The Hall effect includes the transverse (to the flow of current) electric field set up by the charges which accumulate on the edges, to counter the magnetic component of the Lorentz force acting on them to move towards the edges. These charges can be both positive holes and negative electrons in semiconductors. If the concentration of any one kind of carriers is very high as compared to other(n-type and p-type), the electric field set up can be understood from the accumulation of charges (negative for n-type and vice versa) at one edge due to the Lorentz force acting on it. But if the intrinsic semiconductors contain the same density of the positive holes and the electrons, both move in opposite directions to contribute to the current but the lorentz force acting on them is in the same direction. And hence their deflection owing to the magnetic field is in the same direction.

This means, both the positive and negative charges are deflected towards the same edge and hence cannot produce a net charge density and therefore should not be able to set up an countering electric field and the Hall potential. Does this mean that they will continuosly keep being deflected towards one of the edge and keep annihilating at that edge, never setting up a counter electric field?

If the difference in mobility (or the effective mass) makes any difference, what would happen if they have the same mobility and consequently the same effective mass?

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    $\begingroup$ Basically Hall's effect determines whether the material is n-type or p-type, if we have an intrinsic(undoped) semi-conductor( having same mobility for both the carriers) obviously your argument is correct. But am not sure because practically mobilties are different for e's and h's; Further its hard to perform an experiment for your hypothesis. If you find something please post as an answer i would also like to study it:) $\endgroup$
    – user31782
    Commented Jan 2, 2014 at 11:13

2 Answers 2


The Hall voltage is given by the equation

$$V_H = \frac{I~B}{n~e~d}$$

Where $I$ = current, $B$=applied magnetic field, $d$ = width of conductor, $n$ = density of mobile carriers, and $e$ is the charge of an electron.

The sign of the voltage depends on the predominant charge carrier, and it is possible to imagine a material where the positive and negative carriers cancel - this would be a material that doesn't exhibit a Hall effect.

Now an undoped (intrinsic) semiconductor has equal numbers of electrons and holes - because each electron that is thermally excited from its valence band leaves behind one hole. However, their mobility is typically not the same - so an intrinsic semiconductor will have higher mobility electrons, and the Hall effect will have the same polarity as N-type semiconductor. If you added a tiny amount of P-type dopant, you could increase the density of the holes to offset their lower mobility, and neutralize the Hall effect.

In fact, the drift velocity of the charge carriers is given by $v_D=\frac{I}{neA}$ (where $A$ is the cross sectional area). When the charge carrier density is very high, the velocity is very low (e.g. in good metallic conductors); when the carrier density is lower (semiconductors), the velocity is much higher. It is actually possible to move a conductor in a magnetic field at sufficient speed to make the Hall voltage disappear, and this can be used to measure the drift velocity.


We should consider about the mobility of electrons and holes. Intrinsic semiconductors have a negative hall coefficient such as in metals. If we consider the mass action law , we can find the carrier densities when the hall coefficient is equal to zero. http://physics111.lib.berkeley.edu/Physics111/Reprints/SHE/11-Hall_Effect_in_Semiconductors.pdf


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