Find the average drift speed of conduction electrons in a copper wire of cross-sectional area \( 1.0\times10^{-7}\,\mathrm{m^{2}} \) carrying a current of 1.5 A. (Assume the density of conduction electrons to be \( 9\times10^{28}\,\mathrm{m^{-3}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.04\times10^{-3}\,\mathrm{m\,s^{-1}} \)
The current in a conductor is related to the drift speed of its charge carriers by \( I = neAv_d \), where n is the number of electrons per unit volume, A the cross-sectional area and e the electron charge.
\( \displaystyle v_d = \frac{I}{neA} = \frac{1.5}{(9\times10^{28})(1.6\times10^{-19})(1.0\times10^{-7})} \)
\( \displaystyle v_d = \frac{1.5}{1.44\times10^{3}}\,\mathrm{m\,s^{-1}} = 1.04\times10^{-3}\,\mathrm{m\,s^{-1}} \)
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. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 103 kg/m3, and its atomic mass is 63.5 u.”