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 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

• A. $\frac{\pi r^2\omega B}{2R}$
• B. $\frac{(\pi r^2\omega B{)}^2}{8R}$
• C. $\frac{(\pi r\omega B{)}^2}{2R}$
• D. $\frac{(\pi r{\omega }^{2}B{)}^2}{8R}$

Answer 1

The correct option is B

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

The induced voltage across the ends of the wire at an instant of time t is given by
Flux through the coil at an instant is given by $\frac{\pi r^{2}Bcos\left(\omega t\right)}{2}$
Voltage developed is given by $V=\frac{d\varphi }{dt}=\frac{\pi r^{2}Bsin\left(\omega t\right)\omega }{2}$
Current through the coil $i=\frac{V}{R}$
Power delivered in the coil $=i^{2}R=\frac{\left(\pi r^{2}B\omega {)}^2\right(si{n}^2\left(\omega t\right))}{4R}$
Average of $\left(si{n}^2\right(\omega t\left)\right)=\frac{1}{2}$. Hence,
average power $=\frac{(\pi r^{2}B\omega {)}^2}{8R}$