UNIT # 10. JEE-Physics ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT EXERCISE I N B A

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1 J-Physcs UN # OMAGN NDUON & ANANG UN XS. otl chnge n flu otl chrge flown through the col resstnce. esstnce. Webers. N A t W.mJ v t r r... d N S d N e N 5volt d e.8 volt volt d 6 NA ()( ). > mg < mg 5. NA cost e d NA snt em NA e 8 8 e.h t / t ( /.5) 8 5. q NA q NA 6. r d e (r) dr 7. W Q.d Q.d Q ncreses ncreses So from enz's lw urrent n A s clockwse 8. d.dr A.dr sme nd flu lnked wth A ncreses So from enz's lw current n A clockwse 9. P v +Q e () v rv 8. d b. Accordng to enzs lw. Plte wll become postvely chrged. All spokes re n prllel Nv N. W N() N t A M but db d b (d) n d b M n d M d d 9. Work done (.) (5). J. Accordng to enz lw current n loop s shown n fgure due to ths current, mgnetc force m s cted on br mgnetc f s the ccelerton of the br mgnet then Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

2 J-Physcs. mpednce ( ) At low frequency so mpednce At hgh frequency so mpednce m mg g /m < g. + ut () (5 ) 6volt volt. & 9. Here constnt.5, 6mH We cn't chnge or.. () 5 P n A crcut P cos () (5) z (5) ( ) cos z z 5 5 f Hz. or curve tme constnt s more t / t / ( e ) ( e ) t n n (.69).s 5. Averge vlue rms vlue 6.. X c. c fc X tn. Z X 5. f X s cpctve X tn tn Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 7. rms reches t mmum vlue t phse before reches t ts pek vlue t phse So s leds by So s leds by t (t ) (t) A 8. Ac mmeter reds rms vlue dc mmeter reds verge vlue 9. sn t sn t sn t rms tn tn () 5 but / 8 5 s s.5 ms 5 ( ) 5 8. Here esonnce condton oltge cross combnton remns sme 9. tn5 X X f fc

3 J-Physcs 5 5. d q d q c m f f f(f ) esonnt frequency. X () 5 volt. +., 5 A 5 5. rcut wll be cpctve f X > X rcut wll be nductve f X > X 6. edng of voltmeter s 5. or crcut q q cos t, q cos t Accordng to gven condton q q Q 5. ( ) ( +t) 5 5. d ( +t) ntclockwse v (.)(.)() 5 A 6 / 5 / 5 edng of mmeter 7. Here A rd / s v v 5/6 so resonnce condton /.A 6 nd redng of voltmeter () () P cos 9. cos 5.7 tn tn tn () 5 wttless current rms sn A A Z 5 6. d d e, 5 7. Dmenson bsed : heck yourself P P P P U U P volt 8 tn tn tn tn 6 Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

4 J-Physcs 6. sn t + sn(t) cos ( ) rms + cos(t) sn sn t + cos(t) 7.5 XS d. e [where.a d ] e (A cos ) ether,a or should be chnged for nduced e.m.f.. otl chrge trnsferred t s ndependent of tme 6. Old power fctor () New power fctor () 5.A ( ) d d n n() q r. e v e de b vd 6. P Q e v n(b / ) v n(b / ) orce () b df b d b f n b n. A r ( ) / 6. P & Q clockwse d r e 5 / ( ) or e to be mmum d(e) d 5. Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 6. q A + (v) () () () ( 6 ) 8 q q A , 6. e effectve length of the wre frme s e 7. 6 X 8. tn 6 X Also tn 6 s X X seres s n resontng condton A P W 5

5 J-Physcs 9.. P P wth A source P Z Z Z Z H W W ;. Mnmum vlue of mpednce s for X X.6 mh. n D (ct s open crcut) So 5 5 At resonnce () (e / ) r e e r ( ) Ad e 6 d er e r t 5..A e e t / b e t / e n d t ms b d d e b n d d t / d 6. t ( t) d t d ( t) Here 5 / 5 5 / 5 So, here 5 rd/sec () 5 rom () & () : H Het ( t) t t t Het 7. Are n the mgnetc feld s gven by A ; d d v 9 d v t, t 8. No nduced current n the loop therefore No force cts over the loop. Work done on the loop wll be zero 9. quvlent crcut dgrm s s shown e e v r r r v r r r r r Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

