ELECTROMAGNETIC INDUCTION CHAPTER - 38

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Gertrude Watts
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1 . (a) CTOMAGNTIC INDUCTION CHAPT dl MT I M I T 3 (b) BI T MI T M I T (c) d / MI T M I T. at + bt + c s / t Volt (a) a t t Sc b t Volt c [] Wbr (b) d [a., b.4, c.6, t s] at + b volt 3. (a) B.A d mv 3 3 B.A d 4 5 d 4 mv 4 B.A d 4 5 d 4 mv 5 B.A. d 5 d mv (b) mf is not constant in cas of ms and 3 ms as 4 mv and 4 mv. 4. BA.5 (5 ) t 5 5. A mm ; i A, d cm ;. s d BA i A d V. 6. (a) During rmoval, B.A Tsla-m t A cm (ms) 38.

2 d.5 5 5V.5 5 (b) During its rstoration ;.5 Tsla-m ; t.5 s V. (c) During th motion, d 7. 5 (a) 5 V, T.5 s i / A, H i T J (b) 5 V, T.5 s i / A, H i T 5 J (c) Sinc nrgy is a scalar quantity Nt thrmal nrgy dvlopd 5 J + 5 J 5 J. 8. A 5 cm 5 4 m B B sin t. sin(3 t) 6 a) Max mf inducd in th coil d d (BA cos ) d 4 (B sint 5 ) 5 4 d B (sin t) B5 4 cos t cost 5 3 cost max V b) Inducd mf at t (/9) s 5 3 cos t 5 3 cos (3 /9) 5 3 ½.5/ V c) Inducd mf at t /6 s 5 3 cos (3 /6) 5 3 V. 9. B. T A cm 4 m T s B.A V d. mv 3 V A ( ) 4 4 Dt. s, 8 38.

3 BA, BA d BA d BA 3 B 3 4 B 3 B 3 3 5T 4 4. Ara A, sistanc, B Magntic fild BA Ba cos BA d BA BA ; i it BA/. r cm m n turns / cm turns/m i 5 A B ni T n turns r cm m Flux linking pr turn of th scond coil Br B 4 Total flux linking Bn r 4 3 Whn currnt is rvrsd. d d 4 I 4 4 q I C. 3. Spd u Magntic fild B Sid a a) Th prpndicular componnt i.. a sin is to b takn which is r to vlocity. So, l a sin 3 a/. Nt a charg 4 a/ a So, inducd mf BI aub b) Currnt aub wbr,.85 wbr D (.85.35) wbr.5 wbr.5 sc a sin a u 3 a a B B 38.3

4 d.5 v..5 Th inducd currnt is anticlockwis as sn from abov. 5. i v(b l) v B l cos is angl btwn normal to plan and B 9. v B l cos u cm/, B.6 T a) At t sc, distanc movd cm/s cm d.6 ( 5 ) 3 4 V b) At t sc distanc movd cm Th flux linkd dos not chang with tim c) At t sc distanc cm Th loop is moving out of th fild and cm outsid. d da B.6 ( 5 ) 3 4 V d) At t 3 sc Th loop is total outsid and flux linkd. 7. As hat producd is a scalar prop. So, nt hat producd H a + H b + H c + H d 4.5 m a) 3 4 V i Amp. H a (6.7 ) H b H d [sinc mf is inducd for 5 sc] H c (6.7 ) So Total hat H a + H c (6.7 ) J. 8. r cm, 4 db d db. T/, A d db r A i A 4 9. a) S closd S opn nt cm a r b cm B d c

5 d db A i through ad b) db A 5 V V.5 7 A along ad 38.5 i.5 7 A along d a 6 c) Sinc both S and S ar opn, no currnt is passd as circuit is opn i.. i d) Sinc both S and S ar closd, th circuit forms a balancd what ston bridg and no currnt will flow along ad i.. i.. Magntic fild du to th coil () at th cntr of () is B Flux linkd with th scond, B.A () (a.m.f. inducd b) (a x Nia Na a (a x 3 / ) a d Na a 3 / (a x ) 3 / x ) Na a (a 3 / x ) Na a a) For x (a 3 / ) x d 3 / ) ( / )x r. /.v ( / )x r V ( / Na a v r) ( r) di. N 5, B. T ; r. cm. m 6, t. s a) Nd NB.A NBA cos 6 T T (for x /, / x /) 5 (.) V 6.8 V 6.8 b) i.57 A 4 Q it C. n turns, B 4 4 T A 5 cm 5 4 m a) Whn th coil is prpndicular to th fild nba Whn coil gos through half a turn BA cos 8 nba d nba (a Nia x 3 / ) d c a b a () d a () B

