Physics 9702 · Paper 1

I = nAvq & Drift Velocity

The electrons in a lamp wire drift at about 0.01 mm per second. At that speed, a single electron would take half a day to cross a table.
The light still comes on at once, because electrons everywhere in the wire start moving at the same time. This lesson builds the equation that tells us how slow they drift: $I = nAvq$ .

Building I = nAvq

Take a piece of wire of length $L$ and cross-section area $A$ . Suppose every free electron drifts at speed $v$ .
In a time $t$ , every electron inside a length $vt$ of wire passes the end point. That length of wire has volume $A \times vt$ .
The wire holds $n$ free electrons in each cubic metre. $n$ is the . So the number that passes is $n \times Avt$ , and the charge is $nAvtq$ .
Current is charge over time. Divide by $t$ and the equation appears.

I = nAvq

Symbols

$I$ = current (A)
$n$ = number density of charge carriers (\text{m}^{-3})
$A$ = cross-sectional area of the wire (\text{m}^2)
$v$ = mean drift velocity of the carriers (\text{m s}^{-1})
$q$ = charge of one carrier (e for electrons) (C)

$v$ is called the . Each electron actually moves randomly at around $10^{5}$ m/s, colliding with ions. The slow drift is the small overall movement forward on top of that.

How slow the drift really is

Worked example

Drift velocity in a copper wire

A copper wire has cross-section area $5.0 \times 10^{-6}$ m² and carries 1.0 A. Copper has $n = 8.5 \times 10^{28}$ m⁻³. Find the mean drift velocity.

Rearrange: $v = \dfrac{I}{nAe}$ .
Substitute: $v = \dfrac{1.0}{8.5 \times 10^{28} \times 5.0 \times 10^{-6} \times 1.6 \times 10^{-19}}$ .
This gives $v = 1.5 \times 10^{-5}$ m/s, which is 0.015 mm per second.

Answer

v \approx 1.5 \times 10^{-5} \text{ m s}^{-1}

Drag the sliders below to see how $v$ changes. More current needs faster drift. A thinner wire needs faster drift (fewer electrons in each section of wire, so each must move faster). Fewer carriers (smaller n) also needs faster drift.
This is why semiconductors matter here: their n is about a million times smaller than a metal's, so for the same current their electrons drift about a million times faster.

current I1.0 Aarea A1.0 mm²density n8.5

n is in units of 10²⁸ per m³ (copper ≈ 8.5)

v = 0.074 mm/s·time to drift 1 metre ≈ 3.8 hours

The exam question: scaling

The exam rarely asks you to put all four numbers in. It changes two or three of them and asks what happens to the current or the drift velocity. Treat it as pure scaling.

Your turn— tap to reveal the worked answer (9702/12/M/J/24 Q32)

A wire has area A, number density n, current I and drift speed v. A second wire has area 0.5A and number density 2n, and its electrons drift at 2v. What is the current in the second wire?

Answer: multiply the factors in $I = nAvq$ : $2 \times 0.5 \times 2 = 2$ . The current is 2I. (9702/12/M/J/24 Q32)

Common mistake

Areas often come in mm². Convert before you substitute:

1 \text{ mm}^2 = 1 \times 10^{-6} \text{ m}^2

. Writing

10^{-3}

here is a common error, because both the mm and the squaring must be converted.

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