Frequency doubling of an induction-driven rotating loop (${\omega }_{\mathrm{rad}}=2{\omega }_{\mathrm{spin}}$)

Let a conducting loop of area A rotate about z with normal $(\cos \omega t,\sin \omega t,0)$ in a fixed field $B\hat{\mathbf{x}}$. Neglect inductance so $I=(BA\omega /R)\sin \omega t$. Put $m_0=B{A}^2\omega /R$. Its vector magnetic moment isThe 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 $\overline{P}=2{\mu }_0{B}^2{A}^4{\omega }^6/(3\pi c^3{R}^2)$. 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.

For constant self-inductance L, the steady induced current satisfies $L\dot{I}+RI=BA\omega \sin \omega t$, giving amplitude $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$ and phase lag. Multiplication by the rotating loop normal again gives a static moment plus a twice-frequency rotating moment of magnitude $A{I}_0/2$, independently of lag. Therefore replace the resistance-only denominator $R^2$ in the radiation power by $R^2+{\omega }^2{L}^2$. A wire resistance alone is not sufficient for an exact prediction if inductance is appreciable.