Question

There is current of 40 amperes in a wire of ${10}^{-6}{m}^2$ area of cross -section. If the number of free electrons 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

The correct option is B $2.50\times {10}^{-3}m/s$

$Given:$ Current=$40ampers$ ; Area of cross- section=${10}^{-6}{m}^2$ ; number of free electrons per $m^3$=${10}^{29}$

$Solution:$

$V=\left(\frac{I}{nqA}\right)=\frac{40}{{10}^{29}\times 1.6\times {10}^{-19}\times {10}^{-6}}$

$=\frac{40}{1.6}\times {10}^{-29+25}$

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

$So,the$ $correct$ $option:B$