Question: Q. 7. A wheel with 8 metallic spokes each $50 \mathrm{~cm}$ long is rotated with a speed of $120 \mathrm{rev} / \mathrm{min}$ in a plane normal to the horizontal component of the earth’s magnetic field. The earth’s magnetic field at the place is $0.4 \mathrm{G}$ and the angle of dip is $60^{\circ}$. Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased?
C [CBSE O.D. I, II, III 2013]
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Solution:
Ans. $\because$
$$ \begin{aligned} H & =B \cos \theta=0.4 \cos 60^{\circ} \ & =0.2 \mathrm{G}=0.2 \times 10^{-4} \mathrm{~T} \end{aligned} $$
This component is parallel to the plane of the wheel. Thus, the emf induced is
$$ \begin{aligned} & \varepsilon=\frac{1}{2} H l^{2} \omega, \quad \text { where } \omega=2 \pi f \ &=\frac{1}{2} H l^{2}(2 \pi f) \ & \varepsilon=\frac{1}{2} \times 0.2 \times 10^{-4} \times(0.5)^{2} \ & \quad \times \frac{2 \times 3.14 \times 120}{60} \end{aligned} $$
