Capturing $T^{+}$ Particles.

But still if we were to consider the energy term,

$U_B=\cfrac{1}{2}\cfrac{B^2_o}{\mu_o}$

$U_B=\cfrac{1}{2\mu_o}\left\{\cfrac {\mu_o q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}.sin^{ 2 }(\phi ).Z\right\}^2$

$U_B=\cfrac{\mu_o}{2}\left\{\cfrac { q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}\right\}^2sin^{ 4 }(\phi ).Z^2$

where $Z$ has two terms,

$Z= \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }+  \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right\}$

As $x\le 2r_e$ and $\theta_0$ is small, $y=sec(\theta_0)\approx 1$,  $\sqrt { y^{ 2 }-1 }\approx 0$ because $0\le y\le sec(\theta_0)$.

We simplify by letting the first term of $Z$ be zero.

$\int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }=0$

$Z= \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }$

Then we consider, the average of $U_B$ over one period, ie its power,

$\overline{U_B}=\cfrac{1}{T}\int^{T}_{0}{U_B}dt=\cfrac{1}{T}\int^{T}_{0}\cfrac{\mu_o}{2}\left\{\cfrac { q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}\right\}^2sin^{ 4 }(\phi )\left(\int _{ 0 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right)^2 dt$

$\overline{U_B}=\cfrac{\mu_o}{2T}\left\{\cfrac { q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}\right\}^2\int^{T}_{0}sin^{ 4 }(\phi )\left(\int _{ 0 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right)^2 dt$

but

$x^2=r^2_e+r^2_e-2r^2_ecos(\phi)=2r^2_e(1-cos(\phi))$

$x=r_e\sqrt{2(1-cos(\phi))}=2r_{ e }sin(\phi /2)$

$\overline{U_B}=\cfrac{\mu_o}{2T}\left\{\cfrac { q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}\right\}^2\int _{ 0 }^{ T }sin^{ 4 }(\phi )\left(\int^{cos(\theta_0)}_{0}{ e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y }\right)^2 dt$

$\cfrac{d\phi}{dt}=\omega$ is a constant.

$\overline{U_B}=\cfrac{\mu_o}{2T\omega}\left\{\cfrac { q v  }{4\pi r^2 }.\cfrac{ r^{ 3 }_{ e }}{a^3_e}\right\}^2\int _{ 0 }^{2\pi }sin^{ 4 }(\phi )\left(\int^{ cos(\theta_0)}_{0}{ e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { dy }\right)^2 d\phi$

where $y=cos(\theta)$,

$\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2T\omega  } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } }  \right\} ^{ 2 }\int _{ 0 }^{ 2\pi  } sin^{ 4 }(\phi )\left( \int _{ 0 }^{ \theta_0  }{ -e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta) }  }\left\{ \cfrac { sin(\theta ) }{ cos^{ 4 }(\theta ) }  \right\}  } { d\theta  } \right) ^{ 2 }d\phi$

Consider,

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

$h^{ ' }_{ \theta  }=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }\left( +\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos^{ 2 }(\theta ) }  \right) .(-sin(\theta ))$

$h^{ ' }_{ \theta  }=-h.\left( \cfrac { 2r_{ e }sin(\phi /2)sin(\theta ) }{ a_{ e }cos^{ 2 }(\theta ) }  \right)$

$\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2T\omega  } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } }  \right\} ^{ 2 }\cfrac { a^2_{ e } }{ r^2_{ e } } \int _{ 0 }^{ 2\pi  } sin^{ 2 }(\phi )\left( \int _{ 0 }^{ \theta _{ 0 } }{ -e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }\cfrac { r_{ e } }{ a_{ e } } \cfrac { 2sin(\phi /2)sin(\theta ) }{ cos^{ 2 }(\theta ) } \left\{ \cfrac { cos(\phi /2) }{ cos^{ 2 }(\theta ) }  \right\}  } { d\theta  } \right) ^{ 2 }d\phi$

$\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2T\omega  } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } }  \right\} ^{ 2 }\cfrac { a^2_{ e } }{ r^2_{ e } }\cfrac { r }{ r_{ e } }  \int _{ 0 }^{ 2\pi  } sin^{ 2 }(\phi )\left( \int _{ 0 }^{ \theta _{ 0 } }{h^{' }\left\{ \cfrac {cos(\phi /2) }{ cos^{ 2 }(\theta ) }  \right\}  } { d\theta  } \right) ^{ 2 }d\phi$

$\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2T\omega  } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } }  \right\} ^{ 2 }\cfrac { a^2_{ e } }{ r^2_{ e } } \int _{ 0 }^{ 2\pi  } \left( sin(\phi )cos(\phi /2)\int _{ 0 }^{ \theta _{ 0 } }{ h^{ ' }_{ \theta  }\left\{ \cfrac { 1 }{ cos^{ 2 }(\theta ) }  \right\}  } { d\theta  } \right) ^{ 2 }d\phi$--- (1)

Integrating by parts,

$\int _{ 0 }^{ \theta _{ 0 } }{h^{' }\left\{ \cfrac {1 }{ cos^{ 2 }(\theta ) }  \right\}  } { d\theta  }= h\cfrac {1 }{ cos^{ 2 }(\theta ) }+2\int_{0}^{ \theta _{ 0 } } { h\cfrac { sin(\theta ) }{ cos^{ 3 }(\theta ) }  } d\theta$

then, let,

$\Pi=\left( sin(\phi )cos(\phi /2)\int _{ 0 }^{ \theta _{ 0 } }{ h^{ ' }\left\{ \cfrac { 1 }{ cos^{ 2 }(\theta ) }  \right\}  } { d\theta  } \right) ^{ 2 }$

