B Orbits Here To Stay

The expression,

$\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  } } { d\, (cos(\theta )) }$

is an interesting one.  And has solution,

$z=cos(\theta)$

$\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ z^{ 4 } } e^{ -\cfrac { x }{ a_{ e }z }  } } { d\, z}=\cfrac { 1 }{ z^{ 4 } } \left\{ \cfrac { x }{ { a }_{ e } } { E }_{ i }(-\cfrac { x }{ { a }_{ e }z } )+xe^{ -\cfrac { x }{ { a }_{ e }z }  } \right\} +c$

where  $E_i(x)$  is the polyexponential function,

 OR

$\cfrac { a_{ e }(x^{ 2 }+2a_{ e }xz+2a_{ e }^{ 2 }z^{ 2 }) }{ x^{ 3 }z^{ 2 } } e^{ (-\cfrac { x }{ a_{ e }z } ) }$

And so,

$\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  } } { d\, (cos(\theta )) }=\left[\cfrac { a_{ e } }{ x^{ 3 }cos^{ 2 }(\theta ) } \left\{x^{ 2 }+2a_{ e }xcos(\theta )+2a_{ e }^{ 2 }cos^{ 2 }(\theta )\right\}e^{ (-\cfrac { x }{ a_{ e }cos(\theta ) } ) }\right]^{\theta_o}_{0}$

Which has a DC term,

$\cfrac { a_{ e } }{ x^{ 3 } } \left\{ x^{ 2 }+2a_{ e }x+2a_{ e }^{ 2 } \right\} e^{ (-\cfrac { x }{ a_{ e } } ) }$

This term suggests that the $B$ field is always present and the positively charged nucleuii are held permanently in the $B$ orbits.

1/x^4*e^(-a/(2*x)) for 0.5≤a≤5,  1/x^4*e^(-2.5/(2*x)),  and  1/x^5*e^(-2.5/(2*x)) are plotted below.

Since,  $x=cos(\theta)$, $x\le1$ the graph $\cfrac{1}{x^5}\gt\cfrac{1}{x^4}$.  And

$\cfrac{1}{x^4}e^{a/(2x)}$ increases with decreasing $a$ which in our original expression is $x$, the distance between the center of the loop and the orbiting negative charge.