CBSE Class 12 Physics Electromagnetic Induction MCQs Set 04

Answer: (a) 1 × 10⁻³ Wb

Explanation:
1. Φ = BA cos θ, with A = 50 cm² = 50 × 10⁻⁴ m² and θ = 60°
2. Φ = 0.4 × 50 × 10⁻⁴ × 0.5 = 1 × 10⁻³ Wb

Teacher's Note:
1. θ is measured from the normal, so use cos 60°, not cos 30°.
2. 1 cm² = 10⁻⁴ m²; forgetting this gives the 0.1 Wb trap.

Answer: (c) 1 T m²

Explanation:
1. Φ = BA, so the unit of flux is tesla × metre².
2. Hence 1 Wb = 1 T m² (which also equals 1 V s).

Teacher's Note:
1. From Faraday's law, emf = flux/time, so Wb = V × s, not V/s.
2. Tesla is the unit of B; weber is the unit of flux.

Answer: (b) 40 V

Explanation:
1. ε = N \( \frac{\Delta\Phi}{\Delta t} \), with ΔΦ = 5 − 1 = 4 mWb = 4 × 10⁻³ Wb
2. ε = \( \frac{200 \times 4 \times 10^{-3}}{0.02} \) = 40 V

Teacher's Note:
1. Use the change in flux, not the final or initial value.
2. Do not forget to multiply by the number of turns N.

Answer: (d) 4 s to 5 s

Explanation:
1. The induced emf equals the slope of the Φ–t graph.
2. Slope from 0 to 2 s = 4/2 = 2 mWb s⁻¹; from 2 s to 4 s = 0; from 4 s to 5 s = 4/1 = 4 mWb s⁻¹.
3. The steepest part is 4 s to 5 s, so the emf is greatest there.

Teacher's Note:
1. A flat part of the Φ–t graph means zero emf, even though the flux is largest there.
2. A falling flux gives an emf of the opposite sign to a rising flux.

Answer: (c) anticlockwise

Explanation:
1. The flux through the coil towards the magnet's side increases, so by Lenz's law the coil must oppose the approach.
2. To repel a north pole, the near face of the coil must become a north pole.
3. A face shows north polarity when the current in it looks anticlockwise from that side.

Teacher's Note:
1. Rule: anticlockwise current seen from a face means that face is N; clockwise means S.
2. If the magnet were moving away, the current would reverse.

Answer: (a) energy

Explanation:
1. If the induced current aided the change, a magnet once pushed would accelerate on its own, creating energy from nothing.
2. The opposition means work must be done to change the flux, and that work appears as electrical energy.
3. So Lenz's law follows from conservation of energy.

Teacher's Note:
1. Kirchhoff's junction rule comes from conservation of charge, not Lenz's law.
2. The work done against the opposing force is the source of the electrical energy.

Answer: (d) 0.5 V

Explanation:
1. Motional emf ε = Blv
2. ε = 0.25 × 0.5 × 4 = 0.5 V

Teacher's Note:
1. B, l and v must be mutually perpendicular for ε = Blv to apply directly.
2. No closed circuit is needed for an emf; a current needs a closed path.

Answer: (b) 5 V

Explanation:
1. For a rod rotating about one end, ε = \( \frac{1}{2} B \omega l^2 \)
2. ε = \( \frac{1}{2} \) × 0.5 × 20 × 1² = 5 V

Teacher's Note:
1. The factor ½ appears because different parts of the rod move at different speeds; the average speed is ωl/2.
2. Forgetting the ½ gives the 10 V trap.

Answer: (a) 0.63 mH

Explanation:
1. L = \( \frac{\mu_0 N^2 A}{l} \) = \( \frac{4\pi \times 10^{-7} \times (500)^2 \times 10 \times 10^{-4}}{0.5} \)
2. L = \( \frac{4\pi \times 10^{-7} \times 2.5 \times 10^5 \times 10^{-3}}{0.5} \) ≈ 6.28 × 10⁻⁴ H = 0.63 mH

Teacher's Note:
1. L depends on N², so a small error in N makes a big error in L.
2. Convert 10 cm² to 10 × 10⁻⁴ m² and 50 cm to 0.5 m.

