Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 16B / ii

Solution

Let

s = ∣ r - r^{'} ∣

A (r) = μ_{0} (4 π)^{- 1} \int J (r^{'}) / s d V^{'}

r

, so

J (r^{'})

is constant under it. Hence

\nabla \times \frac{J ( r ^{'} )}{s} = \nabla (s^{- 1}) \times J (r^{'}) = - \frac{r - r ^{'}}{s ^{3}} \times J (r^{'}) = \frac{J ( r ^{'} ) \times ( r - r ^{'} )}{s ^{3}} .

Taking the curl through the integral reproduces the Biot-Savart law, proving

B = \nabla \times A

and the requested representation.

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