 Solution

A steady localized electric current density satisfies $\nabla \cdot \mathbf{j}=0$, from charge conservation or the divergence of Ampère's equation. Choose an enclosing region whose boundary lies outside its support; extending the integration region does not change the source integrals. The divergence theorem applied componentwise givesTheir boundary fluxes vanish, proving the localized steady-current moment identitiesFor discontinuous but localized currents the same argument uses the distributional steady conservation law and an outer enclosing surface.

The magnetic vector potential is a vector field with $\mathbf{B}=\nabla \times \mathbf{A}$, consistent with $\nabla \cdot \mathbf{B}=0$. It is defined up to $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$. The steady Maxwell equation $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ givesIn the Coulomb gauge $\nabla \cdot \mathbf{A}=0$, the equation is ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{j}$, whose decaying Green-function solution is the one supplied in the source.

Let $r=|\mathbf{x}|$ and expand its kernel uniformly over the bounded source region:The first term integrates to zero. Put $T_{ki}=\int x_k^{\prime }{j}_i({\mathbf{x}}^{\prime })d{V}^{\prime }$, an antisymmetric tensor by the earlier identity. Define the magnetic dipole moment $m_{\ell }=\frac{1}{2}{ϵ}_{\ell ki}{T}_{ki}$, so $T_{ki}={ϵ}_{ki\ell }{m}_{\ell }$. ConsequentlyorThe displayed dipole term has size $r^{-2}$. If $\mathbf{m}=0$, it vanishes and a higher multipole determines the leading magnetic field instead.

For $r>0$, $\nabla \cdot (\mathbf{x}/r^3)=0$ andTherefore the far magnetic field is