JEE Main 2023 — Current Electricity Question with Solution

Using the formula:

$I=neAv$

where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.

We can isolate $v$ to find:

$v=\frac{I}{neA}$

Substituting the given values:

$v=\frac{3.2\text{A}}{8\times 10^{28}{\text{m}}^{-3}\times 1.6\times 10^{-19}\text{C}\times 2\times 10^{-6}{\text{m}}^2}$

This simplifies to:

$v=\frac{3.2}{16\times 1.6\times 10^3}{\text{ms}}^{-1}=125\times 10^{-6}{\text{ms}}^{-1}$

So, the drift speed of the electrons is indeed $125\times 10^{-6}{\text{ms}}^{-1}$.