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In a Hall generator, charge accumulation on electrodes occurs because there is a generated current, $\bf{J}$, in a magnetic field, $\bf{B}$. This results in charge separation due to a Hall current pushing the electrons by the $\bf{J} \times \bf{B}$ force. As desired, this will produce a net voltage over the electrodes since the electrons will eventually impinge upon an anode presumably. This general Hall effect is displayed below:

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You'll notice the negative charges, which constitute $\bf{J}$, are initially moving straight and then are deflected to hit the walls separated by $w$. Therefore, the trajectory of the charged particle is given by the $\bf{J} \times \bf{B}$ centripetal force and will simply be a gyro-orbit with radius, $\rho = mv / qB$—the charged particle will simply make circles around the magnetic field with this radius. My question is therefore: if the strength of the magnetic field is too large (i.e. high B), then $\rho$ will be very small. But if $\rho$ is too small, the electron may not even hit the sides if $w$ is larger than $\rho$, correct?

If the electrons don't hit the electrodes (or sides in this picture) because their gyro-orbits are too small, how could a Hall MHD generator work? Therefore, is the maximum magnetic field strength, B, set by the condition of sufficiently close spacing between electrodes, i.e. $\rho > w$? Or do we assume the charged particles eventually collide with particles moving straight in the x-direction and continually keep deflecting via their gyro-orbit, and this will move the charged particles closer toward the wall?

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My question is therefore: if the strength of the magnetic field is too large (i.e. high B), then $\rho$ will be very small. But if $\rho$ is too small, the electron may not even hit the sides if $w$ is larger than $\rho$, correct?

The electron velocity in a Hall-effect circuit like this under realistic fields will be determined by the ExB-drift speed, which is given by: $$ \mathbf{V}_{ExB} = \frac{ \mathbf{E} \times \mathbf{B} }{ B^{2} } \tag{1} $$ where $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic field vectors, respectively. One can see from Equation 1 that the drift speed of the electrons will go as $V_{ExB} \propto E/B$. Thus, a smaller ratio will result in a smaller drift.

So long as the applied $\mathbf{E}$ and $\mathbf{B}$ are the only fields and they remain orthogonal to each other, the Hall current is given by: $$ \mathbf{j}_{Hall} = - n_{e} \ e \ \mathbf{V}_{ExB} = - n_{e} \ e \ \frac{ \mathbf{E} \times \mathbf{B} }{ B^{2} } \tag{2} $$ where $n_{e}$ is the number density of electrons undergoing the ExB-drift and $e$ is the fundamental charge. Therefore, one can see that the magnitude of the JxB-force, $\mathbf{F}_{jxB}$, is independent of the magnitude of $\mathbf{B}$. It does, however, depend upon the magnitude of the component of $\mathbf{E}$ orthogonal to $\mathbf{B}$.

Therefore, is the maximum magnetic field strength, B, set by the condition of sufficiently close spacing between electrodes, i.e. $\rho > w$?

Equation 1 alters the particle gyroradius to give: $$ \rho = \frac{ m }{ q \ B } \ \lvert \frac{ \mathbf{E} \times \mathbf{B} }{ B^{2} } \rvert \tag{3} $$ Thus, the particle gyroradius in this setup goes as $\rho \propto E/B^{2}$.

Using Equations 1 and 3, we can show that: $$ \begin{align} \rho & = \frac{ m \ E }{ q \ B^{2} } > w \tag{4a} \\ & \frac{ m \ E }{ q \ w } > B^{2} \tag{4b} \\ & \sqrt{ \frac{ m \ E }{ q \ w } } > B \tag{4c} \end{align} $$ which relates the electrode width to the applied electric and magnetic fields. It's another way of saying that the width requirement depends upon the ratio of $E/B$. In the limit of small $E$ a particle will undergo simple gyro motion orthogonal to $\mathbf{B}$.

Or do we assume the charged particles eventually collide with particles moving straight in the x-direction and continually keep deflecting via their gyro-orbit, and this will move the charged particles closer toward the wall?

In a perfect conductor, no collisions are not an issue. In a semi-conductor, yes, this is an issue and there are numerous articles on computing the mobility of electrons within such materials.

However, these types of collisions generally cause random walk-like trajectories which means there would not be a preference for any given direction. Such collisions tend to be strong in this limit, so they would dominate over any preferential direction caused by the external $\mathbf{E}$ in the limit where $\rho \ll w$.

Hall effect magnetic field probes have designed ranges of sensitivity/response and so are only effective over a defined range of magnetic field strengths. So yes, if you apply a field that is too large (or too small), a Hall effect probe will not work.

Note: All of the above assume classical field limits, i.e., I am not including pulsar-like fields.

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