= Solution
Time-translation symmetry conserves the energy $H$, and axial rotational symmetry makes $\phi$ a <cyclic coordinate>, conserving
$$
\boxed{J=p_\phi=mr^2\dot\phi+\frac{qBr^2}{2}.}
$$
The radial <Euler-Lagrange equation> before elimination is
$$
m\ddot r=mr\dot\phi^2+qBr\dot\phi-4\pi^2k(r-r_0).
$$
Insert $\dot\phi=J/(mr^2)-qB/(2m)$. The cross terms linear in $JqB$ cancel, leaving
$$
\boxed{m\ddot r=\frac{J^2}{mr^3}-\frac{q^2B^2}{4m}r-4\pi^2k(r-r_0).}
$$
Equivalently use the reduced radial potential
$$
U_J(r)=\frac{J^2}{2mr^2}+\frac{q^2B^2r^2}{8m}
+2\pi^2k(r-r_0)^2-\frac{JqB}{2m}.
$$
The last term is constant at fixed $J$ and has no radial derivative, but is part of the full conserved energy. Conservation of $J$ is stated in the rotationally <symmetric gauge>; the conserved mechanical combination can be written in terms of <magnetic flux> in a gauge-independent way.
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