A semicircular wire of radius r rotates in uniform magnetic field B...

The flux linked with coil (made of wire of radius $r$) of area $A and magnetic field$B$, is given by</p><p>$ϕ = BAcos\theta = B\left(\dfrac{\pir^{2}}{2}\right)cos\omegat$</p><p>$ϕ = \dfrac{\piBr^{2}}{2}cos\omegat~ & [∵ A = \dfrac{\pir^{2}}{2}~and&\theta = \omegat$]</p><p>By Faraday's law of electromagnetic induction, induced emf,~$e = - \dfrac{dϕ}{dt}$</p><p>Hence, after differentiating,</p><p>$e = - \dfrac{d}{dt}\left(\dfrac{\piBr^{2}}{2}cos\omegat\right) = \dfrac{\piBr^{2}\omega}{2}sin\omegat$~ (where,$R =$resistance in the circuit)</p><p>$∴~Power&P = \dfrac{e2}{R} = \dfrac{{π}^{2}B^{2}r^{4}{ω}^{2}\sin^{2}\omegat}{4R}$</p><p>$∴~Mean power&\overset{¯}{P} = \dfrac{{π}^{2}B^{2}r^{4}{ω}^{2}}{4R} \cdot \overset{¯}{\sin^{2}\omegat}$</p><p>$= \dfrac{{π}^{2}B^{2}r^{4}{ω}^{2}}{4R} \cdot \dfrac{1}{2}~ & \left[∵ \overset{¯}{\sin^{2}\omegat} = \dfrac{1}{2}\right]

$= \dfrac{\left(\piBr^{2}\omega\right)2}{8R}$