Chapter # 38
Electromagnetic Induction
SOLVED EXAMPLES
Example 38.1 Figure shows a conducting loop placed near a long straight wire carrying a current i as shown. If the current increases 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 positive normal to the loop is going into the plane of the diagram. Also, the same rule shows that the magnetic field at the site of the loop due to the current is also going into the plane of the diagram. Thus, B and dS are along the same direction everywhere so that the flux B dS is positive. If i increases, the magnitude
d is positive. Thus, is negative and hence, the current is negative. The current is, therefore, dt induced in the direction opposite to the arrow.
increases,
Example 38.2 Figure shows a rectangular loop MNOP being pulled out of a magnetic field with a uniform velocity v by applying an external force F. The length MN is equal to l and the total resistance of the loop is R. Find (a) the current in the loop, (b) the magnetic force on the loop, (c) the power delivered by the external force and (e) the thermal power developed in the loop.
Sol.
× ×N × × × × O × × × × × × × × × v× × F × × × × × × × M × × × ×P (a) The emf induced in the loop is due to the motion of the wire MN. The emf is e = vB with the positive end at N and the negative end at M. The current is vB = R R in the clockwise direction (figure). i=
× × × F1 × M ×
N
×
× F2 ×
× ×
×
× v ×
×
×
×
×
×
×
F2
O F P
vB 2 2 (b) The magnetic force on the wire MN is F1 i B . The magnitude is F1 = iB = and is opposite to R the velocity. The forces on the parts of the wire NO and PM, lying in the field, cancel each other. The resultant B2 2 v opposite to the velocity.. R (c) To move the loop at a constant velocity, the resultant force on it should be zero. Thus, one should pull the loop with a force magnetic force on the loop is, therefore, F1 =
vB 2 2 . R (d) The power delivered by the extenal force is F = F1 =
P = Fv =
v 2B 2 2 . R
(e) The thermal power developed is 2
vB v 2B 2 2 R= P = i2R = . R R 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. manishkumarphysics.in Page # 1
Chapter # 38
Electromagnetic Induction
Example 38.3 An average induced emf of 0.20 V appears in a coil when the current in it is changed form 5.0 A in one direction to 5.0 A in the opposite direction in 0.20 s. Find the self-inductance of the coil. Sol.
Average
di {5.0 A ) (5.0 A ) 50 A / s dt 0.20 s
Using
= –L
or,
L=
di dt 0.2 V = L(50 A/s)
0.2V 4.0 mH 50 A / s
Example 38.4 Consider the circuit shown in figure. The sliding contact is being pulled towards right so that the resistance in the circuit is increasing. Its value at the instant shown in 12 W. Will the current be more than 0.50 A or less than it at this instant? 20 mH
r
6V
Sol.
As the sliding contact is being pulled, the current in circuit charnges. An induced emf e = – L across the inductor. The net emf in the circuit is 6 V = L
i
6V L 12
di dt
di is produced dt
di and hence the current is dt
....(i)
at the instant shown. Now the resistance in the circuit is increasing, the current is decreasing and so
di is dt
6V negative. Thus, the numerator of (i) is more than 6V and hence i is greater than 12 = 0.50 A. Example 38.5 In inductor (L = 20 mH), a resistor (R = 100 ) and a battery ( = 10 V) are connected in series. Find (a) the time constant, (b) the maximum current and (c) the time elapsed before the current reaches 99% of the maximum value. Sol. (a) The time constant is
L 20 mH R 100 = 0.20 ms
(b) The maximum current is
10 V i = /R = 100 = 0.10 A. (c) Using or, or, or,
i = i0 (1 – e–t/), 0.99 i0 = i0(1 – e–t/) e–t/ = 0.01 t = – ln(0.01) t = (0.20 ms) ln (100) = 0.92 ms.
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Chapter # 38 Electromagnetic Induction Example 38.6 An inductor (L = 20 mH), a resistor (R = 100 ) and a battery ( = 10 V) are connected in series. After a long time the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit 1 ms after short-circuiting. Sol.
The initial current is i = i0 = /R = The time constant is = L/R =
10 V = 0.10 A. 100
20 mH .20 ms . 100
The current at t = 1 ms is i = i0e–t/ = (0.10 A)e–(1 ms / 0.20 ms) = (0.10 A)e–5 = 6.7 × 10–4 A. Example 38.7 Calculate 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
U
1 2 1 Li (50 10 3 H)(2.0 A )2 0.10 J 2 2
Example 38.8 A solenoid S1 is placed inside another solenoid S2 as shown in figure. The radii of the the inner and the outer solenoids 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.
s2
Sol.
s1
Suppose a current i is passed through the inner solenoid s1. A magnetic field B = s n1i is produced inside s1 whereas the field outside it is zero. The flux through each trun of s2 is Br12 = m0 ir12. The total flux through all the turns in a length of s2 is = (0 n1ir12) n2 = (0 n1n2r12)i. Thus, M = 0 n1n2r12.
QUESTIONS
FOR
SHORT
ANSWER
1.
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 is much 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 if the magnet if the magnet is approaching the loop and attracts the magnet if ir is going away from the loop.
5.
Two circular loops are placed coaxially but separated by a distance. A battery is suddenly connected to 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 the other 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 field inside the box ? Such low-resistivity metals are used to form enclosures which shield objects inside manishkumarphysics.in
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Chapter # 38 Electromagnetic Induction 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 slow down as compared to the nonmetallic ones and hence are separated. Discuss the role of eddy currents in the process.
9.
A pivoted aluminimum bar falls much more slowly through a small region containing a magnetic field than a similar bar of an insulating material . Explain.
10.
A metallic bob A oscillates through the space between the poles of an electromagnet . The oscillates are more quickly damped when the circuit is on , as compared to the case when the circuit is off. Explain.
N
S
A
11.
Two circular loops are placed with their centres separated by a fixed distance. How would you orient the 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 fo the two is greater ?
13.
Consider the energy density in a solenoid at its centre and that near its ends. Which of the two is greater ?
Objective - I 1.
A rod a length l rotates with a small but uniform angular velocity about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is yEckbZ okyh ,d NM+ blds yEc v)Zd ds ifjr% de fdUrq ,d leku dks.kh; osx ls ?kwe jgh gSA ogk¡ ij ?kw.kZu v{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
2.
1 2 Bl 8
(C)
1 2 Bl 2
(D) Bl2
A rod of length l rotates with a uniform angular velocity about its perpendicualr bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is yEckbZ 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 lekUrj bafxr le:i pqEcdh; {ks=k B fo|eku gSA NM+ ds dsUnz rFkk NM+ ds nksuksa fljksa ds chp foHkokUrj gS & (A*) zero
3.
(B*)
(B)
1 Bl2 2
(C) Bl2
(D) Bl2
Consider the situation shown in fig. If the switch is closed and after some time it is opened again, the closed loop will show
fp=k esa iznf'kZr ifjfLFkfr ij fopkj dhft;sA ;fn fLop cUn fd;k tkrk gS vkSj dqN le; i'pkr~ bldks iqu% [kksy fn;k tkrk gS] cUn ywi iznf'kZr djsxk &’
(A) an anticlockwise current-pulse manishkumarphysics.in
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Chapter # 38 Electromagnetic Induction (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 the magnet
rkacs 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 (C) will move with an acceleration g (A) uyh esa :d tk;sxkA (C) g Roj.k ds lkFk xfr djsxkA 6.
(B*) will move with almost contant speed (D) will oscillate (B*) yxHkx fu;r pky ls xfr djsxkA (D) nksy u djsxkA
Fig. shown a horizontal solenoid connected to a battery and a switch. A copper ring is place on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will
fp=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 iFk ij 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 cUn fd;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 on
ifjufydk dh vksj xfr djsxh ;k mlls ijs which 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 ¼/kukRed ;k _.kkRed½ tqM+k gqvk gSA 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 (C) B is true but A is false (A*) A o B nksu ksa gh lR; gSA (C) B lR; gS] fdUrq A vlR; 8.
gS
(B) A is true but B is false (D) Both A and B are false (B) A lR; gS] fdUrq B vlR; (D) A o B nksu ksa xyr gSA
gSA
Consider the situation shown in fig. The wire AB is slid on the fixed rails with a constant velocity. If the wire AB is rep;aced by a semicircular wire, the magnitude of the induced current will fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA rkj AB fLFkj iVfj;ksa ij fu;r osx ls fQly jgk gSA ;fn rkj AB dks
v)Zo`Ùkkdkj rkj ls izfrLFkkfir dj fn;k tk;s rks izsfjr /kkjk dk ifjek.k gksxk &
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Chapter # 38
Electromagnetic Induction
(A) increase (B*) remain the same (C) decrease (D) increase or decrease depending on whether the semicircle bulges towards the resistance or away from it. (A) c
Fig. shows a conducting loop being pulled out of a magnetic field with a speed . Which of the four plots shown in fig. may represnet the power delovered by the pulling agent as a function of the speed . fp=k (a) esa n'kkZ;k x;k gS fd ,d pkyd ywi dks fu;r pky ds lkFk pqEcdh; {ks=k ds ckgj [khapk tk jgk gSA fp=k (b) esa iznf'kZr pkj xzkQksa esa ls dkSulk [khapus okyh ;qfDr&}kjk 'kfä dks pky v ds Qyu esa iznf'kZr djrk gSA
(B*) 10.
Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inseted in between to dirve a current in it. The current changes slightly because of the variation in resistance with temperature. During the period, the two loops
leku f=kT;kvksa okys nks o`Ùkkdkj ywiksa dks dqN varjky ij lek{kr% j[kk x;k gSA igys ywi esa /kkjk izokfgr djus ds fy;s bldks dkVdj blesa ,d cSVjh yxk;h tkrh gSA rki esa ifjorZu ds dkj.k izfrjks/k ifjofrZr gksus ls /kkjk FkksM+h lh ifjofrZr gksrh gSA bl dky esa nksuksa ywi & (A*) attract each other (B) repel each other (C) do not exert any force on each other (D) attract or repel each other depending on the sense of the current (A*) ijLij vkdf"kZr djsaxsA (B) ijLij izfrdf"kZr djsxsaA (C) ,d nwljs ij dksbZ cy ugha yxk;saxsA (D) /kkjk dh fn'kk ds vk/kkj ij ,d nwljs dks vkdf"kZr djsxsa ;k izfrdf"kZr 11.
djsaxsaA
A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is
,d NksVk o`Ùkkdkj pkyd ywi] ,d yEch /kkjkokgh ifjufydk esa j[kk gqvk gSA ywi ds ry esa ifjufydk dh v{k fLFkr gSA ;fn ifjufydk dh v{k fLFkr gSA ;fn ifjufyk esa /kkjk ifjofrZr dh tk;s rks ywi esa izsfjr /kkjk gksxh & (A) clockwise (B) anticlockwise (C*) zero (D) clockwise or anticlockwise depending on whether the resistance in increased or decreased. (A) nf{k.kkorhZ (B) okekorhZ (C*) 'kwU; (D) nf{k.kkorhZ gksxh ;k okekorhZ gksxh ;g bl ij fuHkZj djsxk fd izfrjks/k c<+k;k tkrk gS ;k de fd;k 12.
tkrk gSA
A conducting square loop of side l and resistance R moves in its plane with a uniform velocity u perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop in fig. The current induced in the loop is Hkqtk vkSj R izfrjks/k okyk ,d pkyd ywi blds ry ds vuqfn'k ,d leku osx v los bldh ,d Hkqtk ds yEcor~ xfr'khy gSA ywi ds ry ds yEcor~ fp=k esa n'kk;s vuqlkj ,d le:i pqEcdh; {ks=k B fo|eku gSA ywi esa izsfjr
/kkjk gS &
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Chapter # 38
Electromagnetic Induction
(A) Bl/R clockwise (C) 2Bl/R anticlockwise (A) Bl/R nf{k.kkorhZ (C) 2Bl/R okekorhZ
(B) Bl/R anticlockwise (D*) zero (B) Bl/R okekorhZ (D*) 'kwU;
Objective - II 1.
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 gSApqEcd dh 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*) mÙkj /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) mÙkj /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 potential difference appears across the two ends ,d pky NM+ f;r osx b ds lkFk fdlh pqEcdh; {ks=k esa xfr'khy gSA blds nksuksa fljksa ds chp foHkokrj mRiUu gksxk& (A) if
3.
l
(B) if
B
(C) if l B
(D*) none of these
A conduction loop is placed in a uniform magnetic filed with its plane perpenducular to the field. An emf is induced 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-cizsfjr gksxk] ;fn & (A) it is translated (C*) it is rotated about a diameter (A) bldks LFkkukuarfjr fd;k tk;sA (C*) bldks] O;kl ds ifjr% ?kw.kZr fd;k 4.
tk;sA
(B) it is rotated about its axis (D*) it is deformed (B) bldks bldh v{k ds ifjr% ?kwf.kZr (D*) bldks fo:fir fd;k tk;sA
fd;k tk;sA
A metal sheet is placed in front of a strong magnetic pole. A force needed to
izcy 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]
xq:Ro ds fdlh Hkh izHkko dks ux.; eku fyft'sA 5.
izfr pqEcdRo ;k
A constant current i is maintained in a solenoid. Which of the following quantities will increase if an iron rod is 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;s
rks fuEu esa ls dkSulh jkf'k c<+sxhA
(A*) magnetic field at the centre (C*) self-inductance of the solenoid
(B*) mangetic flux linked with the solenoid (D) rate of Joule heating
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Chapter # 38
Electromagnetic Induction
(A*) dsUnz ij pqecdh; {ks=k (C*) ifjufydkvksa dk LoizsjdRo 6.
(B*) ifjufydk (D) twy Å"ek
ls lEc) pqEcdh; ¶yDl dh nj
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 battery
nks ifjufydkvksa dh T;kferh; cukoV ,d tSlh gS] fdUrq ,d eksVs rkj ls o nwljh irys rkj ls cuk;h x;h gSA nksuksa ifjufydkvksa ds fy;s fuEu jkf'k;ksa esa ls dkSulh fHkUu&fHkUu gS (A) LoizsjdRo (B*) ;fn nksu ksa ls leku /kkjk,a izo kfgr gksa rks twy Å"ek dh nj (C) ;fn nksuksa ls leku /kkjk,a izokfgr gks rks pqEcdh; {ks=k dh ÅtkZ (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 the connection ? (A) current in the circuit (B) magnetic field energy in the inductor (C) power delivered by the battery (D*) emf induced in the inductor fdlh 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; ÅtkZ (C) izsjdRo esa pqEcdh; ÅtkZ (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 end A 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 Å"ek ds dkj.k NM+ xeZ gks tkrh gSA 9.
L, C and R represent the physical quantities inductance, capacitance and resistance combinations have dimensions of frequency ? HkkSfrd jkf'k;ks]a izjs dRo] /kkfjrk ,oa izfrjks/k dks Øe'k% L, C ,oa R }kjk O;Dr fd;k tkrk gSA fuEu esa ls fdl dh foek,a vko`fÙk dh gS (A*)
10.
1 RC
(B*)
R L
(C*)
1 LC
(D) C/L
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. (C*) The current in L just before t = t0 is /R fp=kksa (a) o (b) esa fLopksa dks t = 0 ij can fd;k
x;k
(B*) The change on C long after t = 0 is C. (D) The current in L long after t = t0 is /R gS] vkSj ,d yEcs le; t = t0 ds i'pkr~ iqu% [kksyk
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x;k gS -
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Chapter # 38
Electromagnetic Induction
(A) t = 0 ds rqjar i'pkr~ C ij vkos' k C gSA (C*) t = t0 ds rqjar igys L ls izo kfgr /kkjk /R gSA
(B*) t = 0 ds yEcs le; i'pkr~ C ij vkos' k C gSA (D) t = t0 ds cgqr le; i'pkr~ L ls izokfgr /kkjk /R gSA
WORKED OUT EXAMPLES EXERCISE 1.
Calculate the dimensions of (a)
E . d l.
(c)
d B . The symbols have their usual meanings. dt
(a)
E . d l.
Ans.
(b) vBL and
d B dt 2 –1 –3 M L I T in each case.
(b) vBL rFkk (c)
dh foekvksa dh x.kuk dhft;sA ladsrksa ds lkekU; vFkZ gSA
2.
The flux of magnetic field through a closed conducting loop changes with time according to the equation at2 + 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 vuqlkj ifjofrZr gksrk gSA (a) a, b ,oa c ds S ek=kd fyf[k;sA (b) ;fn a, b ,oa c ds ifjek.k Øe'k% 0.20 , 0.40 ,oa 0.60 gS rks t = 2 lsd.M ij izsfjr fo-ok-cy Kkr dhft;sA Ans: 1.2 Volt
3.
(a) The magnetic field in a region varies as shown in figure. Calculate the average induced emf in a conducting loop of area 2.0 × 10 –3 m 2 placed perpendicular to the field in each of the 10 ms intervals shown. (b) In which intervals is the emf not constant? Neglect the behavior near the ends of 10 ms intervals. (a) fdlh LFkku ij pqEcdh; {ks=k fp=kkuqlkj ifjofrZr gksrk gSaA {ks=k ds yEcor~ j[ks gq, 2.0 × 10 –3 m 2 {ks=kQy okys pkyd 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) fdlh varjky ds fy;s fo-ok-cy fu;r ugha gS\ 10 feyh lsd.M varjky ds lehi O;ogkj dks ux.; eku yhft;sA B(T) 0.03 – –
Ans: 4.
–
–
–
–
0.01 –
10
20
30
40
(a)–2.0 mV, –4.0 mV, 4.0mV, 2.0mV
t(ms)
(b)10 ms to 20 ms and 20 ms to 30 ms.
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`Ùkkdkj 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 manishkumarphysics.in
Page # 9
Chapter # 38 Ans: 7.8 X 10-3 V 5.
