A copper wire of area of cross-section \( 2 \times 10^{-6}\,\mathrm{m^2} \) and electron density \( 10^{28}\,\mathrm{m^{-3}} \) carries a current of 1.6 A. Time taken by an electron to drift from one end of a 1.0 m long wire to its other end will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2 \times 10^{3}\,\mathrm{s} \)
The drift velocity follows from the relation between current and drift velocity, \( I = n e A v_d \). The electron covers the length of the wire at this speed, so the time taken is the length divided by the drift velocity.
\( \displaystyle v_d = \frac{I}{n e A} \)
\( n = 10^{28}\,\mathrm{m^{-3}} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( A = 2\times10^{-6}\,\mathrm{m^2} \)
\( \displaystyle v_d = \frac{1.6}{10^{28} \times 1.6\times10^{-19} \times 2\times10^{-6}} = \frac{1.6}{3.2\times10^{3}} = 5\times10^{-4}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle t = \frac{L}{v_d} = \frac{1.0\,\mathrm{m}}{5\times10^{-4}\,\mathrm{m\,s^{-1}}} = 2\times10^{3}\,\mathrm{s} \)
The drift speed is very small, so an electron takes more than half an hour to cross a 1 m wire, even though the current starts almost at once.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
