Charged circular elastic ring in a uniform magnetic field ($J=m{r}^2\dot{\varphi }+qB{r}^2/2$)

A uniformly charged circular ring hasHere $k$ is the spring constant for extension of its full circumference and the symmetric gauge is used. The canonical angular momentum $J=m{r}^2\dot{\varphi }+qB{r}^2/2$ is conserved. Eliminating the angle gives $m\ddot{r}=J^2/(m{r}^3)-q^2{B}^{2}r/(4m)-4{\pi }^{2}k(r-r_0)$. This model neglects electric self-interaction and noncircular deformation.

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.

With $J=0$, define ${\omega }_s^2=4{\pi }^{2}k/m$ and ${\omega }_c=qB/m$. Then $r_e={\omega }_s^2{r}_0/({\omega }_s^2+{\omega }_c^2/4)$, the radial frequency is $\Omega =\sqrt{{\omega }_s^2+{\omega }_c^2/4}$, and release at $r_0$ gives $r=r_e+(r_0-r_e)\cos \Omega t$ with $\dot{\varphi }=-{\omega }_c/2$. The physical solution requires $r>0$. If $2r_e<r_0$, the formal oscillation reaches zero and cannot be continued as a negative ring radius without a new collapse model.

Although the energy in velocities is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2+2{\pi }^{2}k(r-r_0{)}^2$, the Hamiltonian in canonical variables isThe field therefore enters Hamilton's equations through the relation between canonical and mechanical momentum, even though a magnetic field does no work.