University Physics II · Current, Resistance, and DC Circuits · 6.3

Current Density & Drift Speed

Flip a switch and the lamp lights in nanoseconds, yet the electrons carrying that current creep along at a fraction of a millimetre per second. Current density is the bookkeeping that makes both facts true at once.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A current is charge crossing a surface, and the microscopic count is simple. If a conductor holds n mobile carriers per cubic metre, each of charge q, drifting at average velocity v(d), then exactly the carriers within v(d)Δt of a cross-section pass through it in Δt, so I = n q v(d) A. Divide the area out and what is left belongs to the point: the current density J = n q v(d), a vector field pointing along the conventional current whatever the sign of the carriers, with current recovered as its flux, I = ∫ J · dA.

Writing v(d) = μE turns that into J = σE, Ohm's law stated at a point rather than across a component. The numbers are the surprise. A 1.0 A current in a 1.0 mm² copper wire needs a drift speed of only 7.3 × 10⁻⁵ m s⁻¹, about 26 cm an hour, riding on random electron speeds near 10⁶ m s⁻¹.

Circuits work not because charge travels far but because carriers are already everywhere the field reaches.

Simple definition: Current density J = n q v(d) is current per unit cross-sectional area, a vector along the conventional current. Drift velocity v(d) is the slow average velocity the field adds to the carriers' fast random motion.
Example: A 1.0 mm² copper wire carrying 1.0 A has J = 1.0 × 10⁶ A m⁻² and a drift speed of 7.3 × 10⁻⁵ m s⁻¹, so one electron needs over seven hours to cross a 2.0 m lamp cord.

Microscopic currentI = n q v(d) A

The carriers within v(d)Δt of the face are exactly the ones that cross in Δt.

n in m⁻³, q in C, v(d) in m s⁻¹, A in m²; I in A

Current density vectorJ = n q v(d)

Electrons have q < 0 and drift against E, so the two sign flips leave J along E.

A m⁻²; direction of conventional current for either sign of q

Current as the flux of JI = ∫ J · dA

Tilt the cross-section: area grows by 1/cos θ, J·n̂ falls by cos θ, I is unchanged.

Collapses to I = JA only for uniform J normal to the face

Drift velocity and mobilityv(d) = μ E, μ = |q| τ / m

Accelerate for τ, then lose it in a collision: the average is a creep, not a runaway.

μ in m² V⁻¹ s⁻¹; τ is the mean free time between collisions

Ohm's law at a pointJ = n |q| μ E = σ E, σ = n q² τ / m

Conductivity splits into how many carriers there are and how freely each one moves.

σ in S m⁻¹ = Ω⁻¹ m⁻¹; σ = 1/ρ, and V = IR is its integrated form

Two carrier speciesJ = n₊q₊v₊ + n₋q₋v₋, σ = e(n μ(e) + p μ(h))

Two sign reversals cancel, so anions and cations push J in the same direction.

Electrolytes and semiconductors; opposite charges drifting oppositely add

Counting the carriers

Take a conductor holding n mobile carriers per cubic metre, each of charge q, all drifting at average velocity v(d) along the wire. Mark a cross-section of area A and ask what crosses it in time Δt. Every carrier inside a slab of length v(d)Δt behind the face gets there, and no carrier outside it does, so the slab volume A v(d)Δt contains n A v(d)Δt carriers and delivers ΔQ = n q v(d) A Δt. Divide by Δt: I = n q v(d) A. Each factor is separately knowable. For copper, n comes from density and molar mass on the assumption of one conduction electron per atom: (8960 kg m⁻³ ÷ 0.06355 kg mol⁻¹) × 6.022 × 10²³ mol⁻¹ = 8.5 × 10²⁸ m⁻³. That means n|q| = 1.36 × 10¹⁰ C of mobile charge in every cubic metre of copper. The figure is enormous, which already tells you that the remaining factor, v(d), will have to be very small.

Current density is local; current is not

I = n q v(d) A mixes a material property with a geometric one. Divide the area out and what remains belongs to the point: J = n q v(d), the current density, in A m⁻². Unlike current, it is a vector, and it is defined everywhere inside the conductor rather than only on a chosen surface. Its direction is that of the conventional current, and that survives the sign of the carrier: in a metal q = −e and v(d) points against E, so the two minus signs cancel and J lies along E. In an electrolyte, cations drift with E and anions against it, and both terms of J = n₊q₊v₊ + n₋q₋v₋ add rather than cancel. Working in J rather than I is what lets you describe conductors that are not neat wires — current spreading through a grounding plate, through seawater, or through tissue, where 'the' cross-sectional area is not a well-posed quantity.

