1. The magnetic flux through a coil perpendicular to the plane is varying according to the relation $\phi = (5t^3 + 4t^2 + 2t - 5)$ Weber. The induced current through the coil at $t = 2$ s, if the resistance of the coil is $5 \Omega$ , is:

(a) $15.6$ A
(b) $1.56$ A
(c) $0.78$ A
(d) $156$ A

Answer: (a)
$e = -d\phi/dt = -(15t^2 + 8t + 2)$ . At $t=2$ , $|e| = 15(4) + 8(2) + 2 = 60 + 16 + 2 = 78$ V.
$I = e/R = 78/5 = 15.6$ A.

2. Lenz’s law is a consequence of the law of conservation of:

(a) Charge
(b) Momentum
(c) Energy
(d) Mass

Answer: (c)
The mechanical work done in moving the magnet against the opposing force is converted into electrical energy.

3. A metal ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is:

(a) Equal to $g$
(b) Less than $g$
(c) More than $g$
(d) Zero

Answer: (b)
The induced current opposes the motion (Lenz’s Law), creating an upward magnetic force that retards the fall.

4. The self-inductance $L$ of a solenoid of length $l$ and area of cross-section $A$ , with a fixed number of turns $N$ , increases as:

(a) $l$ and $A$ increase
(b) $l$ decreases and $A$ increases
(c) $l$ increases and $A$ decreases
(d) Both $l$ and $A$ decrease

Answer: (b)
$L = \mu_0 N^2 A / l$ . Thus $L \propto A$ and $L \propto 1/l$ .

5. Eddy currents are produced when:

(a) A metal is kept in a varying magnetic field
(b) A metal is kept in a steady magnetic field
(c) A circular coil is placed in a magnetic field
(d) Through a circular coil, current is passed

Answer: (a)
Changing magnetic flux through a bulk conductor induces circulating currents called eddy currents.

6. Two coils of self-inductances $2$ mH and $8$ mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

(a) $16$ mH
(b) $10$ mH
(c) $6$ mH
(d) $4$ mH

Answer: (d)
For perfect coupling ( $k=1$ ), $M = \sqrt{L_1 L_2} = \sqrt{2 \times 8} = \sqrt{16} = 4$ mH.

7. A conducting rod of length $L$ is moving with a constant velocity $v$ perpendicular to a uniform magnetic field $B$ . The potential difference across its ends will be:

(a) $BLv$
(b) $BLv/2$
(c) $2BLv$
(d) Zero

Answer: (a)
This is the standard formula for motional EMF: $\varepsilon = Bvl$ .

8. The dimension of magnetic flux is:

(a) $[M L^2 T^{-2} A^{-1}]$
(b) $[M L^2 T^{-2} A^{-2}]$
(c) $[M L^2 T^{-1} A^{-2}]$
(d) $[M L^2 T^{-2} A]$

Answer: (a)
$\Phi = B A$ . Since $F = qvB$ , $B = F/qv$ . So $\Phi = (F/qv)A = [MLT^{-2}] / [AT][LT^{-1}] \times [L^2] = [M L^2 T^{-2} A^{-1}]$ .

9. When current in a coil changes from $5$ A to $2$ A in $0.1$ s, an average voltage of $50$ V is produced. The self-inductance of the coil is:

(a) $1.67$ H
(b) $6$ H
(c) $3$ H
(d) $0.67$ H

Answer: (a)
$|e| = L |dI/dt| \Rightarrow 50 = L \times (3/0.1) \Rightarrow 50 = 30L \Rightarrow L = 5/3 \approx 1.67$ H.

10. In an AC generator, a coil with $N$ turns, all of the same area $A$ and total resistance $R$ , rotates with frequency $\omega$ in a magnetic field $B$ . The maximum value of EMF generated in the coil is:

(a) $N.A.B.R.\omega$
(b) $N.A.B$
(c) $N.A.B.\omega$
(d) $N.A.B.R$

Answer: (c)
$\varepsilon = \varepsilon_0 \sin(\omega t)$ , where peak EMF $\varepsilon_0 = NAB\omega$ .

