
It is relatively easy to calculate the magnitude of the Hall voltage U_{Hall} that is induced by the magnetic field B. 


First we note that we must also have an electrical field E parallel to j because it is the driving force for the current. 


Second, we know that a magnetic field at right angles to a current causes a force on the moving carriers, the socalled Lorentz force F_{L}, that is given by 











We have to take the drift velocity v_{D} of the carriers, because the other velocities (and the forces caused by these componentes) cancel to zero on average. The vector product assures that F_{L} is perpendicular to v_{D} and B. 


Note that instead the usual word “electron” the neutral term carrier is used, because in principle an electrical current could also be carried by charged particles other than electrons, e.g. positively charged ions. Remember a simple but important picture given before! 

For the geometry above, the Lorentz force F_{L} has only a component in y – direction and we can use a scalar equation for it. F_{y} is given by 





F_{y} 
= 
– q · v_{D,x} · B_{z} 







We have to be a bit careful: We know that the force is in ydirection, but we do no longer know the sign. It changes if either q, v_{D}, or B_{z} changes direction and we have to be aware of that. 


However, it is important to note that for a fixed current density j_{x} the direction of the Lorentz force is independent of the sign of the charge carriers – the sign of the charge and the sign of the drift velocity just cancel each other. 

With v_{D} = µ · E and µ = mobility of the carriers, we obtain a rather simple equation for the force 





F_{y} 
= 
– q · µ · E_{x} · B_{z} 






This means that the current of carriers will be deflected from a straight line in ydirection. In other words, there is a component of the velocity in ydirection and the surfaces perpendicular to the ydirection will become charged as soon as the current (or the magnetic field) is switched on. The flowlines of the carriers will look like this: 











The charging of the surfaces is unavoidable, because some of the carriers eventually will end up at the surface where they are “stuck”. 


Notice that the sign of the charge for a given surface depends on the sign of the charge of the carriers. Negatively charged electrons (e^{–} in the picture) end up on the surface opposite to posively charged carriers (called h^{+} in the picture). 


Notice, too, that the direction of the force F_{y} is the same for both types of carriers, simply because both q and v_{D} change signs in the force formula 

The surface charge then induces an electrical field E_{y} in ydirection which opposes the Lorentz force; it tries to move the carriers back. 


In equilibrium, the Lorentz force F_{y} and the force from the electrical field E_{y} in ydirection (which is of course simply q · E_{y}) must be equal with opposite signs. We therefore obtain 





– q · E_{y} 
= 
– q · µ · E_{x} · B_{z} 



E_{y} 
= 
sgn(q) · µ · E_{x} · B_{z} 






The Hall voltage U_{Hall} now is simply the field in ydirection multiplied by the dimension d_{y} in ydirection. 


It is clear then that the (easily measured) Hall voltage is a direct measure of the mobility µ of the carriers involved, and that its sign or polarity will change if the sign of the charges flowing changes. 

It is customary to define a Hall coefficient R_{Hall} for a given material. 


This can be done in different, but equivalent ways. In the link we look at a definition that is particularly suited for measurements. Here we use the following definition: 





R_{Hall} 
= 
E_{y}
B_{z} · j_{x} 






In other words, we expect that the Hall voltage E_{y} · d_{y} (with d_{y} = dimension in ydirection) is proportional to the current(density) j and the magnetic field strength B, which are, after all, the main experimental parameters (besides the trivial dimensions of the specimen): 





E_{y} 
= 
R_{Hall} · B_{z} · j_{x} 






The Hall coefficient is a material parameter, indeed, because we will get different numbers for R_{Hall} if we do experiments with identical magnetic fields and current densities, but different materials. The Hall coefficient, as mentioned before, has interesting properties: 


R_{Hall} will change its sign, if the sign of the carriers is changed because then E_{y} changes its sign, too. It thus indicates in the most unambiguous way imaginable if positive or negative charges carry the current. 


R_{Hall} allows to obtain the mobility µ of the carriers, too, as we will see immediately. 

R_{Hall} is easily calculated: Using the equation for E_{y} from above, and the basic equation j_{x} = s · E_{x}, we obtain for negatively charged carriers: 





R_{Hall} 
= – 
µ · E_{x} · B_{z}
s · E_{x} · B_{z} 
= – 
µ
s 
= 
– µ
q · n · µ 
= 
– 1
q · n 






The blue part corresponds to the derivation given in the link; n is (obviously) the carrier concentration. 


If one knows the Hall coefficient or the carrier concentration, the Hall effect can be used to measure magnetic field strengths B ( not so easily done otherwise!). 

Measurements of the Hall coefficient of materials with a known conductivity (something easily measurable) thus give us directly the mobility of the carriers responsible for the conductance. 


The minus sign above is obtained for electrons, i.e. negative charges. 


If positively charged carriers would be involved, the Hall constant would be positive. 


Note that while it is not always easy to measure the numerical value of the Hall voltage and thus of R with good precision, it is the easiest thing in the world to measure the polarity of a voltage. 

Let’s look at a few experimental data: 




Material 
Li 
Cu 
Ag 
Au 
Al 
Be 
In 
Semiconductors
(e.g. Si, Ge, GaAs, InP,…) 
R
(× 10^{–24})
cgs units 
–1,89 
–0,6 
–1,0 
–0,8 
+1,136 
+2,7 
+1,774 
positive or negative values, depending on “doping” 
Comments:
1. the positive values for the metals were measured under somewhat special conditions (low temperatures; single crystals with special orientations), for other conditions negative values can be obtained, too.
2. The units are not important in the case, but multiplying with 9 · 10^{13} yields the value in m^{3}/Coulomb 


