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A copper disc of radius 0.1m is rotated about its natural axis with 10rps in a uniform magnetic field of 0.1 T with its plane perpendicular to the field. The emf induced across the radius of the disc is

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$π \times 10 v$

$2 π \times 10 v$

$π \times 10^{- 2} v$

$2 π \times 10^{- 2} v$

answer is C.

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Detailed Solution

$\begin{aligned} e = \frac{1}{2} B R^{2} w \\ = \frac{1}{2} \times 0.1 \times (0.1)^{2} 10 \times 2 π \\ = 0.1 \times 10^{- 2} \times 10 π = π \times 10^{- 2} \\ R = 0.1 m; B = 0.1 T \\ w = 10 \times 2 π; e = ? \end{aligned}$

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