Imagine a conductor with rectangular cross-section d x t that is parallel to the y-z plane with (conventional) current passing in the positive x direction. With a magnetic field acting in the positive y direction, there is a Lorentz force on the charge carriers given by $F = qv_{d}B$. Fleming’s left hand rule suggests that both positive and negative charge carriers should experience a force to the side of the conductor.

However, this build up of charge creates the Hall electric field that prevents the build-up of further charge. A steady state is reached where:

$$\begin{aligned} F_{B} &= F_{E} \\ Bqv_{d} &= qE \\ Bv_{d} &= \frac{V}{d} \\ B &= \frac{V}{v_{d}d} \\ V &= Bv_{d}d \end{aligned}$$

, where V is the hall voltage,

v_{d} is the drift velocity,

d is the width of the conductor.

Since $I = Anv_{d}q$,

$$v_{d} = \frac{I}{Anq}$$

Combining the equation for V and v_{d},

$$\begin{aligned} V &= Bd \left( \frac{I}{Anq} \right) \\ V &= \frac{BI}{nq} \left( \frac{d}{A} \right) \end{aligned}$$

Since the cross-sectional area A is equal to the product of the width of the sample d and its thickness t, (A = dt)

$$V = \frac{BI}{nqt}$$

The hall effect is used extensively to study conduction in materials, particularly in semiconductors. We have assumed so far that the mobile charge carriers within solids are electrons and this is in agreement with the sign of the Hall voltage for most materials. However, anomalous results can be obtained for metals such as aluminium or indium, and some semiconducting materials behave as if there were positive charge carriers in the materials.