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.
