The SI unit of the Hall coefficient is the cubic metre per coulomb, written m³/C or equivalently m³ A⁻¹ s⁻¹. Its dimensional formula is [M⁰ L³ T⁻¹ A⁻¹]. The Hall coefficient is defined as RH = Ey / (Jx Bz), which for a single type of carrier simplifies to RH = 1/(nq), where n is the carrier density in m⁻³ and q is the charge on one carrier in coulombs.
What is the Hall coefficient?
Send a current through a conductor and place it in a magnetic field at right angles to that current. The moving carriers feel a magnetic force, drift to one face of the slab and pile up there. That build-up creates a transverse electric field, and the voltage across the two side faces is the Hall voltage. Edwin Hall measured this in 1879, before the electron was discovered.
The Hall coefficient is the number that fixes how big that transverse field is:
RH = Ey / (Jx Bz)
- Ey is the Hall electric field, across the sample, in V/m
- Jx is the current density along the sample, in A/m²
- Bz is the magnetic flux density, perpendicular to both, in tesla
In the simple one-carrier (Drude) model this reduces to RH = 1/(nq). For electrons q = −e, so RH = −1/(ne). For holes q = +e, so RH = +1/(pe).
In a laboratory you rarely measure E and J directly. For a slab of thickness t carrying current I with Hall voltage VH, the working formula is RH = VH t / (I B). The width cancels out, which is why only the thickness appears.
SI unit of Hall coefficient
Take the definition apart unit by unit:
RH = (V/m) ÷ [(A/m²) × T]
Write the tesla in base units. Since B = F/(qv), one tesla is 1 kg·s⁻²·A⁻¹, and in electrical terms 1 T = 1 V·s/m². Substituting:
RH = (V/m) ÷ [(A/m²) × (V·s/m²)] = (V/m) × (m⁴)/(A·V·s) = m³/(A·s)
And because 1 coulomb = 1 ampere second, m³/(A·s) is the same thing as m³/C.
The same answer falls out of RH = 1/(nq) in one step: n is measured in m⁻³ and q in coulombs, so 1/(nq) has units of m³/C. Two different routes, one unit, which is a good sign you have the formula right.
Other valid ways to write the same unit
| Written as | Why it is the same | Value of 1 m³/C |
|---|---|---|
| m³ C⁻¹ | Index notation for m³/C | 1 |
| m³ A⁻¹ s⁻¹ | 1 C = 1 A·s | 1 |
| m³/(A·s) | Same as above, written as a fraction | 1 |
| Ω·m/T (ohm metre per tesla) | Ω·m ÷ T = (V·m/A) ÷ (V·s/m²) = m³/(A·s) | 1 |
| V·m/(A·T) | The measurement form VHt/(IB) | 1 |
| cm³/C | Practical unit in older papers | 10⁶ |
| Ω·cm/G (ohm centimetre per gauss) | 1 Ω·cm = 10⁻² Ω·m and 1 G = 10⁻⁴ T | 0.01 |
So 1 Ω·cm/G equals 100 m³/C, and 1 m³/C equals 10⁶ cm³/C. If an old data table quotes RH in cm³/C, divide by 10⁶ to get m³/C.
Dimensional formula of Hall coefficient, derived
Do not memorise this one. Build it from the three quantities in the definition.
