Class 12 Physics Chapter 6: Electromagnetic Induction worksheet with answer key

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.

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.

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.

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.

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.

The ends of the rotating coil are joined to slip rings that turn with it; fixed brushes press on the rings and carry the current to the external circuit without the wires twisting. The generator converts mechanical energy (of whatever turns the armature) into electrical energy.

• slip rings and brushes connect the rotating coil to the external circuit (1)
• mechanical energy to electrical energy (1)

U = ½LI². ΔU = ½ × 0.50 × (6.0² − 2.0²) = 0.25 × 32 = 8.0 J. The source does this work against the back emf while the current rises; it is stored as magnetic energy.

• ΔU = ½L(I₂² − I₁²) = 8.0 J (1)
• work done by the source against the back emf (1)

The self-induced (back) emf opposes any change in the current, growth or decay, just as inertia opposes any change in velocity. In U = ½LI², the inductance L plays the part of mass m and the current I that of velocity v.

• back emf opposes both growth and decay of current, as inertia opposes change of velocity (1)
• L corresponds to m and I to v (1)

(a) F = evB = 1.6 × 10⁻¹⁹ × 4.0 × 0.25 = 1.6 × 10⁻¹⁹ N. For a positive charge, v × B (right × into page) points up the rod towards P; an electron, being negative, is pushed towards Q. (b) Moving a charge q along the rod the force does work W = qvBl, so ε = W/q = Blv = 0.25 × 0.60 × 4.0 = 0.60 V. (c) Electrons collect at Q, so P is at the higher potential (the induced current flows from Q to P inside the rod, anticlockwise round the circuit, opposing the rising flux).

• F = evB = 1.6 × 10⁻¹⁹ N, electron pushed towards Q (1)
• ε = W/q = Blv = 0.60 V (1)
• P at higher potential (1)

(a) Closing the switch makes the coil's current, and the flux through the ring, rise sharply (the iron core makes the flux large). By Lenz's law the induced current in the ring opposes this rise: its field is opposite to the coil's, so the ring is repelled and jumps. (b) Once the current is steady the flux stops changing, the induced current dies away, there is no repulsion and gravity pulls the ring back. (c) With a cut the ring is an open circuit: an emf appears across the cut but no current flows, so there is no force and the ring does not jump.

• rising flux induces a current opposing it, so the ring is repelled (1)
• steady current, no change of flux, no induced current (1)
• cut ring: emf only, no current, so no jump (1)

(a) At time t the area vector makes θ = ωt with B, so Φ_B = BA cos ωt per turn. ε = −N dΦ_B/dt = −NBA d(cos ωt)/dt = NBAω sin ωt. (b) ε is maximum, ε₀ = NBAω, at θ = 90° and 270°, when the plane of the coil is parallel to B: the flux is zero but changing fastest. ε is zero at θ = 0° and 180°, when the plane is perpendicular to B: the flux is maximum and momentarily not changing.

• Φ_B = NBA cos ωt written (1)
• ε = −N dΦ/dt = NBAω sin ωt (1)
• maximum when plane parallel to B, zero when perpendicular, with the rate-of-change reason (1)