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