= Frequency doubling of an induction-driven rotating loop
{title2=$\omega_{\rm rad}=2\omega_{\rm spin}$}
Let a conducting loop of area A rotate about z with normal $(\cos\omega t,\sin\omega t,0)$ in a fixed field $B\widehat{\mathbf x}$. Neglect <inductance> so $I=(BA\omega/R)\sin\omega t$. Put $m_0=BA^2\omega/R$. Its vector magnetic moment is
$$
\mathbf m=\frac{m_0}{2}[\sin(2\omega t)\widehat{\mathbf x}+(1-\cos(2\omega t))\widehat{\mathbf y}].
$$
The static component does not radiate. The remaining component rotates at twice the mechanical <frequency> and has $|\ddot{\mathbf m}|^2=4m_0^2\omega^4$. Thus $\overline P=2\mu_0 B^2A^4\omega^6/(3\pi c^3R^2)$. Treating the same sinusoidal current as a fixed-axis moment underestimates this rotating geometry by a factor of eight. The <energy> source is mechanical rotation, with additional ohmic <dissipation> and <magnetic braking>.
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