INTERACTIVE LEARNING PLANS ELECTROMAGNETIC INDUCTION

Program: II IIT - JEE Achiever II-IIT JEE PHYSICS

Chapter: Electromagnetic Induction ILP # 45 Year 2023 - 24

Chapter/Topic Electromagnetic Induction / Faraday’s laws of electromagnetic induction

Syllabus Electromagnetic Induction − Faraday’s Laws − Lenz law − Motional emf − Induced
electric field.
Concept List Faraday’s Experiments − Faraday’s laws − Lenz law − motional emf − Induced and Static
Electric field
Learning 1. To understand the experiments of Faraday.
Objects 2. To understand Faraday’s laws.
3. To understand Lenz’s law.
4. To understand the concept of motional emf.
5. To understand the distinction between induced and static electric field
Concept map

Electromagnetic Induction
The work of Oersted, Biot, Savart, Laplace and Ampere established that an electric current can produce a
magnetic field. On the basis of the principle of symmetry, physicists strongly felt that a reverse effect, where
a magnetic field can produce an electric current, must also exist. Ten years of search for such an effect,
particularly by Michel Faraday, culminated in the discovery of the phenomenon of electromagnetic induction
in 1930.

Faraday’s experiments
Faraday conducted a series of experiments to find how a magnetic field can produce an electric current. His
experiments can be broadly classified into the following two types.

517
2I2324PILP45(EMI)

1. Coil – Magnet experiment

Consider a solenoid which is connected to a sensitive galvanometer as
shown in the figure. Just by keeping the magnet, anywhere around the
coil, howsoever strong be the magnet, no current is detected by the
galvanometer. However, when the magnet is kept parallel to the axis of
the coil and moved towards or away from it, the galvanometer shows a
deflection. Faster the magnet is moved, larger would be the deflection. The direction of deflection is reversed
when the direction of motion of the magnet is reversed. The deflection is also reversed when the poles of the
magnet are reversed. Instead of moving the magnet, if the coil is moved keeping the magnet stationary,
deflection is produced. If both the magnet and the coil are moved towards each other or moved away from
each other, then also deflection is observed. However if the coil and the magnet are moved with same
velocity in same direction, no deflection is observed. A stronger magnet, wider coil, larger number of turns
in the coil produces larger deflection

Coil-Coil Experiment
In the second set of experiments, Faraday dispensed with the magnet and instead used two coils. the coil
containing a battery and a tap key is called a primary coil. The other containing a sensitive galvanometer is
called secondary coil(s). Whenever the tap key K is closed, there is a
sudden deflection in the galvanometer towards one side. When the key K
is opened there is again a deflection but in the opposite direction.
However when the key K is kept closed, there is no deflection. Only
during making or breaking of the primary circuit, a deflection is
observed in the galvanometer connected in the secondary circuit.

Faraday’s conclusions
From the above set of experiments, Faraday concluded that whenever the magnetic field passing through a
closed circuit changes a current is produced in the circuit. This current is called ‘induced current’, the emf
responsible for it is called ‘induced emf’ and the phenomenon is called ‘electromagnetic induction’. In the
first experiment, the change in the magnetic field is brought about, by moving a magnet. In the second
experiment the growing or decaying current in the primary coil produces a growing or decaying magnetic
field. In both experiments the number of magnetic field lines linked with the secondary circuit changes,
causing an induced emf in it.

Faraday’s laws of electromagnetic Induction

Faraday crystallised his observations into two laws:
First law: Whenever there is a change in the magnetic flux linked with a closed circuit, an emf is induced
in the circuit.
Second law: The magnitude of the induced emf is directly proportional to the rate of change of magnetic
flux.
d
 B
dt

518
 2I2324PILP45(EMI)

Magnetic flux, in analogy with electric flux, is defined as the surface integral of magnetic field over any
arbitrary surface S through which it is passing, i.e., B =  B  dS . If a uniform magnetic field B is passing
S

through an area S , making an angle  with it, then the flux of magnetic field through the area S is given by
d
B = BScos  .    ( BScos  ) . Hence an emf will be induced in a circuit, whether B changes with time
dt
or S changes with time or  changes with time or all these change with time.

