 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/15d/b/solution

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

Take the curl of the integral from part (a). Since

J (r^{'})

is independent of

r

B (r) = \nabla \times A (r) = \frac{μ _{0}}{4 π} \int \nabla (\frac{1}{∣ r - r ^{'} ∣}) \times J (r^{'}) d^{3} r^{'} = \frac{μ _{0}}{4 π} \int \frac{J ( r ^{'} ) \times ( r - r ^{'} )}{∣ r - r ^{'} ∣ ^{3}} d^{3} r^{'} .

 New to topics? Read the docs here!