Class 9 Physics Chapter 6 Electromagnetic Induction Important Questions — Complete Guide with Answers

Electromagnetic Induction is a high-weightage chapter in Class 9 Physics. The 2024-25 CBSE rationalized curriculum emphasizes conceptual clarity over rote memorization, meaning examiners focus on application-based questions rather than definition recall. In recent board papers, Faraday's law and Lenz's law have been tested through numerical problems and diagram-based questions (4–5 marks). Understanding magnetic flux (Φ = B·A·cos θ), induced EMF (ε = −N·dΦ/dt), and the direction of induced current using Lenz's law are non-negotiable skills. Students who can solve 2-mark 'explain with diagram' questions and 5-mark 'prove Faraday's law' problems typically secure 18–20 marks in the electromagnetic section. This guide drills all question patterns in increasing difficulty, mirroring the actual exam blueprint. Questions on self-inductance and mutual inductance also appear in assertion–reason format (1 mark) and short-answer formats (2 marks), making them frequent board-exam fixtures.

**Q1. When a bar magnet is pushed into a coil, the induced current in the coil will:** (a) Be in the same direction always (b) Be zero (c) Oppose the motion of the magnet (d) Increase the magnetic flux through the coil **Answer: (c)** Oppose the motion of the magnet **Explanation:** Lenz's law states that the induced current creates a magnetic field opposing the change in flux. When the magnet approaches, the induced current opposes its motion. --- **Q2. The SI unit of magnetic flux is:** (a) Tesla (T) (b) Ampere (A) (c) Weber (Wb) (d) Henry (H) **Answer: (c)** Weber (Wb) **Explanation:** Magnetic flux Φ = B × A (in tesla × m²) is measured in Weber. 1 Wb = 1 T·m². --- **Q3. A coil has self-inductance L = 2 H. If the current through it changes from 0 to 5 A in 2 seconds, the induced EMF is:** (a) 2.5 V (b) 5 V (c) 10 V (d) 20 V **Answer: (b)** 5 V **Explanation:** Using ε = −L(dI/dt) = 2 × (5−0)/(2) = 5 V (magnitude). --- **Q4. Lenz's law is a consequence of:** (a) Conservation of energy (b) Ohm's law (c) Newton's first law (d) Coulomb's law **Answer: (a)** Conservation of energy **Explanation:** The induced current opposes the change causing it, ensuring energy is not created from nothing—a direct consequence of energy conservation. --- **Q5. Two coils are placed close to each other. If current in coil 1 increases, the induced EMF in coil 2 depends on:** (a) Mutual inductance M and rate of change of current (b) Resistance of coil 2 only (c) Distance between coils only (d) Number of turns in coil 1 only **Answer: (a)** Mutual inductance M and rate of change of current **Explanation:** ε₂ = −M(dI₁/dt), where M depends on geometry, number of turns, and proximity.

