 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/36d/b/solution

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

M

J_{b} = \nabla \times M, K_{b} = M \times n .

Starting from the vector potential of magnetic dipoles and integrating by parts gives

A (x) = \frac{μ _{0}}{4 π} [\int_{V} \frac{\nabla ^{'} \times M ( x ^{'} )}{∣ x - x ^{'} ∣} d^{3} x^{'} + \int_{S} \frac{M ( x ^{'} ) \times n}{∣ x - x ^{'} ∣} d S^{'}],

which has precisely the form generated by those bound currents.

R

ρ

. Ampère's law for the free axial current gives, inside the material,

H (ρ) = \frac{I}{2 π ρ} \hat{ϕ}, M = \frac{B}{μ _{0}} - H = (\frac{μ}{μ _{0}} - 1) \frac{I}{2 π ρ} \hat{ϕ} .

ρ = R

this is tangential to the cylinder. Since

n = \overset{r}{^}

K_{b} = M \times \overset{r}{^} = - (\frac{μ}{μ _{0}} - 1) \frac{I}{2 π R} \overset{z}{^},

and

M (R) = (\frac{μ}{μ _{0}} - 1) \frac{I}{2 π R} \hat{ϕ} .

For

μ > μ_{0}

, the bound surface current therefore runs opposite to the free wire current.

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