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): An induced current is developed in a conductor moved in a direction perpendicular to the magnetic field.
Reason (R): This phenomenon is called the dynamo effect.

Answer: (B)
Both statements are true. The motion induces EMF (Motional EMF), but the reason is just a name, not the physical explanation (Lorentz force).

2. Assertion (A): When a magnet is brought closer to a coil, the induced current in the coil flows in a direction such that the front face of the coil behaves as a North Pole (if North pole is approaching).
Reason (R): Lenz’s law states that induced EMF opposes the change in magnetic flux.

Answer: (A)
Correct. The induced North pole repels the approaching North pole, opposing the motion.

3. Assertion (A): Inductance coils are made of copper.
Reason (R): Induced current is more in wire having less resistance.

Answer: (A)
Low resistance (high conductivity) of Copper minimizes energy loss and maximizes induced effects.

4. Assertion (A): The possibility of an electric bulb fusing is higher at the time of switching ON and OFF.
Reason (R): Inductive effects produce a surge of current at these times.

Answer: (A)
Rapid change in current ( $dI/dt$ ) induces a high back-EMF or forward-EMF (surge) due to self-inductance of the circuit.

5. Assertion (A): Eddy currents are produced in any metallic conductor when magnetic flux is changed around it.
Reason (R): Electric potential determines the flow of charge.

Answer: (B)
Both are true. But potential difference is not the specific explanation for the *cause* of eddy currents (induction).

6. Assertion (A): Two coils are wound on a soft iron core. The mutual inductance is $M$ . If the coils are air cored, the mutual inductance will decrease.
Reason (R): The relative permeability of soft iron is much greater than 1.

Answer: (A)
$M = \mu_r \mu_0 n_1 n_2 A l$ . Removing the core replaces $\mu_r$ (high) with 1, decreasing $M$ .

7. Assertion (A): An EMF can be induced between the two ends of a straight copper wire when it is moved through a uniform magnetic field.
Reason (R): When a conductor moves in a magnetic field, the free electrons experience a Lorentz force.

Answer: (A)
This is the physical origin of Motional EMF.

8. Assertion (A): A spark occurs between the poles of a switch when the switch is opened.
Reason (R): Current flowing in the conductor produces a magnetic field.

Answer: (B)
Both are true. But the spark is due to the high induced EMF ( $L di/dt$ ) trying to maintain the current across the gap (self-induction), not just the presence of a field.

9. Assertion (A): The magnetic flux through a closed surface is always zero.
Reason (R): Gauss law applies in the case of electric flux only.

Answer: (C)
Assertion is True (Gauss Law for Magnetism). Reason is False; Gauss Law exists for both, but the result differs.

10. Assertion (A): Acceleration of a magnet falling through a long solenoid decreases.
Reason (R): The induced current produced in a circuit always flows in such a direction that it opposes the change or the cause that produced it.

Answer: (A)
The opposing magnetic force reduces the net downward force, hence acceleration is less than $g$ .

Part 3: Important Derivations & Theory

1. State Faraday’s law of electromagnetic induction. Derive an expression for motional EMF induced in a conductor moving in a uniform magnetic field. (3 Marks)

Statement: The magnitude of induced EMF is equal to the time rate of change of magnetic flux through the circuit. $\varepsilon = -d\Phi/dt$ .
Derivation:
1. Consider a rectangular loop with movable arm of length $l$ moving with velocity $v$ .
2. Area $A = lx$ . Flux $\Phi = Blx$ .
3. EMF $\varepsilon = -d\Phi/dt = -B l (dx/dt)$ .
4. Since $dx/dt = -v$ , $\varepsilon = B l v$ .

