A Conductor wire having \( 10^{29} \) electrons per \( \mathrm{m^3} \) carries a current of 20 A. If the cross-section of wire is \( 1\,\mathrm{mm^2} \), then drift velocity of the electron will be (\( e = 1.6 \times 10^{-19}\,\mathrm{C} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.25\times 10^{-3}\,\mathrm{m/s} \)
The current in a conductor is related to the drift velocity by \( I = n e A v_d \).
\( \displaystyle v_d = \frac{I}{n e A} \)
\( A = 1\,\mathrm{mm^2} = 1\times10^{-6}\,\mathrm{m^2} \)
\( n = 10^{29}\,\mathrm{m^{-3}} \)
\( \displaystyle v_d = \frac{20}{10^{29} \times 1.6\times10^{-19} \times 10^{-6}} = \frac{20}{1.6\times10^{4}} = 1.25\times10^{-3}\,\mathrm{m\,s^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… Example 3.1 (a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10–7 m2 carrying a current of 1.5 A. … Solution (a) … The drift speed vd is given by Eq. (3.18)
\( \displaystyle v_d = \frac{I}{neA} \)”