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There is a current of 40 ampere in a wire of $10^{- 6} m^{2}$ area of cross-section. If the number of free electron per $m^{3}$ is $10^{29}$ , then the drift velocity will be

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$250 \times 10^{- 3} m / s$

$2.50 \times 10^{- 3} m / s$

$25.0 \times 10^{- 3} m / s$

$1.25 \times 10^{3} m / s$

answer is B.

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Detailed Solution

$V_{d} = \frac{i}{n e A} = \frac{40}{10^{29} \times 10^{- 6} \times 1.6 \times 10^{- 19}}$

$= 2.5 \times 10^{- 3} m / s e c$

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