekuksa dk iz;ksx djus ij Q= 1.2× 10–6 C 21.33 A current flows through a rectangular conductor in the presence of uniform magnetic field B pointing out of the page as shown. Then the potential difference VP – VQ is equal to. (assume charge carriers in the conductor to be positively charged moving with a drift velocity of v) n'k Z;svuqlkj,dvk;rkdkjpkydls]pqEcdh;{ks=kBtksdkxtlsckgjdhvksjbafxrgS]dhmifLFkfresa/k jkizokfgr gksjghgSArksfoHkokUrjVP –VQ cjkcjgS&(pkyddsvUnjvkos'k&okgd/kukRedekus,oaosvuqxeu¼viogu½osx v ls pyrs gSaA) (A) Bvb (B) –Bvb (C*) Bvc (D) –Bvc Sol. |  | = B 𝑙v where l is the edge perpendicular to both B and → i.e.c. ||=B𝑙vtgkalHkqtkgSvFk ZrctksfdBrFk  |  | = B v c. → nksuksdsyEcor~gSA Now by right hand thumb rule magnetic force an a positive charge moving towards right is in down ward direction Hence end P will be positive. nka;sgLrfu;elsnka;hrjQxfrekufdlh/kukRedvkos'kijpqEcdh;cyuhpsdhrjQyxsxkAvr%fcUnqP/kukRed gksxkA  Vp – VQispositive/kukRed gSA   = + Bv𝑙. 21.34 A conducting circular ring and a conducting elliptical ring both are moving pure translationally in a uniform magnetic field of strength B as shown in figure on a horizontal conducting plane then potential difference between top most points of circle and ellipse is : ,dpkydoy;rFk ,dpkydnh?kZo`Ùknksuksa,d,dlekupqEcdh;{ks=kBesafp=k uqlkj'kq)LFk ukUrj.kxfrdjrs gSmudh;gxfr{kSfrtpkydryijgSrkso`Ùkds'kh"kZfcUnqonh?kZo`Ùkds'kh"kZfcUnqdse/;foHkokUrjgksxk& (A) 12 va (B*) 4 vBa (C) 8 vBa (D) 2 vBa Sol. 21.35 A small circular wire loop of radius a is located at the centre of a much larger circular wire loop of radius b as shown above(b>>a). Both loops are coaxial and coplanar. The larger loop carries a time (t) varying current I = I0 cos  t, where 0 and  are constants. The magnetic field generated by the current in the large loop induces in the small loop an emf that is approximately equal to which of the following. ,d a f=kT;k dk NksVk o`Ùkkdkj ywi tks fd ,d cM+h f=kT;k b ds o`Ùkkdkj ywi ds dsUnz ij gS tSlk fd fp=k esa crk;k x;k gS (b > > a)A nksuksa ywi lek{kh; rFkk leryh; gSA cM+s okys ywi esa le; ds lkFk ifjorhZ /kkjk I = I0 cos  t gSA tgk¡ 0 rFkk  fu;r gSA cM+s okys ywi ds dkj.k pqEcdh; {ks=k ds dkj.k NksVs ywi esa fo- ok- cy (emf) izsfjr gksrk gS rks og buesa ls yxHkx fdlds cjkcj gksxk & (A) 00 a2  cos t (B*) 00 a2  sint 2 b 2 b (C) 00 a  sint (D) 00 a  cos t 2 b2 2 b2 Sol. The magnetic filux in inner loop due to current in outer loop is ckgjhywidsdkj.kvkUrfjdywiesapqEcdh;¶yDlgSA   0 2i  2  a2   = BA =  4  . a b  = 0   i0 cos t   e = d dt = 0 a2 i0 2b sin t 21.36 The number of turns, cross-sectional area and length for four solenoids are given in the following table. pkjifjukfydkvksa(solenoids)dhyEckbZ]vuqizLFkdkV{kS=kQyrFk ?ksjksdhla[;kfuEulkj.khesafn;sx;sgSA Solenoid Total Turns Area Length 1 2N 2A 𝑙 2 2N A 𝑙 3 3N 3A 2𝑙 4 2N 2A 𝑙/2 i fj ukfydk dqy ?ksj ksadhl a[ ; k vuqi zLFkdkV dk{ks=kQy yEckbZ 1 2N 2A 𝑙 2 2N A 𝑙 3 3N 3A 2𝑙 4 2N 2A 𝑙/2 The solenoid with maximum self inductance is : ogifjukfydk ftldkvf/kdre LoizsjdRogS & (A) 1 (B) 2 (C) 3 (D*) 4 Sol. (D) L =  n2r2𝑙 A = r2 N n = 𝑙  L = N2 0 𝑙 A By putting the given values, it can be seen that it is maximum for solenoid no.4. fn;sx;sekuksdksj[kusij];gns[k tkldrkgSfd;gifjukfydk4dsfy;svf/kdregSA 21.37 Assume Earth's surface is a conductor with a uniform surface charge density . It rotates about its axis with angular velocity . Suppose the magnetic field due to Sun at Earth at some instant is a uniform field B pointing along earth’s axis. Then the emf developed between the pole and equator of earth due to this field is. (Re = radius of earth) i`Fohlrgdks,dlekui`"Bh;vkos'k?kuRodkpkydekusaA;gviusv{kdsifjr%dks.kh;osxls?kwejghgSAfdlh {k.klw;Zdsdkj.ki`FohijpqEcdh;{ks=k]i`Fohdsv{kdsvuqfn'k,dlekuBdhdYiukdjsaArks/kzqo,oaHkwe/;js[k ds chpbl{ks=kdsdkj.kmRiUufo-ok-cygS (Re =i`Fohdhf=kT;k) – (A*) 1 BR2 (B) BR2 (C) 3 BR2 (D) zero 2 e e 2 e Sol. the equator can be seen as a conducting ring of radiur R revolving with angular velocity  in a perpendicular magnetic field B. v{k a'k,dRef=kT;kdhpkydoy;tksfdpEcdh;{ks=kBdsyEcor~dks.kh;osxlsxfrekugS]dslekuizrhrgksxhA BR2 Potentialdifference. acrossitscenter andperipherydsUnz vksj ifjf/k ds e/; foHkokUrj = e 2 Potential at pole = potential of the axis of earth i.e. potential at point O /kzqoksaijfoHko=i`Fohdhv{kijfoHkovFkZrfcUnqOijfoHko BRe 2  Veqvator v{k – Vpole /kzqo = 2 . 