University Physics II · Magnetic Fields and Magnetic Forces · 8.8

The Hall Effect

Send a current through a conductor in a magnetic field and charge collects on one edge. The voltage it builds measures the field, counts the carriers, and reveals their sign — if you respect the geometry.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put a current-carrying conductor in a magnetic field and every carrier picks up a transverse force qv × B. The carriers cannot leave, so they collect on one edge and leave a matching deficit on the other, and the transverse electric field that builds grows until it exactly cancels the magnetic force. After that transient the carriers travel straight again, and what remains is a steady transverse potential difference: the Hall voltage.

Two lines of algebra turn it into VH = IB/(nqt), where n is the carrier number density, q the carrier charge, and t the thickness measured along B — the width between the contacts cancels completely. That one equation does three jobs. Its polarity gives the sign of the carriers, which no current measurement can.

Its magnitude counts them. And with n fixed by the material it measures B. Everything after that is asking how far a single-carrier, single-drift-speed model can be trusted.

Simple definition: The Hall effect is the transverse voltage across a current-carrying conductor in a magnetic field, appearing once the charge driven to one edge builds an electric field that cancels the magnetic force on the carriers.
Example: Run 5.0 A along a 0.10 mm-thick copper strip in a 1.0 T field and 3.7 μV appears across its edges — small, steady, and with a polarity that says the carriers are negative.

Transverse force balanceq EH = q vd B, so EH = vd B

The carrier charge cancels, so the same balance holds for electrons and for holes.

Steady state; EH in V m⁻¹, vd in m s⁻¹, B in T

Hall voltage across the contactsVH = EH w = vd B w

A uniform transverse field multiplied by the gap it acts across.

w is the contact separation across the strip, in m

Drift speed from the currentI = n q vd w t, so vd = I/(n q w t)

Copper, 5.0 A through 20 mm × 0.10 mm: vd = 1.8 × 10⁻⁴ m s⁻¹.

n in m⁻³, q in C, cross-section A = w t in m²

Hall voltageVH = I B/(n q t)

A thin sample and a low carrier density both raise the signal.

t is the thickness measured along B; the width w cancels

Hall coefficientRH = Ey/(Jx B) = 1/(n q), VH = RH I B/t

Copper: −7.3 × 10⁻¹¹ predicted, about −5.5 × 10⁻¹¹ measured.

Unit m³ C⁻¹; negative for electrons, positive for holes

Hall angle and mobilitytan θH = Ey/Ex = μ B, with μ = |RH| σ

Copper at 1.0 T: μ ≈ 4.4 × 10⁻³, so the total field tilts by only 0.25°.

μ in m² V⁻¹ s⁻¹; σ is the conductivity in S m⁻¹

Carriers get pushed sideways, then stop

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.

From the force balance to VH = I B/(n q t)

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.

Polarity names the sign; size counts the carriers

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 = −7.3 × 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.

Geometry: one length matters, and the contacts must be honest

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.

Why field sensors are semiconductors, not metals

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.

What the single-carrier model assumes

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 and RH = 1/(n q) inherits that approximation — most of why copper's measured RH misses the free-electron prediction. Two carrier types break it 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. Band structure alone can flip it: 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Probe current I5.0 mA

Field B, along the thickness0.50 T

Plate thickness t0.10 mm

Halve the thickness and the bar doubles while the two force arrows stay equal — t is the only length in VH = IB/(nqt) — then cut B and watch the bar fall with it while the sensitivity readout VH/B, which belongs to the plate and not to the field, refuses to move.

DRIFT SPEED vd78.1 m s⁻¹

HALL FIELD EH = vd B39.1 V m⁻¹

HALL VOLTAGE IB/(nqt)0.156 V

SENSITIVITY VH/B0.313 V T⁻¹

Live interpretationDRIFT SPEED vd: 78.1 m s⁻¹. HALL FIELD EH = vd B: 39.1 V m⁻¹. HALL VOLTAGE IB/(nqt): 0.156 V. SENSITIVITY VH/B: 0.313 V T⁻¹

Catch the common trap

Explain before calculating.

A Hall plate carries a fixed current I in a fixed magnetic field B normal to its face, with the Hall contacts well clear of the current contacts. Which single change doubles the Hall voltage?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA Hall probe is an n-type silicon plate 0.50 mm thick with carrier density n = 2.0 × 10²² m⁻³. It carries a steady 3.0 mA and is mounted face-on in an unknown field. The contacts read VH = 1.5 mV. Find B.

VH = I B/(n q t), and B is the only unknown, so rearrange to B = VH n q t/I.
Group the plate's own constants first: n q t = (2.0 × 10²² m⁻³)(1.60 × 10⁻¹⁹ C)(5.0 × 10⁻⁴ m) = 1.6 C m⁻².
B = VH (n q t)/I = (1.5 × 10⁻³ V)(1.6 C m⁻²)/(3.0 × 10⁻³ A) = 0.80 T.
The plate's width was never needed. The only length in the formula is t, the 0.50 mm measured along B — and it only counts as the full 0.50 mm because the plate is face-on, so B is normal to it.

