Drift Velocity Formula: vd = I/(neA), Units and Examples

The drift velocity formula is $v_d = \frac{I}{neA}$ : divide the current by the number density of free electrons, the electron charge and the area of the wire. The result is the small average speed at which electrons move along the conductor when a potential difference is applied, usually a fraction of a millimetre per second.

Inside a metal, free electrons already rush about at very high random speeds, but in every direction, so their average displacement is zero. Switch on an electric field and each electron picks up a small extra push against the field between collisions. That steady sideways creep is the drift. It is a Class 12 topic in current electricity and sits on the electricity formulas list right after the definition of current.

Quick facts

Formula
$v_d = \frac{I}{neA}$
Microscopic form
$v_d = \frac{eE\tau}{m}$ with relaxation time τ
SI unit
metre per second (m/s)
Dimensional formula
[M⁰ L¹ T⁻¹]
Typical size
About 10⁻⁴ m/s in a household copper wire
Direction
Opposite to the electric field for electrons
Course
Class 12, JEE Main, NEET, AP Physics 2

Symbols in the drift velocity formula

Quantities in v_d = I / (neA)
Symbol	Meaning	SI unit
$v_d$	Average drift speed of the free electrons	m/s
$I$	Current through the wire	A
$n$	Free electrons per unit volume (about 8.5 × 10²⁸ m⁻³ for copper)	m⁻³
$e$	Electronic charge, 1.602 × 10⁻¹⁹ C	C
$A$	Area of the wire's cross section	m²

The denominator $neA$ is the amount of mobile charge in each metre of wire. Multiply that by how many metres the charge advances per second and you get coulombs per second, which is the current. So the formula is really $I = neAv_d$ turned around.

Class 12 derivation with relaxation time

An electron of mass m in a field E feels a force eE, so its acceleration is $a = \frac{eE}{m}$ . Between two collisions it travels for an average time τ, called the relaxation time. Its random starting velocities average to zero, so the average velocity gained is

$v_d = a\tau = \frac{eE\tau}{m}$

Now count charge. In a time t the electrons in a length $v_d t$ of wire pass a cross section. That length holds $nAv_d t$ electrons, carrying charge $neAv_d t$ . Dividing by t gives $I = neAv_d$ . Both results appear in board answers, so learn both lines.

The ratio $\mu = \frac{v_d}{E} = \frac{e\tau}{m}$ is the mobility, measured in m²/(V s).

Worked examples

Example 1: copper wire with area in mm²

A copper wire of cross section 1.0 mm² carries 2.0 A. Take n = 8.5 × 10²⁸ m⁻³. Find the drift velocity.

Convert: 1.0 mm² = 1.0 × 10⁻⁶ m².

$v_d = \frac{2.0}{(8.5\times10^{28})(1.602\times10^{-19})(1.0\times10^{-6})} = \frac{2.0}{1.36\times10^{4}}$

v_d ≈ 1.5 × 10⁻⁴ m/s, or about 0.15 mm/s.

Example 2: diameter instead of area

A wire of diameter 1.2 mm carries 1.5 A with n = 8.5 × 10²⁸ m⁻³. Find v_d.

Halve the diameter: r = 0.60 mm, or 6.0 × 10⁻⁴ m, so A = πr² = 1.13 × 10⁻⁶ m².

$v_d = \frac{1.5}{(8.5\times10^{28})(1.602\times10^{-19})(1.13\times10^{-6})} \approx 9.7\times10^{-5}\ \text{m/s}$

Example 3: from the relaxation time

In a metal, τ = 2.5 × 10⁻¹⁴ s and the field is 0.50 V/m. Find v_d. Electron mass is 9.11 × 10⁻³¹ kg.

$v_d = \frac{(1.602\times10^{-19})(0.50)(2.5\times10^{-14})}{9.11\times10^{-31}} \approx 2.2\times10^{-3}\ \text{m/s}$

The mobility here is v_d / E ≈ 4.4 × 10⁻³ m²/(V s).

If drift is so slow, why does the light switch work at once

A drift of 0.15 mm/s means an electron takes well over an hour to crawl one metre. Yet the lamp glows the instant you flip the switch. The reason is that the wire is already full of free electrons. The electric field travels along the circuit at close to the speed of light, and it sets every electron in the wire drifting at almost the same moment. No single electron has to travel from the switch to the bulb.

Do not confuse drift speed with the random thermal speed of electrons, which is of the order of 10⁵ m/s. The drift is a tiny bias laid on top of that motion.

How drift velocity links to other current formulas

Divide $I = neAv_d$ by A and you get $J = nev_d$ , the current density. Put $v_d = \frac{eE\tau}{m}$ into that and compare with $J = \sigma E$ to find the conductivity $\sigma = \frac{ne^2\tau}{m}$ . Its inverse is the resistivity, which explains why resistance rises with temperature: hotter ions shorten τ.

For the same current, a thinner section of wire has a faster drift because A is smaller. The current itself, from the current formula, stays the same along a series circuit.

Common errors in drift velocity questions

Using the diameter as the radius. Halve it, then square, then multiply by π.
Leaving area in mm² or cm². Divide mm² by a million, or cm² by ten thousand, to get m².
Forgetting that v_d depends on E, not on wire length alone. At fixed voltage, a longer wire has a weaker field, so the drift falls.
Stating a big answer. If you get metres per second in a metal wire, a power of ten has gone wrong.

Frequently asked questions

What is the formula of drift velocity?

In terms of current, v_d = I / (neA). In terms of the field, v_d = eEτ / m. Both describe the same average speed of free electrons.

Drift velocity is measured in what unit?

Being a speed, it is measured in m/s, which reduces to [M⁰ L¹ T⁻¹] in base dimensions. Answers in metal wires are usually near 10⁻⁴ m/s.

Does drift velocity depend on temperature?

Yes. Raising the temperature makes the lattice ions vibrate more, so collisions are more frequent and τ gets shorter. At the same field, the drift velocity drops.

What happens to drift velocity if the diameter of the wire is doubled at the same current?

The area becomes four times larger, so the drift velocity falls to one quarter, since v_d is inversely proportional to A.

Is drift velocity the same as the speed of electricity?

No. Drift velocity is the slow average motion of electrons. The signal that starts current flowing everywhere in the circuit moves at nearly the speed of light.