2. Define self-inductance. Derive the expression for the self-inductance of a long solenoid. (3 Marks)

Definition: The property of a coil to oppose the change in current flowing through it by inducing a back EMF.
Derivation:
1. Field inside solenoid $B = \mu_0 n I$ .
2. Flux through one turn $\phi = B A = \mu_0 n I A$ .
3. Total Flux $N\phi = (nl) (\mu_0 n I A) = \mu_0 n^2 A l I$ .
4. Since $N\phi = L I$ , comparing gives $L = \mu_0 n^2 A l$ .

3. Describe the principle and working of an AC generator. Derive the expression for the instantaneous EMF. (5 Marks)

Principle: Based on electromagnetic induction (coil rotating in B-field).
Derivation:
1. Flux $\Phi = NBA \cos(\omega t)$ where $\theta = \omega t$ .
2. $\varepsilon = -d\Phi/dt = -NBA \frac{d}{dt}(\cos \omega t)$ .
3. $\varepsilon = -NBA (-\sin \omega t \cdot \omega) = NBA\omega \sin \omega t$ .
4. Max EMF $\varepsilon_0 = NBA\omega$ . So $\varepsilon = \varepsilon_0 \sin \omega t$ .

Part 4: Numericals

1. A horizontal straight wire 10 m long extending from east to west is falling with a speed of $5.0 \text{ ms}^{-1}$ , at right angles to the horizontal component of the earth’s magnetic field, $0.30 \times 10^{-4} \text{ Wb m}^{-2}$ . What is the instantaneous value of the emf induced in the wire?

Formula: $\varepsilon = B l v$ .
$\varepsilon = (0.30 \times 10^{-4}) \times 10 \times 5.0$ .
$\varepsilon = 1.5 \times 10^{-3} \text{ V} = 1.5 \text{ mV}$ .

2. A circular coil of radius 10 cm, 500 turns and resistance $2 \Omega$ is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through $180^\circ$ in $0.25$ s. Estimate the magnitude of the emf and current induced in the coil. ( $H_E = 3.0 \times 10^{-5}$ T).

Initial flux $\Phi_i = NBA \cos 0^\circ = NBA$ . Final flux $\Phi_f = NBA \cos 180^\circ = -NBA$ .
Change $\Delta \Phi = -2NBA$ .
$\varepsilon = -N \Delta \Phi / \Delta t = 2NBA / t$ .
Area $A = \pi (0.1)^2 = 0.0314 \text{ m}^2$ .
$\varepsilon = (2 \times 500 \times 3 \times 10^{-5} \times 0.0314) / 0.25 = 3.8 \times 10^{-3} \text{ V}$ .
$I = \varepsilon / R = 1.9 \times 10^{-3} \text{ A}$ .

3. 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.

$\varepsilon = -L (dI/dt)$ .
$200 = L \times (5 - 0) / 0.1$ .
$200 = 50 L \Rightarrow L = 4 \text{ H}$ .

Part 5: Case Study

Case Study: Eddy Currents
Foucault (1895) observed that when a mass of metal moves in a magnetic field or is subjected to a changing magnetic field, induced currents are produced in the volume of the metal. These currents flow in closed loops and look like eddies in water. The direction of eddy currents is given by Lenz’s law. They produce significant heating and magnetic damping.

Why does a metallic pendulum oscillating in a magnetic field come to rest quickly?
How can the effect of eddy currents be reduced in transformer cores?
Mention two useful applications of eddy currents.
What is the source of heat generated by eddy currents?

1. Due to **electromagnetic damping**. Eddy currents oppose the motion, dissipating mechanical energy.
2. By using **laminated cores** (thin sheets insulated from each other) to break the path of currents.
3. **Induction Furnace** and **Magnetic Brakes** in trains.
4. The **electrical resistance** of the metal conductor (Joule heating $I^2R$ ).

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Electromagnetic Induction: Question Bank

Electromagnetic Induction: Question Bank

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

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Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

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