Current is the flux of J through a surface

Given J everywhere, the current through a surface is I = ∫ J · dA, with dA the outward normal element. The dot product earns its place. Only the component through the face counts, so tilting a plane cross-section of a wire by θ multiplies its area by 1/cos θ and its normal component of J by cos θ, leaving I unchanged — as it must be, since the same charge per second passes either way. If J is uniform and normal, the integral collapses to I = JA. If it is not, do the integral: a wire of radius R with J(r) = J₀ r/R gives I = ∫₀ᴿ (J₀ r/R)(2πr dr) = (2/3)πR²J₀, so the mean current density is two-thirds of the peak; with R = 1.0 mm and J₀ = 3.0 × 10⁶ A m⁻², I = 6.3 A. Steady flow also forces J₁A₁ = J₂A₂. Charge cannot accumulate, so a wire that narrows carries the same I with proportionally larger J and drift speed.

How slow the drift really is

Put numbers in. A 1.0 mm² copper wire — ordinary lamp cord — carrying 1.0 A has J = 1.0 × 10⁶ A m⁻², so v(d) = J/(n|q|) = 1.0 × 10⁶ / 1.36 × 10¹⁰ = 7.3 × 10⁻⁵ m s⁻¹. That is 0.073 mm s⁻¹, about 26 cm in an hour; an electron entering a 2.0 m cord reaches the lamp 2.7 × 10⁴ s later, more than seven hours. The electrons themselves are not slow at all. A conduction electron in copper moves at roughly the Fermi speed, 1.6 × 10⁶ m s⁻¹, in a direction re-randomised every few tens of femtoseconds; the drift is a bias on that motion of order one part in 10¹⁰. The lamp still lights immediately because the field that sets up the drift propagates along the wire at a good fraction of c — about 10 ns for that 2.0 m cord — and carriers were already present everywhere, filament included. Nothing has to arrive from the switch.

Mobility, and Ohm's law at a point

Why is v(d) proportional to E instead of growing without limit? Between collisions a carrier accelerates at qE/m; each collision randomises its velocity, so on average it restarts from scratch and accumulates v(d) = (|q|E/m)τ over a mean free time τ. Define the mobility μ = v(d)/E = |q|τ/m, in m² V⁻¹ s⁻¹, and the rest follows: J = n|q|μE = σE with σ = n q² τ/m. That is Ohm's law written at a point; V = IR is what it integrates to for a uniform conductor. Copper's measured σ = 5.96 × 10⁷ S m⁻¹ divided by n|q| = 1.36 × 10¹⁰ C m⁻³ gives μ = 4.4 × 10⁻³ m² V⁻¹ s⁻¹, hence τ = μm/|q| = 2.5 × 10⁻¹⁴ s and a mean free path v(F)τ ≈ 40 nm, roughly 150 atomic spacings. The field inside our 1.0 A wire is only J/σ = 0.017 V m⁻¹, and the net forward creep per free flight is v(d)τ ≈ 2 × 10⁻¹⁸ m.

Where this model gives out

The Drude argument behind μ = |q|τ/m is classical and wrong in detail: metal electrons are degenerate, so the speed setting the mean free path is the Fermi speed, not the thermal √(3kT/m) = 1.2 × 10⁵ m s⁻¹ at 300 K. It survives because the derivation needs only 'accelerate, then randomise'. Three limits matter. In a metal n is essentially fixed and temperature acts through τ, so resistivity rises with T; in intrinsic silicon n itself climbs steeply with T and resistivity falls — and with n ≈ 10¹⁶ m⁻³ against copper's 8.5 × 10²⁸, a mobility some 30 times larger still leaves σ smaller by some 2 × 10¹¹. Next, v(d) = μE is linear only at low field: carriers in silicon saturate near 10⁵ m s⁻¹ once they can shed energy to optical phonons. Finally, alternating current crowds into a skin depth δ = √(ρ/(π f μ₀)) — 8.4 mm in copper at 60 Hz, so mains wiring stays near-uniform, but 65 μm at 1.0 MHz, where I = JA fails.

Change one variable at a time

Make the relationship visible.

Interactive model

Current I3.0 A

Narrow cross-section A₂1.0 mm²

Shrink the narrow section with the current held fixed and watch its arrow stretch — the same I crosses both faces, so J and v(d) rise exactly as fast as the area falls.

J, WIDE SECTION1.00 MA m⁻²

J, NARROW SECTION3.00 MA m⁻²

DRIFT SPEED, WIDE73.4 μm s⁻¹

DRIFT SPEED, NARROW220.3 μm s⁻¹

Live interpretationJ, WIDE SECTION: 1.00 MA m⁻². J, NARROW SECTION: 3.00 MA m⁻². DRIFT SPEED, WIDE: 73.4 μm s⁻¹. DRIFT SPEED, NARROW: 220.3 μm s⁻¹

Catch the common trap

Explain before calculating.

