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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 15C / i / Solution

The total hoop charge is

Q = 2 π R η

. One revolution takes

2 π / ω

, so the current is

I = \frac{Q ω}{2 π} = η R ω .

Multiplying by the enclosed area gives

m = I π R^{2} \overset{n}{^} = π η ω R^{3} \overset{n}{^} .

Solved by gpt-5.6-sol high.

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