electromagnetic induction

Chapter # 38 Electromagnetic Induction

Page # 1manishkumarphysics.in

SOLVED EXAMPLESExample 38.1

Figure shows a conducting loop placed near a long straight wire carrying a current i as shown. If the currentincreases continuously, find the direction of the induced current in the loop.

Sol. Let us put an arrow on the loop as shown in the figure. The right-hand thumb rule shows that the positivenormal to the loop is going into the plane of the diagram. Also, the same rule shows that the magnetic fieldat the site of the loop due to the current is also going into the plane of the diagram. Thus, B

and Sd

are

along the same direction everywhere so that the flux SdB

is positive. If i increases, the magnitude

increases, dtd

is positive. Thus, is negative and hence, the current is negative. The current is, therefore,

induced in the direction opposite to the arrow.

Example 38.2Figure shows a rectangular loop MNOP being pulled out of a magnetic field with a uniform velocity v byapplying an external force F. The length MN is equal to l and the total resistance of the loop is R. Find (a) thecurrent in the loop, (b) the magnetic force on the loop, (c) the power delivered by the external force and (e) thethermal power developed in the loop.

v

F

O

PM

N

Sol. (a) The emf induced in the loop is due to the motion of the wireMN. The emf is e = vB with the positive end at N and the negativeend at M. The current is

i =R

=R

vB

v

F

O

PM

N

F1

F2

F2

in the clockwise direction (figure).

(b) The magnetic force on the wire MN is BiF1

. The magnitude is F1 = iB = R

vB 22and is opposite to

the velocity. The forces on the parts of the wire NO and PM, lying in the field, cancel each other. The resultant

magnetic force on the loop is, therefore, F1 = RvB 22

opposite to the velocity..

(c) To move the loop at a constant velocity, the resultant force on it should be zero. Thus, one should pull theloop with a force

F = F1 = RvB 22

.

(d) The power delivered by the extenal force is

P = Fv =RBv 222

.

(e) The thermal power developed is

P = i2R =2

RvB

R =

RBv 222

.

We see that the power delivered by the external force is equal to the thermal power developed in the loop.This is consistent with the principle of conservation of energy.



Example 38.3An average induced emf of 0.20 V appears in a coil when the current in it is changed form 5.0 A in onedirection to 5.0 A in the opposite direction in 0.20 s. Find the self-inductance of the coil.

Sol. Average s/A50s20.0

)A0.5()A0.5{dtdi

Using = L dtdi

0.2 V = L(50 A/s)

or, L = mH0.4s/A50V2.0

Example 38.4Consider the circuit shown in figure. The sliding contact is being pulled towards right so that the resistancein the circuit is increasing. Its value at the instant shown in 12 W. Will the current be more than 0.50 A or lessthan it at this instant?

r

20 mH

6V

Sol. As the sliding contact is being pulled, the current in circuit charnges. An induced emf e = L dtdi

is produced

across the inductor. The net emf in the circuit is 6 V = L dtdi

and hence the current is

12dtdiLV6

i ....(i)

at the instant shown. Now the resistance in the circuit is increasing, the current is decreasing and so dtdi

is

negative. Thus, the numerator of (i) is more than 6V and hence i is greater than 12V6

= 0.50 A.

Example 38.5In inductor (L = 20 mH), a resistor (R = 100 ) and a battery ( = 10 V) are connected in series. Find (a) thetime constant, (b) the maximum current and (c) the time elapsed before the current reaches 99% of themaximum value.

Sol. (a) The time constant is

100mH20

RL

= 0.20 ms

(b) The maximum current is

i = /R = 100V10

= 0.10 A.

(c) Using i = i0 (1 et/),

0.99 i0 = i0(1 et/)

or, et/ = 0.01

or,

t= ln(0.01)

or, t = (0.20 ms) ln (100) = 0.92 ms.



Example 38.6An inductor (L = 20 mH), a resistor (R = 100 ) and a battery ( = 10 V) are connected in series. After a longtime the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit 1 msafter short-circuiting.

Sol. The initial current is i = i0 = /R = 100V10

= 0.10 A.

The time constant is = L/R = ms20.100mH20

.

The current at t = 1 ms isi = i0e

t/

= (0.10 A)e(1 ms / 0.20 ms)= (0.10 A)e5 = 6.7 104 A.

Example 38.7Calculate the energy stored in an inductor of inductance 50 mH when a current of 2.0 A is passed through it.

Solution :The energy stored is

J10.0)A0.2)(H1050(21Li

21U 232

Example 38.8A solenoid S1 is placed inside another solenoid S2 as shown in figure. The radii of the the inner and the outersolenoids are r1 and r2 respectively and thenumbers of turns per unit length are n1 and n2 respectively.Consider a length l of each solenoid. Calculate the mutual inductance between them.

s1s2

Sol. Suppose a current i is passed through the inner solenoid s1. A magnetic field B = s n1i is produced inside s1whereas the field outside it is zero. The flux through each trun of s2 is

Br12 = m0 ir1

2.The total flux through all the turns in a length of s2 is

= (0 n1ir12) n2 = (0 n1n2r1

2)i.Thus, M = 0 n1n2r1

2.

QUESTIONS FOR SHORT ANSWER1. A metallic loop is placed in a uniform magnetic field. Will an emf be induced in the loop ?

2. An inductor is connected to a battery through a switch. Explain why the emf induced in the indutor ismuch larger the switch is opened as comapered to the emf induced when the switch is closed.

3. The coil of a moving-coil galvanometer keeps on oscillating for a long time if it is deflected and released.If the ends of the coil are connected together , the oscillation stops at once , Explain.

4. A short magnet is moved along the axis of a conducting loop. Show that the loop repels the magnet ifthe magnet if the magnet is approaching the loop and attracts the magnet if ir is going away from theloop.

5. Two circular loops are placed coaxially but separated by a distance. A battery is suddenly connectedto one of the loops establishing a current in it. Will there be a current induced in the other loop ? If yes, when does the current start and when does it end ? Do the loops attract each other or do they repel?

6. The battery discussed in the previous question is suddenly disconnected. Is a current induced in theother loop ? If yes , when does it start and when does it end ? Do the loops attract each other or repel?

If the magnetic field outside a copper box is suddenly changed , what happens to the magnetic fieldinside the box ? Such low-resistivity metals are used to form enclosures which shield objects inside



them against varying magnetic fields.

8. Metallic (nonferromagnetic) and nonmetallic particles in a solid waste may be separated as follows.The waste is allowed to slide down an incline over permanent magnets. The metallic particles slowdown as compared to the nonmetallic ones and hence are separated. Discuss the role of eddy currentsin the process.

9. A pivoted aluminimum bar falls much more slowly through a small region containing a magnetic fieldthan a similar bar of an insulating material . Explain.

10. A metallic bob A oscillates through the space between the poles of an electromagnet . The oscillatesare more quickly damped when the circuit is on , as compared to the case when the circuit is off.Explain.

N

SA

11. Two circular loops are placed with their centres separated by a fixed distance. How would you orientthe loops to have (a) the largest mutual inductance (b) the smallest mutual inductance ?

12. Consider the self-inductance per unit length of a solenoid at its centre and that near its ends. Which fothe two is greater ?

13. Consider the energy density in a solenoid at its centre and that near its ends. Which of the two isgreater ?

Objective - I1. A rod a length l rotates with a small but uniform angular velocityabout its perpendicular bisector. Auniform

magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rodand an end isyEckbZ okyh ,d NM+ blds yEc v)Zd ds ifjr% de fdUrq ,d leku dks.kh; osx ls ?kwe jgh gSA ogk ij ?kw.kZuv{k ds lekUrj bafxr le:i pqEcdh; {ks=k B fo|eku gSA NM+ ds dsUnz rFkk blds ,d fljs ds e/; foHkokUrj gS &

(A) zero (B*)18 Bl

2 (C)12 Bl

2 (D) Bl2

2. Arod of length l rotates with a uniform angular velocityabout its perpendicualr bisector.Auniform magneticfield B exists parallel to the axis of rotation. The potential difference between the two ends of the rod isyEckbZ okyh ,d NM+ blds yEc v/kZd ds ifjr% ,oa leku dks.kh; osx ls ?kwe jgh gSA ogk ij ?kw.kZu v{k ds lekUrjbafxr le:i pqEcdh; {ks=k B fo|eku gSA NM+ ds dsUnz rFkk NM+ ds nksuksa fljksa ds chp foHkokUrj gS &

(A*) zero (B)12 Bl

2 (C) Bl2 (D) Bl2

3. Consider the situation shown in fig. If the switch is closed and after some time it is opened again, the closedloop will showfp=k esa iznf'kZr ifjfLFkfr ij fopkj dhft;sA ;fn fLop cUn fd;k tkrk gS vkSj dqN le; i'pkr~ bldks iqu% [kksyfn;k tkrk gS] cUn ywi iznf'kZr djsxk &

(A) an anticlockwise current-pulse



(B) a clockwise current-pulse(C) an anticlockwise current-pulse and then a clockwise current-pulse(D*) a clockwise current-pulse and then an anticlockwise current-pulse(A) ,d okekorhZ /kkjk Lian(B) ,d nf{k.kkorhZ /kkjk Lian(C) ,d okekorhZ /kkjk Lian vkSj fQj ,d nf{k.kkorhZ /kkjk Lian(D*) ,d nf{k.kkorhZ /kkjk Lian vkSj fQj ,d okekorhZ /kkjk Lian

4. Solve the previous question if the closed loop is completely enclosed in the circuit containing the switch.fiNys iz'u dks gy dhft;s] ;fn can ywi iw.kZr;k fLop okys ifjiFk ds vUnj ifjc) gSA(C*)

5. A bar magnet is relased from rest along the axis of a very long, vertical copper tube. After some time themagnetrkacs dh ,d cgqr yEch uyh dh v{k ds vuqfn'k ,d NM+ pqEcd dks fxjk;k tkrk gSA dqN le;i'pkr~ pqEcd &(A) will stop in the tube (B*) will move with almost contant speed(C) will move with an acceleration g (D) will oscillate(A) uyh esa :d tk;sxkA (B*) yxHkx fu;r pky ls xfr djsxkA(C) g Roj.k ds lkFk xfr djsxkA (D) nksyu djsxkA

6. Fig. shown a horizontal solenoid connected to a battery and a switch. A copper ring is place on a frictionlesstrack, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring willfp=k esa iznf'kZr fd;k x;k gS fd ,d {ksfrt ifjufydk ,d cSVjh ,oa ,d fLop ds lkFk tqM+h gqbZ gSA ?k"kZ.k jfgr iFkij rkacs dh ,d oy; bl izdkj j[kh gqbZ gS fd oy; dh v{k] ifjufydk dh v{k ds vuqfn'k gSA tSls gh fLop cUnfd;k tkrk gS rks oy; &

(A) remain stationery fLFkj jgsxhA(B) move towards the solenoid ifjufydk dh vksj xfr djsxhA(C*) move away from the sloenoid ifjufydk ls ijs xfr djsxhA(D) move towards the solenoid or away from it depending onifjufydk dh vksj xfr djsxh ;k mlls ijswhich terminal (positive or neagtive) of the battery is connected to the left end of the solenoid.;g bl ij fuHkzj djsxk fd ifjufydk ds ck;sa fljs ls cSVjh dk dkSulk VfeZuy /kukRed ;k _.kkRed tqM+k gqvkgSA

7. Consider the following statements:(a) An emf can be induced by moving a conductor in a magnetic field(b) An emf can be induced by changing the magnetic field.fuEu dFkuksa ij fopkj dhft;s &(a) fdlh pky dks pqEcdh; {ks=k esa xfr djokdj fo-ok-c- izsfjr fd;k tk ldrk gSA(b) pqEcdh; {ks=k ifjofrZr djds fo-ok-c- izsfjr fd;k tk ldrk gSA(A*) Both A and B are true (B) A is true but B is false(C) B is true but A is false (D) Both A and B are false(A*) Ao B nksuks a gh lR; gSA (B) AlR; gS] fdUrq BvlR; gSA(C) B lR; gS] fdUrq AvlR; gS (D) A o B nksuks a xyr gSA

