A-LevelPhysicsMagnetic fields9702/41 · May/June 2020 · Q910 Marks

(a) A coil of wire is situated in a uniform magnetic field of flux density B. The coil has diameter 3.6 cm and consists of 350 turns of wire, as illustrated in Fig. 9.1. [Figure 9.1] The variation with time t of B is shown in Fig. 9.2. [Figure 9.2]

(a)(i)Show that, for the time t = 0 to time t = 0.20 s, the electromotive force (e.m.f.) induced in the coil is 0.080 V.[2]
(a)(ii)On the axes of Fig. 9.3, show the variation with time t of the induced e.m.f. E for time t = 0 to time t = 0.80 s. [Figure 9.3][4]
(b)A bar magnet is held a small distance above the surface of an aluminium disc by means of a rod, as illustrated in Fig. 9.4. [Figure 9.4] The aluminium disc is supported horizontally and held stationary. The magnet is rotated about a vertical axis at constant speed. Use laws of electromagnetic induction to explain why there is a torque acting on the aluminium disc.[4]

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Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[C1] e.m.f. = (Δ)B × AN / t
[A1] = 45 × 10⁻³ × π× (1.8 × 10⁻²)² × 350 / 0.20 = 0.080 V
[B1] 0 to 0.2 s: straight horizontal line at 0.080 V or -0.080 V
[B1] 0.2 s to 0.4 s: zero
[B1] 0.4 s to 0.8 s: straight horizontal line at 0.040 V or –0.040 V
[B1] opposite polarity to 0 to 0.2 s line
[B1] either disc cuts flux lines (of the magnet) / there is a changing flux in the disc
[B1] (by Faraday's law) e.m.f. is induced in the disc
[B1] e.m.f. causes (eddy) currents in the disc
[B1] current in the magnetic field (of the magnet) causes force on disc

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