Class 12 Physics Notes – Chapter 6: Electromagnetic Induction

1. Introduction to Electromagnetic Induction

The phenomenon of generating electric current in a circuit by changing the magnetic flux linked with it is called Electromagnetic Induction (EMI). This was discovered by Michael Faraday and Joseph Henry in 1831.

2. Magnetic Flux

Magnetic flux (Φ_B) through a surface of area A placed in a uniform magnetic field B is defined as the total number of magnetic field lines crossing the surface normally.

Formula: Φ_B = B · A = BA cosθ

Where θ is the angle between the magnetic field B and the area vector A.

SI Unit: Weber (Wb) or Tesla meter² (T m²). It is a scalar quantity.

3. Faraday’s Laws of Induction

Based on his experiments, Faraday gave two laws:

• First Law: Whenever the magnetic flux linked with a closed circuit changes, an emf (and hence a current) is induced in it, which lasts only so long as the change in flux continues.
• Second Law: The magnitude of the induced emf is equal to the time rate of change of magnetic flux through the circuit.

Formula: ε = – dΦ_B / dt

For a closely wound coil of N turns, the induced emf is: ε = – N (dΦ_B / dt)

4. Lenz’s Law and Conservation of Energy

Lenz’s Law: The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

The negative sign in Faraday’s law equation represents Lenz’s law. Lenz’s law is a direct consequence of the Law of Conservation of Energy. The mechanical work done in moving the magnet against the opposing force is converted into electrical energy (induced current).

5. Motional Electromotive Force

The emf induced across the ends of a conductor due to its motion in a magnetic field is called motional emf.

If a straight conductor of length l moves with velocity v perpendicular to a uniform magnetic field B, the induced emf is:

Formula: ε = B · l · v

If the conductor is moving at an angle θ with the magnetic field, ε = B l v sinθ.

6. Inductance

An electric current can be induced in a coil by flux change produced by another coil in its vicinity or by flux change produced by the same coil.

(i) Mutual Inductance (M)

It is the property of two coils due to which each opposes any change in the current flowing through the other. If current I₁ in coil 1 changes, emf is induced in coil 2.

Φ₂ = M · I₁ ⇒ ε₂ = – M (dI₁ / dt)

Where M is the coefficient of mutual induction. SI unit is Henry (H).

(ii) Self-Inductance (L)

It is the property of a single coil due to which it opposes any change in the current flowing through itself by inducing an emf.

Φ = L · I ⇒ ε = – L (dI / dt)

Where L is the coefficient of self-induction. Self-inductance of a long solenoid is given by: L = μ₀ n² A l (where n is turns per unit length).

Energy Stored in an Inductor

The energy required to build up the current I in an inductor is stored as magnetic potential energy:

U = 1/2 L I²

7. AC Generator

An AC generator converts mechanical energy into electrical energy based on the principle of electromagnetic induction.

When a coil is rotated with a constant angular speed ω in a uniform magnetic field B, the induced emf is given by:

ε = N B A ω sin(ωt)

Here, ε₀ = N B A ω is the maximum (peak) value of the emf.

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Multiple Choice Questions (MCQs)

1. The SI unit of magnetic flux is:
a) Tesla
b) Weber
c) Henry
d) Faraday
Answer: (b) Weber

2. Electromagnetic induction was discovered by:
a) Oersted
b) Ampere
c) Faraday and Henry
d) Maxwell
Answer: (c) Faraday and Henry

3. Lenz’s law is a consequence of the law of conservation of:
a) Charge
b) Momentum
c) Mass
d) Energy
Answer: (d) Energy

4. The formula for motional emf is:
a) ε = B/lv
b) ε = Blv
c) ε = B²lv
d) ε = Bvl²
Answer: (b) ε = Blv

5. The SI unit of self-inductance and mutual inductance is:
a) Weber
b) Tesla
c) Henry
d) Ohm
Answer: (c) Henry

6. The self-inductance of a long solenoid is directly proportional to:
a) The current flowing through it
b) Square of the number of turns
c) Radius of the solenoid
d) None of the above
Answer: (b) Square of the number of turns

