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# 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 ?

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Solution

## 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/secMagnitude of the field $\left(B\right)$= $0.4$ Gauss = $0.4×{10}^{-4}\mathrm{T}$Step 2. Formula used:Induced emf between axle & rim is given by,$e=Bvl$, where all the symbols are described above.Step 3: CalculationWe know, that one revolution = $2\mathrm{\pi }\mathrm{rad}$For two revolutions, angular velocity $\omega =2\left(2\mathrm{\pi }\right)rad=4\mathrm{\pi }\mathrm{rad}}{s}$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$e=\left(0.4×{10}^{-4}\right)×\left(\frac{1}{2}×4\mathrm{\pi }×0.5\right)×0.5=6.28×{10}^{-5}V$Hence, the induced emf between the axle and the rim of the wheel is $6.28×{10}^{-5}V$.  Suggest Corrections  0      Similar questions  Explore more