Explanation

(b) The flux associated with coil of area $A$ and magnetic induction$\ B$ is
$\phi = BA\cos\theta$
$\ \ \ \ = \frac{1}{2}B\pi r^{2}\cos{\omega t\ }\ \ \ \ \ \ \ \ \ \left\lbrack \because A = \frac{1}{2}\pi r^{2} \right\rbrack$
$\therefore e_{induced} = - \frac{d\phi}{dt}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ - \frac{d}{dt}\left( \frac{1}{2}B\pi r^{2}\cos{\omega t} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}B\pi r^{2}\omega\sin{\omega t}$
$\therefore power\ p = \frac{e_{induced}^{2}}{R}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{B^{2}\pi^{2}r^{4}\omega^{2}\sin^{2}{\omega t}}{4R}$
Hence,$P_{mean} = < p >$
$\ \ \ \ \ \ \ \ \ \ \ = \frac{B^{2}\pi^{2}r^{4}\omega^{2}}{4R}.\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \left( \because < \sin{\omega t} > = \frac{1}{2} \right)$
$\ \ \ \ \ \ \ \ \ \ = \frac{\left( B\pi r^{2}\omega \right)^{2}}{8R}$