21.38 A circuit containing capacitors C1 and C2 as shown in the figure are in steady state with key K1 closed. At the instant t = 0, if K1 is opened and K2 is closed then the maximum current in the circuit will be : ,difjiFkesala/kfj=k C1rFk C2fp=kuqlkjLFk;hvoLFk esagSarFk daqthK1cangSAt=0le;ijvxjK1dks[k syfn;k tkrk gS rFkk K2 dks can dj fn;k tkrk gS ] rks ifjiFk esa vf/kdre /kkjk gksxh & [ (A*) 1 A (B) 1 A (C) 2 A (D) Noneof these buesa ls dksbZ ugha Sol. When switch K1 is opended and K2 is closed it becomes L–C circuit so applying energy conservation tcdqathK1dks[k syrsgSvksjK2dscUndjrsgSrks;gL–CifjiFkcutkrkgSvr%ÅtkZlja{k.kls Q2 0 = 2C 1 Li2 ; Q 2 0 = CeqV = C1C2 C1  C2 · V = (20 ×10–6) (20  10–6 )2 6 = 1 (0.2 × 10–3) i2  i = 1 A 2 2  2 10 21.39 In the given circuit having an ideal inductor of inductance L, resistor of resistance R and an ideal cell of emf , the work done by the battery in one time constant after the switch is closed is fn;s x;s ifjiFk esa L izsjdRo dh vkn'kZ izsjd dq.Myh gS o izfrjks/k R gS o fo|qr okgd cy  dk ,d vkn'kZ lsy gSA fn;s x;s ifjiFk esa fLop dks cUn djus ds i'pkr~ ,d le; fu;rkad esa cSVjh }kjk fd;k x;k dk;Z gSA 2 (A) R (B) 2L R2 e (C) 2L R2 (D*) 2L eR2 Sol. i =  (1 eRt / L ) R charge passing through battery cSVjhlsizokfgrvkos'k%  L / R   L L  L q = idt  R (1 eRt / L )dt 0 2L =  R R  e1  1 R  = eR2  Wb = q  = eR2 SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M- II: cgqy lgh fodYi izdkj 21.40 A conducting rod of length 𝑙 is moved at constant velocity ‘v0’ on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct : ,dpkydNM+ftldhyEckbZ 𝑙 gSnkslekUrj?k"kZ.kfofgupkydiVfj;ksadschp‘v0’vpjosxlspyjghgSAfp=k uqlkj tksfdvpjpqEcdh;{ks=kBesatksfdjsydsrydsvfHkyEcor~gS]esaj[k x;kgSAiVfj;ksadks,dizfrjks/kRlstksM+k x;kgSArcfuEuesalsdkSu&lk@lslghgS@gSa: R (A*) The thermal power dissipated in the resistor is equal to rate of work done by external person pulling the rod. (B*) If applied external force is doubled than a part of external power increases the velocity of rod. (C) Lenz’s Law is not satisfied if the rod is accelerated by external force (D*) If resistance R is doubled then power required to maintain the constant velocity v0 becomes half. (A*) izfrjks/k esa Å"eh; 'kfDr cká O;fDr }kjk NM+ dks [khapus esa fd;s x;s dk;Z dh nj ds cjkcj gksxhA (B*) vxjckácydksnqxukfd;ktk,rc]cká'kfDrdkdqNHkkxNM+dhpkydksc<+krkgSA (C) vxjckácy}kjkNM+dksRofjrfd;ktk,rcysatfu;elUrq"VughagksrkgSA (D*) vxj izfrjks/k R dks nqxuk fd;k tk, rc NM+ dks vpj osx v0 ij cuk;s j[kus dh vko';d 'kfDr vk/kh gks tk;sxhA Sol. Rate of work done by external agent is : ckádkjd}kjkdk;Zfd;stkusdhnjgS% dw BL.dx dt = dt dw BL.dx dt = dt = BLv & thermal power dissipated in the resistor = e = (BvL)  =BLv oÅ"eh;ÅtkZtksizfrjks/kesa{k;gqbZgS= e=(BvL) clearly both are equal, hence (A) . nksuksacjkcjgSblfy,(A) If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increases, hence (B) vxjyxk;kx;kcynqxukfd;ktk,rcNM+ij,difj.kkehcyoRoj.kyxsxkAifj.kkekuqlkjosxc<+rkgS]blfy, (B) Since ;  = e R pw¡fd ;  = e On doubling ‘R’, current and hence required power becomes half. ‘R’,dksnqxukdjusij]/k jkvkSjvko';d'kfDrvk/khjgtk,xhA Since, P = BLv Hence (D). pwafd , P = BLv blfy, (D). 21.41 In the figure shown ‘R’ is a fixed conducting fixed ring of negligible resistance and radius ‘a’. PQ is a uniform rod of resistance r. It is hinged at the centre of the ring and rotated about this point in clockwise direction with a uniform angular velocity . There is a uniform magnetic field of strength ‘B’ pointing inwards. ‘r’ is a stationary resistance fp=k uqlkj‘R’,dn`<+pkydNYykgSizfrjks/kux.;vkSjf=kT;kagSAPQizfrjks/krdh,dlekueksVkbZdhNM+gSA bldksNYysdsdsUnzijdCtslsdlkx;kgSvkSjblfcUnqdslkis{knf{k.k orZfn'k esavpjdks.kh;osxls?kqek;ktkrk gS;gk¡,dlekupqEcdh;lkeF;Z‘B’dk{ks=kvUnjdhrjQgSA‘r’,dfLFkjizfrjks/kgSAfuEuesalslghdFkupqfu,- (A) Current through ‘r’ is zero. 2Ba2 (B*) Current through ‘r’ is 5r . (C) Direction of current in external ‘r’ is from centre to circumference. (D*) Direction of current in external ‘r’ is from circumference to centre. (A) ‘r’ ls 'kwU; /kkjk cgrh gS 2Ba2 (B*) ‘r’ ls/kkjk 5r cgrhgS (C)cká‘r’esa/kjkdsUnzlsifjf/kdhfn'k esagksxh (D*) cká ‘r’ esa /kkjk ifjf/k ls dsUnz dh rjQ gksxh Sol. Equivalent circuit : rqY;ifjiFk% Inducedemf izsfjrfo-ok-c-e= Br 2  2  =   ( Radius f=kT;k= a) By nodal equation : nodal uksMylehdj.k}kjk% 2  2   X  e   X  0  4   +     r  = 0  5X = 4e 4e 2Ba2  X = 5 = 5 X 2Ba2 andvkSj  = r = 5r also direction of current in ‘r’ will be towards negative terminal i.e. from rim to origin. Alternatively;by equivalent of cells (figure (ii)) : vkSj'r'esa/k jkdhfn'k _.k RedVfeZuydhrjQgksxh vFk Zr~ifjf/klsewyfcUnqdhrjQlSyksadsrqY;}kjk(figure(ii)): e I = r  r 4 4 e = 5 r 2Ba2 = 5r Br 2 ( e = ) 2 21.42 A circuit consisting of a constant e.m.f. 'E', a self induction 'L' and a resistance 'R' is closed at t = 0. The relation between the current I in the circuit and time t is as shown by curve 'a' in the figure. When one or more of parameters E, R & L are changed, the curve 'b' is obtained.The steady state current is same in both the cases. Then it is possible that: (A*) E & R are kept constant and L is increased. (B) E & R are kept constant and L is decreased (C*) E & R are both halved and L is kept constant (D) E & L are kept constant and R is decreased ,d fu;rfo0ok0cy 'E',,dLoçsjdRo 'L'o ,dizfrjks/k 'R'okysifjiFk dksle; t=0ijcUn fd;ktkrkgSA/kkjk I o le; t ds chp lEcU/k fp=k esa n'kkZ;s x;s oØ 'a' ds vuqlkj çkIr gksrk gSA ;fn E , R o L esa ls ,d ;k ,d ls vf/kd çkpyksa dks ifjofrZr dj fn;k tk, rks oØ 'b' izkIr gskrk gSA nksuksa fLFkfr;ksa esa fu;r /kkjk leku gS rks ;g lEHko gS] fd (A*)E