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Multipole radiation expands radiation from localized time-dependent sources into electric and magnetic multipoles. Electric dipole radiation is the leading electric term, while magnetic dipole radiation and higher electric multipoles contribute when the leading term vanishes or higher precision is needed.

Let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$, and $t_R=t-R/c$. To leading nonzero multipole expansion order, the loop is a time-dependent magnetic dipole withExpanding the retarded electromagnetic potential across the small loop givesThe $R^{-2}$ term is the near magnetic-dipole field and the $R^{-1}$ term is the magnetic dipole radiation field. Substituting the Fourier series for the current givesThe $n=0$ summand vanishes for the stated sine series.

For a localized charge and current distribution, take the retarded vector potentialLet $R=|x|$, $n=x/R$, and $t_R=t-R/c$. If the source size is much smaller than both $R$ and the characteristic radiation wavelength, the leading dipole term iswhere the last identity follows from charge conservation.

Keeping only the $1/R$ terms in the radiation zone gives the transverse fieldsThe radial Poynting vector therefore yieldsUsingor equivalently $\int {\sin }^2\theta d\Omega =8\pi /3$, we recover the electric dipole radiation formulaThis requires a localized, nonrelativistic source of size $d\ll \lambda$, observation distance $R\gg d$ and $R\gg \lambda$, and neglect of higher multipoles and faster-decaying near fields.

For the pulsar, write its magnetic dipole moment as $m(t)$ to distinguish it from the electric dipole above. The magnetic dipole radiation formula differs by a factor $c^{-2}$:Hereand hence

On the slow spin-down timescale, energy conservation gives the magnetic-dipole spin-down equationsoHalf of the initial rotational energy remains when $\Omega ={\Omega }_0/\sqrt{2}$. Thereforewhere $I=2M{\mathcal{R}}^2/5$. Since $E_0=I{\Omega }_0^2/2$, the equivalent expression is

A fixed electrically small loop of area A with peak sinusoidal current $I_0\cos \omega t$ has magnetic moment $m=AI$. The magnetic dipole radiation law gives mean power ${\mu }_0{A}^2{I}_0^2{\omega }^4/(12\pi c^3)$. Defining it as $I_0^2{R}_{\mathrm{rad}}/2$ yields $R_{\mathrm{rad}}={\mu }_0{A}^2{\omega }^4/(6\pi c^3)$. This equivalent resistance accounts for radiation, not heating in the conductor. Rotating the loop changes its vector moment and cannot in general be modelled by the same scalar current alone.