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Electric Current and Resistance · Lesson 2 of 6

Drift Velocity

15 min read4 ideas7 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 4
Free electrons and drift
Maths you need

A gas of free electrons

In a metal each atom releases one or two of its outer electrons to the whole crystal. These free electrons are bound to no atom. They wander through the lattice of fixed positive ions, like the molecules of a gas in a box. Copper has about \(n = 8.5 \times 10^{28}\) free electrons in every cubic metre. That is one for each atom.

The free electrons are never still. At room temperature they move at speeds of the order of \(10^5\,\text{m/s}\), and they change direction at every collision with a vibrating ion. Yet a copper wire lying on a table carries no current. For every electron that crosses a section one way, another crosses it the other way. The net charge carried is zero.

Switching on a field

Now keep a field \(\vec E\) inside the wire, with a cell across its ends. Each electron feels a force \(-e\vec E\), against the field. Between two collisions it gains a little velocity against \(\vec E\). At the next collision its velocity is thrown into a random direction again. So the gain never grows large; it is restarted at every collision.

One electron among the ions, with and without a fieldA free electron, the blue disc, darts among the fixed ions of a metal, the grey dots, and changes direction at every collision; the pale blue line is its path. First chip, no field: the electron wanders at high speed, but after 12 seconds of the slowed-down motion it has got nowhere in particular; the purple arrow from the start to the electron stays short. Second chip, a field E pointing right: the same random flights, but each one bends a little to the left, against the field. The path leans left and the arrow grows steadily: a slow drift laid on a fast random motion, restarted at every collision. The drift is hugely exaggerated here; in copper it is about a billion times slower than the random motion. no field: no net drift
Fig 2.1One free electron among the ions, slowed down enormously. Play it with the field off, then with the field on. The random flights are the same in both. With the field on, each flight bends slightly to the left, and the arrow from the start grows steadily.

Averaged over all the electrons, this small lean is a steady velocity against the field. It is the drift velocity \(\vec v_d\). Fig 2.1 exaggerates it hugely. In copper the drift is about a billion times slower than the random motion. The drift is a tiny steady bias laid on a fast random walk.

Deriving \(\vec v_d = -\dfrac{e\tau}{m}\,\vec E\) step 1 of 4
1

\[ \vec v_i = \vec u_i - \dfrac{e\vec E}{m}\,t_i \]

Why. electron i left its last collision with a random velocity \(\vec u_i\) and has been accelerated by \(-e\vec E/m\) for a time \(t_i\) since.

RememberDrift velocity and mobility
\[ \begin{gathered} v_d = \dfrac{eE\tau}{m} \\ \mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m} \end{gathered} \]
Holds whenfree electrons in a metal; the drift is against \(\vec E\); \(\tau\) is the relaxation time, the average time since the last collision.
Symbols\(v_d\) drift speed, \(E\) field in the conductor, \(\tau\) relaxation time, \(e\) and \(m\) charge and mass of the electron, \(\mu\) mobility
Units\(v_d\) in m/s, \(E\) in V/m, \(\tau\) in s, \(\mu\) in m²/(V s)

The drift speed is proportional to the field, not to how long the field has acted. That is the work of the collisions, and it is the seed of Ohm's law in the next lesson. The drift speed per unit field is the mobility \(\mu = e\tau/m\). A carrier that collides seldom has a long \(\tau\) and a large mobility.

Common mistake

Some books take \(\tau\) as the mean time between two collisions and write \(v_d = eE\tau/2m\). This book, like the NCERT, takes \(\tau\) as the relaxation time and writes \(v_d = eE\tau/m\). Never put a \(\tau\) taken from one convention into the formula of the other.

ExampleDrift speed and mobility

In a metal the relaxation time is \(\tau = 2.25 \times 10^{-14}\,\text{s}\). Find the drift speed of the electrons in a field of \(100\,\text{V/m}\), and their mobility.

Try it on paper first. The solution is one tap away when you are ready.
Checkpoint · select all that applyUnanswered

Inside a wire the field \(\vec E\) points from end P to end Q. Which statements are true?

Checkpoint · in your own wordsUnanswered

The field inside the wire is constant, so each free electron feels a constant force. Why do the electrons not speed up without limit?

Current from the drift speed
Maths you need

From drift speed to current

Now count the charge. Take a wire of section \(A\) with \(n\) free electrons per unit volume, all drifting at \(v_d\). In a time \(t\) each electron drifts a distance \(v_d t\). So every electron within \(v_d t\) of a section, on the upstream side, crosses it in that time. Those electrons fill a cylinder of volume \(A v_d t\).

Counting the charge that crosses a faceA straight wire of cross-section A with its free electrons drifting to the left at the drift speed v_d. A dashed face is drawn across the wire. The shaded cylinder to the right of the face has length v_d t: in a time t every electron inside it, and no other, crosses the face. It holds n A v_d t electrons, so the charge through the face in time t is n e A v_d t, and the current is n e A v_d. face of area A vd vd t in time t: q = n e A vd t
Fig 2.2The electrons in the shaded cylinder, of length \(v_d t\) and section \(A\), are exactly the ones that cross the dashed face in a time \(t\).
Deriving \(I = neAv_d\) step 1 of 4
1

\[ N = n\,A\,v_d\,t \]

Why. the cylinder of Fig 2.2 has volume \(A v_d t\) and holds \(n\) free electrons per unit volume.