Answer: (c) 8 V

Explanation:
1. |ε| = L \( \frac{\Delta I}{\Delta t} \) = 0.2 × \( \frac{5 - 1}{0.1} \)
2. |ε| = 0.2 × 40 = 8 V

Teacher's Note:
1. Use the change in current (4 A), not the final current.
2. The induced emf tries to keep the current flowing, so it acts in the direction of the falling current.

Answer: (d) 0.4 J

Explanation:
1. U = \( \frac{1}{2} L I^2 \) = \( \frac{1}{2} \) × 50 × 10⁻³ × 4²
2. U = 0.025 × 16 = 0.4 J

Teacher's Note:
1. The energy is stored in the magnetic field of the inductor.
2. Leaving out the ½ gives 0.8 J, a common slip.

Answer: (b) the currents flowing in the coils

Explanation:
1. Mutual inductance depends on the geometry of the two coils and on the medium, for example M = μ₀n₁n₂πr₁²l for coaxial solenoids.
2. The current appears on both sides of N₁Φ₁ = MI₂ and cancels out, so M is independent of the currents in the coils.

Teacher's Note:
1. The induced emf depends on the rate of change of current, but M itself does not.
2. Inserting a soft-iron core increases M by a factor μr.

Answer: (c) 50 V

Explanation:
1. ε₀ = NBAω = 100 × 0.5 × 0.02 × 50
2. ε₀ = 50 V

Teacher's Note:
1. Here ω is already given in rad s⁻¹; if a frequency ν is given, use ω = 2πν.
2. The peak occurs when the plane of the coil is parallel to the field.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A steady field linked with a stationary coil gives a constant flux, so dΦ/dt = 0 and no emf is induced. The Assertion is correct.
2. By Faraday's law ε = −dΦ/dt, so the Reason is correct.
3. The Reason directly explains the Assertion.

Teacher's Note:
1. Large flux and large emf are different things.
2. An emf appears only while the flux is changing.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The emf depends only on N and dΦ/dt, not on the material, so the Assertion is correct.
2. I = ε/R is true, so the Reason is correct.
3. The Reason is about the current, while the Assertion is about the emf, so the Reason does not explain the Assertion.

Teacher's Note:
1. The induced current, unlike the emf, does depend on resistance.
2. Even an open coil has an emf induced in it.

Answer: (d) A is false, but R is true.

Explanation:
1. Lenz's law is a consequence of conservation of energy; the opposition makes us do work to produce electrical energy. The Assertion is false.
2. The Reason is the statement of Lenz's law and is true.

Teacher's Note:
1. If the induced current aided the change, energy would be created from nothing.
2. The mechanical work done appears as electrical energy.

Answer: (c) A is true, but R is false.

Explanation:
1. ε = NBAω sin ωt is maximum when ωt = 90°, i.e. when the normal is perpendicular to B and the plane is parallel to B. The Assertion is true.
2. In this position Φ = BA cos 90° = 0, so the flux is zero, not maximum. The Reason is false.

Teacher's Note:
1. Zero flux can go with maximum emf, because the flux is then changing fastest.
2. Maximum flux goes with zero emf.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For coaxial solenoids with a core, M = μ_rμ₀n₁n₂πr₁²l, so M is proportional to μ_r. The Reason is true.
2. Soft iron has a very large μ_r, so M increases. The Assertion is true.
3. The Reason explains the Assertion.

Teacher's Note:
1. This is why Faraday found a larger deflection with an iron rod inside the coils.
2. M depends on geometry and the medium only.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. U = ½LI² is stored in the magnetic field of the inductor, so the Assertion is true.
2. L = μ₀N²A/l, so L ∝ N² and the Reason is true.
3. The dependence of L on N does not explain why energy is stored, so the Reason is not the correct explanation.

Teacher's Note:
1. The energy is stored against the back emf while the current builds up.
2. Choose (b) when both statements are true but unrelated.

