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A wheel of 20 metallic spokes each $40 \mathrm{~cm}$ long is rotated with a speed of $180 \mathrm{rev} / \mathrm{min}$ in a plane normal to the horizontal domponent of earth's magnetic field $\mathrm{H}_{\mathrm{e}}$ at a place. If $\mathrm{H}_{\mathrm{e}}=0.4 \mathrm{G}$ (Gauss) at that place, the induced emf between the axle and the rim of the wheel is
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Verified Answer
The correct answer is:
$192 \pi \times 10^{-7} \mathrm{~V}$
Let induced emf be 'e'. Then
$\begin{aligned}
& \mathrm{e}=\frac{1}{2} \mathrm{~B} \omega \mathrm{L}^2 \\
& =\frac{1}{2} \times 0.4 \times 10^{-4} \times 2 \pi \times 3 \times 0.4^2 \\
& =1.92 \pi \times 10^{-5} \mathrm{~V} \\
& =192 \pi \times 10^{-7} \mathrm{~V}
\end{aligned}$
As all spokes are in parallel. So, overall emf is equal to e $=1.92 \times 10^{-7} \mathrm{~V}$
$\begin{aligned}
& \mathrm{e}=\frac{1}{2} \mathrm{~B} \omega \mathrm{L}^2 \\
& =\frac{1}{2} \times 0.4 \times 10^{-4} \times 2 \pi \times 3 \times 0.4^2 \\
& =1.92 \pi \times 10^{-5} \mathrm{~V} \\
& =192 \pi \times 10^{-7} \mathrm{~V}
\end{aligned}$
As all spokes are in parallel. So, overall emf is equal to e $=1.92 \times 10^{-7} \mathrm{~V}$
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