Hall Effect

The Hall effect is the generation of a voltage between a conductor place in both an electric and magnetic field, and is due to the Lorentz Force.

Let’s start with a simple apparatus of a Current carrying wire, and a Conductor between it. In this explanation, we assume both the wire and the conductor to be perfect conductors, i.e. have a conductivity of $\infty$.

The current flows directly through the conductor, as it would in a wire (because a wire is just a thin cylindrical conductor). It does redistribute itself as per the Current Density formula, but other than that, nothing interesting happens.

Well...

The wire and conductors do generate Magnetic Fields, as magnetic fields form whenever a current is created. They would form concentric-ish ring shapes. Here, however we choose to ignore them. Or assume that the new magnetic field we are about to add is stronger than this induced one.

Now, what happens if we add an external Magnetic Field, perpendicular to the current? A Magnetic Force would be generated, as the magnetic field is perpendicular to the electron flow (current).

Note that electrons are negative charge carriers, and as such the magnetic force is actually the opposite direction to that given by the Right Hand Rule.

This force causes the electrons to ‘bend’, resulting in them building up on an edge. Remember, a charge build-up causes a Voltage difference.

Hall Voltage §

One thing to note is that inside the conductor, the electric field generated can be likened to that generated by a Parallel Plate Capacitor:

$$\vec{E}=\frac{\Delta V}{w}$$

$d$ is the length of the conductor.

To achieve the steady state configuration, where ${\vec{F}}_E={\vec{F}}_B$ is achieved when:

$$q\vec{v}\times \vec{B}=q\vec{E}$$

Assuming magnetic field is perpendicular to the electric field, and setting $q=q_e$ and $v=v_{drift}$

$$\cancel{q_e}v\vec{B}=\cancel{q_e}\vec{E}$$

$$\Delta V=w{v}_{drift}\vec{B}$$

We can rearrange this equation using the formula for Current Density:

$$v_{drift}=\frac{I}{A{\rho }_e{q}_e}\to \frac{I}{{\rho }_e{q}_{e}lw}A=l\times w$$

$$\Delta V_{hall}=\frac{\cancel{w}I\vec{B}}{{\rho }_e{q}_{e}l\cancel{w}}$$

$$\Delta V_{hall}=\frac{IB}{{\rho }_e{q}_{e}l}$$