$\Pi=\left( { h\cfrac { sin(\phi )cos(\phi /2) }{ cos^{ 2 }(\theta ) } +2\cfrac { a_{ e } }{ r_{ e } } \int _{ 0 }^{ \theta _{ 0 } }{ h\cfrac { r_{ e } }{ a_{ e } } \cfrac { 2sin(\phi /2)sin(\theta ) }{ cos^{ 2 }(\theta ) } \cfrac { cos(\phi /2) }{ cos(\theta ) }  }  }d\theta  \right) ^{ 2 }$

$\Pi=\left( { h\cfrac { sin(\phi )cos(\phi /2) }{ cos^{ 2 }(\theta ) } +2\cfrac { a_{ e }cos(\phi /2) }{ r_{ e } } \int _{ 0 }^{ \theta _{ 0 } }{ h^{ ' }_{ \theta  }\cfrac { 1 }{ cos(\theta ) }  } d\theta  } \right) ^{ 2 }$---(2)

$\int _{ 0 }^{ \theta _{ 0 } }{ h^{ ' }_{ \theta  }\cfrac { 1 }{ cos(\theta ) }  } d\theta =h\cfrac { 1 }{ cos(\theta ) } +\int _{ 0 }^{ \theta _{ 0 } }{ h\cfrac { sin(\theta ) }{ cos^{ 2 }(\theta ) }  } d\theta$---(3)

and

$\int _{ 0 }^{ \theta _{ 0 } }{ h\cfrac { sin(\theta ) }{ cos^{ 2 }(\theta ) }  } d\theta =\cfrac { a_{ e } }{ 2r_{ e }sin(\phi /2) } \int _{ 0 }^{ \theta _{ 0 } }{ h\cfrac { r_{ e } }{ a_{ e } } \cfrac { 2sin(\phi /2)sin(\theta ) }{ cos^{ 2 }(\theta ) }  } d\theta$

$=-\cfrac { a_{ e } }{ 2r_{ e }sin(\phi /2) } \int _{ 0 }^{ \theta _{ 0 } }{ h^{ ' }_{ \theta  } } d\theta$

$=-\cfrac { a_{ e } }{ 2r_{ e }sin(\phi /2) } \int _{ 0 }^{ \theta _{ 0 } }{ h^{ ' }_{ \theta  } } d\theta =-\left| \cfrac { a_{ e } }{ 2r_{ e }sin(\phi /2) } h \right| ^{ \theta }_{ 0 }$---(4)

Substituting (4) into (3), (2) and (1), we have,

$\overline { U_{ B } } =\cfrac { \mu _{ o } }{ 2T\omega  } \left\{ \cfrac { qv }{ 4\pi r^{ 2 } } .\cfrac { r^{ 3 }_{ e } }{ a^{ 3 }_{ e } }  \right\} ^{ 2 }\cfrac { a^2_{ e } }{ r^2_{ e } } \int _{ 0 }^{ 2\pi  } \left( { \left| \cfrac { 1 }{ cos^{ 2 }(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }2cos^{ 2 }(\phi /2)sin(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }\\+\left| \cfrac { 2a_{ e } }{ r_{ e }cos(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }cos(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }\\-\left| \cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } cot(\phi /2)e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  } \right| ^{ \theta _{ 0 } }_{ 0 } } \right) ^{ 2 }d\phi$

with the substitution, $h=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

Let $III|_{ 0 }^{ 2\pi  }$ be the integral,

$III|_{ 0 }^{ 2\pi  }=\int _{ 0 }^{ 2\pi  } \left( { \left| \cfrac { 1 }{ cos^{ 2 }(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }2cos^{ 2 }(\phi /2)sin(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }+\left| \cfrac { 2a_{ e } }{ r_{ e }cos(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }cos(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }\\-\left| \cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } cot(\phi /2)e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  } \right| ^{ \theta _{ 0 } }_{ 0 } } \right) ^{ 2 }d\phi$

and consider,

$(a+b+c)^{ 2 }=a^{ 2 }+b^{ 2 }+c^{ 2 }+2ab+2ac+2bc$

let,

$a= \left| \cfrac { 1 }{ cos^{ 2 }(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }2cos^{ 2 }(\phi /2)sin(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }$

$b=\left| \cfrac { 2a_{ e } }{ r_{ e }cos(\theta ) } e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }cos(\phi /2) \right| ^{ \theta _{ 0 } }_{ 0 }$

$c=-\left| \cfrac { a^{ 2 }_{ e } }{ r^{ 2 }_{ e } } cot(\phi /2)e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  } \right| ^{ \theta _{ 0 } }_{ 0 }$

then consider,

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

$h^{ ' }_{ \phi  }=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }\left( -\cfrac { 2r_{ e }cos(\phi /2) }{ a_{ e }cos(\theta ) }  \right) \cfrac { 1 }{ 2 }$

$h^{ ' }_{ \phi  }=-h.\cfrac { r_{ e }cos(\phi /2) }{ a_{ e }cos(\theta ) }$

This is a long tedious way home...


Note:

$\sigma =\cfrac { \pi -\phi  }{ 2 }$

$v=\cfrac { d }{ d\, t } xcos(\pi /2-\sigma )=\cfrac { d }{ d\, t } xsin(\sigma )=\cfrac { d }{ d\, t } xcos(\phi /2)$

$v=cos(\phi /2)\cfrac { dx }{ d\, t } -xsin(\phi /2)\cfrac { 1 }{ 2 } \cfrac { d\phi  }{ d\, t }$

$\left\{ r_{ e }+\cfrac { x }{ 2 } sin(\phi /2) \right\} \cfrac { d\phi  }{ d\, t } =cos(\phi /2)\cfrac { dx }{ d\, t }$