Last updated: 2026-07-13

Hall Voltage

The Hall voltage is the transverse potential difference that appears across a current-carrying conductor when it is placed in a magnetic field perpendicular to the current. The field bends the paths of the moving charge carriers, pushing them toward one edge of the sample until the charge that accumulates there produces an electric field strong enough to balance the deflection. The voltage between the two edges at that balance point is the Hall voltage, and this calculator finds it from the current, the magnetic flux density, the carrier density, and the sample thickness using $V_H = I B / (n q t)$.

Where the formula comes from

A carrier of charge $q$ moving with drift velocity $v_d$ through a magnetic field $B$ feels a Lorentz force $F = q\, v_d\, B$ that pushes it sideways. Charge builds up on one edge of the sample and creates a transverse electric field $E$. Equilibrium is reached when the electric force cancels the magnetic force:

The Hall voltage is that field times the sample width $w$, so $V_H = E\, w = v_d\, B\, w$. The drift velocity is set by the current, $I = n q\, v_d\, (w t)$, where $w t$ is the cross-sectional area, so $v_d = I / (n q\, w t)$. Substituting gives:

The width cancels, leaving the Hall voltage in terms of the current, field, carrier density, charge, and thickness alone.

Formula

The quantity $R_H = 1/(n q)$ is called the Hall coefficient, so the relation is often written $V_H = R_H\, I B / t$. Its sign follows the sign of the dominant carrier.

Worked example

Consider a lightly doped semiconductor strip carrying $I = 1\ \text{A}$ in a field of $B = 0.5\ \text{T}$, with a carrier density $n = 1\times10^{25}\ \text{m}^{-3}$ and a thickness $t = 0.1\ \text{mm} = 1\times10^{-4}\ \text{m}$. Each carrier holds the elementary charge $q = 1.602\times10^{-19}\ \text{C}$:

That is about 3.1 mV — a comfortably measurable signal, which is why devices like this are used as magnetic-field sensors.

Metals versus semiconductors

Run the same numbers for a metal such as copper, with $n = 8.5\times10^{28}\ \text{m}^{-3}$, and the Hall voltage collapses to roughly $3.7\times10^{-7}\ \text{V}$ — less than a microvolt. The difference is entirely due to the carrier density in the denominator: metals pack in tens of thousands of times more free electrons than a lightly doped semiconductor, so each one carries a smaller share of the current and the transverse pile-up is far weaker. This is why practical Hall sensors and Hall probes are made from semiconductors, where the carrier density is low enough to yield a robust output.

Reading the carrier sign

The polarity of the Hall voltage carries information beyond its size. Because negative electrons and positive holes are deflected toward opposite edges of the sample, the sign of the Hall voltage tells you whether the current is dominated by electrons or by holes. Measuring it is one of the standard ways to distinguish n-type from p-type semiconductors, and rearranging the formula as $n = I B / (q t\, V_H)$ turns a Hall-voltage measurement into a direct determination of the carrier density.

Limitations

The formula assumes a single dominant carrier type with a uniform density, a magnetic field exactly perpendicular to the current, and a thin sample so the field is uniform across it. In materials where electrons and holes contribute comparably, the two partly cancel and the simple expression no longer holds. At high fields or low temperatures the Hall response can also depart from the classical result. For a well-defined single-carrier sample in a moderate field, though, $V_H = I B / (n q t)$ is an accurate and widely used relation.