Question 270: Electromagnetic Induction

Question

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

Accepted Answer

As the angular velocity is constant, the linear velocity changes from 0 to $\mathrm{R} \omega$ as we move from the axle to the rim.
$$\begin{aligned}
& 	herefore \quad 	ext { Average velocity } \mathrm{v}_{	ext {avg }}=\frac{\mathrm{R} \omega}{2} \
& \quad 	ext { We know motional e.m.f, } \mathrm{E}_{\mathrm{mot}}=\mathrm{B} / \mathrm{v} \
& 	herefore \quad \mathrm{E}_{\mathrm{mot}}=\frac{\mathrm{B} 	imes \mathrm{R} 	imes \mathrm{R} \omega}{2}=\frac{\mathrm{BR}^2 	imes 2 \pi \mathrm{~F}}{2} \
&={\mathrm{B} \pi \mathrm{R}^2}^2
\end{aligned}$$

Answer

$\frac{1}{2} \mathrm{~B} \pi \mathrm{FR}^2$

Answer

$\mathrm{B} \pi \mathrm{FR}^2$

Answer

$\frac{1}{\mathrm{n}} \mathrm{B} \pi \mathrm{FR}$

Answer

$B \pi \mathrm{FR}^2 n$

Electromagnetic Induction

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

Choose the correct answer: