Still Not

Consider,

$p=\int { \cfrac { \partial \, t }{ \partial x^{ ' } }  } \partial \, E$

$\Delta p=\int _{ E }^{ E+\Delta E }{ \cfrac { \partial \, t }{ \partial x^{ ' } }  } \partial \, E$

As $\Delta E\rightarrow0$,

$\Delta p=\cfrac { \partial \, t }{ \partial x^{ ' } } \Delta E$

$\cfrac { \partial \, p }{ \partial \, E } =\cfrac { \partial \, t }{ \partial x^{ ' } }$

and

$\cfrac { \partial \, E }{ \partial \, p } =\cfrac { \partial \, x^{ ' } }{ \partial \, t } =v$

If the particle is in circular motion in $B$ orbits, the centripetal force is provided by,

$\pi E=\pi\cfrac { F }{ q } =\cfrac {\pi mv^{ 2 } }{ q x}$

then

$\cfrac { \partial \, E }{ \partial \, v } =\cfrac { 2mv }{ q x }$

Let,

$p=\cfrac { 2mv }{ q x}$

such that,

$\cfrac { \partial \, E }{ \partial \, p } =\cfrac { q x}{ 2m } \cfrac { \partial \, E }{ \partial \, v } =v$

Wait a minute, does not  $E$ depends on $x$?  In this case the integral is taken over changing values of $E$. $x$ may or may not change over the integral interval.  It is not a variable that the integral is dependent on.

$\cfrac { \partial \, p }{ \partial \, E } =\cfrac{ 2m }  { q x}\cfrac{ \partial \, v } { \partial \, E } =\cfrac { \partial \, t }{ \partial x^{ ' } }$

and so,

$\int { \cfrac { \partial \, t }{ \partial x^{ ' } }  } \partial \, E=\int{\cfrac{ 2m }  { q x}}{ \partial \, v }=\cfrac{ 2m }  { q x}v$

From the previous post "Maybe Not",

$\cfrac { \partial B^{ 2 } }{ \partial \, t } =-i2\int { \cfrac { \partial \, t }{ \partial x^{ ' } }  } \partial \, E\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } }  \right\}$

$\cfrac { \partial B^{ 2 } }{ \partial \, t } =-i2\cfrac{ 2m }  { q x}v\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } }  \right\}$

$\cfrac { 1 }{ 2\mu _{ o } } \cfrac { \partial B^{ 2 } }{ \partial \, t } =-i\cfrac { 1 }{ 2\mu _{ o } }2\cfrac{ 2m }  { q x}v\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } }  \right\}$

$\cfrac { 1 }{ 2\mu _{ o } } \cfrac { \partial B^{ 2 } }{ \partial \, t } =-i\cfrac{ 2m }  { \mu _{ o }q x}v\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } }  \right\}$

Let $\psi=\cfrac{1}{2\mu_o}B^2=-\cfrac{1}{2}\varepsilon_oE^2$,

$\cfrac{\partial\,\psi}{\partial\,t}=-\cfrac{ 2m }  { \mu _{ o }q x}(iv)\left\{ \cfrac { \partial ^{ 2 }\, V }{ \partial \, x^{ 2 } }  \right\}$

where  $iv$ is along the $x$ direction.  And $x$ is the orbit of the particle.

This equation shows the rate of change in total energy required as a particle subjected to a B field and is in circular motion along the B field line attains a velocity $v$, perpendicular to its orbit at the present orbital radius of $x$.