**Q1. State Faraday's law of electromagnetic induction. Write the mathematical expression.** **Answer:** Faraday's law states: The induced EMF in a coil is equal to the negative rate of change of magnetic flux through it. Mathematical expression: ε = −N(dΦ/dt) Where N = number of turns, dΦ/dt = rate of change of magnetic flux. The negative sign indicates the direction (Lenz's law). --- **Q2. A rectangular coil is moving through a uniform magnetic field. When will the induced EMF be zero?** **Answer:** The induced EMF will be zero when: (i) The coil moves parallel to the magnetic field lines (no flux change), or (ii) The coil moves perpendicular to the field but the velocity is perpendicular to the length of the conductor cutting field lines at zero angle. Induced EMF depends on dΦ/dt. If flux remains constant (Φ = 0 Wb/s), then ε = 0 V. --- **Q3. A circular coil of area 50 cm² is placed perpendicular to a magnetic field of 0.5 T. The field is reduced to zero in 0.1 s. Calculate the magnitude of induced EMF.** **Answer:** Given: A = 50 cm² = 50 × 10⁻⁴ m² = 0.005 m², B₁ = 0.5 T, B₂ = 0 T, Δt = 0.1 s Initial flux: Φ₁ = B₁A = 0.5 × 0.005 = 0.0025 Wb Final flux: Φ₂ = 0 Wb Change in flux: ΔΦ = 0.0025 Wb Induced EMF: |ε| = N × ΔΦ/Δt = 1 × 0.0025/0.1 = **0.025 V or 25 mV** --- **Q4. Distinguish between self-inductance and mutual inductance with one example each.** **Answer:** **Self-inductance (L):** The EMF induced in a coil due to change in its own current. Example: When you switch off a fan, the induced current in its coil opposes the sudden drop in current (creating a small spark at the switch). **Mutual inductance (M):** The EMF induced in one coil due to change in current in a neighboring coil. Example: In a transformer, when AC current changes in the primary coil, it induces EMF in the secondary coil through mutual inductance. Formula: ε_self = −L(dI/dt) and ε_mutual = −M(dI₁/dt) --- **Q5. Why does a copper ring fall slowly through a magnetic field compared to a steel ring?** **Answer:** When a conductor (copper ring) falls through a magnetic field, the change in flux induces an EMF and current in the ring. By Lenz's law, this induced current creates a magnetic field opposing the change, producing an upward magnetic force opposing gravity. This creates an electromagnetic damping effect. Copper has lower resistivity than steel, allowing larger induced currents → stronger opposing force → slower fall. The copper ring reaches terminal velocity quickly due to electromagnetic braking, while steel (higher resistivity) experiences less damping and falls faster.

**Q1. A rectangular coil ABCD of dimensions 20 cm × 10 cm is rotating in a uniform magnetic field of 0.5 T at a frequency of 50 Hz. The coil has 100 turns. Calculate the maximum induced EMF.** **Answer:** Given: Length l = 20 cm = 0.2 m, Width w = 10 cm = 0.1 m, B = 0.5 T, f = 50 Hz, N = 100 turns Area of coil: A = l × w = 0.2 × 0.1 = 0.02 m² Angular frequency: ω = 2πf = 2 × 3.14 × 50 = 314 rad/s Maximum flux through coil: Φₘₐₓ = BA = 0.5 × 0.02 = 0.01 Wb As the coil rotates, flux varies as: Φ(t) = Φₘₐₓ cos(ωt) = 0.01 cos(314t) Induced EMF: ε = −N(dΦ/dt) = −N × (−Φₘₐₓ ω sin(ωt)) Maximum EMF: εₘₐₓ = N × Φₘₐₓ × ω = 100 × 0.01 × 314 = **314 V** --- **Q2. Explain Lenz's law with a diagram. How does it relate to the conservation of energy?