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 -\sigmaWe 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.




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