Coulomb-gauge vector potential of a localized steady current
= Coulomb-gauge vector potential of a localized steady current
{c}
If $\nabla\cdot J=0$ and the current decays enough for boundary terms to vanish, then
$$
A(x)=\frac{\mu_0}{4\pi}\int\frac{J(x')}{|x-x'|}\,d^3x'
$$
satisfies $\nabla\cdot A=0$. Its far field is
$$
A(x)=\frac{\mu_0}{4\pi}\frac{m\times x}{|x|^3}+\cdots.
$$