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Suppose I have a current of both negative and positive charges(I know that there is also current from only negative and only positive charges,I'm not confused) along an infinite wire of square cross-section. Now suppose we put a homogeneous magnetic field normal to the current.

Then a Lorentz force acts on both the positive and negative charges and it is in the same direction for both of them. In other words, negative charges move in the opposite direction of the positive charges. This folows from the equation of the Lorentz force.The negative sign from the negative charge and the negative sign from the velocity vector cancel each other out. Here I am considering the positive velocity vector to be in the direction that the positive charges are moving.

But because the forces act in the same direction AND because negative and positive charges attract each other,then we will have accumulation of both negative AND positive charges near one end of the wire.

So,the question is, will there be a potential difference transverse to the wire like the normal classical Hall effect?

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  • $\begingroup$ Look up the ExB drift in a plasma. There you'll see that the drift is charge-independent, thus if both positive and negative charges are allowed to move freely in uniform, static fields, there will be no net current. $\endgroup$ Commented Mar 1, 2021 at 23:29

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In real-life conductors you always have some of both kind of charge carriers (electrons and holes), so, the main question is, which of those is more abundant?

In fact materials with positive and negative hall coefficients can be found.

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The Hall voltage can indeed be equal to zero if the electrons and holes balance out. You can find the formula in these Hall Effect lab notes by Pengra, Stoltenberg, Dyck, Vilches, eq. 16:

$$R_H = \frac{1}{|q|}\frac{n_h \mu_h^2 - n_e \mu_e^2}{(n_h \mu_h + n_e \mu_e)^2}$$

where $\mu$ is mobility and $n$ is density and $e$ means electron and $h$ means hole and $\pm q$ is the charge of an electron or hole. So if $n_h\mu_h^2 = n_e\mu_e^2$, no Hall voltage.

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  • $\begingroup$ so a am right then? $\endgroup$ Commented Mar 18, 2015 at 9:15

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