Mean Drift Velocity

Spec mapping: OCR H556 Module 4.1 — Charge and current (mean drift velocity equation $I = nAvq$I=nAvq relating macroscopic current to microscopic charge-carrier motion in a conductor; number density $n$n of free carriers; distinction between conductors, semiconductors and insulators in terms of $n$n). Refer to the official OCR H556 specification document for exact wording.

Lesson 1 defined current as $I = dQ/dt$I=dQ/dt, the macroscopic rate of charge flow. That definition is clean and exam-friendly, but it tells you nothing about what is happening inside the wire. How fast are the electrons actually moving? How many are there? Why does a thick copper wire carry more current than a thin one for the same voltage? Why does a thermistor's resistance fall as it warms?

This lesson derives and applies the mean drift velocity equation

$I = nAvq$I=nAvq

which links the macroscopic current $I$I to the microscopic picture: free-carrier number density $n$n, wire cross-sectional area $A$A, mean drift velocity $v$v, and carrier charge $q$q (which for electrons equals $e$e). It is a required OCR H556 calculation skill and appears in exam questions every single year. By the end you should be able to compute drift velocities in metals, semiconductors and ionic solutions, and to explain the famous A-Level surprise that drift velocity is mere millimetres per second while electrical signals travel at near the speed of light.

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A Wire Full of Electrons

A typical metal conductor — copper, say — contains a vast number of free electrons. Each copper atom contributes about one outer electron to a "sea" of delocalised charge carriers that move throughout the crystalline lattice; the remaining positively-charged ion cores stay fixed at their lattice positions. This Drude picture (1900) treats the conductor as a classical gas of electrons inside a fixed background lattice.

In the absence of a potential difference, the free electrons move randomly at very high thermal speeds — about $10^5$105 m s$^{-1}$−1 at room temperature — but with no preferred direction. Their average velocity is zero, so there is no net current. Each electron's velocity vector points in a random direction; vector summing over $10^{29}$1029 such vectors per cubic metre gives almost exactly zero.

When a potential difference is applied across the ends of the wire, an electric field $\vec{E}$E is set up inside the conductor (this happens in roughly $L/c \sim 1$L/c∼1 ns for a metre of wire — see "Signal speed" below). The field exerts a force $\vec{F} = -e\vec{E}$F=−eE on every electron, accelerating it. The random thermal motion continues — but now there is a superimposed slow drift in the direction opposite to $\vec{E}$E (because electrons carry negative charge). This slow average drift is the mean drift velocity, symbol $v$v (or sometimes $v_d$vd​).

The thermal speed dominates the speed of any individual electron at any instant; the drift velocity dominates the average motion when you take time-averages or carrier-averages.

flowchart LR
  subgraph Wire
    e1((e)) --> e2((e)) --> e3((e)) --> e4((e))
  end
  EF["Electric field E (right)"] -.->|"pushes positive charges right"| R[Right]
  EL["Electrons drift LEFT (opposite to E)"] -.-> L[Left]

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Deriving $I = nAvq$I=nAvq

Consider a straight cylindrical wire of cross-sectional area $A$A, containing $n$n free charge carriers per cubic metre, each carrying charge $q$q (equal to $-e$−e for electrons, with magnitude $e$e). Suppose all the carriers drift along the wire at a mean velocity of magnitude $v$v.

We want to count the charge that crosses an imaginary plane perpendicular to the wire in a short time $\Delta t$Δt.

• In time $\Delta t$Δt, each carrier moves a distance $v \Delta t$vΔt along the wire.
• Therefore every carrier currently inside the cylinder of length $v \Delta t$vΔt ending at the plane will cross the plane within $\Delta t$Δt.
• Volume of this cylinder: $V_{\text{cyl}} = A \cdot v \cdot \Delta t$Vcyl​=A⋅v⋅Δt.
• Number of carriers in this volume: $N = n V_{\text{cyl}} = nAv\Delta t$N=nVcyl​=nAvΔt.
• Magnitude of charge crossing the plane: $|\Delta Q| = Nq = (nAv\Delta t)q$∣ΔQ∣=Nq=(nAvΔt)q (taking $q$q here as the magnitude $e$e).

