Chapter 12 Electromagnetic Induction
1. Choose the correct option.
i) A circular coil of 100 turns with a cross-sectional
area (A) of 1 m2 is kept with its plane perpendicular to the
magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(A) 1 Wb
(B) 100 Wb
(C) 50 Wb
(D) 200 Wb
Answer:
(B) 100 Wb
ii) A conductor rod of length (l) is moving with velocity
(v) in a direction normal to a uniform magnetic field (B). What will be the
magnitude of induced emf produced between the ends of the moving conductor?
iii) Two inductor coils with inductance 10 mH and 20 mH
are connected in series. What is the resultant inductance of the combination of
the two coils?
iv) A current through a coil of self inductance 10 mH
increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
(A) 0.1 V
(B) 1 V
(C) 10 V
(D) 0.01 V
Answer:
(A) 0.1 V
v) What is the energy required to build up a current of 1
A in an inductor of 20 mH?
(A) 10 mJ
(B) 20 mJ
(C) 20 J
(D) 10 J
Answer:
(A) 10 mJ
2. Answer in brief.
i) What do you mean by electromagnetic induction? State
Faraday’s law of induction.
Answer:
The phenomenon of production of emf in a conductor or circuit by a changing
magnetic flux through the circuit is called electromagnetic induction.
Faraday’s laws of electromagnetic induction :
(1) First law ; Whenever there is a change in the magnetic flux associated with
a circuit, an emf is induced in the circuit.
(2) Second law : The magnitude of the induced emf is directly proportional to
the time rate of change of magnetic flux through the circuit.
[Note : The phenomenon was discovered in 1830 by Joseph Henry (1797-1878), US
physicist, and independently in 1832 by Michael Faraday (1791 -1867), British
chemchemist and physicist.]
ii) State and explain Lenz’s law in the light of
principle of conservation of energy.
Answer:
Lenz’s law : The direction of the induced current is such as to oppose the
change that produces it.
The change that induces a current may be (i) the motion of a
conductor in a magnetic field or (ii) the change of the magnetic flux through a
stationary circuit.
Explanation : Consider Faraday’s magnet-and- coil
experiment. If the bar magnet is moved towards the coil with its N-pole facing
the coil, as in Fig., the number of magnetic lines of induction (pointing to
the left) through the coil increases. The induced current in the coil sets up a
magnetic field of its own pointing to the right (as given by Amperes right-hand
rule) to oppose the growing flux due to the magnet. Hence, to move the magnet
towards the coil against this repulsive flux of the induced current, we must do
work. The work done shows up as electric energy in the coil.
When the magnet is withdrawn, with its N-pole still facing the coil, the number
of magnetic lines of induction (pointing left) through the coil decreases. The
induced current reverses its direction to supplement the decreasing flux with
its own, as shown in Fig.. Facing the coil along the magnet, the induced
current is in the clockwise sense. The electric energy in the coil comes from
the work done to withdraw the magnet, now against an attractive force. Thus, we
see that Lenz’s law is a consequence of the law of conservation of energy.
[Note : The above law was discovered by Heinrich Friedrich
Emil Lenz (1804-65), Russian physicist.]
iii) What are eddy currents? State applications of eddy
currents.
Answer:
Whenever a conductor or a part of it is moved in a magnetic field “cutting”
magnetic field lines, or placed in a changing magnetic field, the free
electrons in the bulk of the metal start circulating in closed paths equivalent
to current-carrying loops. These loop currents resemble eddies in a fluid
stream and are hence called eddy or Foucault currents [after Jean Bernard Leon
Foucault (1819-68), French physicist, who first detected them].
Applications :
(1) Dead-beat galvanometer : A pivoted moving-coil galvanometer used for
measuring current has the coil wound on a light aluminium frame. The rotation
of the metal frame in magnetic field produces eddy currents in the frame which
opposes the rotation and the coil is brought to rest quickly. This makes the
galvanometer dead-beat.