Lenz’s law
Faraday’s laws only tell about the cause of induced emf and its magnitude. In which of the two possible
directions the induced emf is produced in a circuit? This information is given by Lenz’s law. On the basis of
the law of conservation of energy Lenz arrived at a law which gives the direction of induced emf. This law,
known as Lenz’s law can be stated as follows.
Lenz’s law: The induced emf in a circuit is setup in such a direction that it can oppose the changing
magnetic flux associated with the circuit.
d d
   − B ( or )  = −k B . The proportionality constant.
dt dt
d
K = 1 in the S.I. system. Hence  = − B . To understand Lenz’s law consider the following illustrations.
dt

Illustrations

1. In the figure shown, let the magnet with its north pole towards the coil move towards the coil. The
induced emf will be set up in such a direction, that it produces an induced current which will produce a
magnetic field which will repel the magnet. The end of the solenoid
facing the magnet develops N-polarity so that it can repel the magnet.
The work done against this repulsive force is stored as the magnetic
energy in the coil. If the magnet is moved away from the coil, the
nearer end of the solenoid will behave like a south pole, attracting the
magnet.
2. Consider a circular coil through which a magnetic field is
increasing with time. Then the induced current in the coil,
according to Lenz’s law, will flow in the anticlockwise direction,
so that it produces a magnetic field which opposes the applied
magnetic field. Similarly if the applied magnetic field is
decreasing, the induced current will flow in the clock-wise
direction trying to increase it.

Static electric field and induced electric field

The electric field produced by changing magnetic field, which produces the induced current is called
‘induced electric field’. This field is intrinsically different from the static electric field produced by electric
charges. The two fields are different in the following two respects.

519
2I2324PILP45(EMI)

(i) As we know the static electric field is radial. The static electric field lines radially diverge from a
positive charge and converge onto a negative charge. But the induced electric field lines are circular and
form closed loops.

(ii) Static electric field is a conservative field. Along any arbitrary closed loop C in a static electric field
 E  dl = 0 . But induced electric field is a non-conservative field. Along any arbitrary closed loop C in
an induced electric field is non-zero given by,
F 1 W
C E  dl = C q  dl = q C F  dl = q =  , induced emf.

Motional emf
The emf induced in a conductor moving in a magnetic field is called motional emf.
To see the origin of this emf consider a conductor of length l moving with a velocity
v in a magnetic field B. Let us consider a simple situation, where l , v and B are
mutually perpendicular. As the conductor moves with a velocity v, the free
electrons inside the metal also share this velocity, over and above their random
thermal velocities. Due to this velocity v, the free electrons experience a magnetic
force FM in the downward direction and tend to collect at the lower end Q of the
conductor. Due to the motion of the electrons in the direction PQ, the end P gets
positively charged and the end Q, negatively charged. This charge separation builds
up an electric field in the conductor. Due to this electric field, the free electrons start experiencing an upward
force FE. As this process continues, a stage will be reached, when the upward electric force just balance the
downward magnetic force. In this situation,
FM = FB  evB = eE  E = vB
The emf developed between the two ends of the conductor is given by  = El = vBl.
(a) If v, B and l are not mutually perpendicular, then  = v  B  l . Thus a conductor moving in a
magnetic field behaves like a cell with an emf,  = v  B  l .
(b) If a conductor of length l but arbitrary shape moves in a magnetic field, then the motional emf
produced in it is the same as that produced in a straight conductor of length l, running from one
end of the arbitrary conductor to another.

520
 2I2324PILP45(EMI)

(c) Consider a conductor moving over an infinitely long

conducting frame as shown in the figure. Then the moving
conductor acting as a source of emf  = Bvl, sends a current
through the circuit. If R is the resistance of the circuit, the
 Bvl
current is given by I = = .
R R

Now, since the current conductor PQ is moving in a magnetic field, it experiences a magnetic force
 Bvl  B2 vl 2
FM = BIl = B   l = acting towards left. To keep the rod moving towards the right with a uniform
 R  R
velocity v, we have to apply an external force Fext to balance this magnetic force. The power required to keep
the rod moving is given by
 B2 vl 2  B2 v 2 l 2
Pext = (Fext)v =   v = … (1)
 R  R
The power dissipated as thermal energy in the circuit is given by
2
 Bvl  B2 v2l 2
Pdis = I R = 
2
 R= … (2)
 R  R
From equations (1) and (2), we see that the power expended in pulling the conductor is completely dissipated
as heat in the circuit. The law of conservation of energy is not violated in the production of motional emf.

