1. Magnetic flux through the rotating semicircle

The wire forms a semicircular loop of radius $r$, so its area is $A=\frac{1}{2}\pi r^2.$

The loop rotates with angular frequency $\omega$ about its diameter. The axis of rotation is perpendicular to the magnetic field, so the angle between the loop's area vector and the magnetic field changes as $\theta=\omega t.$

Hence the magnetic flux through the loop is $\Phi = BA\cos\theta = B\left(\frac{1}{2}\pi r^2\right)\cos(\omega t).$

2. Induced emf

By Faraday's law, $e = -\frac{d\Phi}{dt}.$ So, $e = -\frac{d}{dt}\left[B\left(\frac{1}{2}\pi r^2\right)\cos(\omega t)\right]$ $e = B\left(\frac{1}{2}\pi r^2\right)\omega \sin(\omega t).$

Thus the peak emf is $e_0 = \frac{1}{2}B\pi r^2\omega.$

3. Instantaneous power

The circuit has total resistance $R$, so instantaneous power generated is $P = \frac{e^2}{R} = \frac{e_0^2\sin^2(\omega t)}{R}.$

4. Mean power over one full period

Over one complete cycle, $\langle \sin^2(\omega t) \rangle = \frac{1}{2}.$

Therefore, $P_{\text{mean}} = \frac{e_0^2}{2R}.$

Substitute $e_0$: $P_{\text{mean}} = \frac{1}{2R}\left(\frac{1}{2}B\pi r^2\omega\right)^2$ $P_{\text{mean}} = \frac{(B\pi r^2\omega)^2}{8R}.$

5. Compare with options

This matches: $\boxed{\frac{(B\pi r^2\omega)^2}{8R}}$ which is Option B.