= Solution
The azimuth is a <cyclic coordinate>, so its <canonical momentum> is conserved:
$$
\boxed{p_\theta=\frac{\partial L}{\partial\dot\theta}=\gamma m\rho^2\dot\theta+q\rho A_\theta},\qquad
\gamma=\frac1{\sqrt{1-\mathbf v^2/c^2}}.
$$
Initially the <velocity> is parallel to the axis in a field-free region, so $p_\theta=0$. Conservation of <canonical angular momentum across a solenoid fringe> therefore gives, inside the uniform field,
$$
\boxed{\dot\theta=-\frac{qB}{2\gamma m}\equiv-\Omega}.
$$
This holds away from the coordinate singularity $\rho=0$; the Cartesian orbit will be smooth there.
Time independence also conserves the <energy> <Hamiltonian>,
$$
E=\sum_i\dot q_i\frac{\partial L}{\partial\dot q_i}-L
=\boxed{\gamma mc^2}.
$$
The velocity-linear magnetic term cancels in this <Legendre transform>. Thus $\gamma=\gamma_v=(1-v^2/c^2)^{-1/2}$ is constant, as expected because a static magnetic <Lorentz force> does no work. Mechanical angular <momentum> can change even though $p_\theta$ and the total speed are conserved.
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