oRfu;r j[kdj Ldksc<+k;k tkrkgSA (B)EoRfu;r j[kdj Ldks?kVk;k tkrkgSA (C*)EoRnksuksavk/ksdjfn;sx;sgSrFk L fLFkjj[k tkrkgSA(D)EoLdksfu;rj[kdjRdks?kVk;ktkrkgSA Sol. (A, C) Since curve 'b' is below curve 'a' hence it is posssible that E & R are kept constant and L is increased or E & R are both halved and L is kept constant Sol. (A,C) D;ksfd xzkQ 'b' xzkQ 'a' ls uhpsgSA blfy,;g laHko gksldrk gSfd EoRfu;rgks rFkkLc<+rkjgs ;k EoRnksuksavk/ksgksrFk Ldksfu;rj[k tk;sA 21.43 In the circuit diagram shown n'kZ;sx;sifjiFkesa (A*) time constant is L/R (B) time constant is 2L/R (C) steady state current in inductor is 2/R (D*) steady state current in inductor is /R (A*)le; fu;rkad L/RgSA (B) le; fu;rkad 2L/RgSA (C)izsjddq.MyhesaLFk;h/kjk2/RgSA (D)izsjddq.MyhesaLFk;h/kjk/RgSA Sol. (A, D) At steady state no resistance will be offered by inductor hence, time constant is L/R and steady state current in inductor is E/R. Sol. (A,D)LFk ;hvoLFk esaizsjd}kjkdksbZHkhizfrjks/kughfn;ktkrkgSAvr%le;fu;rkadL/RrFk LFk ;hvoLFk esaizsjdesa/kjkE/RgSA 21.44 A conducting loop rotates with constant angular velocity about its fixed diameter in a uniform magnetic field. Whose direction is perpendicular to that fixed diameter. ,dpkydywifu;rdks.kh;osxls bldsO;kldsifjr%,dlekupqEcdh;{ks=kesa?kw.kZudjrkgSAfuEupqEcdh;{ks=k dh fn'kk fLFkj O;kl ds yEcor~ gS & (A*) The emf will be maximum at the moment when flux is zero. (B*) The emf will be '0' at the moment when flux is maximum. (C*) The emf will be maximum at the moment when plane of the loop is parallel to the magnetic field (D*) The phase difference between the flux and the emf is /2 (A*) fo0ok0cyml{k.kvf/kdregksxktc¶yDl'kwU;gSA (B*) fo0ok0cyml{k.k'kwU;gksxktc¶yDlvf/kdregSA (C*) fo0ok0cyml{k.kvf/kdregksxktcywidkrypq0{ks=kdslekUrjgSA (D*) ¶yDl rFkk fo0ok0cy ds e/; dykUrj /2 gSA Sol. |e| = Ba  sint  = Ba cost ; * |e| is maximum when t = /2 so  is zero. * |e| is zero then t = 0 So  = is maximum. Sol. |e| = Ba  sint  = Ba cost ; * |e| vf/kdre gS tc t = /2 vr%'kwU;gSA * |e| 'kwU; gS rc t = 0 vr%= vf/kdregSA 21.45 An ideal inductor, (having initial current zero) a resistor and an ideal battery are connected in series at time t = 0. At any time t, the battery supplies energy at the rate PB, the resistor dissipates energy at the rate PR and the inductor stores energy at the rate PL. ,d vkn'kZ izsjdRo ('kwU; izkjfEHkd /kkjk), ,d izfrjks/k rFkk ,d vkn'kZ cSVjh dks t = 0 ij Js.khØe esa tksM+rs gSaA fdlh le; t ij] cSVjh PB nj ls ÅtkZ iznku djrh gS] izfrjks/k PR nj ls ÅtkZ [kpZ djrk gS rFkk izsjdRo PL nj ls ÅtkZ lafpr djrk gS & (A*) PB = PR + PL for all times t. (B) PR < PL for all times t. (C*) PR < PL only near the starting of the circuit. (D) PR > PL only near the starting of the circuit. (A*) lHkh le; t ij PB = PR + PL (B) lHkh le; t ij PR < PL (C*) ifjiFkds'kq:gksusdsrqjUrcknPR PL Sol. By principal of energy conservation. PB = PR + PL Near the starting of the circuit PR = i R and P 2 di = L i di dt . As dt has greater value at the starting of the circuit, PL > PR gy% ÅtkZlaj{k.kdsfl)kUrls& PB = PR + PL di ifjiFk'kq:gksusdsle;& PR = i R rFkkP = L i dt . 2 di ifjiFk 'kq: gksus ds le; dt dk eku T;knk gksxk & PL >PR SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 21.46 STATEMENT-1 : Two coaxial conducting rings of different radii are placed in space. The mutual inductance of both the rings is maximum if the rings are also coplanar. STATEMENT-2 : For two coaxial conducting rings of different radii, the magnitude of magnetic flux in one ring due to current in other ring is maximum when both rings are coplanar. oDrO;-1 : fHkUu&fHkUu f=kT;kvksa dh nks lek{kh; pkyd oy; ,d LFkku esa fLFkr gSaA nksuksa oy;ksa dk vU;ksU; izsjdRo vf/kdre gksxk ;fn oy; leryh; Hkh gSaA oDrO;-2: fHkUuf=kT;kvksadhnkslek{kh;pkydoy;ksadsfy,],doy;esa]nwljhoy;dh/k jkdsdkj.kpqEcdh; ¶yDl dk ifjek.k vf/kdre gksrk gS tc nksuksa oy; leryh; gksaA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA (C) Statement-1 is True, Statement-2 is False oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D) Statement-1 is False, Statement-2 is True. oDrO; -1 vlR; gS , oDrO;-2 lR; gSA Sol. x It is obvious that flux linkage in one ring due to current in other coaxial ring is maximum when x = 0 (as shown) or the rings are also coplanar. Hence under this condition their mutual induction is maximum. ;gLi"VgSfd,doy;esanwljhlek{kh;oy;dh/k jkdsdkj.klEc)¶yDlvf/kdregksxktcx=0(tSlkfdfn[k ;k x;kgS );koy;leryh;HkhgksAblfy,blizfrcU/kdsvUrxZr]budkvU;ksU;izsjdRovf/kdregksxkA 21.47 STATEMENT-1 : A resistance R is connected between the two ends of the parallel smooth conducting rails. A conducting rod lies on these fixed horizontal rails and a uniform constant magnetic field B exists perpendicular to the plane of the rails as shown in the figure. If the rod is given a velocity v and released as shown in figure, it will stop after some time. The total work done by magnetic field is negative . oDrO;-1 : ,d izfrjks/k R dks nks lekUrj fpduh pkyd iVfj;ksa ds chp tksM+k tkrk gSA ,dpkyd NM+ mu fLFkj