RememberCurrent and drift speed
\[ \begin{gathered} I = neAv_d \\ J = nev_d \end{gathered} \]
Holds whenany conductor with one kind of carrier, \(n\) per unit volume, all drifting at \(v_d\); for electrons \(\vec J\) points opposite to \(\vec v_d\)
Symbols\(I\) current, \(J\) current density, \(n\) number of free electrons per unit volume, \(e\) their charge in size, \(A\) section, \(v_d\) drift speed
Units\(n\) in m⁻³, \(A\) in m², \(v_d\) in m/s, \(I\) in A
ExampleHow slowly do the electrons drift?

A copper wire of section \(1\,\text{mm}^2\) carries \(1.7\,\text{A}\). Copper has \(8.5 \times 10^{28}\) free electrons per m³. Find the drift speed, and the time an electron takes to drift \(1\,\text{m}\) along the wire.

Reveal one step at a time and finish the rest yourself.
Checkpoint · numericUnanswered

A metal wire of section \(5\,\text{mm}^2\) carries \(10\,\text{A}\). The metal has \(8 \times 10^{28}\) free electrons per m³. Find the drift speed, in millimetres per second.

mm/s± 3 %
CheckpointUnanswered

The same current flows through a copper wire of half the diameter. How does the drift speed change?

Why the bulb lights at once

Why the bulb lights at once

At \(1.25 \times 10^{-4}\,\text{m/s}\), an electron would need \(5/(1.25 \times 10^{-4}) = 40\,000\,\text{s}\) to cross a \(5\,\text{m}\) room. That is about \(11\) hours. Yet a bulb lights the instant the switch closes. The electrons from the switch never need to reach the bulb.

The wire is already full of free electrons along its whole length, the filament included. Closing the switch creates the field along the circuit at nearly the speed of light. The electrons already inside the filament start to drift almost at once, together with those near the switch.

Intuition

A garden pipe already full of water gives water at the far end the moment the tap opens. No drop has travelled from the tap to the end. The push travelled, and the water already in the pipe moved.

Property
What it does
Its size in a copper wire
Random motion of a free electron
carries no current
about \(10^5\,\text{m/s}\)
Drift of the free electrons
carries the current
about \(10^{-4}\,\text{m/s}\)
Spread of the field along the circuit
switches the current on everywhere
nearly \(3 \times 10^8\,\text{m/s}\)
Checkpoint · in your own wordsUnanswered

The drift speed in a wire is of the order of \(0.1\,\text{mm/s}\). Why does a bulb light as soon as the switch is closed?

Drift through a narrowing wire
Predict firstLocked

A copper wire carrying a steady current narrows to half its cross-section. How does the drift speed in the narrow part compare with that in the wide part?

Electrons drifting through a narrowing wireA copper wire carrying 1.7 amperes narrows from part 1, of cross-section 2 square millimetres, to part 2, whose cross-section is set by a slider as a fraction of part 1. Press play: the free electrons, the blue dots, drift to the left, against the field E. They move slowly in the wide part and faster in the narrow part, because the same number must pass every section each second. The readouts give the drift speed in each part and the ratio of the drift speeds, which is also the ratio of the current densities and of the fields. At half the section the electrons drift twice as fast, 0.125 against 0.063 millimetre per second. The drift is shown hugely speeded up.E I I I = 1.7 A through both parts vd = 0.063 mm/s vd = 0.125 mm/s v2/v1 = J2/J1 = E2/E1 = 2.00

In the steady state, charge cannot pile up anywhere inside the wire. So the current \(I\) is the same through every section, wide or narrow. Then \(J = I/A\) is larger where \(A\) is smaller, and so is \(v_d = J/ne\). The field must be larger there too, because \(v_d = eE\tau/m\) with the same \(\tau\).

In the figure above, part 1 has section \(2\,\text{mm}^2\) and part 2 has \(1\,\text{mm}^2\), with \(1.7\,\text{A}\) through both. The current is in the ratio \(1 : 1\). The drift speed, the current density and the field are all in the ratio \(1 : 2\). The drift is \(0.063\,\text{mm/s}\) in part 1 and \(0.125\,\text{mm/s}\) in part 2.

Checkpoint · select all that applyUnanswered

A steady current \(I\) flows through a wire whose part 1 has section \(2A\) and part 2 has section \(A\). Which ratios, part 1 to part 2, are correct?

Story · Paul Drude, 1900
Electrons as a gas in a box

In 1900 the German physicist Paul Drude had just heard of the electron, discovered three years before. He pictured the electrons in a metal as a gas, flying freely between collisions with the fixed ions. From that picture alone he derived Ohm's law. He also explained why good conductors of electricity are good conductors of heat. His model needed quantum mechanics for its details, but its drift and its relaxation time are still how engineers think about wires.

Exam traps
  1. Taking the drift speed to be the speed of the free electrons.
    Free electrons move at about \(10^5\,\text{m/s}\) at random; the drift is only their small average velocity, about \(10^{-4}\,\text{m/s}\).
  2. Drawing the drift along the field.
    Electrons carry negative charge, so they drift against \(\vec E\) and against the current.
  3. Halving the drift speed when the diameter is halved.
    The area goes as the diameter squared: half the diameter, four times the drift.
  4. Using \(eE\tau/2m\) with the relaxation time of this book.
    With \(\tau\) the relaxation time, \(v_d = eE\tau/m\).