8. Consider the situation shown in fig. The wire AB is slid on the fixed rails with a constant velocity. If the wireAB is rep;aced by a semicircular wire, the magnitude of the induced current willfp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA rkj AB fLFkj iVfj;ksa ij fu;r osx ls fQly jgk gSA ;fn rkj ABdksv)Zo`kkdkj rkj ls izfrLFkkfir dj fn;k tk;s rks izsfjr /kkjk dk ifjek.k gksxk &



(A) increase (B*) remain the same(C) decrease(D) increase or decrease depending on whether the semicircle bulges towards the resistance or away fromit.(A) c



(A) Bl/R clockwise (B) Bl/R anticlockwise(C) 2Bl/R anticlockwise (D*) zero(A) Bl/R nf{k.kkorhZ (B) Bl/R okekorhZ(C) 2Bl/R okekorhZ (D*) 'kwU;

Objective - II1. A bar magnet is moved along the axis of a copper ring placed far away from the magnet, an anticlockwise

current is found to be induced in the ring. Which of the following may be true ?,d NM+ pqEcd dks] blls cgqr vf/kd nwj fLFkr j[kh gqbZ rkacs dh oy; dh v{k ds vuqfn'k xfr djok;h tkrh gSApqEcddh vksj ls ns[kus ij] oy; esa izsfjr /kkjk dh fn'kkokekorhZ izsf{kr gksrh gSA fuEu esa ls dkSulk lR; gks ldrk gS &(A) The south pole faces the ring and the magnet moves towards it(B*) The north pole faces the ring asnd the magnet moves towards it(C*) The south pole faces the ring and the magnet moves away from it.(D) The north pole faces the ring and the manget moves away from it.(A) nf{k.k /kzqo] oy; dh vksj gS rFkk pqEcd bldh vksj vk jgk gSA(B*) mkj /kzqo] oy; dh vksj gS rFkk pqEcd bldh vksj vk jgk gSA(C*) nf{k.k /kzqo] oy; dh vksj gS rFkk pqEcd blls nwj tk jgk gSA(D) mkj /kzqo] oy; dh vksj gS rFkk pqEcd bls nwj tk jgk gSA

2. A conducting rod is moved with a constant velocity u in a magnetic field. A potential difference. A potentialdifference appears across the two ends,d pky NM+ f;r osx bds lkFk fdlh pqEcdh; {ks=k esa xfr'khy gSA blds nksuksa fljksa ds chp foHkokrj mRiUu gksxk&

(A) if

l (B) if

B (C) if

l

B (D*) none of these

3. A conduction loop is placed in a uniform magnetic filed with its plane perpenducular to the field. An emf isinduced in the loop if,d pkyd ywi dks le:i pqEcdh; {ks=k esa bl izdkj j[kk x;kgS fd bldk ry {ks=k ds yEcor~ gSA ywi esa fo-ok-c-izsfjr gksxk] ;fn &(A) it is translated (B) it is rotated about its axis(C*) it is rotated about a diameter (D*) it is deformed(A) bldks LFkkukuarfjr fd;k tk;sA (B) bldks bldh v{k ds ifjr% ?kwf.kZr fd;k tk;sA(C*) bldks] O;kl ds ifjr% ?kw.kZr fd;k tk;sA (D*) bldks fo:fir fd;k tk;sA

4. A metal sheet is placed in front of a strong magnetic pole. A force needed toizcy pqEcdh; /kzqo ds lkeus /kkrq dh IysV j[kh tkrh gSA ,d cy dh vko';drk gksxh &(A*) hold the sheet there if the metal is magnetic(B) hold the sheed there if the metal is nonmagnetic(C*) move the sheet away from the pole with uniform velocity if the metal is magnetic(D*) move the sheet away from the pole with uniform velocity if the metal is nonmagnetic.

Negative any effect if oaramagnetism, diamagnetism and gravity.(A*) ;fn /kkrq pqEcdh; gS] rks IysV dks ogk idM+dj j[kus ds fy;sA(B) ;fn /kkrq vpqEcdh; gS] rks IysV dks ogk idM+dj j[kus ds fy;sA(C*) ;fn /kkrq pqEcdh; gS rks 'khV dks ,d leku osx ls /kzqo ls nwj ys tkus ds fy;sA(D*) ;fn /kkrq vpqEcdh; gS rks 'khV dks ,d leku osx ls /kzqo ls nwj ys tkus ds fy;s vuqpqEcdRo] izfr pqEcdRo ;kxq:Ro ds fdlh Hkh izHkko dks ux.; eku fyft'sA

5. A constant current i is maintained in a solenoid. Which of the following quantities will increase if an iron rodis inserted in the soleniod along axis ?,d ifjufydk esa fu;r /kkjk v LFkkfir SA ;fn ifjufydk esa bldh v{k ds vuqfn'k yksgs dh NM+ izfo"V djok;h tk;srks fuEu esa ls dkSulh jkf'k c



(A*)dsUnz ij pqecdh; {ks=k (B*) ifjufydk ls lEc) pqEcdh; yDl(C*) ifjufydkvksa dk LoizsjdRo (D) twy "ek dh nj

6. Two solenoids have identical geometrical construction but one is made of thick wire and the other of thin wire.Which of the following quantities are different for the two solenoids ?(A) self-inductance (B*) rate of Joule heating if the same current goes through them(C) magnetic field energy if the same current goes through them(D*) time constant if one solenoid is connected to one battery and the other is connected to another batterynks ifjufydkvksa dh T;kferh; cukoV ,d tSlh gS] fdUrq ,d eksVs rkj ls o nwljh irys rkj ls cuk;h x;h gSA nksuksaifjufydkvksa ds fy;s fuEu jkf'k;ksa esa ls dkSulh fHkUu&fHkUu gS -(A) LoizsjdRo (B*) ;fn nksuks a ls leku /kkjk,a izokfgr gksa rks twy "ek dh nj(C) ;fn nksuksa ls leku /kkjk,a izokfgr gks rks pqEcdh; {ks=k dh tkZ(D*) ;fn ,d ifjufydk dk ,d cSVjh ls tksM+k tk;s rFkk nwljh dks] bldh cSVjh ls tksM+k tk;s rks le; fLFkjkad

7. An LR circuit with a battery is connected at t =0. Which of the following quantities is not zero just after theconnection ?(A) current in the circuit (B) magnetic field energy in the inductor(C) power delivered by the battery (D*) emf induced in the inductorfdlh LR ifjiFk esa t =0 ij cSVjh yxk;h x;h gSA la;kstu ds rqjar i'pkr~ fuEu jkf'k;ksa esa ls dkSulh v'kwU; gS -(A) ifjiFk esa /kkjk (B) izsjdRo esa pqEcdh; tkZ(C) izsjdRo esa pqEcdh; tkZ (D*) izsjdRo esa izsfjr fo-ok-cy

8. A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in fig.,d NM+ AB ,d leku osx v ds lkFk fp=k esa n'kkZ;s vuqlkj le:i pqEcdh; {ks=k esa xfr'khy gS -

(A) The rod becomes electrically charged(B*) The endA becomes positively charged(C) The end B become positibely charged(D) The rod becomes hot because of Joule heating(A) NM+ fo|qr vkosf'kr gks tkrh gSA(B*)A fljk /kukosf'kr gks tkrk gSA(C) B fljk /kukosf'kr gks tkrk gSA(D) twy "ek ds dkj.k NM+ xeZ gks tkrh gSA

9. L, C and R represent the physical quantities inductance, capacitance and resistance combinations havedimensions of frequency ?HkkSfrd jkf'k;ksa] izsjdRo] /kkfjrk ,oa izfrjks/k dks e'k% L, C ,oa R }kjk O;Dr fd;k tkrk gSA fuEu esa ls fdl dh foek,avkof`k dh gS -

(A*)1

RC (B*)RL (C*)

1LC (D) C/L

10. The switches in fig. and are closed at t = 0 and reopened after a long time at t = t0.

(A) The change on C just after t = 0 is C. (B*) The change on C long after t = 0 is C.(C*) The current in L just before t = t0 is /R (D) The current in L long after t = t0 is /Rfp=kksa (a) o (b) esa fLopksa dks t = 0 ij can fd;k x;k gS] vkSj ,d yEcs le; t = t0ds i'pkr~ iqu% [kksyk x;k gS -



(A) t = 0ds rqjar i'pkr~ C ij vkos'k C gSA (B*) t = 0 ds yEcs le; i'pkr~ C ij vkos'k C gSA(C*) t = t0 ds rqjar igys L ls izokfgr /kkjk /R gSA (D) t = t0ds cgqr le; i'pkr~ L ls izokfgr /kkjk /R gSA

WORKED OUT EXAMPLES

EXERCISE1. Calculate the dimensions of

(a) .ld.E

(b) vBL and

(c) dtd B . The symbols have their usual meanings.

(a) .ld.E

(b) vBL rFkk (c) dtd B dh foekvksa dh x.kuk dhft;sA ladsrksa ds lkekU; vFkZ gSA

Ans. M L2 I1 T3 in each case.

2. The flux of magnetic field through a closed conducting loop changes with time according to the equationat2 + bt + c. (a) Write the SI units of a, b, and c. (b) If the magnitudes of a, b and c are 0.20 , 0.40,0.60 respectively, find the induced emf at t = 2 s.,d can pkyd ywi ls xqtjus okys pqEcdh; {ks=k dk yDl le; ds lkFk lehdj.k at2 + bt + c ds vuqlkjifjofrZr gksrk gSA (a) a, b ,oa c ds S ek=kd fyf[k;sA (b) ;fn a, b ,oa c ds ifjek.k e'k% 0.20 , 0.40 ,oa 0.60gS rks t = 2 lsd.M ij izsfjr fo-ok-cy Kkr dhft;sAAns: 1.2 Volt

3. (a) The magnetic field in a region varies as shown in figure. Calculate the average induced emf in aconducting loop of area 2.0 10 3 m 2 placed perpendicular to the field in each of the 10 ms intervalsshown. (b) In which intervals is the emf not constant? Neglect the behavior near the ends of 10 msintervals.(a) fdlh LFkku ij pqEcdh; {ks=k fp=kkuqlkj ifjofrZr gksrk gSaA {ks=k ds yEcor~ j[ks gq, 2.0 10 3 m2 {ks=kQy okyspkyd ywi esa] n'kkZ;s x;s izR;sd 10 feyh lsd.M varjky ds fy;s izsfjr vkSlr fo-ok-cy dh x.kuk dhft;sA (b) fdlhvarjky ds fy;s fo-ok-cy fu;r ugha gS\ 10 feyh lsd.M varjky ds lehi O;ogkj dks ux.; eku yhft;sA

0.01

0.03

B(T)

10 20 30 40 t(ms)

Ans: (a)2.0 mV, 4.0 mV, 4.0mV, 2.0mV (b)10 ms to 20 ms and 20 ms to 30 ms.4. A conducting circular loop having a radius of 5.0 cm, is placed perpendicular to a magnetic field of 0.50

T .It is removed from the field in 0.50 s. Find the average emf produced in the loop during this time.5.0 lseh f=kT;k okyk ,d pkyd o`kkdkj ywi] 0.50 T pqEcdh; {ks=k ds yEcor~ j[kk x;k gSA bldks 0.50 lsd.M esa{ks=k ds ckgj fudky fy;k tkrk gSA bl le;karj esa ywi esa izsfjr vkSlr fo-ok-c- Kkr dhft;sA



Ans: 7.8 X 10-3 V5. A conducting circular loop of area 1 mm2 is placed 20 cm from it. The straight wire at a distance of 20

cm from it. The straight wire at a distance of 20 cm from it. The straight wire carries an electric currentwhich changes from 10A to zero in 0.1 s. Find the average emf induced in the loop in 0.1s. Find theaverage emf induced in the loop in 0.1s.1 feeh2 {ks=kQy okys pkyd o`kkdkj ywi dks blls 20 lseh nwj fLFkr yEcs ,oa lh/ks rkj ds leryh; j[kk x;k gSAlh/ks rkj ls izokfgr /kkjk 0.1 lsd.M esa 10 Als 'kwU; rd ifjofrZr gks tkrh gSA ywi esa 0.1 lsd.M esa izsfjr vkSlrfo-ok-c- Kkr dhft;sA

