 Solution

Every element of the uniformly distributed mass has speed squared ${\dot{r}}^2+r^2{\dot{\varphi }}^2$, so the total kinetic energy is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2$. The extension of the ring's full circumference is $2\pi (r-r_0)$, and the stated spring constant therefore givesThe magnetic coupling for each charge element is $dq\mathbf{A}\cdot \mathbf{v}$. With the symmetric gauge $\mathbf{A}=(Br/2)\hat{\mathbf{\varphi }}$ and tangential speed $r\dot{\varphi }$, integration over the ring gives $qB{r}^2\dot{\varphi }/2$. ThusThis idealization neglects charge self-energy, radiation and deformation away from a circular ring. The elastic constant refers to extension of the whole circumference; replacing it by a radial spring constant without the factor $4{\pi }^2$ changes the problem.