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What is electric current, and how is it related to the drift of charge carriers?

Charge and current: electric charge and the elementary charge, current as the rate of flow of charge, conservation of charge at junctions, and the mean drift velocity equation.

A focused answer to the OCR H556 content on charge and current, covering electric charge and the elementary charge, current as the rate of flow of charge, the conservation of charge at junctions, and the mean drift velocity equation that links current to carrier number density.

Generated by Claude Opus 4.811 min answer

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

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to define electric charge and the elementary charge, define current as the rate of flow of charge, apply the conservation of charge at a junction, and use the mean drift velocity equation I=nAveI = nAve to link current to the motion of charge carriers.

The answer

Charge and the elementary charge

Current as the rate of flow of charge

Conservation of charge

Mean drift velocity

In metals nn is very large (around 1028 m−310^{28}\ \text{m}^{-3}), so the drift velocity is tiny, often a fraction of a millimetre per second, even though the electric field (and the lamp lighting) responds almost instantly.

Examples in context

Fuse ratings depend on the heating effect of current, which is greater in a thin element where the drift velocity (and collisions) are higher. Semiconductor devices have a much smaller carrier density than metals, so for the same current the drift velocity is much larger. Electroplating and electrolysis use Q=ItQ = It to relate the charge passed to the mass of material deposited. The near-instant response of circuits, despite the slow drift, explains why a light comes on the moment the switch is closed.

Try this

Q1. State what is meant by an electric current. [1 mark]

  • Cue. The rate of flow of charge, I=ΔQΔtI = \frac{\Delta Q}{\Delta t}.

Q2. A current of 2.0 A2.0\ \text{A} flows for 30 s30\ \text{s}. Find the charge that passes. [1 mark]

  • Cue. Q=It=2.0×30=60 CQ = It = 2.0 \times 30 = 60\ \text{C}.

Q3. Explain, using I=nAveI = nAve, why current is the same at every point in a series circuit. [2 marks]

  • Cue. Charge is conserved, so the same charge per second flows through each component; the drift velocity adjusts to the local area and carrier density.

Exam-style practice questions

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

OCR 20184 marksA copper wire of cross-sectional area 1.5×10−6 m21.5 \times 10^{-6}\ \text{m}^2 carries a current of 3.2 A3.2\ \text{A}. Copper has a free-electron number density of 8.5×1028 m−38.5 \times 10^{28}\ \text{m}^{-3}. Calculate the mean drift velocity of the electrons. Take e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}.
Show worked answer →

Rearrange I=nAveI = nAve for the drift velocity: v=InAev = \frac{I}{nAe}.

Substitute: v=3.2(8.5×1028)(1.5×10−6)(1.6×10−19)v = \frac{3.2}{(8.5 \times 10^{28})(1.5 \times 10^{-6})(1.6 \times 10^{-19})}.

Denominator: (8.5×1028)(1.5×10−6)=1.275×1023(8.5 \times 10^{28})(1.5 \times 10^{-6}) = 1.275 \times 10^{23}; times 1.6×10−19=2.04×1041.6 \times 10^{-19} = 2.04 \times 10^{4}.

So v=3.22.04×104=1.6×10−4 m s−1v = \frac{3.2}{2.04 \times 10^{4}} = 1.6 \times 10^{-4}\ \text{m s}^{-1}. Markers reward rearranging the equation, correct substitution, and the value about 1.6×10−4 m s−11.6 \times 10^{-4}\ \text{m s}^{-1} (showing drift is surprisingly slow).

OCR 20213 marksA battery delivers a steady current of 0.45 A0.45\ \text{A} for 2.02.0 hours. Calculate the total charge that flows and the number of electrons this represents. Take e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}.
Show worked answer →

Charge: Q=It=0.45×(2.0×3600)=0.45×7200=3240 CQ = It = 0.45 \times (2.0 \times 3600) = 0.45 \times 7200 = 3240\ \text{C}, about 3.2×103 C3.2 \times 10^{3}\ \text{C}.

Number of electrons: N=Qe=32401.6×10−19=2.0×1022N = \frac{Q}{e} = \frac{3240}{1.6 \times 10^{-19}} = 2.0 \times 10^{22}.

Markers reward converting hours to seconds, the charge about 3.2×103 C3.2 \times 10^{3}\ \text{C}, and dividing by the elementary charge for the number of electrons.

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