Irreducible Set of Problems

1. A coil of mean area 500 cm2 and having 1000 turns is held perpendicular to a uniform field of 4  10–5 T.
th
1
The coil is turned through 180 in of a second. The average emf induced in the coil is
10
(A) 20 mV (B) 40 mV (C) 2 mV (D) 4 mV
Ans (B)
 −  (BA cos180 − BA cos0) 2NBA
 = −N  2 1  = −N =
 t  t t
2(1000)(4  10−5 )(500  10−4 )
= = 0.04 V
0.1

2. A conducting circular loop is placed in a uniform magnetic field of 0.04T, with its plane perpendicular to
the magnetic field. The radius of the loop starts shrinking at a rate 2 mms –1. The induced emf in the loop
when the radius is 2 cm is
(A) 4.8  v (B) 0.8 v (C) 1.6  v (D) 3.2  v
Ans (D)
d d dA d dr
 = − = − (BA) = − B = − B ( r 2 ) = − B2r
dt dt dt dt dt
 2  10 
−3
= −(0.04)( )(2)(2  10−2 )  −  = 3.2V
 1 

521
2I2324PILP45(EMI)

3. A wire in the form of a circular loop of radius 10 cm lies in a plane normal to a magnetic field of 100T.
If the wire is pulled to take a square shape in the same plane in 0.1 s, the average emf induced in the loop
is
(A) 9 V (B) 12 V (C) 6.75 V (D) 5 V
Ans (C)
2r  2 r 2
2

Circle → square; r 2 →    A = r 2
−
 4  4
BA Br    100(3.14)(0.1)  3.14  31.4  0.86
2 2
= = 1 −  = 1 −  6.75V
t t  4 0.1  4  4

4. A square wire loop with 2.0 m sides is ⊥r to a uniform magnetic field, with half the area of the loop in the
field as shown in the figure. The loop contains a 20.0 V battery with negligible internal resistance. If the
magnitude of the field varies with time as to B = 0.042 – 0.870 t, what is the total e.m.f in the circuit?
Solution
L2
 = BA = B
2
d d  L2 
E = − = − B  B
dt dt  2 
L2 d
=− (0.042 − 0.87t)
2 dt
20.0V
L2 22
= − ( −0.87) =  0.87 = 1.74V
2 2
Since B is out of the page and decreasing, the e.m.f induced tends to increase it. Hence its flows
anticlock wise. Then the e.m.f of the battery adds to induced e.m.f.
 Total emf = 20.0 + 1.74 = 21. 74 V
5. Two centric circular loops made of wire with resistance / unit length 10 –4  / m, have diameters 0.2 m
and 2m. A time varying potential difference (4 + 2.5 t) V is applied to the larger loop. Calculate the
current in the smaller loop
Solution
4 + 2.5t 4 + 2.5t
I= =
Resistacne (2 R)10−4 I I
 2I 0 4 + 2.5t r
B0 = 0 =  −4
O
4 R 2R (2 R)10 R
 4 + 2.5t 2
 = BA = 0 2 r
4r 10−4
d 0 r 2 2.5
= =
dt 4R 2 10−4
  0 r 2 2.5 1
I = −4
= = 1.25A
(2 r)  10 4R 10 2 r  10−4
2 −4

522
 2I2324PILP45(EMI)

6. A rectangular frame ABCD, made of a uniform metal wire, has a straight connection A E B
between E and F made of the same wire, as shown in the figure AEFD is a square of B B
side 1m, and EB = FC = 0.5m. The entire circuit is placed in steadily increasing,  
uniform magnetic field directed into the plane of paper and normal to it. The rate of
change of magnetic field is 1 Ts–1. The resistance / unit length of the wire is D F C
1 m–1 . Find the magnitude and direction of the current in the segment AE, BE and EF
Solution
1 0.5 
d dB A E B
AEFD   = − = −A = −1  1 = −1V
dt dt 1 1 1
dB
EBCF   = −A = −0.5  1 = −0.5V i1– i2 i 2
dt
i1 0.5 V
D F C
1 0.5 