6 J-Physcs. e v where ll the three should be perpendculr to ech other e ( v) e j k [( j) ] (j k)(8k) e volt. As cpctor blocks the current there wll be no current n the crcut HKD. e effectve length for the gven dgrm s ( ) l eff + e. y flemngs left hnd rule equvlent crcut dgrm s s shown current wll be P to Q n the dsc 7. Het () O Het produced energy stored n nductor u u Joule?, A H, P ohm me (). sec 9. As current n the crcut s gven by. d 5 e t P Q tsec 5. n loop D current s ndependent of the tme s current wll remn ' ' ll the tme X X cos Z. cos t 6 s voltge cross nductor s more crcut behves lke n nductve crcut s current lgs voltge by n mount. cos t 6 rcut equvlent cn be consdered s t current t ny tme [ e ]. or ( ) crcut cos 5 5 X or ( ) crcut Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 y super mposng the two current the net current n t wll be t / e e t / 6. Decy of current n crcut s gven by d t d ( e ) 7 cos 6 X n power fctor s.e. rectnce s zero X X. cos 8 eff

7 J-Physcs. rms snt nd current n the more thn > >. f nd P moves towrds Q, then ccordng to enz lw current n opposte s s nduced n Q. Sme s nd Q moves towrds P when, re n opposte drecton then the cols repels ech others. 5. No emf wll be nduced n ny drecton of ts moton 6. We know qq m cost q m cos + t, cos t Mmum current q m Mmum chrge on ( + ) q m Mmum chrge on. hrge on the cpctor Q (v ) As ll the qunttes re constnt dq so dq Hence. eplce the nduced emf's n the rngs by cell e (r) () r e (r) () r e +e 8r. e d d d [A cos ] d [ Are of squre d, ] d d d ( ) 6. hrgng or determnng eq eq Dschrgng 7. Sudden ncrese n the e.m.f. cuse the sprk n the nductor 8. n bsence of whole emf of goes on lmp nd lmp wll glow wth full brghtness nstntneously but n presence of some emf s nduced n. oltge drop on P decrese nd brghtness. 9. M, A. Z X 5 X () 6 Z ncreses 8 tmes current decreses by /8. s the nduced current of, so the nture of become should becomes opposte of e negtve nergy per unt volume r rdr r rdr r dr 6 8. urrent through nd would be equl fter tme f when q Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

8 J-Physcs q 5. he electrc feld force due to vrble mgnetc feld d d q e d d q q e t / c, q e t / Accelerton d e m q e t / q e / e t / e t / t / e e t/ n t n tn t / nstntneous voltge cross cpctor snt, X fc 5 5 sn t X X rms X nduced current 5 6. e nd N A As wre s fed n ( ) 5 t ; N r N' r Nr N'N 5 8. rms (N) r / / / / ' t Averge vlue Are tme (n ntervl) Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 X 9 tn X So phse dfference between nd X tn 5. me constnt of crcut s rom opton (A) + 6. () c 5 c & c 9 c 9 + c c c 5X c X c 5 9

9 J-Physcs XS Mtch the column sn( t) sn t ( A ) Phse dfference men [for hlf cycle] ( ) men men otl Are otl tme / men / cos Z Z ; X tn X ( D ) men otl Are otl tme ( ) Z Z Z X 5 X X () 5 ( ) Averge power cos 5 Mtch the column Pek vlue current n the crcut s gven by f wll be less current wll be mmum Slope of (v/s) t grph gves tme constnt s ( A ) Grph nd denotes for ' ' nd slope s more for therefore, represent nd () represent, ( ) Smlrly nd denotes for ' ' nd slopes s more for '' Mtch the column ( A ) rms men for full cycle men for hlf cycle ( ) men for hlf cycle r men t / 6 t rms otl Are otl tme men for full cycle men men / / hlf cycle / Mtch the column () t t t t t e 8 6A 8 A 6 8 A 6 Mtch the column ( A ) tn 6 X tn ( ) X tn s ( ) X tn s ( D ) X X tn s ( ) X 5 tn Mtch the column (A) d 5 5 Antclockwse () d zero Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

