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An electron revolves in a circular orbit a plane perpendicular to a uniform magnetic field of induction $6 \times 10^{- 9} w b / m^{2}$ . Then the time period of revolution is

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$5.96 \times 10^{- 9} \sec$

$5.96 \times 10^{- 3} \sec$

$5.96 \times 10^{- 5} \sec$

$5.96 \times 10^{- 11} \sec$

answer is B.

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Detailed Solution

$T = \frac{2 π m}{q B} = \frac{2 \times 3.14 \times 9.1 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 6 \times 10^{- 9}} = 5.95 \times 10^{- 3} \sec$ .

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