Chapter - 16_Electromagnetic Induction

303

CHAPTER-16ELECTROMAGNETIC INDUCTION

FILL IN THE BLANKS

1. A uniformly wound solenoidal coil of self-inductance 1.8 104 H and resistance 6W is broken up into two identicalcoils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The time constantfor the current in the circuit is ..... s and the steady state current through the battery is ..... A. (1989; 2M)

2. In a striaght conducting wire, a constant current is flowing from left to right due to a source of emf. When the sourceis switched off, the direction of the direction of the induced current in the wire will be .... (1993; 1M)

3. The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A and is decreasingat a rate of 103. As then VB VA = ...V (1997C; 1M)

15VA 1W 5 mH B

TRUE/FALSE1. A coil of metal wire is kept stationary in a non-uniform magnetic field. An emf is induced in the coil. (1986; 3M)2. A conducting rod AB moves parallel to the x-axis (see fig.) in a uniform magnetic field pointing in the positive z-

direction. The end A of the rod gets positively charged. (1987; 2M)

B

y

o x

A

2. A thin semicircular conducting ring of radius R isfalling with its plane vertical in a horizontal magnetic

induction Bur

. At the position MNQ the speed of thering is v and the potential difference across the ringis : (1986; 2M)

B

N

v

QM(a) zero(b) BvpR2/2 and M is at higher potential(c) pBRv and Q is at higher potential(d) 2RBv and Q is at higher potential

OBJECTIVE QUESTIONSOnly One option is correct :1. A conducting square loop of side L and resistance R

moves in its plane with a uniform velocity vperpendicular to one of its sides. a magnetic inductionB, constant in time and space, pointing perpendicularto and into the plane of the loop exists everywhere.

++

++ + +

+ ++ +

+++

+

+

+++ +

+ ++

+

+

+

++

+

+

+

V

B

The current induced in the loop is : (1989; 2M)(a) BLv/R clockwise(b) BLv/R anticlockwise(c) 2BLv/R anticlockwise(d) zero

304

3. A metal rod moves at a constant velocity in a directionperpendicular to its length. A constant uniformmagnetic field exists in space in a direction perpendicularto the rod as well as its velocity. Select the correctstatement (s) from he following : (1988; 2M)(a) The entire rod is at the same electric potential(b) There is an electric field in the rod.(c) The electric potential is highest at the centre of

the rod and decrease towards its ends(d) The electric potential is lowest at the centre of the

rod and increases towards its ends

4. A small square loop of wire of side l is placed insidea large square loop of wire of side L ( L > > l). Theloops are coplanar and their centres coincide. Themutual inductance of the system is proportional to:

(1998; 2M)(a) l/L (b) l2/L(c) L/l (d) L2/l

5. Two identical circular loops of metal wire are lying ona table without touching each other. Loop A carries acurrent which increases with time. In response, theloop B : (1999; 2M)(a) remains stationary(b) is attracted by the loop A(c) is repelled by the loop A(d) rotates about its CM, with CM fixed

6. A coil of inductance 8.4 mH and resistance 6W isconnected to a 12 V battery. The current in the coil is1 A at approximately the time : (1999; 2M)(a) 500 s (b) 20 s(c) 35 ms (d) 1 ms

7. A uniform but time varying magnetic field B (t) existsin a circular region of radius a and is directed into theplane of the paper as shown. The magnitude of theinduced electric field at point P at a distance r from thecentre of the circular region : (2000; 2M)

a

B(t)

rP

(a) is zero (b) decreases as 1/r(c) increases as r (d) decreases as 1/r2

8. A coil of wire having finite inductance and resistancehas a conducting ring placed co-axially within it. Thencoil is connected to a battey at time t = 0, so that a timedependent current I1 (t) starts flowing through the coil.

