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A wire shaped as a semi-circle of radius a rotates about an axis ${OO}^{^{|}}$ with an angular velocity $ω$ in a uniform magnetic field of induction B. The rotation axis is perpendicular to the field direction. The total resistance of the circuit is equal to R. Neglecting the magnetic field of the induced current, find the mean amount of thermal power being generated in the loop during a rotation period.

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\frac{{(π ω aB)}^{2}}{R}

{(π ω a^{2} B)}^{2} /R

\frac{1}{2} {\frac{(π ω a^{2} B)}{R}}^{2}

\frac{1}{8} \frac{{(π ω a^{2} B)}^{2}}{R}

answer is D.

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Detailed Solution

flux

ϕ = BA \cos θ \Rightarrow ϕ = BA \cos (ω t)

Induced emf

$E = \frac{- d ϕ}{dt}$

$E = - B \frac{π a^{2}}{2} \frac{d}{dt} [\cos ω t]$

$= \frac{B π a^{2}}{2} ω tsin ω t$

$Power P = \frac{E^{2}}{R} = (\frac{B π ω a^{2}}{2}) \frac{\sin^{2} ω t}{R}$

$P_{avg} = \frac{\int_{0}^{T} P dt}{\int dt}$

$= \frac{I}{T} \int_{0}^{T} (\frac{B ω π a^{2}}{2}) \frac{\sin^{2} ω t}{R}$

$= \frac{(B ω π a^{2})}{4 RT} \int_{0}^{T} [\frac{1 - \cos 2 ω t}{2}] dt$

$= \frac{1}{8} \frac{{(B ω π a^{2})}^{2}}{R}$

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