tM+or~ (fixed) {kSfrt iVfj;ksa ij j[kh gS rFkk ,d leku fu;r pqEcdh; {ks=k B iVfj;ksa ds ry ds yEcor~ fp=kkuqlkj fo|eku gSA ;fn NM+ dks fp=kkuqlkj osx v nsdj NksM+ fn;k tkrk gS rks ;g dqN le; ckn :d tk;sxhA pqEcdh; {ks=k }kjk fd;k x;k dqy dk;Z _.kkRed gSA STATEMENT-2 : If force acts opposite to direction of velocity its work done is negative. oDrO;-2: ;fncyosxdsfoijhrfn'k esadk;ZjrgksrkgSrksbldkdk;Z_.k RedgksrkgSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True (A) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ; (D*) oDrO;-1 vlR; gS] oDrO;-2 lR; gS Sol: Magnetic field cannot do work, hence statement-1 is false. pqEcdh;{ks=kdk;Zughdjldrkvr%oDrO;-1xyrgSA 21.48 STATEMENT–1 : Consider the arrangement shown below. A smooth conducting rod, CD, is lying on a smooth U-shaped conducting wire making good electrical contact with it. The U-shape conducting wire is fixed and lies in horizontal plane. There is a uniform and constant magnetic field B in vertical direction (perpendicular to plane of page in figure). If the magnetic field strength is decreased, the rod moves towards right. oDrO;–1 : fuEu O;oLFkk ij fopkj dhft,A ,d pkyd NM+ CD tks fd ,d fpduh U-vkdkj ds pkyd rkj ij blds lkFk vPNs fo|qr lEidZ cukrs gq, j[kh gSA U-vkdkj dk pkyd rkj tM+or~ (fixed) gS rFkk ;g {kSfrt ry esa gSA ogk¡ ,dleku o fu;r pqEcdh; {ks=k B tks fd Å/okZ/kj fn'kk (dkxt ds ry ds yEcor~) gSA ;fn pqEcdh; {ks=k dks ?kVk;k tkrk gS rks NM+ nka;h rjQ xfr djrh gSA . STATEMENT–2 : In the situation of statement-1, the direction in which the rod will slide is that which tends to maintain constant flux through the loop. Providing a larger loop area counteracts the decrease in magnetic flux. So the rod moves to the right independent of the fact that the direction of magnetic field is into the page or out of the page. oDrO;–2: fn;sx;s oDrO;-1dhfLFkfresaNM+ dhxfrblrjg lsgksrhgSfdywils ikfjr¶yDlfu;rjgus dh izo`fÙkjguhpkfg,AcM+kywi{ks=kQycudjpqEcdh;{ks=kds¶yDldhdehdks{kfriwfrZ(conteracts)gksrhgSA vr%NM+nka;h rjQxfr djrhgS rFkk;gxfrdhfn'kkblckrlsLorU=kgSfdpkgspqEcdh;{ks=kdkxtdsry ds vUnj dh rjQ gS ;k ckgj dh rjQ gSA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. (A*) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA (B) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k ugha gSA (C) oDrO; -1lR; gS, oDrO;-2vlR; gSA (D) oDrO; -1vlR; gS, oDrO;-2lR; gSA Sol. Obiviously statement 2 is correct explanation of statement-1 Li"V oDrO;-2, oDrO;-1 dh lgh O;k[;k gSA 21.49 STATEMENT–1 : No electric current will be present within a region having uniform and constant magnetic field. → STATEMENT–2 : Within a region of uniform and constant magnetic field B , the path integral of → → → → magnetic field B  d𝑙 along any closed path is zero. Hence from Ampere circuital law B  d𝑙  oI (where the given terms have usual meaning), no current can be present within a region having uniform and constant magnetic field. oDrO;–1: ,dlekuo fu;r pqEcdh; {ks=k okys {ks=k (region) dsvUnjdksbZHkhfo|qr/k jkmifLFkrughagksxhA → → → oDrO;–2: ,dlekuo fu;rpqEcdh; {ks=kB okys {ks=k ds vUnjfdlh Hkh cUn (closed) iFkds vuqfn'k B  d𝑙 → → dkeku'kwU;gSAvr%,fEi;jifjiFkfu;e B  d𝑙  oI (tgk¡lHkhinksadslkekU;vFkZgS) lsfu;ro,dleku pqEcdh;{ks=kdsLFk udsvUnjdksbZHkh/k jkmifLFkrughjgldrhgSA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. (A*) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO; -1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k ugha gSA (C) oDrO; -1lR; gS, oDrO;-2vlR; gSA (D) oDrO; -1vlR; gS, oDrO;-2lR; gSA → → Sol. (A) B  d𝑙 along any closed path within a uniform magnetic field is always zero. Hence the closed path can be chosen of any size, even very small size enclosing a very small area. Hence we can prove that net current through each area of infinitesimally small size within region of uniform magnetic field is zero. Hence we can say no current (rather than no net current) flows through region of uniform magnetic field. Hence statement -2 is correct explanation of statement-1. → → B  d𝑙 ,dle:ipqEcdh;{ks=kesafdlhHkhcUniFkdsvuqfn'kges'k 'kwU;gksrkgSAvr%cUniFkfdlhHkhvkdkj dkysldrsgS]tcHkhcgqrNksVsvkdkjdkcUnHkkxysaxsrkscgqrNksVk{ks=kQygksxkAvr%gefl)djldrs gSfdle:ipqEcdh;{ks=kesavfrlw{eNksVsizR;sdvkdkjdsfy,dqy/k jk'kwU;gSvr%gedgldrsgSfdle:i pqEcdh;{ks=kdsifjljlsdksbZ/kjk¼vFkZr~dksbZifj.keh/kjk½ughizokfgrgksxhA vr% dFku-2,dFku-1dh lgh O;k[;k gSA 21.50 Statement-1 : Electric field produced by changing magnetic field is nonconservative. oDrO;-1 : ifjorhZ pqEcdh; {ks=k }kjk mRiUu fo|qr {ks=k vlaj{kh gksrk gSA → Statement-2: For the electric field E induced by a changing magnetic field which has closed lines of → → force, E.d𝑙  0 → oDrO;-2 :ifjorhZpqEcdh;{ks=kftldhcyjs[k ,sacUniFkgSadsdkj.kizsfjrfo|qr{ks=k E gksrks , → → E.d𝑙  0 [3 min.] (A) Statement -1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C*) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C*) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D) oDrO;-1vlR; gS] oDrO;-2lR; gS SECTION - IV : COMPREHENSION TYPE [k.M- IV:cks/ku