# Is there a limit on the magnetic field in a Hall generator?

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:

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?

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.