Drift Velocity

In one sentence: Electrons in a copper wire drift along at a fraction of a millimeter per second, even though the electric signal races along the wire.

$v_d = \frac{I}{n q A}$

Read it out loud: “v sub d equals I divided by the quantity n times q times A. Read v sub d as v sub d, the drift velocity, the average speed of the electrons moving along the wire. I is the current, in amperes. n is the carrier density, the number of free charge carriers in each cubic meter. q is the charge of one carrier. A is the cross-sectional area, the area of a slice cut straight across the wire.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $8.5 \times 10^{28}$ and $10^{-19}$: Very large and very small numbers are written as a short number times a power of ten. The raised exponent says how many times to multiply by 10. So $10^{28}$ is a 1 followed by 28 zeros, and $8.5 \times 10^{28}$ is 8.5 with the decimal point moved 28 places to the right. A negative exponent means divide by 10 instead. $10^{-19}$ is a decimal point, then 18 zeros, then a 1, a very small number. To multiply powers, add the exponents, and to divide, subtract them.
Electric charge, the coulomb (C) and the elementary charge $e$: Electric charge is a property of some particles that makes them push and pull on one another. Its unit is the coulomb, written C. An electron carries a tiny charge of about $1.602 \times 10^{-19}$ C, called the elementary charge, written $e$ . A coulomb is a large amount of charge: it equals about $6.24 \times 10^{18}$ electrons.
Current, the ampere (A) and the charge per second: Current is how much charge passes a point each second. The unit of current is the ampere, written A. One ampere means one coulomb of charge passes a point every second. So a current of 1 A moves 1 C of charge past a point each second. The formula turns this flow of charge into the speed at which the carriers move inside the wire.
Charge carriers and free electrons: A metal contains many electrons that are not tied to any one atom. These free electrons can drift through the metal when a voltage pushes on them. Any particle that carries charge through a material is called a charge carrier. Each carrier has a charge $q$ , and the number of carriers in each cubic meter is the carrier density $n$ .
Cubic meters, square meters and the unit $\text{m}^{-3}$: A cubic meter, written $\text{m}^3$ , is a box one meter long, wide and high. A square meter, written $\text{m}^2$ , is one meter by one meter, the unit of area. The unit $\text{m}^{-3}$ means one over a cubic meter, so a density in $\text{m}^{-3}$ counts the particles that fill one cubic meter. The negative exponent is a way to write division.
The prefix milli (m), as in millimeters: A millimeter, written mm, is one thousandth of a meter, so 1 mm = 0.001 m. The prefix milli always means divide by 1000. The formula needs meters, so a wire diameter given in millimeters must be converted to meters before any area is found. The radius is half the diameter.

What each symbol means

Symbol	Meaning
$v_d$	The drift velocity, the average speed of the electrons moving along the wire in one direction, in meters per second (m/s). Electrons also move in random directions at far higher speeds, but only the small net drift makes the current. In the example, the drift is about 0.09 mm/s.
$I$	The current through the wire, in amperes (A), where one ampere is one coulomb per second. A small LED indicator may draw about 20 milliamperes, which is 0.02 A. A 12 W lamp on a 12 V supply draws $12\ \text{W} / 12\ \text{V} = 1$ A.
$n$	The carrier density, the number of free charge carriers in each cubic meter, in per cubic meter ( $\text{m}^{-3}$ ). Copper has about $8.5 \times 10^{28}$ free electrons in each $\text{m}^3$ . That is a huge count, which is why a small current is made of an enormous number of moving electrons.
$q$	The charge of each carrier, measured in coulombs (C). For the free electrons in copper, its size is the elementary charge, $e = 1.602176634 \times 10^{-19}$ C. The minus sign of an electron’s charge is left out here, because the formula needs only the size of the charge.
$A$	The cross-sectional area of the wire, the area of a slice cut straight across it, measured in square meters ( $\text{m}^2$ ). A round wire 1.0 mm across has an area of about $7.85 \times 10^{-7}\ \text{m}^2$ , which is less than one square millimeter.

