Current, charge flow and drift velocity

Key idea: Current measures charge flow rate. The drift model connects that circuit-scale quantity to the slow average motion of many charge carriers.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 1 of 3

Check your understanding

By the end of this lesson, you should be able to

Use I = Q/t for steady or average current.
Derive I = nAvq from a carrier model.
Distinguish carrier drift speed from the much faster establishment of an electric field around a circuit.

Learn the idea

Big question: If electrons drift slowly, why does a lamp respond almost as soon as the switch closes?

Define current at a chosen cross-section

Current is charge crossing a chosen section per unit time: I = Q/t for a steady or average current. Conventional current follows the direction positive charge would move; electron drift in a metal is opposite.

Current is not used up by a component. In a steady series circuit, charge does not build up inside a lamp, so the same charge per second enters and leaves it.

Check your understanding: A current of 2 A enters a lamp steadily. How much current leaves it?

2 A. Energy is transferred in the lamp, but charge flow is conserved.

Build the drift equation

In time t, carriers with drift speed v move distance vt. The cylinder of carriers crossing area A has volume Avt, contains nAvt carriers and carries charge nAvtq. Dividing by time gives I = nAvq, with q as carrier-charge magnitude.

Drift can be slow because the carrier density is enormous. The lamp responds quickly because an electric field is established around the whole circuit; it does not wait for one electron to travel from switch to lamp.

Check your understanding: If wire area doubles while current, n and q stay fixed, what happens to drift speed?

It halves because v = I/(nAq).

Key ideas

Current is not used up by a component.
Convert mm² to m² before using the drift equation.
State whether q is charge magnitude when discussing electrons.

Relationships to know

I = Q/t
Q = nAvtq
I = nAvq

Follow the reasoning

Worked example

Find drift speed with a careful area conversion

Question: A 1.8 A current flows in a metal wire of diameter 0.80 mm. The electron number density is 8.5 × 10²⁸ m⁻³. Find drift speed using e = 1.60 × 10⁻¹⁹ C.

Step 1: Convert radius and area
Why: The equation requires area in m², and diameter must first be halved.
Working: r = 0.40 mm = 4.0 × 10⁻⁴ m; A = πr² = 5.03 × 10⁻⁷ m².
Step 2: Rearrange the carrier equation
Why: All contributing carriers are represented by n, A and q.
Working: v = I/(nAq).
Step 3: Substitute
Why: Keeping powers of ten visible avoids confusing mm with mm².
Working: v = 1.8/[(8.5 × 10²⁸)(5.03 × 10⁻⁷)(1.60 × 10⁻¹⁹)] = 2.63 × 10⁻⁴ m s⁻¹.

Answer: The electron drift speed is about 2.6 × 10⁻⁴ m s⁻¹, opposite to conventional current.

Check: The very small speed is plausible because about 10²⁹ carriers occupy each cubic metre.

Now try it with support

Practise with support

A steady current of 0.80 A flows for 3.0 min. Find the charge that passes.

Hints

Convert minutes to seconds.
Rearrange I = Q/t.

View the guided answer

t = 180 s, so Q = It = 0.80(180) = 144 C.

Your turn

Practise independently

Find the drift speed in a 1.0 mm² wire carrying 2.0 A if n = 8.5 × 10²⁸ m⁻³ and q = 1.60 × 10⁻¹⁹ C.

Check your answer

A = 1.0 mm² = 1.0 × 10⁻⁶ m². v = I/(nAq) = 2.0/[(8.5 × 10²⁸)(1.0 × 10⁻⁶)(1.60 × 10⁻¹⁹)] = 1.47 × 10⁻⁴ m s⁻¹.

Common mistakes and exam guidance

Watch out for

Using 1 mm² = 10⁻³ m².
Equating drift speed with the speed at which the circuit responds.

In an exam

Show the carrier-volume step when asked to derive I = nAvq.
Include the area conversion explicitly; it is a frequent source of powers-of-ten errors.

Put the ideas together

Exam-style practice [6 marks]

A copper wire of cross-sectional area 1.2 mm² carries 3.0 A. Take n = 8.5 × 10²⁸ m⁻³ and e = 1.60 × 10⁻¹⁹ C. Calculate electron drift speed and the charge passing in 4.0 min. Explain why the circuit can respond much faster than the drift time along the wire.

Plan before you answer

Convert mm² and minutes separately.
Use I = nAvq and Q = It.
Distinguish field establishment from carrier drift.

View the marking points and model answer

Marking points

Uses A = 1.2 × 10⁻⁶ m².
Obtains v ≈ 1.84 × 10⁻⁴ m s⁻¹.
Uses t = 240 s.
Obtains Q = 720 C.
States the electric field/signal is established around the circuit rapidly.
States individual electrons need not travel from source to component.

Model answer

v = 3.0/[(8.5 × 10²⁸)(1.2 × 10⁻⁶)(1.60 × 10⁻¹⁹)] = 1.84 × 10⁻⁴ m s⁻¹. In 4.0 min, Q = It = 3.0(240) = 720 C. Closing the switch establishes an electric field around the circuit rapidly, so nearby carriers everywhere begin drifting; no electron has to cross the whole circuit first.

Finish from memory

Three-question recap

Define electric current.
Check
Rate of flow of charge through a cross-section.
How does electron drift direction compare with conventional current?
Check
It is opposite.
What does n mean in I = nAvq?
Check
Number of mobile charge carriers per unit volume.

Try this next

Move from charge per second to energy per coulomb in the potential-difference lesson.

Continue to Potential difference and electromotive force Return to Currents

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027