# Drift velocity in a conductor

by guardaiinalto
Tags: conductor, drift, velocity
 P: 1 Dear all, I have a question on the derivation of the drift velocity in a conductor, and was wondering if somebody could help me with it: drift velocity is the average velocity which a free charge moving in a conductor has due to the influence of an electric field applied to the conductor. In a metal, the free charge will be an electron. As they move through the conductor, electrons will frequently bump into ions. If τ is the mean free time of the electron, i.e. the time between successive collisions, then between two collisions, the action of an external electric field will make the electron accelerate by (E*e/m)*τ, where E is the strength of the field (and this strength is constant), e the charge of an electron, an m the mass of the electron. Now, two sources I have consulted say that this quantity (E*e/m)*τ is equal to the magnitude of the drift velocity in the conductor. This appears to me to be false, since the quantity expresses the maximum speed gained by the electron, i.e. the speed it has just before it collides with the next ion. But drift velocity is supposed to be the average velocity of the electron due to the field, so I think its magnitude should be just one-half of this quantity. Any help would be much appreciated. Guardaiinalto
 Mentor P: 11,997 Wouldn't it be the average magnitude since you are using T as the mean free time in your equation? If you average your time between collisions, then whatever you base off that time should already be an average I would think. Does that sound like it makes any sense?
 Sci Advisor Thanks P: 2,539 Drift velocity in a conductor If you have a normal conductor, i.e., not a superconductor, then the usual classical model is that you have charges, freely movable around as in metals, where the conduction electrons are not bound to the ions of the lattice and thus free to move. "Free" is of course not really true, because there are always some defects by the lattice vibrations of the ions due to temperature and other defects in the crystal lattice of the metal and thus the conduction electrons will suffer some friction moving around. Now putting an electric field across a conductor the motion of a fluid cell of conduction electrons is given by (in the non-relativistic limit!) $$n \dot{\vec{v}}=- n e \vec{E} - \alpha \vec{v}.$$ Here $n$ is the density of conduction electrons (particles per cubic metres), $e \approx 1.6 \cdot 10^{-19} \text{As}$ the elementary charge (=charge of a proton), $\alpha$ the friction constant. In the stationary limit the right-hand side must be 0, i.e., the accelerating electric force must compensate precisely the friction force. Then $\vec{v}=\text{const}$ and thus $$\vec{v}=-\frac{n e}{\alpha} \vec{E}.$$ The current density is $\vec{j}=-n e \vec{v}$ and thus $$\vec{j}=\frac{n^2 e^2}{\alpha} \vec{E}.$$ This means the conductance of the metal is given in microscopic terms by $$\sigma=\frac{n^2 e^2}{\alpha}.$$ Independently from this simple classical microscopic picture of conductance, the drift velocity is easy calculated from the current through the conductor, which is given by $$I=A |\vec{j}|=A n e |\vec{v}|\; \Rightarrow \; |\vec{v}|=\frac{I}{A n e}.$$ Here, $A$ is the cross-sectional area of a not too thick wire, so that we can assume the current density to be homogeneous across the wire. Take $I=1 \; \text{A}$ and a copper wire with cross-sectional area of $1 \; \text{mm}^2=10^{-6} \text{m^2}$. For copper, in Wikipedia you find the value $n=8.5 \cdot 10^{28}/\text{m}^3.$ There you also find an explanation for how this value is obtained from the density and the atomic weight. This leads to $$v=\frac{1 \text{A}}{10^{-6} \text{m}^2 8.5 \cdot 10^{28}/\text{m}^3 \cdot 1.6 \cdot 10^{-19} \text{A s}} \approx 7.3 \cdot 10^{-2} \frac{\text{mm}}{\text{s}}.$$ The drift velocity in every-day applications is thus very very small.