KVL to EADFE  (–1) i1 – 1(i1) + 1 – 1 (i1) –1(i1 – i2) = 0  4i1 – i2 = 1 ... (1)
KVL to EBCFE  (0.5)i2 + (1)i2 – 0.5 + (0.5) i2 – 1 (i1 – i2) = 0  3i2 – i1 = 0.5 ... (2)
3.5 3 0.5 1
Solving (i) and (ii) we get i1 = ;i 2 = ;i1 − i 2 = = A
11 11 11 22
7. A wire bent as a parabola y = kx2 is located in a magnetic field as shown in the figure. At time t = 0, a
connector starts sliding up from the apex of the parabola with a constant acceleration a. The emf induced
in the loop thus formed as a function of y is
8a y
(A) By
k
6a B B
(B) By
k
a
(C) 2By
k
x
2a
(D) By
k
Ans (A)
d = B  ds = B(2x)dy
y
But y = kx 2  x = B
k B

 y  d y dy y dy
d = B  2 dy   = 2B = 2B v x x
 k  dt k dt k
y
d y y 2y  1 2 2y 
= 2B at = 2B a  y = at  t = 
dt k k a  2 a 
8a
 = By
k

8. A metal bar of length 1.0 m falls from rest under gravity while horizontal with its ends pointing towards
magnetic east and west. What is the potential difference between its ends when it has fallen by 10 m?
(Given, BH = 3  10–5 T; Bv = 2  10–5 T)

523
2I2324PILP45(EMI)

Solution
BH = 3  10–5 Wb m–2
v 2 = u2 + 2gh  v = 2gh = 2  9.8 10 = 196 = 14ms−1.
e = BH vl = 3  10−5  14  1 = 4.2  10−4 V

9. A square frame with side l and a long straight wire carrying current I are located in the same plane as
shown in the figure. Then frame moves towards right with a constant velocity v. Find the e.m.f induced
in the frame as a function of distance x
Solution
e.m.f is induced only in the sides KN and ML and in the same direction N l M
0 2I 0 Ivl I
e KN = Bvl = vl =
4 x 2 x x
v
0 2I 0 Ivl

e LM = B vl = vl =
4 x + l 2  (x + l)
K L
 1 1   0 Ivl 2
net e.m.f = 0 Ivl  − =
2  x x + l  2  x(x + l )
B0 ˆ
10. A magnetic field in a region is given by B = y k , where L is a fixed length. A conducting rod of
L
length L lies along the y-axis between (0, 0, 0) and (0, L, 0). If the rod moves with a constant velocity
v0 î , find the emf induced between the ends of the rod.
Solution y
The motional emf induced in the element dy is given by  B
B Bv
d = Bv0dy = 0 yv0dy = 0 0 ydy dy
L L
L 2
Bv B v L B0 v0 L
 = 0 0  ydy = 0 0 = v0
L 0 L 2 2
x

11. Two parallel vertical metallic rails AB and CD are separated by a distance of 1 m. They are connected at
the two ends by resistances R1 and R2 as shown in the figure. A horizontal metallic bar L of mass 0.2 kg
slides without friction, vertically down under the action of gravity. There is a uniform magnetic field of
0.6 T. It is observed that when the bar achieves the terminal velocity, the powers dissipated in R 1 and R2
are 0.76 W and 1.2 W respectively. Find the terminal velocity of the bar and values of R1 and R2.
Solution
The bar acquires the terminal velocity vT when mg = Bil
R1
mg ( 0.2 )( 9.8)  
i= = = 3.27 A . But i1 + i2 = 3.27 … (1)
Bl ( 0.6 ) (1) i1
i1 P2 0.76
P1 = i1, P2 = i2  = = = 0.63 … (2) 
i1 + i 2

i 2 P1 1.2
From Eqn. (1) and (2), we get i2 = 2A and i1 = 1.27 A i2
R2
P
From, P2 = i2   = 2 = 0.6 V  
i2

524
 2I2324PILP45(EMI)

 0.6
When v = vT,  = BvTL  vT = = = 1 m s −1
BL ( )
0.6 1 ( )
 0.6  0.6
R1 = = = 0.47  and R 2 = = = 0.3 
i1 1.27 i2 2

12. Two metal bars are fixed vertically and are connected on the top by a capacitor of capacitance C. A
sliding conductor of length l and mass m slides with its ends in contact with the bars. If the conductor is
released from rest, find the displacement of the conductor x(t) as a function of time.
Solution C
dv dq dv
mg − BIl = ma = m . But q = C = CBlv  I = = CBl
dt dt dt B  B
 dv  dv 
m + B2l 2 C = mg
dv
 mg − B  CBl  l = m 
 dt  dt dt
dv mg mgt 2
 a= = ; x ( t ) = 1 at 2 =
dt ( m + B2l 2 C ) 2 2 ( m + B2 l 2 C ) B  B