10 J-Physcs () d 5 5 lockwse omprehens on. M s sme s tht of (D) d 5 Antclockwse k m s sme s tht of ompreshenson. f f v 6 v. 5 v 5 m/s. e v 8 5 v. r v omprehens on. y pplyng K... or AD 8 6 d y pplyng K... for AG...() ( ) k s sme s tht of omprehens on. q q sn ()...() mk dq q cos () 5 q cos (t + ) y dvdng () nd () we get 5 tn (t + ) tn () from bove equton sn (t + ) Q Q sn (t + ) Q Q 6. (t + ) t t sn 5 Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p () Substtutng ts vlue n equton () 8 d t d e t e t t n n() e t + 8 6e t. d omprehens on 5. ( + ) s current n both drecton re ddtve n nture whle ( ) s current n both the loops re n opposte drecton.. y lenz lw n both the loop re n clock wse drecton therefore t flours from b to nd d to c n both the loop.. Agn by lenz lw current n both drecton should be clockwse but t s not possble therefore t s clockwse n bgger loop nd n ntclockwse n smller loop s e.m.f. due to bgger one s greter thn smller one. >

11 J-Physcs. XS (A) Due to current n no chnge n of. So no nduced current n. ut due to n A, s chngng n becuse A s movng towrds. s chngng (ncresng) n. So ccordng to enz's lw drecton of nduced current n wll be such tht t wll try to decrese the n so current wll be opposte n drecton n thn A.. Q P they produced mgnetc feld whch s norml to the plne of the pper but pont upwrds,.e. towrds the reder. hs requres tht current nd flow n the drectons shown n the fgure. Snce the resstnce per unt length s m, the resstnce of wres AD, A, D nd re ech nd those of wres, nd re.5 ech. Applyng Krchoff's loop rule to loop (ADA), e or e volt...() Applyng Krchhoff's loop rule to loop (), we hve e.5 ( ).5 or e.5 volt...() Solvng equton () nd (), we get 7 nd A.. S e v v A D O e ( ) e. ms efer to fgure lectromotve force (emf) s nduced n crcuts nd due to chnge of mgnetc flu thredng the crcuts becuse mgnetc feld s chngng wth tme. As the res enclosed by the crcuts remn unchnged, the mgntude of the nduced emf s gven by d d Ad e A Are enclosed by crcut s A AD A m m m. herefore, the emf nduced n crcut s e m s m s Are enclosed by crcut s A.5 m m.5 m nduced emf n crcut s e.5 m s.5 et nd be the nduced currents n crcuts nd respectvely. rom enz's lw, the drectons of these currents must be such tht they oppose the ncrese n currents. n other words, the drectons of the current n crcuts nd must be such tht eferrng to fgure, the current n segment A 7 A n the drecton from to A, the current n segment A n the drecton from to nd the current n segment. q HG A n the drecton from to. NA K J HG q NA ( ) KJ 7 5. nduced emf n col em d Mutul nductnce of system M NN A, N N,A M n N ( ) where () n N d herefore e nn( ) n N ( ) cost 6. () lockwse, ()Antclockwse, () AntclockwseA, (v) lockwse A, ( snt) 7. nduced emf n the loop v v ( ) v NM ( ) O QP (v) () HG b gk J v Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

12 J-Physcs 7 5 (. ).. 8. () Mmum urrent m () () e m 5 µ NA A 5 lu s mmum when plne of col s t 9 to the mgnetc feld.lu s zero when plne of col s t to the mgnetc feld. Yes t wll work becuse relted to col contnuous n chnge. 9 nduced emf n prmry col p d d ( + t) volt nduced emf n secondry col s N s s p HG N N s p N p KJ p 5 HG 5 K J () volt. Here d d ut + Hence () urrent n HG KJ H G K J () urrent n. () At t ct s open crcut HG KJ H G K J Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65. () A Power n 7 A P A e t / ().6A 5 eq (A).6 A () P.5 W sec e e t / t / e t / t / n tn t.69 t.8 sec nergy stored n : H H.5 J.6 7 So A () After some tme ct s short crcut 6.5A 6. () After ms wve s repeted so tme () perod s ms. f Hz Averge vlue Are/tme perod ( / ) ( ) b. rms cos t sn t g 5 volt cos t sn t sn t cos t HG K J H G K J b g 5. (). 68 () 6 ( ) 6. () mpednce Z 5. s ()

