\(B=-i\cfrac{\partial E}{\partial x^{'}}\)
For \(E_p\) at a distance \(a\) from the center of the circle, The distance \(R\) from the negative charge at an angle \(\theta\) on the circumference of the circular path of radius \(r\),
\(R^2 =a^2+r^2-2a.r.cos{(\theta)}\)
and
\(E_p = \cfrac{q}{4\pi \varepsilon_o R^2}\), so
\(E_p = \cfrac{q}{4\pi \varepsilon_o (a^2+r^2-2a.r.cos{(\omega.t)})}\)
where both \(a\) and \(r\) are constants and \(\theta = \omega.t\) where \(\omega\) is the angular velocity of the negative charge.
Consider, the tangential and normal component of \(E_p\),
\(E_{pt} = \cfrac{q}{4\pi \varepsilon_o (a^2+r^2-2a.r.cos{(\theta)})}.cos{(\theta)}\)
\(E_{pn} = \cfrac{q}{4\pi \varepsilon_o (a^2+r^2-2a.r.cos{(\theta)})}.sin{(\theta)}\)
For \(\theta\) very small,
\(\triangle x^{'}= Rcos{(\theta)}=cos(\theta)\sqrt{a^2+r^2-2a.r.cos{(\theta)}}\)
\(\cfrac{\partial \theta}{\partial x^{'}}=-\cfrac{\sqrt { { a }^{ 2 }+{ r }^{ 2 }-2a.r.cos(\theta) }}{sin(\theta )(({ a }^{ 2 }+{ r }^{ 2 })-3a.r.{ cos }(\theta ))}\)
Therefore,
\(|B|=\cfrac{\partial E}{\partial x^{'}}=\cfrac{\partial E}{\partial \theta}\cfrac{\partial \theta}{\partial x^{'}}=\cfrac{\partial E_{pt}}{\partial \theta}\cfrac{\partial \theta}{\partial x^{'}}\)
Since the normal component does not effect the \(x^{'}\) direction.
\(\cfrac{\partial E_{pt}}{\partial \theta}=-\cfrac{({ a }^{ 2 }+{ r }^{ 2 })sin(\theta)}{({ a }^{ 2 }+{ r }^{ 2 }-2a.r.cos(\theta))^{ 2 }}.\cfrac{q}{4\pi \varepsilon_o}\)
\(|B|=-\cfrac{({ a }^{ 2 }+{ r }^{ 2 })sin(\theta)}{({ a }^{ 2 }+{ r }^{ 2 }-2a.r.cos(\theta))^{ 2 }}*-\cfrac{\sqrt { { a }^{ 2 }-2a.r.cos(\theta)+{ r }^{ 2 } }}{sin(\theta )(({ a }^{ 2 }+{ r }^{ 2 })-3a.r.{ cos }(\theta ))}.\cfrac{q}{4\pi \varepsilon_o}\)
\(|B|=\cfrac{({ a }^{ 2 }+{ r }^{ 2 })}{({ a }^{ 2 }+{ r }^{ 2 }-2a.r.cos(\theta))^{ 3/2 }}*\cfrac{1}{(({ a }^{ 2 }+{ r }^{ 2 })-3a.r.{ cos }(\theta ))}.\cfrac{q}{4\pi \varepsilon_o}\)
If we define \(R_s=\cfrac{a.r}{{ a }^{ 2 }+{ r }^{ 2 }}\)
\(|B|=\cfrac{q}{4\pi \varepsilon_o ({ a }^{ 2 }+{ r }^{ 2 })^{3/2}}*\cfrac{1}{(1-2.R_s.cos(\theta))^{ 3/2 }}*\cfrac{1}{(1-3.R_s.{ cos }(\theta ))}\)
When \(r\) is small compared to \(a\), \(a\) is also the perpendicular distance to the axis of travel,
\(|B|=\cfrac{1}{4\pi \varepsilon_o }.\cfrac{q}{a^3}=\cfrac{1}{4\pi \varepsilon_o }.q/a^3\)
At this point, we know that \(B\) curves around the direction of the moving charge but Gauss's Equation in either integral or differntial form, gives no indication of this fact. Compare this expression with Ampere's Law for B-field around a current carrying wire, we understand why current density, \(J\) was defined and used.
