A bicycle wheel of radius ' $R$ ' has ' $n$ ' spokes. It is rotating at the rate of ' $F$ ' r.p.m.…
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
$\frac{1}{2}{\mathrm{~B} \pi \mathrm{FR}^2}$
$\quad \mathrm{B} \pi \mathrm{FR}^2$
$\frac{1}{\mathrm{n}} \mathrm{B} \pi \mathrm{FR}$
$\mathrm{B} \pi \mathrm{FR}^2 \mathrm{n}$
Solution
As the angular velocity is constant, the linear velocity changes from 0 to $R \omega$ as we move from the axle to the rim.
$\therefore \quad$ Average velocity $\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{R} \omega}{2}$
We know motional e.m.f, $\mathrm{E}_{\mathrm{mot}}=\mathrm{B} / \mathrm{v}$
$\begin{aligned}
\therefore \quad \mathrm{E}_{\mathrm{mot}} & =\frac{\mathrm{B} \times \mathrm{R} \times \mathrm{R} \omega}{2}=\frac{\mathrm{BR}^2 \times 2 \pi \mathrm{~F}}{2} \\
& =\mathrm{B} \pi \mathrm{FR}^2
\end{aligned}$
^