Mean Drift Velocity — OCR A Level Physics
Module 4 · Electrons, waves and photons

Mean Drift Velocity

Specification: OCR A H556  |  Section: 4.1.2  |  Focus: mean drift velocity of charge carriers, the equation I = nAve, and the distinction between conductors, semiconductors and insulators in terms of the number density of charge carriers.

By the end of this topic you should be able to…
  • define mean drift velocity and explain why electrons in a wire move slowly even when the current is large
  • derive and use the equation I = nAve to calculate current or drift velocity
  • explain the physical meaning of each term in the equation
  • distinguish between conductors, semiconductors and insulators based on the number density n of charge carriers
  • apply the equation to real exam-style problems involving wires, semiconductors and different materials

Big idea: current is not about how fast individual electrons zoom along. It is about how many charge carriers pass a point each second. A huge number of very slow electrons can give a large current.

Part 1

What is mean drift velocity?

When a potential difference is applied across a conductor, the free electrons inside experience a force and start to drift in one direction. However, they do not move in straight lines. They constantly collide with the vibrating ions in the lattice.

The mean drift velocity v is the average velocity of the charge carriers along the conductor, taking all these collisions into account.

Common misconception

Electrons in a wire do not travel at the speed of light. Their drift velocity is typically only a few millimetres per second. The electrical signal (the electric field) travels very quickly, but the charge carriers themselves move slowly.

Why is drift velocity so small?

  • electrons have very large random thermal velocities (around 106 m s−1)
  • they constantly collide with the positive ions in the metal lattice
  • after each collision they are scattered in random directions
  • the applied electric field only adds a tiny extra drift component on top of the random motion

Thermal motion

Random, very fast, in all directions. Averages to zero net movement.

Drift motion

Slow, directed, superimposed on the thermal motion. Responsible for the net current.

Part 2

Deriving the equation I = nAve

Consider a wire of cross-sectional area A containing free charge carriers each with charge e. The number of free charge carriers per unit volume is n (the number density). The charge carriers move with a mean drift velocity v along the wire.

Electron path through a metal lattice

Electron drift through lattice positive ions in lattice net drift electron
Electrons collide repeatedly with the lattice ions. The zigzag path shows the random thermal motion, while the dashed arrow shows the small net drift in the direction of conventional current.

Derivation of I = nAve

I = nAve derivation v L = v × t A volume = A × L = A × v × t
In time t, all charge carriers in the shaded volume sweep past the end of the wire. The number of carriers is n × A × v × t, each carrying charge e.
  1. In a time t, each charge carrier travels a distance L = v × t.
  2. The volume of wire swept out in time t is volume = A × L = A × v × t.
  3. The number of charge carriers in this volume is number = n × volume = n × A × v × t.
  4. The total charge passing a point in time t is Q = number × e = n × A × v × t × e.
  5. Current is charge per unit time: I = Q / t = n × A × v × e.
Key equationI = n A v e
SymbolQuantitySI unitTypical value (copper)
IcurrentA (ampere)depends on circuit
nnumber density of charge carriersm−3~8.5 × 1028 m−3
Across-sectional aream2~1 mm² = 1 × 10−6 m²
vmean drift velocitym s−1~10−4 m s−1
eelementary chargeC (coulomb)1.60 × 10−19 C
Part 3

Conductors, semiconductors and insulators

The value of n varies enormously between different materials. This is the main reason conductors, semiconductors and insulators behave so differently.

Material typen (m−3)ExamplesCharge carriers
Conductor~1028 – 1029copper, aluminium, silverfree electrons from the outer shells of atoms
Semiconductor~1016 – 1019silicon, germaniumelectrons and holes; increases sharply with temperature
Insulator~0 (effectively)rubber, glass, plasticalmost no free charge carriers
Exam tip

For the same current I and cross-sectional area A, a material with a smaller n must have a larger drift velocity v. This is why drift velocities in semiconductors are often much larger than in metals.

Why does n matter?

  • Conductors: huge n means even a tiny v gives a large current. Electrons are abundant.
  • Semiconductors: n is much smaller, so v must be larger to carry the same current. Heating or doping increases n dramatically.
  • Insulators: n ≈ 0, so effectively no current can flow even with a large potential difference.
Part 4