6 Th coil undrgos 3 rv, in min 3 rad/min rad/sc rad is swpt in sc. / rad is swpt / / sc d nba b) nba, nba ( 36 ) d / c) i q i 5 4 / 5 5 C. 3. r cm. m 4, N 8, B H 3 5 T N(B.A) NBA Cos 8 or NBA whr d NBA 6 4 wbr i Q d V C V d db.a cos 4. mf B A sin BA sin (dq/ th rat of chang of angl btwn arc vctor and B ) a) mf maximum BA volt. b) Sinc th inducd mf changs its dirction vry tim, so for th avrag mf t 5. H t B A i sin t B A t ( cost) B A sin t t minut B A sin 8 / / r 4 B J. 9 9

7 6. BA, 4 d (.) V 7. l cm. m v cm/s. m/s B. T a) F q v B N b) q qvb V/m This is cratd du to th inducd mf. c) Motional mf Bvl... 3 V 8. l m, B. T, v m/s, Blv..4 V 9. l m, v 3 7 m/s, B 3 T Motional mf Bvl V 3. v 8 km/h 5 m/s B. 4 T, m Bvl. I V Th voltmtr will rcord mv. 3. a) Zro as th componnts of ab ar xactly opposit to that of bc. So thy cancl ach othr. Bcaus vlocity should b prpndicular to th lngth. b) Bv l Bv (bc) +v at C c) as th vlocity is not prpndicular to th lngth. d) Bv (bc) positiv at a. i.. th componnt of ab along th prpndicular dirction. 3. a) Componnt of lngth moving prpndicular to V is B v b) Componnt of lngth prpndicular to vlocity a b c V V 33. l cm. m ; 6 ; B T V cm/s. m/s Bvl sin6 [As w hav to tak that componnt of lngth vctor which is r to th vlocity vctor].. 3 / V. 34. a) Th.m.f. is highst btwn diamtr r to th vlocity. Bcaus hr lngth r to vlocity is highst. max VB b) Th lngth prpndicular to vlocity is lowst as th diamtr is paralll to th vlocity min. 6 v 38.7

8 35. F magntic ilb This forc producs an acclration of th wir. But sinc th vlocity is givn to b constant. Hnc nt forc acting on th wir must b zro. 36. Bvl sistanc r total lngth r (l + vt) 4(l + vt) i 37. Bvl i Bv r( vt) Bv r( vt) a) F ilb b) Just aftr t Bv B v B r( vt) r( vt) Bv B v F i B B r r F B v B v 4r r( vt) l l + vt T l/v 38. a) Whn th spd is V mf Blv sistanc r + r Bv Currnt r b) Forc acting on th wir ilb BvB B v r r Acclration on th wir B v m( r) B v c) v v + at v t [forc is opposit to vlocity] m( r) d) a dx v B x m( r) dv B v v dx m( r) dvm( r) B m( r)v x B 39.., B. T, l 3 cm.3 m B 8 cm.8 m a) F ilb 3. 5 N B v l v +v v i l v l V b c a d F

9 (.) (.8) 5 v v 5 m/s b) mf vbl V c) sistanc pr unit lngth sistanc of part ad/cb Bv V ab i. 8 d) sistanc of cd V 3.6 V V i 4 3 V 4. l cm m v cm/s m/s B H 3 5 T i A 6 A. i B v v B v tan i v B B v H Bv B cos v cos 4. I Bv cos 5 Tsla (dip) tan (/3) Bv cos B F ilb Now, F mg sin [Forc du to gravity which pulls downwards] B v cos Now, mg sin B mg sin v cos 4. a) Th wirs constitut paralll mf. Nt mf B v Nt rsistanc 9 Nt currnt 4. ma. b) Whn both th wirs mov towards opposit dirctions thn not mf Nt currnt B 4cm a l l Cos, vcos v b B 9 BT 38.9

10 43. P P 4cm 9 Q Q BT P P a) No currnt will pass as circuit is incomplt. b) As circuit is complt VP Q B v V 3 i 3 A ma. 44. B T, V 5 I m/, a) Whn th switch is thrown to th middl rail Bvl 5 3 Currnt in th rsistor / 3 4. ma b) Th switch is thrown to th lowr rail Bvl 5 4 Currnt 45. Initial currnt passing i Hnc initial mf ir mf du to motion of ab Blv Nt mf ir Blv Nt rsistanc r 4 Hnc currnt passing 46. Forc on th wir ilb ib Acclration m 4. ma ir Bv r ibt Vlocity m 47. Givn Blv mg () Whn wir is rplacd w hav mg Blv ma [whr a acclration] mg Bv a m Now, s ut at mg Bv t [ s l] m t 4ml mg Bv 4ml mg mg / g. [from ()] 38. cm cm Q Q P P Q Q 5cm/s S d a g c i g b B g B b a

