Question

A current of $0.43A$ is flowing through a wire of thickness $2cm$. If there are ${10}^{18}$ electrons per $c{m}^3$ of the wire, calculate the drift velocity of electrons in $m{s}^{-1}$. Take the electron charge as $1.6\times {10}^{-19}C$.

• A. $4.277\times {10}^{-3}m{s}^{-1}$
• B. $8.554\times {10}^{-3}m{s}^{-1}$
• C. $1.283\times {10}^{-4}m{s}^{-1}$
• D. $3.697\times {10}^{-3}m{s}^{-1}$

Answer 1

The correct option is B $8.554\times {10}^{-3}m{s}^{-1}$

Current $i$ is related to the drift velocity $v_D$ of electrons as:
$i=neA{v}_D$
where, $n$ is the number of free electrons per $m^3$, $e$ is the electron charge and $A$ is the cross sectional area of the conductor.
$⟹v_D=\frac{i}{neA}$

Given: $i=0.43A$, $n={10}^{18}c{m}^{-3}$, $e=1.6\times {10}^{-19}C$, wire thickness, $d=2cm$.
Area of cross section of wire, $A=\pi (\frac{d}{2}{)}^2=\pi (\frac{2}{2}{)}^2$ = $\pi c{m}^2$

$⟹v_D=\frac{0.43}{{10}^{18}\times 1.6\times {10}^{-19}\times \pi }$
$⟹v_D=0.8554cm{s}^{-1}$
$\therefore$ Drift velocity, $v_D=8.554\times {10}^{-3}m{s}^{-1}$