Electron Orbit B Field II

This post is wrong.  Please refer to the post "Not To Be taken Too Seriously, Please" dated 14 May 2016. 

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.

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,

$e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }=e^{ -\cfrac { r_{ e }\sqrt { 2(1-cos(\phi )) }  }{ a_{ e }cos(\theta ) }  }=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

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

$\omega=\cfrac{d\,\phi}{d\,t}$

$B_o$  becomes,

$B_{ o }=\cfrac {\mu _{ o }\varepsilon _{ o }E_{ o } }{ a_{ e } } cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t }x_{ \bot  }  \cfrac {cos(\theta_o) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }$

$B_{ o }=\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })$

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  $E$.