The Hall Effect: Voltage, Carrier Density and the Sign of the Charge

Q: Why do electrons and holes pile up on the same face?

Because the force qv × B has the sign of the charge in it twice. A negative carrier drifts opposite to the conventional current, so both q and v change sign together, and the product does not change at all. Both are pushed the same way. The diagram holds the red force arrows still while you switch carrier type, which is the point.

Q: Why are Hall probes made of semiconductors rather than metals?

Because VH goes as 1/n. A metal has around 1029 carriers per cubic metre and gives a fraction of a microvolt; a doped semiconductor has around 1022 and gives volts from the same current and field. There is a row above that computes exactly that comparison for your own numbers.

Put a current-carrying slab in a magnetic field and a voltage appears across it, sideways. Its size tells you how many charge carriers the material has; its SIGN tells you whether they are negative or positive — which is the only straightforward way anyone has of finding that out, and it is why we know some semiconductors conduct by holes.

Last updated 11 Nov 2026

Hall voltage, carrier density and carrier sign

V_H = BI/(ntq)

Current through the slab (A)The Hall voltage is proportional to the current, so a Hall probe is usually driven from a constant-current source rather than a constant voltage. Three amperes through a thin foil is a lot; real probes use small currents through semiconductors instead, for the reason the rows below make obvious.

Magnetic flux density through the slab’s face (T)The field must be perpendicular to the broad face of the slab, going through it — into the page in the diagram. A field in the plane of the slab produces no Hall voltage at all. Cambridge 9702 examines the use of a Hall probe to measure flux density under outcome 20.3.4, and this is why: once you have calibrated n and t, V_H reads B directly.

Thickness of the slab, along the field (mm)The dimension parallel to the magnetic field — the one the field has to travel through. This is the t in V_H = BI/(ntq), and it is the one students pick wrong. A thinner slab gives a bigger Hall voltage, which is why Hall probes are made from very thin films.

Width of the slab, across the current (mm)The dimension across which you measure the Hall voltage. It does not appear in V_H = BI/(ntq) at all — a wider slab has a proportionally weaker Hall field over a proportionally longer distance, and the two cancel exactly. It is needed here only for the drift velocity and the current density rows.

Carrier density (×10²⁸ per m³)Copper is about 8.5 × 10²⁸ m⁻³, roughly one free electron per atom. A doped semiconductor is ten million times lower — enter 0.000001 for 10²² m⁻³ and watch the Hall voltage go from microvolts to volts. That single contrast is why every real Hall sensor is a semiconductor.

What carries the currentThis is the input the whole page exists for. Both kinds of carrier are pushed to the SAME face by the magnetic force — a negative one because it drifts backwards, a positive one because it drifts forwards, and the two reversals cancel. So the face that charges up does not change; the sign of the charge on it does, and with it the sign of the Hall voltage. Switch this and watch the diagram.

A slab carrying current to the right, with the magnetic field going into the page through its broad face. The red arrows are the magnetic force on the carriers, and it pushes them towards the top face whichever sign they have, so the top face is always the one that charges up. What changes when you switch between electrons and holes is the SIGN on each face, and the direction the carriers themselves drift. Switch the carrier type and watch both reverse while the forces stay where they are.

-0.5507µVExample

3.0 A through a 0.20 mm thick copper strip, 5.0 mm wide, in 0.50 T

The sideways voltage, and what each part of it tells you

V_H = BI/(ntq) · V_H = Bv_dw · R_H = 1/nq · v_d = I/(nAq) · at balance: qE_H = qv_dB

t: the thickness along the field, which Cambridge 9702 states explicitly in outcome 20.3.3: “derive and use the expression V_H = BI /(ntq), where t = thickness”. Choosing the wrong one of the slab’s three dimensions is the commonest error on this topic, and nothing in the answer looks wrong when you do
n: the number of mobile charge carriers per cubic metre. This is what the Hall effect measures that nothing else easily does: rearrange to n = BI/(V_Htq) and a voltmeter has counted the free electrons in a lump of metal. About 8.5 × 10²⁸ per m³ for copper, about one per atom
the sign: both kinds of carrier are deflected to the SAME face, because reversing the sign of the charge also reverses its drift direction and the two cancel in qv × B. So the face that charges up never changes — only the sign of the charge on it, and therefore the sign of V_H. That is how we know some semiconductors conduct by positive holes
the equilibrium: charge piles up on one face until its electric field pushes back as hard as the magnetic force pulls: qE_H = qv_dB, so E_H = v_dB and V_H = v_dBw. That is a second route to the same answer and there is a row above checking the two agree
why the width cancels: V_H = v_dBw looks as though a wider slab gives a bigger voltage, but v_d = I/(nAq) and A contains the width, so the two w’s cancel and leave BI/(ntq). Only the thickness survives

