Radial escape in an inverse-radius axial magnetic field (${\dot{r}}^2=v_{\perp }^2-(\ell /r-c{)}^2,c=q{B}_0/m$)

For an axial magnetic field $\mathbf{B}=(B_0/r){\mathbf{e}}_z$, let $L=m{r}^2\dot{\theta }$. The Lorentz force gives $\dot{L}=-q{B}_0\dot{r}$, so $\ell =(L+q{B}_{0}r)/m$ 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 $|q{B}_0|/m$; motion parallel to the magnetic field cannot provide transverse escape energy.