Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 36C / 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 becomesWith the four-potential $A^a=(\phi /c,\mathbf{A})$, the Lorenz gauge ${\partial }_a{A}^a=0$ converts Maxwell equations to $\square A^a={\mu }_0{j}^a$, where $\square =c^{-2}{\partial }_t^2-{\nabla }^2$. Choosing no incoming homogeneous radiation, the retarded electromagnetic potential isCurrent 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 $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$ and $t_R=t-R/c$. ThenIn the small-loop magnetic-dipole approximation, expand $|\mathbf{x}-{\mathbf{x}}^{\prime }|=R-\hat{\mathbf{R}}\cdot {\mathbf{x}}^{\prime }+\cdots$. The constant term integrates to zero because $\oint d{\mathbf{\ell }}^{\prime }=0$. The first nonzero term isFor the oriented circular loop the last integral is $\pi r^2\mathbf{n}\times \hat{\mathbf{R}}$. Thus with magnetic dipole moment $\mathbf{m}(t)=\pi r^{2}I(t)\mathbf{n}$,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 $r/R\ll 1$, the expansion of the retarded current requires $\omega r/c\ll 1$.

If only $r/R\ll 1$ 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 insteadwhere $\cos \theta =\mathbf{n}\cdot \hat{\mathbf{R}}$ 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.