13 J-Physcs Power fctor cos cos HG K J (lggng) 7. ; f Hz, 5 tn X X tn X f.59 H 5.9 mh. 8. () X s resstor nd Y s cpctor () Snce the current n the two devces s the sme (.5A t volt) When nd re n seres cross the sme voltge then X. 5 ( ) ( ) rms rms X.5A O H Pt ( cos ) t J HG Use cos Z K J HG 5 5 ()Wttless current rms sn 5. mpednce of crcut HG 5. A X sn 5 KJ X Z (X X ) (5) ( ) otl current n crcut Z 5 KJ 5 Z (edng of mmeter) oltmeter connect cross nd so redng of voltmeter Now X X So redng of voltmeter. Power dsspton 9 A 5 9. () resstor () nductor. () At resonnce condton X X cos.. Z Z Z...() () cos Z cos, Z Z, Z (X ) No, t s lwys zero.. () mpednce of the crcut Z ( X X ) ( ) ( 7 ) 5 current flow n ckt. rms rms 5 / Z 5 Het developed HG 5 A rms t 5 8 joule KJ X 5 X Put ll these vlue n eq () P loss ( ). Mutul nductnce M s p 5 wtt r ( ) ( r ) 5.. v k. (v v j) k. v y k v y y r Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

14 J-Physcs 6. eff 6 6. dr XS () 6 v eff A eff N 7. mg t v 6 A.8 t / [ e ] mg v 6 t hrge pssed through the bttery t / e e ( / e) ( ) t e 5 A 6 t / r () e de rdr r () e t r e Now torque requred for power loses r P P r r orque requred to move the rod n crculr moton gnst grvttonl feld r mg cos mgr cos t he totl torque r + mgr cos t Drecton of torque : clockwse. H 5 Mgnetc energy he current n the crcut for one forth of mgnetc energy t ut e e t t Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65. where t / e t n hrge flown t / n e tn t / (e ) e e t/5 t n 5 t / 5 t 5 n s. efer to fgure. z y v m mg 5

15 J-Physcs et v be the velocty of the rod long the postve drecton t n nstnt of tme nd let the mgnetc feld ct perpendculr to the tble long the postve y drecton. he emf nduced n the rod s e v. herefore, the nduced current s e v he rod of length crryng current n mgnetc feld wll eperence force...() long the negtve drecton. Snce the rod s mssless, ths force wll lso be equl to the tenson n the strng ctng long the postve drecton,.e. et be the ccelerton of mss m movng n the downwrd drecton, then m net force ctng on m mg mg or g m...() Due to, the flu through the loop s M A nduced emf n the loop s e d d M v d d v d d d d M nduced current nt he loop s M v e M v Mgnetc moment of the loop s M re enclosed by the loop M v Usng (), () nd (), we hve g m g m v g v m...() () he rod wll cqure termnl velocty v t when. Puttng nd v v t n eq. () we hve g v t m mg or v t () When the velocty of the rod s hlf the termnl velocty.e. when v then from equton (), we hve g v t / m. efer to fgure oop g Mgnet v v t mg m mg z g g g he mgnetc feld t dstnce on the s of mgnetc of length nd dpole moment M s gven by >>, we hve y M M 6 5. Potentl energy U M M cos 8 U M M M M U 7 du M 8 d hs force s cused by the movng mgnet. rom enz's lw, ths force opposes the moton of the mgnet. O A P v 5.. or the crcut + + A v Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

16 J-Physcs 6. ntl current through for swtch n poston nergy stored n nductor Het developed cross 7. r r d or n electron d 8. M r k f e erk m m m M(kt) () s decresng wth so nduced current wll try to ncrese () (enz's lw). So drecton of current ntclockwse () or upper hlf re Ad e.87 e.7 olt otl emf n crcut volts. e v, de (d)vd Q km 9. d d(.t) (.) orque on col q q m Angulr velocty ttned q t m rd/sec. or crcut q d c q Q sn(t + ) At t, q Q Q Q sn / t q Q sn. A Kt nduced current wll be clockwse n A : e n D : v vd e n v e vd e n So net ee e v n e v n orce, d (d) d) Work done dw d.. W d W d Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 hrge q At t; At So t k q...87 t q 6. D W A n M M 7