izdkj COMPREHENSION # 1 A fan operates at 200 volt (DC) consuming 1000 W when running at full speed. It's internal wiring has resistance 1 . When the fan runs at full speed, its speed becomes constant. This is because the torque due to magnetic field inside the fan is balanced by the torque due to air resistance on the blades of the fan and torque due to friction between the fixed part and the shaft of the fan. The electrical power going into the fan is spent (i) in the internal resistance as heat, call it P1 (ii) in doing work against internal friction and air resistance producing heat, sound etc., call it P2. When the coil of fan rotates, an emf is also induced in the coil. This opposes the external emf applied to send the current into the fan. This emf is called back-emf, call it 'e'. Answer the following questions when the fan is running at full speed. 200 oksYV(fn"V½ij dk;Zjr,d ia[kk]1000W dkmiHkksxdjrkgS] tcfd;g viuhiwjhpkyls pyrkgSA blds vkUrfjdrkjksadkizfrjks/k1gSAtc;giwjhpkylsxfrdjrkgS]rksbldhpkyfu;rgkstkrhgSAbldkdkj.k ;ggSfdpqEcdh;{ks=kdsdkj.kdk;Zjrcyk?kw.kZ]ia[kqfM+;ksaijyxusokysok;qizfrjks/khcyk?kw.kZ,oaia[ksds'kW¶V (shaft) dh NM+ fLFkj Hkkx ds chp yxus okys ?k"kZ.k cyk?kw.kZ ls larqfyr gks tkrk gSA ia[ks dks nh xbZ fo|qr 'kfDr [kpZ gksrh gS & (i) vkUrfjd izfrjks/k esa m"ek ds :i esa P1 (ii) vkUrfjd ?k"kZ.k ,oa ok;q izfrjks/ k ds fo:) dk;Z djus esa tSls gok] /ofu bR;kfn esa bls P2 ekusA tc ia[ks dh dq.Myh ?kwerh gS rks dq.Myh esa ,d fo-ok-cy- Hkh izsfjr gksrk gSA ;g ckg~; fo-ok- cy dk ia[ks dks /kkjk Hkstus dk fojks/k djrk gSA bls i'p fo-ok-cy dgrs gS] 'e' dgsaA. fuEu iz'uksa dk mRrj nks ;fn ia[kk iwjh pky ls pyrk gS & 21.51 The current flowing into the fan and the value of back emf 'e' is : ia[ksesaizokfgr/k jkrFk i'pfo-ok-cyedkekugS& (A) 200 A, 5 volt (B) 5 A, 200 volt (C*) 5 A, 195 volt (D) 1 A , 0 volt 21.52 The value of power 'P1' is 'kfDrP1dkifjek.kgS& (A) 1000 W (B) 975 W (C*) 25 W (D) 200 W 21.53 The value of power 'P2 is 'kfDrP2dkifjek.kgS& (A) 10000 W (B*) 975 W (C) 25 W (D) 200 W Sol. 51 to 53 1000 The fan is operating at 200 V, consuming 1000 W, then  = 200 = 5A But as coil resistance is 1 then power dissipated by internal resistance heat is P If V is net emf across coil then V 2 = 2R = 25W = 25 W V = 5 volt R Net emf = source emf – back emf V = V – e  e = 195 V The work done P2 = 1000 – 25 = 975 W. 1000 gy% ia[kk200 V ij 1000 W miHkksx djrk gS] vr%  = 200 = 5A ijUrqdq.Myhizfrjks/k1gSrksvkUrfjdizfrjks/k}kjk[kir'kfDrP =2R=25W ;fnV dq.Myhdsifjr%ifj.kkehfo-ok-cygSrks& V 2 = 25 W V = 5 volt R usVfo-ok-cy =L=k srfo-ok-cy–i'pfo-ok-cy V = Vs – e  e = 195 V fd;k x;k dk;Z P2 = 1000 – 25 = 975 W. COMPREHENSION # 2 Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 1/m. Position of the conducting rod at t = 0 is shown. A time t dependent magnetic field B = 2t Tesla is switched on at t = 0. fp=k esa ux.; izfrjks/k dh ,d pkyd NM+ n'kkZ;h xbZ gS tks 1/m izfrjks/k ds rkj ls cuh U-vkdkj dh fpduh iVfj;ksa ij ljd ldrh gSA t = 0 ij pkyd NM+ dh fLFkfr fn[kkbZ xbZ gSA le; t = 0 ij B = 2t Vslyk dk le; ij fuHkZj pqEcdh; {ks=k pkyw djrs gSa & 𝑙= 40cm Conducting Rod 𝑙= 40cm pky d NM 21.54 The current in the loop at t = 0 due to induced emf is t=0ijywiesaizsfjrfo-ok-cydsdkj.k/k jkgS& (A*) 0.16 A, clockwise (B) 0.08 A, clockwise (C) 0.08 A, anticlockwise (D) zero (A*) 0.16A, nf{k.kkorZ (B)0.08A,nf{k.k orhZ (C) 0.08A, okekorhZ (D)'kwU; dB Sol. dt = 2T/s E = – AdB = – 800 × 10–4 m2 × 2 = – 0.16 V dt 0.16 i = 1 = 0.16A, clockwise (nf{k.kkorZ). 21.55 At t = 0, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed 5 cm/s by some external means. The rod moves perpendicular to the rails. At t = 2s, induced emf has magnitude. le;t=0ij]tcpqEcdh;{ks=kpkywdjrsgSa]rkspkydNM+ck;harjQfLFkjpky5lseh-@ls-lsckádkjd}kjk xfrekugksrhgSANM+]iVfj;ksadsyEcor~xfrekugSAle;t=2ls-ij]izsfjrfo-ok-cydkifjek.kgS& (A) 0.12 V (B*) 0.08 V (C) 0.04 V (D) 0.02 V dB Sol. At t = 2s B = 4T; dt dB t = 2 s ij & B = 4T; dt A = 20 × 30 cm2 = 2T/s = 2T/s = 600 × 10–4 m2; dA = –(5 × 20) cm2/s dt = – 100 × 10–4 m2/s E = – d d(BA)  = – = – BdA  AdB  dt  dt   dt dt  = – [4×(–100 × 10–4) + 600 × 10–4 × 2] = – [–0.04 + 0.120] = – 0.08 v Alternative : (fodYi %)  = BA = 2t x 0.2 (0.4 – vt) = 0.16t – 0.4 vt2 d E = – dt at t = 2s t = 2s ij = 0.8 vt – 0.16 E = – 0.08 V 21.56 Following situation of the previous question, the magnitude of the force required to move the conducting rod at constant speed 5 cm/s at the same instant t = 2s, is equal to mijksäiz'udhfLFkfresa]leku{k.kt=2sls-ijpkydNM+dksfLFkjpky5lseh-@ls-lsxfrekuj[kusdsfy, vko';d cy dk ifjek.k gS & (A) 0.16 N (B) 0.12 N (C*) 0.08 N (D) 0.06 N Sol. At t = 2s, length of the wire = (2 × 30 cm) + 20 cm = 0.8 m Resistance of the wire = 0.8  Current through the rod = 0.08 0.8 1 = 10 A 1 Force on the wire = il B = 10 × (0.2) × 4 = 0.08 N Same force is applied on the rod in opposite direction to make net force zero. gy% t = 2s ij rkj dh yEckbZ = (2 × 30 cm) + 20 cm = 0.8 m rkjdkizfrjks/k =0.8 NM+lsizokfgr/kkjk= 0.08 = 1 10 A 1 rkj ij cy = il B = 10 × (0.2) × 4 = 0.08 N NM+ijlekucyfoijhrfn'k esaifj.k ehcy'kwU;djusdsfy,yxkrsgSaA COMPREHENSION A train of mass 100 tons (1 ton = 1000 kg) runs on a meter gauge track(distance between the two rails is 1 m.) The coefficient of friction between the rails and the train is 0.045. The train is powered by an electric engine of 90% efficiency. The train is moving with uniform speed of 72 Kmph at its highest speed limit. Horizontal and vertical component of earth’s magnetic field are BH = 10–5T and BV = 2 × 10– 5 T. Assume the body of the train and rails to be perfectly conducting. 