 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,Time independence and the cyclic angular coordinate give two constants:The three Euler-Lagrange equations areThe supplied angular velocity is $\dot{\varphi }=-2eg(z_0)/m$. Set $r=r_0$, $z=z_0$ and use this constant angular velocity. The radial right-hand side is $m{r}_0({\dot{\varphi }}^2+2eg\dot{\varphi }/m)=0$, the angular momentum is constant, and the vertical equation vanishes when $g^{\prime }(z_0)=0$. These equations and the initial data therefore giveFor nonzero radius and charge this is the requested circular orbit.