Electromagnetic Induction
A conducting circular loop of area 1 mm 2 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 current which changes from 10A to zero in 0.1 s. Find the average emf induced in the loop in 0.1s. Find the average emf induced in the loop in 0.1s. 1 feeh2 {ks=kQy okys pkyd o`Ùkkdkj ywi dks blls 20 lseh nwj fLFkr yEcs ,oa lh/ks rkj ds leryh; j[kk x;k gSA lh/ks rkj ls izokfgr /kkjk 0.1 lsd.M esa 10 A ls 'kwU; rd ifjofrZr gks tkrh gSA ywi esa 0.1 lsd.M esa izsfjr vkSlr
fo-ok-c- Kkr dhft;sA 6.
A square- shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicular to a 1.0 T magnetic field. It is removed from the magnetic field in 0.25 s and restored in its original place in 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 vkÑfr okyh ,d dq.Myh dh Hkqtk,a 50 lseh yEch gS vkSj blesa 50 Qsjs gSaA bldks 1.0 T pqEcdh; {ks=k ds yEcor~ 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% blds
iwoZ 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 esa Ans: (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 with uniform 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 ,d leku osx ls xfr djrh gSA dq.Myh esa mRiUu Å"ek ÅtkZ Kkr dhft;s % (a) bldks gVkrs gq,A (b) bldks iqu% j[krs gq, vkSj (c) bldh xfr esaA Ans: (a) 25 J (b) 25 J (c) 50 J
8.
A conducting loop of area 5.0 cm 2 is placed in a magnetic field which varies sinusoidally with time as B = B0 sin t where B0 = 0.20 T and = 300 s –1. The normal to the coil makes an angle of 60º with the field. Find the (a) the maximum emf induced in the coil, (b) the emf induced at = (/900)s and (c) the emf 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 lkFk T;koØh; :i esa ifjofrZr gks jgk gS] tgk¡ B0 = 0.20 T vkSj = 300 izfr lsd.M gSA dq.Myh ij vfHkyEc] {ks=k ds lkFk 60º dks.k cukrk gSA Kkr dhft;s (a) dq.Myh esa izsfjr vf/kdre fo-ok-cy (b) = (/900) lsd.M ij izsfjr fook-cy vkSj (c) t = (/600) lsd.M ij izsfjr fo-ok-cy Ans. (a) 0.015 V, (b) 7.5 × 10–3 V, (c) zero
9.
Figure shows a conducting squares loop placed parallel to the pole-faces of a ring magnet. The polefaces have an area of 1 cm 2 each and the field between the poles is 0.10 T. The wires making the loop are all outside the magnetic field. If the magnet is removed in 1.0 s, what is the average emf induced in the loop?
fp=k esa n'kkZ;k x;k gS fd ,d oxkZdkj pkyd ywi ,d fjax&pqEcd ds /kzoq &Qydksa ds lekUrj j[kk x;k gSA /kzoq &Qydksa esa 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=k ds 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 180º about a diagonal in 0.20 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. manishkumarphysics.in
Page # 10
Chapter # 38
Electromagnetic Induction
2.0 lseh
Hkqtkvksa okyk ,d oxkZdkj pkyd ywi blds ,d fod.kZ ds ifjr% 0.20 lsd.M esa 180º dks.k ls ?kqek fn;k tkrk 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;sA Ans. 5.0 T 11.
A conducting loop of face area A and resistance R is placed perpendicular to a magnetic field B. The loop is withdrawn completely from the field. Find the charge which flows through any cross section of the wire in the process. Note that it is independent of the shape of the loop as well as the away it is withdrawn. Qyd {ks=kQy A o izfrjks/k R okyk ,d pkyd ywi pqEcdh; {ks=k B ds yEcor~ j[kk gqvk gSA ywi dks pqEcdh; {ks=k
ls iw.kZr;k gVk fy;k tkrk gSA bl izfØ;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 vkÑfr ,oa lkFk gh bldks ckgj gVkus ds rjhds ij fuHkZj ugha djrk gSA Ans. 12.
BA/R
A long solenoid of radius 2 cm has 100 turns/ cm and carries a current of 5A. A coil of radius 1 cm having 100 turns and a total resistance of 20 is placed inside the solenoid coaxially .The coil is connected is reversed galvanometer. If the current in the solenoid is reversed in direction find the charge 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 1 lseh 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 mRØfer dj nh tk;s rks /kkjkekih ls izokfgr vkos'k Kkr dhft;sA Ans. 13.
2 × 10–4 C
Figure shows a metallic square frame of edge a in a vertical plane. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They starts pulling the corners at t = 0 and displace the corners at a uniform speed u. (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60º. (b) Find the induced current in the frame at this instant if the total resistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line. fp=k esa m/okZ/kj ry esa fLFkr ,d oxkZdkj /kkfRod Ýse iznf'kZr dh x;h gS] ftldh Hkqtk a gSA ogk¡ ij fp=k esa ry ds yEcor~ le:i pqEcdh; {ks=k B fo|eku gSA nks yM+ds oxkZdkj Ýse oxkZdkj Ýse ds foijhr dksuksa dks [khap jgs gSa ftlls fd ;g lekUrj prqHkZqtkdkj gks tk;sA os dksuksa dks t = 0 ij [khapuk izkjEHk djrs gSa vkSj dksuksa dks ,d leku pky u ls foLFkkfir djrs gSaA (a) Ýse esa izsfjr fo-ok-c- ml {k.k ij Kkr dhft;s tc bu dksuksa ds dks.k 60º rd de gks tkrs gSaA (b) ;fn Ýse dk dqy izfrjks/k R gS] rks bl {k.k ij Ýse esa izsfjr /kkjk Kkr dhft;sA (c) oxZ ds
lh/kh js[kk rd fo:fir gksus rd fdlh ,d Hkqtk ls izokfgr dqy vkos'k Kkr dhft;sA
///////////////////////////////////////////////////// 14.
The north pole of a magnet is brought down along the axis of a horizontal circular coil (figure ). As a result 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 as you look into the coil from the side of the magnet?
,d pqEcd dk mÙkjh /kzqo fp=k esa n'kkZ;s vuqlkj {kSfrt dq.Myh dh v{k ds vuqfn'k uhps yk;k tkrk gSA blds ifj.kkeLo:i dq.Myh ls xqtjus okyk ¶yDl vk/ks lsd.M esa 0.35 oscj ls .85 oscj rd ifjofrZr gksrk gSA bl dky esa 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\
manishkumarphysics.in
Page # 11
Chapter # 38
Ans: 15.
Electromagnetic Induction
1.0 V ,anticlockwise
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 loop
fdlh ry esa jgus ds fy;s cká rkj dk ,d ywi blds ry esa dqN dks.kh; osx ls ?kwf.kZr gks jgk gSA bl {ks=k esa ,d le:i pqEcdh; {ks=k fo|eku gSA ywi esa izsfjr fo-ok-c- Kkr dhft;sA Ans. 16.
zero
Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s The front 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 bldk lkeus 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 × 10–4 V (b) zero (c) 3 × 10–4 V (d) zero
17.
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 Å"ek Kkr dhft;sA Ans: 2 × 10–4 J
18
A uniform magnetic field B exist in a cylindrical region or radius 10 cm as shown in figure. A uniform wire of length 80 cm and resistance 4.0 is bent into a square frame and is placed with one side along a diameters of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s find the 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 vkSj 4.0 izfrjks/k okys ,d le:i rki dks oxkZdkj Ýse ds :i esa eksM+dj] bl izdkj j[kk x;k gS fd bldh ,d Hkqtk csyukdkj {ks=k ds O;kl ds vuqfn'k gSA ;fn pqEcdh; {ks=k 0.010 VsLyk@ls dh fu;r nj ls c<+ jgk gks rks Ýse esa izsfjr
/kkjk Kkr dhft;sA
Ans. 19
3.9 × 10–5 A
The magnetic field in the cylindrical region shown in figure increases at a constant rate of 20.0 m T/s. Each side of the square loop abcd and defa has a length of 1.00 cm and resistance of 4.00 . Find the current (magnitude and sense ) in the wire ad if (a) the switch S1 is closed but S2 is open (b) S1 is open but S2 is closed (c) both S1 and S2 are open and (d) both S1 and S2 are closed. fp=k esa iznf'kZr csyukdkj {ks=k esa pqEcdh; {ks=k 20.0 feyh
[email protected] dh fu;r nj ls c<+ jgk gSA oxkZdkj ywi manishkumarphysics.in
Page # 12
Chapter # 38
Electromagnetic Induction
abcd vkSj defa dh izR;sd Hkqtk dh yEckbZ 1.00 lseh vkSj izfrjks/k 4.00 gSA ad rkj esa /kkjk ¼ifjek.k ,oa fn'kk½ Kkr dhft;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 gksaA vkSj (d) S1 ,oa S2 nksuksa gh can gksaA [6 min.] [M.Bank(07-08)_HCV_Ch.38_Ex._19]
Ans. 20.
