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Drift Velocity & Current

Key idea: Derive and use I = nAvq to link current, drift velocity, number density and cross-sectional area in exam-style A Level Physics questions and calculations.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Relate current to charge flow, number density and drift velocity.

1. Definitions (Must Know)

A. Electric current, I

Electric current, I, is the rate of flow of charge:

I = (Δ Q)/(Δ t)

Unit: ampere (A), where 1 A = 1 C s⁻¹.

B. Drift velocity, v

Drift velocity, v, is the average velocity of charge carriers along a conductor due to an electric field.

Unit: m s⁻¹.

C. Number density, n

Number density of charge carriers, n, is the number of charge carriers per unit volume.

Unit: m⁻³.

D. Charge per carrier, q, and cross-sectional area, A

  • Charge per carrier, q (C): for an electron, q = -1.60 × 10⁻¹⁹ C.
  • Cross-sectional area, A (m²): area perpendicular to the current.

2. Key Ideas (What Earns Marks)

  • For a uniform conductor with carriers moving with drift speed v: I = nAvq

  • In calculations, it’s usually safest to use magnitudes:

    • I = nAv|q|
    • and state the direction separately (electron drift is opposite to conventional current).

  • Unit check:

    • (m⁻³)(m²)(m s⁻¹)(C)
    • simplifies to C s⁻¹ = A
Direction reminder

Conventional current is defined in the direction of positive charge flow. In metals, the charge carriers are electrons, so electron drift is opposite to the current direction.

Exam pitfall: electron flow vs conventional current

Electron drift direction is opposite to conventional current. Keep this distinction explicit when drawing directions in metallic conductors.

3. Detailed Explanations

A. Derivation of I = nAvq (exam-ready)

Consider a conductor with cross-sectional area A and charge carriers of number density n moving with drift speed v.

In time Δ t, the carriers move a distance vΔ t, so the volume swept out is:

Δ V = A(vΔ t)

Number of carriers in that volume is:

N = nΔ V = nA(vΔ t)

Total charge passing the cross-section in time Δ t is:

Δ Q = Nq = nA(vΔ t)q

So the current is:

I = (Δ Q)/(Δ t); = (nA(vΔ t)q)/(Δ t); = nAvq

B. What this tells you physically

  • Large current can come from:
    • large n (many carriers per m³),
    • large A (thicker wire),
    • large v (faster drift),
    • large |q| (bigger charge per carrier).
  • In metals, n is huge, so the drift speed is usually very small.

4. Common Mistakes

  • Using A in mm² without converting to m².
  • Using n in cm⁻³ without converting to m⁻³.
  • Forgetting electron charge is negative (then mixing up directions).
  • Confusing drift speed with the signal/field propagation speed.

5. Exam Tips

  • If the question says “drift velocity”, it usually wants:
    • v = I/nA|q|
  • If you’re given diameter d of a wire:
    • A = π(d/2)² (convert d to metres first)
  • Always quote v with a direction statement if needed (e.g. “electron drift to the left”).

6. Worked Examples

Modelled example 1

Find drift speed in a copper wire

Core

Problem

A 2.0 A current flows through area 1.0 × 10⁻⁶ m² in copper with n = 8.5 × 10²⁸ m⁻³. Find electron drift-speed magnitude; use |q| = 1.60 × 10⁻¹⁹ C.
Study the worked solution

  1. Rearrange the carrier equation

    Method

    Use v = I/(nA|q|).

    Reason

    Current is charge crossing per second from nAv carriers per second.

    Working

    v = 2.0/((8.5 × 10²⁸)(1.0 × 10⁻⁶)(1.60 × 10⁻¹⁹))

  2. Evaluate and interpret

    Method

    Obtain 1.5 × 10⁻⁴ m s⁻¹.

    Reason

    The enormous carrier density permits a substantial current at small drift speed.

    Working

    v = 1.5 × 10⁻⁴ m s⁻¹.

Guided practice 2

Find current from drift speed

About 5 min

Problem

For n = 5.0 × 10²⁸ m⁻³, A = 2.0 × 10⁻⁶ m², v = 2.0 × 10⁻⁴ m s⁻¹ and |q| = 1.60 × 10⁻¹⁹ C, find current magnitude.

Try this before viewing the solution

Unit: A

Hints

Hint 1: group powers of ten carefully
First calculate nA = 1.0 × 10²³ m⁻¹.
View solution step by step

  1. Count carriers crossing

    Method

    Form nAv.

