 Solution

For static fields, the Maxwell equations give $\nabla \times B={\mu }_{0}J$ and $\nabla \cdot B=0$. Write $B=\nabla \times A$. A gauge transformation $A\mapsto A+\nabla \chi$ leaves $B$ unchanged; choose $\Delta \chi =-\nabla \cdot A$ to impose the Coulomb gauge. ThenFor a localized conserved current with the potential chosen to vanish at infinity, the supplied fundamental solution gives the magnetic vector potentialIts divergence vanishes by integration by parts and $\nabla \cdot J=0$. For a filamentary loop the current integral becomes $I{\oint }_{L}dr$, soFor $R=|x|$ much larger than the fixed loop size, expand $|x-r{|}^{-1}=R^{-1}+(x\cdot r)R^{-3}+O(R^{-3})$. The leading constant term integrates to zero because $\oint dr=0$. ThereforeThe vector triple-product identity gives $x\times (r\times dr)=r(x\cdot dr)-(x\cdot r)dr$. Integrating the derivative of $(x\cdot r)r$ around the closed loop shows $\oint r(x\cdot dr)=-\oint (x\cdot r)dr$. ConsequentlyEquivalently, $A={\mu }_{0}m\times x/(4\pi R^3)+O(R^{-3})$, where $m=(I/2)\oint r\times dr$ is the magnetic dipole moment. The leading displayed potential is of order $R^{-2}$; the remainder is smaller by a factor $R^{-1}$.