11 48. a) mf dvlopd Bdv (whn it attains a spd v) Currnt Bdv Bd v Forc This forc opposs th givn forc Bd v Nt F F Bd v F F B d v Nt acclration m b) Vlocity bcoms constant whn acclration is. F B d v m m F m B d v m F V B d c) Vlocity at lin t a dv v dv t F l B v m ln [F l B v] l B v tl B l n(f l B v) m v t m t B t ln(f l B v) ln(f) m l B t l B v m F l B v F l B t m l B vt F v m Fvm v v( ) l B 49. Nt mf Bvl Bv I from b to a r F I B Bv B lb ( Bv ) towards right. r r Aftr som tim whn Bvl, Thn th wir movs constant vlocity v Hnc v / Bl. t F d a b 38.

12 5. a) Whn th spd of wir is V mf dvlopd B V Bv b) Inducd currnt is th wir (from b to a) c) Down ward acclration of th wir mg F m du to th currnt B V mg - i B/m g m d) t th wir start moving with constant vlocity. Thn acclration B v m m V m g gm B a b dv ) a dv mg B m v / dv mg B v / m v mdv B v mg m B t B v log(mg v t m B v log log mg log(mg) t B B v mg tb log mg m B v tb log mg m B v mg ( v tb m B /m B v ) mg mg B / m B gt / Vm v v ( ) m v m mg B 38.

13 ds f) v ds v s vm ( t gt / vm ) Vm Vm t g gt / vm V / vm m V t gt m g Vm Vmt g gt / vm V g m g) d mgs mg ds mgv ( m gt / vm) dh BV i B v B gt / vm Vm( ) Aftr stady stat i.. T d mgs mgv m dh B B mg Vm Vm mgv m B Hnc aftr stady stat d H d mgs 5. l.3 m, B. 5 T, rpm v v rad/s mf Blv V V V. 5. V at a distanc r/ r From th cntr B.3 B r Blv B r Br 53. B.4 T, rad/, r r 5 cm.5 m Considring a rod of lngth.5 m affixd at th cntr and rotating with th sam..5 v.5 Blv I ma It lavs from th cntr. 3 V B v 38.3

14 B 54. B ykˆ ngth of rod on y-axis V V î Considring a small lngth by of th rod d B V dy d B y V dy B d V ydy BV ydy B V y BV BV 55. In this cas B varis Hnc considring a small lmnt at cntr of rod of lngth dx at a dist x from th wir. B i x So, d i x vxdx d iv xt / xt / iv x / iv x ln ln x / x x dx iv [ln (x + l/) ln(x - l/)] x 56. a) mf producd du to th currnt carrying wir iv x ln x t currnt producd in th rod i iv x ln x Forc on th wir considring a small portion dx at a distanc x df i B l df i df i F iv x i ln x x v x dx ln x x v x ln x x t / xt / dx dx x i v x x ln ln x x v i x ln x ln x b) Currnt ln x 38.4 i x i x V l dx l

15 c) at of hat dvlopd i iv x iv x ln x x d) Powr dvlopd in rat of hat dvlopd i iv x ln x 57. Considring an lmnt dx at a dist x from th wir. W hav a) B.A. d i adx x a ia ab dx ia d ln{ a / b} b x d d ia b) ln[ a / b] a d ln[ a / n] i sin t ai cos t ln[ a / b] aicos t c) i ln[ a / b] r r H i rt aicos t ln( a / b) r t r a i ln [ a / b] r 4 r 5a i ln [ a / b] r [ t ] 58. a) Using Faraday law Considr a unit lngth dx at a distanc x B i x Ara of strip b dx d i dx x al a i bdx x al i dx ib a l b log x a mf a d d ib a l log a ib a va (a l)v (whr da/ V) a l a i b x a dx i l b x a dx v 38.5