17 J-Physcs 7. m m, m + m m X X m m m nd m m m m m t,, 6H + d + t (+5t) + 6(5) 9.() et v be the velocty of the rod MN t n nstnt of tme t when t s t dstnce from. hen, the nduced emf t tht nstnt s e vd. Snce s the resstnce per unt length of ech wre, the totl resstnce n seres wth t tht nstnt s +. hus the totl resstnce of the crcut t tme t +. Hence the current n the crcut s nduced emf vd totl resstnce constnt (gven) v...() d he mgnetc force ctng on the rod s m d drected to the left. he net force ctng on the rod s net m d...() net s the force eperenced by the rod MN t tme t when ts velocty s v t dstnce from. rom Newton's lw net m m dv m dv d d Now d v. Usng (), we hve net mv d d mv d (,, nd d re constnts) m (+) [Use equton ()] d Usng ths n equton (), we get m d + (+)...() d () Work done n tme d. herefore, work done per second.e., power s P d (d) d Usng quton (), we hve P d Het produced per second s Q (+) to Q P d Usng () n (), we get Q d v d...() P d m. et the mgnetc feld be perpendculr to the plne of rls nd nwrds. f be the termnl velocty of the rls, then potentl dfference cross nd would be wth t lower potentl nd t hgher potentl. he equvlent crcut s shown n fgure (). n fgure (). () () A e...() e D Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 8

18 J-Physcs Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 e...() Power dsspted n s.76 wtt herefore () e.76 wtt...() Smlrly () e. wtt...() Now the totl current n br s + (rom to )...(5) Under equlbrum condton, mgnetc force ( m ) on br weght ( g ) of br.e., m g or mg...(6) rom equton (6) mg A m g or.7 A Multplyng equton (5) by e, we get e e + e (.75 +.) wtt (rom equton nd ).96 wtt e.96 volt.96 or e.6.7 ut snce e.6 e.6. m/s. m/s Hence, termnl velocty of br s. m/s. Power n s.76 wtt e e Smlrly.. efer to fgure () e..6.. O y z G H onsder smll element of the loop of wh dy nd sde. he re of the element da dy. Snce the re vector nd the mgnetc feld vector pont n the sme drecton,.e. the ngle between the norml to the plne of the loop nd the mgnetc dy () feld s zero, the mgnetc flu through the element d da cos da cos da he totl mgnetc flu lnked wth the loop s da depends upon y, we hve y y y dy y y y ˆk ydy y y y [(y+) y ]...() nduced emf s d d e [(y+) y ] dy dy y y dy e dy he nduced current s...() Snce the mgnetc feld ponts n the postve z drecton, t follows from enz's lw tht the drecton of the nduced current wll be long the negtve z s. hus the current n the loop wll flow n the counterclockwse drecton s shown n fgure. O y (y+) y G he orentz force ctng on current element d of length d n mgnetc feld s gven by H d Now, the forces ctng on sdes G nd H of the loop re equl nd opposte. Hence they cncel ech other. he forces ctng on sdes nd GH re n opposte drectons but ther mgntudes re dfferent snce depends upon y. hus, force ctng on sde s mg ˆ y k ˆ ˆ yj Smlrly, force ctng on sde GH s y ˆ GH ˆ k y ˆj herefore, the net force ctng on the loop s net + GH ˆ yj y ˆj ˆ j Substtutng the epresson for from equton (), 9

19 J-Physcs () we get net dy ĵ he negtve sgn ĵ ndctes tht the force cts long the negtve y drecton. As force net s drected long the negtve y s nd the grvttonl force mg s long the postve y drecton, the totl force on the loop n the downwrd (postve y) drecton s mg dy where dy v, the speed wth whch the loop s fllng downwrds. rom newton's second lw, the equton of moton of the loop s mss ccelerton or m dv m dv mg v. A. XS (A) n prllel D s H he equvlent crcut of the conductor wll be A v or dv g v m or dv g kv...() evl where k m errngng epresson (), we hve dv g kv ntergrtng, we hve v dv g kv whch gves g kt v e k t or log e k gm g kv g t t ep m hs s the requred epresson for the speed v of the loop s functon of tme t. When the loop cqures termnl velocty v no force cts on t. Hence ts ccelerton dv/ s zero. Usng ths n equton (), we hve g kv t or v t g k mg. flu A; d (d) (da) d d b d b d n b b n D. he core of trnsformer s lmnted so s to reduce the energy loss due to eddy currents. 6. A A t.5, e 8 e d A A H 7. Number of turns n esstnce of col esstnce of glvnometer nduced current e t estnce of crcut nw nw t 5 W W n 5t Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

