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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-1/16b/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 16B / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Use the identity

\nabla \times B = \nabla (\nabla \cdot A) - \nabla^{2} A

. The Laplacian identity for the Coulomb kernel gives, component by component,

- \nabla^{2} A = - \frac{μ _{0}}{4 π} \int J (r^{'}) \nabla^{2} (s^{- 1}) d V^{'} = μ_{0} J (r) .

(1)

Also

\nabla (s^{- 1}) = - \nabla^{'} (s^{- 1})

, so integration by parts yields

\nabla \cdot A = - \frac{μ _{0}}{4 π} \int J (r^{'}) \cdot \nabla^{'} (s^{- 1}) d V^{'} = \frac{μ _{0}}{4 π} \int \frac{\nabla ^{'} \cdot J ( r ^{'} )}{s} d V^{'} .

(2)

Here the surface integral at infinity vanishes under the assumed localization of the current. As usual, this assumption means sufficient decay for these integrals and the boundary limit, rather than merely pointwise convergence to zero. Combining the two terms gives

\nabla \times B = μ_{0} J + \frac{μ _{0}}{4 π} \nabla \int \frac{\nabla ^{'} \cdot J ( r ^{'} )}{∣ r - r ^{'} ∣} d V^{'} .

(3)

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