= Radial escape in an inverse-radius axial magnetic field
{title2=$\dot r^2=v_\perp^2-(\ell/r-c)^2,\quad c=qB_0/m$}
For an axial <magnetic field> $\mathbf B=(B_0/r)\mathbf e_z$, let $L=mr^2\dot\theta$. The <Lorentz force> gives $\dot L=-qB_0\dot r$, so $\ell=(L+qB_0r)/m$ is constant. Since <magnetic forces do no work> and $\dot z$ is constant, the transverse speed $v_\perp$ is constant and $\dot r^2=v_\perp^2-(\ell/r-c)^2$. If $c\ell>0$ and $0<v_\perp<|c|$, two finite radial <turning points> confine the motion. If $v_\perp>|c|$, an outward branch escapes to infinite radius. At equality a marginal escaping branch exists when $c\ell>0$, but $\ell=0$ gives a circular orbit instead. The transverse <escape velocity> threshold is $|qB_0|/m$; motion parallel to the <magnetic field> cannot provide transverse escape energy.
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