Question

A copper wire has a cross sectional area of $1m{m}^2$. The electron density of copper is $8.5\times {10}^{28}{m}^3$. Calculate the mean drift velocity of the electrons through the wire when the current is 3 A.

Answer 1

Given:
i= 3A
n= $8.5\times {10}^{28}{m}^3$
A= $1m{m}^2$ = $1\times {10}^{-6}{m}^2$
e= $1.6\times {10}^{-19}C$
We know
$i=neA{v}_d$
$v_d=\frac{1}{neAi}$

$v_d=\frac{1}{8.5\times {10}^{28}\times 1.6\times {10}^{-19}\times 1\times {10}^{-6}\times 3}$

$v_d=0.0245\times {10}^{-3}m{s}^{-1}$

$v_d=0.0245mm{s}^{-1}$