Question

A wheel with $10$ metallic spokes each $0.5m$ long is rotated with a speed of $120rev/min$ in a plane normal to the horizontal component of earth's magnetic field $B_h$ at a place. If $B_h=0.4G$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that $1G={10}^{-4}T$?

Answer 1

Given,
A wheel with $10$ metallic spokes.
Length of metallic spoke $=0.5m$
Angular speed of the wheel $=120rev/min$
Earth's magnetic field, $B=0.4G$
The area covered by an angle $4\theta$.
$A=\pi r^2\frac{\theta }{2\pi }=\frac{r^2}{2}\theta$

The induces emf,
$E=\frac{\varphi }{dt}=\frac{d\left(BA\right)}{dt}$

$=B\frac{r^2}{2}\frac{d\varphi }{dt}=B\frac{r^2}{2}\omega$

$v=120rev/min=2rev/s$

$\omega =2\pi r=4\pi rad/s$

$r=0.5m$

$B=0.4G=0.4\times {10}^{-4}T$

Therefore,
Induced emf is given by,
$E=\frac{1}{2}B{r}^2\omega$

$E=\frac{2\pi \times 2\times 0.4\times {10}^{-4}\times \left({0.5}^2\right)}{2}$

$E=6.28\times {10}^{-5}V$
Each spoke will act as a parallel source of emf.
Hence, the emf will be $0.628mV$.