Question

Number of electrons in a conductor of area of cross section $0.4m^2$ is ${10}^{23}{m}^{-3}$. If the drift velocity of the electrons is $4\times {10}^{-5}m{s}^{-1}$, what is the current flowing through the conductor in ampere? (Take electron charge as $1.6\times {10}^{-19}C$)

• A. 0.256

Answer 1

The correct option is A 0.256
We know, current $i=neA{v}_D$, where,
$n$ is the number of electrons, $e$ is the charge of the electron, $A$ is the area of cross section of the conductor and $v_D$ is the drift velocity.
Given: $n={10}^{23}{m}^{-3}$, $v_D=4\times {10}^{-5}m{s}^{-1}$ and $A=0.4m^2$ and $e=1.6\times {10}^{-}19C$

$\therefore i={10}^{23}\times 1.6\times {10}^{-19}\times$ $0.4\times 4\times {10}^{-5}=0.256A$