Question

In a uniform magnetic field of induction $B$, a wire in the form of semicircle of radius $r$ rotates about the diameter of the circle with angular velocity $\omega$. 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. $\frac{(B\pi r^2\omega {)}^2}{8R}$
• C. $\frac{(B\pi r\omega {)}^2}{2R}$
• D. $\frac{(B\pi r{\omega }^2{)}^2}{8R}$

Answer 1

The correct option is B $\frac{(B\pi r^2\omega {)}^2}{8R}$

Flux $\varphi =BA\cos \theta =B\times \left(\frac{1}{2}\pi r^2\right)\cos \theta$
$|\epsilon |=|\frac{d\varphi }{dt}|=\frac{1}{2}\pi B{r}^2\frac{d\left(\cos \theta \right)}{dt}$
$|\epsilon |=\frac{1}{2}\pi B{r}^2\left(\sin \theta \right)\frac{d\theta }{dt}$
$|\epsilon |=\frac{1}{2}\pi B{r}^2\left(\sin \omega t\right)\omega$
Power: $P=\frac{|\epsilon {|}^2}{R}=\frac{(\frac{1}{2}\pi B{r}^2\omega \sin \omega t{)}^2}{R}$
Mean Power $P_{mean}=\frac{{\int }_0^T\frac{|\epsilon {|}^2}{R}dt}{{\int }_0^{T}dt}=\frac{{\pi }^2{B}^2{r}^4{\omega }^2{\int }_0^T{\sin }^2\omega tdt}{4TR}$
$=\frac{{\pi }^2{B}^2{r}^4{\omega }^2}{4TR}\frac{T}{2}=\frac{{\pi }^2{B}^2{r}^4{\omega }^2}{8R}$