= Solution
In <Lagrangian mechanics>, the <energy balance for an autonomous Lagrangian> uses $E=\sum_i\dot q_i\,\partial L/\partial\dot q_i-L$. For the stationary <vector potential> the velocity-linear magnetic term cancels in this expression, leaving <kinetic energy>. In <cylindrical coordinates>,
$$
L=\frac m2(\dot r^2+r^2\dot\phi^2+\dot z^2)+er^2g(z)\dot\phi.
$$
Time independence and the cyclic angular coordinate give two constants:
$$
\boxed{E=\frac m2(\dot r^2+r^2\dot\phi^2+\dot z^2),\qquad
p_\phi=mr^2\dot\phi+er^2g(z).}
$$
The three <Euler-Lagrange equations> are
$$
m\ddot r=mr\dot\phi^2+2erg\dot\phi,\qquad
\frac d{dt}(mr^2\dot\phi+er^2g)=0,\qquad
m\ddot z=er^2g'\dot\phi.
$$
The supplied angular velocity is $\dot\phi=-2eg(z_0)/m$. Set $r=r_0$, $z=z_0$ and use this constant angular velocity. The radial right-hand side is $mr_0(\dot\phi^2+2eg\dot\phi/m)=0$, the <angular momentum> is constant, and the vertical equation vanishes when $g'(z_0)=0$. These equations and the initial data therefore give
$$
\boxed{r(t)=r_0,\quad z(t)=z_0,\quad
\phi(t)=\phi_0-\frac{2eg(z_0)}m t.}
$$
For nonzero radius and charge this is the requested circular orbit.
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