Answer: (b) 1 V

Explanation:
1. ε = Blv, with l = 40 cm = 0.4 m
2. ε = 0.5 × 0.4 × 5 = 1 V

Teacher's Note:
1. Here l is the separation between the rails, i.e. the length of rod inside the circuit.
2. Remember to convert 40 cm to 0.4 m.

Answer: (d) 2 m² s⁻¹

Explanation:
1. Area = lx, so \( \frac{dA}{dt} \) = l \( \frac{dx}{dt} \) = lv
2. \( \frac{dA}{dt} \) = 0.4 × 5 = 2 m² s⁻¹

Teacher's Note:
1. Multiplying this by B gives dΦ/dt = 1 Wb s⁻¹, which equals the emf in volts.
2. This shows that ε = Blv is just Faraday's law for a growing area.

Answer: (a) doubles

Explanation:
1. ε = Blv, so ε is directly proportional to v.
2. Doubling v doubles the emf, to 2 V.

Teacher's Note:
1. Only the speed has changed; B and l are the same.
2. A faster rod sweeps out area faster, so the flux changes faster.

Answer: (c) 0.5 V

Explanation:
1. Area A = 0.4 × 0.5 = 0.2 m²; ΔΦ = A ΔB = 0.2 × 0.5 = 0.1 Wb
2. ε = \( \frac{\Delta\Phi}{\Delta t} \) = \( \frac{0.1}{0.2} \) = 0.5 V

Teacher's Note:
1. An emf is induced whenever the flux changes, whether B, A or θ changes.
2. Here there is no motion, so the emf comes from the changing field, not from a Lorentz force on moving charges.

Answer: (b) from Q to P

Explanation:
1. As the rod moves away from the bulb, the area and hence the flux into the page increase.
2. By Lenz's law, the induced current must produce a field out of the page inside the loop, i.e. flow anticlockwise as seen from above.
3. With the bulb on the left and the rod on the right, anticlockwise from above means the current flows up the rod, away from the student: from Q (near rail) to P (far rail).

Teacher's Note:
1. Fleming's right-hand rule gives the same direction: thumb along v, forefinger into the page.
2. Lenz's law always works: decide whether the flux is rising or falling, then oppose it.

Answer: (c) 0.39 mH

Explanation:
1. For coaxial solenoids, M = μ₀n₂N₁πr₁², where N₁ = n₁l is the total number of turns of the inner coil.
2. M = 4π × 10⁻⁷ × 2500 × 400 × π × (0.01)² = 4π² × 10⁻⁵ H
3. M ≈ 3.95 × 10⁻⁴ H ≈ 0.39 mH

Teacher's Note:
1. Use the radius of the inner solenoid, because the flux of the outer one passes only through the inner coil's area.
2. n₁l is simply the total number of turns of the inner coil.

Answer: (a) 20 mV

Explanation:
1. |ε| = M \( \frac{dI}{dt} \) = 3.95 × 10⁻⁴ × 50
2. |ε| ≈ 0.0197 V ≈ 20 mV

Teacher's Note:
1. The emf depends on how fast the current changes, not on the current itself.
2. A steadier change gives a steady voltmeter reading.

Answer: (d) zero

Explanation:
1. With a steady current the flux through the inner coil is constant.
2. dI/dt = 0, so the induced emf is zero.

Teacher's Note:
1. This matches Faraday's observation: a steady current in one coil gives no deflection in the other.
2. Multiplying M by the current (3 A) gives a flux linkage, not an emf.

Answer: (b) twice as large

Explanation:
1. M = μ₀n₂N₁πr₁² is proportional to N₁.
2. Doubling N₁ makes M twice as large.

Teacher's Note:
1. M is proportional to the product of the turns of the two coils, not to the square of either.
2. Self-inductance, by contrast, goes as N².

Answer: (a) only a changing current produces a changing flux in the phone's coil

Explanation:
1. An emf is induced in the phone's coil only if the flux through it changes.
2. An alternating current in the pad coil changes continuously, so it keeps the flux in the phone's coil changing and induces an emf continuously.
3. A steady current also produces a magnetic field, but its flux would not change.

Teacher's Note:
1. M depends on the coils and the medium, not on the type of current.
2. Alternating current does heat the coils; that is why chargers get warm.