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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
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Verified Answer
The correct answer is:
$\frac{\left(B \pi r^2 \omega\right)^2}{2 R}$
$\frac{\left(B \pi r^2 \omega\right)^2}{2 R}$
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)
$$
\therefore\langle\mathrm{P}\rangle=\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}} \text {. }
$$
$$
\begin{aligned}
& \therefore \varepsilon_{\text {ind }}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{1}{2} \mathrm{~B} \pi \mathrm{r}^2 \omega \sin \omega \mathrm{t} \\
& \therefore \mathrm{P}=\frac{\varepsilon_{\text {ind }}^2}{\mathrm{R}}=\frac{\mathrm{B}^2 \pi^2 \mathrm{r}^4 \omega^2 \sin ^2 \omega \mathrm{t}}{4 \mathrm{R}}
\end{aligned}
$$
Now, $ < \sin ^2 \omega t>=1 / 2$ (mean value)
$$
\therefore\langle\mathrm{P}\rangle=\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}} \text {. }
$$
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