Question

A wheel with $10$ metallic spokes each $0.5\text{m}$ long is rotated with a speed of $120\text{rev/min}$ in a plane normal to the earth's magnetic field at the place. If the magnitude of the magnetic field is $0.4\text{G}$, what is the induced emf between the axle and the rim of the wheel ?

• A. $1.256\times {10}^{-3}\text{V}$
• B. $6.28\times {10}^{-4}\text{V}$
• C. $1.256\times {10}^{-4}\text{V}$
• D. $6.28\times {10}^{-5}\text{V}$

Answer 1

The correct option is D $6.28\times {10}^{-5}\text{V}$

Given,

Length of each spoke $l=0.5\text{m}$

Frequency of rotation, $f=120\text{rev/min}=2\text{rev/s}$

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

We know that, $\omega =2\pi f$

$\Rightarrow \omega =2\pi \times 2=4\pi \text{rad/s}$

$\therefore$ Emf induced across $1$ spoke,

$E=\frac{B\omega l^2}{2}$

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

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

All the spokes are connected in parallel combinations.
Therefore, their net emf is the same as the emf across the individual spoke.

Hence, $\left(\text{D}\right)$ is the correct answer.

Note: Induced emf does not depend upon the number of spokes present in the wheel.