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A wheel with \( 10 \) spokes each of length ' \( L \) ' \( m \) is rotated with a uniform angular velocity ' \( \omega \) ' in a
plane normal to the magnetic field 'B'. The emf induced between the axle and the rim of the
wheel.
Options:
plane normal to the magnetic field 'B'. The emf induced between the axle and the rim of the
wheel.
Solution:
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Verified Answer
The correct answer is:
\( \frac{1}{2} \omega B L^{2} \)
Given, number of spokes \( =10 \); length of each spoke \( =L ; \) angular velocity \( =\omega \); angular field = B
Therefore, emf induced between axle and rim is given as
\( E=B l v_{\mathrm{avg}}=\frac{B l v}{2}=B L \frac{r \omega}{2} \)
Now, here \( r=L \)
\( \Rightarrow E=\frac{B L^{2} \omega}{2} \)
Thus, emf induced between the axle and rim of the wheel is \( \frac{1}{2} B L^{2} \omega \)
Therefore, emf induced between axle and rim is given as
\( E=B l v_{\mathrm{avg}}=\frac{B l v}{2}=B L \frac{r \omega}{2} \)
Now, here \( r=L \)
\( \Rightarrow E=\frac{B L^{2} \omega}{2} \)
Thus, emf induced between the axle and rim of the wheel is \( \frac{1}{2} B L^{2} \omega \)
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