The Hall effect

A current is carriers drifting at an average speed vd along the strip. Switch on a field B perpendicular to that drift and every carrier picks up a transverse magnetic force q vd × B. Nothing lets them leave the conductor, so they collect on one edge and leave the opposite edge with a matching deficit. Those two sheets of surface charge build a transverse electric field EH across the strip, and it grows until the electric force cancels the magnetic one: q EH = q vd B, so EH = vd B. The charge q drops out, which is why the balance is identical for electrons and for holes. After the transient the carriers travel straight down the strip again with no net sideways force, the interior stays neutral, and what is left is a steady transverse potential difference — the Hall voltage. Note what never happens: the magnetic force does no work, and in the steady state neither does EH, because the carriers no longer move across the strip.

EH is uniform across the strip, so the voltage between contacts a distance w apart is VH = EH w = vd B w. Now eliminate vd, which is not a quantity you set. The current through a cross-section A = w t is I = n q vd w t, so vd = I/(n q w t) and VH = I B/(n q t). The width has cancelled exactly: widen the strip at fixed current and the drift speed falls in the same proportion as the gap grows, leaving the product unchanged. The only length still standing is t, the thickness measured along B. Put numbers on it. Copper has roughly one free electron per atom, n = 8.5 × 10²⁸ m⁻³. A strip 20 mm wide and 0.10 mm thick carrying 5.0 A gives vd = 5.0/((8.5 × 10²⁸)(1.60 × 10⁻¹⁹)(2.0 × 10⁻⁶)) = 1.8 × 10⁻⁴ m s⁻¹, so in B = 1.0 T, EH = 1.8 × 10⁻⁴ V m⁻¹ and VH = 3.7 μV. Metals give microvolts, and the reason is entirely the enormous n in the denominator.

Fix the current and the field. Positive carriers drift along the current; negative carriers drift against it. Reversing the sign of q also reverses vd, so q vd × B flips twice and points the same way — both species are driven to the same edge. What differs is the charge deposited there, so the polarity of VH reverses. An ammeter cannot tell you the sign of the carriers; a Hall probe can, and that is how conduction in p-type semiconductors was established as behaving like positive charge in motion. Rearranged, the same measurement counts them: n = I B/(q t VH). Report it as the Hall coefficient RH = 1/(n q), which is what an experiment actually returns. One free electron per copper atom predicts RH = −7.4 × 10⁻¹¹ m³ C⁻¹; the measured room-temperature value is about −5.5 × 10⁻¹¹ m³ C⁻¹, implying some 1.3 carriers per atom. Right order of magnitude, and not much better than that.

Three geometric facts follow from VH = I B/(n q t). First, t is measured along B, not along the current, so a Hall plate is mounted face-on to the field and reads only the normal component: tilt it by 10° and it under-reads by 1.5%. Second, the width cancels only if the current contacts are far away. They are metal, they hold the transverse field near zero locally, and a short sample has its Hall voltage partly shorted out; a length-to-width ratio of about 3 or more keeps the geometry factor near 1. Third, and most awkward, the two Hall contacts must face each other. Offset them by δ along the current path and they also straddle an ordinary resistive drop I ρ δ/(w t). For the copper strip above with δ = 0.10 mm that offset is 4.2 μV — larger than the 3.7 μV Hall signal itself. It does not reverse with B, so measure at +B and −B and take VH = [V(+B) − V(−B)]/2.

VH goes as 1/n, so the useful material is the one with few carriers. A doped semiconductor at n = 1.0 × 10²¹ m⁻³ is 8.5 × 10⁷ times more dilute than copper. Take a plate of it 0.10 mm thick, drive 5.0 mA through it and place it in 0.50 T: VH = (5.0 × 10⁻³)(0.50)/((1.0 × 10²¹)(1.60 × 10⁻¹⁹)(1.0 × 10⁻⁴)) = 0.16 V. That is a signal you can read without a nanovoltmeter, which is why gaussmeter probes, clamp current sensors, brushless-motor commutation and crank-position pickups are all semiconductor Hall plates. Two practical consequences of VH ∝ I B. Drive the plate from a current source, not a voltage source, or the temperature-dependent resistivity rides straight into the reading. And respect the calibration range: a doped semiconductor holds n nearly constant across its extrinsic plateau, but heat it until intrinsic carriers take over and the sensitivity collapses.

Every line above used one carrier species, one charge, and one shared drift speed. Real carriers have a spread of speeds and scattering times, so vd is a Drude average — but in a metal that spread very nearly cancels out of RH = 1/(n q), which is why the free-electron estimate lands as close as it does. What pushes copper's measured RH off the prediction is band structure: its Fermi surface is not the sphere the free-electron model assumes. Two carrier types break the model harder. In an intrinsic semiconductor or a semimetal, electrons and holes deflect to the same edge and deposit opposite charge, so the contributions partly cancel: RH = (p μh² − n μe²)/(e (p μh + n μe)²). That can be small, it can change sign with temperature, and no single measurement separates p from n. The same band structure can flip the sign outright: aluminium, zinc and beryllium conduct by electrons yet return a positive RH. And in a two-dimensional electron gas at low temperature and high field, VH/I stops tracking B/(n q t) and locks onto plateaus at h/(ν e²), with h/e² = 25812.807 Ω — the resistance standard.