Worked examples

Worked example 1
A copper wire has a cross-sectional area of 2.0 mm² and carries a current of 5.0 A. Calculate the mean drift velocity of the free electrons in the wire. n for copper = 8.5 × 1028 m−3, e = 1.60 × 10−19 C.
Rearrange I = nAve for v: v = I / (n A e)
Convert A to m²: 2.0 mm² = 2.0 × 10−6 m²
Substitute: v = 5.0 / (8.5 × 1028 × 2.0 × 10−6 × 1.60 × 10−19)
v = 5.0 / (2.72 × 104)
v = 1.8 × 10−4 m s−1 (about 0.18 mm per second)
Worked example 2
A semiconductor has n = 1.0 × 1017 m−3. If the drift velocity is 5.0 m s−1 and the cross-sectional area is 1.0 mm², find the current.
Use I = n A v e
Convert A: 1.0 mm² = 1.0 × 10−6 m²
I = (1.0 × 1017) × (1.0 × 10−6) × (5.0) × (1.60 × 10−19)
I = 8.0 × 10−8 A = 80 nA
Worked example 3
Explain why the drift velocity in a semiconductor is typically much larger than in a metal for the same current and cross-sectional area.
From I = nAve, rearranging gives v = I / (nAe)
For the same I, A and e, v is inversely proportional to n
Semiconductors have n around 1016 – 1019 m−3, much smaller than metals (~1028 m−3)
Therefore v must be larger in semiconductors to carry the same current
Interactive

Drift velocity explorer

Change n, A and v to see how current I changes
Current I = 2.72 A
Charge carriers per second = 1.7 × 10¹⁹
Typical for household wiring

Try selecting the semiconductor option. Notice how the same current requires a much larger drift velocity when n is small.

Part 5

Common misconceptions

Thermal velocity vs drift velocity

Thermal vs drift velocity thermal velocity ~ 10⁶ m s⁻¹ drift velocity ~ 10⁻⁴ m s⁻¹ Drift velocity is about 10¹⁰ times smaller than thermal velocity (not to scale)
The drift velocity is tiny compared with the random thermal velocity. Electrons move very fast in random directions, but the net drift along the wire is extremely slow.

"Electrons travel at the speed of light"

No. The electric field propagates quickly, but drift velocity is only ~10−4 m s−1.

"Larger current means faster electrons"

Not necessarily. Current depends on n, A and e as well as v. A larger area or larger n can give more current at the same v.

"Insulators have no electrons"

They have electrons, but they are not free to move. n ≈ 0 for conduction.

"I = nAve only works for metals"

It works for any material with mobile charge carriers, including semiconductors and electrolytes. For ions, replace e with the ion charge q.

Knowledge

Knowledge Check

1
Define mean drift velocity.
2 marks
  • The average velocity of charge carriers along a conductor
  • Taking into account their random thermal motion and collisions
2
State the equation relating current to number density, cross-sectional area, drift velocity and charge.
1 mark
  • I = nAve
3
Explain why n is much smaller in semiconductors than in metals.
2 marks
  • In semiconductors, only a small fraction of electrons are free to move
  • Most electrons are bound in covalent bonds; only those excited across the band gap are free
4
A wire carries a current of 3.0 A. If the drift velocity is 1.5 × 10−4 m s−1 and the cross-sectional area is 4.0 mm², calculate n. e = 1.60 × 10−19 C.
3 marks
  • n = I / (A v e)
  • A = 4.0 × 10−6 m²
  • n = 3.0 / (4.0 × 10−6 × 1.5 × 10−4 × 1.60 × 10−19) = 3.1 × 1028 m−3
Exam Practice

Exam-Style Questions

1
A copper wire of diameter 0.80 mm carries a current of 2.5 A. Calculate the mean drift velocity of the electrons. n = 8.5 × 1028 m−3, e = 1.60 × 10−19 C. [3 marks] generated exam-style
3 marks
  • A = πr² = π × (0.40 × 10−3)² = 5.03 × 10−7 m²
  • v = I / (n A e) = 2.5 / (8.5 × 1028 × 5.03 × 10−7 × 1.60 × 10−19)
  • v = 3.7 × 10−4 m s−1
2
A silicon semiconductor has n = 1.5 × 1016 m−3. When the drift velocity is 80 m s−1, the current is 1.9 × 10−7 A. Calculate the cross-sectional area of the semiconductor. e = 1.60 × 10−19 C. [2 marks] generated exam-style
2 marks
  • A = I / (n v e) = 1.9 × 10−7 / (1.5 × 1016 × 80 × 1.60 × 10−19)
  • A = 9.9 × 10−7 m² (about 1 mm²)
3
Explain, in terms of n, why metals are much better conductors than semiconductors at room temperature. [2 marks] generated exam-style
2 marks
  • Metals have n ~ 1028 m−3 whereas semiconductors have n ~ 1016 m−3
  • Metals have many more free charge carriers per unit volume, so can carry much larger currents for the same v and A
Summary

Topic Summary

Mean Drift Velocity

Definition

Average velocity of charge carriers along a conductor, after accounting for collisions and random thermal motion.

Key equation

I = n A v e

Current depends on number density, area, drift velocity and charge.

Materials

Conductors: n ~ 1028 m−3

Semiconductors: n ~ 1016 m−3

Insulators: n ≈ 0

Key insight

Drift velocity is tiny (~mm/s) but current can be large because n is enormous in metals.

4.1.2(a) mean drift velocity 4.1.2(b) I = nAve 4.1.2(c) conductors, semiconductors, insulators