** **Answer:** **Lenz's Law:** The direction of induced current is such that it opposes the change in magnetic flux that produced it. **Explanation:** When a magnet is pushed toward a coil: (i) Magnetic flux through coil increases (into the page). (ii) By Lenz's law, induced current opposes this increase. (iii) Induced current creates a magnetic field opposing the approaching magnet (repulsive force). (iv) You must do work against this repulsive force to push the magnet. **Energy Conservation:** The mechanical work done against the magnetic force is converted into electrical energy (induced current) and then into heat (I²R losses) in the coil's resistance. No energy is created or destroyed—it only transforms. If induced current aided the change (by Lenz's law), it would create energy from nothing, violating conservation. **Diagram:** [Magnet approaching coil from left; coil develops induced current shown as circulating arrows; induced field repels magnet] --- **Q3. Two identical coils are placed coaxially with separation d. When current I₁ = 2 A in coil 1 is switched off in 0.05 s, an EMF of 0.4 V is induced in coil 2. Calculate the mutual inductance between them.** **Answer:** Given: I₁ = 2 A (initial), Final current = 0 A, ΔI₁ = 2 A, Δt = 0.05 s, Induced EMF in coil 2: ε₂ = 0.4 V From Faraday's law for mutual inductance: ε₂ = −M(dI₁/dt) Taking magnitude: 0.4 = M × (ΔI₁/Δt) 0.4 = M × (2/0.05) 0.4 = M × 40 M = 0.4/40 = **0.01 H or 10 mH** This mutual inductance indicates the coils are coupled (share magnetic flux) with weak coupling due to the small M value. --- **Q4. A solenoid has 500 turns and cross-sectional area 10 cm². When current through it changes from 2 A to 5 A in 0.1 s, calculate the self-induced EMF. What does the negative sign in Faraday's law indicate?** **Answer:** Given: N = 500 turns, A = 10 cm² = 10 × 10⁻⁴ m² = 0.001 m², ΔI = 5 − 2 = 3 A, Δt = 0.1 s Self-inductance of solenoid: L = μ₀(N²A/l) For a solenoid, if we assume length l (not given, we use the general relationship): Induced EMF: ε = −L(dI/dt) = −L × (3/0.1) = −30L If L ≈ 0.01 H (typical for such a solenoid), then: ε = −30 × 0.01 = **−0.3 V** The magnitude of self-induced EMF = 0.3 V **Negative Sign:** The negative sign indicates that the self-induced EMF opposes the change in current. Since current is increasing, the induced EMF acts to reduce it, trying to maintain the original magnetic state (Lenz's law).

**Q1. Derive Faraday's law of electromagnetic induction. Explain the terms in the equation ε = −N(dΦ/dt).** **Solution:** **Derivation of Faraday's Law:** Consider a coil with N turns placed in a time-varying magnetic field. **(i) Single Loop (N = 1):** When magnetic flux Φ through a single loop changes, an EMF is induced. Experimental observations show that the induced EMF is directly proportional to the rate of change of flux: ε ∝ −dΦ/dt Proportionality constant = 1 (in SI units): ε = −dΦ/dt ... (for one turn) **(ii) Multiple Turns (N turns):** Each turn links the same changing flux. The total induced EMF is the sum of EMFs in all turns: ε = −N(dΦ/dt) ... (Faraday's Law for N turns) **Explanation of Terms:** • **ε (Induced EMF):** The potential difference generated across the coil due to change in magnetic flux, measured in volts (V). • **N:** Number of turns in the coil. More turns mean more cutting of field lines, thus larger induced EMF. • **dΦ/dt:** Rate of change of magnetic flux with respect to time (Wb/s = V). Faster change → larger induced EMF. • **Φ (Magnetic Flux):** Total magnetic field passing through the coil, given by Φ = B·A·cos θ, where B is magnetic field, A is area, and θ is angle between B and normal to the area. • **Negative Sign:** Indicates Lenz's law—the induced EMF opposes the change causing it. If flux increases, ε is negative (opposes increase); if flux decreases, ε is positive (opposes decrease). **Example:** A coil of 50 turns has area 0.02 m². When the magnetic field through it increases from 0.1 T to 0.3 T in 0.04 s: ΔΦ = (0.3 − 0.1) × 0.02 = 0.004 Wb ε = −50 × (0.004/0.04) = −50 × 0.1 = **−5 V** (magnitude = 5 V) --- **Q2. Explain Lenz's law with a detailed example. Show how it prevents violation of the conservation of energy.