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A copper disc of diameter 20 cm makes 1200 r.p.m. about its natural axis kept parallel to a uniform magnetic field of 10^-2 T. The potential difference between the centre and edge of the disc is

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$6.28 \times 10^{- 3} V$

$62.8 \times 10^{- 3} V$

$0.628 \times 10^{- 3} V$

$0.628 V$

answer is A.

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Detailed Solution

$e = \frac{1}{2} B R^{2} ω$ where $ω = 2 π f$

$e = \frac{1}{2} \times 10^{- 2} \times 0.1 \times 0.1 \times 2 π \times \frac{1200}{60} = 6.28 \times 10^{- 3} V$

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