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Current Electricity: JEE Main Physics Question with Solution

A current of

2\text{ A}

flows through a wire of cross-sectional area

25.0\mathrm{mm}^2

. The number of free electrons in a cubic meter are

2.0 \times 10^{28}

. The drift velocity of the electrons is

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

(given, charge on electron

= 1.6 \times 10^{-19}\mathrm{C}

)

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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

Quick checks

Students also ask

Why is

1\text{ mm}^2 = 10^{-6}\text{ m}^2

and not

10^{-3}\text{ m}^2

?

$1\text{ mm} = 10^{-3}\text{ m}$ , so squaring both sides gives $(1\text{ mm})^2 = (10^{-3}\text{ m})^2 = 10^{-6}\text{ m}^2$ .

Answer

The drift velocity of electrons is $25 \times 10^{-6}\text{ m/s}$ , so the numerical value to fill in is 25.

Why each option works or fails

Step-by-step solution

given: Current $I = 2\text{ A}$ , Area of cross-section $A = 25.0\text{ mm}^2 = 25.0 \times 10^{-6}\text{ m}^2$ , Electron number density $n = 2.0 \times 10^{28}\text{ m}^{-3}$ , Elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ .
find: Value of the pre-factor in the drift velocity $v_d = \underline{\quad} \times 10^{-6}\text{ m s}^{-1}$ .
visualise: Electrons drift along a conductor of uniform cross-section under an electric field, carrying current $I = n e A v_d$ .
strategise: Use the direct relation between macroscopic current and microscopic drift velocity: $I = n e A v_d \implies v_d = \frac{I}{n e A}$ .
execute: Substitute the values: $v_d = \frac{2}{2.0 \times 10^{28} \times 1.6 \times 10^{-19} \times 25.0 \times 10^{-6}}$ $v_d = \frac{2}{80 \times 10^{3}} = \frac{1}{40000} = 2.5 \times 10^{-5}\text{ m s}^{-1} = 25 \times 10^{-6}\text{ m s}^{-1}$ Thus, the integer to enter is 25.
verify: Drift velocities in typical metallic conductors carrying a few amperes are on the order of $10^{-4}$ to $10^{-5}\text{ m s}^{-1}$ . $2.5 \times 10^{-5}\text{ m s}^{-1}$ is physically standard and consistent with SI units.