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Current Electricity: JEE Main Physics Question with Solution

The number density of free electrons in copper is nearly

8 \times 10^{28}\text{ m}^{-3}

. A copper wire has its area of cross section

= 2 \times 10^{-6}\text{ m}^2

and is carrying a current of

3.2\text{ A}

. The drift speed of the electrons is

\underline{\quad\quad\quad\quad} \times 10^{-6}\text{ ms}^{-1}

.

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Question type: Numerical
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

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Students also ask

Is the electronic charge

e

given in the stem?

No, $e = 1.6 \times 10^{-19}\text{ C}$ is a standard fundamental physical constant that candidates are expected to know.

Answer

The drift speed of the electrons is 125 × 10⁻⁶ m s⁻¹.

Why each option works or fails

Step-by-step solution

given: $n = 8 \times 10^{28}\text{ m}^{-3}$ , $A = 2 \times 10^{-6}\text{ m}^2$ , $I = 3.2\text{ A}$ , and fundamental electronic charge $= 2 \times 10^{-6}\text{ m}^2$ .
find: Drift speed $v_d$ expressed in units of $10^{-6}\text{ ms}^{-1}$ .
visualise: Electrons drift through the cross-sectional area $A$ of the copper conductor under an electric field, giving rise to an electric current $I = n e A v_d$ .
strategise: Use the microscopic formula for current: $I = n e A v_d \implies v_d = \frac{I}{n e A}$ .
execute: $v_d = \frac{3.2}{(8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2 \times 10^{-6})} = \frac{3.2}{25.6 \times 10^3} = \frac{1}{8 \times 10^3} = 0.125 \times 10^{-3} = 125 \times 10^{-6}\text{ m/s}$
verify: Drift velocities in typical copper wires carrying a few amperes are of the order of $0.1\text{ mm/s}$ to $1\text{ mm/s}$ ( $10^{-4}-10^{-3}\text{ m/s}$ ). $125 \times 10^{-6}\text{ m/s} = 0.125\text{ mm/s}$ , which is physically consistent.