Worked example

3.0 A through a 0.20 mm thick copper strip, 5.0 mm wide, in 0.50 T
GET THE THICKNESS RIGHT, AND INTO METRES. t is the dimension along the field: 0.20 mm = 2.0 × 10⁻⁴ m. The 5.0 mm width is a decoy for this formula and is used only for the drift-speed rows.
BUILD THE DENOMINATOR IN TWO STEPS. n·t = 8.5 × 10²⁸ × 2.0 × 10⁻⁴ = 1.7 × 10²⁵ m⁻²; then × e = 1.7 × 10²⁵ × 1.602 × 10⁻¹⁹ = 2.724 × 10⁶.
DIVIDE. V_H = BI/(ntq) = (0.50 × 3.0)/(2.724 × 10⁶) = 1.5/2.724 × 10⁶ = 5.507 × 10⁻⁷ V = 0.5507 µV. Half a microvolt, from three amperes in half a tesla. That is how little a metal gives you.
PUT THE SIGN ON. Copper conducts by electrons, so the carriers are negative, the top face in the diagram collects negative charge and the headline is −0.5507 µV. Switch the page to holes and the magnitude is identical while the sign flips — which is the entire measurement.
CHECK IT THE OTHER WAY. The cross-section is 0.20 × 5.0 mm = 1.0 mm² = 10⁻⁶ m², so v_d = I/(nAq) = 3.0/(8.5 × 10²⁸ × 10⁻⁶ × 1.602 × 10⁻¹⁹) = 0.220 mm/s. Then V_H = Bv_dw = 0.50 × 2.203 × 10⁻⁴ × 5.0 × 10⁻³ = 0.5507 µV. Same number, different route.
NOW SEE WHY PROBES ARE SEMICONDUCTORS. Keep everything else and drop the carrier density to 10²² m⁻³, a typical doped semiconductor. V_H becomes 4.68 V — eight and a half million times larger, and now trivially easy to measure. The Hall voltage goes as 1/n, and that is the whole design argument.

The same 3.0 A in 0.50 T through a 0.20 mm slab

The last two rows are the measurement that matters: identical in every respect except the sign, and the sign is what tells you which kind of doping you have. The first two show why you would never build a sensor out of a metal — the signal is a fraction of a microvolt, buried under thermal noise and under the much larger voltage along the strip. The aluminium figure assumes three free electrons per atom and should be treated as an order of magnitude, not a measurement.
Material	Carriers per m³	Carrier sign	Hall voltage
Copper	8.5 × 10²⁸	negative (electrons)	−0.55 µV
Aluminium, roughly	1.8 × 10²⁹	negative (electrons)	−0.26 µV
n-type semiconductor	10²²	negative (electrons)	−4.68 V
p-type semiconductor	10²²	positive (holes)	+4.68 V

The one experiment that tells you the sign of the charge carriers

A current tells you how much charge is moving, and nothing about what sign it is. Electrons drifting left and positive carriers drifting right are the same current, and no ammeter can separate them. The Hall effect can, and that is the whole reason it is on your syllabus rather than being a curiosity about a sideways voltage.

Here is why the sign survives. The magnetic force on a carrier is qv × B. Change the sign of q and the carrier also drifts the other way, so v changes sign too — and a product with both factors reversed is unchanged. Both kinds of carrier are therefore pushed to the same face. The diagram shows that: switch between electrons and holes and the red force arrows do not move. What does change is the sign of the charge piling up on that face, and therefore the sign of the voltage you measure.

The build-up stops when the electric field it creates fights back hard enough. Charge accumulates until qE_H = qv_dB, so E_H = v_dB and V_H = v_dBw. Substituting v_d = I/(nAq) and cancelling the width leaves Cambridge’s own form, V_H = BI/(ntq) — with t the thickness along the field and the width gone entirely. This page does the calculation both ways and prints a check row so you can see them agree.

Where this goes wrong is almost always the t. A slab has three dimensions and only one of them belongs in this formula: the one the field passes through. Pick the width instead and the answer is wrong by the ratio of the two, with nothing to warn you. The second trap is the size of the thing: for copper the Hall voltage is half a microvolt, which is smaller than the thermal EMFs in your wiring, so a real Hall probe uses a doped semiconductor whose carrier density is millions of times lower and whose Hall voltage is millions of times larger. Drop the carrier density on this page to 0.000001 and watch microvolts turn into volts. For the force this all rests on see the magnetic force on a moving charge, and for how a current relates to drift speed in the first place, drift velocity and current density.

Frequently asked questions

Why do electrons and holes pile up on the same face?

Because the force qv × B has the sign of the charge in it twice. A negative carrier drifts opposite to the conventional current, so both q and v change sign together, and the product does not change at all. Both are pushed the same way. The diagram holds the red force arrows still while you switch carrier type, which is the point.

Then what does change when the carriers are positive?

