Continuing from the post "Electron Orbit B Field", we replace
\(E=\cfrac{e}{4\pi\varepsilon_o d^2}\) with,
\(E=E_oe^{-\cfrac{x}{a_{e}}}\),
where \(a_e\) is the radius of the electron, and
\(E_o=\cfrac{e}{4\pi\varepsilon_o a^2_e}\)
This exponential form of \(E\) was developed analogously, using exponential gravity as guide.
Total \(E\) over flat circular x-section when the charge is at a distance \(x\) from the loop,
\(E_A=\int_{0}^{r} {2\pi xtan(\theta). E_oe^{-\cfrac{x}{a_{e}cos(\theta)} } }d\,r\)
\(E_{ A }=2\pi E_{ o }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r\)
\(\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r\)
Cosinder the term,
\(\Pi=\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r\)
\( \Pi=\int _{ 0 }^{ r }{ \cfrac { d\, x }{ d\, t } tan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }} \\+xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }\cfrac { -1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } \\+xsec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } e^{-\cfrac{x}{a_{e}cos(\theta)} }d\, r\)
\( \Pi=\int _{ 0 }^{ r }{ e^{-\cfrac{x}{a_{e}cos(\theta)} }\left\{ { \cfrac { d\, x }{ d\, t } tan(\theta ) }-xtan(\theta )\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } +xsec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } \right\} } d\, r\)
Since,
\( \cfrac { r }{ x } =tan(\theta )\)
\( -\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } \)
\(xsec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t }=-\cfrac { r }{ x } \cfrac { d\, x }{ d\, t }=-\cfrac { d\, x }{ d\, t }tan(\theta) \)
Substitute into the expression for \(\Pi\),
\(\Pi=\int _{ 0 }^{ r }{ e^{-\cfrac{x}{a_{e}cos(\theta)} }\left\{ { \cfrac { d\, x }{ d\, t } tan(\theta ) }-xtan(\theta )\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } -\cfrac { d\, x }{ d\, t }tan(\theta ) \right\} } d\, r\)
\( \Pi=-\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ r }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r \)
and
\( \cfrac { r }{ x } =tan(\theta )\)
\( \cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ dr } \)
\( dr=xsec^{ 2 }(\theta ){ d\, \theta }\)
\( \Pi=-\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ r }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } }xsec^{ 2 }(\theta ){ d\, \theta } \)
Substitute \(\Pi\) back into \(E_A\),
\(\cfrac { \partial \, E_{ A } }{ \partial t } =-2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }} xsec^{ 2 }(\theta ){ d\, \theta }\)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =-2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ sec^{ 4 (\theta )}e^{-\cfrac{x}{a_{e}cos(\theta)} } } sin(\theta ){ d\, \theta }\)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac{1}{cos^{ 4 }(\theta )}e^{-\cfrac{x}{a_{e}cos(\theta)} } } { d\, (cos(\theta )) }\) --- (*)
This however, is not the final solution. We have to consider the fact that the charge is in orbit of radius \(r_e\) and the loop defining \(B\) is centered at the circumference along a radial line.
\( -\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } \)
\(xsec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t }=-\cfrac { r }{ x } \cfrac { d\, x }{ d\, t }=-\cfrac { d\, x }{ d\, t }tan(\theta) \)
Substitute into the expression for \(\Pi\),
\(\Pi=\int _{ 0 }^{ r }{ e^{-\cfrac{x}{a_{e}cos(\theta)} }\left\{ { \cfrac { d\, x }{ d\, t } tan(\theta ) }-xtan(\theta )\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } -\cfrac { d\, x }{ d\, t }tan(\theta ) \right\} } d\, r\)
\( \Pi=-\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ r }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r \)
and
\( \cfrac { r }{ x } =tan(\theta )\)
\( \cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ dr } \)
\( dr=xsec^{ 2 }(\theta ){ d\, \theta }\)
\( \Pi=-\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ r }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } }xsec^{ 2 }(\theta ){ d\, \theta } \)
Substitute \(\Pi\) back into \(E_A\),
\(\cfrac { \partial \, E_{ A } }{ \partial t } =-2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }} xsec^{ 2 }(\theta ){ d\, \theta }\)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =-2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ sec^{ 4 (\theta )}e^{-\cfrac{x}{a_{e}cos(\theta)} } } sin(\theta ){ d\, \theta }\)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac{1}{cos^{ 4 }(\theta )}e^{-\cfrac{x}{a_{e}cos(\theta)} } } { d\, (cos(\theta )) }\) --- (*)
This however, is not the final solution. We have to consider the fact that the charge is in orbit of radius \(r_e\) and the loop defining \(B\) is centered at the circumference along a radial line.
We are concerned with change in \(E\) perpendicular to the loop.
