Question

There is a current of 40 amperes 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. $2.5\times {10}^{-3}m/s$
• C. $25.0\times {10}^{-3}m/s$
• D. $250\times {10}^{-3}m/s$

Answer 1

The correct option is B $2.5\times {10}^{-3}m/s$

$V=\left(\frac{I}{nqA}\right)=\frac{40}{{10}^{29}\times 1.6\times {10}^{-19}\times {10}^{-6}}$
$=\frac{40}{1.6}\times {10}^{-29+25}$
$V=2.5\times {10}^{-2}m/s$