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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-b/10/d/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 10 / d / Solution by

Codex  0 2026-10-20

The regular hexagon has area

A = 6 (3 a^{2} /4) = 33 a^{2} /2

. Choose

t = 0

with its positive normal parallel to the imposed field. The external magnetic flux is

Φ = B_{0} A cos ω t

, so Faraday's law of induction gives

E = - \frac{d Φ}{d t} = B_{0} A ω sin ω t .

(1)

Taking the stated resistance to determine the loop response,

I (t) = \frac{3 3 B _{0} a ^{2} ω}{2 R} sin ω t .

(2)

A different initial orientation adds a phase. The sign is with respect to the chosen loop orientation and obeys Lenz's law; reversal of that orientation reverses the signed current. This expression neglects self-inductance, as in the resistive-loop model.

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