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In a uniform magnetic field of induction B a wire in the form of a semi-circle of radius r rotates about the diameter of the circle with angular frequency $ω$ . 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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$\frac{{(B π r ω)}^{2}}{8 R}$

$\frac{{(B π r^{2} ω)}^{2}}{8 R}$

$\frac{B π r^{2} ω}{2 R}$

$\frac{{(B π r ω)}^{2}}{2 R}$

answer is B.

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Detailed Solution

$\in = \frac{1}{2} B π r^{2} ω \sin ω t ρ = \frac{\in^{2}}{R} = \frac{B^{2} π^{2} r^{4} ω^{2}}{4 R} \sin^{2} ω t = \frac{{(B π r^{2} ω)}^{2}}{8 R}$

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