100Vu(1Vu=1000fdxzk)dhjsyxkM+hehVjxstiVfj;ksaij¼nksuksaiVfj;ksadschpdhnwjh=1m.)nkSM+jghgSAiVfj;ksa rFk jsyxkM+hdse/;?k"kZ.kxq.kad0.045gSAbljsyxkM+hdks90%n{krkokysfo|qrbatu}kjk[khapktkrkgSAjsyxkM+h bldhmPprepkylhekijle:ipky72Kmphlsxfr'khygSAi`FohdspqEcdh;{ks=kds{kSfrtrFk Å/okZ/kj?kVd Øe'k%BH =10–5TrFkkBV =2× 10–5TgSAekukjsyxkM+hdsiqtsZrFk iVfj;k¡pkydgSaA 21.57 The electrical power consumed by the train is - jsyxkM+h}kjkiz;qDrdhxbZfo|qr'kfDrgS- (A) 1.11 MW (B*) 1 MW (C) 0.50 MW (D) 0.90 MW µmg.v Sol. Power = fractional efficiency µmg.v 'kfDr = n{krk vuqkri 0.45 102  103 10  20 = 0.9 = 106 watt 21.58 The potential difference between the two rails is - nksuksaiVfj;ksadse/;foHkokUrjgS- (A) 100 µV (B) 200 µV (C*) 400 µV (D) 800 µV Sol.  = B V𝑙 = BV .V𝑙 = 2 × 10–5 × 20 × 1 = 40 × 10–5 Volts 21.59 If now a resistor of 10–3  is attached of between the two rails, the extra units of energy (electricity) consumed during a 324 km run of the train is (1 unit of power = 1 kW hour) (assume the speed of train to remain unchanged) ;fn vc 10–3  ds izfrjks/k dks nksuksa iVfj;ksa ds e/; tksM+ nsa rks jsyxkM+h ds 324 km pyus ij vfrfjDr [kpZ dh xbZ ÅtkZ ¼fo|qr½ dh ;wfuV gSA (1 ;wfuV 'kfDr = 1 kW ?kaVk) (ekuk jsyxkM+h dh pky vifjofrZr jgrh gSA) (A) 8 × 10–4 KW hour (B) 8 × 10–5 KW hour (C) 8 × 10–6 KW hour (D*) 8 × 10–7 KW hour Sol. Time taken = 324 = 72 9 hour 2 2 (40 10 5 )2 extra power engine = Power dissipated in resistor = = R 103 = 16 × 10–5 watt. For this the extra power consumed by the train engine will be 16  105 0.9 watt.  energy consumed = 16 10 5 0.9 watt × 9 hour 2 = 8 × 10–7 KW hour Sol. fy;kx;kle;= 324 = 9 ?kaVsa 72 2 2 (40 10 5 )2 baftudhvfrfjDr'kfDr=izfrjks/kesa[kpZ'kfDr= R = 103 =16×10–5 okWV. blvfrfjDr'kfDrdsfy,jsyxkM+hdsbatu}kjk[kpZdhxbZ'kfDr= 16  105 0.9 okWV iz;qDrÅtkZ= 16 10 5 0.9 okWV× 9 ?kaVsa = 8 × 10–7 KW hour SECTION - V : MATRIX - MATCH TYPE [k.M- V:eSfVªDl&lqesy izdkj 21.60 The figure shows a metallic solid block, placed in a way so that its faces are parallel to the coordinate axes. Edge lengths along axis x, y and z are a, b and c respectively. The block is in a region of uniform magnetic field of magnitude 30mT. One of the edge length of the block is 25 cm. The block is moved at 4 m/s parallel to each axis and in turn, the resulting potential difference V that appears across the block is measured. When the motion is parallel to the y axis, V = 24 mV; with the motion parallel to the z axis, V = 36 mV; with the motion parallel to the x axis, V = 0. Using the given information, correctly match the dimensions of the block with the values given. fp=kkuqlkj Bksl /kkfRod fi.M bl izdkj j[kk x;k gS] ftlls blds i`"B] funsZ'kkad v{kksa ds lekUrj gSaA v{kksa x, y rFkk z ds vuqfn'k Hkqtkvksa dh yEckbZ Øe'k% a, b rFkk c gSaA fi.M 30 mT ds ifjek.k okysa ,d leku pqEcdh; {ks=k ds {ks=k esa j[kk gSA fi.M dh ,d Hkqtk dh yEckbZ 25 lseh- gSA fi.M dks Øe ls izR;sd v{k ds vuqfn'k 4 eh-@ls- ds osx ls pykrs gSa] fi.M ds ifjr% mRiUu ifj.kkeh foHkokUrj V ekik tkrk gSA tc xfr y v{k ds vuqfn'k gS] rks V = 24 mV gS] tc xfr z-v{k ds vuqfn'k gS] rks V = 36 mV gS] tc xfr x-v{k ds vuqfn'k gS] rks V = 0 gSA nh xbZ lwpuk ds vk/kkj ij] fi.M dh ekiksa dks fn;s x;s ifjek.kksa ds lkFk feykb;s & y x z (A) (B) (C) a b c (p) (q) (r) 20 cm 24 cm25 cm bc (D) a Ans. (A) r, (B) s , (C) p, (D) q (s) 30 cm (t) 26 cm Sol. Magnetic field is along x axis because when the cube is moved along x-axis, there is no motional emf as → →  0 . When the block is moved along y axis, force on the electrons is in direction v x B –(j × i) = k Therefore electric field will be created along z-axis. Now, c . v. B = 24mV  c = 20 cm similarly b vB = 36 mV  b = 30 cm  a = 25 cm gy% pqEcdh;{ks=kx-v{kdsvuqfn'kgSD;ksafdtcfi.Mdksx-v{kdsvuqfn'kxfrdjkrsgSa]rksmlesadksbZxfrdfo-ok- cymRiUuughagksrkD;ksafd → →  0 Atcfi.Mdksy-v{kdsvuqfn'kxfrdjkrsgS]rksbysDVªkWuijcydhfn'k v x B –(j × i) = k vr%fo|qr{ks=kz-v{kdsvuqfn'kmRiUugksrkgSA vc c . v. B = 24mV  c = 20 cm blhizdkj  b vB = 36 mV a = 25 cm  b = 30 cm 21.61 Column-I gives situations involving a charged particle which may be realised under the condition given in column-II. Match the situations in column-I with the condition in column-II. LrEHk-Iesa,dvkosf'krd.klslEcfU/krfLFkfr;k¡nhxbZgSftUgsaLrEHk-IIesafn;sx;sizfrcU/k sadsvUrxZreqDrfd;k tk ldrk gSA LrEHk-I dh fLFkfr;ksa dks LrEHk-II ds izfrcU/kksa ls lqesfyr dhft,A Column I Column II (A) Increase in speed of a charged