Class 12 Physics Chapter 6: Electromagnetic Induction MCQs with answers
10 multiple-choice questions on Electromagnetic Induction, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (b) V s
From Faraday's law ε = −dΦ_B/dt, flux = emf × time, so 1 Wb = 1 V s. It is also 1 T m², not T m⁻².
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Answer: (c) swings to the right, as before
Only the relative motion of magnet and coil matters. Moving the coil towards the north pole is the same relative motion as moving the north pole towards the coil, so the deflection is the same in direction and size.
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Answer: (d) a momentary deflection when the key is pressed, then none while it is held
Pressing the key makes the current in P, and so the flux through Q, rise: an emf is induced for that moment. While the key is held the current is steady, the flux through Q does not change, and the induced current is zero.
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Answer: (a) 4.0 V
ε = N ΔΦ/Δt = 200 × (5.0 − 1.0) × 10⁻³ / 0.20 = 4.0 V. 6.0 V adds the two fluxes instead of subtracting; 0.020 V leaves out N; 0.80 V forgets to divide by the time.
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Answer: (c) clockwise
Only relative motion matters, so the flux through the ring decreases as for a receding magnet. To oppose the decrease the ring's near face becomes a south pole, attracting the magnet; a current that is clockwise as seen from that face makes it a south pole.
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Answer: (d) acts in the same direction as the current, opposing its decrease
The self-induced (back) emf ε = −L dI/dt opposes any change in the current. When the current falls, the emf tries to keep it flowing, so it acts in the direction of the current.
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Answer: (b) [M L² T⁻² A⁻²]
L = NΦ_B / I, so [L] = [flux]/[current] = [M L² T⁻² A⁻¹]/[A] = [M L² T⁻² A⁻²]. [M L² T⁻² A⁻¹] is flux, [M L² T⁻³ A⁻¹] is emf and [M L² T⁻³ A⁻²] is resistance.
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Answer: (a) the plane of the coil is perpendicular to the magnetic field
ε = NBAω sin ωt, with θ = ωt the angle between B and the area vector. ε = 0 at θ = 0° and 180°, when the area vector is along B, that is, the plane of the coil is perpendicular to B; the flux is then at its extreme and momentarily not changing.
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Answer: (d) Rotating the loop about the axis through its centre that is parallel to the field
Φ_B = BA cos θ changes if B, A or θ changes. Turning the loop about its own normal (which is along B) changes none of them, so the flux stays constant and no emf is induced.
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Answer: (b) 0.80 H
L = μ_r μ₀ n² A l, so L is multiplied by μ_r: 400 × 2.0 × 10⁻³ H = 0.80 H. 0.0050 H divides by μ_r; 0.040 H uses √μ_r; 320 H uses μ_r².
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