(a) 1.25 × 10–7 A, a to d (b) 1.25 × 10–7 A, d to a. (c) zero (d) zero
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 circular loop of radius a’ and resistance r’ is placed coaxially with the coil. The centre of the loop is at a distance 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 right at 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`Ùkkdkj dq.Myh iznf'kZr dh x;h gS] bldks ,d /kkjk fu;a=kd }kjk fo-ok-c- dh cSVjh ls 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`Ùkkdkj ywi] dq.Myh ds lkFk lek{kr% j[kk gqvk gSA ywi dk dsUnz] dq.Myh ds dsUnz ls x nwjh ij fLFkr gSA izkjEHk es]a /kkjk fu;a=kd dk 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`Ùkkdkj ywi esa izsfjr fo-ok-cy ml {k.k ij Kkr dhft;s tc % (a) foliZd fQlyuk izkjEHk djrk gS vkSj (b) foliZd vk/kh yEckbZ
rd f[kld pqdk gksA
Ans. 21.
0Na2a' 2 Rv 2L(a 2 x 2 )3 / 2 (R'r )2
where R’ = R for part (a) and R/2 for part (b)
A circular coil of radius 2.00 cm has 50 turns .A uniform magnetic field B = 0.200 T exists in the space in a direction parallel to the axis of the loop The coil is now rotated about a diameter through an angle of 60° The operation takes 0.100s. (a) Find the average of emf induced in the coil .(b) If the coil a is closed one with the two ends joined together and has a resistance of 4.00 calculate the net charge crossing a cross- section of the wire of the coil. 2.00 lseh f=kT;k okyh o`Ùkkdkj dq.Myh esa 50 Qsjs gSaA ywi dh v{k ds lekUrj fn'kk esa ,d le:i pqEcdh; {ks=k B = 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 izfØ;k esa 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 fljs ijLij tqM+s gq, vkSj bldk izfrjks/k 4.00 gS] rks dq.Myh ds rkj ds fdlh vuqizLFk dkV ls izokfgr dqy vkos'k dh
x.kuk dhft;sA Ans. 22.
(a) 6.28 × 10–3 V (b) 1.57 × 10–3 C
A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 × 10 – 4 about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25cm 2 and its resistance is 4.0 . Find (a) the average emf developed in half a turn from 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).
manishkumarphysics.in
Page # 13
Chapter # 38
Electromagnetic Induction
,d dq.Myh esa 100 Qsjs gS]a 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 Kkr dhft;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'k Ans. 23.
(a) 2.0 × 10–3 V (b) zero (c) 5.0 × 10–5 C
A coil of radius 10 cm and resistance 40 has 1000 turns It is placed with its plane vertical and its axis paralleled to the magnetic meridian The coil is connected to a galvanometer and is rotated about if the horizontal 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;ksÙkj ds lekUrj ,oa ry m/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.k ls ?kqek;k tkrk gSA ;fn i`Foh ds pqEcdh; {ks=k dk {kSfrt ?kVd BH = 3.0 × 10 – 5 VsLyk gS rks /kkjkekih ls izokfgr
vkos'k Kkr dhft;sA Ans. 24.
4.7 × 10–5 C
A circular coil of one turns of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute A uniform magnetic field revolutions per minute A uniform magnetic field B = 0.010 rotation Find a direction perpendicular to the axis of rotation Find (a) the maximum emf induced. (b) the average emf induced in the coil over a long period and (c) the average of the squares of emf induced over a long period. ,d Qsjs okyh o`Ùkkdkj dq.Myh dh f=kT;k 5.0 lseh gS] bldks O;kl ds ifjr% 80 pDdj@fefuV dh fu;r dks.kh; pky ls ?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 Kkr dhft;s % (a) vf/kdre izsfjr fo-ok-cy (b) yEcs dky esa dq.Myh esa izfs jr vkSlr fo-ok-cy rFkk (c) yEcs dky esa izfs jr
fo-ok-cy ds oxZ dk vkSlr Ans. 25.
(a) 6.6 × 10–4 V (b) zero (c) 2.2 × 10–7 V2
Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Neglecting the 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 dk
izfrjks/k ux.; ekudj ifjiFk esa ,d fefuV esa mRiUu Å"ek Kkr dhft;sA Ans. 26.
1.3 × 10–7 J
Figure shows a circular wheel of radius 10.0 cm whose upper half shown dark in the figure is made of iron and the lower half of wood the two junctions are joined by an iron rod A uniform magnetic field B of magnitude 2.00 × 10–4 exists in the space above the central line as suggested by the figure The wheel is set into pure rolling on the horizontal surface it takes 2.00 seconds for the iron part to come down and the wooden part to go up find the average emf induced during this period
fp=k esa 10.0 lseh f=kT;k dk o`Ùkkdkj ifg;k n'kkZ;k x;k gS] ftldk Åijh vk/kk Hkkx tks xgjk gSA yksgs dk cuk gS vkSj uhps 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[kk ds Åij okys Hkkx esa le:i pqEcdh; {ks=k B fo|eku gS] ftldk ifjek.k 2.00 × 10–4 VsLyk gSA ifg;k {kSfrt lrg ij 'kq) yksVuh xfr dj jgk gSA ;fn yksgs okys Hkkx dks uhps vkus esa vkSj ydM+h okys Hkkx dks Åij tkus esa 2.00 lsd.M yxrs gSa rks bl dky esa izsfjr vkSlr fo-ok-cy Kkr dhft;sA Ans. 27.
1.57 × 10–6 V
A 20 cm long conducting rod is set pure translation with a uniform velocity of 10 cm/s perpendicular to its lengths. A uniform magnetic filed of magnitude 0.10 T exists in a direction perpendicular to the plane of motion . (a) Find the average magnetic force on the free electrons of the rod. (b) for what electronic field inside the rod the electronic force on a free electron will balance the magnetic force ? How is this electronic 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.k xfr'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 bysDVªkWuksa ij vkSlr pqEcdh; cy dk ifjek.k Kkr dhft;sA (b) NM+ ds vUnj fdrus fo|qr {ks=k ds fy;s eqDr bysDVªkWuksa 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;sA Ans.
(a) 1.6 × 10–21 N (b) 1.0 × 10–2 V/m (c) 2.0 × 10–3 V manishkumarphysics.in
Page # 14
Chapter # 38 Electromagnetic Induction 28. A metallic metre sticks moves with a velocity of 2 m/ s in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2 T Find the emf induced between the ends of 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;sA Ans: 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 space craft 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 xfr ds ry ds yEcor~ B = 3 × 10– 10 VsLyk pqEcdh; {ks=k fo|eku gSA varfj{k ;ku dks pkyd ekurs gq, bldh pkSM+kbZ
ds chp izsfjr fo-ok-cy dh x.kuk dhft;sA Ans: 30.
0.9 V
The two rails of a railway track insulated from each other and from the ground are connected to a millivoltmeter. What will be the reading of the milli-voltmeter when a train travels on the track at a speed of 180 km/h.? The vertical component of earth’s magnetic field is 0.2 × 10 – 4 T and the rails are separated by 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 Vªsu bl iFk ij 180 fdeh@?kaVk pky ls xfr'khy gks rks feyh oksYVehVj dk ikB~;kad fdruk gksxk\ i`Foh ds pqEcdh; {ks=k ds m/oZ?kVd dk eku 0.2 × 10 – 4 VsLyk rFkk iVfj;ksa ds chp dh nwjh 1eh- gSA Ans: 31.
1 mV
A right angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in 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 pky v 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 c B
v
a
Ans:
b
(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 constant velocity v .A uniform magnetic field B exist in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if (a) the velocity is perpendicular to the diameter joining free ends (b) the velocity is parallel to this diameter. rkacs dk ,d rkj v)Zo`Ùkkdkj vkÑfr esa eksM+k x;k gS] bldh f=kT;k r gS vkSj ;g blds ry esa fu;r osx v ls foLFkkfir gks 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 Kkr dhft;s] ;fn (a) osx] eqDr fljksa dks tksM+us okys O;kl ds yEcor~ gS] (b) osx] bl O;kl ds lekUrj gSA Ans: (a) 2rvB (b) zero 'kwU;
33.
A wire of length 10 cm translates in a direction making an angle of 60º with its length. The plane of motion is perpendicular to a uniform magnetic field of 1.0 T that exists in the space Find the emf induced between the ends of the rod if the speed of translation is 20 cm/s. 10 lseh yEck rkj bldh yEckbZ ls 60º dks.k cukrs gq, LFkkukarfjr gks jgk gSA bldh xfr dk ry ml LFkku ij fo|eku 1.0 VsLyk rhozrk ds pqEcdh; {ks=k ds yEcor~ gSA ;fn LFkkukarj.k dh pky 20 lseh@ls gS] rks NM+ ds fljksa ds chp izfs jr
fo-ok-cy Kkr dhft;sA
manishkumarphysics.in
Page # 15
Chapter # 38 Ans: 17 × 10–3V 34.
