Question

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

• A. $ 1.25\times 10^{3}\,m/s $
• B. $ 2.50\times 10^{-3}\,m/s $
• C. $ 25.0\times 10^{-3}\,m/s$
• D. $ 250\times 10^{-3}\,m/s $

Answer 1

Answer: (B)

The relation between drift velocity $v_{d}$ and current $i$ are related as given by
$i=A n e v_{d},$ So, $v_{d}=\frac{i}{A n e}$
$v_{d}=\frac{40}{10^{-6} \times 1.6 \times 10^{-19} \times 10^{29}}$
$=\frac{40}{16} \times 10^{-3} m / s$
$=2.5 \times 10^{-3} m / s$