Question

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

• A. 0 V
• B. 0.628 mV
• C. 0.628 $\mu$V
• D. 62.8 $\mu$V

Answer 1

The correct option is B 0.628 mV

Given,

A wheel with $410$ 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,

$E=\frac{d\varphi }{dt}=\frac{d\left(BA\right)}{dt}$

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

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

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

$r=0.5m$

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

Therefore,

Induced emf is given by,

$E=\frac{1}{2}B{r}^2\omega$

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

$=6.28\times {10}^{-5}V$

Each spoke will act as a parallel source of emf.

Hence, the emf will be $0.628mV$