# The Hall Effect

Suppose you have a conductor.  We’ve already seen that it contains individual charges.  It is easy to measure the current flowing through the conductor.   But, it is not necessarily easy to tell if a positive current corresponds to positive charges moving forwards, or negative charges moving backwards.  It might be possible to find out using magnetism.

Magnetism exerts a force on particles proportional to their charge.  The total force on a particle of charge $q$ and velocity $\vec{v}$ from an electromagnetic field $\vec{E}, \vec{B}$ is:

$\vec{F} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$

So supposing you look at a current $\vec{J} = J\hat{x}$ moving in the $\hat{x}$ direction, with a magnetic field $\vec{B}=B\hat{y}$ in the y direction.  Since no charges are piling up just yet, $\vert \vec{E} \vert = 0$.  So, the force on a free particle of charge $q$ is $\vec{F} = q\vec{v} \times \vec{B} = qvB\left( - \hat{z} \right)$.

We’ve been writing $q$ as positive, but there’s no need for it to be anything other than zero.  Let’s look at the positive and negative cases separately, and see what kind of behavior we find.

When $q$ is negative, the velocity of the particles is in the same direction as the current.  So, the velocity coefficient $v$ is also positive.  Since we’re assuming that $\vec{B}$ is positive (and we will get the opposite result if it is pointing in the opposite direction), we can conclude that a force $F = qvB\left( - \hat{z} \right)$ will push the particle in the $-\hat{z}$ direction.

When $q$ is positive, the velocity of the particles is in the opposite direction as current, making the velocity coefficient $v$ negative.  Since the charge is also negative, a force $F = qvB \left( -\hat{z} \right)$ direction will also push a negative charge towards the $- \hat{z}$ direction.

No matter what sign charge we use, charges will pile up on the $-\hat{z}$ wall of the conductor.  There, they will act like a plate with surface charge density, which we will denote with $\sigma$.  Since the overall material is electrically neutral, there will be a corresponding pileup of not-electrons on the other side of the conductor, corresponding to a surface charge density $-\sigma$We know that the electric field from one such plate is $\vec{E}_{\perp} = \frac{\sigma}{2 \epsilon_0}$, so from two plates we expect a total electric field of magnitude $\frac{\sigma}{\epsilon_0}$.

We could run the current and let the charge build up until the charges reach a steady state.  In this configuration, the electric force from the charged walls of the conductor exactly cancels out the magnetic force from the external magnetic field.

$F_\perp = q \left( E_\perp + \vec{v} \times \vec{B} \right)$ should be zero, so we would expect $\frac{\sigma}{\epsilon_0} = \vec{F}_{E} = \vec{F}_{B} = vB$.  We don’t know $v$, however.  If we know the density of charged particles $\rho$, and we know the current $J$, then we can find the velocity of each particle, $v = J \frac{1}{\rho q}$.

$\sigma = \frac{\epsilon_0 J B}{\rho q}$.

The potential across the plates can then be calculated.