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Step-by-Step Solution
1. Understanding the Problem
We have a wire shaped as a semicircle of radius $r$, rotating with angular frequency $\omega$ about its diameter. The axis of rotation is perpendicular to a uniform magnetic field $B$. The total resistance of the circuit is $R$. We want to find the mean (average) power generated per period of rotation.
2. Determining the Magnetic Flux
The area of a full circle of radius $r$ is $\pi r^2$, so for a semicircle, the area is $\frac{\pi r^2}{2}.$ If $\overrightarrow{B}$ is uniform and perpendicular to the axis of rotation, the magnetic flux at an instant is:
$\phi = B \, A \cos(\omega t),$
where $A = \frac{\pi r^2}{2}$ is the area of the semicircle and $\omega t$ is the angle the semicircle’s plane makes with respect to the magnetic field direction.
3. Induced EMF
The induced emf $\varepsilon$ is given by the time derivative of the negative flux:
$\varepsilon = - \frac{d\phi}{dt}
= - \frac{d}{dt}\bigl(B A \cos(\omega t)\bigr)
= \omega B A \sin(\omega t).$
4. Induced Current
Since the total circuit resistance is $R$, the instantaneous current $i$ is:
$i = \frac{\varepsilon}{R}
= \frac{\omega B A}{R}\,\sin(\omega t).$
5. Instantaneous Power
The instantaneous power $P_{\text{inst}}$ delivered is given by:
$P_{\text{inst}} = i^2 R
= \left(\frac{\omega B A}{R}\sin(\omega t)\right)^2 R
= \frac{(\omega B A)^2}{R}\,\sin^2(\omega t).$
6. Average Power Over One Period
The average power $P_{\text{avg}}$ is the time average of the instantaneous power over one full period $T = \frac{2\pi}{\omega}$:
$P_{\text{avg}}
= \frac{1}{T} \int_{0}^{T} P_{\text{inst}} \, dt
= \frac{1}{T} \int_{0}^{T} \frac{(\omega B A)^2}{R} \sin^2(\omega t)\, dt
= \frac{(\omega B A)^2}{R} \cdot \frac{1}{T} \int_{0}^{T} \sin^2(\omega t)\, dt.
$
We use the standard result $\frac{1}{T}\int_{0}^{T} \sin^2(\omega t)\, dt = \frac{1}{2}.$ Hence,
$P_{\text{avg}}
= \frac{(\omega B A)^2}{R} \times \frac{1}{2}
= \frac{(\omega B A)^2}{2R}.
$
7. Substituting the Semicircle’s Area
Recall $A = \frac{\pi r^2}{2}.$ Therefore,
$\omega B A = \omega B \left(\frac{\pi r^2}{2}\right)
= \frac{\omega B \pi r^2}{2}.
$
Substitute this into the average power expression:
$P_{\text{avg}}
= \frac{\left(\frac{\omega B \pi r^2}{2}\right)^2}{2R}
= \frac{\left(\omega B \pi r^2\right)^2}{8R}.
$
8. Final Answer
The mean power generated per period of rotation is:
$\displaystyle \frac{\bigl(B \,\pi \,r^2\,\omega \bigr)^2}{8\,R}.\!
$
