• # question_answer 25) In a uniform magnetic field of induction B, a wire in the form of a semi-circle of radius r rotates about the diameter of the circle with 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: A) $\frac{{{(B\pi r\omega )}^{2}}}{2R}$                  B) $\frac{B\pi {{r}^{2}}\omega }{2R}$C) $\frac{B\pi {{r}^{2}}\omega }{8R}$                    D) $\frac{{{(B\pi r{{\omega }^{2}})}^{2}}}{8R}$

$\Phi =B\frac{\pi {{r}^{2}}}{2}\cos \omega t$ $\therefore$${{e}_{ind}}=-\frac{d\Phi }{dt}=\frac{1}{2}B\pi {{r}^{2}}\omega \sin \omega t$ $\therefore$$P=\frac{e_{ind}^{2}}{2}=\frac{{{B}^{2}}{{\pi }^{2}}{{\omega }^{2}}{{\sin }^{2}}\omega t}{4R}$ Now $<{{\sin }^{2}}\omega t>=\frac{1}{2}$$\therefore$$<P>=\frac{{{(B\pi {{r}^{2}}\omega )}^{2}}}{8R}$ Hence, the correction option is [b].