 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/36b/c/solution

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

M

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

H = B / μ_{0} - M

obeys

\nabla \times H = J_{free},

H

isolates free currents while

M

accounts for bound currents. For a linear isotropic material,

B = μ H

H = \frac{I}{2 π r} e_{ϕ}

in both media. Therefore

M_{i} = (\frac{μ _{i}}{μ _{0}} - 1) \frac{I}{2 π r} e_{ϕ} .

r = R

K_{b} = (M_{1} - M_{2}) \times e_{r} = - \frac{I ( μ _{1} - μ _{2} )}{2 π R μ _{0}} e_{z} .

Hence its magnitude is

∣ K_{b} ∣ = \frac{I ∣ μ _{1} - μ _{2} ∣}{2 π R μ _{0}},

+ e_{z}

it points along

- e_{z}

when

μ_{1} > μ_{2}

, reversing when

μ_{1} < μ_{2}

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