A copper conductor (n = 8.5 × 10²⁸ m⁻³) carries a steady 3.0 A. It narrows from a 3.0 mm² section, where J = 1.0 × 10⁶ A m⁻² and v(d) = 7.3 × 10⁻⁵ m s⁻¹, into a 1.0 mm² section. What holds in the narrow section?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA copper wire of cross-sectional area 2.5 mm² carries 4.0 A. Copper has n = 8.5 × 10²⁸ conduction electrons per cubic metre. Find the current density and the drift speed.

Work in SI: A = 2.5 mm² = 2.5 × 10⁻⁶ m².
J is uniform and normal to the face, so the flux integral collapses to J = I/A = 4.0 A ÷ 2.5 × 10⁻⁶ m² = 1.6 × 10⁶ A m⁻².
Mobile charge per cubic metre: n|q| = 8.5 × 10²⁸ m⁻³ × 1.60 × 10⁻¹⁹ C = 1.36 × 10¹⁰ C m⁻³.
J = n|q|v(d) rearranges to v(d) = J/(n|q|) = 1.6 × 10⁶ ÷ 1.36 × 10¹⁰ = 1.2 × 10⁻⁴ m s⁻¹ — about 0.12 mm each second.

AnswerJ = 1.6 × 10⁶ A m⁻²; v(d) = 1.2 × 10⁻⁴ m s⁻¹

MediumA steady 6.0 A runs through an aluminium busbar (n = 1.8 × 10²⁹ m⁻³) that steps down from a 12 mm² section to a 3.0 mm² section. Find J and v(d) in each section.

Steady flow forbids charge piling up at the step, so the same I = 6.0 A crosses both faces; only J and v(d) change.
Wide: J₁ = 6.0 A ÷ 12 × 10⁻⁶ m² = 5.0 × 10⁵ A m⁻². Narrow: J₂ = 6.0 A ÷ 3.0 × 10⁻⁶ m² = 2.0 × 10⁶ A m⁻² — four times larger, which is J₁A₁ = J₂A₂.
The metal is the same on both sides, so n|q| = 1.8 × 10²⁹ m⁻³ × 1.60 × 10⁻¹⁹ C = 2.88 × 10¹⁰ C m⁻³ throughout.
v(d) = J/(n|q|): 5.0 × 10⁵ ÷ 2.88 × 10¹⁰ = 1.7 × 10⁻⁵ m s⁻¹ in the wide part, 2.0 × 10⁶ ÷ 2.88 × 10¹⁰ = 6.9 × 10⁻⁵ m s⁻¹ in the narrow part — also four times larger, since n is fixed and v(d) tracks J.

AnswerJ₁ = 5.0 × 10⁵ A m⁻² with v(d)₁ = 1.7 × 10⁻⁵ m s⁻¹; J₂ = 2.0 × 10⁶ A m⁻² with v(d)₂ = 6.9 × 10⁻⁵ m s⁻¹

HardAn n-type silicon bar has n = 1.0 × 10²² m⁻³ and electron mobility μ = 0.135 m² V⁻¹ s⁻¹. It is 2.0 mm long with a 0.50 mm² cross-section and carries 8.0 mA. Find J, the field inside it, the drift speed, and the voltage across it.

J = I/A = 8.0 × 10⁻³ A ÷ 0.50 × 10⁻⁶ m² = 1.6 × 10⁴ A m⁻².
Build the conductivity from the microscopic pieces: σ = n|q|μ = 1.0 × 10²² × 1.60 × 10⁻¹⁹ × 0.135 = 216 S m⁻¹, i.e. 2.2 × 10² S m⁻¹.
Ohm's law at a point, J = σE, gives E = J/σ = 1.6 × 10⁴ ÷ 216 = 74 V m⁻¹.
v(d) = μE = 0.135 × 74 = 10 m s⁻¹; check it independently against v(d) = J/(n|q|) = 1.6 × 10⁴ ÷ 1.60 × 10³ = 10.0 m s⁻¹.
The bar is uniform, so the field is uniform and V = EL = 74.0 V m⁻¹ × 2.0 × 10⁻³ m = 0.148 V, giving V ≈ 0.15 V and R = V/I = 18.5 Ω.
Copper at this same J would drift at 1.6 × 10⁴ ÷ 1.36 × 10¹⁰ = 1.2 × 10⁻⁶ m s⁻¹. Silicon's carrier density is some seven orders of magnitude smaller, so its carriers must crawl that much faster to deliver the same current density.

AnswerJ = 1.6 × 10⁴ A m⁻², E = 74 V m⁻¹, v(d) = 10 m s⁻¹, V = 0.15 V