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A conductor shaped as a semicircle of radius a rotates about an axis 00’ with an angular velocity $ω$ in a uniform magnetic field of induction B (see figure). The rotation axis is perpendicular to the field direction. The total resistance of the circuit is equal to R. Neglecting the magnetic field of induced current , if the mean amount of thermal power being generated in the loop during a rotation period is $\frac{B^{2} ω^{2} π^{2} a^{2}}{x R} .$ Find the value of $x$

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Detailed Solution

The area of the loop , $A = \frac{π a^{2}}{2}$ .

If $θ$ is the angle between magnetic field and normal to the loop at any time t, then

$θ = ω t,$

And flux $ϕ = B A \cos ω t$

$\begin{aligned} = B (\frac{π a^{2}}{2}) \cos ω t \\ e = - \frac{d ϕ}{d t} \\ = B (\frac{π a^{2}}{2}) ω \sin ω t \end{aligned}$

Induced current $i = \frac{e}{R} = (\frac{B ω π a^{2}}{2 R}) \sin ω t$

The instantaneous power generated

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

Thus mean power $P_{mean} = \frac{\int_{0}^{T} P d t}{T}$

$\begin{aligned} = \frac{{(\frac{B ω π a^{2}}{2})}^{2} \frac{1}{R} \int_{0}^{T} \sin^{2} ω t d t}{T} \\ = {(\frac{B ω π a^{2}}{2})}^{2} \frac{1}{2 R} \end{aligned}$

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