6. A square- shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicularto a 1.0 T magnetic field. It is removed from the magnetic field in 0.25 s and restored in its original placein the next 0.25 s. Find the magnitude of the average emf induced in the coil during(a) its removal, (b) its restoration and(c) its motion (during removal and restoration).oxkZdkj vkfr okyh ,d dq.Myh dh Hkqtk,a 50 lseh yEch gS vkSj blesa 50 Qsjs gSaA bldks 1.0 T pqEcdh; {ks=k dsyEcor~ j[kk x;k gSA bldks 0.25 lsd.M esa pqEcdh; {ks=k ls gVk fy;k tkrk gS] vkSj vxys 0.25 lsd.M esa iqu% bldsiwoZ LFkku ij j[k fn;k tkrk gSA ywi esa izsfjr vkSlr fo-ok-cy Kkr dhft;s %(a) bldks gVkrs gq,A (b) bldks iqu% j[krs gq,A (c) bldh xfr esaAns: (a) 50 V (b) 50 V (c) zero 'kwU;

7. Suppose the resistance of the coil in the previous problem is 25 Assume that the coil moves withuniform velocity during its removal and restoration .Find the thermal energy developed in the coil during(a) its removal, (b) its restoration and (c) its motion.ekuk fd fiNys iz'u esa dq.Myh dk izfrjks/k 25 gSA ;g eku fyft;s fd gVkrs gq, ,oa iqu% j[krs gq, dq.Myh ,dleku osx ls xfr djrh gSA dq.Myh esa mRiUu "ek tkZ Kkr dhft;s % (a) bldks gVkrs gq,A (b) bldks iqu% j[krsgq, vkSj (c) bldh xfr esaAAns: (a) 25 J (b) 25 J (c) 50 J

8. A conducting loop of area 5.0 cm2 is placed in a magnetic field which varies sinusoidally with time asB = B0 sin t where B0 = 0.20 T and = 300 s

1. The normal to the coil makes an angle of 60 with thefield. Find the (a) the maximum emf induced in the coil, (b) the emf induced at = (/900)s and (c) theemf induced at t = (/600)s.5.0 lseh2 {ks=kQy okyk pkyd ywi] ,d pqEcdh; {ks=k esa fLFkr gS tks fd B = B0 sin t ds vuqlkj le; ds lkFkT;koh; :i esa ifjofrZr gks jgk gS] tgk B0 = 0.20 T vkSj = 300 izfr lsd.M gSA dq.Myh ij vfHkyEc] {ks=k dslkFk 60 dks.k cukrk gSA Kkr dhft;s (a) dq.Myh esa izsfjr vf/kdre fo-ok-cy (b) = (/900) lsd.M ij izsfjr fo-ok-cy vkSj (c) t = (/600) lsd.M ij izsfjr fo-ok-cyAns. (a) 0.015 V, (b) 7.5 103 V, (c) zero

9. Figure shows a conducting squares loop placed parallel to the pole-faces of a ring magnet. The pole-faces have an area of 1 cm 2 each and the field between the poles is 0.10 T. The wires making the loopare all outside the magnetic field. If the magnet is removed in 1.0 s, what is the average emf induced inthe loop?fp=k esa n'kkZ;k x;k gS fd ,d oxkZdkj pkyd ywi ,d fjax&pqEcd ds /kzqo&Qydksa ds lekUrj j[kk x;k gSA /kzqo&Qydksaesa izR;sd dk {ks=kQy 1 lseh2 gS] vkSj /kzqoksa ds chp {ks=k 0.10 T gSA ywi fufeZr djus okyk lEiw.kZ rkj pqEcdh; {ks=kds ckgj gSA ;fn pqEcd dks 1.0 lsd.M esa gVk fy;k tkrk gS] rks ywi esa izsfjr vkSlr fo-ok-cy fdruk gS\

Ans. 10 V.

10. A conducting square loop having edges of length 2.0 cm is rotated through 180 about a diagonal in0.20 s. A magnetic field B exists in the region which is perpendicular to the loop in its initial position.If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field.



2.0 lseh Hkqtkvksa okyk ,d oxkZdkj pkyd ywi blds ,d fod.kZ ds ifjr% 0.20 lsd.M esa 180 dks.k ls ?kqek fn;ktkrk gSA ogk ij ,d pqEcdh; {ks=k B fo|eku gS] ftldh fn'kk ywi izkjfEHkd fLFkfr esa blds ry ds yEcor~ gSA ;fn?kw.kZu esa izsfjr vkSlr fo-ok-cy 20 mV gS] rks pqEcdh; {ks=k dk ifjek.k Kkr dhft;sAAns. 5.0 T

11. A conducting loop of face area A and resistance R is placed perpendicular to a magnetic field B. Theloop is withdrawn completely from the field. Find the charge which flows through any cross section ofthe wire in the process. Note that it is independent of the shape of the loop as well as the away it iswithdrawn.Qyd {ks=kQy A o izfrjks/k R okyk ,d pkyd ywi pqEcdh; {ks=k Bds yEcor~ j[kk gqvk gSA ywi dks pqEcdh; {ks=kls iw.kZr;k gVk fy;k tkrk gSA bl izf;k esa rkj ds fdlh Hkh vuqizLFk dkV ls izokfgr gksus okyk vkos'k Kkr dhft;sA/;ku jf[k;s fd ;g ywi dh vkfr ,oa lkFk gh bldks ckgj gVkus ds rjhds ij fuHkZj ugha djrk gSAAns. BA/R

12. A long solenoid of radius 2 cm has 100 turns/ cm and carries a current of 5A. A coil of radius 1 cmhaving 100 turns and a total resistance of 20 is placed inside the solenoid coaxially .The coil isconnected is reversed galvanometer. If the current in the solenoid is reversed in direction find thecharge flown through the galvanometer.,d yEch ifjufydk dh f=kT;k 2 lseh f=kT;k ,oa blesa 100 Qsjs@lseh gS rFkk blls 5A /kkjk izokfgr gks jgh gSA 1lseh f=kT;k vkSj 100 Qsjksa rFkk dqy izfrjks/k 20 okyh ,d dq.Myh ifjufydk esa lek{kr% j[kh gqbZ gSA dq.Myh ls,d /kkjkekih tqM+k gqvk gSA ;fn ifjufydk esa /kkjk dh fn'kk mRfer dj nh tk;s rks /kkjkekih ls izokfgr vkos'k Kkrdhft;sAAns. 2 104 C

13. Figure shows a metallic square frame of edge a in a vertical plane. A uniform magnetic field B exists inthe space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners ofthe square to deform it into a rhombus. They starts pulling the corners at t = 0 and displace the cornersat a uniform speed u. (a) Find the induced emf in the frame at the instant when the angles at thesecorners reduce to 60. (b) Find the induced current in the frame at this instant if the total resistance ofthe frame is R. (c) Find the total charge which flows through a side of the frame by the time the squareis deformed into a straight line.fp=k esa m/okZ/kj ry esa fLFkr ,d oxkZdkj /kkfRod se iznf'kZr dh x;h gS] ftldh Hkqtk a gSA ogk ij fp=k esa ryds yEcor~ le:i pqEcdh; {ks=k B fo|eku gSA nks yM+ds oxkZdkj se oxkZdkj se ds foijhr dksuksa dks [khap jgs gSaftlls fd ;g lekUrj prqHkZqtkdkj gks tk;sA os dksuksa dks t = 0 ij [khapuk izkjEHk djrs gSa vkSj dksuksa dks ,d lekupky u ls foLFkkfir djrs gSaA (a) se esa izsfjr fo-ok-c- ml {k.k ij Kkr dhft;s tc bu dksuks a ds dks.k 60 rdde gks tkrs gSaA (b) ;fn se dk dqy izfrjks/k R gS] rks bl {k.k ij se esa izsfjr /kkjk Kkr dhft;sA (c) oxZ dslh/kh js[kk rd fo:fir gksus rd fdlh ,d Hkqtk ls izokfgr dqy vkos'k Kkr dhft;sA

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14. The north pole of a magnet is brought down along the axis of a horizontal circular coil (figure ). As aresult the flux through the coil changes from 0.35 weber to .85 weber in an interval of half a second.Find the average emf induced during this period Is the induced current clockwise or anticlockwise asyou look into the coil from the side of the magnet?,d pqEcd dk mkjh /kz qo fp=k esa n'kkZ;s vuqlkj {kSfrt dq.Myh dh v{k ds vuqfn'k uhps yk;k tkrk gSA bldsifj.kkeLo:i dq.Myh ls xqtjus okyk yDl vk/ks lsd.M esa 0.35 oscj ls .85 oscj rd ifjofrZr gksrk gSA bl dkyesa izsfjr vkSlr fo-ok-cy Kkr dhft;sA ;fn vki pqEcd dh vksj ls dq.Myh esa ns[ksaxs rks izsfjr /kkjk nf{k.kkorhZ gksxh;k okekorhZ\



Ans: 1.0 V ,anticlockwise15. A wire - loop confined in a plane is related in its own plane with some angular velocity. A uniform

magnetic field exist in the region. Find the emf induced in the loopfdlh ry esa jgus ds fy;s ck rkj dk ,d ywi blds ry esa dqN dks.kh; osx ls ?kwf.kZr gks jgk gSA bl {ks=k esa ,dle:i pqEcdh; {ks=k fo|eku gSA ywi esa izsfjr fo-ok-c- Kkr dhft;sAAns. zero

16. Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s Thefront edge enters the 20 cm wide magnetic field at t = 0 Find the emf induced in the loop at(a) t = 2s, (b) t = 10s, (c) t = 22s and (d) t = 30s.fp=k esa iznf'kZr 5 lseh Hkqtk okyk oxkZdkj ywi nk;ha vksj 1 lseh@ls fu;r pky ls xfr'khy fd;k tk jgk gSA bldklkeus okyk fdukjk t = 0 ij 20 lseh pkSM+s pqEcdh; {ks=k esa izos'k djrk gSA ywi esa izsfjr fo-ok-cy (a) t = 2ls,(b) t = 10 ls, (c) t = 22 ls- rFkk (d) t = 30 ls- ij Kkr dhft;sA

Ans: (a) 3 104 V (b) zero (c) 3 104 V (d) zero17. Find the total heat produced in the loop of the previous problem during the interval 0 to 30 s if the

resistance of the loop is 4.5 m fiNys iz'u esa ;fn ywi dk izfrjks/k 4.5 m gS] rks 0 ls 30 ls- le;kUrjky esa mRiUu dqy "ek Kkr dhft;sAAns: 2 104 J

18 A uniform magnetic field B exist in a cylindrical region or radius 10 cm as shown in figure. A uniformwire of length 80 cm and resistance 4.0 is bent into a square frame and is placed with one side alonga diameters of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s findthe current induced in the frame.fp=k esa n'kkZ;s vuqlkj 10 lseh f=kT;k ds csyukdkj {ks=k esa le:i pqEcdh; {ks=k B fo|eku gSA 80 lseh yEcs vkSj4.0 izfrjks/k okys ,d le:i rki dks oxkZdkj se ds :i esa eksM+dj] bl izdkj j[kk x;k gS fd bldh ,d Hkqtkcsyukdkj {ks=k ds O;kl ds vuqfn'k gSA ;fn pqEcdh; {ks=k 0.010 VsLyk@ls dh fu;r nj ls c



abcd vkSj defa dh izR;sd Hkqtk dh yEckbZ 1.00 lseh vkSj izfrjks/k 4.00 gSA ad rkj esa /kkjk ifjek.k ,oa fn'kk Kkrdhft;s] ;fn (a) fLop S1 can gks fdUrq S2 [kqyk gS] (b) S1 [kqyk gks fdUrq S2 can gks] (c) S1 ,oa S2 nksuksa gh [kqys gksaAvkSj (d) S1 ,oa S2 nksuks a gh can gksaA [6 min.] [M.Bank(07-08)_HCV_Ch.38_Ex._19]

Ans. (a) 1.25 107 A, a to d (b) 1.25 107 A, d to a. (c) zero (d) zero20. Figure shows a circular coil of N turns and radius a, connected to a battery of emf through a rheostat.