** **Solution:** **Statement of Lenz's Law:** The direction of the induced current is always such that it opposes the change in magnetic flux that caused it. **Detailed Example: Magnet Approaching a Coil** *Step 1: Changing Flux* When a north pole of a magnet approaches a coil from the left: - Magnetic flux through the coil increases (field lines enter the coil). - Change in flux: dΦ/dt > 0 (into the page). *Step 2: Finding Induced Current Direction* By Lenz's law, induced current opposes this increase: - Induced current must create a field opposing the increasing external field. - Using right-hand rule: if external field is into the page and increasing, induced current flows counterclockwise (viewed from the approaching magnet) to create an outward field. - This makes the coil's left face a north pole (repelling the approaching magnet). *Step 3: Mechanical Consequence* - Repulsive force pushes back on the approaching magnet, slowing it down. - You must do mechanical work W = F·d against this repulsive force. **Energy Conservation Proof:** *Without Lenz's law (hypothetically):* - If induced current aided the flux change, it would strengthen the external field. - No work needed; magnet accelerates freely. - But electrical energy is generated in the coil (I²R dissipation)—**energy created from nothing!** This violates conservation of energy. *With Lenz's law (actual case):* - Mechanical work done against repulsion = W - Induced current = I, coil resistance = R - Joule heating in coil = I²R·Δt (over time Δt) - Energy balance: W = I²R·Δt (mechanical energy converts to electrical energy, then heat) - **Energy is conserved—no creation or loss.** **Mathematical Illustration:** Induced EMF: ε = −dΦ/dt = −N·A·dB/dt Induced current: I = ε/R = −(N·A/R)·dB/dt Joule heating: P = I²R = (N²A²/R)·(dB/dt)² Mechanical power needed: P_mech = F·v (where v is magnet's velocity) When magnet is pushed slowly (quasi-static process), P_mech = P (energy balance confirmed) --- **Q3. A rectangular coil of 200 turns, each of area 0.005 m², is rotated about an axis perpendicular to a uniform magnetic field of 0.2 T. The coil rotates at 50 Hz. Derive the expression for instantaneous induced EMF and calculate its peak value.** **Solution:** **Given:** N = 200 turns A = 0.005 m² B = 0.2 T (uniform, perpendicular to rotation axis) f = 50 Hz (frequency) **Derivation of Instantaneous EMF Expression:** *Step 1: Magnetic Flux as Function of Time* At t = 0, let the coil plane be perpendicular to B (maximum flux position): Φ(t) = BA cos(ωt) where ω = 2πf = 2π × 50 = 100π rad/s *Step 2: Apply Faraday's Law* Induced EMF: ε = −N(dΦ/dt) ε = −N·d[BA cos(ωt)]/dt ε = −N·BA·(−ω sin(ωt)) ε = NBAω sin(ωt) **Instantaneous EMF Expression:** **ε(t) = NBAω sin(ωt)** or **ε(t) = ε₀ sin(ωt)**, where ε₀ = NBAω *Step 3: Calculate Peak (Maximum) EMF* ε₀ = N × B × A × ω ε₀ = 200 × 0.2 × 0.005 × 100π ε₀ = 200 × 0.2 × 0.005 × 314.16 ε₀ = 0.2 × 314.16 ε₀ = **62.83 V ≈ 63 V** **Physical Meaning:** - Maximum EMF is generated when the coil plane is parallel to B (sin(ωt) = 1), i.e., when the rate of flux change is maximum. - At any time t, ε(t) varies sinusoidally between −63 V and +63 V. - RMS EMF for AC: ε_rms = ε₀/√2 = 63/1.414 ≈ **44.5 V** (used in AC circuit calculations)

**Case-Study: Electromagnetic Braking in Electric Trains** Modern electric trains use electromagnetic brakes to slow down safely without friction-based wear. A conducting disk (100 turns, area 0.04 m²) rotates inside a uniform magnetic field of 0.6 T. When the train needs to brake, the disk's rotation speed decreases from 200 rpm to 50 rpm in 8 seconds. **Questions:** **(i) Calculate the average rate of change of magnetic flux through the disk as it decelerates.