By the definition $I = \Delta Q / \Delta t$I=ΔQ/Δt:

$I = \frac{nAv\Delta t \cdot q}{\Delta t} = nAvq$I=ΔtnAvΔt⋅q​=nAvq

Hence the mean drift velocity equation:

$I = nAvq$I=nAvq

where:

• $I$I is current (A).
• $n$n is the number density of free charge carriers (m$^{-3}$−3).
• $A$A is the cross-sectional area of the conductor (m$^2$2).
• $v$v is the mean drift velocity of the carriers (m s$^{-1}$−1).
• $q$q is the magnitude of each carrier's charge ($= e = 1.60 \times 10^{-19}$=e=1.60×10−19 C for electrons).

OCR sometimes writes this as $I = nAve$I=nAve when the carrier is the electron. Both notations are accepted in mark schemes.

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Worked Example 1 — Copper Wire

A copper wire of cross-sectional area $1.0$1.0 mm$^2$2 carries a current of $2.0$2.0 A. The number density of free electrons in copper is $n = 8.5 \times 10^{28}$n=8.5×1028 m$^{-3}$−3. Calculate the mean drift velocity of the electrons.

Step 1 — Convert units. $A = 1.0$A=1.0 mm$^2 = 1.0 \times 10^{-6}$2=1.0×10−6 m$^2$2. (Remember $1$1 mm $= 10^{-3}$=10−3 m so $1$1 mm$^2 = 10^{-6}$2=10−6 m$^2$2. A factor-of-1000 error here is one of the most common slips at A-Level.)

Step 2 — Rearrange for $v$v.

$v = \frac{I}{nAq} = \frac{2.0}{(8.5 \times 10^{28})(1.0 \times 10^{-6})(1.60 \times 10^{-19})}$v=nAqI​=(8.5×1028)(1.0×10−6)(1.60×10−19)2.0​

Denominator: $(8.5 \times 10^{28}) \times (1.0 \times 10^{-6}) \times (1.60 \times 10^{-19}) = 8.5 \times 1.60 \times 10^{28-6-19} = 13.6 \times 10^{3} = 1.36 \times 10^4$(8.5×1028)×(1.0×10−6)×(1.60×10−19)=8.5×1.60×1028−6−19=13.6×103=1.36×104 A m$^{-1}$−1.

$v = \frac{2.0}{1.36 \times 10^4} \approx 1.5 \times 10^{-4} \text{ m s}^{-1}$v=1.36×1042.0​≈1.5×10−4 m s−1

That is about 0.15 mm per second. Less than a millimetre per second! The "classic A-Level surprise" — electrons in a copper wire carrying a perfectly ordinary household current of $2$2 A drift along at the speed of an inch-worm. How can a light bulb switch on "instantly"?

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Signal Speed vs Drift Speed — the Critical Distinction

A bulb does not wait for a particular electron to travel from the battery to the filament. When you flip the switch, two things happen:

1. The electric field inside the wire propagates along the conductor at close to the speed of light (more precisely, at the phase velocity of an electromagnetic wave in the dielectric surrounding the wire — typically $\sim 0.6c$∼0.6c to $\sim 0.9c$∼0.9c in everyday cables). This takes only a few nanoseconds per metre.
2. Every electron in the circuit — including those already in the filament — begins to drift simultaneously, the moment the field reaches them.

So the bulb lights because the field propagates near $c$c, not because the electrons travel from battery to bulb. The electrons already inside the filament start drifting within nanoseconds; their drift speed is small (mm/s), but they are everywhere in the circuit, and the current flows immediately.