20 8. he flu ssocted wth col NAcost r cos t e d r sn t P P nst. v e r sn t P sn t Hence P P v v P v r 8 sn t r r r 8. t 5t +5 d t 5 e d 5 t e (t) 5. mh; ; A J-Physcs ong tme fter, current n the crcut s A On short crcutng e e A e t o 9. he emf developed cross the ends of the pvoted rod s e 5 rd,., m sec Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 e. 5. stedy A ; t e e t t n e t.s /. he vertcl rm of the both tubes wll becomes bttery of emf lv. lv lv he emf nduced n the crcut s lv.. MS vlue of electrc feld 7 N/ Pek vlue of electrc feld 7 N/ the verge totl energy densty of electromgnetc wve u ; v On solvng we get u v.58 6 J/m 5. nductnce of the col H esstnce of the col 5 As - s connected cross bttery, hence nture of current tht wll flow through the crcut wll be trnsent current,.e., ( e t/ ) 5 where s nd A 5 5 hence ( e / ) ; ( e )A

21 J-Physcs 6. N N A M. H 7. quvlent resstnce cross (by short crcutng bttery) eq then current through ( e t/ ) where H, 6A 6( e 5t )... () p.d. cross s d d [6( e 5t )] e 5t hs queston contns Sttement nd Sttement. Of the four choces gven fter the sttements, choose the one tht best descrbes the two sttements. 8. At t nductor behves s broken wre then...5 Mutul nductnce so flu through trgger col ( X ) /. ( X ) / 7 (.) ( ) ( ) ((.) (.5) ) / ANANG UN. On drwng the mpednce trngle; we get t t nductor behves s conductng wre / 9. rcut cn be reduced s e v v e v /. e v (5 5 ) () (.5).5 m.. wb/m W e v 5. H m. Due to conductng nture of Al eddy currents re produced X Z X he power fctor cos Z 6. et Q m Q m Q...() [Gven tht energy stored n cpctor nergy stored n conductor]. Accordng to the conservton of energy we know tht Q m Q Q m Q Q Q Q m [from eq. ()] Q Q m Q Q m 7. A D meter mesure the verge vlue nd the verge vlue of A over one full cycle s zero. Hence, D meters cn't mesure A. Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

22 J-Physcs 8. oltge cross combnton oltge cross combnton he resonnt frequency r or On squrng both sdes, we get ; ;. n order to trnsfer mmum power the genertor should work t resonnt frequency,.e., should be such so tht f f f 5 6. Power fctor cos Z cos f resstnce wll be the prt of crcut, phse dfference between voltge nd current cn not be. 5. Gven tht sn(t) nd \ sn(t /) 6. he phse dfference between nd s. Power dsspted n n A crcut P rms rms cos So, power dsspted for ths stuton where phse dfference between voltge nd current s be zero. X tn tn X X X X tn X X X ondton for resonnce So P cos P W 7. nergy s shred eqully between nd t t, so where t 8 8 wll. he voltge cross X At resonnce, current. Also t resonnce X X oltge esstnce ; Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 X 6 X. 5. Mgnetc flu ssocted wth rottng col NAcost NAcost d (NA)sn(t) ee snt Mmum vlue of emf generted n col NA