particle (p) Electric field uniform in space and constant with time (B) Exert a force on an electron initially at rest (q) Magnetic field uniform in space and constant with time. (C) Move a charged particle in a circle with (r) Magnetic field uniform in space uniform speed but varying with time. (D) Accelerate a moving charged particle (s) Magnetic field non-uniform in space but constant with time (t) Electric field non–uniform in space but constant with time. LrEHk-I LrEHk-II (A) ,dvkosf'krd.kdhpkyesao`f) (p) vkdk'kesa,dlekurFkkle;dslkFkfu;rfo|qr{ks=k (B)izkjEHkesafojkeijfLFkr,dbysDVªkWu (q) vkdk'kesa,dlekurFk le;dslkFkfu;rpqEcdh;{ks=k ij ,d cy vkjksfir djuk (C) ,d vkosf'kr d.k dh o`Ùk esa ,dleku (r) vkdk'k esa ,dleku ysfdu le; ds lkFk ifjorhZ pky ls xfr pqEcdh; {ks=k (D) xfr djrs gq;s vkosf'kr d.k dk Rofjr gksuk (s) vkdk'k esa le; ds lkFk fu;r rFkk vleku pqEcdh; {ks=k (t) vkdk'k esa fo|qr {ks=k vleku gS ysfdu le; ds lkFk fu;r gSA Ans. (A) p, r,t (B) p, r , t (C) q,s (D) p, q, r, s, t Sol. (A) Speed of charged particle cannot be changed by magnetic force because magnetic force does no work on charged particle. Only electric field in case (p), (t) and induced electric field in case (r) can change speed of charged particle. (B) Magnetic field cannot exert a force on charged particle at rest. Only electric field in case (p), (t) and induced electric field in case (r) can exert force on charge initially at rest. In case (r) after the charge starts moving even the magnetic field can exert force on charge. (C) A charged particle can move in a circle within uniform speed due to uniform and constant magnetic field in case . Even within a region of non uniform magnitic field, at all points on a circle field may be uniform for example on any circle coaxial with a current carring ring (D) A moving charged particle is accelerated by electric field and also accelerated by magnetic field (provided v is not parallel to B). (A)vkosf'krd.kdhpkypqEcdh;cy}kjkifjofrZrughgksldrhgSApw¡fdpqEcdh;cyvkosf'krd.kijdksbZdk;Zugh djrkgSAdsoyfLFkfr(p),(t)esafo|qr{ks=krFk fLFkfr(r)esasizsfjrfo|qr{ks=kvkosf'krd.kdhpkycnyldrsgSA (B)fojkeijvkosf'krd.kijpqEcdh;{ks=kcyvkjksfirughdjldrkgSAdsoyfLFkfr(p),(t)esafo|qr{ks=krFk fLFkfr (r)esaiszfjrfo|qr{ks=kizkjEHkesafojkeijfLFkrvkos'kijcyvkjksfirdjldrkgSAfLFkfr(r)esavkos'kdsxfrizkjEHkdjus dscknHkhpqEcdh;{ks=kvkos'kijcyvkjksfirdjldrkgSA (C) fLFkfr(q)esa,dlekurFk fu;rpqEcdh;{ks=kdsdkj.k,dvkosf'krd.k,dlekupkyls,do`ÙkesaxfrdjldrkgSA vle:ipqEcdh;{ks=kesa,do`ÙkdslHkhfcUnqvksijfo|qr{ks=kle:igksldrkgSAmnkgj.kdsfy,,d/k jkokghoy; dsladsUnzh;o`ÙkijA (D) ,dxfrdjrkgqvkvkosf'krd.kfo|qr{ks=k}kjkRofjrgksrkgSrFk pqEcdh;{ks=k}kjkHkhRofjrgksrkgS(;gfn;ktk; fd v,Bds lekUrjugh gSA) 21.62 A square loop of conducting wire is placed near a long straight current carrying wire as shown. Match the statements in column-I with the corresponding results in column-II. pkydrkjdk,doxkZdkjywi,dyEcslh/ks/k jkokghrkjdsutnhdfp=k uqlkjj[k tkrkgSALrEHk-Iesafn;sx;s dFkuksadksLrEHk-IIesamudsifj.k eksalslqesfyrdhft;s& Column-I Column-II (A) If the magnitude of current I is increased (p) Induced current in the loop will be clock wise (B) If the magnitude of current I is decreased (q) Induced current in the loop will be anticlockwise (C) If the loop is moved away from the wire (r) wire will attract the loop (D) If the loop is moved towards the wire (s) wire will repel the loop (t) Torque about centre of mass of loop is zero due to magnetie force LrEHk-I LrEHk-II (A);fn/k jkIdkifjek.kc<+k;ktkrkgSA (p)ywiesaizsfjr/kjknf{k.korZgksxh (B);fn/kkjkIdkifjek.k?kVk;ktkrkgSA (q)ywiesaizsfjr/kjkokekorZgksxh (C);fnywidksrkjlsnwjgVk;ktkrkgSA (r)rkj]ywidksvkdf"kZrdjsxk (D) ;fn ywi dks rkj dh vksj pyk;k tkrk gSA (s) rkj] ywi dks izfrdf"kZr djsxk (t) pqEcdh; oy ds dkj.k ywi ds nzO;eku dsUnz ds pkjks rjQ cyk?kw.kZ 'kwU; gksxkA Ans (A) q,s (B) p,r (C) p,r (D) q,s Sol. (A) Due to current carrying wire, the magnetic field in loop will be inwards w.r.t. the paper. As current is increased, magnetic flux associated with loop increases. So a current will be induced so as to de- crease magnetic flux inside the loop. Hence Induced current in the loop will be anticlockwise. The current in left side of loop shall be downwards and hence repelled by wire. The current in right side of loop is upwards and hence attracted by wire. Since left side of loop is nearer to wire, repulsive force will dominate. Hence wire will repel the loop (B) Options in (B) will be oppsite of that in (A) (C) When the loop is moved away from wire, magnetic flux decreases in the loop. Hence the options for this case shall be same as in (B) (D) When the loop is moved towards the wire, magnetic flux increases in the loop. Hence the op- tions for this case shall be same as in (A) (A)/k jkokghrkjdsdkj.kywiesapqEcdh;{ks=kistdsrydsvUnjgksxkAvr%/k jkc<+uslslEcfU/kr¶yDlHkh c<+sxkAvr%/k jkblizdkjizsfjrgksxhfd;gywids¶yDldks?kVk;svr%ywiesaizsfjr/k jkokekorZgksxhywids ck;sarjQ/k jkÅ/okZ/kjuhpsdhrjQgksxhvr%rkj}kjkizfrdf"kZrgksxkAywidsnka;srjQ/k