A-Level Magnetic flux, the induction laws of Faraday and Lenz, and power transformers: worked solution

Question

The 0.72 m diameter disc in the 0.017 T field cuts about 6.9 × 10⁻³ Wb per revolution and spins at 25 revolutions per second. Find the magnitude of the e.m.f. induced between axle and rim. [3]

Principle: Faraday's law says the magnitude of the induced e.m.f. equals the rate at which flux is cut, |E| = ΔΦ / Δt, where ΔΦ is the flux cut in time Δt. Here a radius of the spinning disc sweeps across the field, cutting flux, and the e.m.f. this produces appears between the axle at the centre and the rim at the edge. Step 1 - identify the flux cut in one revolution. In one complete turn a radius sweeps out the whole area of the disc, so it cuts the flux passing through the whole disc: ΔΦ = 6.9 × 10⁻³ Wb per revolution (given). (Check that this is consistent: the disc's radius is half its diameter, r = 0.72/2 = 0.36 m, so its area is πr² = π × 0.36² = 0.407 m², and ΔΦ = B A = 0.017 T × 0.407 m² = 6.9 × 10⁻³ Wb, which agrees.) Step 2 - find the time taken for one revolution. The disc makes f = 25 revolutions each second, so the time for one revolution is Δt = 1 / f = 1 / 25 s = 0.040 s Step 3 - apply Faraday's law. |E| = ΔΦ / Δt |E| = 6.9 × 10⁻³ Wb / 0.040 s |E| = 0.1725 V (Equivalently, and more quickly: e.m.f. = flux cut per revolution × revolutions per second = 6.9 × 10⁻³ × 25 = 0.17 V, which is the same calculation written in one line.) The e.m.f. induced between the axle and the rim is about 0.17 V.