Physics - EMI/AC Question with Solution | TestHub

The flux associated with coil of area$A$and magnetic induction$B$is

$\mathrm{\varphi }=BA\cos \mathrm{\theta }$

$=\frac{1}{2}B\pi r^2\cos \omega t \left[\because A=\frac{1}{2}\pi r^2\right]$

$\therefore e_{\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}=-\frac{d\mathrm{\varphi }}{dt}$

$= -\frac{d}{dt}\left(\frac{1}{2}B\pi r^2\cos \omega t\right)$

$=\frac{1}{2}B\pi r^2\omega \sin \omega t$

$\therefore \mathrm{p}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r} p=\frac{e_{\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}}^2}{R}$

$= \frac{B^2{\pi }^2{r}^4{\omega }^2{\sin }^2\omega t}{4R}$

Hence,$P_{\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{n}}=<p>$

$=\frac{B^2{\pi }^2{r}^4{\omega }^2}{4R}.\frac{1}{2} \left(\because <\sin \omega t>=\frac{1}{2}\right)$

$=\frac{{\left(B\pi r^2\omega \right)}^2}{8R}$