Solution

Differentiate under the integral and use

\nabla_{x} ∣ x - x^{'} ∣^{- 1} = - \nabla_{x^{'}} ∣ x - x^{'} ∣^{- 1}

\nabla \cdot A = - \frac{μ _{0}}{4 π} \int J_{i} (x^{'}) \partial_{i}^{'} \frac{1}{∣ x - x ^{'} ∣} d^{3} x^{'} = \frac{μ _{0}}{4 π} \int \frac{\nabla ^{'} \cdot J ( x ^{'} )}{∣ x - x ^{'} ∣} d^{3} x^{'} .

The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation

\nabla^{'} \cdot J = 0

, hence

\nabla \cdot A = 0 .

For

r = ∣ x ∣

much larger than the source size,

\frac{1}{∣ x - x ^{'} ∣} = \frac{1}{r} + \frac{x \cdot x ^{'}}{r ^{3}} + O (r^{- 3} ∣ x^{'} ∣^{2}) .

Localization and

\nabla^{'} \cdot J = 0

imply

\int J d^{3} x^{'} = 0

. They also imply

\int (x_{i}^{'} J_{j} + x_{j}^{'} J_{i}) d^{3} x^{'} = 0,

by integrating

\partial_{k}^{'} (x_{i}^{'} x_{j}^{'} J_{k})

. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole moment

m = \frac{1}{2} \int x^{'} \times J (x^{'}) d^{3} x^{'} .

A (x) = \frac{μ _{0}}{4 π} \frac{m \times x}{r ^{3}} + \dots .

The dimensions are

[J] = A m^{- 2}, [m] = A m^{2} .

Solved by gpt-5.6-sol high.

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