23 J-Physcs XS -. A motonl emf, e v s nduced n the rod. Or we cn sy potentl dfference s nduced between the two ends of the rod A, wth A t hgher potentl nd t lower potentl. Due to ths potentl dfference, there s n electrc feld n the rod. A v. Mgnetc feld produced by current n lrge squre loop t ts centre. sy K Mgnetc flu lnked wth smller loop,.s K ( ) herefore, the mutul nductnce M K M. or understndng, let us ssume tht the two loops re lyng n the plne of pper s shown. he current n loop wll produce * mgnetc feld n loop. herefore, ncrese n current n loop wll produce n nduced current n loop whch produce mgnetc feld pssng through t.e., nduced current n loop wll lso be clokwse s shown n the fgure. 5. nd 8.. s 6 nd A (gven) Substtutng these vlues n equton (), we get t.97 s t.97 ms t ms.d d d S (r) d for r > nduced electrc feld r or r < ; (r) r d At r, d herefore, vrton of wth r(dstnce from centre)wll be s follows : d r r d r d r r r 6. he equtons of (t), (t) nd (t) wll tke the followng forms : (t) K ( e k t ) current growth n crcut (t) K ( e k t ) (t) (t); N n cse of solenod col nd N (t) K e k t n cse of crculr col.e., r Perpendculr to pper outwrds Perpendculr to pper nwrds he loops wll now repel ech other s the currents t the nerest nd frthest ponts of the two loops flow n the opposte drectons.. he current tme ( t) equton n crcut s gven by [Growth of current n crcut] where ( e t/ ) A 6...() e d d (t) nd e : e M herefore the product (t) (t) K 5 e k t ( e k t ). he vlue of ths product s zero t t nd t. herefore, the product wll pss through mmum vlue (K : K : K : K nd K 5 re postve constnts nd M s the mutul nductnce between the col nd the rng). he correspondng grph wll be s follows : (t) t (t) t (t)(t) t Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65

24 J-Physcs 7. lectrc feld wll be nduced n both AD nd. 8. When current flows n ny of the cols, the flu lnked wth the other col wll be mmum n the frst cse. herefore, mutul nductnce wll be mmum n cse (). 9. When swtch S s closed mgnetc feld lnes pssng through Q ncreses n the drecton from rght to left. So, ccordng to enz's lw nduced current n Q.e., Q wll flow n such drecton, so tht the mgnetc feld lnes due to Q psses from left to rght through Q. hs s possble when Q flows n ntclockwse drecton s seen by. Opposte s the cse when swtch S s opened.e., Q wll be clockwse s seen by.. Power P e Here, e nduced emf d where NA e NA d Also, r where resstnce, r rdus, length N r P P P. As the current leds the emf e by, t s n crcut. tn or tn X. When resstvty s low current nduced wll be more; therefore mpulsve force on the rng wll lso be more nd t jumps to hgher levels. [ut for ths mss should be ether less or equl to the other] omprehens on#. hrge on cpctor t tme t s : Here, q q ( e t/ ) q nd t q ( e / ) ( e ). rom conservton of energy, m m e. omprng the osclltons wth norml SHM we get, d Q Q Here, Q d Q Subjectve. Mgnetc feld () vres wth tme (t) s shown n fgure. ().8 d / s nduced emf n the col due to chnge n mgnetc flu pssng through t, t(s) Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 As rd/s he product of should be s.. Polrty of emf wll be opposte n the two cse whle enterng nd whle levng the col. Only n opton (b) polrty s chngng.. n unform mgnetc feld, chnge n mgnetc flu s zero. herefore, nduced current wll be zero.. As re of outer loop s bgger therefore emf nduced n outer loop s domnnt nd therefore ccordng to lenz lw current n outer loop s Antclockwse nd nner loop s clockwse M Q. lectrosttc nd grvttonl feld do not mke closed loops. 5 e d NA d Here, A Are of col 5 m N Number of turns Substtutng the vlues, we get e () (5 ) () herefore, current pssng through the col e ( esstnce of col.6 ).5 A.6 Note tht from to. s nd from.s to.6s, mgnetc feld pssng through the col ncreses, whle durng the tme.s to.s nd from.6s to.8s mgnetc feld pssng through the col decreses. herefore, drecton of current through

