A copper wire of length $L=4.0,\text{m}$, cross‑sectional area $A=0.8,{\text{mm}}^2$, and free‑electron number density $n=8.5\times 10^{28},{\text{m}}^{-3}$ carries a current $I=5.0,\text{A}$. (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 area to metres squared: $0.8,{\text{mm}}^2=0.8\times 10^{-6},{\text{m}}^2=8.0\times 10^{-7},{\text{m}}^2$.
2. Use the drift velocity formula $v_d=\frac{I}{nAe}$, where $e=1.602\times 10^{-19},\text{C}$ is the electron charge.
3. Substitute the values: $v_d=\frac{5.0}{(8.5\times 10^{28})(8.0\times 10^{-7})(1.602\times 10^{-19})}$.
4. Calculate the denominator stepwise: $(8.5\times 10^{28})(8.0\times 10^{-7})=6.8\times 10^{22}$; multiply by $1.602\times 10^{-19}$ gives $1.089\times 10^4$.
5. Divide the numerator by the denominator: $v_d=5.0/1.089\times 10^4\approx 4.59\times 10^{-4},\text{m/s}$.
6. Thus the drift speed is about $4.6\times 10^{-4},\text{m/s}$, which is very small.
7. Explain the smallness: Electrons in a metal are constantly scattering with lattice ions, so each individual electron moves only a little before its direction is changed. The collective effect of all electrons is a slow drift. The electric field that causes the drift propagates through the metal at a speed close to the speed of light, so the electrical signal travels almost instantaneously while the individual electrons lag behind.

Concept tested: Electron drift velocity in conductors

Published 19 September 2026 · How this solution was prepared and checked