Why do conductivities and resistivities change with magnetic field strength?

[SchroedingersLion]

Greetings,

assume we have a 2-dimensional system in the x-y-plane. An electric field is applied in x-direction, a magnetic field is applied in z-direction. As is well-known, the charge carriers get pushed in the y-direction due to the Lorentz-force until the Hall field is strong enough to counteract this motion. In steady state, there will thus be no current in y-direction.

The magneto-conductivity terms are given by
$$J_x = \sigma_{xx} * E_x + \sigma_{xy} *E_y $$
and
$$J_y= \sigma_{yx} * E_x + \sigma_{yy} *E_y
$$

The magneto-resistivities (given by inversion of the $\sigma$ matrix) are given by
$$E_x = \rho_{xx} * J_x + \rho_{xy} *J_y $$
and
$$E_y= \rho_{yx} * J_x + \rho_{yy} *J_y
$$

It holds that $\sigma_{xx}=\sigma_{yy}$ and $\sigma_{xy}=-\sigma_{yx}$. Same for $\rho_{ij}$.

A plot of these quantities w.r.t strength of the magnetic field is attached.

I am trying to understand them qualitatively.

It makes sense that $\sigma_{ii}$ decrease with increasing B, as more and more charge carriers are bound to create the Hall field $E_y$, meaning $E_x$ needs to be stronger and stronger to keep the current flowing.
On the same line, one can argue that $\sigma_{xy}$ has to decrease from 0 into the negative numbers, as the Hall-field removes charge carriers from their motion in x-direction.
First question: Why does this effect saturate? Why is there an extremum in $\sigma_{ij}$?

Now to the resistivities:
At B=0 (and in an isotropic system), the resistivity $\rho$ is simply the inverse of the conductivity $\sigma$. Here, however, $\rho_{xx}$ stays constant even though $\sigma_{xx}$ goes to zero.
I was trying to explain it like this: The conductivity describes the strength of the current that gets created by a field. The resistivity gives the resistance against a current that is already flowing. In other words, $\sigma_{xx}=0$ means that no current can flow, whereas $\rho_{xx}>0$ means that a current WOULD experience a resistance if it could flow.
Second question: Does this make sense?

Third question: Why does $\rho_{yx}$ (or, as in the figure, $-\rho_{xy}$) increase with B?
Is it because the growing B-field increases $E_y$ which, again, draws away charges from their motion in x-direction, effectively increasing the resistance in x-direction?

It's funny, I did all the maths to find expressions for the different matrix components, but it is harder to me to understand it intuitively.
SL

 

[trurle]

First question: Why does this effect saturate? Why is there an extremum in $\sigma_{ij}$?

I never encountered Hall effect saturating at practical sensors. Readout circuitry may saturate though. The extreme on your plot is likely due transition from Hall effect to cyclotron resonance, which is possible but require very high magnetic field.

Second question: Does this make sense?

Partially. sigma_{xx}=0 mean no average current. Single electrons will still move in random directions after scattering, contributing to resistance in that direction even if sigma_{xx}=0

Third question: Why does $\rho_{yx}$ (or, as in the figure, $-\rho_{xy}$) increase with B?
Is it because the growing B-field increases $E_y$ which, again, draws away charges from their motion in x-direction, effectively increasing the resistance in x-direction?

Correct.
In other words: Electrons after scattering (and losing part of velocity), would accelerate in average at angle to X direction, due Hall field. The movement direction is re-aligned with X axis by Lorenz force as electron gathering speed, making paths of each electron between scatterings curved, therefore increasing probability of scattering per unit of X-axis travel (hence increasing resistivity).

 

[SchroedingersLion]

Thank you for the answer trurle.

I never encountered Hall effect saturating at practical sensors. Readout circuitry may saturate though. The extreme on your plot is likely due transition from Hall effect to cyclotron resonance, which is possible but require very high magnetic field.

We are not talking about practical sensors, just physics :)
So, at a certain B strength, instead of reaching the boundaries in y direction, the electrons will be localized in their cyclotron gyration since the cyclotron radius R got too small. This would explain the maximum in $\sigma_{yx}$. At very high B fields, the Lorentz-force will thus not be able to transfer as many electrons to the edges in y-direction, meaning $\sigma_{yx}$ goes to zero.

At the same B field $\sigma_{xx}$ has a point of inflection. It decreases less quickly. I would say this is because due to the smaller R, electrons remain longer at the vicinity of the x-axis and have thus again a higher chance of being scattered in x-direction, which might slow down the initial decrease in $\sigma_{xx}$.
You agree?

Correct.
In other words: Electrons after scattering (and losing part of velocity), would accelerate in average at angle to X direction, due Hall field. The movement direction is re-aligned with X axis by Lorenz force as electron gathering speed, making paths of each electron between scatterings curved, therefore increasing probability of scattering per unit of X-axis travel (hence increasing resistivity).

This is the other way around, right? So, the speed between collisions is typically higher than the drift velocity of electrons. That means that between collisions, the Lorentz-force on the electrons is higher than the electrostatic force from the Hall field. Thus, electrons want to leave the x-axis to enter circular motion. Scattering events happen, and, on average, the Hall field compensates the Lorentz-force.
However, due to their tendency to leave the x-axis, they will scatter more often per distance traveled in x-direction?