The rheostat has a total length L and resistance R. The resistance of the coil is r is A small circularloop of radius a and resistance r is placed coaxially with the coil. The centre of the loop is at adistance x from the centre of coil . The In the beginning the sliding contact of the rheostat is at the left.end and then sliding contact of the rheostat is at the left end and then onwards it is moved towards rightat a constant speed sym Find the emf induced in the small circular loop at the instant(a) the contact begins to slide and (b) it has sild through half the length of the rheostat.fp=k esa N Qsjksa o a f=kT;k okyh o`kkdkj dq.Myh iznf'kZr dh x;h gS] bldks ,d /kkjk fu;a=kd }kjk fo-ok-c- dh cSVjhls tksM+k x;k gSA /kkjk fu;a=kd dh dqy yEckbZ L ,oa izfrjks/k R gSA r f=kT;k ,oa a izfrjks/k okyk ,d NksVk o`kkdkjywi] dq.Myh ds lkFk lek{kr% j[kk gqvk gSA ywi dk dsUnz] dq.Myh ds dsUnz ls x nwjh ij fLFkr gSA izkjEHk esa] /kkjk fu;a=kddk foliZd ck;sa fljs ij gS vkSj blds i'pkr~ bldks nk;ha vksj fu;r pky ls f[kldk;k tkrk gSA NksVs o`kkdkj ywiesa izsfjr fo-ok-cy ml {k.k ij Kkr dhft;s tc % (a) foliZd fQlyuk izkjEHk djrk gS vkSj (b) foliZd vk/kh yEckbZrd f[kld pqdk gksA

Ans. 22/32222

0

)r'R()xa(L2Rv'aNa

where R = R for part (a) and R/2 for part (b)

21. A circular coil of radius 2.00 cm has 50 turns .A uniform magnetic field B = 0.200 T exists in the spacein a direction parallel to the axis of the loop The coil is now rotated about a diameter through an angleof 60 The operation takes 0.100s. (a) Find the average of emf induced in the coil .(b) If the coil a isclosed one with the two ends joined together and has a resistance of 4.00 calculate the net chargecrossing a cross- section of the wire of the coil.2.00 lseh f=kT;k okyh o`kkdkj dq.Myh esa 50 Qsjs gSaA ywi dh v{k ds lekUrj fn'kk esa ,d le:i pqEcdh; {ks=kB = 0.200 VsLyk fo|eku gSA vc dq.Myh dks blds fdlh O;kl ds ifjr% 60 dks.k ls ?kqek;k tkrk gSA bl izf;kesa 0.100 lsd.M yxrs gSaA (a) dq.Myh esa izsfjr vkSlr fo-ok-cy Kkr dhft;sA (b) ;fn dq.Myh can gSA nksuksa fljsijLij tqM+s gq, vkSj bldk izfrjks/k 4.00 gS] rks dq.Myh ds rkj ds fdlh vuqizLFk dkV ls izokfgr dqy vkos'k dhx.kuk dhft;sAAns. (a) 6.28 103 V (b) 1.57 103 C

22. A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 10 4 about a diameterwhich is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. Thearea of the coil is 25cm 2 and its resistance is 4.0 . Find (a) the average emf developed in half a turnfrom position where the coil is perpendicular to the magnetic field (b) the average emf on a full turn and(C) the net charge displaced in part (a).



,d dq.Myh esa 100 Qsjs gSa] bldks le:i pqEcdh; {ks=k esa fdlh O;kl ds ifjr% ?kwf.kZr fd;k tkrk gS] tks {ks=k ds yEcor~gSA ?kw.kZu dk dks.kh; osx 300 pDdj izfr fefuV gSA dq.Myh dk {ks=kQy 25 lseh2 ,oa bldk izfrjks/k 4.0 gSA Kkrdhft;s % (a) tc dq.Myh ds pqEcdh; {ks=k ds yEcor~ fLFkfr ls vk/kk pDdj yxkus ij mRiUu vkSlr fo-ok-cyA(b) iwjs pDdj esa vkSlr fo-ok-cy (C) Hkkx (a) esa foLFkkfir dqy vkos'kAns. (a) 2.0 103 V (b) zero (c) 5.0 105 C

23. A coil of radius 10 cm and resistance 40 has 1000 turns It is placed with its plane vertical and its axisparalleled to the magnetic meridian The coil is connected to a galvanometer and is rotated about if thehorizontal component of the earth magnetic field is BH = 3.0 10

5 T.10 lseh f=kT;k ,oa 40 izfrjks/k okyh dq.Myh esa 1000 Qsjs gSaA bldh v{k pqEcdh; ;kE;kskj ds lekUrj ,oa rym/okZ/kj fLFkfr esa j[kk x;k gSA dq.Myh ls,d /kkjkekih tqM+k gqvk gS vkSj bldks blds m/oZ O;kl ds ifjr% 180 dks.kls ?kqek;k tkrk gSA ;fn i`Foh ds pqEcdh; {ks=k dk {kSfrt ?kVd BH = 3.0 10 5 VsLyk gS rks /kkjkekih ls izokfgrvkos'k Kkr dhft;sAAns. 4.7 105 C

24. A circular coil of one turns of radius 5.0 cm is rotated about a diameter with a constant angular speedof 80 revolutions per minute A uniform magnetic field revolutions per minute A uniform magnetic fieldB = 0.010 rotation Find a direction perpendicular to the axis of rotation Find (a) the maximum emfinduced. (b) the average emf induced in the coil over a long period and (c) the average of the squaresof emf induced over a long period.,d Qsjs okyh o`kkdkj dq.Myh dh f=kT;k 5.0 lseh gS] bldks O;kl ds ifjr% 80 pDdj@fefuV dh fu;r dks.kh; pkyls ?kwf.kZr fd;k tk jgk gSA ?kw.kZu v{k ds yEcor~ fn'kk esa le:i pqEcdh; {ks=k B = 0.010 VsLyk fo|eku gSA Kkrdhft;s % (a) vf/kdre izsfjr fo-ok-cy (b) yEcs dky esa dq.Myh esa izsfjr vkSlr fo-ok-cy rFkk (c) yEcs dky esa izsfjrfo-ok-cy ds oxZ dk vkSlrAns. (a) 6.6 104 V (b) zero (c) 2.2 107 V2

25. Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Neglectingthe resistance of the coil find the heat produce in the circuit in one minute.eku yhft;s fd fiNys iz'u esa of.kZr dq.Myh ds fljksa dks 100 izfrjks/k ls tksM+ fn;k x;k gS] dq.Myh dkizfrjks/k ux.; ekudj ifjiFk esa ,d fefuV esa mRiUu "ek Kkr dhft;sAAns. 1.3 107 J

26. Figure shows a circular wheel of radius 10.0 cm whose upper half showndark in the figure is made of iron and the lower half of wood the twojunctions are joined by an iron rod A uniform magnetic field B ofmagnitude 2.00 104 exists in the space above the central line assuggested by the figure The wheel is set into pure rolling on thehorizontal surface it takes 2.00 seconds for the iron part to come downand the wooden part to go up find the average emf induced during thisperiod

fp=k esa 10.0 lseh f=kT;k dk o`kkdkj ifg;k n'kkZ;k x;k gS] ftldk ijh vk/kk Hkkx tks xgjk gSA yksgs dk cuk gS vkSjuhps okyk Hkkx ydM+h dk gSA nksuksa laf/k;ksa dks yksgs dh NM+ ls tksM+k x;k gSA tSlkfd fp=k esa iznf'kZr gS] dsfUnz; js[kkds ij okys Hkkx esa le:i pqEcdh; {ks=k B fo|eku gS] ftldk ifjek.k 2.00 104 VsLyk gSA ifg;k {kSfrt lrgij 'kq) yksVuh xfr dj jgk gSA ;fn yksgs okys Hkkx dks uhps vkus esa vkSj ydM+h okys Hkkx dks ij tkus esa 2.00lsd.M yxrs gSa rks bl dky esa izsfjr vkSlr fo-ok-cy Kkr dhft;sAAns. 1.57 106 V

27. A 20 cm long conducting rod is set pure translation with a uniform velocity of 10 cm/s perpendicular toits lengths. A uniform magnetic filed of magnitude 0.10 T exists in a direction perpendicular to the planeof motion . (a) Find the average magnetic force on the free electrons of the rod. (b) for what electronicfield inside the rod the electronic force on a free electron will balance the magnetic force ? How is thiselectronic field created ? (c) Find the motional emf between the ends of the rod.20 lseh yEch ,d pkyd NM+ dks bldh yEckbZ ds yEcor~ 10 lseh@ls ds ,d leku osx ds lkFk 'kq) LFkkukarj.kxfr'khy fd;k tkrk gSA xfr ds ry ds yEcor~ fn'kk esa 0.10 VsLyk ifjek.k dk le:i pqEcdh; {ks=k fo|eku gSA(a) NM+ ds eqDr bysDVkWuksa ij vkSlr pqEcdh; cy dk ifjek.k Kkr dhft;sA (b) NM+ ds vUnj fdrus fo|qr {ks=k dsfy;s eqDr bysDVkWuksa ij pqEcdh; cy larqfyr gks tk;sxk\ ;g fo|qr {ks=k fdl izdkj mRiUu fd;k tk;sxk\ (c) NM+ds fljksa ij xfrd fo-ok-cy Kkr dhft;sAAns. (a) 1.6 1021 N (b) 1.0 102 V/m (c) 2.0 103 V



28. A metallic metre sticks moves with a velocity of 2 m/ s in a direction perpendicular to its length andperpendicular to a uniform magnetic field of magnitude 0.2 T Find the emf induced between the endsof the stick.,d /kkfRod ehVj NM+ bldh yEckbZ ds yEcor~ fn'kk esa rFkk 0.2 VsLyk ifjek.k ds le:i pqEcdh; {ks=k ds yEcor~2 eh@ls osx ds lkFk xfr'khy gSA NM+ ds fljksa ds chp izsfjr fo-ok-cy Kkr dhft;sAAns: 0.4 V

29. A 10m wide spacecraft moves through the interstellar space at a speed 3 10 7 m/s magnetic field B =3 10 10 T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor , calculate the emf induced across its width.10 eh pkSM+k varfj{k ;ku varj rkjdh; vkdk'k esa 3 10 7 eh@ls pky ls xfr dj jgk gSA varfj{k esa ;ku dh xfrds ry ds yEcor~ B = 3 10 10 VsLyk pqEcdh; {ks=k fo|eku gSA varfj{k ;ku dks pkyd ekurs gq, bldh pkSM+kbZds chp izsfjr fo-ok-cy dh x.kuk dhft;sAAns: 0.9 V

30. The two rails of a railway track insulated from each other and from the ground are connected to a milli-voltmeter. What will be the reading of the milli-voltmeter when a train travels on the track at a speed of180 km/h.? The vertical component of earths magnetic field is 0.2 10 4 T and the rails are separatedby 1m,d jsy iFk dh nks iVfj;k ijLij ,oa tehu ds lkFk dqpkfyr gSa] buds lkFk ,d feyh oksYVehVj tksM+k x;k gSA tc,d Vsu bl iFk ij 180 fdeh@?kaVk pky ls xfr'khy gks rks feyh oksYVehVj dk ikB~;kad fdruk gksxk\ i`Foh dspqEcdh; {ks=k ds m/oZ?kVd dk eku 0.2 10 4 VsLyk rFkk iVfj;ksa ds chp dh nwjh 1eh- gSAAns: 1 mV

31. A right angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shownin the figure. A uniform magnetic field B exists in the perpendicular direction. Find the emf induced(a) in the loop abc, (b) in the segment bc, (c) in the segment ac and(d) in the segment ab.,d ledks.k f=kHkqt abc, /kkfRod rkj ls cuk gqvk gS vkSj blds ry ds vuqfn'k fp=k esa n'kkZ;s vuqlkj ,d leku pkyv ds lkFk xfr'khy gSA yEcor~ fn'kk esa le:i pqEcdh; {ks=k B fo|eku gSA izsfjr fo-ok-cy Kkr dhft;s %(a) ywi abc esa , (b) Hkkx bc esa , (c) Hkkx ac esa vkSj (d) Hkkx ab esa

v

b

B

a

c

Ans: (a) zero, (b) vB (bc), positive at c, (c) zero(d) vB (bc), positive at a

32. A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constantvelocity v .A uniform magnetic field B exist in the direction perpendicular to the plane of the wire. Findthe emf induced between the ends of the wire if (a) the velocity is perpendicular to the diameter joiningfree ends (b) the velocity is parallel to this diameter.rkacs dk ,d rkj v)Zo`kkdkj vkfr esa eksM+k x;k gS] bldh f=kT;k r gS vkSj ;g blds ry esa fu;r osx v ls foLFkkfirgks jgk gSA rkj ds ry ds yEcor~ fn'kk esa le:i pqEcdh; {ks=k B fo|eku gSA rkj ds fljksa ij izsfjr fo-ok-cy Kkrdhft;s] ;fn (a) osx] eqDr fljksa dks tksM+us okys O;kl ds yEcor~ gS] (b) osx] bl O;kl ds lekUrj gSAAns: (a) 2rvB (b) zero 'kwU;