jkÅijdhrjQgksxh vr%rkj}kjkvkdf"kZrgksxkpwafdywidkcka;kHk xrkjdsiklgSAvr%izfrd"khZcyT;knkizHk ohgksxkAvr%rkj ywidksizfrdf"kZrdjsxkA (B)(B)esafodYi(A) dsfoijhrgksxsaA (C)tcywirkjlsnwjtkrkgSrksywiesapqEcdh;¶yDl?kVrkgSAvr%blfLFkfresafodYi(B)dslekugksxkA (D) tcywi rkjdh rjQxfr djsxkrks ¶yDlc<+sxk vr% bl fLFkfresa fodYi(A) dsleku gksaxsA SECTION - VI : INTEGER TYPE [k.M-VI:iw.kkZadizdkj SHORT SUBJECTIVE y?kqÙkjkRediz'u 21.63 In the figure, a long thin wire carrying a varying current i = i sin t lies at a distance y above one edge of a rectangular wire loop of length L and width W lying in the x-z plane. What emf is induced in the loop. fp=kesa/k jki=i0 sint dk,dyEckiryk/k jkokghrkj]x–zryesafLFkrLyEckbZoWpkSMk+bZds,dvk;rkdkj ywidh,ddksjdsÅijynwjhijj[k gqvkgSAywiesafdrukfo0ok0c0izsfjrgksrkgS\ Sol. Figure - 1 Figure - 2 µ0 i The magnetic field at point P (figure -2) is B = 2 µ0 i PfcUnqijpqEcdh;{ks=k(fp=k-2)gksxk B = 2 µ0 i The magnetic flux through the shaded strip in figure -1 is d = (W dz) 2 sin  fp=k1esaNk;kafdr{ks=kesapqEcdh;¶yDldgksxk =(Wdz) µ0 z i sin where tgk¡ sin =  total magnetic flux through rectangular loop is  vk;rkdkjywilsdqypqEcdh;¶yDlgksxk L µ i sint W z dz  y2  L2   =  0 0 = µ0 W ln   i sint 2 y2  z2 0 4  y2  d µ0  L2  y2   induced emf in the loop is e = dt = 4 i W  cost ln  2    d µ  L2  y2   ywi esa izsfjr fo-ok-c- e= = 0 i Wcostln  2  0i0Wcos t  L2  dt 4 0   Ans : 𝑙n  1 4  y2  21.64 The magnetic field of a cylindrical magnet that has a pole-face radius 2.8 cm can be varied sinusoidally 60 between minimum value 16.8 T and maximum value 17.2 T at a frequency of  Hz. Cross section of the magnetic field created by the magnet is shown. At a radial distance of 2cm from the axis find the amplitude of the electric field (in mN/C) induced by the magnetic field variation. ,d csyukdkj pqEcd dk pqEcdh; {ks=k ¼/kzqo ds ry dh f=kT;k2.8cm½ U;wure eku 16.8TvkSj vf/kdre eku 60 17.2Tdschp  Hzvko`frlssineQyuds:ilscnyjgkgSpqEcd}kjkmRiUupqEcdh;{ks=kdkvuqizLFkdkV{ks=kfp=k esafn[k;kgSAv{kls2cmnwjhijpqEcdh;{ks=kdsifjorZu}kjkmRiUuizsfjrfo|qr{ks=kdkvk;keKkrdjksa(mN/Cesa)& Ans. 240 mN/C → → dB Sol. E . d𝑙 = – A . dt As B = 17 + (0.2) sin (t + ) ; E (2r) = –r2(0.2) . cos (t + ) E = – r (0.2) . cos (t + ) 2 Magnitude of the amplitude = r (0.2) .  = 240 mN/C 2 vk;ke dk ifjek.k = r (0.2) .  = 240 mN/C LONG SUBJECTIVE fucU/k Rediz'u 21.65 In the figure shown ABCDEFGH is a square conducting frame of side 2m and resistance 1 /m. A uniform magnetic field B is applied perpendicular to the plane and pointing inwards . It increases with time at a constant rate of 10 T/s . Find the rate at which heat in watt is produced in the circuit , AB = BC = CD = BH . fp=k uqlkjABCDEFGH,d2ehVjdhHkqtkrFk 1/eh-çfrjks/kokykoxkZdkjÝsegSAblij,dlekupqEcdh;{ks=k B rydsyEcor~vUnjdhvksjvkjksfirfd;ktkrkgSA;gle;dslkFkfu;rnj10T/slsc<+rkgSAifjiFkesamRiUu Å"ek dh nj okWV esa Kkr djksA AB = BC = CD = BH . Ans. 40 Sol. The induced emf in loop ABHFG ywiABHEGesaizsfjrfo-ok-cy = d (BA) = A dB = 2  10 = 20 V dt dt The induced emf in loop BCDH & DEFH ywiBCDHvkSjDEFHesaizsfjrfo-ok-c- = 1  10 = 10 volt . KVL in top left loop is , = 1  10 = 10 volt . KVL Åij ds ck;sa ywi esa yxkus ij, 10  (y  z) + (x  y)  2y = 0  x  4y + z =  10  (1) KVL in right loop : 10  2x  (x  y)  (x  z) = 0 KVL , nk;sa ywi esa : 10  2x  (x  y)  (x  z) = 0   4 x + y + z =  10  (2) By equation (1) & (2) it is seen that x  y = 0  no current in DH lehdj.k(1) vkSj(2) ls];gizkIrfd;ktkldrkgSfdx y=0  DHesadksbZ/k jkugha { This can also be seen by symmetry } {blsleferhlsns[kktkldrkgS} This makes solution very simple now the circuit is , bllsgyvklkugkstk,xk] Assume ekuk vB = 0 , & vF = v , then v  20 4 + v  20 4 + v  0 2 = 0  v = 0  no current in FB FBesadksbZ/kjkugh.  circuit is :  ifjiFk gS %  rate of heat production = (40)2/8 = 200 watt . Ans. Note : If you are not able to observe the symmetry or decide x  y = 0 , then write kVL in the lower loop . It will be x + y  6 z =  20  (3) Solving (1) , (2) and (3) you will get x = +5 , y = 5 , z = 5 A . Heat rate will be , 10 x + 10 y + 20 z = 40  5 = 200 W  Å"ek ds mRiUu gksus dh nj = (40)2/8 = 200 watt . mÙkj uksV% vxjvki leferhdks r; ugha djldrs rc x  y = 0 , rc KVL fupys ywi esa x + y  6 z =  20  (3) (1) , (2) vkSj (3) dks gy djus ij vki ik;saxs x = +5 , y = 5 , z = 5A. Å"ek dh nj gkssxh, 10 x + 10 y + 20 z = 40  5 = 200 W Method II   H = 2R =  R where R = 2 × 4 × 1 = 8  and  = 4 × 10 = 40 21.66 A long coaxial cable consists of two thin walled conducting cylinders with inner radius 2cm and outer radius 8cm. The inner cylinder carries a steady current 1A, and the outer cylinder provides the return path for that current. The current produces a magnetic field between the two cylinders. Find the energy stored in the magnetic field for length 1 m of the cable. Express answer in nJ (use 𝑙n2 = 0.7). ,dyEchlek{kh;dscyftlesanksiryhnhokjksaokyspkydcsyuftudhf=kT;k¼vkUrfjd½2lsehvkSjckgjh8lsehgSA vUnjdkcsyu1AfLFkj/k jkizokfgrdjrkgSvkSjckgjhcsyubldkokilhdkiFkiznkudjrkgSA/k jknksuksacsyuksads