Skip to content

Physics · Ch 2 — Current Electricity

Drift Velocity

2.1.2

Drift Velocity

With no electric field applied, a conductor's free electrons undergo a purely random zig-zag motion, repeatedly colliding with the lattice's positive ions; on average, exactly as many electrons travel in any one direction as travel in the opposite direction, so there is no net flow and no current, exactly as described in section 2.1. Once a battery is connected and a potential difference is set up across the conductor, an electric field E⃗\vec E appears inside it. This field exerts a force on every free electron, and because the electron's charge is negative, the resulting acceleration a⃗\vec a points opposite to E⃗\vec E:

a⃗=−eE⃗m(since F⃗=−eE⃗)(2.3)\vec a = -\dfrac{e\vec E}{m}\qquad\text{(since } \vec F=-e\vec E\text{)}\qquad (2.3)

Between one collision and the next, each electron accelerates in this direction; the mean free time τ\tau is the average time interval between successive collisions. Averaging over very many electrons and collisions, the electric field superimposes a slow, steady net motion on top of the electrons' underlying random zig-zag -- this net average velocity is called the drift velocity, v⃗d=a⃗ τ\vec v_d = \vec a\,\tau, i.e.

v⃗d=−eτmE⃗(2.4)\vec v_d = -\dfrac{e\tau}{m}\vec E \qquad (2.4)

which is conventionally written as v⃗d=−μE⃗\vec v_d = -\mu \vec E (2.5), where μ=eτ/m\mu = e\tau/m is called the mobility of the electron -- defined as the magnitude of the drift velocity produced per unit applied electric field, μ=vd/E\mu = v_d/E (2.6). The SI unit of mobility is m2V−1s−1\text{m}^2\text{V}^{-1}\text{s}^{-1}. …

Figure 2.4Zig-zag motion and drift velocity

What this figure shows. A conductor is drawn with several negative charge symbols following jagged, angular paths that repeatedly change direction -- representing the electrons' frequent random collisions with the lattice's positive ions. Superimposed on this zig-zag is a straight arrow labelled vdv_d pointing to the left, opposite to a second arrow labelled E (the applied electric field) pointing to the right, showing that although each electron's actual path is a chaotic zig-zag, its net displacement over many collisions is a slow, …

Misc Example 2.2Acceleration of an electron in a given field

Worked out. The question gives an electric field of magnitude 570 N/C applied in a copper wire and asks for the acceleration experienced by an electron. Using Newton's second law with the electric force, F=ma=eEF = ma = eE, so a=eE/ma = eE/m. Substituting the electron's charge e=1.6×10−19e = 1.6\times10^{-19} C and mass m=9.11×10−31m = 9.11\times10^{-31} kg along with E=570E = 570 N/C gives a=(570)(1.6×10−19)/(9.11×10−31)≈1.001×1014 m/s2a = (570)(1.6\times10^{-19})/(9.11\times10^{-31}) \approx 1.001\times10^{14}\ \text{m/s}^2. This enormous acceleration -- roughly 101310^{13} times g -- is exactly why a real electron's motion is dominated by its frequent collisions with the lattice rather than by smooth, unimpeded acceleration; it only accelerates freely for the b …