Current and Drift Velocity

In a copper wire carrying a current, electrons drift through the lattice at a snail's pace — much slower than your phone scrolls. Yet the signal travels at nearly the speed of light. How do we reconcile these facts?

Concept

Electric current $I$ is the rate of flow of charge across a cross-section: $I = \frac{dQ}{dt}.$ Unit: ampere ($A = C/s$). Conventional current flows in the direction of positive charge motion — opposite to actual electron motion in a metal.

Drift velocity. Free electrons in a metal move randomly with thermal speeds $\sim 10^6\,m/s$. With an applied field $\vec E$, they superimpose a small directed velocity $\vec v_d$ opposite to $\vec E$.

The drift velocity is given by $v_d = \frac{eE\tau}{m},$ where $\tau$ is the average time between collisions (relaxation time).

The current is then $I = n e A v_d,$ with $n$ = number density of free electrons, $A$ = cross-section area.

Typical magnitudes: $n \approx 10^{29}\,m^{-3}$ for copper, giving $v_d \sim 10^{-4}\,m/s$ for $1\,A$ in a $1\,mm^2$ wire — a glacial $0.1\,mm/s$.

Derivation

An electron in field $\vec E$ feels force $-e\vec E$, giving acceleration $\vec a = -e\vec E/m$. Between collisions it picks up velocity $\vec a\tau$. After a collision its velocity is reset to a random thermal value, which averages to zero across many electrons. The net average drift is $\vec v_d = -\frac{e\vec E\tau}{m}.$

For the magnitude, $v_d = eE\tau/m$.

To get current, consider a conductor of cross-section $A$. In time $dt$, all electrons within distance $v_d\,dt$ of the cross-section cross it. Number of such electrons: $n A v_d\,dt$. Charge transported: $neAv_d\,dt$. So $I = \frac{dQ}{dt} = neAv_d.$

Worked Example

A copper wire of cross-section $1\,mm^2$ carries $5\,A$. Electron density $n = 8.5\times 10^{28}\,m^{-3}$.

$v_d = \frac{I}{neA} = \frac{5}{(8.5\times 10^{28})(1.6\times 10^{-19})(10^{-6})} \approx 3.7\times 10^{-4}\,m/s.$

About $0.4\,mm/s$. So why does turning on the switch produce immediate light? Because the electric field propagates at nearly $c$ through the wire, simultaneously nudging electrons everywhere.

If $\tau = 2\times 10^{-14}\,s$ and $E = 0.05\,V/m$: $v_d = \frac{(1.6\times 10^{-19})(0.05)(2\times 10^{-14})}{9.1\times 10^{-31}} \approx 1.76\times 10^{-4}\,m/s,$ consistent with the above.

Common Confusions

• Drift velocity is tiny, but the field propagates at $c$. Electrons everywhere start moving almost simultaneously.
• Current is a scalar with sign convention (conventional direction). It is not a true vector — though current density $\vec J$ is.
• Direction of conventional current is opposite to electron drift in a metal.
• $n$ is per cubic metre, $A$ in $m^2$, $v_d$ in $m/s$ — keep SI consistent.

Key Takeaways

• $I = dQ/dt$; unit ampere.
• $v_d = eE\tau/m$; typically $10^{-4}\,m/s$.
• $I = neAv_d$.
• Direction of conventional current = direction of positive charge flow = opposite to electron drift.
• Drift is slow, but the field-driven response is essentially instantaneous.