= Solution
Use SI units, $x^a=(ct,\mathbf x)$ and metric signature $(+,-,-,-)$. The <four-current> is $j^a=(c\rho,\mathbf j)$. Charge density and current transform together under Lorentz transformations, and the <continuity equation> becomes
$$
\boxed{\partial_aj^a=\partial_t\rho+\nabla\cdot\mathbf j=0.}
$$
With the <four-potential> $A^a=(\varphi/c,\mathbf A)$, the <Lorenz gauge> $\partial_aA^a=0$ converts <Maxwell equations> to $\Box A^a=\mu_0j^a$, where $\Box=c^{-2}\partial_t^2-\nabla^2$. Choosing no incoming homogeneous radiation, the <retarded electromagnetic potential> is
$$
\boxed{A^a(\mathbf x,t)=\frac{\mu_0}{4\pi}\int\frac{j^a(\mathbf x',t-|\mathbf x-\mathbf x'|/c)}{|\mathbf x-\mathbf x'|}\,d^3x'.}
$$
Current conservation ensures this solution obeys the Lorenz gauge; a homogeneous solution may be added when additional field boundary data are specified.
For a thin loop, let $R=|\mathbf x|$, $\widehat{\mathbf R}=\mathbf x/R$ and $t_R=t-R/c$. Then
$$
\mathbf A=\frac{\mu_0}{4\pi}\oint\frac{I(t-|\mathbf x-\mathbf x'|/c)}{|\mathbf x-\mathbf x'|}\,d\boldsymbol\ell'.
$$
In the small-loop magnetic-dipole approximation, expand $|\mathbf x-\mathbf x'|=R-\widehat{\mathbf R}\cdot\mathbf x'+\cdots$. The constant term integrates to zero because $\oint d\boldsymbol\ell'=0$. The first nonzero term is
$$
\frac{\mu_0}{4\pi}\left[\frac{I(t_R)}{R^2}+\frac{\dot I(t_R)}{cR}\right]\oint(\widehat{\mathbf R}\cdot\mathbf x')\,d\boldsymbol\ell'.
$$
For the oriented circular loop the last integral is $\pi r^2\mathbf n\times\widehat{\mathbf R}$. Thus with <magnetic dipole moment> $\mathbf m(t)=\pi r^2I(t)\mathbf n$,
$$
\boxed{\mathbf A(\mathbf x,t)=\frac{\mu_0I_0r^2}{4}\left[\frac{\sin(\omega t_R)}{R^2}+\frac{\omega\cos(\omega t_R)}{cR}\right]\mathbf n\times\widehat{\mathbf R}.}
$$
The $R^{-2}$ term is the near dipole potential and the $R^{-1}$ term its radiation potential. This is the leading small-$r$ result at fixed $R,\omega$; besides $r/R\ll1$, the expansion of the retarded current requires $\omega r/c\ll1$.
If only $r/R\ll1$ is imposed and the loop need not be small compared with a wavelength, keep the retarded phase around the loop. For fixed $r,\omega$ and $R\to\infty$, the leading far-field expression is instead
$$
\mathbf A=\frac{\mu_0I_0r}{2R}J_1\left(\frac{\omega r}{c}\sin\theta\right)\cos(\omega t_R)\frac{\mathbf n\times\widehat{\mathbf R}}{\sin\theta}+O(R^{-2}),
$$
where $\cos\theta=\mathbf n\cdot\widehat{\mathbf R}$ and $J_1$ is the <Bessel function of the first kind>. Integrating the sinusoidal phase on a circle gives this Bessel factor; $J_1(z)\sim z/2$ recovers the radiation part of the boxed dipole result.
Back to article page