33. A wire of length 10 cm translates in a direction making an angle of 60 with its length. The plane ofmotion is perpendicular to a uniform magnetic field of 1.0 T that exists in the space Find the emfinduced between the ends of the rod if the speed of translation is 20 cm/s.10 lseh yEck rkj bldh yEckbZ ls 60 dks.k cukrs gq, LFkkukarfjr gks jgk gSA bldh xfr dk ry ml LFkku ij fo|eku1.0 VsLyk rhozrk ds pqEcdh; {ks=k ds yEcor~ gSA ;fn LFkkukarj.k dh pky 20 lseh@ls gS] rks NM+ ds fljksa ds chp izsfjrfo-ok-cy Kkr dhft;sA



Ans: 17 103V34. A circular copper-ring of radius r translates in its plane with a constant velocity v. A uniform magnetic field B

exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametricallyopposite points on the ring. (a) Between which pair of points is the emf? (b) Between which pair of points isthe emf minimum? What is the value of this minimum emf?r f=kT;k okyh rkacs dh oy; blds ry esa fu;r osx vls LFkkukarfjr gks jgh gSA oy; ds ry ds yEcor~ le:i pqEcdh;{ks=k B fo|eku gSA oy; ij O;klr% vfHkeq[k fcUnqvksa ds fofHkUu ;qXeksa ij fopkj dhft;sA (a) fcUnqvksa ds fdl ;qXeds fy, fo-ok-c- vf/kdre gksxk\ bl vf/kdre fo-ok-cy dk eku fdruk gksxk\ (b) fcUnqvksa ds fdl ;qXe ds fy;s fo-ok-cy U;wure gksxk\ bl U;wure fo-ok-cy dk eku fdruk gksxk\Ans. (a) at the ends of the diameter perpendicular to the velocity, 2rvB (b) at the ends of the diameterparallel to the velocity, zero.

35. Figure shows a wire sliding on two parallel, conducting rails placed at a separation . A magnetic field Bexists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire movingat a constant velocity v?fp=k esa n'kkZ;k x;k gS fd nwjh ij fLFkr nks lekUrj pkyd iVfj;ksa ds ij ,d rkj fQly jgk gSA iVfj;ks dsry ds yEcor~ fn'kk esa pqEcdh; {ks=k B fo|eku gSA rkj dks fu;r osx vls xfr'khy j[kus ds fy;s fdruk cy vko';dgS ?

v

Ans. zero36. Figure shows a long U-shaped wire of width placed in a perpendicular magnetic field B.A wire of length is

slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unitlength. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuitdiagram, showing the induced emf as a battery. Calculate the current in the circuit.fp=k esa n'kkZ;k x;k gS fd pkSM+kbZ okyk U-vkfr dk rkj yEcor~ pqEcdh; {ks=k B esa j[kk gqvk gSA U-vkkdj okysrkkj ij ,d yEckbZ okyk rkj nk;ha vksj fu;r osx vds lkFk xfr'khy gSA leLr rkjksa dh ,dkad yEckbZ dkizfrjks/k r gSA t = 0, ij] f[kldus okyk rkj U-vkdkj okys rkj ds ck;sa fdukjs ds lehi gSA izsfjr fo-ok-cy dks cSBjhds :i esa n'kkZrs gq, ,d rqY; ifjiFk cukb;sA ifjiFk esa /kkjk dh x.kuk dhft;sA

v

Ans. (a) )vt(r2vB

(b) /v

37. Consider the situation of the previous problem. (a) Calculate the force needed to keep the sliding wire movingwith a constant velocity v. (b) If the force needed just after t = 0 is F0, find the time at which the force neededwill be F0/2.fiNys iz'u esa of.kZr fLFkfr ij fopkj dhft;sA (a) f[kldus okys rkj dks fu;r osx v ls xfr'khy j[kus ds fy;svko';d cy dh x.kuk dhft;sA (b) ;fn t = 0 lsd.M ds i'pkr~ vko';d cy F0gS rks og le; Kkr dhft;sA ftlij vko';d cy F0/2 gks tk;sxkA

Ans. (a) )vt(r2vB 22

(b) /v.

38. Consider the situation shown in figure. The wire PQ has mass m, resistance r and can slide on the smooth,horizontal parallel rails separated by a distance . The resistance of the rails is negligible. Auniform magneticfield B exists in the rectangular region and a resistance R connects the rails outside the field with a speed v0.Find (a) the current in the loop at an instant when teh speed of the wire PQ is v, (b) the acceleration of the wireat this instant, (c) the velocity v as a function of x and (d) the maximum distance the wire will move,



fp=k esa n'kkZ;h x;h fLFkfr ij fopkj dhft;sA rkj PQdk nzO;eku m, izfrjks/k r gS rFkk ;g l nwjh ij fLFkr nks fpduh,oa lekUrj {kSfrt iVfj;ksa ij fQly ldrk gSA iVfj;ksa dk izfrjks/k ux.; gSA ,d vk;rkdkj ifj{ks=k esa le:ipqEcdh; {ks=k B fo|eku gS vkSj {ks=k ds ckgj iVfj;ksa ls ,d izfrjks/k R tksM+k x;k gSA t = 0 ij] rkj PQdks v0pkyls nk;ha vksj /kdsyk tkrk gSA Kkr dhft;s % (a) ftl {k.k ij PQdh pky v gS] ywi esa izokfgr /kkjk (b) bl {k.k ijrkj dk Roj.k, (c) xds Qyu :i esa osx vkSj (d) og vf/kdre nwjh tks rkj r; djsxkA

P

Q

Ans. (a)rR

vB

(b) )rR(m

vB 22

towards left (c) v = v0 )rR(m

xB 22

(d) 22

0

B)rR(mv

.

39. A rectangular frame of wire abcd has dimensions 32 cm 8.0 cm and a total resistance of 2.0 . It is pulledout of a magnetic field B = 0.020 T by applying a force of 3.2 105 N (figure). It is found that the frame moveswith constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential differencebetween the points a and b and (d) the potential difference between the points c and d.rkj ds ,d vk;rkdkj se abcddh foek,a 32 lseh 8.0 lseh gS vkSj bldk dqy izfrjks/k 2.0 gSA bldks 3.2 105 U;wVu cy yxkdkj pqEcdh; {ks=k B = 0.020 VsLyk ds ckgj [khapk tkrk gSA fp=k ;g izsf{kr gksrk gS fd sefu;r pky ls xfr djrh gSA Kkr dhft;s % (a) ;g fu;r pky, (b)ywi esa izsfjr fo-ok-c- (c) fcUnqvksa a o bds chpfoHkokUrj] rFkk (d) fcUnqvksa c o dds chp foHkokarj

Ans. (a) 25 m/s (b) 4.0 102 V (c) 3.6 102 V (d) 4.0 103 V.40. Figure shows a metallic wire of resistance 0.20 W sliding on a horizontal, U-shaped metallic rail. The

separation between the parallel arms is 20 cm. An electric current of 2.0A passes through the wire when itis slid at a rate of 20 cm/s. If the horizontal component of the earths magnetic field is 3.0 x 105 T, calculatethe dip at the place.fp=k esa iznf'kZr gS fd 0.20 izfrjks/k okyk /kkfRod rkj] {kSfrt ,oa U-vkfr dh /kkfRod iVjh ij fQly jgk gSAlekUrj Hkqtkvksa ds chp 20lseh varjky gSA tc rkj 20 lseh@ls dh nj ls fQlyrk gS rks rkj ls 2.0A /kkjk izokfgrgksrh gSA ;fn i`Foh ds pqEcdh; {ks=k ds {kSfrt ?kVd dk eku 3.0 x 105 T gS rks bl LFkku ij ufr Kkr dhft;sA

Ans. tan1 (1/3)41. A wire ab of length , mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the

bottom as shown in figure. The plane of the rails makes an angle with the horizontal. A vertical magnetic

field B exists in the region. If the wire slides on the rails at a constant speed v, show that B =

22 cosv

sinRmg

,d rkj ab ftldh yEckbZ , nzO;eku m vkSj izfrjks/k R gS] fp=kkuqlkj uhps ls tqM+h gqbZ nks eksVh iVfj;ksa ij fQlyrkgSA iVfj;ksa dk ry {kSfrt ls dks.k curk gSA ;gk ij ,d m/okZ/kj pqEcdh; {ks=k B fo|eku gSA ;fn rkj iVfj;ksa



ij fu;r pky v ls fQlyrk gS] rks O;Dr dhft;s fd B =

22 cosv

sinRmg

ba

B

Ans.42. Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same

speed 5 cm/s. Find the electric current in the 19 resistor if (a) both the wires move towards right and (b) ifP1Q1 moves towards left but P2Q2 moves towards right.fp=k esa n'kkZ;h x;h fLFkfr ij fopkj dhft;sA rkj P1Q1vkSj P2Q2dks iVfj;ksa ij leku pky 5 lseh@ls ds lkFkfQlyk;k tkrk gSA 19 izfrjks/k ls izokfgr /kkjk Kkr dhft;s] ;fn % (a) nksuksa rkj nk;ha vksj xfr djsa] rFkk (b) P1Q1ck;ha vksj xfr djs fdUrq P2Q2 nk;ha vksj xfr djsA

19 2 2 4 cm

P1 P2

Q1 Q2 B=1.0 T

Ans. (a) 0.1 mA (b) zero43. Suppose the 19 resistor of the previous problem is disconnected. Find the current through P2Q2 in the two

situations (a) and (b) of that problem.fiNys iz'u esa] ekukfd 19 izfrjks/k gVk fn;k tkrk gS rks bl iz'u dh fLFkfr;ksa (a) o (b)ds P2Q2 ls izokfgr /kkjkKkr dhft;sAAns. (a) zero 'kwU; (b) 1 mA

44. Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on thethree rails with a constant speed of 5 cm/s. Find the current in the 10 resistor when the switch S is thrownto (a) the middle rail (b) bottom rail.fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA rkj PQdk izfrjks/k ux.; gS vkSj bldks rhu iVfj;ksa ij 5 lseh@ls dh fu;rpky ls fQlyk;k tkrk gSA 10 izfrjks/k esa izokfgr /kkjk ml fLFkfr esa Kkr dhft;s tcfd fLop S : (a) chp okyhiVjh dh vksj gSA (b) uhps okyh iVjh dh vksj gSA

SB=1.0T

Q

P

2 cm

2 cm

10

Ans. (a) 0.1 mA (b) 0.2 mA45. The current generator g, shown in figure, sends a constant current i through the circuit. The wire cd is fixed

and ab is made to slide on the smooth, thick rails with a constant velocity v towards right. Each of thesewires has resistance r. Find the current through the wire cd.fp=k esa iznf'kZr /kkjk lzksr g , ifjiFk esa fu;r /kkjk i izokfgr djrk gSA rkj cd fLFkj gS vkSj abdks nk;ha vksj fu;rpky vls eksVh ,oa fpduh iVfj;ksa ij fQlyk;k tkrk gSA buesa ls izR;sd rkj dk izfrjks/k r gSA rkj cd ls izokfgr/kkjk Kkr dhft;sA



g v

a

b

d

c

i

Ans.r2

vBir

46. The current generator g, shown in figure, sends a constant current i through the circuit. The wire ab has alength and mass m and can slide on the smooth, horizontal rails connected to g. The entire system lies ina vertical magnetic field B. Find the velocity of the wire as a function of time.fp=k esa iznf'kZr /kkjk lzksr g, ifjiFk esa fu;r /kkjk i izokfgr djrk gSA rkj abdh yEckbZ vkSj nzO;eku m gS rFkk;g g dks tksM+us okyh fpduh ,oa {kSfrt iVfj;ksa ij fQly ldrk gSA lEiw.kZ fudk; m/okZ/kj pqEcdh; {ks=k B esa fo|ekugSA rkj dk osx le; ds Qyu :i esa Kkr dhft;sA

g

a

b

d

cAns. iBt/m, away from the generator

47. The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontalmagnetic field B that is perpendicular to the plane of the rails (figure). It is found that the wire stays inequilibrium. If the wire ab is replaced by another wire of double its mass, how long will it take in falling througha distance equal to its length?fiNys iz'u esa of.kZr iVfj;ksa ,oa rkj okyk fudk; {kSfrt pqEcdh; {ks=k B esa m/okZ/kj j[kk tkrk gS] pqEcdh; {ks=k iVfj;ksads ry ds yEcor~ gSA fp=k nsf[k;s ;g ik;k x;k fd rkj lkE;koLFkk esa :dk jgrk gSA ;fn rkj abdks ,d vU;rkj ls izfrLFkkfir dj fn;k tkrk gSA ftldk nzO;eku blls nqxuk gS] ;g bldh yEckbZ ds cjkcj nwjh rd fxjus esafdruk le; yxk;sxk\

g

b a

Ans. g/2 .

