Electric current measures how much charge crosses a surface per unit time. To connect that macroscopic flow to the motion of individual charge carriers, introduce the current density $\mathbf J$ and the drift velocity $\mathbf v_d$.

Current density

For a uniform current passing perpendicularly through cross-sectional area $A$,

$$\boxed{J=\frac{I}{A}}.$$

The SI unit is

$$\mathrm{A/m^2}.$$

More generally, $\mathbf J$ is a vector field whose direction is the direction of conventional positive-charge flow. The current crossing an oriented surface is

$$I=\int_S \mathbf J\cdot d\mathbf A.$$

For a uniform $\mathbf J$ normal to a flat area, this reduces to $I=JA$.

From moving carriers to current

Suppose a conductor contains mobile carriers with

• number density $n$: number of carriers per unit volume;
• charge $q$ per carrier;
• average drift velocity $\mathbf v_d$.

During a short time $dt$, carriers drifting through a cross section move an average distance

$$v_d,dt.$$

The volume swept through area $A$ is

$$A v_d,dt,$$

so the number of carriers crossing is

$$nA v_d,dt.$$

The corresponding charge is

$$dQ=nqA v_d,dt$$

with sign determined by $q$ and the chosen direction. Therefore

$$I=\frac{dQ}{dt}=nqA v_d$$

in one-dimensional signed form.

In vector form,

$$\boxed{\mathbf J=nq\mathbf v_d}.$$

If the carriers are electrons, $q=-e$, so their drift velocity points opposite to conventional current density.

Drift is usually much slower than microscopic motion

Charge carriers in matter can have large microscopic velocities because of thermal or quantum motion. Current is associated with the small net drift produced by an applied field, not with the full random microscopic speed.

A small drift speed can still produce a large current because an ordinary conductor contains an enormous number of carriers per unit volume.

Worked example

A wire has cross-sectional area

$$A=1.0,\mathrm{mm^2}=1.0\times10^{-6},\mathrm{m^2}$$

and carries current

$$I=2.0,\mathrm A.$$

The current-density magnitude is

$$J=\frac{2.0}{1.0\times10^{-6}} =2.0\times10^6,\mathrm{A/m^2}.$$

Suppose the mobile carrier density is

$$n=8.5\times10^{28},\mathrm{m^{-3}}$$

and each carrier has charge magnitude

$$e=1.60\times10^{-19},\mathrm C.$$

The drift-speed magnitude is

$$v_d=\frac{J}{ne} =\frac{2.0\times10^6}{(8.5\times10^{28})(1.60\times10^{-19})} \approx1.47\times10^{-4},\mathrm{m/s}.$$

Thus

$$\boxed{v_d\approx0.15,\mathrm{mm/s}}.$$

The electrical response of a circuit can begin far faster than this carrier drift because changes in the electromagnetic field propagate through the circuit; individual electrons do not need to travel from the source to the load before the load responds.

Current density therefore connects circuit current to a spatial description of moving charge and provides the natural local quantity for conductivity and Maxwell's equations.