Electromagnetic Induction
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 diametrically opposite points on the ring. (a) Between which pair of points is the emf? (b) Between which pair of points is the emf minimum? What is the value of this minimum emf? r f=kT;k okyh rkacs dh oy; blds ry esa fu;r osx v ls 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 ;qXe ds 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 diameter parallel to the velocity, zero. 35.
Figure shows a wire sliding on two parallel, conducting rails placed at a separation . A magnetic field B exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity v? fp=k esa n'kkZ;k x;k gS fd nwjh ij fLFkr nks lekUrj pkyd iVfj;ksa ds Åij ,d rkj fQly jgk gSA iVfj;ks ds ry ds yEcor~ fn'kk esa pqEcdh; {ks=k B fo|eku gSA rkj dks fu;r osx v ls xfr'khy j[kus ds fy;s fdruk cy vko';d gS ? ×
×
×
×
×
×
×
×
Ans. 36.
×
×
×
×
×
×
v
×
×
×
×
×
×
×
×
×
×
×
×
×
×
zero
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 unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, 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-vkÑfr dk rkj yEcor~ pqEcdh; {ks=k B esa j[kk gqvk gSA U-vkkdj okys rkkj ij ,d yEckbZ okyk rkj nk;ha vksj fu;r osx v ds lkFk xfr'khy gSA leLr rkjksa dh ,dkad yEckbZ dk izfrjks/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 cSBjh
ds :i esa n'kkZrs gq, ,d rqY; ifjiFk cukb;sA ifjiFk esa /kkjk dh x.kuk dhft;sA
Ans. 37.
×
×
×
×
×
×
×
×
×
×
×
×
×
v ×
×
×
×
×
×
×
×
×
×
×
×
×
×
Bv (b) /v 2r( vt )
Consider the situation of the previous problem. (a) Calculate the force needed to keep the sliding wire moving with a constant velocity v. (b) If the force needed just after t = 0 is F0, find the time at which the force needed will 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;s vko';d cy dh x.kuk dhft;sA (b) ;fn t = 0 lsd.M ds i'pkr~ vko';d cy F0 gS rks og le; Kkr dhft;sA ftl ij vko';d cy F0/2 gks tk;sxkA Ans.
38.
(a)
×
B2 2 v (a) 2 r( vt )
(b) /v.
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. A uniform magnetic field 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 wire at this instant, (c) the velocity v as a function of x and (d) the maximum distance the wire will move, manishkumarphysics.in
Page # 16
Chapter # 38
Electromagnetic Induction
fp=k esa n'kkZ;h x;h fLFkfr ij fopkj dhft;sA rkj PQ dk 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:i pqEcdh; {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 PQ dks v0 pky ls nk;ha vksj /kdsyk tkrk gSA Kkr dhft;s % (a) ftl {k.k ij PQ dh pky v gS] ywi esa izokfgr /kkjk (b) bl {k.k ij rkj dk Roj.k, (c) x ds Qyu :i esa osx vkSj (d) og vf/kdre nwjh tks rkj r; djsxkA
Ans. 39.
×
×
×
×
×
×
×
×
×
×
×
×
×
× Q×
×
×
mv 0 (R r ) B2 2 v B2 2 x B v (b) towards left (c) v = v0 – (d) . B2 2 m (R r ) m (R r ) Rr
(a) 25 m/s (b) 4.0 × 10–2 V (c) 3.6 × 10–2 V (d) 4.0 × 10–3 V.
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 2.0µA passes through the wire when it is slid at a rate of 20 cm/s. If the horizontal component of the earth’s magnetic field is 3.0 x 10–5 T, calculate the dip at the place. fp=k esa iznf'kZr gS fd 0.20 izfrjks/k okyk /kkfRod rkj] {kSfrt ,oa U-vkÑfr dh /kkfRod iVjh ij fQly jgk gSA lekUrj Hkqtkvksa ds chp 20 lseh varjky gSA tc rkj 20 lseh@ls dh nj ls fQlyrk gS rks rkj ls 2.0µA /kkjk izokfgr gksrh gSA ;fn i`Foh ds pqEcdh; {ks=k ds {kSfrt ?kVd dk eku 3.0 x 10–5 T gS rks bl LFkku ij ufr Kkr dhft;sA
Ans. 41.
× P×
A rectangular frame of wire abcd has dimensions 32 cm × 8.0 cm and a total resistance of 2.0 . It is pulled out of a magnetic field B = 0.020 T by applying a force of 3.2 × 10–5 N (figure). It is found that the frame moves with constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the points a and b and (d) the potential difference between the points c and d. rkj ds ,d vk;rkdkj Ýse abcd dh foek,a 32 lseh × 8.0 lseh gS vkSj bldk dqy izfrjks/k 2.0 gSA bldks 3.2 × 10–5 U;wVu cy yxkdkj pqEcdh; {ks=k B = 0.020 VsLyk ds ckgj [khapk tkrk gSA ¼fp=k½ ;g izsf{kr gksrk gS fd Ýse fu;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 b ds chp foHkokUrj] rFkk (d) fcUnqvksa c o d ds chp foHkokarj
Ans. 40.
(a)
×
tan–1 (1/3)
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 =
mg R sin v 2 cos 2
,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 fQlyrk gSA 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 manishkumarphysics.in
Page # 17
Chapter # 38
Electromagnetic Induction
mg R sin
ij fu;r pky v ls fQlyrk gS] rks O;Dr dhft;s fd B = a
v 2 cos 2
b
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) if P1Q1 moves towards left but P2Q2 moves towards right. fp=k esa n'kkZ;h x;h fLFkfr ij fopkj dhft;sA rkj P1Q1 vkSj P2Q2 dks iVfj;ksa ij leku pky 5 lseh@ls ds lkFk fQlyk;k tkrk gSA 19 izfrjks/k ls izokfgr /kkjk Kkr dhft;s] ;fn % (a) nksuksa rkj nk;ha vksj xfr djsa] rFkk (b) P1Q1 ck;ha vksj xfr djs fdUrq P2Q2 nk;ha vksj xfr djsA
× P1 × P2 ×
× × 4 cm ×
× 2 ×
× Ans. 43.
×
× Q1 × Q2 ×
×
× 19 ×
×
× × B=1.0 T
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 /kkjk Ans.
(a) zero
'kwU; (b) 1 mA
Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm/s. Find the current in the 10 resistor when the switch S is thrown to (a) the middle rail (b) bottom rail. fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA rkj PQ dk izfrjks/k ux.; gS vkSj bldks rhu iVfj;ksa ij 5 lseh@ls dh fu;r pky ls fQlyk;k tkrk gSA 10 izfrjks/k esa izokfgr /kkjk ml fLFkfr esa Kkr dhft;s tcfd fLop S : (a) chp okyh iVjh dh vksj gSA (b) uhps okyh iVjh dh vksj gSA
Ans. 45.
× 2 ×
×
(a) 0.1 mA (b) zero
Kkr dhft;sA 44.
×
×
10 ×
×
P×
×
×
×
2 cm ×
×
×
×
×
×
2 cm ×
×
×
×
×
S × × B=1.0T
×
Q×
×
×
×
×
×
×
× ×
×
(a) 0.1 mA (b) 0.2 mA
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 these wires 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 ab dks nk;ha vksj fu;r pky v ls 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
manishkumarphysics.in
Page # 18
Chapter # 38
Electromagnetic Induction
× d × a×
× i
× ×
×
×
× c × b×
×
v×
×
×
×
ir Bv 2r
Ans. 46.
g
×
The current generator g, shown in figure, sends a constant current i through the circuit. The wire ab has a length and mass m and can slide on the smooth, horizontal rails connected to g. The entire system lies in a 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 ab dh yEckbZ vkSj nzO;eku m gS rFkk ;g g dks tksMu+ s okyh fpduh ,oa {kSfrt iVfj;ksa ij fQly ldrk gSA lEiw.kZ fudk; m/okZ/kj pqEcdh; {ks=k B esa fo|eku
gSA rkj dk osx le; ds Qyu :i esa Kkr dhft;sA ×
×d × a×
×
×
×
×
×
×c × b×
×
×
× g × × Ans. 47.
×
iBt/m, away from the generator
The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontal magnetic field B that is perpendicular to the plane of the rails (figure). It is found that the wire stays in equilibrium. If the wire ab is replaced by another wire of double its mass, how long will it take in falling through a 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;ksa ds ry ds yEcor~ gSA ¼fp=k nsf[k;s½ ;g ik;k x;k fd rkj lkE;koLFkk esa :dk jgrk gSA ;fn rkj ab dks ,d vU;
rkj ls izfrLFkkfir dj fn;k tkrk gSA ftldk nzO;eku blls nqxuk gS] ;g bldh yEckbZ ds cjkcj nwjh rd fxjus esa fdruk le; yxk;sxk\ g
b
a
Ans.
2 /g .
48.