48. The rectangular wire-frame, shown in figure, has uniform magnetic field B exists to the left of the frame. Aconstant force F starts pushing the frame into the magnetic field at t = 0. (a) Find the acceleration of theframe when its speed has increased to . (b) Show that after some time the frame will move with a constantvelocity till the whole frame enters into the magnetic field. Find this velocity 0. (c) Show that the velocity at

time t is given by )e1(vv 0mv/Ft0 .

fp=k esa iznf'kZr rkj ds vk;rkdkj se dh pkSM+kbZ d, nzO;eku m , izfrjks/k RvkSj yEckbZ cgqr vf/kd gSA t = 0 ij ,dfu;r cy F,se dks pqEcdh; {ks=k esa /kdsyuk izkjEHk djrk gSA (a)se dh pky v rd c



Fd

Ans. (a)mR

BvRF 22(b) 22B

RF

49. Figure shows a smooth pair of thick metallic rails connected across a battery of emf having a negligibleinternal resistance. A wire ab of length and resistance r can slide smoothly on the rails. The entire systemlies in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire isgiven a small velocity towards right. (a) Find the current in it at this instant. What is the direction of thecurrent ? (b) What is the force acting on the wire at this instant ? (c) Show that after some time the wire abwill slide with a constant velocity. Find this velocity.fp=k esa iznf'kZr gS fd eksVh ,oa fpduh /kkfRod iVfj;ksa ds ;qXe ds lkFk fo-ok-cy ,oa ux.; vkarfjd izfrjks/k okyhcSVjh tqM+h gqbZ gSA yEckbZ ,oa r izfrjks/k okyk rkj iVfj;ksa ij lqxerkiwoZd fQly ldrk gSA lEiw.kZ fudk; {kSfrtry esa fLFkr gS] tgk ij le:i ,oa m/oZ pqEcdh; {ks=k O;kIr gSA fdlh {k.k t ij] rkj dks nk;ha vksj vYi osx iznkufd;k tkrk gSA (a) bl {k.k ij /kkjk Kkr dhft;sA bl /kkjk dh fn'kk fd/kj gksxh\ (b) bl {k.k ij rkj ij yx jgkcy fdruk gS\ (c) O;Dr dhft;s fd dqN le; i'pkr~ rkj ab fu;r osx ls fQlysxkA ;g osx Kkr dhft;sA

a

b

Ans. (a)r1

(E vB), from b to a (b)rB

(E vB) towards right (c)B

E.

50. A conducting wire ab of length , resistance r and mass m starts sliding at t = 0 down a smooth, vertical,thick pair of connected rails as shown in figure. A uniform magnetic field B exists in the space in a directionperpendicular to the plane of the rails. (a) Write the induced emf in the loop at an instant t when the speed ofthe wire is . (b) What would be the magnitude and direction of the induced current in the wire ? (c) Find thedownward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with aconstant velocity. Find this velocity m. (e) Find the velocity of the wire as a function of time. (f) Find thedisplacement of the wire as a function of time. (g) Show that the rate of heat developed in the wire is equal tothe rate at which the gravitational potential energy is decreased after steady state is reached.yEckbZ, r izfrjks/k vkSj m nzO;eku okyk ,d pkyd rkj t = 0 ij fp=k esa iznf'kZr nks tqM+h gqbZ m/okZ/kj ,oa eksVh rFkkfpduh iVfj;ksa ij fQlyuk izkjEHk djrk gSA bl LFkku ij iVfj;ksa ds ry ds yEcor~ fn'kk esa le:i pqEcdh; {ks=kB fo|eku gSA (a) fdlh {k.k t ij tc rkj dk osx gS] izsfjr fo-ok-cy fyf[k;sA (b) rkj esa izsfjr /kkjk dh fn'kk ,oaifjek.k fdruk gksxk\ (c) bl {k.k ij rkj dk uhps dh vksj Roj.k Kkr dhft;sA (d) i;kZIr le; ds i'pkr~ rkj fu;rpky ls xfr djrk gSA ;g fu;r pky mKkr dhft;sA (e) le; ds Qyu :i esa rkj dk osx Kkr dhft;sA (f) le;ds Qyu :i esa rkj dk foLFkkiu Kkr dhft;sA (g) O;Dr dhft;s fd LFkk;h voLFkk izkIr djus ds i'pkr~ rkj esa mRiUu"ek dk eku xq:Roh; fLFkfrt tkZ esa deh ds cjkcj gSA

a b



Ans. (a) vB (b)r

vB, b to a (c) g

mrB 22

v (d) 22Bmgr

(e) vm(1 m

v/gte ) (f) vmt gv 2m (1 mv/gte )

51. A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speedof 100 revolutions per minute. If the length of each spoke is 30.0 cm and the horizontal component of theearths magnetic field is 2.0 x 105 T, find the emf induced between the axis and the outer end of a spoke.Neglect centripetal force acting on the free electrons of the spoke.,d lkbfdy blds LVs.M ij iwoZ&if'pe fn'kk esa [kM+h gqbZ gS vkSj bldk fiNyh ifg;k 100 pDdj@fefuV dks.kh;pky ls ?kwe jgk gSA ;fn izR;sd LikWd dh yEckbZ 30.0 lseh gS vkSj iF`oh ds pqEcdh; {ks=k dk {kSfrt ?kVd2.0 x 105 T gS rks v{k ,oa LikWd ds ckgjh fljs ds chp izsfjr foHkokarj Kkr dhft;sA LikWd ds bysDVkstu ij yxusokyk vfHkdsUnzh; cy ux.; eku yhft;sAAns. 9.4 106 V

52. A conducting disc of radius r rotates with a small but constant angular velocity about its axis. A uniformmagnetic field B exists parallel to the axis of rotation. Find the motional emf between the centre and theperiphery of the disc.r f=kT;k okyh pkyd pdrh bldh v{k ds ifjr% vYi fdUrq fu;r dks.kh; osx ls ?kwe jgh gSA ,d le:i pqEcdh;{ks=k B bldh ?kw.kZu v{k ds lekUrj fo|eku gSA dsUnz ,oa pdrh ds chp xfrd fo-ok-cy Kkr dhft;sA

Ans. Br21 2

53. Figure shows a conducting disc rotating about its axis in a perpendicular magnetic field B. A resistor ofresistance R is connected between the centre and the rim. Calculate the current in the resistor. Does it enterthe disc or leave it at the centre ? The radius of the disc is 5.0 cm, angular speed = 10 rad/s, B = 0.40 Tand R = 10 .yEcor~] pqEcdh; {ks=k B esa viuh v{k ds ifjr% ?kwf.kZr pkyd pdrh fp=k esa n'kkZ;h x;h gSA dsUnz ,oa ifjf/k ds chp,d izfrjks/k RtksM+k x;k gSA izfrjks/k esa izokfgr /kkjk Kkr dhft;sA D;k ;g dsUnz ij pdrh ls ckgj fudy jgh gS;k vUnj izfo"V gks jgh gS\ pdrh dh f=kT;k 5.0 lseh] dks.kh; pky = 10 js/ls, B = 0.40 T vkSj R = 10 . gSA

R

Ans. 0.5 mA, leaves

54. The magnetic field in a region is given by B =

LBk 0

y where L is a fixed length. A conducting rod of length L

lies along the Y-axis between the origin and the point (0, L, 0). If the rod moves with a velocity ivv 0

, find

the emf induced between the ends of the rod.

fdlh LFkku ij pqEcdh; {ks=k B

=L

Bk 0

y }kjk O;Dr fd;k tkrk gS] tgk L ,d fu;r yEckbZ gSA LyEckbZ dh ,d

pkyd NM+ ewy fcUnq ,oa fcUnq (0, L, 0)ds chp Y-v{k ds vuqfn'k j[kh gqbZ gSA ;fn NM+ ivv 0

osx ls xfr'khygS] ds NM+ ds fljksa ds chp izsfjr fo-ok-c- Kkr dhft;sA

Ans.2vB 00

55. Figure shows a straight, long wire carrying a current i and a rod of length coplanar with the wire andperpendicular to it. The rod moves with a constant velocity v in a direction parallel to the wire. The distanceof the wire from the centre of the rod is x. Find the motional emf induced in the rod.,d yEck ,oa lh/kk rkj] ftlls i /kkjk izokfgr gks jgh gS vkSj rkj ds leryh; rFkk blds yEcor~ yEckbZ dh NM+dks fp=k esa iznf'kZr fd;k x;k gSA NM+ rkj ds lekurj fn'kk esa fu;r osx ls xfr'khy gSA NM+ ds dsUnz ls rkj dhnwjh x gSA NM+ esa izsfjr xfrd fo-ok-c- Kkr dhft;sA



x

i

v

Ans.

2iv0 ln

x2x2

56. Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair ofthick metallic rails laid parallel to the wire. At one end the rails are connected by resistor of resistance R. (a)What force is needed to keep the rod sliding at a constant speed v ? (b) In this situation what is the currentin the resistance R ? (c) Find the rate of heat developed in the resistor. (d) Find the power delivered by theexternal agent exerting the force on the rod.fiNys iz'u tSlh gh fLFkfr ij fopkj dhft;s dsoy varj ;g gS fd NM+ ds fljs&rkj ds lekUrj fLFkr eksVh ,oa pkydiVfj;ksa ij fQly jgs gSaA iVfj;ksa ds ,d fljs ij izfrjks/k R tqM+k gqvk gSA (a) NM+ dks fu;r pky vls fQlyrk gqvkj[kus ds fy;s fdrus cy dh vko';drk gksxh\ (b) bl fLFkfr esa izfrjks/k R ls fdruh /kkjk izokfgr gksxh (c)izfrjks/kesa "ek mRiUu gksus dh nj fdruh gksxh\ (d) NM+ ij cy yxkus ds fy;s ck ;qfDr }kjk iznk 'kfDr Kkr dhft;sA

Ans. (a)2

0

x2x2ln

2i

Rv

(b)

x2x2ln

R2iv0 (c)

20

x2x2ln

2iv

R1

(d) same as (c)

57. Figure shows a square frame of wire having a total resistance r placed complanarly with a long, straight wire.The wire carries a current i given by i = i0 sin t. Find (a) the flux of the magnetic field through the square

frame, (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to

20.

fp=k esa iznf'kZr gS fd r izfrjks/k okyk rkj dk ,d oxkZdkj se ,d yEcs ,oa lh/ks rkj ds leryh; j[kk gqvk gSA rkjls i /kkjk izokfgr gks jgh gS] tks i = i0 sin t }kjk O;Dr dh tkrh gSA Kkr dhft;s % (a) oxkZdkj se ls xqtjus okyk

pqEcdh; yDl (b)se esa izsfjr fo-ok-c- vkSj (c) le;karjky 0 ls

20ds fy;s se esa mRiUu "ekA

a

b

i

58. A rectangular metallic loop of length and width b is placed complanary with a long wire carrying a current i(figure). The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and theloop. Calculate the emf induced in the loop when the rear end of the loop is at a distance a from the wire.Solve by using Faradays law for the flux through the loop and also by replacing different segments withequivalent batteries.yEckbZ vkSj bpkSM+kbZ okyk ,d vk;rkdkj /kkfRod ywi ,d yEcs ,oa lh/ks rkj ds leryh; j[kk gqvk gS] rkj ls i /kkjk izokfgrgks jgh gSA ywi dks] ywi ,oa rkj ds ry esa rFkk rkj ds yEcor~ v osx ls xfr'khy fd;k tkrk gSA tc ywi dk fiNyk fljkrkj ls a nwjh ij gks] rc ywi esa izsfjr fo-ok-c- dh x.kuk dhft;sA ywi ls xqtjus okys yDl ds fy;s QSjkMs dk fu;e izLrqrdjrs gq, rFkk lkFk gh fofHkUu [k.Mksa dks lerqY; cSVfj;ksa ls izfrLFkkfir djrs gq,] gy dhft;sA

b

a

i

v



Ans. )a(a2vbi

59. Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. Athick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre Oand a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middlepoint of the rod OAperpendicularly, so that the rod rotates clockwise at a uniform angular velocity. Find theforce.fp=k esa iznf'kZr gS fd a f=kT;k okyk o`kkdkj pkyd ywi yEcor~ le:i pqEcdh; {ks=k B esa j[kk gqvk gSA /kkrq dh,d eksVh NM+ dk nwljk fljk ywi ds Aij Li'kZdj jgk gSA dsUnz O ,oa ywi ij ,d fLFkj fcUnq Cdks R izfrjks/k okysrkj OC ls la;ksftr fd;k x;k gSA NM+OAds e/; fcUnq ij ,d cy yEcor~ yxk;k tkrk gS] ftlls NM+ nf{k.kkorhZfn'kk esa ,d leku dks.kh; osx ls ?kwerh gSA cy Kkr dhft;sA

FO A

C

Ans.R2Ba 23

to the right of OA in the figure.

60. Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C haszero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA ismade to rotate with a uniform angular speed as shown in the figure. Find the current in the rod when

90AOC .fiNys iz'u ds fp=k esa of.kZr fLFkfr ij fopkj dhft;sA eku yhft;s fd la;kstu rkj O o Cdk izfrjks/k 'kwU; gS] fdUrqo`kkdkj ywi dk izfrjks/k R gS vkSj ;g bldh yEckbZ ij ,d leku :i ls forfjr gSA NM+ OAdks fp=kkuqlkj fu;rdks.kh; osxls ?kqek;k tkrk gSA tc 90AOC gks rks NM+ esa /kkjk Kkr dhft;sA

Ans.R

Ba38 2

61. Consider a variation of the previous problem (figure). Suppose the circular loop lies in a vertical plane. The rodhas a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has aresistance R. The rod is made to rotate with a uniform angular velocity in the clockwise direction byapplying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force whenthe rod makes an angle with the vertical.fiNys iz'u esa ,d ifjorZu ij fopkj dhft;sA eku yhft;s fd o`kkdkj ywi m/oZ ry esa fLFkr gSA NM+ dk nzO;ekum gSA NM+ ,oa ywi dk izfrjks/k ux.; gS] fdUrq la;kstu rkj O o Cdk izfrjks/k R gSA OAds e/; fcUnq ij] NM+ dsyEcor~ ,d cy yxkdj NM+ dks ,d leku dks.kh; osx ls nf{k.kkorhZ fn'kk esa ?kwf.kZr fd;k tkrk gSA tc NM+ m/oZ ls dks.k cukrh gS] rks bl cy dk ifjek.k Kkr dhft;sA

Ans.R2Ba 23

mg sin

62. Figure shows a situation similar to the previous problem. All parameters are the same except that a batteryof emf and a variable resistance R are connected between O and C. The connecting wires have zeroresistance. No external force is applied on the rod (except gravity, forces by the magnetic field and by thepivot). In what way should the resistance R be changed so that the rod may rotate with uniform angularvelocity in the clockwise direction ? Express your answer in terms of the given quantities and the angle made by the rod OA with the horizontal.fp=k esa Bhd fiNys iz'u tSlh gh fLFkfr iznf'kZr dh x;h gSA lkjh jkf'k;k oSlh gh gSA dsoy O o Cds chp esa fo-



ok-cy okyh ,d cSVjh vkSj ,d izfrjks/k RtksM+s x;s gSaA la;kstu rkjksa dk izfrjks/k 'kwU; gSA NM+ ij dksbZ ck cy xq:Ro]dhyd ,oa pqEcdh; {ks=k ds cyksa ds vykok ugha yxk;k tk jgk gSA Rdks fdl izdkj ifjofrZr fd;k tk;s ftlls fdNM+ dks nf{k.kkorhZ fn'kk esa ,d leku dks.kh; osx ls ?kqek;k tk lds\ vius mkj dks nh x;h jkf'k;ksa rFkk NM+ OA}kjk {kSfrt ls cuk;s x;s dks.k ds inksa esa O;Dr dhft;sA

FO A

C

R

Ans.

cosmg2)Ba2(aB 2

63. A wire of mass m and length can slide freely on a pair of smooth, vertical rails (figure). A magnetic field Bexists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the topend by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.m nzO;eku ,oa yEckbZ okyk ,d rkj] nks m/oZ ,oa fpduh iVfj;ksa ij eqDr :i ls fQly ldrk gSA iVfj;ksa ds ryds yEcor~ fn'kk esa ,d pqEcdh; {ks=k B fo|eku gSA iVfj;ksa dks 'kh"kZ ij C /kkfjrk okys la/kkfj=k ls tksM+k x;k gSA fdlhHkh izfrjks/k dks xkS.k ekurs gq, rkj dk Roj.k Kkr dhft;sA

C

Ans. 22CBmmg

64. A uniform magnetic field B exists in a cylindrical region, shown dotted in figure. The magnetic field increases

at a constant rate dtdB

. Consider a circle of radius r coaxial with the cylindrical region. (a) Find the magnitude

of the electric field E at a point on the circumference of the circle. (b) Consider a point P on the side of thesquare circumscribing the circle. Show that the component of the induced electric field at P along ba is thesame as the magnitude found in part (a).

,d csyukdkj {ks=k esa fp=k esa js[kkafdr n'kkZ;k x;k gS le:i pqEcdh; {ks=k B fo|eku gSA pqEcdh; {ks=k esa dtdB

dh

fu;r nj ls of`) gks jgh gSA csyukdkj {ks=k ds lek{kr% r f=kT;k ds o`k ij fopkj dhft;sA (a) o`k dh ifjf/k ij fLFkrfcUnq ij fo|qr {ks=k Edk ifjek.k Kkr dhft;sA (b) o`k dks ifjc) djus okys oxZ dh Hkqtk ij fLFkr fcUnq P ij izsfjrfo|qr {ks=k dk bads vuqfn'k ?kVd Hkkx (a) esa izkIr ifjek.k ds cjkcj gSA



d c

ba

P

Ans. (a) dtdB

2r

65. The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A/s. The solenoid has 2000 turns/mand its radius in 6.0 cm. (a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding withthe axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds. (b) Find theelectric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a pointoutside the solenoid at a distance 8.0 cm from its axis.,d vkn'kZ yEch ifjufydk esa /kkjk 0.01 ,Eih@ls- dh ,d leku nj ls c



Ans. (a) 0.17 A (b) 0.03 J

72. A coil of resistance 40 is connected across a 4.0 V battery, 0.10 s after the battery is connected, thecurrent in the coil is 63 mA. Find the inductance of the coil.40 izfrjks/k okyh ,d dq.Myh 4.0 V cSVjh ls tksM+h x;h gSA cSVjh tksM+us ds 0.10 lsd.M i'pkr~ dq.Myh esa /kkjk63 mA gSA dq.Myh dk izsjdRo Kkr dhft;sAAns. 4.0 H

73. An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 resistorand a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switchedon.5.0 H izsjdRo ,oa ux.; izfrjks/k okyh dq.Myh dks 100ds izfrjks/k rFkk 2.0 V fo-ok-c- dh cSVjh ds lkFk Js.kheesa tksM+k x;k gSA fLop pkyw djus ds 20 ms i'pkr~ ifjiFk esa izfrjks/k ds fljksa ij foHkokUrj Kkr dhft;sAAns. 0.66 V

74. The time constant of an LR circuit is 40 ms. The circuit is connected at t = 0 and the steady-state current isfound to be 2.0 A. Find the current at (a) t = 10 ms (b) t = 20 ms, (c) t = 100 ms and (d) t = 1 s.,d LR ifjiFk dk le;&fLFkjkad 40 feyh lsd.M gSA t = 0 ij ifjiFk iwjk fd;k tkrk gS vkSj 2.0A LFkk;h /kkjk izkIrgksrh gSA (a) t = 10 feyh lsd.M (b) t = 20 feyh lsd.M , (c) t = 100 feyh lsd.M vkSj (d) t = 1 lsd.M ij /kkjk Kkrdhft;sAAns. (a) 0.44 A (b) 0.79 A (c) 1.8 A and (d) 2.0 A

75. An LR circuit has L = 1.0 H and R = 20 . It is connected across an emf of 2.0 V at t = 0. Find di/dt at (a) t= 100 ms, (b) t = 200 ms and (c) t = 1.0 s.,d LR ifjiFk esa L = 1.0 H ,oa R = 20gSA bldks t = 0 ij 2.0 V fo-ok-c- ds fljksa ij tksM+k x;k gSA (a) t = 100feyh lsd.M , (b) t = 200 feyh lsd.M rFkk (c) t = 1.0 lsd.M] ij di/dtdk eku Kkr dhft;sAAns. (a) 0.44 A (b) 0.79 A (c) 1.8 A and (d) 2.0 A

76. What are the values of the self-induced emf in the circuit of the previous problem at the times indicatedtherein?fiNys iz'u esa fn;s x;s le;ksa ij ifjiFk esa Lo&izsfjr fo-ok-c- ds eku D;k gS\Ans. (a) 0.27 A/s (b) 0.036 V (c) 4.1 109 V

77. An inductor-coil of inductance 20 mH having resistance 10 is joined to an ideal battery of emf 5.0 V. Findthe rate of change of the induced emf at (a) t = 0, (b) t = 10 ms and (c) t = 1.0 s.20 mH izsjdRo ,oa 10 izfrjks/k okyh ,d izsj.k dq.Myh dks 5.0 V fo-ok-c- dh vkn'kZ cSVjh ls tksM+k x;k gSA(a) t = 0, (b) t = 10 feyh lsd.M vkSj (c) t = 1.0 lsd.M ij izsfjr fo-ok-c- esa ifjorZu dh nj Kkr dhft;sAAns. (a) 2.5 10 3 V/s (b) 17 V/s and (c) 0.00 V/s

78. An LR circuit contains an inductor of 500 mH, a resistor of 25.0 and emf of 5.00 V in series. Find thepotential difference across the resistor at t = (a) 20.0 ms, (b) 100 ms and (c) 1.00 s.,d LR ifjiFk esa 500 mH izsjdRo] 25.0 izfrjks/k vkSj 5.00 V fo-ok-c- Js.khe esa la;ksftr gSaA izfrjks/k ds fljksa ijfoHkokarj t = (a) 20.0 feyh lsd.M , (b) 100 feyh lsd.M rFkk (c) 1.00 lsd.M ij Kkr dhft;sAAns. (a) 3.16 V (b) 4.97 V and (c) 5.00 V

79. An inductor-coil of resistance 10 and inductance 120 mH is connected across a battery of emf 6 V andinternal resistance 2W. Find the charge which flows through the inductor in (a) 10 ms, (b) 20 ms and (c) 100ms after the connections are made.10 izfrjks/k vkSj 120 mH izsjdRo okyh ,d izsj.k dq.Myh dks 6 V fo-ok-c- vkSj 2 vkarfjd izfrjks/k okyh cSVjhds fljksa ls tksM+k x;k gSA la;kstu iwjk gksus ds i'pkr~ t = (a) 10 feyh lsd.M, (b) 20 feyh lsd.M vkSj (c) 100 feyhlsd.M] i'pkr~ izsj.k dq.Myh ls izokfgr vkos'k Kkr dhft;sAAns. (a) 1.8 mC (b) 5.7 mC and (c) 45 mC