If I2 (t) is the current induced in the ring and B (t) isthe magnetic field at the axis of the coil due to I1 (t)then as a function of time (t > 0), the product I2 (t) B(t) : (1997; 1M)(a) increases with time(b) decreases with time(c) does not vary with time(d) passes through a maximum

9. A metallic square loop ABCD is moving in its ownplane with velocity v in a uniform magnetic fieldperpendicular to its plane as shown in the figure.Electric field is induced : (2001; 2M)

V

CD

A B

(a) in AD, but not in BC(b) in BC, but not in AD(c) neither in AD nor in BC(d) in both AD and BC

10. Two circular coils can be arranged in any of the threesituations shown in the figure. Their mutual inductancewill be : (2001; S)

(a) (b) (c)(a) maximum in situation (a)(b) maximum in situation (b)(c) maximum in situation (c)(d) the same in all situation

11. As shown in the figure, P and Q are two coaxilconducting loops separated by some distance. Whenthe switch S is closed, a clockwise current Ip flows inP (as seen by E) and an induced current IQ, flows inQ. The switch remains closed for a long time. When S

is opened, a current 2QI flows in Q. Then the direction

1QI and 2QI (as seen by E) are : (2002; 2M)

S

EQP

Battery

305

(a) respectively clockwise and anticlock wise(b) both clockwise(c) both anticlockwise(d) respectively anticlockwisee and clockwise

12. A short-circuited coil is placed in a time varyingmagnetic field. Electrical power is dissipated due to thecurrent induced in the coil. If the number of turns wereto be quadrupled and the wire radius halved, theelectrical power dissipated would be : (2002; 2M)(a) halved (b) the same(c) doubled (d) quadrupled

13. When an AC source of emf e = E0 sin (100t) isconnected across a circuit, the phase differencebetween the emf e and the current i in the circuit is

observed to be 4p

, as shown in the diagram. If the

circuit consists possibly only of R-C or R-L or L-C inseries, find the relationship between the two elements:

(2003; 2M)

t

ei

(a) R = 1kW , C = 10 F(b) R = 1kW , L = 1H(c) R = 1kW , C = 10 F(d) R = 1kW , L = 1H

14. The variation of induced emf (e) with time (t) in a coilif a short bar magnet is moved along its axis with aconstant velocity is best represented as: (2004; 2M)

(a)

e

t(b)

e

t

(c)

e

t(d)

e

t

15. An infinitely long cylinder is kept parallel to an uniformmagnetic field B directed along positive z-axis. Thedirection of induced current as seen from the z-axis willbe : (2005; 2M)(a) clockwise of the +ve z-axis(b) anticlockwise of the +ve z-axis(c) zero(d) along the magnetic field

16. The figure shows certain wire segments joined togetherto form a coplanar loop. The loop is placed in aperpendicular magnetic field in the direction going intothe plane of the figure. The magnitude of the fieldincreases with time. I1 and I2 are the currents in thesegments ab and cd. Then, (2009; M)

c d

a b

(a) I1 > I2(b) I1 < I2(c) I1 is in the direction ba and I2 is in the direction

cd(d) I1 is in the direction ab and I2 is in the direction

dc

OBJECTIVE QUESTIONS

More than one options are correct?1. Two different coils have self-inductances L1 = 8 mH

are L2 = 2 mH. The current in one coil is increased ata constant rate. The current in the second coil is alsoincreased at the same constant rate. at a certain instantof time, the power given to the two coils is the same.At that time, the current, the induced voltage and theenergy stored in the first coil are i1, V1 and W1respectively. Corresponding values for the second coilat the same instant are i2, V2 and W2 respectively.Then : (1994; 2M)

(a)1

2

14

=ii (b)

1

24=

ii

(c)1

2

14

=WW (d)

1

24=

VV

2. A field line is shown in the figure. This field cannotrepresent. (2006; 2M)

306

(a) Magnetic field(b) Electrostatic field(c) Induced electric field(d) Gravitational field

3. Two metallic rings A andB, identical in shape andsize but having different

resistivities A? and

,? B are kept on top of twoidentical solenoidsas shown in the figure.

A B

When current I is switched on in the both the solenoidsin identical manner, the rings A and B jump to heights

Ah and ,Bh respectively, with .BA hh > The possiblerelation(s) between their resistivities and their masses

Am and Bm is (are) (2009; M)

(a) BA ?? > and BA mm =

(b) BA ?? < and BA mm =

(c) BA ?? > and BA mm >

(d) BA ?? < and BA mm e1 i

Net emf, e = e2 e1e = B0av

\ i = 0

=B ave

R Rand direction of current will be counter-clockwise.

(b) Total lorentz force on the loop :

E

H GV

yx

y

B

F

We have seen in part (a) that induced current passingthrough the loop (when its speed is v) is

i = 0B av

RNow magnetic force on EH and FG are equal inmagnitude and in opposite directions, hence theycancel each other and produce no force on the loop.

