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?

 A: 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.
A: 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.