16 ib a vl a l a ibvl (a l)a Th vlocity of AB and CD crats th mf. sinc th mf du to AD and BC ar qual and opposit to ach othr. B AB i a ngth b, vlocity v. B CD i (a l).m.f. CD ngth b, vlocity v. Nt mf 59. Bvl i Ba.m.f. AB i bv a ibv (a l) i bv a B a a ibv (a l) ibvl a(a l) Ba B a F ilb a B towards right of OA. 6. Th rsistancs r/4 and 3r/4 ar in paralll. r / 4 BVl i 3r r / 4 3r 6 a Ba B a Ba Ba 3r /6 6. W know Ba 6 8 Ba 3r 3 r B a F i B Componnt of mg along F mg sin. Nt forc 3 a mg sin. B 3 i C O B O B A a mg A 3r/4 l O B O F C D b A r/4 A F mg sin 6. mf [from prvious problm] B a / Ba Ba Currnt mg cos ilb [Nt forc acting on th rod is O] mg cos Ba a B (Ba )ab. mg cos C O A ilb C O mg mg cos 38.6

17 63. t th rod has a vlocity v at any instant, Thn, at th point, Blv Now, q c potntial c CBlv Currnt I dq d CBlv dv CBl CBla (whr a acclration) From figur, forc du to magntic fild and gravity ar opposit to ach othr. So, mg IlB ma mg CBla lb ma ma + CB l a mg a(m + CB l mg ) mg a m CB 64. a) Work don pr unit tst charg. dl ( lctric fild). dl d dl r r db r db r r db db A b) Whn th squar is considrd, dl r 4 db (r) db 4r 8r r db Th lctric fild at th point p has th sam valu as (a). l mg P r B v 65. di. A/s di For s. A/s n turn/m, 6. cm.6 m r cm. m a) BA d na di [A 4 ] d or, for s.785. d b).dl 38.7

18 c) d.785 dl d d dl 8. 7 V/m di n A 4 7. (.6) d / 4 r V V di I I.5 (.5) 5A. s V di. (.6) 8 (5/.) 5 / 5 4/.4 Hnry. d wbr di n, I 4A, n V/m or, or, 68. d di d n 8 4 H. N A 7 di 4 (4) ( ). 8 4 (4) V. 69. W know i i ( - t/r ) a) 9 t / i i ( r ).9 t/r t/r. Taking ln from both sids ln t/r ln. t.3 t/r.3 b) 99 t / i i ( r ) t/r. ln t/r ln. or, t/r 4.6 or t/r t / r c) i i( ) t/r. ln t/r ln. t/r 6.9 t/r

19 7. i A, 4V, H 4 i i H,, mf 4. V, t. S i 4,. a) i i ( t/ ) 4.. /.7 A b) i (.7).89.3 J. 7. 4, 4V, t., i 63 ma i i ( t/ ) /4 (. 4/ ) 63 3 ( 4/ ) 63 ( 4/ ).63 4/ 4/.37 4/ ln (.37) H 4 H H,, mf. V t ms 3 s s i now i i ( t/ ) 5 i 5 / 5 i ( ) V i V ms i A a) t ms i i ( t/ ) ( /4 ) ( /4 ) (.7788) (.) A.44 A.44 A b) t ms i i ( t/ ) ( /4 ) ( / ) (.66).7869 A.79 A c) t ms i i ( t/ ) ( /4 ) ( /4 ) (.8).835 A.8 A d) t s i i ( t/ ) ( ( 5 ) A 3 / 4 ) ( /4 ) 38.9

20 75.. H,, mf. V.5 i. A i i ( t ) i i t di di t / (ix / ) i / t/. So, di. a) t ms./. 5.7 A.5 di. b) t ms./ A.5 di. c) t s / A a) For first cas at t ms di.7 di Inducd mf.7.7 V b) For th scond cas at t ms di.36 di Inducd mf c) For th third cas at t s di 4. 9 V di Inducd mf V 4. 9 V 77. mh; 5. V, 3 5, i i i ( t/ ) i i i t / i i i t / di d 5 a) / i 3 b) 5 3 di i 5 t / t ms 3 s V/s. d 5. / V/ 38.

21 c) For t s d di 5 3 / mh, 5, 5 V a) t ms i i ( t/ t / ) ( ). V/s / ( ) 5 5 (.3678).64 5 Potntial diffrnc i V 3.6 V. b) t ms i i ( t/ t / ) ( ) / ( 5 ) 5 5 (.67) Potntial diffrnc i V. c) t sc i i ( t/ t / ) ( ) / ( 5 ) 5 5 /5 A 5 Potntial diffrnc i (/5 5) V 5 V. 79. mh. H, mf 6, r i i ( t/ ) Now, dq i i ( t/ ) t / Q dq i( ) t t t / t / i i t ( ) t / t / i [t ( )] i [t ] Now, i A.. a) t. s So, Q.5[. +../..] C.8 mc 38.

