= Solution
Every element of the uniformly distributed mass has speed squared $\dot r^2+r^2\dot\phi^2$, so the total <kinetic energy> is $m(\dot r^2+r^2\dot\phi^2)/2$. The extension of the ring's full circumference is $2\pi(r-r_0)$, and the stated <spring constant> therefore gives
$$
V=\frac k2[2\pi(r-r_0)]^2=2\pi^2k(r-r_0)^2.
$$
The magnetic coupling for each charge element is $dq\,\mathbf A\cdot\mathbf v$. With the <symmetric gauge> $\mathbf A=(Br/2)\hat{\boldsymbol\phi}$ and tangential speed $r\dot\phi$, integration over the ring gives $qBr^2\dot\phi/2$. Thus
$$
\boxed{L=\frac m2\dot r^2+\frac{mr^2}{2}\dot\phi^2
-2\pi^2k(r-r_0)^2+\frac{qBr^2}{2}\dot\phi.}
$$
This 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.
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