** **Solution:** Initial angular velocity: ω₁ = 200 rpm = 200 × 2π/60 = 20.94 rad/s Final angular velocity: ω₂ = 50 rpm = 50 × 2π/60 = 5.24 rad/s At maximum rotation (ω₁), flux varies as Φₘₐₓ = BA = 0.6 × 0.04 = 0.024 Wb As rotation slows, the rate of flux change decreases. During deceleration: Average angular deceleration: α = (ω₂ − ω₁)/(Δt) = (5.24 − 20.94)/8 = −1.96 rad/s² At the start of braking, rate of flux change: |dΦ/dt|_max = BA × ω₁ = 0.024 × 20.94 = 0.503 Wb/s At the end of braking: |dΦ/dt|_min = BA × ω₂ = 0.024 × 5.24 = 0.126 Wb/s **Average rate of change:** |dΦ/dt|_avg ≈ (0.503 + 0.126)/2 ≈ **0.315 Wb/s** **(ii) Calculate the average induced EMF and current during braking (assume disk resistance = 5 Ω).** **Solution:** Average induced EMF (magnitude): ε_avg = N × (dΦ/dt)_avg = 100 × 0.315 = **31.5 V** Average induced current: I_avg = ε_avg / R = 31.5 / 5 = **6.3 A** **(iii) Explain why this braking method is more efficient than friction brakes. What happens to the electrical energy dissipated?** **Solution:** **Why Electromagnetic Braking is More Efficient:** 1. **No Friction Loss:** Friction brakes convert kinetic energy entirely into heat in brake pads and wheels, wasting energy. Electromagnetic brakes can recover some energy. 2. **Regenerative Braking:** The induced current (6.3 A) flowing through the disk creates a magnetic force opposing rotation. This force does negative work, slowing the train. In advanced systems, this electrical energy is fed back to the power grid, recovering 30−40% of kinetic energy. 3. **Even Braking:** Electromagnetic force is proportional to velocity (I ∝ dΦ/dt ∝ ω). As the train slows, braking force decreases smoothly, preventing sudden jerks (unlike friction brakes). 4. **Longer Component Life:** No mechanical wear on brake pads, reducing maintenance costs. **Where Does the Electrical Energy Go?** The electrical energy dissipated during braking: E = I²R·Δt Part of it is: - Dissipated as heat (I²R losses in disk resistance): E_heat = I²R·Δt ≈ 6.3² × 5 × 8 ≈ 1587 J - Some is transmitted back to the grid (in regenerative systems): E_recovered ≈ 20−40% of kinetic energy lost - Small amount radiated as electromagnetic waves **Energy Balance:** Initial kinetic energy of disk ≈ ½Iω₁² (where I is moment of inertia) Final kinetic energy ≈ ½Iω₂² Energy dissipated = Initial − Final ≈ Joule heat + Mechanical resistance losses This case-study demonstrates how Faraday's law and Lenz's law are directly applied in modern technology, converting motion safely and efficiently.

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**Magnetic Flux:** Φ = B·A·cos θ (in Weber, Wb) Φ = B·A (when field is perpendicular to area) **Faraday's Law:** ε = −N(dΦ/dt) [Induced EMF] For uniform B and rotating coil: ε = NBAω sin(ωt) [Peak = NBAω] **Induced Current (Ohm's Law):** I = ε/R = −(1/R)·N(dΦ/dt) **Self-Inductance (L):** ε = −L(dI/dt) For solenoid: L = μ₀(N²A/l), where l is length, N is turns, A is area **Mutual Inductance (M):** ε₂ = −M(dI₁/dt) [EMF in coil 2 due to current change in coil 1] **Direction of Induced Current (Lenz's Law):** 1. Find direction of change in flux (increasing/decreasing, into/out of page). 2. Induced current opposes this change → creates field opposite to the change. 3. Use right-hand rule: curl fingers in direction of induced current, thumb points in direction of induced field. **Motional EMF (Conductor Moving in Magnetic Field):** ε = Blv, where l is length of conductor, v is perpendicular velocity to field. **Energy Dissipated in Induced Current:** P = I²R = (1/R)·[N(dΦ/dt)]² [Joule heating] **Important Constants:** μ₀ (permeability of free space) = 4π × 10⁻⁷ H/m 1 Weber = 1 Tesla·m² 1 Henry = 1 Volt·second/Ampere