From the equation for the Hall effect:

$$\Delta V_H= \frac{I B}{n q t}$$

[Where $I$ is the current, $B$ the electric field magnitude, $n$ the density of charge carriers, $q$ the charge per charge carrier, and $t$ is the thickness of the conductor/semiconductor]

I can see that the more free charge there is (nq), the smaller the Hall voltage. My question is, why would that be the case? Intuitively I would have thought that more charge carriers might result in more charge accumulated on the top and bottom surfaces of the conductor/semiconductor and a higher Hall voltage, rather than a lower Hall voltage.

  • $\begingroup$ $B$ the electric field magnitude: You probably mean $B$ the magnetic field magnitude. $\endgroup$ – Thomas Fritsch Aug 28 at 9:52

Suppose you have higher $n$ (i.e. more free charges per volume). Then these many charges need to flow only slowly (i.e. with smaller velocity $\mathbf{v}$) in order to make the same current $I$.

Now the Lorentz-force ($\mathbf{F}=q\mathbf{v}\times\mathbf{B}$) is smaller when you have low velocity $\mathbf{v}$. And therefore you get a smaller Hall-voltage between the two edges.

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  • $\begingroup$ Thanks a lot - that perfectly answers my question :) $\endgroup$ – PhysMs Aug 29 at 13:09

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