The rectangular wire-frame, shown in figure, has uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t = 0. (a) Find the acceleration of the frame when its speed has increased to . (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity 0. (c) Show that the velocity at time t is given by v v 0 (1 e Ft / mv 0 ) .
fp=k esa iznf'kZr rkj ds vk;rkdkj Ýse dh pkSM+kbZ d, nzO;eku m , izfrjks/k R vkSj yEckbZ cgqr vf/kd gSA t = 0 ij ,d fu;r cy F, Ýse dks pqEcdh; {ks=k esa /kdsyuk izkjEHk djrk gSA (a) Ýse dh pky v rd c<+us ij bldk Roj.k Kkr dhft;sA (b) O;Dr dhft;s fd dqN le; i'pkr~ lEiw.kZ Ýse ds pqEcdh; {ks=k esa izfo"V gksus rd Ýse fu;r osx ls xfr djsxkA ;g osx 0 Kkr dhft;sA (c) O;Dr dhft;s fd le; t ij osx dk eku v v 0 (1 e Ft / mv 0 ) gksxkA
manishkumarphysics.in
Page # 19
Chapter # 38
Ans. 49.
(a)
×
×
×
×
×
×
d ×
×
×
×
×
×
×
×
×
F
RF RF v 2B 2 (b) 2 2 B mR
Figure shows a smooth pair of thick metallic rails connected across a battery of emf having a negligible internal resistance. A wire ab of length and resistance r can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire is given a small velocity towards right. (a) Find the current in it at this instant. What is the direction of the current ? (b) What is the force acting on the wire at this instant ? (c) Show that after some time the wire ab will 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 okyh cSVjh tqM+h gqbZ gSA yEckbZ ,oa r izfrjks/k okyk rkj iVfj;ksa ij lqxerkiwoZd fQly ldrk gSA lEiw.kZ fudk; {kSfrt ry 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 iznku fd;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 jgk cy fdruk gS\ (c) O;Dr dhft;s fd dqN le; i'pkr~ rkj ab fu;r osx ls fQlysxkA ;g osx Kkr dhft;sA
Ans. 50.
Electromagnetic Induction
(a)
× ×
× ×
× ×
× ×
a × ×
× ×
×
×
×
×
×
×
×
×
×
×
b ×
×
1 B E (E – vB), from b to a (b) (E – vB) towards right (c) . r r B
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 direction perpendicular to the plane of the rails. (a) Write the induced emf in the loop at an instant t when the speed of the wire is . (b) What would be the magnitude and direction of the induced current in the wire ? (c) Find the downward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity m. (e) Find the velocity of the wire as a function of time. (f) Find the displacement of the wire as a function of time. (g) Show that the rate of heat developed in the wire is equal to the 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 rFkk
fpduh iVfj;ksa ij fQlyuk izkjEHk djrk gSA bl LFkku ij iVfj;ksa ds ry ds yEcor~ fn'kk esa le:i pqEcdh; {ks=k B 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 ,oa ifjek.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;r pky ls xfr djrk gSA ;g fu;r pky m Kkr 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 Å"ek dk eku xq:Roh; fLFkfrt ÅtkZ esa deh ds cjkcj gSA ×
×
×
×
×
×
×
×
×
a ×
×
b ×
× × × manishkumarphysics.in ×
×
×
Page # 20
Chapter # 38
Electromagnetic Induction 2
Ans.
51.
vm mgr vB B2 2 (a) vB (b) , b to a (c) g – v (d) 2 2 (e) vm(1 – e gt / v m ) (f) vmt – (1 – e gt / v m ) g r B mr
A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of 100 revolutions per minute. If the length of each spoke is 30.0 cm and the horizontal component of the earth’s magnetic field is 2.0 x 10–5 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 i`Foh ds pqEcdh; {ks=k dk {kSfrt ?kVd 2.0 x 10–5 T gS rks v{k ,oa LikWd ds ckgjh fljs ds chp izsfjr foHkokarj Kkr dhft;sA LikWd ds bysDVªkstu ij yxus
okyk vfHkdsUnzh; cy ux.; eku yhft;sA Ans. 52.
A conducting disc of radius r rotates with a small but constant angular velocity about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the centre and the periphery 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.
53.
9.4 × 10–6 V
1 2 r B 2
Figure shows a conducting disc rotating about its axis in a perpendicular magnetic field B. A resistor of resistance R is connected between the centre and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the centre ? The radius of the disc is 5.0 cm, angular speed = 10 rad/s, B = 0.40 T and 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 R tksM+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
54.
×
×
×
×
×
× Ans.
×
×
×
×
×
0.5 mA, leaves
B 0 The magnetic field in a region is given by B = k y where L is a fixed length. A conducting rod of length L L lies along the Y-axis between the origin and the point (0, L, 0). If the rod moves with a velocity v v 0 i , find the emf induced between the ends of the rod.
B
fdlh LFkku ij pqEcdh; {ks=k B = k
0
L
y }kjk
O;Dr fd;k tkrk gS] tgk¡ L ,d fu;r yEckbZ gSA L yEckbZ 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+ v v 0 i osx ls xfr'khy gS] ds NM+ ds fljksa ds chp izsfjr fo-ok-c- Kkr dhft;sA Ans. 55.
B0 v 0 2
Figure shows a straight, long wire carrying a current i and a rod of length coplanar with the wire and perpendicular to it. The rod moves with a constant velocity v in a direction parallel to the wire. The distance of 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 dh nwjh x gSA NM+ esa izsfjr xfrd fo-ok-c- Kkr dhft;sA manishkumarphysics.in
Page # 21
Chapter # 38
Electromagnetic Induction v i
x
Ans. 56.
2x 0 iv ln 2 2x
Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick 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 current in the resistance R ? (c) Find the rate of heat developed in the resistor. (d) Find the power delivered by the external 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 pkyd iVfj;ksa ij fQly jgs gSaA iVfj;ksa ds ,d fljs ij izfrjks/k R tqM+k gqvk gSA (a) NM+ dks fu;r pky v ls fQlyrk gqvk j[kus ds fy;s fdrus cy dh vko';drk gksxh\ (b) bl fLFkfr esa izfrjks/k R ls fdruh /kkjk izokfgr gksxh (c)izfrjks/k esa Å"ek mRiUu gksus dh nj fdruh gksxh\ (d) NM+ ij cy yxkus ds fy;s cká ;qfDr }kjk iznÙk 'kfDr Kkr dhft;sA 2
Ans. 57.
2
iv 2x v 0i 2x 1 0iv 2x ln ln (b) 0 ln (d) same as (c) (a) (c) 2R 2x R 2 2 x R 2 2x
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 Ýse ,d yEcs ,oa lh/ks rkj ds leryh; j[kk gqvk gSA rkj ls i /kkjk izokfgr gks jgh gS] tks i = i0 sin t }kjk O;Dr dh tkrh gSA Kkr dhft;s % (a) oxkZdkj Ýse ls xqtjus okyk pqEcdh; ¶yDl (b) Ýse esa izsfjr fo-ok-c- vkSj (c) le;karjky 0 ls
20
ds fy;s Ýse esa mRiUu Å"ekA
a
i b 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 the loop. 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 Faraday’s law for the flux through the loop and also by replacing different segments with equivalent batteries. yEckbZ vkSj b pkSMk+ bZ okyk ,d vk;rkdkj /kkfRod ywi ,d yEcs ,oa lh/ks rkj ds leryh; j[kk gqvk gS] rkj ls i /kkjk izokfgr gks 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 fljk rkj ls a nwjh ij gks] rc ywi esa izfs jr fo-ok-c- dh x.kuk dhft;sA ywi ls xqtjus okys ¶yDl ds fy;s QSjkMs dk fu;e izLrqr
djrs gq, rFkk lkFk gh fofHkUu [k.Mksa dks lerqY; cSVfj;ksa ls izfrLFkkfir djrs gq,] gy dhft;sA
i b
v
a manishkumarphysics.in
Page # 22
Chapter # 38 Ans. 59.
Electromagnetic Induction
ivb 2a(a )
Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity . Find the force. fp=k esa iznf'kZr gS fd a f=kT;k okyk o`Ùkkdkj pkyd ywi yEcor~ le:i pqEcdh; {ks=k B esa j[kk gqvk gSA /kkrq dh ,d eksVh NM+ dk nwljk fljk ywi ds A ij Li'kZdj jgk gSA dsUnz O ,oa ywi ij ,d fLFkj fcUnq C dks R izfrjks/k okys rkj OC ls la;ksftr fd;k x;k gSA NM+ OA ds e/; fcUnq ij ,d cy yEcor~ yxk;k tkrk gS] ftlls NM+ nf{k.kkorhZ fn'kk esa ,d leku dks.kh; osx ls ?kwerh gSA cy Kkr dhft;sA
×
×
×
×
×
×
×
×
×O
×
×
×
×
×
×
×
×
×
C
× A ×
F
Ans.
60.
a 3B 2 to the right of OA in the figure. 2R
Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA is made to rotate with a uniform angular speed as shown in the figure. Find the current in the rod when
AOC 90 .
fiNys iz'u ds fp=k esa of.kZr fLFkfr ij fopkj dhft;sA eku yhft;s fd la;kstu rkj O o C dk izfrjks/k 'kwU; gS] fdUrq o`Ùkkdkj ywi dk izfrjks/k R gS vkSj ;g bldh yEckbZ ij ,d leku :i ls forfjr gSA NM+ OA dks fp=kkuqlkj fu;r dks.kh; osx ls ?kqek;k tkrk gSA tc AOC 90 gks rks NM+ esa /kkjk Kkr dhft;sA Ans. 61.