FEF =0 0( )

B av B ya

R a (downwards)

(F = ilB)

=20B avyR

and FGH =0

B avR (a)

0 ( )+

B y aa (upwards)

=

20 ( )

+

B avy a

R

FGH > FEF\ Net Lorentz force on the loop

= FGH FEF =2 20=

B a vR

(upwards)

Fur

=2 20 j$

B a vR

(c) Net force on the loop will beF= weight Lorentz force (downwards)

or F = mg 2 20

B a vR

or

dvm

dt = mg 2 20

B a vv

R

\dvdt

= g2 20

=

B av g Kv

mR

where K =2 20B amR

= constant

ordv

g Kv= dt

or 0 v dv

g Kv =

1

0dt

This equation gives (1 )= Ktg

v eK

vT=g/K

v

O t

Here K =

2 20

B amR

i.e., speed of the loop is increasing exponentially withtime t. Its terminal velocity will be

VT = 2 20

=

g mgRK B a at t

15. Magnetic field (B) varies with time (t) as shown infigure.

322

0 0.2 0.4

0.8

B(T)

0.6 0.8 t(s)Induced emf in the coil to change in magnetic fluxpassing through it,

e =fd

dt=

dBNA

dt

Here, A = Area of coil = 5 103m2

N = Number of turns = 100Substituting the values, we get

e = (100) (5 103)(4)V= 2V

Therefore, current passing through the coil

=e

iR

(R = Resistance of coil = 1.6W)

or i =2

1.6 = 1.25 A

Note that from 0 to 0.2 s and from 0.4 to 0.6s, magneticfield passing through the coil increase, while duringthe time 0.2s to 0.4s and from 0.6s to 0.8s magneticfield passing through the coil decrease. Therefore,direction of current through the coil in these two timeintervals will be opposite to each other. The variationof current (i) with time (t) will be as follows :

t(s)

i(A)

+ 1.25

0 0.2 0.4 0.6 0.8

1.25

Power dissipated in the coil isP = i2 R = (1.25)2 (1.6) W = 2.5 W

Power is independent of the direction of current throughthe coil. Therefore, power (P) versus time (t) graph forfirst two cycles will be as follows :

2.5

0 0.8t(s)

P(watt)

Total heat obtained in 12,000 cycles will beH = P. t = (2.5) (12000) (0.4) = 12000J

This heat is usd in raising the temperature of the coiland the water. Let q be the final temperature. Then

H = mwsw (q 30) + mcSc (q 30)Here mw = mass of water = 0.5 kgSw = specific heat of water = 4200 J/kg Kmc = mass of coil = 0.06 kgand Sc = specific heat of coil = 500 J/kg-KSubstituting the values, we get1200 = (0.5) (4200) (q 30) + (0.06) (500) (q 30)or q = 35.6C

16. (a) Given R1 = R2 = 2W , E = 12 Vand L = 400 mH = 0.4 H. Two parts of the circuit arein parallel with the applied battery. So, the uppercircuit can be broken as :

E

SR1

R2

L

E

S

R1+

E

S R2

L

(a) (b)Now refer figure (b) :This is a simple L-R circuit, whose time constant

tL = L/R2 =0.4

0.22

= s

and steady state current i0 = E/R2 = 12/2 = 6ATherefore, if switch S is closed at time t = 0, thencurrent in the circuit at any time t will be given by

i (t) = i0 ( ) /1 tLtei (t) = 6 ( ) / 0.21 te

= 6 ( )51 te = i (say)Therefore; potential drop across L at any time t is :

V = di

Ldt = L (30e

5t) = (0.4) (30)e5t

or V = 12 e5t volt(b) The steady state current in L or R2 is

i = 6ANow, as soon as the switch is opened, current in R1is reduced to zero immediately. But in L and R2 itdecreases exponentially. The situation is as follows :Refer figure (e) :

323

R1

i

R2

L

t = t

R1R2

L

1

Ei 6A

R= =

i=6A0

Steady state condition

i=0

i0

t = 0S is open

(c) (d) (e)Refer figure (e)Time constant of this circuit would be

tL' =

1 2

0.4(2 2)

=+ +L

R R = 0.1s

\ Current through R1 at any time t is

i = ' /0tLti e = 6et/0.1 or i = 6e 1 0t A

Direction of current in R1 is as shown in figure orclcokwise.

