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(A) The angular velocity $\omega$ in radians per second is given by $\omega = 2 \pi f$,where $f$ is the frequency in revolutions per second. Since $F$

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$\frac{1}{2} B \pi F R^2$

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(A) The angular velocity $\omega$ in radians per second is given by $\omega = 2 \pi f$,where $f$ is the frequency in revolutions per second. Since $F$ is in r.p.m.,$f = \frac{F}{60}$.Thus,$\omega = 2 \pi \frac{F}{60} = \frac{\pi F}{30}$.Consider a small element $dr$ at a distance $r$ from the center. The motional e.m.f. $dE$ induced across this element is $dE = B v dr = B (r \omega) dr$.Integrating from $r = 0$ to $r = R$:$E = \int_{0}^{R} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_{0}^{R} = \frac{1}{2} B \omega R^2$.Substituting $\omega = \frac{2 \pi F}{60}$:$E = \frac{1}{2} B \left( \frac{2 \pi F}{60} \right) R^2 = \frac{B \pi F R^2}{60}$.Note: If $F$ is treated as revolutions per second (rps) in the context of standard physics problems of this type,the answer is $\frac{1}{2} B \pi F R^2$. Given the options,the intended answer is $A$.

$A$ bicycle wheel of radius $R$ has $n$ spokes. It is rotating at the rate of $F$ r.p.m. perpendicular to the horizontal component of earth's magnetic field $\vec{B}$. The e.m.f. induced between the rim and the centre of the wheel is

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