 Turning radius of an initially stationary magnetic ring

For negligible natural radius and release from rest at $r_1$, $J=qB{r}_1^2/2$. The effective radial potential is a sum of $r^{-2}$ and $r^2$ terms, whose two turning radii have squared product $J^2/(m^2{\Omega }^2)$. HenceThese formulas assume $qB\ne 0$. At exactly zero field the ideal ring collapses to zero radius instead of making a regular positive-radius turning point.