Physics (AS & A Level) · Oct / Nov 2016 · Paper 43 · Question 9
9702
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9(a)[2]
(induced) e.m.f. proportional to rate of change of (magnetic) flux (linkage) [2]
9(b)[1]
flux linkage = BAN = π × 10–3 × 2.8 × π × (1.6 × 10–2) 2 × 85 = 6.0 × 10–4 Wb [1]
9(c)[2]
e.m.f. = ∆NΦ / ∆t e.m.f. = (6.0 × 10–4 × 2) / 0.30 e.m.f. = 4.0 mV [2]
9(d)[4]
sketch: E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0 s, 1.6 s → 2.0 s E = 4 mV for t = 0.3 s → 0.6 s (either polarity) E = 2 mV for t = 1.0 s → 1.6 s with opposite polarity [4]
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