Electron Orbit B Field..BLEWUP

What is the  $B$  field established by an electron in orbit.

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!!