In a uniform magnetic field of induction $B$ a wire in the form of semicircle of radius $r$ rotates about…

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 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
  1. $\frac{B \pi r^2 \omega}{2 R}$
  2. $\frac{\left(B \pi r^2 \omega\right)^2}{2 R}$
  3. $\frac{(\mathrm{B} \pi \mathrm{r} \omega)^2}{2 \mathrm{R}}$
  4. $\frac{\left(B \pi r \omega^2\right)^2}{8 R}$

Solution

Magnetic flux $=B A \cos \theta=B \cdot \frac{\pi r^2}{2} \cos \omega t$ $ \begin{aligned} & \therefore \varepsilon_{\text {ind }}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{1}{2} \mathrm{~B} \pi \mathrm{r}^2 \omega \sin \omega \mathrm{t} \\ & \therefore \mathrm{P}=\frac{\varepsilon_{\text {ind }}^2}{\mathrm{R}}=\frac{\mathrm{B}^2 \pi^2 \mathrm{r}^4 \omega^2 \sin ^2 \omega \mathrm{t}}{4 \mathrm{R}} \end{aligned} $ Now, $ < \sin ^2 \omega t>=1 / 2$ (mean value) $ \therefore\langle\mathrm{P}\rangle=\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}} \text {. } $

Asked in: JEE Main 2004

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