22 b) t ms. s So, Q.5[. +../..] C 5.6 mc c) t ms. s So, Q.5[. +../..].45 C 45 mc 8. 7 mh, l m, A mm 6 m, f cu.7 8 -m 8 fcu A 8.7 i sc m sc / 5 ms.5 i t /. 6 a) i( ) t /.5 t /.5 ln t/.5 ln / t ms 35 ms. b) P i ( t. / Maximum powr ) So, ( t / ) t/ t/ t ln t 5.75 ms 6. ms. 8. Maximum currnt In stady stat magntic fild nrgy stord Th fourth of stady stat nrgy On half of stady nrgy t / ( ) 8 Again t / 4 8 t / t ln t ln 4 t / ( ) 38.

23 t / t n n So, t t n H,, 4 V a) Tim constant 4.4 s. b) i.63 i Now,.63 i i ( t/ ) t/ n t/ In.37 t/.994 t s. c) i i ( t/ ) 4.4 /.4 ( ) A. Powr dlivrd VI d) Powr dissipatd in Joul hating I (.58) i i ( t/ ) ni n i ( t/ ) B B ( I/ ) B B / ( ).8 ( / ) /. n( / ) n(.) / mf circuit a) dq i i ( t/ ) i ( I. ) [ /] Q t / dq i t t t i [t ( /) ( I/ ) t ] i [t / ( I/ )] Q / [t / ( I/ )] b) Similarly as w know work don VI I i [t / ( I/ )] c) H [t / ( I/ )] t t t / i ( ) t ( B) / t / ( ) 38.3

24 t / t t / t / t / t 3 t x x t (x 4x 3) t d) i t / ( ) [x t/ ] ( x) ) Total nrgy usd as hat as stord in magntic fild T x 4x 3 x x r t x t ( x) nrgy drawn from battry. (Hnc consrvation of nrgy holds good). 86. H,, V, t ms a) l l ( t/ ) 3 /. ( ). (.3678) A. b) Powr dlivrd by th battry VI I ( t/ ) ( t / ) 3 ( / ). ( ).64 mw. c) Powr disspitd in hating th rsistor I t / [i ( )] (6.3 ma) ma. d) at at which nrgy is stord in th magntic fild d/ (/ I ] 4 t / t / I ( ) ( ) (.35) mw. 38.4

25 87. A. H ; B. H ; a) t. s, A., B /. i A i ( t/ ). i B i ( t/ ) B. b) t ms. s i A i ( t/ ). ( ).644. ( / ) ia i /.( ) / i B.( ) ia i.644 c) t s B / i A.( ) / i B.( ) ia i a) For discharging circuit i i t/ 89. B./ (/)./ ln (/) ln (./ ).693././ b) 4 H, i /.4 4/ 4 / Cas - I Cas - II In this cas thr is no rsistor in th circuit. So, th nrgy stord du to th inductor bfor and aftr rmoval of battry rmains sam. i.. V V i So, th currnt will also rmain sam. Thus charg flowing through th conductor is th sam. 38.5

26 9. a) Th inductor dos not work in DC. Whn th switch is closd th currnt chargs so at first inductor works. But aftr a long tim th currnt flowing is constant. Thus ffct of inductanc vanishs. ( ) i nt b) Whn th switch is opnd th rsistors ar in sris. nt 9. i. A, r cm, n turn/m Magntic nrgy stord. B V Whr B Magntic fild, V Volum of Solnoid. n i r h J. 9. nrgy dnsity B Total nrgy stord ( ) 93. I 4. A, V mm 3, d cm. m B i r Now magntic nrgy stord [h m] B V ( i/ r) i V V 4r 8 4 J. B 7 9 i 4 6 V 4 r J.55 4 J 94. M.5 H di A s di.5.5 V V 38.6

27 95. W know d di M From th qustion, i a di d (i sin t) i cos t d aicost n[ a /b] di Now, M aicost or, n[ a / b] M icost M a n[ a /b] 96. mf inducd di V x r N di / (a x ) Na a V 3 / (a x ) ( /x r) (from qustion ) a a 3 / 97. Solnoid I : a 4 cm ; n 4/. m ; cm. m Solnoid II : a 8 cm ; n /. m ; cm. m B n i. lt th currnt through outr solnoid b i. i b n B.A n n i a 4 i 4. d Now M 4 64 di 7 4 di / 64 4 H H. 98. a) B Flux producd du to first coil n i Flux linkd with th scond n i NA n i N mf dvlopd [As Mdi/] di ( nin ) di nn nn icos t. 38.7

SAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS

SAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS It is not possibl to find flu through biggr loop dirctly So w will find cofficint of mutual inductanc btwn two loops and thn find th flu through biggr loop Also rmmbr M = M ( ) ( ) EDT- (JEE) SOLUTIONS