Current Density and Drift Velocity

Current is charge per second through a wire. Inside a copper wire, what is actually moving, and how fast? The answer is surprising: the electrons creep along far more slowly than a snail, even though the lamp at the end of the wire lights the instant you flip the switch.

Key idea

$$I = nqv_dA, \qquad J = \frac{I}{A} = nqv_d,$$ where $n$ is the number of free charges per cubic meter, $q$ the charge of each, $v_d$ their average drift velocity and $A$ the wire's cross section. $J$ (A/m²) is the current density.

Why it works

Picture a cross section of the wire. In a time $\Delta t$, every carrier within a distance $v_d\Delta t$ upstream of it gets through. Those carriers fill a cylinder of volume $Av_d\Delta t$, which holds $nAv_d\Delta t$ of them (a number density, like the charge densities you met earlier). Each carries $q$, so $\Delta Q = nqAv_d\Delta t$, and dividing by $\Delta t$ gives $I = nqv_dA$.

Current in a simple circuit. The moving dots show the direction of the current, not the speed of real electrons, which drift far more slowly.

Copper has about one free electron per atom, so $n \approx 8.5 \times 10^{28}\ \text{m}^{-3}$. With so many carriers, a tiny drift carries a large current. The electrons are not sitting still otherwise: they zip about randomly at around $10^{6}\ \text{m/s}$, but in every direction at once, so the random part carries no net charge. The field adds a slow, steady drift on top.

Worked examples

Worked example: How fast do the electrons drift?

A copper wire of diameter $1.0\ \text{mm}$ carries $2.0\ \text{A}$. Find the drift speed.

$A = \pi r^2 = \pi(0.50 \times 10^{-3}\ \text{m})^2 = 7.85 \times 10^{-7}\ \text{m²}$.
$$v_d = \frac{I}{neA} = \frac{2.0}{(8.5 \times 10^{28})(1.60 \times 10^{-19})(7.85 \times 10^{-7})} = 1.87 \times 10^{-4}\ \text{m/s}$$
That is about $0.19\ \text{mm/s}$.

Answer: $v_d \approx 0.19\ \text{mm/s}$

Worked example: A long trip

At that speed, how long does one electron take to travel $1.0\ \text{m}$ along the wire?

$t = \dfrac{1.0\ \text{m}}{1.87 \times 10^{-4}\ \text{m/s}} = 5340\ \text{s}$, about $1.5$ hours.

Answer: about $1.5\ \text{h}$

Worked example: Current density and the field inside

In the same wire, find $J$, and the electric field that drives the current (copper's resistivity is about $1.7 \times 10^{-8}\ \Omega\cdot\text{m}$).

$J = I/A = 2.55 \times 10^{6}\ \text{A/m²}$.
Ohm's law, written per unit area and length, is $E = \rho J$: $E = (1.7 \times 10^{-8})(2.55 \times 10^{6}) = 0.0433\ \text{V/m}$.
A field of a few hundredths of a volt per meter is enough. It is not zero, as it would be in electrostatics, because charges are moving and losing energy to the metal.

Answer: $J \approx 2.55 \times 10^{6}\ \text{A/m²}$, $E \approx 0.0433\ \text{V/m}$

Common mistake

"The electrons race through the wire at nearly the speed of light." What travels that fast is the field: when you close a switch, the push spreads along the wire in nanoseconds, and electrons everywhere in the circuit, including those already in the lamp, start drifting at once. Each electron creeps along at a fraction of a millimeter per second. Check with the numbers above: at $0.19\ \text{mm/s}$, an electron would take over an hour to cross a room.

Current Density and Drift Velocity

Worked examples

Practice

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