A metal wire of length $L=3.0,\text{m}$ and cross‑sectional area $A=0.5,{\text{mm}}^2$ carries a current $I=4.0,\text{A}$. The number density of free electrons is $n=9\times 10^{28},{\text{m}}^{-3}$. (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 area to square metres: $A=0.5\times 10^{-6},{\text{m}}^2$. Compute drift speed using $v_d=\frac{I}{neA}$, where $e=1.6\times 10^{-19},\text{C}$. Substitute the values: $v_d=\frac{4}{9\times 10^{28}\times 1.6\times 10^{-19}\times 0.5\times 10^{-6}}\approx 5.6\times 10^{-4},\text{m/s}$. Explain that the drift speed is small because the current is carried by a huge number of electrons; each electron moves only a fraction of a millimetre per second, whereas the electrical signal travels at a speed close to $c$ through the electromagnetic field.
2. Explain that the electric signal is a propagating disturbance in the electric field, not a physical movement of electrons, so its speed is determined by the dielectric properties of the wire and is much larger than the drift speed.

Concept tested: Electron drift speed

Published 15 September 2026 · How this solution was prepared and checked