 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-314/4/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The representation

B_{p} = \nabla ψ \times \nabla ϕ

gives

B_{r} = - \frac{1}{r} \frac{\partial ψ}{\partial z}, B_{z} = \frac{1}{r} \frac{\partial ψ}{\partial r} .

Hence

B \cdot \nabla ψ = 0

\int_{0}^{r} 2 π r^{'} B_{z} (r^{'}, z) d r^{'} = 2 π [ψ (r, z) - ψ (0, z)] .

Thus $2 π ψ$ is the enclosed poloidal magnetic flux up to an additive axial reference.

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