The formula piece by piece

$I$: Start with the current, the charge that passes a point each second. This is the flow that the whole wire must carry. More current means more charge must move each second, and the formula shows how that is shared between the speed of the carriers and how many carriers there are.
$n q A$: The charge packed into each meter of wire. $n$ counts the carriers in each cubic meter, $q$ gives each carrier’s charge, and $A$ is the area of the slice. Multiplying them gives coulombs per meter, written C/m, which tells how much charge fills each meter of the wire.
$\frac{I}{n q A}$: The current divided by the charge per meter. An ampere is one coulomb per second, so the division is coulombs per second divided by coulombs per meter. The coulombs cancel and leave meters per second, a speed, which is the right unit for drift velocity.
$v_d = \frac{I}{n q A}$: The drift velocity equals the current divided by the carrier density, times the charge on each carrier, times the cross-sectional area, which gives the average speed of the carriers along the wire in meters per second.

What the formula is doing, and why

Current is the amount of charge that passes a point each second. Inside a copper wire, that charge is carried by free electrons. The formula $v_d = \frac{I}{nqA}$ gives the average speed at which those electrons drift along the wire in one direction. This speed is called the drift velocity. It is a speed with a direction, along the wire.

Why divide by $nqA$ ? The current is the charge per meter multiplied by the speed: $I = nqA\,v_d$ . Rearranging gives $v_d = \frac{I}{nqA}$ . The more charge packed into each meter, the slower the charge has to move to deliver the same current. More carriers per cubic meter spread the charge over more electrons, and a wider wire, with a bigger area $A$ , gives the carriers more room. Both make each carrier move more slowly.

Why multiply by $q$ ? The charge $q$ is what each carrier carries. A carrier with twice the charge moves the same current at half the speed, because each one delivers twice as much charge per trip. That is why the formula needs $q$ in the bottom. For electrons in a metal, $q$ has the size of the elementary charge $e$ .

Why is the drift so slow, but the light turns on at once? Electrons in a metal move in every direction at speeds far faster than the drift. Their random motions cancel out, so the net flow is only the slow drift, and that drift carries the current. The electric field that pushes the electrons travels along the wire at a large fraction of the speed of light. So every electron starts moving almost at once, even though each one moves only a tiny distance.

When the formula needs care. It assumes a steady current and one kind of carrier, with the same charge and density everywhere in the wire. In some materials, such as salt water or semiconductors, more than one kind of carrier carries current. Each kind then needs its own density, charge and drift speed, and the formula must be applied to each kind in turn.

An everyday picture

Think of a packed hallway where people shuffle forward. The current is the number of people who pass a doorway each second. If the hallway is crowded, each person can only move a tiny step at a time. The people push forward slowly, but a long line of people still delivers a steady flow at the far end. A narrower hallway means people must shuffle faster to keep the same flow, and a wider one lets them move more slowly.

Worked example: How fast electrons move in a 1 A copper wire

A copper wire with a diameter of 1.0 mm carries a steady current of 1 A. That is the current a 12 W LED lamp on a 12 V supply draws, since $12\ \text{W} / 12\ \text{V} = 1\ \text{A}$ . Copper has about $8.5 \times 10^{28}$ free electrons in each cubic meter. Find the drift velocity in meters per second and in millimeters per second. Then find how long one electron takes to drift 1 m.

