Charge flow, current and drift velocity
Key idea: H2 Physics lessons on drift velocity, electrical energy, sinusoidal supplies and half-wave rectification.
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The core idea
Build the idea
Learn the idea
Big question: How can slow carrier drift produce an immediate circuit current?
Current is charge flow rate, I = ΔQ/Δt. In a conductor I = nAqvᵈ, where n is carrier number density, A cross-sectional area, q carrier charge magnitude and vᵈ mean drift speed. The electric field is established around the circuit rapidly even though individual electrons drift slowly opposite conventional current.
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 to keep
- Do not confuse carrier density with total number of carriers.
- Use cross-sectional area perpendicular to the flow.
- Conventional current follows positive-charge direction.
See 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.
Another worked model
Question
Derive I = nAvq for carriers crossing a wire section in time Δt.
Check the worked solution
In Δt, carriers within length vΔt cross area A. Their number is nAvΔt and charge magnitude is ΔQ = nAvqΔt. Therefore I = ΔQ/Δt = nAvq. For electrons, conventional current is opposite their drift direction.
Use a hint if needed
Practise with support
Try this
The carrier number density doubles at fixed A, q and current. State the drift-speed factor.
Hint: Hold every named quantity except n and v fixed.
Check your answer
From I = nAvq, v is inversely proportional to n, so it halves.
Now work without the hint
Practise independently
Your turn
Explain the meanings and directions in I = nAvq, including why slow drift does not imply a slow circuit response.
Check your answer
n is carrier number per unit volume, A is perpendicular area, v is mean drift speed and q is carrier-charge magnitude. Electron drift is opposite conventional current. Drift is slow, while the electric-field change that establishes current propagates through the circuit much faster.
Avoid these traps
Common mistakes
Common mistake
Current is the amount of charge in a wire.
What is wrong with this reasoning?
Show better thinking
Current is the rate of charge flow: I = ΔQ/Δt.
Common mistake
Electron drift and conventional current point in the same direction.
What is wrong with this reasoning?
Show better thinking
Because electrons are negative, their drift direction is opposite conventional current.
Write for the examiner
Exam guidance
Write every quantity with SI units; microscopic-current questions often hide powers of ten in area or density.
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.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
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.
Come back in three days
Check what stayed with you
Recall question 1
Define electric current.
Check the answer
Rate of flow of charge through a cross-section.
Recall question 2
How does electron drift direction compare with conventional current?
Check the answer
It is opposite.
Recall question 3
What does n mean in I = nAvq?
Check the answer
Number of mobile charge carriers per unit volume.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 15 states no explicit exclusions. q in I = nAvq is treated as carrier-charge magnitude when calculating current magnitude; electron drift is opposite conventional current. E.m.f. is energy supplied per unit charge, while p.d. is energy transferred from electrical form per unit charge. The peak/√2 and half-maximum-power results are restricted to sinusoidal waveforms and a resistive load; a single ideal diode gives unsmoothed half-wave rectification.
- GCE A-Level H2 PhysicsTopic 15(a) / Topic 15(b) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027