From Maxwell,
\( \oint { B } \,dr=\varepsilon _{ o }\mu _{ o }\cfrac { \partial \, }{ \partial t } \left\{ \oint { E } \,dA \right\} \)
\(E=\cfrac{e}{4\pi\varepsilon_o d^2}\)
Total \(E\) over flat circular x-section when the charge is at a distance \(x\) from the loop,
\(E_A=\int_{0}^{r} {2\pi xtan(\theta).\cfrac{e}{4\pi\varepsilon_o}\left(\cfrac{cos(\theta)}{x}\right)^2}d\,r\)
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { \partial \, }{ \partial t } \left\{ \int _{ 0 }^{ r }{ sin(\theta )cos(\theta ).\cfrac { 1 }{ x } } d\, r \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } \cfrac { \partial \, }{ \partial t } \left\{ \int _{ 0 }^{ r }{ sin(2\theta )\cfrac { 1 }{ x } } d\, r \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } \left\{ \int _{ 0 }^{ r }{ 2cos(2\theta )\cfrac { 1 }{ x } \cfrac { d\, \theta }{ d\, t } } -sin(2\theta )\cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } d\, r \right\}\)
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \left\{ \int _{ 0 }^{ r }{ (2cos^{ 2 }(\theta )-1)\cfrac { 1 }{ x } \cfrac { d\, \theta }{ d\, t } } -sin(\theta )cos(\theta )\cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } d\, r \right\} \)
From geometry,
\( \cfrac { r }{ x } =tan(\theta )\)
\( -\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ d\, t } \)
\( \cfrac { d\, \theta }{ d\, t } =-\cfrac { rcos^{ 2 }(\theta ) }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } \)
and
\( \cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta }{ dr } \)
\( dr=xsec^{ 2 }(\theta ){ d\, \theta }\)
So we have,
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \left\{ \int _{ 0 }^{ r }{ -(2cos^{ 2 }(\theta )-1)\cfrac { 1 }{ x } \cfrac { rcos^{ 2 }(\theta ) }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } } -sin(\theta )cos(\theta )\cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } d\, r \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ r }{ -(2cos^{ 2 }(\theta )-1)tan(\theta )cos^{ 2 }(\theta )-sin(\theta )cos(\theta ) } d\, r \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =-\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ r }{ 2cos^{ 2 }(\theta )sin(\theta )cos(\theta ) } d\, r \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =-\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ \theta _{ o } }{ cos^{ 2 }(\theta )sin(2\theta ) } sec^{ 2 }(\theta )d\, \theta \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =-\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ \theta _{ o } }{ sin(2\theta ) } d\, \theta \right\} \)
\( \cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } \cfrac { 1 }{ x } \cfrac { d\, x }{ d\, t } \left\{ cos(2\theta _{ o })-1 \right\} \) --- (*)
We have also 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.
\(\sigma=\cfrac{\pi-\phi}{2}=\cfrac{\pi}{2}-\cfrac{\phi}{2}\)
\(sin(\sigma)=sin(\cfrac{\pi}{2}-\cfrac{\phi}{2})=cos({\phi}/{2})\)
\(E_\bot=Esin(\sigma)=Ecos(\phi/2)\)
\(x_\bot=xsin(\sigma)=xcos(\phi/2)\)
We are concerned with the change in \(E\) perpendicular to the smaller orbit in the loop. From (*)
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { d\, x }{ d\, t } \cfrac { 1 }{ x } \left\{ cos(2\theta _{ o })-1 \right\} \)
So the equation (*) becomes,
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } sin(\sigma )\cfrac { d\, x_{ \bot } }{ d\, t } \cfrac { 1 }{ x } \left\{ cos(2\theta _{ o })-1 \right\} \)
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } }cos(\phi/2)\cfrac { d\, x_{ \bot } }{ d\, t }\cfrac { 1 }{ x } \left\{ cos(2\theta _{ o })-1 \right\} \) --- (1)
The perpendicular \(E\) field component is given by \(E_\bot=Ecos(\phi/2)\), not by changing \(x\) to \(x_\bot\).
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.
\(\cfrac { dx_{ \bot } }{ dt } =r_{ e }cos(\phi )\cfrac { d\phi }{ dt } \)
\(\cfrac { dx_{ \bot } }{ dt } \cfrac { 1 }{ x } =r_{ e }\cfrac { 1 }{ 2r_{ e }sin(\phi /2) } cos(\phi )\cfrac { d\phi }{ dt } \)
\(\cfrac { dx_{ \bot } }{ dt } \cfrac { 1 }{ x } =\cfrac { 1 }{ 2 } \cfrac { cos(\phi ) }{ sin(\phi /2) } \cfrac { d\phi }{ dt } \)
From (1),
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } }cos(\phi/2)\cfrac { d\, x_{ \bot } }{ d\, t }\cfrac { 1 }{ x } \left\{ cos(2\theta _{ o })-1 \right\} \)
\(\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } cos(\phi /2)\cfrac { 1 }{ 2 } \cfrac { cos(\phi ) }{ sin(\phi /2) } \cfrac { d\phi }{ dt } \left\{ cos(2\theta _{ o })-1 \right\} \)
We know at this point that \(B\) is going to blowup when integrated over \(\phi\). This suggests that \(\cfrac{1}{x^2}\) is not approprite here.
Assuming that \(B\) is constant around the loop,
\(\oint{B}d\,r=2\pi r_pB_o\)
where \(r_p\) is the radius of the \(B\) orbit.
\(2\pi r_pB_o=\mu _{ o }\varepsilon _{ o }\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { \mu _{ o }e }{ 4 } cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \cfrac { d\phi }{ dt } \left\{ cos(2\theta _{ o })-1 \right\}\)
\(B_{ o }=\cfrac { \mu _{ o }e }{ 8\pi r_{ p } } cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \cfrac { d\phi }{ dt } \left\{ cos(2\theta _{ o })-1 \right\}\)
This is a time varying \(B\) field, changing as the electron travels at \(\omega=\cfrac{d\phi}{d\,t}\) around its orbit. We can average \(B^2\) over one period to obtain the energy in needed to establish the orbit.
\(\cfrac { 1 }{ T } \int _{ 0 }^{ T }{ B^{ 2 }_{ o } } d\, t=\cfrac { 1 }{ T } \int _{ 0 }^{ T }{ \left( \cfrac { \mu _{ o }e }{ 8\pi r_{ p } } sin^{ 2 }(\theta _{ o }) \right) ^{ 2 }\left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \right) ^{ 2 }\cfrac { d\phi }{ dt } \cfrac { d\phi }{ dt } }\)
\(=\cfrac { 1 }{ T } \left( \cfrac { \mu _{ o }e }{ 8\pi r_{ p } } sin^{ 2 }(\theta _{ o }) \right) ^{ 2 }\int _{ 0 }^{ 2\pi }{ \left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \right) ^{ 2 }\omega } d\phi \)
as \(\omega=\cfrac{d\phi}{d\,t}\).
Consider,
\(\int _{ 0 }^{ 2\pi}{ \left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \right) ^{ 2 }\omega } d\phi \)
KABOOM!!
as \(\omega=\cfrac{d\phi}{d\,t}\).
Consider,
\(\int _{ 0 }^{ 2\pi}{ \left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \right) ^{ 2 }\omega } d\phi \)
KABOOM!!