17. (a) Applying Kirchhoff's second law :

0f =d diiR Ldt dt or

f = +d diiR Ldt dt

This is the desired relation between i, didt

and fddt

(b) Eq. (1) can be written asd f = iRdt + Ldi

Integrating we getD f = R.Dq + Li1

Dq = 1Df LiR R

...(2)

Here Df = ff fi =

0

0

0 0

2 2

=

= p

x x

x x

Ildx

x

=0 02pI l

In (2)

So, from Eq. (2) charge flown through the resistanceupto time t = T, when current is i1 is

Dq =0 0

11

( 2 ) 2

p

I lIn Li

R

(c) This is the case of current decay in an L-R circuit.Thus,

i = /0tLti e ...(3)

Here, 1 0 1, , (2 )4= = = =

ii i i t T T T and

tL =LR

Substituting these values in eq. (3), we get :

tL =LR

In(4)=

T

18. (a) For an elemental strip of thickness dx at a distancex from left wire, net magnetic field (due to both wires)

a

aI i

3a

dxx

B = 0 0 2 2 3

+p p

I Ix a x

(outwards)

=0 1 1

2 3 + p

Ix a x

Magnetic flux in this strip,

df = BdS =0 1 1

2 3 + p

Iadx

x a x

\ Total flux f 2

= fa

ad

=2

0 1 12 3

+ p a

a

Iadx

x a x or

f =0pIa

In (2)

f =0 In(2)

pIa

(I0 sin wt) ...(1)

Magnitude of induced emf,

e = fd

dt =0 0 In(2)w

paI

coswt = e0 coswt

where e0 =0 0 In(2)w

paI

Charge stored in teh capacitor,q = Ce = Ce0 coswt ...(2)

and current in the loop

i =dqdt = Cwe0 sinwt ..(3)

imax = Cwe0 =2

0 0 In(2)wp

aI

(b) Magnetic flux passing through the square loopf sin wt [From eq. (1)]

i.e., magnetic field passing through the loop isincreasing at t = 0. Hence, the induced current willproduce magnetic field (from Lenz's law). Or the

324

current in the circuit at t = 0 will be clockwise (ornegative as per the given convention). Therefore,charge on upper plate could be written, as

q = q0 cos wt [From Eq. (2)]

Here, q0 = Ce0 0 0 In(2)w

paI

The corresponding q - t graph is shown in figure.

i+

q

q0

q0

T4

T2

3T4

Tt

19. After a long time, resistance across an inductor be-comes zero while resistance across capacitor becomesinfinite. Hence, net external resistance.

Rnet =2

2

+R R

=34R

Current through the batteries,

i =

1 2

234

+ +

ER

r r

Given that potential across the terminals of cell A iszero.\ E ir1 = 0

or 11 2

2 03 / 4

= + +

EE rR r r

Solving this equation, we get, 1 24 ( )3

=R r r

20. Inductive reactanceXL = wL = (50) (2p) (35 10

3) = 11W

Impedence Z 2 2= + LR X

= 2 2(11) (11)+ = 11 2W

Given Vrms = 220VHence, amplitude of voltage

v0 = rmsV2 220 2= V

or i0 = 20 A

Phase difference f = tan1

LXR

= tan1 1111 4

p = In L-R circuit voltage leads the current. Hence, instan-

taneous current in the circuit is,i = (20A) sin(wt p/4)

Corresponding i - t graph is shown in figure.

5T/8T/2

i =20 sin ( t /4) pw

T/4T/8

T 9T/8 t10 2-

O20

V,I

V= 220 2 sin tw

21. Out side the solenoid net magnetic field zero. It can beassumed only inside the solenoid and equal to 0nl.

Induced )(BAdtd

dtde -=f-=

)( 20 anIdtd pmf-=

or |e| = (0 npa2) (I0 w cos wt)

Resistance of the cylindrical vessel

R =(2 )r r p=l R

s Ld

\ Induced current i 2

0 0 cos| |2

w w= =

rLdna I te

R R

ASSERATION AND REASON

1. The induced current in the ring will interact withhorizontal component of magnetic field and both willrepel each other.This repulsion will balance the weight of ring.Hence, option (a) is correct.

COMPREHENSION

1. Change on capacitor at time t is :q = q0 (1 e

t/t)Here, q0 = CV and t = 2t\ q = CV (1 e2t /t)

= CV (1 e2)

2. From conservaion of energy,

2max

12

LI = 212

CV

\LCVI =max

3. Comparing the LC oscillation with normal SHM we get,

2

2d Q

dt= w2Q

Here, w2 =1

LC

\ Q = LC2

2d Q

dt