(2) Electric brakes : When a conducting plate is pushed into
a magnetic field, or pulled out, very quickly, the interaction between the eddy
currents in the moving conductor and the field retards the motion. This
property of eddy currents is used as a method of braking in vehicles.
iv) If the copper disc of a pendulum swings between the
poles of a magnet, the pendulum comes to rest very quickly. Explain the reason.
What happens to the mechanical energy of the pendulum?
Answer:
As the copper disc enters and leaves the magnetic field, the changing
magnetic flux through it induces eddy current in the disc. In both cases,
Fleming’s right hand rule shows that opposing magnetic force damps the motion.
After a few swings, the mechanical energy becomes zero and the motion comes to
a stop.
Joule heating due to the eddy current warms up the disc.
Thus, the mechanical energy of the pendulum is transformed into thermal energy.
v) Explain why the inductance of two coils connected in
parallel is less than the inductance of either coil.
Answer:
Assuming that their mutual inductance can be ignored, the equivalent inductance
of a parallel combination of two coils is given by
Hence, the equivalent inductance is less than the inductance
of either coil.
Question 3.
In a Faraday disc dynamo, a metal disc of radius R rotates with an angular
velocity ω about an axis perpendicular to the plane of the disc and passing
through its centre. The disc is placed in a magnetic field B acting
perpendicular to the plane of the disc. Determine the induced emf between the
rim and the axis of the disc.
Answer:
Question 4.
A horizontal wire 20 m long extending from east to west is falling with a
velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 × 10-4 T.
What is the value of induced emf in the wire?
Answer:
Data : l = 20 m, v = 10 m/s. B = 5 × 10-5 T
The magnitude of the induced emf,
|e| = Blv = (5 × 10-5)(20)(10) = 10-2V = 10 mV
Question 5.
A metal disc is made to spin at 20 revolutions per second about an axis passing
through its centre and normal to its plane. The disc has a radius of 30 cm and
spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of
rotation. Calculate
(a) The area swept out per second by the radius of the disc,
(b) The flux cut per second by a radius of the disc,
(c) The induced emf in the disc.
Answer:
Data: R = 0.3m, f = 20 rps, B = 0.2T
(a) The area swept out per unit time by a given radius = (the frequency of
rotations) × (the area swept out per rotation) = f(πr2)
= (20)(3.142 × 0.09) = 5.656 m2
Question 6.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in
one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage
with the other coil?
Answer:
Data: M = 1.5 H, I1i = 0, I1f = 10A, ∆f =
0.2s
The flux linked per unit turn with the second coil due to current I1 in
the first coil is
Φ21 = MI1
Therefore, the change in the flux due to change in I1 is
∆21 =M(∆I1) = M(I1f – I1i)
= 1.5 (10 – 0)
= 15 Wb
[Note: The rate of change of flux linkage is M(∆I1/∆t) = 15/0.2 = 75
Wb/s] .
Question 7.
A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm2
and 150 turns (Nc) is wound tightly around the centre of the
solenoid. If a current of 3.0A flows through the solenoid, calculate :
(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the direction of the current in the
solenoid is reversed in a time of 0.5 s. (µ0 = 4π × 10-7 H/m)
Answer:
Data: n = 1.5 × 103 m , A = 25 × 10-4 m2,
Nc = 150, I = 3A, ∆t = 0.5s,
µ0 = 4π × 10-7 H/m
(a) Magnetic flux density inside the solenoid,
B = u0 nI = (4π × 10-7)(1500)(3)
= 5.656 × 10-3 T = 5.656 mT
(b) Flux per unit turn through the coils of the solenoid, Φm =
BA
Since the coil C is wound tightly over the solenoid, the flux linkage of C is
NCΦm = NCBA = (150)(5.656 × 10-3)(25
× 10-4)
= 2.121 × 10-3 Wb = 2.121 mWb
Question 8.
A search coil having 2000 turns with area 1.5 cm2 is placed in
a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time
of 0.2 second. Calculate the induced emf across the search coil.
