Hall Effect

From the previous post $B$ generated by a moving charge/dipole, has a $E$ component perpendicular to its direction along the perpendicular to the axis of of travel of the charge/dipole.  This normal component is given by,

$E_{pn} = \cfrac{q}{4\pi \varepsilon_o (a^2+r^2-2a.r.cos{(\theta)})}.sin{(\theta)}$

where $a$ is the perpendicular distance to the axis of travel, $q$ is the negative charge of the dipole, $r$ is the radius of the circular path on which the negative charge is rotating and $\theta$ is the location of the negative charge on the circular path.

Since,

$sin(\theta)=\cfrac{R}{a}=\cfrac{\sqrt{(a^2+r^2-2a.r.cos{(\theta)})}}{a}$

$\therefore E_{pn} = \cfrac{q}{4\pi \varepsilon_o.a.\sqrt{(a^2+r^2-2a.r.cos{(\theta)})}}$

when r is small,

$E_{pn} = \cfrac{q}{4\pi \varepsilon_o.a^2}=B.a$,   since $B=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{q}{a^3}$

It seem that $E_{pn}$ has a bigger effect that $B$.

For a conductor placed perpendicular to the B-field, this normal E-field component $E_{pn}$ acts across the conductor producing a voltage difference across the surface of the conductor.  $E_{pn}$ could be responsible for Hall Effect.