# In a uniform magnetic field of induction 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 Option 1) $\frac{B\pi r^{2}\omega }{2R}\;$ Option 2) $\; \; \frac{(B\pi r^{2}\omega)}{8R}^{2}\; \;$ Option 3) $\; \frac{(B\pi r\omega)}{2R}^{2}\; \;$ Option 4) $\; \frac{(B\pi r\omega^{2})}{8R}^{2}$

As we learnt in

Motional E.m.f due to rotational motion -

Semicircle conducting loop

$\varepsilon = \frac{1}{2}Bwr^{2}$

$i = \frac{Bwr^{2}}{2R}$

$=\frac{(B\pi r^{2}\omega )^{2}}{4R}\times \frac{1}{2}=\frac{(B\pi r^{2}\omega )^{2}}{8R}$

At any instant magnetic flux associated with the coil is

$\phi =B.\frac{\pi r^{2}}{2}coswt$

induced current =$\frac{\varepsilon }{R}$

$= - \frac{d\phi }{dt}.\frac{1}{R}=\frac{B\pi\:r^{2}\omega sin\omega t}{2R}$

Instantaneous power = I2R

$=\left ( \frac{B^{2}\pi^2\:r^{4}\omega^2 sin^2\omega t}{4R^2}\right )R$

Average power

$= \frac{B^{2}\pi^2\:r^{4}\omega^2} {4R}$

$= \frac{B^{2}\pi^2\:r^{4}\omega^2} {8R}$

$\because =\frac{1}{2}$

Option 1)

$\frac{B\pi r^{2}\omega }{2R}\;$

This option is incorrect

Option 2)

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

This option is correct

Option 3)

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

This option is incorrect

Option 4)

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

This option is incorrect

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