Explain Hall effect.
Consider a slab of material wherein there is a current density J resulting by an applied electric field Ex in the x- direction. The electrons will drift along with an average velocity V‾_{x} in the x direction; while a magnetic field of flux density B_{z} (wb/m^{2}) is superposed upon the applied electric field in the Z direction the electrons will experience a Lorentz force perpendicular to V‾_{x }and to B_{z}; the magnitude of this force will be specified by B_{z} (µx)e.
Thus the electrons are driven towards one face in the sample resulting in an excess of electrons near one face and a deficiency of electrons near the other face. These charges will in turn create a counteracting electric field E_{y} in the y-direction. E_{y} would build up, till this is of sufficient magnitude to compensate the lorentz force exerted on the electrons because of the magnetic field we may hence write e E_{y} =B_{z} e( V‾_{x} ). In steady state, a Hall voltage, V_{H}, is thereby put in the y-direction specified by
V_{H} = E_{y} .a =B_{z} (V‾_{x}), a
The current density in the sample is specified with:
J_{x} = N. e (V‾_{x})
Here N=number of conduction electrons/m³
The current density can be computed from the total current and the cross section (a x b) of the sample.
Therefore,
J_{x} = I/a x b = N. e (V‾_{x})
I = N. e (V‾_{x}) a x b ....................................(1)
V_{H} = B_{Z} (V‾_{x}) . a ......................................(2)
Eliminating (V‾_{x}) from equating (1) and (2) we get:
V_{H} = I/(N.e.b)
= (1/N.e) ((B_{Z}.I)/b)
The ratio + (1/N.e) = Ey/(J_{X}.B_{Z}) should be constant.
Another variable that is frequently used to explain the Hall Effect is the ratio of the currents J_{Y} to J_{X}.
This is termed as the Hall angle and is denoted with θ θ = JY / JX = σ EY / JX
= σ RH BZ
= µ H BX
Here µ H is termed as the Hall mobility. The hall angle is equal to the average no. of radians traversed through a particle between collisions.