8 a 2B 3 R
Consider a variation of the previous problem (figure). Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle with the vertical.
fiNys iz'u esa ,d ifjorZu ij fopkj dhft;sA eku yhft;s fd o`Ùkkdkj ywi m/oZ ry esa fLFkr gSA NM+ dk nzO;eku m gSA NM+ ,oa ywi dk izfrjks/k ux.; gS] fdUrq la;kstu rkj O o C dk izfrjks/k R gSA OA ds e/; fcUnq ij] NM+ ds yEcor~ ,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. 62.
a 3B 2 – mg sin 2R
Figure shows a situation similar to the previous problem. All parameters are the same except that a battery of emf and a variable resistance R are connected between O and C. The connecting wires have zero resistance. No external force is applied on the rod (except gravity, forces by the magnetic field and by the pivot). In what way should the resistance R be changed so that the rod may rotate with uniform angular velocity 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 C ds chp esa fomanishkumarphysics.in
Page # 23
Chapter # 38
Electromagnetic Induction
ok-cy okyh ,d cSVjh vkSj ,d izfrjks/k R tksM+s x;s gSAa la;kstu rkjksa dk izfrjks/k 'kwU; gSA NM+ ij dksbZ cká cy ¼xq:Ro] dhyd ,oa pqEcdh; {ks=k ds cyksa ds vykok½ ugha yxk;k tk jgk gSA R dks fdl izdkj ifjofrZr fd;k tk;s ftlls fd NM+ dks nf{k.kkorhZ fn'kk esa ,d leku dks.kh; osx ls ?kqek;k tk lds\ vius mÙkj 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
×
×
×
×
×
×
× C ×
×
×O
×
×
×
×
×
×
×
×
×
R
A × F
Ans. 63.
aB(2 a 2B) 2mg cos
A wire of mass m and length can slide freely on a pair of smooth, vertical rails (figure). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end 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 ry ds 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 fdlh
Hkh izfrjks/k dks xkS.k ekurs gq, rkj dk Roj.k Kkr dhft;sA
Ans. 64.
×
×
C×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
mg m CB2 2
A uniform magnetic field B exists in a cylindrical region, shown dotted in figure. The magnetic field increases dB . Consider a circle of radius r coaxial with the cylindrical region. (a) Find the magnitude dt of the electric field E at a point on the circumference of the circle. (b) Consider a point P on the side of the square circumscribing the circle. Show that the component of the induced electric field at P along ba is the same as the magnitude found in part (a).
at a constant rate
,d csyukdkj {ks=k esa ¼fp=k esa js[kkafdr n'kkZ;k x;k gS½ le:i pqEcdh; {ks=k B fo|eku gSA pqEcdh; {ks=k esa
dB dt
dh
fu;r nj ls o`f) gks jgh gSA csyukdkj {ks=k ds lek{kr% r f=kT;k ds o`Ùk ij fopkj dhft;sA (a) o`Ùk dh ifjf/k ij fLFkr fcUnq ij fo|qr {ks=k E dk ifjek.k Kkr dhft;sA (b) o`Ùk dks ifjc) djus okys oxZ dh Hkqtk ij fLFkr fcUnq P ij izsfjr fo|qr {ks=k dk ba ds vuqfn'k ?kVd Hkkx (a) esa izkIr ifjek.k ds cjkcj gSA
manishkumarphysics.in
Page # 24
Chapter # 38
Electromagnetic Induction
×
× a
× × ×
×
× d ×
65.
(a)
× × c×
×
×
Ans.
× × P × b × × × ×
× ×
×
×
×
×
r dB 2 dt
The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A/s. The solenoid has 2000 turns/m and its radius in 6.0 cm. (a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds. (b) Find the electric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a point outside 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<+ jgh gSA ifjufydk esa 2000 Qsjs@eh- gS ,oa bldh f=kT;k 6.0 lseh gSA (a) ifjufydk ds vUnj 1.0 lseh f=kT;k ds o`Ùk ij fopkj dhft;sA ftldh v{k ifjufydk dh v{k ls lEikfrr gsA o`Ùk ls xqtjus okys ¶yDl esa 2.0 lsd.M esa gksus okyk ifjorZu Kkr dhft;sA (b) o`Ùk dh ifjf/k ij fLFkr fdlh fcUnq ij izsfjr fo|qr {ks=k Kkr dhft;sA (c) ifjufydk ds ckgj ,oa bldh v{k ls 8.0 lseh nwjh
ij fLFkr fcUnq ij izsfjr fo|qr {ks=k Kkr dhft;sA Ans.
(a) 1.6 × 10–8 weber (b) 1.2 × 10–27 V/m (c) 5.6 × 10–7 V/m
66.
An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor. tc fdlh izsj.k dq.Myh esa ,d fn'kk esa izokfgr 2.5 A /kkjk dks 0.1 lsd.M esa ifjofrZr djds brus gh eku dh /kkjk foifjr fn'kk esa izokfgr dh tkrh gS] rks vkSlr fo-ok-c- 20 V izsfjr gksrk gSA izsfjr dq.Myh dk Loizjs dRo Kkr dhft;sA Ans. 0.4 H
67.
A magnetic flux of 8 x 10–4 weber is linked with each turn of a 200 turn coil when there is an electric current of 4A in it. Calculate the self-inductance of the coil. 200 Qsjksa okyh dq.Myh esa 4 A /kkjk izo kfgr djus ij blds izR;sd Qsjs ls lEc) pqEcdh; ¶yDl 8 x 10–4 oscj gSA
dq.Myh ds LoizsjdRo dh x.kuk dhft;sA Ans. 68.
4 × 10–2 H
The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A/s. Find the emf induced in it. 240 Qsjksa] 12 lseh yEckbZ ,oa 2 lseh f=kT;k okyh ifjufydk esa /kkjk 0.8 A/s dh nj ls ifjofrZr gks jgh gSA blesa izsfjr
fo-ok-c- Kkr dhft;sA Ans.
6 × 10–4 V
69.
Find the value of t/ for which the current in an L/R circuit builds up to (a) 90%, (b) 99% and (c) 99.9% of the steady-state value t/ dk og eku Kkr dhft;s] ftlds fy;s ,d L/R ifjiFk esa /kkjk dk eku LFkk;h fLFkfr dk (a) 90%, (b) 99% vkSj (c) 99.9% gks tk;sA Ans. 2.3, 4.6, 6.9.
70.
An inductor-coil carries a steady-state current of 2.0 A when connected across an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of the circuit. ,d izjs d dq.Myh dks 4.0 fo-ok-c- okyh vkn'kZ cSVjh ls tksM+us ij blesa LFkk;h voLFkk esa 2.0 A /kkjk izokfgr gksrh gsA ;fn izsjdRo 1.0 H gS] rks ifjiFk dk le;&fLFkjkad Kkr dhft;sA Ans. 0.50 s
71.
A coil having inductance 2.0 H and resistance 20 is connected to a battery of emf 4.0 V. Find (a) the current at the instant 0.20 s after the connection is made and (b) the magnetic field energy at this instant. 2.0 H izsjdRo ,oa 20 izfrjks/k okyh dq.Myh dks 4.0 V fo-ok-cy dh cSVjh ls la;ksftr fd;k x;k gSA Kkr dhft;s% (a) la;kstu djus ds 0.20 lsd.M i'pkr~ /kkjk] vkSj (b) bl {k.k ij pqEcdh; {ks=k dh ÅtkZA manishkumarphysics.in
Page # 25
Chapter # 38 Ans. (a) 0.17 A (b) 0.03 J
Electromagnetic Induction
72.
A coil of resistance 40 is connected across a 4.0 V battery, 0.10 s after the battery is connected, the current 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 /kkjk 63 mA gSA dq.Myh dk izsjdRo Kkr dhft;sA Ans. 4.0 H
73.
An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switched on. 5.0 H izsjdRo ,oa ux.; izfrjks/k okyh dq.Myh dks 100 ds izfrjks/k rFkk 2.0 V fo-ok-c- dh cSVjh ds lkFk Js.khØe esa tksM+k x;k gSA fLop pkyw djus ds 20 ms i'pkr~ ifjiFk esa izfrjks/k ds fljksa ij foHkokUrj Kkr dhft;sA Ans. 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 is found 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 2.0 A LFkk;h /kkjk izkIr gksrh 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 Kkr
dhft;sA Ans.
(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 = 20 gSA bldks t = 0 ij 2.0 V fo-ok-c- ds fljksa ij tksM+k x;k gSA (a) t = 100 feyh lsd.M , (b) t = 200 feyh lsd.M rFkk (c) t = 1.0 lsd.M] ij di/dt dk eku Kkr dhft;sA Ans. (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 indicated therein?
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 × 10–9 V
77.