A water-pipe analogy: imagine a hose-pipe completely full of water. When you turn on the tap, water flows out the far end immediately — not because the water in the tap has travelled the length of the hose, but because the water already at the far end has been pushed out by the pressure wave (which travels at the speed of sound in water, $\sim 1500$∼1500 m s$^{-1}$−1 — much faster than the water itself, which trickles out at perhaps 1 m s$^{-1}$−1).

Charge in a wire behaves identically. The "pressure wave" of the electric field travels near $c$c; the charges themselves drift at $\sim 10^{-4}$∼10−4 m s$^{-1}$−1. Both speeds are true; they describe different things.

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Worked Example 2 — Varying Cross-Section

A wire carries a current of $1.5$1.5 A and narrows from cross-sectional area $A_1 = 2.0$A1​=2.0 mm$^2$2 to $A_2 = 0.50$A2​=0.50 mm$^2$2. The charge-carrier density is $n = 1.0 \times 10^{29}$n=1.0×1029 m$^{-3}$−3. Calculate the drift velocity in each section.

Thick section ($A_1 = 2.0 \times 10^{-6}$A1​=2.0×10−6 m$^2$2):

$v_1 = \frac{I}{nA_1 q} = \frac{1.5}{(1.0 \times 10^{29})(2.0 \times 10^{-6})(1.60 \times 10^{-19})} = \frac{1.5}{3.2 \times 10^4} = 4.7 \times 10^{-5} \text{ m s}^{-1}$v1​=nA1​qI​=(1.0×1029)(2.0×10−6)(1.60×10−19)1.5​=3.2×1041.5​=4.7×10−5 m s−1

Thin section ($A_2 = 0.50 \times 10^{-6}$A2​=0.50×10−6 m$^2$2):

$v_2 = \frac{I}{nA_2 q} = \frac{1.5}{(1.0 \times 10^{29})(0.50 \times 10^{-6})(1.60 \times 10^{-19})} = \frac{1.5}{8.0 \times 10^3} = 1.9 \times 10^{-4} \text{ m s}^{-1}$v2​=nA2​qI​=(1.0×1029)(0.50×10−6)(1.60×10−19)1.5​=8.0×1031.5​=1.9×10−4 m s−1

The drift velocity is four times larger in the thin section, because the same number of electrons per second must squeeze through a quarter of the area. This is identical in form to water speeding up when a river narrows or air speeding up in a Venturi tube — the flux (charge per second) is conserved, so the flow speed must vary inversely with cross-section.

Conceptual point: Whenever the geometry changes along a series wire, the current $I$I is the same everywhere (Kirchhoff's first law — lesson 3). The drift velocity is not. Students who memorise "current is the same in series" sometimes leap to "drift velocity is the same in series" — that is wrong.

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Worked Example 3 — Same wire, two materials joined

A copper wire ($n_{\text{Cu}} = 8.5 \times 10^{28}$nCu​=8.5×1028 m$^{-3}$−3) of cross-section $1.0$1.0 mm$^2$2 is joined to an aluminium wire ($n_{\text{Al}} = 6.0 \times 10^{28}$nAl​=6.0×1028 m$^{-3}$−3) of the same cross-section. A current of $5.0$5.0 A flows. Calculate the drift velocity in each material.

Copper: $v_{\text{Cu}} = 5.0 / [(8.5 \times 10^{28})(1.0 \times 10^{-6})(1.60 \times 10^{-19})] = 5.0 / 1.36 \times 10^4 = 3.7 \times 10^{-4}$vCu​=5.0/[(8.5×1028)(1.0×10−6)(1.60×10−19)]=5.0/1.36×104=3.7×10−4 m s$^{-1}$−1.

Aluminium: $v_{\text{Al}} = 5.0 / [(6.0 \times 10^{28})(1.0 \times 10^{-6})(1.60 \times 10^{-19})] = 5.0 / 9.6 \times 10^3 = 5.2 \times 10^{-4}$vAl​=5.0/[(6.0×1028)(1.0×10−6)(1.60×10−19)]=5.0/9.6×103=5.2×10−4 m s$^{-1}$−1.

