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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(b) ${V_d} = \frac{i}{{neA}} = \frac{{40}}{{{{10}^{29}} \times {{10}^{ - 6}} \times 1.6 \times {{10}^{ - 19}}}}$

$=$ $2.5 \times {10^{ - 3}}\,m/sec$.

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