| Quantity | SI unit | Dimensions |
|---|---|---|
| Electric field E | V/m = N/C | [M¹ L¹ T⁻³ A⁻¹] |
| Current density J | A/m² | [M⁰ L⁻² T⁰ A¹] |
| Magnetic flux density B | T = kg·s⁻²·A⁻¹ | [M¹ L⁰ T⁻² A⁻¹] |
Multiply the two in the denominator first:
[J][B] = [L⁻² A] × [M T⁻² A⁻¹] = [M¹ L⁻² T⁻² A⁰]
Now divide:
[RH] = [M L T⁻³ A⁻¹] ÷ [M L⁻² T⁻² A⁰]
Subtract the exponents one base at a time. Mass: 1 − 1 = 0. Length: 1 − (−2) = 3. Time: −3 − (−2) = −1. Current: −1 − 0 = −1. That gives
[RH] = [M⁰ L³ T⁻¹ A⁻¹]
Cross-check it against 1/(nq). Carrier density n has dimensions [L⁻³] and charge q has dimensions [A¹ T¹], so 1/(nq) is [L³ A⁻¹ T⁻¹]. The two derivations agree, and L³ T⁻¹ A⁻¹ read back as a unit is exactly m³/(A·s) = m³/C. Note the mass exponent is zero, so the Hall coefficient is dimensionally independent of mass.
Solved example with units carried through
Problem. A semiconductor slab of thickness t = 0.50 mm carries a current I = 10 mA. A magnetic field B = 0.40 T is applied perpendicular to the slab, and a Hall voltage VH = +6.0 mV is measured across it. Find the Hall coefficient, the carrier concentration and the carrier type.
Step 1: convert everything to SI.
t = 0.50 mm = 5.0 × 10⁻⁴ m, I = 10 mA = 1.0 × 10⁻² A, VH = 6.0 mV = 6.0 × 10⁻³ V, B = 0.40 T.
Step 2: apply RH = VH t / (I B).
RH = (6.0 × 10⁻³ V × 5.0 × 10⁻⁴ m) / (1.0 × 10⁻² A × 0.40 T)
RH = (3.0 × 10⁻⁶ V·m) / (4.0 × 10⁻³ A·T) = 7.5 × 10⁻⁴ V·m·A⁻¹·T⁻¹
Step 3: reduce the unit. V·m/(A·T) = (V·m/A) ÷ T = Ω·m/T = m³/C, from the table above. So
RH = +7.5 × 10⁻⁴ m³/C
Step 4: carrier concentration. From RH = 1/(nq) with q = 1.602 × 10⁻¹⁹ C,
n = 1/(RH q) = 1 / (7.5 × 10⁻⁴ m³/C × 1.602 × 10⁻¹⁹ C) = 1 / (1.20 × 10⁻²² m³) ≈ 8.3 × 10²¹ m⁻³
The coulombs cancel and m³ inverts to m⁻³, which is the correct unit for a number density. Step 5: RH came out positive, so the carriers are holes and the sample is p-type.
A useful by-product: Hall mobility is μH = |RH| σ, where σ is conductivity in S/m. The units work out as m³/C × A/(V·m) = m²/(V·s), the standard mobility unit.
What the sign of the Hall coefficient tells you
The sign is the most useful single piece of information the Hall effect gives, because it identifies the majority carrier. The reason is that the magnetic force qv × B pushes electrons and holes to the same side of the slab (they travel in opposite directions and carry opposite charge), but the charge that accumulates there has the opposite sign in the two cases. That flips the polarity of the Hall voltage.
| Sign of RH | Majority carrier | Material type | Formula |
|---|---|---|---|
| Negative | Electrons | n-type semiconductor, most metals | RH = −1/(n e) |
| Positive | Holes | p-type semiconductor | RH = +1/(p e) |
Doping a silicon wafer with phosphorus (group 15) adds electrons and gives a negative RH. Doping with boron (group 13) adds holes and gives a positive one. This is exactly how a fabrication lab confirms which way a wafer was doped, and how strongly.
Two honest caveats. Aluminium and a few other metals show a positive Hall coefficient at high fields, about +1.0 × 10⁻¹⁰ m³/C, which the free-electron model cannot explain; it takes band theory and a hole-like Fermi surface. And when both electrons and holes conduct at once, as in an intrinsic semiconductor, RH depends on both densities and both mobilities, so the simple 1/(nq) form no longer applies.