\(\sigma=\cfrac{\pi-\phi}{2}=\cfrac{\pi}{2}-\cfrac{\phi}{2}\)
\(E_{\bot}=Esin(\sigma)=Esin(\cfrac{\pi}{2}-\cfrac{\phi}{2})=Ecos(\phi/2)\)
Consider,
\(x^2=r^2_e+r^2_e-2r^2_ecos(\phi)=2r^2_e(1-cos(\phi))\),
\(x=r_e\sqrt{2(1-cos(\phi))}=2r_{ e }sin(\phi /2)\)
\(\cfrac { dx }{ dt }=r_{ e }cos(\phi/2)\cfrac{d\,\phi}{d\,t}\)
and,
\(x_\bot=xcos(\phi/2)\)
\(\cfrac { dx_\bot }{ dt }=\cfrac { dx }{ dt }cos(\phi/2)-xsin(\phi/2).\cfrac{1}{2}\cfrac { d\phi }{ dt } \)
\(\cfrac { dx_{ \bot } }{ dt } =\left( r_{ e }cos^{ 2 }(\phi /2)-xsin(\phi /2).\cfrac { 1 }{ 2 } \right) \cfrac { d\phi }{ dt } \)
Substitute \(x\) into the above.
\(\cfrac { dx_{ \bot } }{ dt } =r_{ e }cos(\phi )\cfrac { d\phi }{ dt } \)
Consider the term,\(x^2=r^2_e+r^2_e-2r^2_ecos(\phi)=2r^2_e(1-cos(\phi))\),
\(x=r_e\sqrt{2(1-cos(\phi))}=2r_{ e }sin(\phi /2)\)
\(\cfrac { dx }{ dt }=r_{ e }cos(\phi/2)\cfrac{d\,\phi}{d\,t}\)
and,
\(x_\bot=xcos(\phi/2)\)
\(\cfrac { dx_\bot }{ dt }=\cfrac { dx }{ dt }cos(\phi/2)-xsin(\phi/2).\cfrac{1}{2}\cfrac { d\phi }{ dt } \)
\(\cfrac { dx_{ \bot } }{ dt } =\left( r_{ e }cos^{ 2 }(\phi /2)-xsin(\phi /2).\cfrac { 1 }{ 2 } \right) \cfrac { d\phi }{ dt } \)
Substitute \(x\) into the above.
\(\cfrac { dx_{ \bot } }{ dt } =r_{ e }cos(\phi )\cfrac { d\phi }{ dt } \)
\(e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } }=e^{ -\cfrac { r_{ e }\sqrt { 2(1-cos(\phi )) } }{ a_{ e }cos(\theta ) } }=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }\)
Let
\(h(\phi,\theta)=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) } }\)
which has both \(\phi\) and \(\theta\) dependence.
From (*),
\(\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac{1}{cos^{ 4 }(\theta )}h(\phi,\theta) } { d\, (cos(\theta )) } \)
Adjusted for circular orbit,
\(\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }cos(\phi/2 )\cfrac { d\, x_{ \bot } }{ d\, t } \cfrac { x^{ 2 }_{ \bot } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }\)
The direction of \(E\) is adjusted by \(cos(\phi/2)\). The the distance term in the exponential remains as \(x\) as we see from the above diagram.
Assuming that \(B\) is constant around the loop,
\(\oint{B}d\,r=2\pi rB_o\)
\(2\pi rB_o=\mu _{ o }\varepsilon _{ o }2\pi E_{ o }cos(\phi /2)\cfrac { d\, x_{ \bot } }{ d\, t } \cfrac { x^{ 2 }_{ \bot } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
This is wrong. \(r\ne x_\bot tan(\theta _{ o })\) refer to post "Not To Be taken Too Seriously, Please" dated 14 May 2016
\( x_\bot tan(\theta _{ o })B_{ o }=\cfrac {\mu _{ o }\varepsilon _{ o } E_{ o } }{ a_{ e } }cos(\phi/2 )\cfrac { d\, x_{ \bot } }{ d\, t } x^{ 2 }_{ \bot } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
\(B_{ o }=\cfrac { \mu _{ o }\varepsilon _{ o }E_{ o } }{ a_{ e } }cos(\phi/2 )\cfrac { d\, x_{ \bot } }{ d\, t } x_{ \bot } \cfrac {cos(\theta_o) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }\)
Consider,
\(\cfrac { d\, x_{ \bot } }{ d\, t } x_{ \bot }=2r^{ 2 }_{ e }cos(\phi /2)sin(\phi/2)cos(\phi )\cfrac { d\phi }{ dt } \)
Let
\(f(\phi)=cos(\phi /2).2cos(\phi /2)sin(\phi/2)cos(\phi )=\cfrac{1}{2}sin(2\phi)cos(\phi /2)\)
such that,
\(cos(\phi /2)\cfrac { d\, x_{ \bot } }{ d\, t } x_{ \bot } =r^2_ef(\phi)\cfrac { d\phi }{ dt } \)
And we let,
\(g(\phi,\theta_o)= \cfrac { cos(\theta _{ o }) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
and
\(\omega=\cfrac{d\,\phi}{d\,t}\)
\(B_{ o }=\cfrac {\mu _{ o }\varepsilon _{ o }E_{ o } }{ a_{ e } } cos(\phi /2)\cfrac { d\, x_{ \bot } }{ d\, t }x_{ \bot } \cfrac {cos(\theta_o) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }\)
\(B_{ o }=\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })\)
where the term \(g(\phi,\theta_o)\) share the variable \(\phi\) with \(f(\phi)\) through \(h(\phi,\theta_o)\) and the two are not independent.
A plot of \(f(\phi)\) is shown below. \(f(\phi)\) is periodic and bounded.
This shows that the used of the exponential form is appropriate for \(E\).
