A copper conductor of length $2\text{m}$ and cross‑sectional area $2{\text{mm}}^2$ carries a current of $5\text{A}$. The number density of free electrons in copper is $n=8.5\times 10^{28}{\text{m}}^{-3}$. (i) Find the drift speed of the electrons. (ii) Explain why the drift speed is so small compared to the speed of an electric signal in the wire.

Step-by-step solution

1. Convert the cross‑sectional area to square metres: $2{\text{mm}}^2=2\times 10^{-6}{\text{m}}^2$.
2. Current is related to drift speed by $I=neA{v}_d$, where $e=1.6\times 10^{-19}\text{C}$.
3. Re‑arrange for drift speed: $v_d=\frac{I}{neA}$.
4. Substitute the numbers: $v_d=\frac{5}{(8.5\times 10^{28})(1.6\times 10^{-19})(2\times 10^{-6})}$.
5. Calculate the denominator: $(8.5\times 10^{28})(1.6\times 10^{-19})(2\times 10^{-6})=2.72\times 10^4$.
6. Thus $v_d=\frac{5}{2.72\times 10^4}=1.84\times 10^{-4}\text{m/s}$.
7. Explanation: The electric signal is a change in the electric field which propagates at a speed close to that of light (~$3\times 10^8\text{m/s}$). Only a very small average displacement of electrons (the drift speed) is required to sustain the current, so individual electrons move much slower.

Concept tested: Drift velocity of electrons in a conductor

Published 8 September 2026 · How this solution was prepared and checked