Typical values: metals vs semiconductors
Because RH = 1/(nq), a large carrier density gives a tiny Hall coefficient. Metals have roughly 10²⁸ to 10²⁹ carriers per cubic metre, doped semiconductors around 10²⁰ to 10²³, so the coefficients differ by seven to nine orders of magnitude.
| Material | Carrier density n (m⁻³) | RH (m³/C) | Carrier |
|---|---|---|---|
| Copper | 8.5 × 10²⁸ | about −5.5 × 10⁻¹¹ (measured) | Electrons |
| Silver | 5.9 × 10²⁸ | about −9 × 10⁻¹¹ | Electrons |
| Sodium | 2.65 × 10²⁸ | about −2.5 × 10⁻¹⁰ | Electrons |
| Aluminium (high field) | – | about +1.0 × 10⁻¹⁰ | Hole-like |
| Doped semiconductor | 1 × 10²² | 6.2 × 10⁻⁴ (from 1/nq) | Either sign |
| Lightly doped semiconductor | 1 × 10²⁰ | 6.2 × 10⁻² (from 1/nq) | Either sign |
Values for the metals are measured room-temperature figures and differ somewhat from the free-electron estimate 1/(ne); sodium agrees well, copper does not. The semiconductor rows are calculated straight from 1/(nq) with q = 1.602 × 10⁻¹⁹ C, so you can reproduce them.
Student tip for the lab: because RH is so small in metals, the Hall voltage in a copper foil is in microvolts and needs a sensitive amplifier. In a germanium slab it is in millivolts and a normal multimeter will read it. That is why every undergraduate Hall effect experiment uses a semiconductor sample.
Mistakes to avoid
- Using the width instead of the thickness. In RH = VHt/(IB), t is the dimension along the magnetic field, not the distance between the Hall probes. The probe separation cancels out.
- Leaving millimetres in the formula. Convert t to metres and the current to amperes before substituting, or the power of ten will be wrong by three or six.
- Quoting the unit as m³/A. The seconds are part of it: m³ A⁻¹ s⁻¹, which equals m³/C.
- Dropping the sign. RH is a signed quantity. A magnitude alone loses the carrier type.
- Confusing RH with Hall resistance. Hall resistance Rxy = VH/I is measured in ohms. The Hall coefficient is a material property in m³/C.
References
- Edwin Hall and the 1879 discovery of the Hall effect, Johns Hopkins University.
- NCERT Physics – Current Electricity and Magnetism, Class 12, for the force on a moving charge.
- NPTEL – Semiconductor Physics lectures on carrier transport and Hall measurements.
FAQs
What is the SI unit of Hall coefficient?
The SI unit of the Hall coefficient is the cubic metre per coulomb, m³/C. It can also be written m³ C⁻¹, m³ A⁻¹ s⁻¹ or ohm metre per tesla, and all four are numerically identical.
What is the dimensional formula of Hall coefficient?
The dimensional formula is [M⁰ L³ T⁻¹ A⁻¹]. It comes from dividing the dimensions of electric field [M L T⁻³ A⁻¹] by those of current density [L⁻² A] times magnetic flux density [M T⁻² A⁻¹], and the mass exponent cancels to zero.
What is the CGS unit of Hall coefficient?
In practical CGS work the Hall coefficient is quoted in cm³/C or in ohm centimetre per gauss. One m³/C equals 10⁶ cm³/C, and one ohm centimetre per gauss equals 100 m³/C.
What does a negative Hall coefficient mean?
A negative Hall coefficient means the majority charge carriers are electrons, so the sample is an n-type semiconductor or an ordinary metal. A positive value means holes carry the current, which indicates a p-type semiconductor.
Why is the Hall coefficient larger in semiconductors than in metals?
Because RH = 1/(nq) is inversely proportional to carrier density. A metal has about 10²⁸ carriers per cubic metre and a doped semiconductor about 10²², so the semiconductor’s Hall coefficient is millions of times bigger and its Hall voltage is far easier to measure.