13. Two infinitely long parallel wires carrying currents I = I0 sin t in opposite directions are placed at a
distance 3a apart as shown in the figure. A square loop of side a of negligible resistance with a capacitor
of capacitance C is placed in the plane of wires. Find the maximum current in the square loop
Solution
At a distance x from the wire 1, the magnetic field due to both wires will be in the same direction
I 0I  I 1 1 
B= 0 + = 0  +
2 x 2 (3a − x) 2   x 3a − x 
Consider a strip of thickness dx at distance x. The flux through this strip is given by
 I 1 1 
d = B  ds = 0  + adx
2  x (3a − x) 
0 IQ  1 1   Ia
2a
=  + dx = 0 l n x − l n(3a − x)0
2a dx
x
 
2 a  x (3a − x)  2
a a
2a
 0 Ia   x    0ia  (I sin t)a
= l n  = l n2 = 0 0 l n2
2    3a − x   0 2 2
d  0 I 0a cos cot
= = l n2
dt 2
3a
C0 I0a l n 2
Q = C = cos t
2
dQ C0 I 0a2l n 2
I= = sin t
dt 2
C0 I 0a2
I max = l n2
2

14. A long straight rod translates with a uniform speed v on a V shaped conductor in a normal magnetic
field. If the resistance / unit length of all the conductors is , find the induced current
Solution
Let x be the position of the rod at any instant t
l 2x
= x tan 30 l =
2 3

525
2I2324PILP45(EMI)

1 1 2x 2 x 2
Area, A = lx = =  
2 2 3 3
Bx 2
d 2Bx dx 2Bx
 = BA = ; = − = − =− v
3 dt 3 dt 3 60
l
v v v x
2Bx 2Bx 2x B
 3 3 3 Bv
i= = = = =
R total [l + 2x sec30]  2x 2   1 2  3  B 
 + 2x   2x  +
 3 3  3 3 
Aliter: The motional emf in the moving rod,  = Bvl
 Bvl Bv
Induced current = = =
R 3l 3
15. A 10 A current carrying straight conductor is placed parallel to an U-shaped horizontal rails at a distance
of 1 cm from its nearer side. A rod AB moves with a constant velocity 5 ms–1 I
calculate 1 cm A
(a) The emf in the rod r
(b) current in the conducting loop assuming that the rails are of negligible 9 cm dr
resistance and the resistance of the rod is 0.4  x
(c) the rate at which thermal energy is developed in the rod B
(d) the force that must be applied by an external agent to maintain its motion
(e) the rate at which this external agent does work on the rod
Solution
.i
(a) consider a small element of length x and thickness dr. The field normal to this strip is B = . Area
2r
i
of the strip is xdr. The flux through it is d = 0 xdr  The flux through the entire area enclosed
2r
 ix dr  ix
10
between the rails and the rod is  =  d = 0  = 0 ln 10
2 1 r 2
d  il n10 dx 0i(2  303)log10 4  3.14  10−7  10  2.303  1  5
=− =− 0 = V  = = 2.303  10−5 V
dt 2 dt 2 2  3.14
 2.3  10−5
(b) il = = 6  10−5 A
R 0.4
Since  is increasing,  tries to decrease it . Hence i flows in the clock wise direction
(c) thermal energy generated in the rod, Pth = il2 R = ( 6 10−5 ) 0.4
2

= 14.4  10–10 W = 1.44  10–9 W.

(d) since the rod is moving with uniform velocity, the net force on it must be zero. The force exerted by
the external agent on the rod must be equal and opposite to the magnetic force acting on the rod
i  i i dr 10
 i i dr 0i i l 10
dFm = Bil dr = 0 il dr = 0 l  Fm =  0 l = ln
2r 2 r 1
2 r 2 1
4  10−7  10  6  10−5  2.303
= = 27.6  10−11 N. towards right
2
= 2.8  10–10 N
(e) rate at which work is done by external agent
P = Fm v = 27.6  10−11  5 = 138  10−11 = 1.38  10−9 W = 1.4  10−9 W = Pth

526