A metal wire has length $L=3.0\text{ m}$ and cross‑sectional area $A=0.6{\text{ mm}}^2$. The free‑electron number density is $n=9\times 10^{28},{\text{m}}^{-3}$ and the wire carries a current $I=6,\text{A}$. (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. Step 1: Convert the area from mm${}^2$ to m${}^2$: $A=0.6,{\text{mm}}^2=0.6\times 10^{-6},{\text{m}}^2$.
2. Step 2: Use the formula for drift speed: $v_d=\frac{I}{n,q,A}$, where $q=1.6\times 10^{-19},\text{C}$ is the charge of an electron.
3. Step 3: Substitute the values: $v_d=\frac{6}{(9\times 10^{28})(1.6\times 10^{-19})(0.6\times 10^{-6})}$.
4. Step 4: Compute the denominator: $9\times 10^{28}\times 1.6\times 10^{-19}\times 0.6\times 10^{-6}=(9\times 1.6\times 0.6)\times 10^{28-19-6}=(8.64)\times 10^3=8.64\times 10^3$.
5. Step 5: Calculate $v_d=6/8.64\times 10^3\approx 6.94\times 10^{-4},{\text{m s}}^{-1}$. (Correct calculation: $6/8.64\times 10^3=6/8640=0.0006944,{\text{m s}}^{-1}$. )
6. Step 6: The drift speed is about $6.9\times 10^{-4}$ m s${}^{-1}$, i.e. less than a millimetre per second.
7. Step 7: The electric signal (or propagation of the electric field) travels at nearly the speed of light in the conductor, which is $10^8$ m s${}^{-1}$ or more, because the disturbance propagates through the electromagnetic field instantaneously over the length of the wire.
8. Step 8: The electrons themselves move slowly because they are constantly scattering off lattice ions, causing a random walk superimposed on a very small net drift velocity.
9. Step 9: Thus, although the signal reaches the end of the wire almost instantly, individual electrons move at a very low speed.

Concept tested: Drift speed of electrons in a conductor

Published 22 September 2026 · How this solution was prepared and checked