# Why (intuitively) do more charge carriers result in a smaller Hall effect?

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.

• $B$ the electric field magnitude: You probably mean $B$ the magnetic field magnitude. – 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.