Drift Velocity, Current Density & Conductivity

PHYCurrent, Drift & Resistance

This subtopic develops the microscopic picture of conduction: free electrons drift slowly under an applied field, and the macroscopic current is the sum of their motion through the cross-section. The central relations connect current to drift velocity, current density to field via conductivity, and total charge to the number of electrons.

ISC favours finding $v_d$ (and noticing it is tiny, $\sim 10^{-4}$ m/s) and computing $J$ , $E$ and $\sigma$ from a steady current.

Drift velocity

v_d = \dfrac{I}{n A e}

I

current (A),

n

free-electron density (

m^{-3}

A

area (

m^2

e=1.6\times10^{-19}

C. Drift speed is of order

10^{-4}

m/s even for ordinary currents.

Drift velocity from relaxation time

v_d = \dfrac{eE\tau}{m}, \qquad \sigma = \dfrac{ne^2\tau}{m}

\tau

average relaxation time between collisions,

m

electron mass,

n

free-electron density. Links the microscopic (relaxation-time) picture of conduction to the macroscopic

v_d=I/nAe

relation above.

Derivation

Between collisions a free electron accelerates under the field $E$ with $a=eE/m$ . Averaged over the electron population, each free flight starts from zero drift velocity and gains $a\tau$ by the next collision, where $\tau$ is the average time between collisions:
$v_d = a\tau = \dfrac{eE}{m}\tau$
Equate this to the macroscopic drift velocity from $J=nev_d$ (i.e. $v_d=J/ne$ ):
$\dfrac{eE\tau}{m} = \dfrac{J}{ne}$
Solve for $J$ and compare with Ohm's law in local form $J=\sigma E$ :
$J = \dfrac{ne^2\tau}{m}\,E \quad\Rightarrow\quad \sigma = \dfrac{ne^2\tau}{m}$

Current density

J = \dfrac{I}{A} = n e \, v_d = \sigma E

J

current density (

A/m^2

E

electric field inside the conductor (V/m),

\sigma

conductivity (

S/m

J

is a vector along the field.

Conductivity and field

\sigma = \dfrac{1}{\rho}, \qquad E = \rho J

\sigma

conductivity (

\Omega^{-1}\,m^{-1}

\rho

resistivity (

\Omega \cdot m

). The microscopic Ohm's law

J=\sigma E

is the local form of

V=IR

Charge and electron count

q = I t = N e

q

total charge (C) crossing a section in time

t

(s),

N

number of electrons,

e

electronic charge. So

N = It/e

Drift velocity is extremely small ( $\sim 10^{-4}$ m/s); the electric field, not the electrons, propagates at near light speed to start the current almost instantly.
Current density $J=I/A$ is intensive and a vector; current $I$ is a scalar flux through the area.
The microscopic Ohm's law $J=\sigma E$ holds point-by-point; $E$ inside a current-carrying wire equals $\rho J = \rho I/A$ .
Use $\sigma = 1/\rho$ to switch between conductivity and resistivity; conductivity has unit $S/m$ (siemens per metre).
Total charge in time $t$ is $q=It$ for steady current; the electron count follows from $N=q/e$ .
Drift velocity is inversely proportional to area for fixed current — a thinner wire has faster-drifting electrons.
Relaxation time links the two pictures: $v_d = (eE/m)\tau$ and $\sigma = ne^2\tau/m$ .

Where the marks go

Confusing current $I$ (A) with current density $J$ ( $A/m^2$ ) by forgetting to divide by area.
Forgetting the huge $n$ ( $\sim 10^{28}\,m^{-3}$ ) in $v_d=I/(nAe)$ , which makes $v_d$ come out absurdly large.
Using $\rho$ where $\sigma$ is needed; remember $\sigma=1/\rho$ , not $\sigma=\rho$ .
Computing electron count from current alone without multiplying current by time first ( $N=It/e$ ).

How the board asks it

Numerical $v_d=I/(nAe)$
A copper wire of cross-sectional area $1.0\times10^{-6}\,m^2$ carries a steady current of $1.5\,A$ . If the free-electron density is $8.5\times10^{28}\,m^{-3}$ , calculate the drift velocity of the conduction electrons.
Numerical $J=I/A$ and $E=\rho J$
A wire of radius $0.5\,mm$ and resistivity $1.7\times10^{-8}\,\Omega\,m$ carries a current of $2\,A$ . Determine the current density $J$ in the wire and the electric field $E$ set up inside it.
Derive / proverelaxation time, $v_d=eE\tau/m$ , $\sigma=ne^2\tau/m$
Obtain an expression for the drift velocity of free electrons in a conductor in terms of the relaxation time $\tau$ , and hence derive the relation $\sigma=ne^2\tau/m$ for the conductivity.
Give reasonssmallness of drift velocity vs near-instant field propagation
The drift velocity of electrons in a conductor is only about $10^{-4}\,m/s$ , yet a bulb glows almost immediately when the switch is turned on. Explain why.
Define / statecurrent density, microscopic Ohm's law $J=\sigma E$
Define current density and state its SI unit. Write the microscopic form of Ohm's law $J=\sigma E$ and name the constant $\sigma$ .
Give reasons $v_d=I/(nAe)$ , inverse dependence on area at fixed current
A current-carrying wire is thinner at one section than at another. Explain in which section the electrons drift faster, justifying your answer using the dependence of drift velocity on cross-sectional area for a fixed current.

Practise this topic

NextResistance & Resistivity

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.