An inductor-coil of inductance 20 mH having resistance 10 is joined to an ideal battery of emf 5.0 V. Find the 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;sA Ans. (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 the potential 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.khØe esa la;ksftr gSaA izfrjks/k ds fljksa ij foHkokarj t = (a) 20.0 feyh lsd.M , (b) 100 feyh lsd.M rFkk (c) 1.00 lsd.M ij Kkr dhft;sA Ans. (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 and internal resistance 2W. Find the charge which flows through the inductor in (a) 10 ms, (b) 20 ms and (c) 100 ms 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 cSVjh ds 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 feyh
lsd.M] i'pkr~ izsj.k dq.Myh ls izokfgr vkos'k Kkr dhft;sA Ans. 80.
(a) 1.8 mC (b) 5.7 mC and (c) 45 mC
An inductor-coil of inductance 17 mH is constructed from a copper wire of length 100 m and cross-sectional area 1 mm2. Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper = 1.7 x 10–8 -m. 100 eh- yEcs ,oa 1 mm2 vuqizLFk dkV {ks=kQy okys rkacs ds rkj ls 17 mH izsjdRo okyh izsj.k dq.Myh cuk;h x;h
gSA ;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 rkacs dh izfrjks/kdrk = 1.7 x 10–8 - ehAns.
10 ms
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Chapter # 38 Electromagnetic Induction 81. An LR circuit having a time constant of 50 ms in connected with an ideal battery of emf . Find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its 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) Å"ek ds :i esa O;rhr 'kfDr dk vf/kdre dk eku izkIr djus rd rFkk (c) ifjiFk esa pqEcdh; {ks=k ÅtkZ dk vf/kdre dk vk/kk eku izkIr djus rd Ans. (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 for the magnetic energy stored in the circuit to change from one fourth of the steady-state value to half of the steady-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 ÅtkZ dk
eku LFkk;h voLFkk ds ,d pkSFkkbZ ls LFkk;h voLFkk ds vk/ks rd ifjofrZr gksus esa yxk le; Kkr dhft;sA Ans.
ln
1 2 2
83.
A solenoid having inductance 4.0 H and resistance 10 is connected to a 4.0 V battery at t = 0. Find (a) the time constant (b) the time elapsed before the current reaches 0.63 of its steady-state value, (c) the power delivered 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 V dh 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 }kjk iznÙk 'kfDr vkSj (d) bl {k.k ij twy Å"ek ds :i esa O;f;r 'kfDrA Ans. (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 when the 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 eku blds vf/kdre eku dk 0.80 xquk gks tkrk gSA ifjiFk dk izfrjks/k Kkr dhft;sA Ans. 160
85.
An LR circuit with emf is connected at t = 0. (a) Find the charge Q which flows through the battery during 0 to t. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time t. (e) Verify that the results in the three parts 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'k Q Kkr dhft;sA (b) bl dky esa cSVjh }kjk fd;k x;k dk;Z Kkr dhft;sA (c) bl dky esa mRiUu Å"ek Kkr dhft;sA (d) le; t ij ifjiFk esa lafpr pqEcdh; ÅtkZ Kkr dhft;sA (e) ;g tkafp;s fd mDr rhu Hkkxksa esa izkIr ifj.kke ÅtkZ laj{k.k fu;e
dh iqf"V djrs gSaA Ans.
(d)
(a)
L 2 L 2 L (3 4x x 2 ) t (1 x ) (b) t (1 x ) (c) t R R R R R 2 R
L 2 (1 x )2 where x – e Rt / t 2R
86.
An inductor of inductance 2.00 H is joined in series with a resistor of resistance 200 and a battery of emf 2.00 V. At t = 10 ms, find (a) the current in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field. 2.00 H izsjdRo okyh izjs .k dq.Myh dks 200 ds izfrjks/k rFkk 2.00 V fo-ok-c- dh cSVjh ds lkFk Js.khØe esa tksM+k x;k gSA t = 10 feyh lsd.M ij] Kkr dhft;s % (a) ifjiFk esa /kkjk (b) cSVjh }kjk iznÙk 'kfDr, (c) izfrjks/k dks xeZ djus esa O;f;r 'kfDr rFkk (d) pqEcdh; {ks=k esa lafpr dh tk jgh ÅtkZ dh nj Ans. (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 . Each coil 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 time t. Find the ratio iA/iB at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1 s. nks dq.Mfy;ksa A rFkk B ds izsjdRo Øe'k% 1.0 H ,oa 2.0 H gSA izR;sd dq.Myh dk izfrjks/k 10 gSA izR;sd dq.Myh dks t = 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 iA vkSj iB gSAa vuqikr iA/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;sA Ans. (a) 1.6 (b) 1.4 (c) 1.0 manishkumarphysics.in
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Chapter # 38 Electromagnetic Induction 88. The current in a discharging LR circuit without the battery drops from 2.0 A to 1.0 A in 0.10 s. (a) Find the time constant 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 A ls 1.0 A rd de gks tkrk gSA (a) ifjiFk dk 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.
A constant current exists in an inductor-coil connected to a battery. The coil is short-circuited and the battery is removed. Show that the total charge flown through the coil after the short-circuiting is the same as that which 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 vkSj cSVjh gVk yh tkrh gSA O;Dr dhft;s fd y?kqifFkr djus ds i'pkr~ izokfgr vkos'k dk eku mruk gh gS] ftruk fd y?kqifFkr djus ls ,d le;kUrjky iwoZ izokfgr gqvk FkkA 90.
Consider the circuit shown in figure. (a) Find the current through the battery a long time after the switch S is closed. (b) Suppose the switch is again opened 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 S dks 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) ,d
R2
R1
L
le;&fLFkjkad ds i'pkr~ izsjdRo ls izokfgr /kkjk Kkr dhft;sA Ans. 91.
(a)
(R1 R 2 ) L (b) (c) R1R 2 R1 R 2 R1e
S
A current of 1.0 A is established in a tightly wound solenoid of radius 2 cm having 1000 turns/metre. Find the magnetic energy stored in each metre of the solenoid. dl dj yisVh x;h 2 lseh f=kT;k o 1000 Qsjs@ehVj okyh ifjufydk esa 1.0 A /kkjk izokfgr gks jgh gSA ifjufydk
ds izR;sd ehVj esa lafpr pqEcdh; ÅtkZ Kkr dhft;sA Ans. 92.
7.9 × 10–4 J
Consider a small cube of volume 1 mm3 at the centre of a circular loop of radius 10 cm carrying a current of 4A. Find the magnetic energy stored inside the cube. 10 lseh f=kT;k ,oa 4A /kkjk izokg okys o`Ùkkdkj ywi ds dsUnz ij 1 feeh3 vk;ru okys ?ku ij fopkj dhft;sA ?ku ds
vUnj lafpr pqEcdh; ÅtkZ Kkr dhft;sA Ans. 93.
8 × 10–14 J
A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of 1 mm3 at a distance of 10.0 cm from the wire. ,d yEcs rkj ls 4.00 A /kkjk izokfgr gks jgh gSA rkj ls 10.00 lseh nwjh ij fLFkr 1.00 mm3 vk;ru esa lafpr pqEcdh;
{ks=k ÅtkZ Kkr dhft;sA Ans. 94.
2.55 × 10–14 J
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 gksrh
gS] rks nwljh dq.Myh esa fdruk fo-ok-c- izsfjr gksxk\ Ans. 95.
2.5 V
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
i b
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Chapter # 38 Ans. 96.
Electromagnetic Induction 0 a a ln 1 2 b
Find the mutual inductance between the circular coil and the loop shown in figure.
fp=k esa iznf'kZr o`Ùkkdkj dq.Myh vkSj ywi ds chp vU;ksU; izsjdRo Kkr dhft;sA
Ans. 97.
N
0 a 2a'12 2(a 2 x 2 )3 / 2
A solenoid of length 20 cm, area of cross-section 4.0 cm2 and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-sectional area 8.0 cm 2 and length 10 cm. Find the mutual inductance between the solenoids. 20 lseh yEch] 4.0 lseh2 vuqizL Fk dkV {ks=kQy vkSj 4000 Qsjksa okyh ifjufydk] ,d vU; ifjufydk ds vUnj j[kh gqbZ gSA ftldh yEckbZ 10 lseh vuqizLFk dkVj {ks=kQy 8 lseh2 vkSj Qsjksa dh la[;k 2000 gSA ifjufydkvksa ds chp
vU;ksU; izsjdRo Kkr dhft;sA Ans. 98.
2.0 × 10–2 H
The current in a long solenoid of radius R and having n turns per unit length is given by i = i0 sin t. A coil having N turns is wound around it neat the centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid and the coil. R f=kT;k ,oa izfr ,dkad yEckbZ esa n Qsjksa okyh ,d yEch ifjufydk esa izokfgr /kkjk i = i0 sin t }kjk O;Dr dh tkrh gSA blds dsUnz ds lehi N Qsjksa 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; izsjdRoA Ans. (a) 0i0 nNR2 cos t (b) 0nNR2
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