\(|B|=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{q}{a^3}\)
If we consider the superposition of many moving charges,
\(B=\sum^{all-vectors}_i{\cfrac{1}{4\pi \varepsilon_o }.\cfrac{q_i}{a^3_i}}\)The effects of moving charges before and after the point in consideration is zero. \(B\) is entirely due to a thin sheet of charges immediately below that point, at the foot of the normal, \(n\).
\(B=\cfrac{1}{4\pi \varepsilon_o }.\sum^{all-vectors}_i\cfrac{{q_{sheet_i}}}{{a^3}}\)
If \(a\) is large,
\(B=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{1}{{a^3}}\sum^{all}_i{q_{sheet_i}}\)
This sheet of charge \(J_A\) is given by,
\(J_A=\sum^{all}_i{q_{sheet_i}}=\cfrac{I}{v}\)
where \(I\) is the current and \(v\) is the average velocity of the charges. \(J_A\) would be the current density. This expression for \(J_A\) is also valid for conductors where the charges runs on the surface. The B-field is then given by,
\(B=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{1}{{a^3}}.\cfrac{I}{v}\)
The \({1}/{a^3}\) dependence is at odd with Ampere's Law for B-field around a current carrying wire which has a \(\frac{1}{a}\) factor.
For the normal \(E_{pn}\) component, using similar simplifications,
\(E_{pn} =B.a=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{1}{{a^2}}.\cfrac{I}{v}\)
This means, if Hall Effect were used to obtain \(B\) we would find \(E_{pn}\) has a \(1/a^2\) dependence, assuming that the dipole model for free charge is right, that \(E_{pn}\) is responsible for Hall Effect.