The drift velocity is about $40\%$40% higher in aluminium because it has fewer free electrons per unit volume (only one per atom, with a smaller atom density than copper). The current is the same in both wires (Kirchhoff-1 at the junction) — the speed of each carrier adjusts to compensate for the different carrier density.

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Worked Example 4 — Semiconductor

A sample of pure silicon at room temperature has a free-electron density of $n = 1.0 \times 10^{16}$n=1.0×1016 m$^{-3}$−3. A current of $1.0$1.0 $\mu$μA flows through a cross-section of $1.0$1.0 mm$^2$2. Calculate the drift velocity.

$v = \frac{I}{nAq} = \frac{1.0 \times 10^{-6}}{(1.0 \times 10^{16})(1.0 \times 10^{-6})(1.60 \times 10^{-19})}$v=nAqI​=(1.0×1016)(1.0×10−6)(1.60×10−19)1.0×10−6​

Denominator: $10^{16} \times 10^{-6} \times 1.60 \times 10^{-19} = 1.60 \times 10^{-9}$1016×10−6×1.60×10−19=1.60×10−9.

$v = \frac{1.0 \times 10^{-6}}{1.60 \times 10^{-9}} \approx 625 \text{ m s}^{-1}$v=1.60×10−91.0×10−6​≈625 m s−1

Now the electrons are zipping along at hundreds of metres per second — a factor of about $10^6$106 faster than in copper — because $n$n is about $10^{13}$1013 times smaller. This enormous drift speed is the reason semiconductors can be driven to electrical breakdown with surprisingly modest voltages, and the reason transistors have to dissipate heat aggressively.

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Charge Carrier Density Across Materials

The astonishing range of electrical conductivity in different materials comes almost entirely from the range of $n$n, the charge-carrier density. Conductivity $\sigma \propto n$σ∝n at fixed temperature and carrier mobility.

Material	Type	$n$n (m$^{-3}$−3)	Comment
Copper	Metal	$\sim 10^{29}$∼1029	One free electron per atom
Aluminium	Metal	$\sim 6 \times 10^{28}$∼6×1028	One free electron per atom
Iron	Metal	$\sim 1.7 \times 10^{29}$∼1.7×1029	Up to two free electrons per atom (partial $d$d-band)
Sea water (ions)	Electrolyte	$\sim 10^{26}$∼1026	Mostly Na$^+$+, Cl$^-$−
Doped silicon	Semiconductor	$\sim 10^{22}$∼1022	Donor / acceptor atoms (parts per million)
Intrinsic silicon	Semiconductor	$\sim 10^{16}$∼1016	Thermally excited carriers only
Glass, rubber, dry wood	Insulator	$\lesssim 10^6$≲106	Effectively no free carriers

For a given $I$I and $A$A, the drift velocity scales as $v \propto 1/n$v∝1/n. So semiconductors carry the same current with drift velocities $10^4$104–$10^{12}$1012 times higher than metals; this is why they dissipate so much more heat per unit current and why their I-V characteristics (lesson 6) are so strongly temperature-dependent.

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Metals vs Semiconductors — Temperature Dependence

The microscopic picture also explains the opposite temperature behaviour of metallic resistance versus semiconductor resistance.

Metals: $n$n is essentially constant with temperature (it's just one outer electron per atom). But as temperature rises, the lattice ions vibrate more vigorously, scattering the drifting electrons more often. The electrons' mean free path shortens, $v$v falls, and so does $I$I at fixed voltage. Therefore $R = V/I$R=V/I rises with temperature.

Semiconductors: $n$n rises exponentially with temperature (more electrons gain enough thermal energy to jump the band gap into the conduction band). The increase in $n$n vastly outweighs the lattice-scattering reduction in carrier mobility. The net effect: at fixed voltage, $I$I rises with temperature, and so $R = V/I$R=V/I falls. This is the operating principle of an NTC thermistor (lesson 6).