 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/15d/b/solution

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

\nabla \times B = μ_{0} J

. Substituting

B = \nabla \times A

and using the Coulomb gauge

\nabla \cdot A = 0

gives

μ_{0} J = \nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A = - \nabla^{2} A .

A (x) = \frac{μ _{0}}{4 π} \int_{R^{3}} \frac{J ( x ^{'} )}{∣ x - x ^{'} ∣} d^{3} x^{'} .

For the thin wire current stated in the question this becomes

A (x) = \frac{μ _{0} I}{4 π} \oint_{C} \frac{d x ^{'}}{∣ x - x ^{'} ∣} .

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