The purpose of this post is to review how the fields of an electrostatic dipole \(\boldsymbol{\pi}\) and magnetostatic dipole \(\boldsymbol{\mu}\) arise. For the electrostatic dipole, “fields” means the electrostatic potential \(\phi\) and by extension the electrostatic field \(\textbf E=-\partial\phi/\partial\textbf x\) while for the magnetostatic dipole, “fields” means the magnetostatic potential \(\textbf A\) and by extension the magnetostatic field \(\textbf B=\partial/\partial\textbf x\times\textbf A\).
Because one is working in the regime of electrostatics, Coulomb’s law is valid:
\[\phi(\textbf x)=\frac{1}{4\pi\varepsilon_0}\iiint_{\textbf x’\in\textbf R^3}\frac{\rho(\textbf x’)}{|\textbf x-\textbf x’|}d^3\textbf x’\]
An electrostatic dipole consists of two stationary charges \(\pm q\) separated by \(\Delta\textbf x\); arbitrarily placing the charge \(-q\) at the origin \(\textbf x’=\textbf 0\) implies that \(\rho(\textbf x’)=q(\delta^3(\textbf x’+\Delta\textbf x)-\delta^3(\textbf x’))\). Substituting this into Coulomb’s law picks out:
\[\phi(\textbf x)=\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{|\textbf x+\Delta\textbf x|}-\frac{1}{|\textbf x|}\right)\]
As \(\Delta\textbf x\to\textbf 0\), one has by definition the directional derivative of \(1/|\textbf x|\) along \(\Delta\textbf x\):
\[\frac{1}{|\textbf x+\Delta\textbf x|}-\frac{1}{|\textbf x|}\to\Delta\textbf x\cdot\frac{\partial}{\partial\textbf x}\frac{1}{|\textbf x|}=-\Delta\textbf x\cdot\hat{\textbf x}/|\textbf x|^2\]
Using the definition \(\boldsymbol{\pi}:=-q\Delta\textbf x\) of the electrostatic dipole in this case, the electrostatic potential reduces to:
\[\phi(\textbf x)=\frac{\boldsymbol{\pi}\cdot\hat{\textbf x}}{4\pi\varepsilon_0|\textbf x|^2}\]
The electrostatic field then follows:
\[\textbf E(\textbf x)=-\frac{1}{4\pi\varepsilon_0}\frac{\partial}{\partial\textbf x}\frac{\boldsymbol{\pi}\cdot\textbf x}{|\textbf x|^3}=\frac{3(\boldsymbol{\pi}\cdot\hat{\textbf x})\hat{\textbf x}-\boldsymbol{\pi}}{4\pi\varepsilon_0|\textbf x|^3}\]
Meanwhile, because one is working in the regime of magnetostatics, the Biot-Savart law (in the Coulomb gauge) is valid:
\[\textbf A(\textbf x)=\frac{\mu_0}{4\pi}\iiint_{\textbf x’\in\textbf R^3}\frac{\textbf J(\textbf x’)}{|\textbf x-\textbf x’|}d^3\textbf x’\]
A magnetostatic dipole consists of a steady current loop \(I\) enclosing an area \(\textbf S\), with \(\textbf J(\textbf x’)d^3\textbf x’\mapsto Id\textbf x’\):
\[\textbf A(\textbf x)=\frac{\mu_0 I}{4\pi}\oint_{\textbf x’\in S^1}\frac{d\textbf x’}{|\textbf x-\textbf x’|}\]
As \(\textbf S\to\textbf 0\), one has (still looking for a simple way to see this):
\[\oint_{\textbf x’\in S^1}\frac{d\textbf x’}{|\textbf x-\textbf x’|}\to\frac{\partial}{\partial\textbf x}\frac{1}{|\textbf x|}\times\textbf S\]
Using the definition \(\boldsymbol{\mu}:=I\textbf S\) of the magnetostatic dipole, the magnetostatic vector potential reduces to:
\[\textbf A(\textbf x)=\frac{\mu_0\boldsymbol{\mu}\times\hat{\textbf x}}{4\pi|\textbf x|^2}\]
From which the magnetostatic field is:
\[\textbf B(\textbf x)=\frac{\mu_0(3(\boldsymbol{\mu}\cdot\hat{\textbf x})\hat{\textbf x}-\boldsymbol{\mu})}{4\pi|\textbf x|^3}\]