AnswerB = 0.80 T

MediumA copper strip 25 mm wide and 0.050 mm thick carries 8.0 A in a 0.90 T field normal to its face. The Hall contacts read 8.5 μV, and the edge the carriers pile onto is the negative one. Find the carrier density, the Hall coefficient with its sign, the number of carriers per copper atom (atomic density 8.5 × 10²⁸ m⁻³), and the drift speed.

Invert VH = I B/(n q t) for the quantity the experiment does not set: n = I B/(q t VH).
n = (8.0 A)(0.90 T)/[(1.60 × 10⁻¹⁹ C)(5.0 × 10⁻⁵ m)(8.5 × 10⁻⁶ V)] = 7.2/(6.8 × 10⁻²⁹) = 1.06 × 10²⁹ m⁻³.
RH = 1/(n q) = 1/[(1.06 × 10²⁹)(1.60 × 10⁻¹⁹)] = 5.90 × 10⁻¹¹ m³ C⁻¹ in magnitude; the negative edge is the one the carriers reached, so the carriers are negative and RH = −5.90 × 10⁻¹¹ m³ C⁻¹. No ammeter reading could have told you that sign.
Per atom: (1.06 × 10²⁹)/(8.5 × 10²⁸) = 1.25 electrons per copper atom. The free-electron count of 1 is the right order and not much better than that.
Drift speed from I = n q vd w t: vd = I/(n q w t) = 8.0/[(1.69 × 10¹⁰ C m⁻³)(0.025 m)(5.0 × 10⁻⁵ m)] = 8.0/(2.12 × 10⁴) = 3.8 × 10⁻⁴ m s⁻¹.
Check it back through the force balance: EH = vd B = (3.78 × 10⁻⁴)(0.90) = 3.40 × 10⁻⁴ V m⁻¹, and VH = EH w = (3.40 × 10⁻⁴)(0.025) = 8.5 μV — the reading we started from.

Answern = 1.06 × 10²⁹ m⁻³, RH = −5.90 × 10⁻¹¹ m³ C⁻¹ (carriers negative), 1.25 carriers per atom, vd = 3.8 × 10⁻⁴ m s⁻¹

HardA copper strip 20 mm wide and 0.10 mm thick carries 5.0 A in a 1.0 T field, with n = 8.5 × 10²⁸ m⁻³ and ρ = 1.68 × 10⁻⁸ Ω m. The two Hall contacts are misaligned by δ = 0.10 mm along the current. Find the Hall voltage, the offset the misalignment adds, the readings at +B and at −B, and how tightly δ would have to be held to keep the offset under 10% of VH.

Hall signal: VH = I B/(n q t) = (5.0)(1.0)/[(8.5 × 10²⁸)(1.60 × 10⁻¹⁹)(1.0 × 10⁻⁴)] = 5.0/(1.36 × 10⁶) = 3.68 μV.
Offset contacts also straddle a length δ of plain resistor: Voff = I ρ δ/(w t) = (5.0)(1.68 × 10⁻⁸)(1.0 × 10⁻⁴)/[(0.020)(1.0 × 10⁻⁴)] = (8.4 × 10⁻¹²)/(2.0 × 10⁻⁶) = 4.20 μV — bigger than the signal it is contaminating.
Voff does not care about the sign of B, but VH does. At +B the meter reads 3.68 + 4.20 = 7.88 μV; reverse the field and it reads −3.68 + 4.20 = 0.52 μV.
Half the difference strips the offset out: (7.88 − 0.52)/2 = 3.68 μV, the true VH. Half the sum returns (7.88 + 0.52)/2 = 4.20 μV, the offset itself — one field-reversed pair delivers both.
Trusting the +B reading alone would report 7.88 μV, a factor 7.88/3.68 = 2.14 too large: a 114% error, and one that no amount of averaging would reveal.
For Voff ≤ 0.10 VH: δ ≤ 0.10 VH w t/(I ρ) = (0.10)(3.68 × 10⁻⁶)(2.0 × 10⁻⁶)/[(5.0)(1.68 × 10⁻⁸)] = (7.35 × 10⁻¹³)/(8.4 × 10⁻⁸) = 8.8 μm. Nobody places contacts to 9 μm by hand, which is why field reversal is standard practice rather than a refinement.

AnswerVH = 3.68 μV and Voff = 4.20 μV; the meter reads 7.88 μV at +B and 0.52 μV at −B, whose half-difference recovers 3.68 μV and half-sum recovers 4.20 μV; a single-polarity reading is 114% high, and the offset stays under 10% only if δ ≤ 8.8 μm.