# Direction of Hall current

A Hall current arises when electric currents transverse to a magnetic field exist. In this figure (found online), the Faraday current appears to be going upwards due to a magnetic field pointing into the page, but is this correct?

I'm largely just trying to understand the origin and implications of the Hall current here: The Faraday current density, $${\bf{J_F}} = \sigma \bf{v} \times \bf{B}$$, is upwards since $$\bf{v}$$ is to the right and $$\bf{B}$$ into the page. (I realize the conductivity tensor is anisotropic due to the Hall effect, but I'm trying to understand how the Hall effect itself arises here.) Thus the force density from the Hall effect is $$\bf{F_H} = \bf{J_F} \times \bf{B}$$, but this current associated to the Hall effect is pointing to the left, correct?

To understand the Hall effect better: Why are the external current links (orange) connected in this configuration necessary? Shouldn't only the electrodes connected by the light blue arrows be externally connected in a circuit since that's the direction of net current flow? Also, if the Hall parameter is sufficiently large, wouldn't the Faraday current go to 0?

In MHD theory, the generalised Ohm law is used to determine the current density. For situations encountered in MHD generators the law is given by :

Where σ is the conductivity of the fluid, and β is the Hall parameter. Please note that this equation has some omitted terms which are insignificant for typical situations encountered in MHD generators. The cross product term to the left is what you refer to as the Faraday current, while the cross product term in the right hand side is the Hall current.

The Faraday current term gives a perpendicular current to both the velocity and the applied magnetic field. The Hall current term gives a current that is perpendicular to the Faraday current and the magnetic field (which means it is parallel to the velocity as the figure shows). In that sense, the overall current is titled as it is the vectorial addition of both current components, with the tilt being a result of the Hall current.

A good explanation of the theory behind MHD generators can be found in this paper and this presentation.

In terms of the blue and the orange currents, they are the same with the only difference being that blue current is the induced current in the plasma while the orange current is the flowing current in the circuit. So it is the same current plotted with two colours and directions just to show that the current circulates the circuit.

EDIT: In MHD generators the power delivered to the load as the earlier presentation shows is given by:

Where all parameters were defined earlier, except the loading factor K which is defined by:

Such that RL is the Load resistance and RG is the generator resistance. The factor K(1-K) arises naturally in the power equation as a result of Kirchoff's Current Law, when it is applied in the plasma and in the external circuit. The full derivation of this expression can be found in pages 21-22 of this PhD thesis.

• Thank you for the links—they are excellent and are helping. 1. In the presentation you posted, slide 5 (bottom left), is there any reason for short circuiting the electrodes? Why not just leave the inner electrodes unconnected and simply connect the pairs at either end as already done? 2. For the Hall generator on slide 5, since the Hall current is in the direction of the initial fluid flow, shouldn't the electrodes be facing with their normal parallel to the Hall current? 3. Why is it optimal for $R_g = R_L$? I understand $P = E \cdot J = \sigma u^2 B^2$, but where does $K(1-K)$ come from? – Mathews24 Sep 30 '18 at 23:52
• @Mathews24 1. The short—circuited pairs are the ones facing each other (not the adjacent) to maximise the collected Hall current from all faces 2. If you put the electrodes facing the Hall current, the plasma won't flow because it crashes into them, so all currents will be zero. 3. The factor K(1-K) means that if the load and generator resistances match, all the absorbed Hall current will be used to generate power in the load. If not, a portion of the Hall current will accumulate on the electrode creating and electric field limiting further current from arriving. (No space left for more words) – Gotaquestion Oct 1 '18 at 17:49
• Is it necessary to short circuit the electrodes directly? I suppose diagonal generators would be preferred if using the net current. But if the electrode plates are not perpendicular to the current, how does charge accumulate on them to produce the voltage? Is not the Hall current density in the center of the channel essentially doing nothing then if it's too far away from the electrodes? Do you mind just expanding upon 3 in the main answer? I feel this is essentially one of the core concepts I am missing as I don't quite see the direct derivation nor understand $K(1-K)$ as you explain it. – Mathews24 Oct 2 '18 at 5:47
• As far as I could tell for 3, $K = 0.5$ was optimal since the Hall current is based on the Faraday current itself first forming and trying to maximize the balance of both of them. Does the expression $K(1-K)$ arise in Faraday generators? Perhaps a simple derivation of this quantity will help close this question. – Mathews24 Oct 2 '18 at 5:52
• @Mathews24 The answer is updated to address your 3rd point. For the first point I am not 100% sure but from most diagrams of MHD generators you can find on the internet, you find that they have an increasing cross section in the area where the electrodes are put. I assume that is done to be make the angle between the resultant current and the electrode as close as possible to 90. – Gotaquestion Oct 2 '18 at 17:10

The diagram is correct. Remember that the Hall effect current direction depends on the type of charge carrier, regardless of the current convention that is used. This is how the Hall effect can be used to determine what is the majority charge carrier in a material.

In a plasma, both electrons and ions are charge carriers, but they have very different mobilities, so the net Hall effect current is not zero. The physics can be a bit complicated, especially with the way electrons and ions interact with each other, but the net Hall effect current is in the direction shown in that picture. Refer to this article for some experimental measurements on Hall effect in plasmas.

I agree that they should provided a lot more explanation as to where that current direction is coming from, as it's not immediately obvious.

As to why the electrodes are connected in series, it's just to provide high-voltage, low-current power output. They would be designed so that the currents are matched. To connect them in parallel, you would need to design it so that the voltages are matched instead, and that may be more difficult in this setup.

• I understand the constant current in the diagram, but what makes you say this configuration is high-voltage, low-current? Also, what exactly does it mean to 'match' voltages in an MHD generator? If the currents are all matched in each of the external current links, and they each have the same constant external load applied, are not the voltages also matched inherently since $V = IR$? – Mathews24 Oct 1 '18 at 6:26
• Beautiful explanation and this makes more sense upon a second read. To clarify, a Hall current would not exist then in an electron-positron plasma? It appears the $J \times B$ force accelerates the different charge carriers, but their mobilities vary in typical large ion-electron plasmas, which causes a net current. – Mathews24 Oct 8 '18 at 23:35
• If the positive and negative charge carriers in a material behave the same (e.g. neither were a majority carrier), then yes, there is no net hall current. – Al Nejati Oct 8 '18 at 23:43
• Interesting, this paper appears to verify that via simulation! So if $J$ was composed of an equal number of electrons and positrons moving in opposite directions (to thus produce a net current), there would be no net hall current. But if $J$ was composed of only moving electrons (and the positrons were stationary for some reason), there would be a net Hall current present, correct? – Mathews24 Oct 8 '18 at 23:49
• Yes that is correct. – Al Nejati Oct 9 '18 at 0:45