Electric current and drift velocity

A lamp that lights at once

Press the switch of a lamp. The bulb glows at once, although the other end of the wire may be far away. Yet the electrons inside a wire creep along slower than a snail.

How can both be true? To answer, we must look inside a metal.

Current

A current is a flow of charge. If a charge $Q$ crosses a section of a wire in a time $t$, the steady current is

The unit is the ampere (A), which is one coulomb each second. Charge comes in lumps. One electron carries $e = 1.6 \times 10^{-19}$ C, so the number of electrons that cross is $N = Q/e$. A current of $1$ A means $6.25 \times 10^{18}$ electrons each second.

By an old habit, the conventional current is drawn from the positive terminal of a battery, round the circuit, to the negative terminal. In a metal the carriers are electrons, so they really move the other way.

Inside a metal

A metal is a lattice of positive ions with free electrons moving among them. In copper each atom gives about one free electron, so every cubic metre holds $n = 8.5 \times 10^{28}$ of them.

With no field, the free electrons move at random, at about a million metres each second. They hit the ions again and again, and they change direction at every hit. For every electron going one way, another goes the other way. So nothing flows, and the current is zero.

The drift

Now join the wire to a battery. A field $E$ appears along the wire. It pushes each electron with a force $eE$, against the field. So each electron gets an acceleration $a = eE/m$, where $m = 9.1 \times 10^{-31}$ kg is the mass of an electron.

An electron speeds up only until its next hit. Then it loses its extra speed and starts again. The average time between two hits is the relaxation time $\tau$. For copper $\tau$ is about $2.5 \times 10^{-14}$ s.

So on average each electron has an extra velocity $a\tau$, against the field. This is the drift velocity:

The random jumps still cancel. What is left is a slow, steady drift. The picture follows one electron. It zigzags, but on the whole it moves one way.

The drift speed in each unit of field is the mobility, $\mu = v_d/E = e\tau/m$. Its unit is m²/(V s). For copper,

The current in terms of the drift

Take a wire of cross section $A$. In a time $t$, every free electron within a distance $v_dt$ of a section crosses it. They fill a volume $Av_dt$, so there are $nAv_dt$ of them. Their charge is $nAv_dt \times e$. Dividing by $t$:

The current for each unit of area is the current density, $J = I/A = nev_d$. Its unit is A/m².

A worked example

A copper wire of cross section $2.5\ \text{mm}^2$ carries $10$ A. How fast do the electrons drift?

1. The area. $2.5\ \text{mm}^2 = 2.5 \times 10^{-6}\ \text{m}^2$.
2. The drift speed. We want $v_d$, so $v_d = \dfrac{I}{nAe}$. The bottom is $8.5 \times 10^{28} \times 2.5 \times 10^{-6} \times 1.6 \times 10^{-19} = 3.4 \times 10^4\ \text{C/m}$. So $v_d = \dfrac{10\ \text{A}}{3.4 \times 10^4\ \text{C/m}} = 2.9 \times 10^{-4}\ \text{m/s}$.
3. In words. That is $0.29$ mm each second. An electron needs nearly an hour to drift one metre.

Slide the current up to $30$ A. Does the drift speed ever reach even $1$ cm each second? In the thinnest wire it gets only to about $4$ mm each second, and that wire would be very hot.

Why does the lamp light at once?

The wire is full of free electrons everywhere, in the filament too. When the switch closes, the field spreads along the wire at nearly the speed of light, about $3 \times 10^8$ m/s. Every electron in the wire starts to drift at almost the same moment.

Think of a pipe full of water. Push the water in at one end, and water comes out at the other end at once. The drops that come out are not the ones you pushed in. It is the same in a wire: the electrons that light the filament were already in it.

Practise with new numbers

Try it in Python

The program finds the drift speed in a copper wire for four currents. It then finds how long an electron needs to drift across a room.

Common mistakes

• Thinking the drift speed is the speed of the signal. The field travels at nearly the speed of light. The electrons drift less than a millimetre each second.
• Thinking electrons are used up. They are not. The same electrons go round the circuit, and the number that enters a bulb equals the number that leaves.
• Mixing mm² and m². $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$.
• Using the diameter in place of the area. The area of a round wire is $\pi d^2/4$.
• Drawing the current the wrong way for electrons. The electrons drift against the conventional current.

Remember these

• $I = Q/t$. The number of electrons is $N = Q/e$, with $e = 1.6 \times 10^{-19}$ C.
• Without a field, the random motions cancel and there is no current.
• Drift velocity: $v_d = \dfrac{eE}{m}\tau$. Mobility: $\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m}$.
• $I = nAev_d$ and $J = nev_d$.
• In copper the drift speed is a fraction of a millimetre each second, but the field spreads at nearly the speed of light.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A current of $3.2$ A flows for $5.0$ s. How much charge crosses a section of the wire, and about how many electrons?

2. Two copper wires carry the same current. Wire B has half the cross section of wire A. How do the drift speeds compare?

3. In a field of $0.20$ V/m, copper electrons have a drift speed of $0.88$ mm/s. What is their mobility, in m²/(V s)?

4. A wire carries a conventional current to the east. Which way do the electrons drift, and which way does the field point?