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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field $H_E$ at a place. If $H_E$ = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1 G = $10^{–4}$ T.

$\begin{array}{1 1} 6.28 \times 10^{-5} V \\ 3.14 \times 10^{-5}V \\ 1.57 \times 10^{-5}V \\ 4.71 \times 10^{-5} V \end{array}$
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  • The number of spokes is immaterial because the emf’s across the spokes are in parallel.
Induced emf $\varepsilon = \large\frac{1}{2}$$ \omega B R^2 \rightarrow $$\varepsilon = \large\frac{1}{2}$$ 4\;\pi \times 0.4 \times 10^{-4} \times 0.5^2 = 6.28 \times 10^{-5} V$
answered Mar 10, 2014 by balaji.thirumalai
 

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