
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 


