Variation of Current In The Hall Effect Experiment

Question

As a 12th Grader, our class came across a very simple setup of the Hall Effect Experiment during our course on Electromagnetism. The entire idea and eventual steady state conditions all make perfect sense, but the only thing that bothers me is the effect of current BEFORE the steady state is achieved.

Googling the question never gave results that discussed the situation prior to the achievement of steady state, nor did my teachers give a satisfying answer.

The more I thought about it, 2 contradicting thoughts popped up over and over again -

• If the charges accumulated to one side of the conductor in the presence of the magnetic field, it would imply that more electrons entered the conductor than those that exited out of it leading to a decrease in the current in the circuit.
• If the above were true, Ohm's law would be disobeyed as neither the potential difference driving the circuit nor the resistor of the conductor changed which should imply that the current must remain constant.

I would also like some insight into the work done by the battery to maintain the current if the 2nd thought is correct.

Mind you, I'm well aware of the fact that I may lack a lot of the background needed to understand what's really going on here, let alone justify this effect. Considering the fact that the very idea of relativity and dependence of magnetic force on frame of observation left huge holes in my understanding of Physics, I really hope I can fill out some of these and come to satisfying conclusions.

Thank you for spending your time to help a curious one out :)

Answer 1

The lateral accumulation of charge is not incompatible with a constant current. Intuitively, this is because you have a deficit of charge on one side and an accumulation on the other by the same amount. You therefore do not need to appeal to an exterior source of charge to generate the Hall voltage, it can be described as an internal rearrangement of charges of the conductor. While you don’t need Ohm’s law for the Hall effect, it is a necessary ingredient to understand the transient behavior.

I’ll describe a simplified model of the transient behavior. Say that the current density and electric in the conductor is uniform (assuming a rectangular geometry guarantees that they will stay so). The longitudinal component is fixed by the imposed current. The vertical component is free to vary, and in the stationary regime, it should be zero. The transient regime is merely an exponential relaxation to the stationary solution. In fact, this is a standard situation in electronics, it is an RC circuit. The capacitance is between the lateral edges of the conductor and the resistance is given by the finite conductivity of the conductor.

Doing the math, the relaxation time for a homogeneous rectangular conductor is: $$ \tau=\frac{\epsilon_0}{\sigma} $$ with $\sigma$ the conductivity of the metal. If initially the sides of the conductor were not charged as you turn on the longitudinal current density by imposing the current, the transverse current would immediately turn on as well and relax back to zero. At the same time, the transverse electric field and the lateral charge would saturate at the same rate.

Note that the finite conductivity is crucial to describe the transience. Without it the relaxation time is zero, so the stationary state sets in immediately. For a more complicated geometry, the calculation would be a lot more involved as the current density and electric field would carry a spatial dependence.

Hope this helps.

Answer to comment

Due to the finite resistivity, the battery would beed to supply a constant power which will be dissipated. This injected/dissipated power is independent of the applied magnetic field, in particular, it is the same as when there is no magnetic field. However, you are correct that you need to supply extra energy during transient to separate the charges in order to generate the Hall voltage.

This is the same as charging a capacitor and is therefore given by the same formula $E=\frac{1}{2}QV$, with $V$ the Hall voltage and $Q$ the capacitors charge. You can view the “capacitor” to be conductor, the armatures being its lateral edges. To calculate the capacitance, in the case of rectangular geometry, the same formula applies: $$ C=\epsilon_0 \frac{A}{d} $$ with $A$ the area of the lateral faces and $d$ the distance between the two.