Find the cross-sectional area. The radius is half the diameter, so $r = 0.5\ \text{mm}$ . A millimeter is $10^{-3}$ m, so $r = 5 \times 10^{-4}$ m. The area of a circle is $\pi r^2$ , and squaring a power of ten doubles its exponent: $(10^{-4})^2 = 10^{-8}$ , while $5^2 = 25$ . So $A = \pi \times 25 \times 10^{-8}\ \text{m}^2 = \pi \times 2.5 \times 10^{-7}\ \text{m}^2 \approx 7.85 \times 10^{-7}\ \text{m}^2$ .
Multiply $n$ , $q$ and $A$ to find the charge in each meter. First multiply $n$ and $q$ . The front numbers give $8.5 \times 1.602 \approx 13.62$ , and the powers of ten add, $10^{28} \times 10^{-19} = 10^{9}$ . So $nq \approx 13.62 \times 10^{9} = 1.362 \times 10^{10}\ \text{C/m}^3$ . Then multiply by the area: $nqA \approx 1.362 \times 10^{10} \times 7.854 \times 10^{-7} \approx 10.70 \times 10^{3} = 1.07 \times 10^{4}\ \text{C/m}$ .
Divide the current by that charge per meter. The formula is $v_d = \frac{I}{nqA} = \frac{1\ \text{A}}{1.07 \times 10^{4}\ \text{C/m}}$ . An ampere is one coulomb per second, so the coulombs cancel and leave meters per second. Dividing 1 by 1.07 gives about 0.935, and dividing by $10^{4}$ moves the decimal point four places to the left. So $v_d \approx 0.935 \times 10^{-4} = 9.35 \times 10^{-5}\ \text{m/s}$ .
Convert the speed to millimeters per second. A meter is 1000 millimeters, so a speed in meters per second is multiplied by 1000 to get millimeters per second. Multiplying the powers adds the exponents: $9.35 \times 10^{-5} \times 10^{3} = 9.35 \times 10^{-2}$ . So the drift velocity is $0.0935$ mm/s, which is about 0.09 mm/s, about a tenth of a millimeter every second.
Find the time for one electron to drift 1 meter. Time equals distance divided by speed: $t = \frac{1\ \text{m}}{9.35 \times 10^{-5}\ \text{m/s}}$ . Dividing 1 by 9.35 gives about 0.107, and dividing by $10^{-5}$ is the same as multiplying by $10^{5}$ . So $t \approx 0.107 \times 10^{5} = 1.07 \times 10^{4}$ seconds, about 10,700 seconds. There are 3600 seconds in an hour, so $10\,700 / 3600 \approx 2.97$ hours, close to 3 hours.
Check the result by comparing with walking. A walking pace is about 1.4 m/s. Dividing it by the drift speed, $1.4 / (9.35 \times 10^{-5}) \approx 15\,000$ , shows that the drift is about 15,000 times slower than a walk. Yet the current is large, because about $6.7 \times 10^{22}$ electrons pass each meter of wire, a number of billions of billions moving together to make one ampere.

Answer: The electrons drift at about 0.09 mm/s ( $9.35 \times 10^{-5}$ m/s), so one electron would take roughly 3 hours to drift 1 m. The electric signal still reaches the far end of the wire almost at once, because it travels much faster than the electrons drift.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Thinking the electrons move at the speed of the current or at the speed of the signal. The drift is slow, and the signal that tells a circuit to switch on travels much faster, at a large fraction of the speed of light.
Mixing up the units of $n$ . Copper has about $8.5 \times 10^{28}$ free electrons per $\text{m}^3$ , which is $8.5 \times 10^{22}$ per $\text{cm}^3$ . Since one cubic meter holds a million cubic centimeters, using the per-cubic-centimeter number in a formula that uses cubic meters makes the answer about a million times too large.
Using the diameter as the radius when finding the area. The area is $\pi r^2$ , so convert the diameter to a radius first by halving it. Using the diameter directly would make the area four times too large.
Assuming the formula gives the speed of random motion. The electrons move in random directions at much higher speeds, and only the small net drift makes up the current.

Where you will meet it

Explaining why a light switch turns a lamp on at once, even though the electrons in the wire move slowly. The signal travels fast, and the electrons do not need to travel far.
The Hall effect, where measuring a small voltage across a conductor in a magnetic field gives the carrier density $n$ and the sign of the carrier charge.
Checking the current density, $I/A$ , in wire design. It is the current per square meter of wire, and it is a key limit on how hot a wire can get.
Modeling how current flows in semiconductors, where more than one kind of charge carrier can carry the current, each with its own density and speed.

Questions people ask

How fast do electrons move in a wire?

Usually a fraction of a millimeter per second for ordinary currents. The exact value depends on the current, the size of the wire and the material.

If electrons are so slow, why does a light turn on instantly?

The electric field that pushes the electrons travels along the wire at a large fraction of the speed of light. Every electron starts moving almost at once, even though each one moves only a tiny distance.

Does drift velocity depend on the length of the wire?

No. The formula has no length in it. A longer wire takes a longer time for the drift to cover the distance, but the drift speed itself is the same.

Why is the formula written with $q$ and not $e$?

The letter $q$ stands for the charge of each carrier. For free electrons in a metal, $q$ has the size of $e$ . For charged atoms moving through salt water, the charge is different, so the letter $q$ lets the same formula cover both.

Charge and Current: Electric current is how much charge passes a point each second. The same charge moving in less time means a bigger current.
Ohm's Law: Ohm’s law says the push on a wire, called voltage, equals the flow through it, called current, times how hard the wire resists that flow, called resistance.
Resistivity: A wire’s resistance grows with its length, shrinks with its thickness, and depends on what the wire is made of.
Kirchhoff's Current Law: Kirchhoff’s current law says that the electricity flowing into a junction has to flow out again, because nothing piles up or disappears at a junction.