More Whacko, Jacko

From the post "The Whacko Part",

$|B|=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{q}{a^3}$

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.