Question on Hall Effect and magnetic force

In summary: Hall voltage can be used to characterize the type of doping materialIn summary, the Hall effect is a phenomenon in which a magnetic field and a current in a conductor result in a perpendicular electric field. This effect is caused by the Lorentz force acting on charge carriers. The direction of the Hall voltage can determine the type of charge carriers present in the conductor, with the voltage being in opposite directions for positive and negative charge carriers. This can be used to characterize the type of material present in doped semiconductors.
  • #1
dainceptionman_02
18
4
so with a Hall Voltage, you have positive current traveling upwards in a wire in the +y-direction and a magnetic field into the screen in the -z-direction. the right hand rule has positive charge deflecting to the left. now if you look at the drift velocity of electrons moving downward in the -y-direction, the negative of the right hand rule has the electrons deflecting to the left. if both positive and negative charge deflect to the left, then why is it assumed that there is a net negative charge on the left hand side of the wire and a positive charge on the right causing a Hall Voltage with an electric field pointing from the right to left?

with magnetic forces, the force is perpendicular to the direction of the field. if this is so, then why do permanent magnets stick together or repel in a direction that seems parallel to the direction of the field. same with solenoids or whatever creates a constant magnetic field that uses a magnet to pick up cars in the dump lot.
 
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  • #2
The force on a charge carrier is given by the Lorentz force
$$\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}).$$
Now if you have a DC, then ##\vec{F}=0##, i.e., the electric field is given by
$$\vec{E}_{\text{perp}}=-\vec{v} \times \vec{B}.$$
Now you have the driving voltage such that
$$\vec{j}=q v \vec{e}_y, \quad q v>0.$$
If now ##q>0## you have thus ##v>0## and thus (according to your description)
$$\vec{E}_{\perp}=-v B \vec{e}_y \times (-\vec{e}_z)=v B \vec{e}_x,$$
i.e., the potential is
$$V_{\text{H}}=-v B x.$$
If ##q<0##, then ##v<0## and thus
$$\vec{E}_{\text{perp}}=-v B \vec{e}_x,$$
i.e., the Hall voltage is
$$V_{\text{H}}=+v B x,$$
i.e., it's in the opposite direction as if the charge carriers are positive. Thus, with the Hall effect you can check, whether the conducting particles are positive or negative. For usual metallic conductors these are electrons and thus negatively charged. In some semiconductors the conduction is due to the motion of positively charged "quasiparticles", i.e., "missing electrons"/"holes", and for them the Hall voltage is opposite than in metallic conductors.
 
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  • #3
If you have both types of carriers they have different drift velocities (different mobilities) so the Hall voltages don't cancel out even if they have the same concentration. In doped semiconductors they have both different concentrations and different mobilities
 
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