The sign of the charge that arrives on that face, and therefore the sign of the Hall voltage. The page’s headline goes from negative to positive and the band under it says which carrier that implies. In the diagram the shaded faces swap colour and the plus and minus signs swap over, and the carriers’ own drift arrow reverses.

Which dimension is t?

The one along the magnetic field — the thickness the field passes through, not the width across which you measure the voltage. Cambridge 9702 spells it out in the outcome itself: “where t = thickness”. The width cancels out of V_H = BI/(ntq) entirely, which you can check on this page by changing the width and watching the Hall voltage sit still.

Why are Hall probes made of semiconductors rather than metals?

Because V_H goes as 1/n. A metal has around 10²⁹ carriers per cubic metre and gives a fraction of a microvolt; a doped semiconductor has around 10²² and gives volts from the same current and field. There is a row above that computes exactly that comparison for your own numbers.

Can I use this to measure a magnetic field?

That is the standard use, and Cambridge examines it as outcome 20.3.4. Once n and t are known for a particular probe, V_H is proportional to B, and the sensitivity row above tells you the microvolts per tesla. The row below it says what field a one-microvolt voltmeter could just resolve — which for a metal strip is most of a tesla, and for a semiconductor is a fraction of a microtesla.

Related calculators

References

Syllabus: Cambridge International AS & A Level Physics 9702, for examination in 2025, 2026 and 2027, Cambridge Assessment International Education, cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf (read 11 November 2026). Cited for outcomes a Cambridge candidate is examined on that the Indian board words differently or omits. 20.3.3: “understand the origin of the Hall voltage and derive and use the expression V_H = BI /(ntq), where t = thickness”. 20.3.4: “understand the use of a Hall probe to measure magnetic flux density”. 20.5.2: “recall and use Φ = BA”. 20.5.3: “understand and use the concept of magnetic flux linkage”. 20.5.4 asks candidates to explain experiments demonstrating “that the induced e.m.f. is in such a direction as to oppose the change producing it”. 20.5.5: “recall and use Faraday’s and Lenz’s laws of electromagnetic induction”. Note that induction is topic 20.5 here and not a topic of its own; topic 21 is alternating currents. The copy read was retrieved in two parts and the last 19,825 characters were not read, so nothing is claimed about what is on the data-and-formulae sheet.
University Physics Volume 2, chapters 11 and 14, OpenStax, Rice University (openstax.org/books/university-physics-volume-2, key-equations pages read 11 November 2026). An openly licensed university text, cited here because the coaching sites that dominate searches for this topic are not citable and mostly copy one another. Chapter 11 gives the radius of a charged particle’s path as r = mv/(qB), its period as T = 2πm/(qB), the Hall potential as V = IBl/(neA), the drift velocity in crossed fields as v_d = E/B, the magnetic dipole moment as μ = NIA and the torque as τ = μ × B. Chapter 14 gives L_solenoid = μ₀N²A/l, NΦ = LI, ε = −L dI/dt, M = N₂Φ₂₁/I₁ = N₁Φ₁₂/I₂ and U = ½LI². Neither key-equations page lists a cyclotron frequency or a field energy density, so those two are stated here from the definitions and not attributed to this source.
Fundamental Physical Constants — Frequently used constants (2022 CODATA wallet card), NIST Committee on Data for Science and Technology, physics.nist.gov/cuu/pdf/wallet_2022.pdf (read 11 November 2026). Every constant on this page comes from here, and the thing worth noticing is which numbers have a bracket after them. The elementary charge is printed 1.602 176 634 × 10⁻¹⁹ C and the speed of light 299 792 458 m s⁻¹, both with no bracket at all, because since 2019 they are exact by definition — the coulomb and the metre are now defined from them. But the vacuum electric permittivity is printed 8.854 187 8188(14) × 10⁻¹² F m⁻¹ and the vacuum magnetic permeability 1.256 637 061 27(20) × 10⁻⁶ N A⁻²: both now carry an uncertainty of about 1.6 parts in 10¹⁰, where before 2019 μ₀ was exactly 4π × 10⁻⁷ and ε₀ exactly 1/μ₀c². Almost every school book still prints the old exact forms. At three significant figures nothing changes; which constants are definitions changed completely.
Annual Syllabus, Class XII, Physics, Code 42, Session 2025-26, Central Board of Secondary Education, edustud.nic.in/edu/SYLLABUS_2025_26/12/12_Physics_EM.pdf (read 11 November 2026). The board’s own document, cited and not endorsed by it. Unit I Chapter 1 asks for “Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside)”. Unit III Chapter 4 asks for “force on a moving charge in uniform magnetic and electric fields”, “torque experienced by a current loop in uniform magnetic field” and “Current loop as a magnetic dipole and its magnetic dipole moment”. Unit IV Chapter 6 is one line and covers three of this batch’s records: “Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction”. The Hall effect is not in this document.