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A wheel of radius 0.5m has 20spokes. It is rotating at the rate of 180 revolution per minute, perpendicular to the horizontal component of earth’s magnetic field of $0.4 x 10^{- 4}$ T. The emf induced between the rim and the centre of wheel will be nearly

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$6 π \times 10^{- 4} V$

$6 π \times 10^{- 3} V$

$3 π \times 10^{- 5} V$

$3 π \times 10^{- 6} V$

answer is A.

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Detailed Solution

$\begin{array}{l} E = \frac{1}{2} B l^{2} w \\ = \frac{1}{2} \times 4 \times 10^{- 5} \times 25 \times 10^{- 2} \times \frac{180 \times 2 π}{60} \end{array}$

$= 3 π \times 10^{- 5} V$

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