A wire shaped as a semicircle of radius o rotates about an axis OO' with an angular velocity $ω$ in a uniform magnetic field of induction B [Fig. (5)]. The rotation
axis is perpendicular to the field direction. The total resistance of the circuit is equal to fl. Neglect the magnetic field of induced current. The mean amount of thermal power being generated into the loop during a rotation period will be

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$\frac{1}{2 R} {(\frac{{πωa}^{2} B}{2})}^{2}$

$\frac{1}{2 R} (\frac{{πωa}^{2} B}{2})$

$\frac{1}{2 R} (\frac{{πωaB}^{2}}{2})$

$\frac{1}{2 R} (\frac{{πω}^{2} aB}{2})$

answer is A.

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Detailed Solution

$\begin{array}{l} Φ = B \cdot A = B (\frac{{πα}^{2}}{2}) \cos ωt \\ e = - \frac{dΦ}{dt} = B (\frac{{πa}^{2}}{2}) ωsin ωt \end{array}$
Induced current $i = \frac{e}{R} = \frac{{Bπa}^{2}}{2 R} ωsin ωt$
At any moment t, the thermal power generated in the circuit
$P_{t} = e \times i = {(\frac{{Bπa}^{2} ω}{2})}^{2} \frac{1}{R} \sin^{2} ωt$
Mean power (P) = $\frac{{(\frac{{Bπa}^{2} ω}{2})}^{2} \frac{1}{R} \int_{0}^{T} \sin^{2} ωt}{\int_{0}^{T} dt}$
$= \frac{1}{2 R} {(\frac{{πωa}^{2} B}{2})}^{2}$