I Question on Hall Effect and magnetic force

AI Thread Summary
The discussion clarifies the Hall effect, explaining how positive and negative charge carriers deflect in a magnetic field, leading to a net charge separation that creates a Hall voltage. It emphasizes that while both positive and negative charges deflect to the left, the resulting electric field points from right to left due to the nature of charge carriers in conductors. The Lorentz force equation is used to illustrate how the force on charge carriers is affected by both electric and magnetic fields. The conversation also notes that in metallic conductors, electrons are the primary charge carriers, while semiconductors can have both electrons and "holes," leading to different Hall voltages. Ultimately, the Hall effect can be used to determine the type of charge carriers in a material based on the observed voltage direction.
dainceptionman_02
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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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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.
 
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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