Not To Be taken Too Seriously, Please

There was a reason why this configuration, with $x$ through the center of the loop, was chosen

instead of taking the tangential component at angle $\phi$.

Because it is a charge moving away from a circular disc.  The $E$ field of this charge passes through the area of the disc.  As the charge perform circular motion, this $E$ field though the disc changes.  The component of this field perpendicular to the area of the disc, induces a $B$ field along the perimeter of the disc.  The $B$ field is given by Maxwell's Law of electromagnetism.

$\nabla \times B=\mu_o\varepsilon_o\cfrac{\partial\,E}{\partial\,t}$

So, a changing $E$ field through a disc, at a distance $x$ from a the electron is,

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r$

The normal component when this charge has traveled through an arc $\phi$ is,

$\cfrac { \partial \, E_{ A } }{ \partial t } sin(\sigma) =2\pi E_{ o }\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r=2\pi E_{ o }\Pi$

but,

$\sigma=\cfrac{\pi-\phi}{2}=\cfrac{\pi}{2}-\cfrac{\phi}{2}$

$sin(\sigma)=sin(\cfrac{\pi}{2}-\cfrac{\phi}{2})=cos(\phi/2)$

and

$\Pi=\cfrac { x^{ 2 } }{ a_{ e } }\cfrac { dx }{ dt } \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 } -1}  \right\}  } { d\, y }+  \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right\}$

So,

$\cfrac { \partial \, E_{ A } }{ \partial t } cos(\phi/2) =2\pi E_{ o }cos(\phi/2)\cfrac { x^{ 2 } }{ a_{ e } }\cfrac { dx }{ dt } \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }+  \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right\}$

and

$\nabla\times B=\oint{B}d\,r=2\pi rB_o$

where $B=B_o$ is a constant around the loop.

$2\pi rB_o=\mu_o\varepsilon_o2\pi E_{ o }cos(\phi/2)\cfrac { x^{ 2 } }{ a_{ e } }\cfrac { dx }{ dt }\\ \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }+  \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right\}$

Since,

$x^2=r^2_e+r^2_e-2r^2_ecos(\phi)=2r^2_e(1-cos(\phi))$

$x=r_e\sqrt{2(1-cos(\phi))}=2r_{ e }sin(\phi /2)$

$\cfrac { dx }{ dt }=r_{ e }cos(\phi/2)\cfrac{d\,\phi}{d\,t}$

$x^{ 2 } \cfrac { dx }{ dt }=4r^3_{ e }sin^2(\phi /2)cos(\phi/2)\cfrac{d\,\phi}{d\,t}$

and let $Z= \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }+  \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right\}$

$\cfrac{d\,\phi}{d\,t}=\omega$    and    $E_o=\cfrac{q}{4\pi\varepsilon_o a^2_e}$

We have,

$B_o=\cfrac {\mu_o q \omega }{4\pi a^3_{ e } r }.cos(\phi/2)4r^3_{ e }sin^2(\phi /2)cos(\phi/2).Z$

$v=r\omega$  where $v$ is the speed of the electron,

$B_{ o }=\cfrac {\mu_o q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}.sin^{ 2 }(\phi ).Z$

The term $\cfrac {\mu_o q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}.Z$.  Given a radial distance $r$ from the orbit of the electron producing $B$, this term is a constant.  $r_e$ is the electron orbital radius, $a_e$ is the radius of the electron.  $q$ is the electron charge.

Although $sin^2(x)$ is much less interesting than $\cfrac{1}{2}sin(2\phi)cos(\phi /2)$.  Ths post "Electron Orbit B Field II" dated 17 Oct 2016 is wrong.