7. The energy stored in an inductor of inductance L carrying current I is given by:
a) LI²
b) 1/2 LI²
c) 1/2 L²I
d) LI
Answer: (b) 1/2 LI²

8. An AC generator works on the principle of:
a) Electrostatic induction
b) Magnetic effect of current
c) Electromagnetic induction
d) Heating effect of current
Answer: (c) Electromagnetic induction

9. According to Faraday’s law, the induced emf is proportional to:
a) Magnetic flux
b) Rate of change of magnetic flux
c) Magnetic field
d) Area of the coil
Answer: (b) Rate of change of magnetic flux

10. The negative sign in the formula ε = – dΦ/dt is explained by:
a) Gauss’s Law
b) Ampere’s Law
c) Lenz’s Law
d) Biot-Savart Law
Answer: (c) Lenz’s Law

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Textbook Exercise Solutions

(Selected Important Questions from NCERT)

Q 6.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Solution:
Number of turns per unit length, n = 15 turns/cm = 1500 turns/m.
Area of loop, A = 2.0 cm² = 2.0 × 10^-4 m².
Rate of change of current, dI/dt = (4.0 A – 2.0 A) / 0.1 s = 20 A/s.
Magnetic field inside solenoid, B = μ₀ n I.
Flux through the loop, Φ = B · A = (μ₀ n I) · A.
Induced emf, ε = dΦ / dt = μ₀ n A (dI/dt)
ε = (4π × 10^-7) × 1500 × (2.0 × 10^-4) × 20
ε = 4 × 3.14 × 10^-7 × 1500 × 2 × 10^-4 × 20 = 7.54 × 10^-6 V.

Q 6.4: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Solution:
Magnetic field B = 0.3 T. Velocity v = 1 cm/s = 0.01 m/s.
(a) When velocity is normal to the longer side (length l = 8 cm = 0.08 m):
Induced emf ε = Blv = 0.3 × 0.08 × 0.01 = 2.4 × 10^-4 V.
Time t = distance / velocity = width / v = 2 cm / 1 cm/s = 2 seconds.

(b) When velocity is normal to the shorter side (length l = 2 cm = 0.02 m):
Induced emf ε = Blv = 0.3 × 0.02 × 0.01 = 0.6 × 10^-4 V.
Time t = distance / velocity = length / v = 8 cm / 1 cm/s = 8 seconds.

Q 6.5: A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Solution:
Length of rod (R) = 1.0 m. Angular frequency (ω) = 400 rad/s. Magnetic field (B) = 0.5 T.
The emf developed across the rotating rod is given by:
ε = 1/2 B R² ω
ε = 1/2 × 0.5 × (1.0)² × 400
ε = 0.25 × 400 = 100 V.

Q 6.6: A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s^-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^-4 Wb m^-2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Solution:
Given: l = 10 m, v = 5.0 m/s, B = 0.30 × 10^-4 T.
(a) Induced emf ε = Blv = 0.30 × 10^-4 × 10 × 5.0 = 1.5 × 10^-3 V (or 1.5 mV).
(b) By Fleming’s Right-Hand Rule, the direction of induced current (and emf) is from West to East.
(c) Inside a source of emf, current flows from lower to higher potential. So, the East end of the wire is at a higher electrical potential.

Q 6.7: Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Solution:
Initial current I₁ = 5.0 A, Final current I₂ = 0.0 A. Time dt = 0.1 s.
Change in current dI = I₂ – I₁ = 0.0 – 5.0 = – 5.0 A.
Induced emf ε = 200 V.
We know, ε = – L (dI / dt)
200 = – L × (- 5.0 / 0.1)
200 = L × 50
L = 200 / 50 = 4.0 H (Henry).

Q 6.8: A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Solution:
Mutual inductance M = 1.5 H.
Change in current ΔI = 20 A – 0 A = 20 A.
We know that flux linkage Φ = M · I.
Therefore, change of flux linkage ΔΦ = M · ΔI
ΔΦ = 1.5 × 20 = 30 Wb (Weber).