80. An inductor-coil of inductance 17 mH is constructed from a copper wire of length 100 m and cross-sectionalarea 1 mm2. Calculate the time constant of the circuit if this inductor is joined across an ideal battery. Theresistivity of copper = 1.7 x 108 -m.100 eh- yEcs ,oa 1 mm2vuqizLFk dkV {ks=kQy okys rkacs ds rkj ls 17 mH izsjdRo okyh izsj.k dq.Myh cuk;h x;hgSA ;fn bl izsj.k dq.Myh dks ,d vkn'kZ cSVjh ds fljksa ls tksM+k tk;s rks ifjiFk dk le; fLFkjkad Kkr dhft;sA rkacsdh izfrjks/kdrk = 1.7 x 108 - eh-Ans. 10 ms



81. An LR circuit having a time constant of 50 ms in connected with an ideal battery of emf . Find the timeelapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches halfits maximum value and (c) the magnetic field energy stored in the circuit reaches half its maximum value.,d LR ifjiFk] ftldk le; fLFkjkad 50 feyh lsd.M gS] fo-ok-c- dh vkn'kZ cSVjh ls tksM+k x;k gSA O;rhr gqvk le;Kkr dhft;s % (a) /kkjk ds vf/kdre dk vk/kk eku izkIr djus rd, (b)"ek ds :i esa O;rhr 'kfDr dk vf/kdre dkeku izkIr djus rd rFkk (c) ifjiFk esa pqEcdh; {ks=k tkZ dk vf/kdre dk vk/kk eku izkIr djus rdAns. (a) 35 ms (b) 61 ms (c) 61 ms

82. A coil having an inductance L and a resistance R is connected to a battery of emf . Find the time taken forthe magnetic energy stored in the circuit to change from one fourth of the steady-state value to half of thesteady-state value.L izsjdRo ,oa R izfrjks/k okyh dq.Myh fo-ok-c- dh cSVjh ls tksM+h x;h gSA ifjiFk esa lafpr pqEcdh; {ks=k tkZ dkeku LFkk;h voLFkk ds ,d pkSFkkbZ ls LFkk;h voLFkk ds vk/ks rd ifjofrZr gksus esa yxk le; Kkr dhft;sA

Ans. ln 221

83. A solenoid having inductance 4.0 H and resistance 10 is connected to a 4.0 V battery at t = 0. Find (a) thetime constant (b) the time elapsed before the current reaches 0.63 of its steady-state value, (c) the powerdelivered by the battery at this instant and (d) the power dissipated in Joule heating at this instant.4.0 H izsjdRo ,oa 10 izfrjks/k okyh ,d ifjufydk dks t = 0 ij 4.0 Vdh cSVjh ls tksM+k x;k gSA Kkr dhft;s %(a) le; fLFkjkad (b) /kkjk dk eku LFkk;h voLFkk eku dk 0.63 xquk gksus esa O;rhr le; (c) bl {k.k ij cSVjh }kjkiznk 'kfDr vkSj (d) bl {k.k ij twy "ek ds :i esa O;f;r 'kfDrAAns. (a) 0.40 s (b) 0.40 s (c) 1.0 W and (d) 0.64 W

84. The magnetic field at a point inside a 2.0 mH inductor-coil becomes 0.80 of its maximum value in 20 s whenthe inductor is joined to a battery. Find the resistance of the circuit.cSVjh ls tksM+us ds 20 s i'pkr~ 2.0 mH izsjdRo okyh dq.Myh ds vUnj fLFkr fdlh fcUnq ij pqEcdh; {ks=k dk ekublds vf/kdre eku dk 0.80 xquk gks tkrk gSA ifjiFk dk izfrjks/k Kkr dhft;sAAns. 160

85. An LR circuit with emf is connected at t = 0. (a) Find the charge Q which flows through the battery during0 to t. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during thisperiod. (d) Find the magnetic field energy stored in the circuit at time t. (e) Verify that the results in the threeparts above are consistent with energy conservation.,d LR ifjiFk fo-ok-c- ds lkFk t = 0 ij tksM+k x;k gSA (a) 0 to t rd cSVjh ls izokfgr vkos'kQ Kkr dhft;sA(b) bl dky esa cSVjh }kjk fd;k x;k dk;Z Kkr dhft;sA (c) bl dky esa mRiUu "ek Kkr dhft;sA (d) le; t ijifjiFk esa lafpr pqEcdh; tkZ Kkr dhft;sA (e) ;g tkafp;s fd mDr rhu Hkkxksa esa izkIr ifj.kke tkZ laj{k.k fu;edh iqf"V djrs gSaA

Ans. (a)

)x1(

RLt

R (b)

)x1(

RLt

R

2

(c)

)xx43(

R2Lt

R2

2

(d) 22

)x1(R2

L

where x t/Rte

86. An inductor of inductance 2.00 H is joined in series with a resistor of resistance 200 and a battery of emf2.00 V. At t = 10 ms, find (a) the current in the circuit, (b) the power delivered by the battery, (c) the powerdissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field.2.00 H izsjdRo okyh izsj.k dq.Myh dks 200ds izfrjks/k rFkk 2.00 V fo-ok-c- dh cSVjh ds lkFk Js.khe esa tksM+k x;kgSA t = 10 feyh lsd.M ij] Kkr dhft;s % (a) ifjiFk esa /kkjk (b) cSVjh }kjk iznk 'kfDr, (c) izfrjks/k dks xeZ djus esaO;f;r 'kfDr rFkk (d) pqEcdh; {ks=k esa lafpr dh tk jgh tkZ dh njAns. (a) 6.3 mA (b) 12.6 mW (c) 8.0 mW and (d) 4.6 mW.

87. Two coils A and B have inductances 1.0 H and 2.0 H respectively. The resistance of each coil is 10 . Eachcoil is connected to an ideal battery of emf 2.0 V at t = 0. Let iA and iB be the currents in the two circuit at timet. Find the ratio iA/iB at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1 s.nks dq.Mfy;ksa ArFkk Bds izsjdRo e'k% 1.0 H ,oa 2.0 H gSA izR;sd dq.Myh dk izfrjks/k 10gSA izR;sd dq.Myh dkst = 0 ij 2.0 V fo-ok-c- dh vkn'kZ cSVjh ls tksM+k x;k gSA ekuk fd nksuksa ifjiFkksa esa le; t ij /kkjk,a iAvkSj iB gSaA vuqikriA/iB dk eku % (a) t = 100 feyh lsd.M , (b) t = 200 feyh lsd.M rFkk (c) t = 1 lsd.M] le; ij Kkr dhft;sAAns. (a) 1.6 (b) 1.4 (c) 1.0



88. The current in a discharging LR circuit without the battery drops from 2.0A to 1.0A in 0.10 s. (a) Find the timeconstant of the circuit. (b) If the inductance of the circuit is 4.0 H, what is its resistance?fcuk cSVjh okys LR foltZu ifjiFk esa 0.10 lsd.M esa /kkjk dk eku 2.0 Als 1.0 Ard de gks tkrk gSA (a) ifjiFkdk le;&fLFkjkad Kkr dhft;sA (b) ;fn ifjiFk dk izsjdRo 4.0 H gS] rks izfrjks/k fdruk gS\Ans. (a) 0.14 s (b) 28

89. Aconstant current exists in an inductor-coil connected to a battery. The coil is short-circuited and the batteryis removed. Show that the total charge flown through the coil after the short-circuiting is the same as thatwhich flows in one time constant just before the short-circuiting.cSVjh ls tqM+h gqbZ ,d izsj.k dq.Myh esa LFkk;h /kkjk izokfgr gks jgh gSA dq.Myh dks y?kqifFkr dj fn;k tkrk gS vkSjcSVjh gVk yh tkrh gSA O;Dr dhft;s fd y?kqifFkr djus ds i'pkr~ izokfgr vkos'k dk eku mruk gh gS] ftruk fdy?kqifFkr djus ls ,d le;kUrjky iwoZ izokfgr gqvk FkkA

90. Consider the circuit shown in figure. (a) Find the current through the batterya long time after the switch S is closed. (b) Suppose the switch is againopened at t = 0. What is the time constant of the discharging circuit? (c)Find the current through the inductor after one time constant.fp=k esa iznf'kZr ifjiFk ij fopkj dhft;sA (a) fLop Sdks can djus ds yEcs le;i'pkr~ cSVjh ls izokfgr /kkjk Kkr dhft;sA (b) eku yhft;s fd t = 0 ij fLop iqu%[kksy fn;k tkrk gSA foltZu ifjiFk dk le;&fLFkjkad fdruk gS\ (c) ,dle;&fLFkjkad ds i'pkr~ izsjdRo ls izokfgr /kkjk Kkr dhft;sA

R2

R1 L

SAns. (a)

21

21

RR)RR(

(b)21 RR

L (c) eR1

91. A current of 1.0A is established in a tightly wound solenoid of radius 2 cm having 1000 turns/metre. Find themagnetic energy stored in each metre of the solenoid.dl dj yisVh x;h 2 lseh f=kT;k o 1000Qsjs@ehVj okyh ifjufydk esa 1.0A /kkjk izokfgr gks jgh gSA ifjufydkds izR;sd ehVj esa lafpr pqEcdh; tkZ Kkr dhft;sAAns. 7.9 104 J

92. Consider a small cube of volume 1 mm3 at the centre of a circular loop of radius 10 cm carrying a current of4A. Find the magnetic energy stored inside the cube.10 lseh f=kT;k ,oa 4A /kkjk izokg okys o`kkdkj ywi ds dsUnz ij 1 feeh3vk;ru okys ?ku ij fopkj dhft;sA ?ku dsvUnj lafpr pqEcdh; tkZ Kkr dhft;sAAns. 8 1014 J

93. A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of 1 mm3at a distance of 10.0 cm from the wire.,d yEcs rkj ls 4.00A /kkjk izokfgr gks jgh gSA rkj ls 10.00lseh nwjh ij fLFkr 1.00 mm3vk;ru esa lafpr pqEcdh;{ks=k tkZ Kkr dhft;sAAns. 2.55 1014 J

94. The mutual inductance between two coils is 2.5 H. It the current in one coil is changed at the rate of 1 A/s,what will be the emf induced in the other coil?nks dq.Mfy;ksa ds chp vU;ksU; izsjdRo 2.5 H gSA ;fn ,d dq.Myh esa /kkjk 1 ,[email protected] dh nj ls ifjofrZr gksrhgS] rks nwljh dq.Myh esa fdruk fo-ok-c- izsfjr gksxk\Ans. 2.5 V

95. Find the mutual inductances between the straight wire and the square loop of figure.fp=k esa iznf'kZr lh/ks rkj ,oa oxkZdkj ywi ds chp vU;ksU; izsjdRo Kkr dhft;sA

a

b

i



Ans.

ba1ln

2a0

96. Find the mutual inductance between the circular coil and the loop shown in figure.fp=k esa iznf'kZr o`kkdkj dq.Myh vkSj ywi ds chp vU;ksU; izsjdRo Kkr dhft;sA

Ans. 2/322122

0

)xa(2'aa

N

97. A solenoid of length 20 cm, area of cross-section 4.0 cm2 and having 4000 turns is placed inside anothersolenoid of 2000 turns having a cross-sectional area 8.0 cm 2 and length 10 cm. Find the mutual inductancebetween the solenoids.20 lseh yEch] 4.0 lseh2vuqizLFk dkV {ks=kQy vkSj 4000Qsjksa okyh ifjufydk] ,d vU; ifjufydk ds vUnj j[khgqbZ gSA ftldh yEckbZ 10 lseh vuqizLFk dkVj {ks=kQy 8 lseh2vkSj Qsjks a dh la[;k 2000 gSA ifjufydkvksa ds chpvU;ksU; izsjdRo Kkr dhft;sAAns. 2.0 102 H

98. The current in a long solenoid of radius R and having n turns per unit length is given by i = i0 sin t. A coilhaving N turns is wound around it neat the centre. Find (a) the induced emf in the coil and (b) the mutualinductance between the solenoid and the coil.R f=kT;k ,oa izfr ,dkad yEckbZ esa nQsjksa okyh ,d yEch ifjufydk esa izokfgr /kkjk i = i0 sint }kjk O;Dr dh tkrhgSA blds dsUnz ds lehi NQsjksa okyh ,d dq.Myh yisVh x;h gSA Kkr dhft;s % (a)dq.Myh esa izsfjr fo-ok-c- rFkk(b) ifjufydk ,oa dq.Myh ds chp vU;ksU; izsjdRoAAns. (a) 0i0 nNR2 cos t (b) 0nNR2

electromagnetic induction

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