25 J-Physcs the col n these two tme ntervls wll be opposte to ech other. the vrton of current () wth tme (t) wll be s follows : +.5 (A) t(s) Power dsspted n the col s P (.5) (.6)W.5W Power s ndependent of the drecton of current through the col. herefore, power (P) versus tme (t) grph for frst two cycles wll be s follows : P(wtt) hs s smple crcut, whose tme constnt. /.s nd stedy stte current / / 6A herefore, f swtch S s closed t tme t, then current n the ccut t ny tme t wll be gven by (t) ( e t/ ) (t) 6 ( e t/. ) 6( e 5t ) (sy) herefore, potentl drop cross t ny tme t s: d (e ) (.)()e e 5t volt 5 t 5 t (b) he stedy stte current n or s 6A Now, s soon s the swtch s opened, current n s reduced to zero mmedtely. ut n nd t decreses eponentlly. he stuton s s follows: 6A.5.8 t(s) otl het obtned n, cycles wll be H P.t (.5) () (.) J hs het s used n rsng the temperture of the col nd the wter. et be the fnl temperture. hen H m w S w ( ) + m c S c ( ) Here m w mss of wter.5 kg S w specfc het of wter J/kg K m c mss of col.6 kg nd S c specfc het of col 5J/kg K Substtutng the vlues, we get (.5) () ( ) + (.6) (5) ( ) 5.6. () Gven, nd mh. H. wo prts of the crcut re n prllel wth the ppled bttery. So, the upper crcut cn be broken s : S S () Now refer fgure (b) : + S (b) 6A Stedy stte condton (c) efer fgure (e) : t S s open (d) me constnt of ths crcut would be. '.s ( ) urrent through t ny tme t s e t/ ' 6e t/. 6e t A Drecton of current n s s shown n fgure or clockwse.. () Applyng Krchhoff' second lw : d d d tt (e) d...() hs s the desred relton between, d () equton () cn be wrtten s d + d ntegrtng we get,. q + q Here, f...() d nd d. n() So, from quton () chrge flown through the resstnce upto tme t, when current s, s Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 6

26 J-Physcs q n() () hs s the cse of current decy n n crcut. hus, t / e...() Here,,, t ( ) nd Substtutng these vlues n equton (), we get :. () or element strp of thckness d t dstnce from left wre, net mgnetc feld (due to both wres) (outwrds) Mgnetc flu n ths strp, d ds totl flu n d d d d n() n() ( sn t) Mgntude of nduced emf, d n() e cos t e cos t...() Here, q e n() he correspondng q t grph s shown n fgures. + q q q t 5. After long tme, resstnce cross n nductor becomes zero whle resstnce cross cpctor becomes nfnte. Hence, net eternl resstnce, net urrent through the btteres, r r Gven tht potentl cross the termnls of cell A s zero. r r / r r Solvng ths equton, we get (r r ) 6. nductve rectnce X (5) () (5 ) mpednce Z X () () Gven v rms Hence, mpltude of voltge Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 where e n() hrge stored n the cpctor, q e e cos t...() nd current n the loop m e n() dq e sn t...() () Mgnetc flu pssng through the squre loop sn t [rom equton ()].e., * mgnetc feld pssng through the loop s ncresng t t. Hence, the nduced current wll produce mgnetc feld (from enz's lw). Or the current n the crcut t t wll be clockwse (or negtve s per the gven conventon). herefore, chrge on upper plte could be wrtten s, q +q cos t [rom equton ()] v v rms Ampltude of current v Z A X Phse dfference tn tn n crcut voltge leds the current. Hence, nstntneous current n the crcut s, (A) sn (t /) orrespondng t grph s shown n fgure., O v sn t sn ( t /) /8 / / 5/8 9/8 t 7

27 J-Physcs 7. Out sde the solenod net mgnetc feld zero. t cn be ssumed only nsde the solenod nd equl to n. nduced e d d d (A ) ( n / ) or e ( n ) ( cost) ( ) esstnce of the cylndrcl vessel s d e dn cos t nduced current 8. hs s problem of osclltons. hrge stored n the cpctor osclltes smple hrmonclly s Q Q sn (t+ ) Here, Q m. vlue of Q et t t, QQ then 6 ( H)(5. ) s dq Q(t) Q cos t...() (t) Q sn t...() nd d(t) () Q Q cos (t)...() Q At cos t t d. nduce electrc feld Q torque on chrge ˆk +Z Q d by d Q kˆ hnge mgnetc dpole moment Q ˆk. Mgntude of nduced electrc feld d At cos(t), from equton () :. d (. )( s ) A/s () Q or Q when cos(t).e. t,... At ths tme (t) Q sn t (t) [sn sn ] () (t) Q sn t Mmum vlue of s Q m Q (. ) ( s ) m.a (v) rom energy conservton Q m m Q ( ) m 6 Q (. )(5. )( ) Q.7.A step up for step up trnsformer, olt for step down trnsformere N N. urrent n trnsmsson lne Power 6 oltge, 5A step down esstnce of lne. 8 Power loss n lne (5) 8 8 KW percentge of power dsspton n durng trnsmsson 8 % 6 Node-6\:\Dt\\Kot\J-Advnced\SMP\Phy\Soluton\Unt-9 & \.M & A.p65 8