 Solution

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

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

A^{2} = \frac{μ _{0} ρ u _{p}^{2}}{B _{p}^{2}} = \frac{μ _{0} k ^{2}}{ρ},

where

u_{p} = k B_{p} / ρ

. Thus, along a given field line,

ρ = \frac{μ _{0} k ^{2}}{A ^{2}};

The azimuthal velocity representation gives

u_{ϕ} = r ω + \frac{k B _{ϕ}}{ρ} .

Substitution into

ℓ = r u_{ϕ} - r B_{ϕ} / (μ_{0} k)

yields

ℓ = r^{2} ω + \frac{r B _{ϕ}}{μ _{0} k} (A^{2} - 1) .

A^{2} = 1

. Smoothness requires the numerator of the solution for

B_{ϕ}

to vanish there, so

ℓ = ω r_{a}^{2} .

Solving away from the surface then gives

B_{ϕ} = \frac{μ _{0} kω}{r} \frac{r ^{2} - r _{a}^{2}}{1 - A ^{2}} .

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