In a uniform magnetic field of induction $$B$$ a wire in the form of a semicircle of radius $$r$$ rotates about the diameter of the circle with an angular frequency $$\omega .$$ The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $$R,$$ the mean power generated per period of rotation is

Solution :
$$\eqalign{ & \phi = \vec B \cdot \vec A;\phi = BA\cos \omega t \cr & \varepsilon = - \frac{{d\phi }}{{dt}} = \omega BA\sin \omega t;\,i = \frac{{\omega BA}}{R}\sin \omega t \cr & {P_{{\text{inst}}}} = {i^2}R = {\left( {\frac{{\omega BA}}{R}} \right)^2} \times R{\sin ^2}\omega t \cr & {P_{{\text{avg}}}} = \frac{{\int\limits_0^T {{P_{{\text{inst}}}} \times dt} }}{{\int\limits_0^T {dt} }} = \frac{{{{\left( {\omega BA} \right)}^2}}}{R}\frac{{\int\limits_0^T {{{\sin }^2}\omega tdt} }}{{\int\limits_0^T {dt} }} = \frac{1}{2}\frac{{{{\left( {\omega BA} \right)}^2}}}{R} \cr & \therefore {P_{{\text{avg}}}} = \frac{{{{\left( {\omega B\pi {r^2}} \right)}^2}}}{{8R}}\,\,\left[ {A = \frac{{\pi {r^2}}}{2}} \right] \cr} $$