Answer:
Question 9.
An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field
of 6 × 10-5 T at a speed of 400 m/s. Calculate the emf
generated between the tips of the wings of the aircraft.
Answer:
Data : l = 50 m, B = 6 × 10-5T, v = 400 m/s
The magnitude of the induced emf,
|e| = Blv = (6 × 10-5)(400)(50) = 1.2V
Question 10.
A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B
about an axis passing through its ends. If the frequency of rotation of the
wire is f, calculate the amplitude of the alternating emf induced in the wire.
Answer:
Question 11.
Calculate the value of induced emf between the ends of an axle of a railway
carriage 1.75 m long traveling on level ground with a uniform velocity of 50 km
per hour. The vertical component of Earth’s magnetic field (Bv) is
given to be 5 × 10-5T.
Answer:
Question 12.
The value of mutual inductance of two coils is 10 mH. If the current in one of
the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in
the other coil.
Answer:
Question 13.
An emf of 96.0 mV is induced in the windings of a coil when the current in a
nearby coil is increasing at the rate of 1.20 A/s. What is the mutual
inductance (M) of the two coils?
Answer:
Question 14.
A long solenoid of length l, cross-sectional area A and having N1 turns
(primary coil) has a small coil of N2 turns (secondary coil)
wound about its centre. Determine the Mutual inductance (M) of the two coils.
Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid
passes through the outer coil C. For a steady current Is through the solenoid,
the uniform magnetic field inside the solenoid is
Question 15.
The primary and secondary coil of a transformer each have an inductance of 200
× 10-6H. The mutual inductance (M) between the windings is 4 × 10-6H.
What percentage of the flux from one coil reaches the other?
Answer:
Question 16.
A toroidal ring, having 100 turns per cm of a thin wire is wound on a
nonmagnetic metal rod of length 1 m and diameter 1 cm. If the permeability of
bar is equal to that of free space (µ0), calculate the magnetic
field inside the bar (B) when the current (i) circulating through the turns is
1 A. Also determine the self-inductance (L) of the coil.
Answer:
Questions and Answers
Do you know (Textbook Page No. 274)
Question 1.
If a wire without any current is kept in a magnetic field, then it experiences
no force as shown in figure (a). But when the wire is carrying a current into
the plane of the paper in the magnetic field, a force will be exerted on the
wire towards the left as shown in the figure (b). The field will be
strengthened on the right side of the wire where the lines of force are in the
same direction as that of the magnetic field and weakened on the left side
where the field lines are in opposite direction to that of the applied magnetic
field. For a wire carrying a current out of the plane of the paper, the force
will act to the right as shown in figure (c).
Answer:
Use your brain power (Textbook Page No. 282)
Question 1.
It can be shown that the mutual potential energy of two circuits is W = MI1I2.
Therefore, the mutual inductance (M) may also be defined as the mutual
potential energy (W) of two circuits corresponding to unit current flowing in
each circuit.
Answer:
Mutual inductance of two magnetically linked coils equals the potential
energy for unit currents in the coils.
1 H = 1 T∙m2/A (= 1 V∙s/A = 1 Ω ∙ s = 1 J/A2)
Use your brain power (Textbook Page No. 284)
Question 1.
Do you know (Textbook Page No. 285)
Question 1.
The flux rule is the terminology that Feynman used to refer to the law relating
magnetic flux to emf. (RP Feynman, Feynman lectures on Physics, Vol II)
Answer:
Modern applications of Faraday’s law of induction :
- Electric
generators and motors
- Dynamos
in vehicles
- Transformers
- Induction
furnaces (industrial), induction cooking stoves (domestic)
- Radio
communication
- Magnetic
flow meters and energy meters
- Metal
detectors at security checks .
- Magnetic
hard disk and tape, storage and retrieval
- Graphics
tablets
- ATM
Credit/debit cards, ATM and point-of-sale (POS) machines
- Pacemakers
Faraday’s second law of electromagnetic induction is
referred by some as the “flux rule”.