Question

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

Answer 1

Step 1. Given data:

Number of metallic spokes $\left(N\right)$ = $10$

Spoke's length or radius of the wheel, $\left(r\right)$ or $\left(L\right)$ = $0.5m$

Wheel's rotation $\left(f\right)$ = $120$ rev/min = $2$ rev/sec

Magnitude of the field $\left(B\right)$= $0.4$ Gauss = $0.4\times {10}^{-4}\mathrm{T}$

Step 2. Formula used:

Induced emf between axle & rim is given by,

$e=Bvl$, where all the symbols are described above.

Step 3: Calculation

We know, that one revolution = $2\mathrm{\pi }\mathrm{rad}$

For two revolutions, angular velocity $\omega =2\left(2\mathrm{\pi }\right)rad=4\mathrm{\pi }\raisebox{1ex}{$\mathrm{rad}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Velocity of spoke at rim = $\omega r$

Velocity of spoke at axle = $0$

Average velocity $\left(v\right)$= $\frac{\omega r+0}{2}=\frac{\omega r}{2}$

Step 3. Calculations:

Putting all the given values in the above equation, we get

$e=\left(0.4\times {10}^{-4}\right)\times \left(\frac{1}{2}\times 4\mathrm{\pi }\times 0.5\right)\times 0.5=6.28\times {10}^{-5}V$

Hence, the induced emf between the axle and the rim of the wheel is $6.28\times {10}^{-5}V$.