In order to make a neuron fire via an external electrode, the electrode must input some electrons in the extracellular space(just outside of the neuron). Imagine a straight conducting wire touching a neuron and other end free in the air. If you move a strong magnet near the wire, such that the wire cuts the magnetic flux, then voltage will be generated at the two ends of the wire. In this case will the neuron fire? or equivalently, will there be flow of electrons to the neuron? Note that the wire is in open circuit i.e. other end is in the air. If there will not be any flow of electrons, can some simple modification on this setup will make it happen?


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Your goal is to move the electrons in a neuron with a magnetic field. If we assume a homogeneous magnetic field $\mathbf{B}$, then this will generate an electrodynamic force


that acts on a charge $q$ which moves with the velocity $\mathbf{v}$.

If we move the magnet from orthogonal to the direction of the neuron-axis, then this is equivalent to moving the electors in the reverse orthogonal direction.

Let us assume that $\mathbf{v}=-v\mathbf{e}_y$ points in the negative $y$-direction. The magnetic field is pointing in the negative $z$-direction $\mathbf{B}=-B\mathbf{e}_z$ and we look at one electron with the charge $q=-1.609\cdot 10^{-19} \text{C}$. Then using $\mathbf{F}=q\mathbf{v}\times\mathbf{B}=-qvB\mathbf{e}_x$ (see picture).

The problem is that your electron will try to stay on a circular trajectory with radius $r$. This is obvious because your force is orthogonal to your velocity. The radius can be determined by assuming an equilibrium of the electrodynamic force and the centripetal force (both with magnitudes).


If you would rotate the neuron-axis according to this knowledge, then you can move the electron from one side to the other side. But it is quite complicated compared to a standard electric field. Another more theoretical way would be to use a very small magnetic field so that the radius becomes very large, but that will lead to very slow accelerations because $\mathbf{F}$, will be very small.

enter image description here


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