The Whacko Part

In the previous derivation $B$ and $E$ are in the plane perpendicular to the direction of travel, and in the case of a free charge/dipole.

$B=-i\cfrac{\partial E}{\partial x^{'}}$

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

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

$E_p = \cfrac{q}{4\pi \varepsilon_o (a^2+r^2-2a.r.cos{(\omega.t)})}$

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.