= Solution
For the imposed <electric current> $I=I_0\cos\omega t$, <Faraday's law> gives the self-induced <electromotive force>
$$
\boxed{\mathcal E_{\rm self}=-L\frac{dI}{dt}=\omega LI_0\sin\omega t.}
$$
The externally applied terminal <voltage>, with the <passive sign convention>, has the opposite sign:
$$
v_L=L\frac{dI}{dt}=-\omega LI_0\sin\omega t.
$$
Thus the terminal <voltage> leads the <electric current> by $\pi/2$. The instantaneous input <power> is
$$
p=v_LI=-\frac12\omega LI_0^2\sin2\omega t=\frac{d}{dt}\left(\frac12LI^2\right),\qquad \boxed{\langle p\rangle=0.}
$$
An ideal <inductor> alternately stores <magnetic energy> and returns it to the source. Real winding <electrical resistance> would add mean <Joule heating> $I_0^2R_w/2$; the ideal zero result assumes that loss is neglected.
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