Ch-38 (Electro Magnetic Induction)

Chapter # 38

1.

[1]

Objective - I

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

Sol.

Electromagnetic Induction

(B*)

1 2 Bl 8

(C)

1 2 Bl 2

(D) Bl2

B Take a small element dx at a distance of ‘x’ centre x

 

2.

l 2

0

0

Bwx2 d  Bxdx  2



dx 1 Bwl2 8

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 baf xr le:i pqEcdh; {ks=k B fo|eku gSA NM+ ds dsUnz rFkk NM+ ds nksuksa fljksa ds chp foHkokUrj gS & (A*) zero

Sol.

l 2

(B)

1 Bl2 2

(C) Bl2

(D) Bl2

D 1 Bwl2 8 So the potential difference between the two ends of therod is zero.

Emf at both end is same =

3.

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 &’

Sol.

(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 D When the switch is closed than a clock wise current pulse generated (Because initially current flow the terminal to negative terminal). Due to Mutual Induction, current is generated in the loop. If circuit is open after some time. Dut to loop an anticlock wise current pulse generated in the circuit.

manishkumarphysics.in

+ -

Chapter # 38 Electromagnetic Induction [2] 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

Sol. 5.

(C*) C An anticlock wise current-pulse generated and then a clock-wise current pulse. 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 &

Sol.

(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 B

(B*) will move with almost contant speed (D) will oscillate (B*) yxHkx fu;r pky ls xfr djsxkA (D) nksyu djsxkA

Copper Tube

M After sometime the Magnet will move with almost contant speed. 6.

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. Sol.

;g bl ij fuHkzj djsxk fd ifjufydk ds ck;sa fljs ls cSVjh dk dkSulk VfeZuy ¼/kukRed ;k _.kkRed½ tqM+k gqvk gSA C

N i 'LR' circuit

I

F

W

E S

di dt Current flow in the CKt is clock wise direction, due to Mutual Induction current flow in the loop anti clockwise direction. The net force applied on the loop in east direction. So we can say that the ring will move away from the solenoid. e  L

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.


Chapter # 38


[3]

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) pqE cdh; {ks=k ifjofrZr djds fo-ok-c- izsf jr fd;k tk ldrk gSA

Sol.

(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*) A o B nksuksa gh lR; gSA (B) A lR; gS] fdUrq B vlR; gSA (C) B lR; gS] fdUrq A vlR; gS (D) A o B nksuksa xyr gSA A  An emf con be induced by moving a condcutor in a magnetic field  = Bvl  An emf can be induced by charging the magnetic field. 

8.

d dt

  flux

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 &

Sol.

(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
x

x

x

v x

A

x

x

R

 x

x

B

R

x

E =Bvl If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will be same. Because it is depend on the velcoty & lenght of the wire. 9.

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*)


Chapter # 38 Sol. B


  Bvl 2

2

2 22

v  B v l   R R R Here P × v2 Power 

10.

[4]

P  Power v  velocity B  Magnetic field

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 &

Sol.

11.

(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 dksb Z cy ugha yxk;saxsA (D) /kkjk dh fn'kk ds vk/kkj ij ,d nwljs dks vkdf"kZr djsxsa ;k izfrdf"kZr djsaxsaA A Due to Mutual induction, current is generated in second loop and that causes the two loops attract each other.

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 &

Sol. 12.

(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 tkrk gSA C

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 yw i ds ry ds yEcor~ fp=k es a n'kk;s vuq l kj ,d le:i pq E cdh; {ks = k B fo|eku gS A yw i es a iz s f jr

/kkjk gS &

Sol.

(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; D Induced emf is AB is Bvl and Induced emf is DC is also Bvl. Net emf in the closed circuit (loop) is zero. So induced current in the loop is zero.


Chapter # 38

1.


[5]

Objective - II

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 &

Sol.

2.

(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 pqE cd 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 BC  The north pole faces the ring and the magnet moves towards it  The south Pole faces the ring and the Magnet moves away from it.

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&



Sol.

3.











(A) if   l  (B) if   B  (C) if l  B  D Potential difference appears across the two ends = Bvl v B , v l , 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-c- izsfjr gksxk] ;fn &

Sol.

4.

(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 CD  An emf is induced in the loop is it is rotated about a diameter.  An emf is induced in the loop if it is deformed. 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 pqE cdh; gS rks 'khV dks ,d leku osx ls /kzqo ls nwj ys tkus ds fy;sA (D*) ;fn /kkrq vpqE cdh; gS rks 'khV dks ,d leku osx ls /kzqo ls nwj ys tkus ds fy;s vuqpqE cdRo] izf r pqE cdRo ;k xq:Ro ds Sol. 5.

fdlh Hkh izHkko dks ux.; eku fyft'sA ACD

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


Chapter # 38

Sol. 6.

Sol.


(A*) dsUnz ij pqecdh; {ks=k (C*) ifjufydkvksa dk LoizsjdRo ABC

[6]

(B*) ifjufydk ls lEc) pqEcdh; ¶yDl (D) twy Å"ek 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) Loizsj dRo (B*) ;fn nksuksa 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 BD

l A - Crossectional Area A Thick wire “A” is large than thin wire. Rthick wire < Rthin wire R 

  7.

Sol. 8.

Sol.

L R Power dissipatedin Heating = I2R time constant   

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) izsj dRo esa pqE cdh; ÅtkZ (C) izsjdRo esa pqEcdh; ÅtkZ (D*) izsjdRo esa izsf jr fo-ok-cy D 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 B The end ‘A’ becomes, positively charged. Because magnetic field exerts an average Force    F0  qv  B on each free electron where q = 1.6 ×1019C is the charge on the electron. This Force is towards AB and hence the free electrons will move towareds B. Negative charge is accumulated at ‘B’ and positive charge appears at A.

9.

L, C and R represent the physical quantities inductance, capacitance and resistance combinations have dimensions of frequency ? HkkSfrd jkf'k;ksa] izsjdRo] /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 -


Chapter # 38 (A*) Sol.

ABC 


1 RC

(B*)

R L

1 LC

(D) C/L

Time constant  = RC in RC circuit frequency 

1 1   RC

....(i)



Time constant in LR circuit is  



1 R   L eq. (i) & (ii) multiply  frequency 

frequency 

frequency 

10.

(C*)

[7]

L RC

....(i)

1 LC

1 2C

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 /R fp=kksa (a) o (b) esa fLopksa dks t = 0 ij can fd;k x;k gS] vkSj ,d yEcs le; t = t0 ds i'pkr~ iqu% [kksyk x;k gS -

Sol.

(A) t = 0 ds rqj ar i'pkr~ C ij vkos'k C gSA (C*) t = t0 ds rqj ar igys L ls izo kfgr /kkjk /R gSA BC

C

R

'RC' circuit  

(B*) t = 0 ds yEcs le; i'pkr~ C ij vkos'k C gSA (D) t = t0 ds cgqr le; i'pkr~ L ls izo kfgr /kkjk /R gSA

L

R

'LR' circuit

A long time after capacitor is fully charged is equal to Q = CV = C Q = C (1-e-t/) The current in ‘L’ just before t = t0 is i= /R (1-e-t/) = /R


Ch-38 (Electro Magnetic Induction)

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