= Solution
Orient the ring's area normal with positive external <magnetic flux>. <Faraday's law> gives the applied <electromotive force>
$$
\mathcal E_{\rm ext}=-\frac{d}{dt}(AB_0\cos\omega t)
=AB_0\omega\sin\omega t.
$$
The ring's <self-inductance> opposes changes in its <electric current>. With positive <electrical resistance> $R$, <Kirchhoff's voltage law> therefore gives $L\dot I+RI=\mathcal E_{\rm ext}$. Define $\phi=\tan^{-1}(\omega L/R)$. The periodic steady state is
$$
\boxed{I(t)=\frac{AB_0\omega}{\sqrt{R^2+\omega^2L^2}}\sin(\omega t-\phi),\qquad
\langle P\rangle=\frac{R A^2B_0^2\omega^2}{2(R^2+\omega^2L^2)}.}
$$
The second result follows from <Joule heating>, $P=RI^2$, and $\langle\sin^2\rangle=1/2$. For unspecified initial conditions, add a transient $I_h=C\exp(-Rt/L)$; the displayed time average is over the periodic steady state after that transient has decayed. The <inductor> temporarily stores and returns <energy>, with no net storage change over one complete cycle.
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