    Reason

    This gives the number of carriers crossing each second.

    Working

    (1.0 × 10²³)(2.0 × 10⁻⁴).

  2. Convert to charge rate

    Method

    Multiply by |q|.

    Reason

    Current is charge transferred per second.

    Working

    I = nAv|q| = 3.2 A

Common misconception 3

Count electrons from beam current

Find and correct the mistake

Learner claim

An electron beam has current 20.0 μA. A learner calculates It = 1.20 × 10⁻³ and reports that 1.20 × 10⁻³ electrons reach the screen in one minute. Diagnose the first error and find the number of electrons. Use e = 1.602 × 10⁻¹⁹ C.

Try this before viewing the solution

First error

View solution step by step

  1. Interpret current multiplied by time

    Method

    Calculate the magnitude of charge delivered in 60.0 s.

    Reason

    Current is the rate of charge flow, so Q = It.

    Working

    Q = (20.0 × 10⁻⁶)(60.0) = 1.20 × 10⁻³ C

  2. Convert charge to particle count

    Method

    Divide by the magnitude of the charge on one electron.

    Reason

    N electrons carry charge magnitude Ne.

    Working

    N = Q/e = (1.20 × 10⁻³)/(1.602 × 10⁻¹⁹) = 7.49 × 10¹⁵

Examiner practice 4

Use diameter to find drift speed

4 marks

Examination question

A copper wire of diameter 1.0 mm carries 1.5 A. Use n = 8.5 × 10²⁸ m⁻³ and |q| = 1.60 × 10⁻¹⁹ C to find drift speed. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Convert diameter

    1 mark

    Method

    Write d = 1.0 × 10⁻³ m.

    Reason

    Area must be in square metres.

    Working

    r = 5.0 × 10⁻⁴ m.

  2. Find area

    1 mark

    Method

    Use A = π(d/2)².

    Reason

    The conductor cross-section is circular.

    Working

    A = 7.85 × 10⁻⁷ m².

  3. Rearrange

    1 mark

    Method

    Use v = I/(nA|q|).

    Reason

    Drift speed is the unknown in the carrier equation.

    Working

    v = 1.5/(nA|q|).

  4. Evaluate

    1 mark

    Method

    Obtain 1.40 × 10⁻⁴ m s⁻¹.

    Reason

    All inputs now use SI units.

    Working

    v = 1.40 × 10⁻⁴ m s⁻¹.

Challenge 5

Time for an electron to drift 1 m (order of magnitude)

Minimal support

Scale-and-interpret transfer

At drift speed 2.0 × 10⁻⁴ m s⁻¹, estimate the time for an electron to drift 1.0 m and explain why a lamp still responds rapidly.

Try this before viewing the solution

Hints

Hint 1: separate particle drift from field propagation
Calculate t = s/v, then distinguish individual electron motion from establishment of the circuit’s electric field.
View solution step by step

  1. Estimate transit time

    Method

    Divide distance by drift speed.

    Reason

    The drift speed is the mean carrier progress along the wire.

    Working

    t = 1.0/(2.0 × 10⁻⁴) = 5.0 × 10³ s ≈ 1.4 h

  2. Interpret circuit response

    Method

    State that the electric-field change propagates through the circuit far faster than carrier drift.

    Reason

    Local electrons throughout the wire begin drifting; one source electron need not reach the lamp.

    Working

    Slow drift is compatible with rapid switching response.

7. Mind Stretchers

Mind stretcher 1: Why drift speed increases when the wire is thinnerExtension

For the same material (same n and |q|) and the same current I, show how v depends on A.

Show Answer

From I = nAv|q|: v = I/nA|q|

So v ∝ 1/A. A smaller cross-sectional area means a larger drift speed is needed to carry the same current.

Mind stretcher 2: Why can current be large when drift speed is small?Extension

Explain (using I = nAv|q|) how a wire can carry a large current even though the drift speed is tiny.

Show Answer

In metals, the number density n of free electrons is extremely large (typically around 10²⁸ m⁻³).

So even if v is small, the product nAv|q| can still be large because n (and often A) is large.

Mind stretcher 3: Optional (Enrichment)Extension

A. Drift speed vs signal speed (extra context)

In metals, drift speeds are typically very small (often ∼ 10⁻⁴ m s⁻¹), but electrical signals propagate through a circuit much faster because the electric field is established through the conductor almost immediately. This is why a light turns on quickly even though individual electrons drift slowly.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027