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Unit VIII · Electric current (lesson 17)
12 min readConcept sheetSection A MCQ (1 mark); Section B short answer (3 marks): define drift velocity and derive I = nAev_d, or show that drift leads to Ohm's law; also a NEET favourite

Electric current, free electrons and drift velocity

Current is the rate of flow of charge, I = Δq/Δt. Free electrons drift slowly against the field, giving I = nAev_d.

Electric currentI = Δq/Δt (average), I = dq/dt (at an instant); SI unit ampere, 1 A = 1 C s⁻¹
Drift velocityv_d = eEτ/m, opposite to E for electrons; about 10⁻⁴ m s⁻¹, while random speeds are about 10⁶ m s⁻¹
Current and driftI = nAev_d; n = free electrons per m³, A = area of cross-section, e = 1.6 × 10⁻¹⁹ C
Resistivity from driftρ = m/(ne²τ); τ = average time between collisions, m = mass of an electron

Free and bound electrons, and current

In an atom, electrons are held to the nucleus by electric attraction. The outermost (valence) electrons are held most weakly. In a metal they leave their atoms and move freely through the solid: these are free electrons. The inner electrons stay bound to their atoms. A potential difference across a wire sets up an electric field inside it, and the free electrons move against the field. This flow of charge is an electric current: the rate of flow of charge through a cross-section, I = Δq/Δt, or I = dq/dt if it changes. Its unit is the ampere: 1 A = 1 C s⁻¹. By convention, current flows the way a positive charge would move, which is opposite to the electron flow. In semiconductors, both electrons and holes carry current.

Derivation: drift velocity v_d = eEτ/m

A wire has free electrons, each of charge e and mass m. A field E acts inside it.

  1. 1
    With no field, the free electrons dash about at random at about 10⁶ m s⁻¹, colliding with the atoms. Their average velocity is zero, so there is no current.
  2. 2
    With a field, each electron feels a force eE, opposite to E. Its acceleration is a = eE/m.
  3. 3
    Between two collisions it gains extra velocity against the field. Each collision throws away this gain and randomises its direction again.
  4. 4
    Let τ be the average time between collisions. The average extra velocity is then aτ.
  5. 5★
    So, the drift velocity is v_d = eEτ/m, opposite to E. It is only about 10⁻⁴ m s⁻¹.
  6. 6
    The energy the electrons hand over in collisions makes the atoms vibrate more, so the wire heats up.

★ key step

Derivation: current I = nAev_d

Take a wire of area of cross-section A with n free electrons per unit volume, all drifting with speed v_d.

  1. 1
    In a time Δt, each electron moves a distance v_d Δt along the wire.
  2. 2
    So every electron within a length v_d Δt of the area A crosses it in that time. The volume of this slice is A v_d Δt.
  3. 3
    Number of electrons crossing = n A v_d Δt. Charge crossing = e n A v_d Δt.
  4. 4
    Current = charge ÷ time: I = e n A v_d Δt ÷ Δt.
  5. 5★
    So, I = nAev_d: the current is proportional to the drift velocity.

★ key step

Derivation: Ohm's law from the drift of electrons

  1. 1
    Put v_d = eEτ/m into I = nAev_d: I = nAe(eEτ/m) = (ne²τA/m)E.
  2. 2
    For a wire of length l with a potential difference V across it, E = V/l.
  3. 3
    So I = (ne²τA/m)(V/l), which gives V/I = ml/(ne²τA).
  4. 4★
    For a given wire at a fixed temperature, n, τ, m, l and A are all constant, so V/I is constant. This is Ohm's law, with R = ml/(ne²τA).
  5. 5
    Compare with R = ρl/A: the resistivity is ρ = m/(ne²τ), and the conductivity is σ = 1/ρ = ne²τ/m.

★ key step

How to draw the drift diagram

  • Draw a cylinder for the wire. Shade a slice of length v_d Δt next to one cross-section, and label its area A.
  • Draw a long arrow along the wire and label it E. Draw the current arrow I in the same direction.
  • Draw a short arrow the other way and label it v_d (electrons).
  • To show random motion, draw the zigzag path of one electron that slowly shifts opposite to E.

Worked example: drift speed in a copper wire

Problem

A copper wire of area of cross-section 2.0 mm² carries a current of 3.4 A. Copper has 8.5 × 10²⁸ free electrons per m³. Find the drift speed of the electrons. How long does an electron take to drift 1 m? (e = 1.6 × 10⁻¹⁹ C)

Working
  1. Given: I = 3.4 A, A = 2.0 mm² = 2.0 × 10⁻⁶ m², n = 8.5 × 10²⁸ m⁻³.
  2. Formula: I = nAev_d, so v_d = I/(nAe).
  3. neA = 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁶ = 27.2 × 10³ = 2.72 × 10⁴ C m⁻¹.
  4. v_d = 3.4 A ÷ (2.72 × 10⁴ C m⁻¹) = 1.25 × 10⁻⁴ m s⁻¹.
  5. Time to drift 1 m: t = 1 m ÷ (1.25 × 10⁻⁴ m s⁻¹) = 8.0 × 10³ s, a little over 2 hours.
Answer

v_d = 1.25 × 10⁻⁴ m s⁻¹, about 1.3 × 10⁻⁴ m s⁻¹ (0.13 mm per second); about 8.0 × 10³ s to drift 1 m

In real life

A ceiling fan starts the moment you press the switch, even though electrons drift less than a millimetre per second. The electric field travels along the wiring almost at the speed of light and sets all the free electrons moving together.

Exam angle

1 marks: “What is the SI unit of electric current?” The ampere (A): 1 A = 1 C s⁻¹.

2 marks: “2.5 × 10¹⁸ electrons pass through a cross-section of a wire in 2 s. Find the current. (e = 1.6 × 10⁻¹⁹ C)” q = Ne = 2.5 × 10¹⁸ × 1.6 × 10⁻¹⁹ = 0.40 C. I = q/t = 0.40 ÷ 2 = 0.20 A, flowing opposite to the electrons.

3 marks: “Define drift velocity. Derive the relation between the current in a wire and the drift velocity of its electrons.” Drift velocity is the average velocity with which free electrons move against the field. In time Δt, the electrons within a length v_dΔt cross area A: their number is nAv_dΔt and their charge nAev_dΔt. So I = nAev_d.

Remember it

I = nAev_d, read as 'nave' with A: number density, Area, electron charge, drift velocity. Slow drift, fast signal.