Question

In a uniform magnetic field B a wire in the form of a semicircle 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. $\frac{B\pi r^{2}\omega}{2R}$
• B. $\frac{\left( B\pi r^{2}\omega \right)^{2}}{8R}$
• C. $\frac{(B\pi r\omega)^{2}}{2R}$
• D. $\frac{\left( B\pi r\omega^{2} \right)^{2}}{8R}$

Answer 1

Answer: (B)

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

Answer 2

$\varphi = B\frac{\pi r^{2}}{2}\cos\omega t$

$\therefore\varepsilon = - \frac{d\varphi}{dt} = \frac{1}{2}B\pi r^{2}\omega\sin\omega t$

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

$\therefore < P > = \frac{(B\pi r^{2}\omega^{2})}{8R}$ $\left( \because < \sin^{2}\omega t > = \frac{1}{2} \right)$