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The local form of conservation of electric charge relates charge density to current density. Taking the divergence of the Ampère-Maxwell equation and substituting Gauss's law proves the equation, using that the divergence of a curl is zero.

In a homogeneous Ohmic conductor, the charge continuity equation and Gauss's law imply exponential decay of the bulk charge density, $\rho (t)=\rho (0)e^{-\sigma t/ϵ}$. For an isolated finite body, bulk charge moves to its boundary; the accumulating surface charge density preserves total electric charge. The formula uses a constant permittivity $ϵ$ and the corresponding free-charge convention, or ${ϵ}_0$ for total charge in the vacuum form of Gauss's law.

For the aligned magnetic dipole field,Since ${\mathbf{e}}_z=\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta }$,The corotation velocity is $\mathbf{u}=\Omega r\sin \theta {\mathbf{e}}_{\varphi }$. The current density is therefore the advected charge density,

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is