A current of 0.50\ A flows for 2.0\ minutes. Find the charge that passes. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A copper wire of cross-sectional area 1.5 × 10^-6\ m^2 carries a current of 4.0\ A. The number density of free electrons in copper is 8.5 × 10^28\ m^-3. Calculate the drift velocity of the electrons. Take e = 1.6 × 10^-19\ C.

Rearrange I = nAvq for the drift velocity: v = InAq. v = 4.0(8.5 × 10^28)(1.5 × 10^-6)(1.6 × 10^-19). Denominator = 8.5 × 1.5 × 1.6 × 10^28-6-19 = 20.4 × 10^3 = 2.04 × 10^4. So v = 4.02.04 × 10^4 = 1.96 × 10^-4\ m s^-1, about 0.2\ mm s^-1. Markers reward rearranging I = nAvq, correct powers of ten, and the drift velocity about 2 × 10^-4\ m s^-1.

EnglandPhysicsSyllabus dot point

What is an electric current at the level of charge carriers, and how does the drift of electrons carry it through a conductor?

Conduction of electricity: electric current as the rate of flow of charge, the equation I = nAvq for charge carriers, drift velocity, and the distinction between conductors, semiconductors and insulators.

A focused answer to the Eduqas A-Level Physics Component 2 conduction content, covering electric current as the rate of flow of charge, the charge-carrier equation I = nAvq, drift velocity, and how the number density of free carriers distinguishes conductors, semiconductors and insulators.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Electric current is the rate of flow of charge, $I = \frac{\Delta Q}{\Delta t}$ , measured in amperes ( $1\ \text{A} = 1\ \text{C s}^{-1}$ ). In a conductor the current is carried by drifting charge carriers, described by $I = nAvq$ , where $n$ is the number density of carriers, $A$ the cross-sectional area, $v$ the mean drift velocity and $q$ the charge on each carrier. The drift velocity is tiny (well under a millimetre per second) even though the electric field, and so the signal, travels near the speed of light. A material is a conductor, semiconductor or insulator according to the number density $n$ of free carriers: metals have a huge $n$ , insulators almost none, and semiconductors lie in between with $n$ that rises sharply with temperature.

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What this dot point is asking

Eduqas wants you to define electric current as the rate of flow of charge, derive and use the charge-carrier equation $I = nAvq$ , explain what drift velocity is and why it is so small, and use the number density of free charge carriers to distinguish conductors, semiconductors and insulators.

The answer

Current as the rate of flow of charge

The charge-carrier equation

Drift velocity

Conductors, semiconductors and insulators

Charge carriers in a thin and a thick wire

A wire narrows from a cross-sectional area of $2.0\ \text{mm}^2$ to $0.50\ \text{mm}^2$ along its length, carrying a steady current. The drift velocity in the thick part is $1.0 \times 10^{-4}\ \text{m s}^{-1}$ . Find the drift velocity in the thin part.

step 1: Use conservation of current

The same current flows through both sections, and $n$ and $q$ are the same material, so $nA_1 v_1 q = nA_2 v_2 q$ , which gives $A_1 v_1 = A_2 v_2$ .

step 2: Rearrange for the thin-section drift velocity

$v_2 = v_1 \dfrac{A_1}{A_2} = (1.0 \times 10^{-4}) \times \dfrac{2.0}{0.50} = (1.0 \times 10^{-4}) \times 4.0$ .

step 3: Evaluate

$v_2 = 4.0 \times 10^{-4}\ \text{m s}^{-1}$ . The carriers drift four times faster where the wire is four times narrower.

Examples in context

The charge-carrier model underpins the design of wires and busbars (sized for current density to limit heating), and the behaviour of thermistors and light-dependent resistors used as sensors. Understanding that semiconductors have a temperature-sensitive carrier density is the basis of the entire electronics industry, from diodes and transistors to the silicon chips in every computer and phone.

Try this

Q1. Define electric current and state its unit. [2 marks]

Cue. The rate of flow of charge, $I = \frac{\Delta Q}{\Delta t}$ , measured in amperes (coulombs per second).

Q2. A current of $0.50\ \text{A}$ flows for $2.0\ \text{minutes}$ . Find the charge that passes. [2 marks]

Cue. $Q = It = 0.50 \times 120 = 60\ \text{C}$ .

Q3. State why the drift velocity in a semiconductor is much greater than in a metal carrying the same current. [1 mark]

Cue. A semiconductor has a far smaller number density $n$ of free carriers, so from $I = nAvq$ the drift velocity $v$ must be larger.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20194 marksA copper wire of cross-sectional area

1.5 \times 10^{-6}\ \text{m}^2

carries a current of

4.0\ \text{A}

. The number density of free electrons in copper is

8.5 \times 10^{28}\ \text{m}^{-3}

. Calculate the drift velocity of the electrons. Take

e = 1.6 \times 10^{-19}\ \text{C}

Show worked answer →

Rearrange $I = nAvq$ for the drift velocity: $v = \dfrac{I}{nAq}$ .

$v = \dfrac{4.0}{(8.5 \times 10^{28})(1.5 \times 10^{-6})(1.6 \times 10^{-19})}$ .

Denominator $= 8.5 \times 1.5 \times 1.6 \times 10^{28-6-19} = 20.4 \times 10^{3} = 2.04 \times 10^{4}$ .

So $v = \dfrac{4.0}{2.04 \times 10^{4}} = 1.96 \times 10^{-4}\ \text{m s}^{-1}$ , about $0.2\ \text{mm s}^{-1}$ .

Markers reward rearranging $I = nAvq$ , correct powers of ten, and the drift velocity about $2 \times 10^{-4}\ \text{m s}^{-1}$ .

Eduqas 20213 marksA metal wire and a semiconductor of the same dimensions carry the same current. Using the equation

I = nAvq

, explain why the drift velocity of the charge carriers is far greater in the semiconductor.

Show worked answer →

For the same current $I$ , area $A$ and carrier charge $q$ , the equation $I = nAvq$ rearranges to $v = \dfrac{I}{nAq}$ , so the drift velocity is inversely proportional to the number density $n$ of free charge carriers.

A semiconductor has a much smaller number density of free carriers than a metal (perhaps a factor of $10^{6}$ or more), so to carry the same current each carrier must move much faster: a far greater drift velocity.

Markers reward $v \propto \frac{1}{n}$ from $I = nAvq$ , stating the semiconductor has fewer free carriers, and concluding the drift velocity must be larger.

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)