Hall Effect — Physica

Equivalent forms

R_H = \frac{1}{nq}

E_H = v_d B

V_H = R_H \frac{I B}{t}

One transverse voltage simultaneously delivers the carriers' sign, their density, and (in 2D, quantized) a window into topological physics — a remarkably information-dense measurement.

Symbol	Name	SI unit	Dimension	Range
$V_H$	Hall voltageoutput Transverse voltage across the strip	V	M·L²·T⁻³·I⁻¹	0 – 0.001
$I$	Current Longitudinal current through the strip	A	I	0.1 – 10
$B$	Magnetic field Field perpendicular to the strip	T	M·T⁻²·I⁻¹	0 – 2
$n$	Carrier density Charge carriers per unit volume	m⁻³	L⁻³	1e+22 – 1e+29
$t$	Thickness Strip thickness along B	m	L	0.00001 – 0.01

Unit systems

Symbol	SI	CGS	Imperial
$V_H$	V	statV	V
$I$	A	statA	A
$B$	T	G	T
$n$	m⁻³	cm⁻³	m⁻³
$t$	m	cm	in

Where it holds

Classical formula valid for a single carrier type with t much larger than the mean free path. Multi-carrier conductors need a weighted R_H; ultra-thin cold samples in strong fields show the quantum (integer/fractional) Hall plateaus instead.

Dimensional analysis

$IB/(nqt) = A\cdot T / (m^{-3} \cdot A\cdot s \cdot m)$ = $(A\cdot kg\cdot s^{-2}\cdot A^{-1})/(m^{-2}\cdot A\cdot s) = V \checkmark$

Discovery

Edwin Hall · 1879

As a graduate student, Hall tested whether a magnet pushes the current or the wire itself. He found a measurable transverse voltage across a thin gold leaf — proof the force acts on the moving charges. The sign later revealed that some materials conduct via positive 'holes'.

Try this

How does your phone know which way is up, and how do engineers count electrons they can't see? Both lean on a sideways voltage discovered in 1879.

A copper strip 1 mm thick carries 1 A in a 0.5 T perpendicular field. With n = 8.5×10²⁸ m⁻³, find the Hall voltage across its width.

Research status: stable

Real-world applications

Position/angle and current sensors (brushless motors, wheel speed)
Smartphone magnetometers / digital compasses
Characterizing semiconductor carrier type and density
Resistance metrology via the quantum Hall effect

Common misconceptions

The Hall voltage is along the current — it's perpendicular to both I and B
Only electrons matter — the sign of V_H distinguishes electron vs hole conduction
Thicker samples give bigger signal — $V_H \propto 1/t$ , so thin films are more sensitive

Experimental verification

A calibrated Hall probe in a known field returns

V_H = IB/(nqt)

to high precision; reversing B flips its sign. The quantum Hall effect (von Klitzing, 1980) gives

R_H = h/e^{2}

plateaus so exact they define the resistance standard.

Derivation

In steady state the transverse electric force balances the magnetic force:

qE_H = qv_d B

, so

E_H = v_d B

.

The transverse voltage is

V_H = E_H w = v_d B w

.

With

v_d = I/(nqA) = I/(nq\cdot wt)

, the width w cancels:

V_H = (I/(nq\cdot wt))\cdot B\cdot w = IB/(nqt)

.

Limiting cases

B = 0

⟶

V_H = 0

No magnetic deflection means no transverse charge separation.

n \to

small (semiconductor)⟶ V_H largeFew carriers must each carry more current, so the deflection and voltage are big — why Hall sensors use semiconductors.

t \to

thin (2D)⟶

V_H \propto 1/t

, quantizesIn a clean 2D electron gas at high B and low T, R_H jumps in steps of

h/(e^{2}\nu )

— the quantum Hall effect.

What if…

What if the carriers are positive holes?

They drift the same way as conventional current but deflect to the opposite edge — V_H reverses sign, revealing p-type conduction.

What if you use a semiconductor with n a million times smaller?

V_H grows a million-fold to easily-measured millivolts — the basis of practical Hall sensors.

What if the sample is 2D, very cold, and B is huge?

R_H locks onto quantized plateaus $h/(\nu e^{2})$ , independent of material details — the quantum Hall effect.

1

Copper strip Hall voltage

Given ·

I:: 1
B:: 0.5
n:: 8.5e+28
q:: 1.602176634e-19
t:: 0.001

Find · V_H

Steps

Denominator $nqt = 8.5\times 10^{28} \times 1.602\times 10^{-19} \times 10^{-3} \approx 1.36\times 10^{7}$
$V_H = 0.5 / 1.36\times 10^{7} \approx 3.7\times 10^{-7} V$
Sub-microvolt — metals make poor Hall sensors; semiconductors (small n) are far better

Result ·

V_H = (1\times 0.5)/(8.5\times 10^{28} \times 1.602\times 10^{-19} \times 10^{-3}) \approx 3.7\times 10^{-7} V

Lorentz Force Law→Drift Velocity→Magnetic Force on a Current-Carrying Wire→