 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-1/15c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

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}{^} .

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