= Solution
Put $\boldsymbol\pi=\boldsymbol p-e\boldsymbol A/c=m\dot{\boldsymbol q}$. From <Hamilton's equations>,
$$
\dot q_i=\frac{\pi_i}{m},\qquad \dot p_i=\frac e{mc}\pi_j\partial_iA_j.
$$
For the stated time-independent <vector potential>, differentiating $\pi_i$ gives $m\ddot q_i=(e/c)\dot q_j(\partial_iA_j-\partial_jA_i)$, or
$$
\boxed{m\ddot{\boldsymbol q}=\frac ec\dot{\boldsymbol q}\times\boldsymbol B,\qquad\boldsymbol B=\nabla\times\boldsymbol A.}
$$
Use the <Poisson bracket> convention $[F,G]=\sum_i(F_{q_i}G_{p_i}-F_{p_i}G_{q_i})$. Then direct differentiation gives
$$
\boxed{[\pi_i,q_j]=-\delta_{ij},\qquad[\pi_i,\pi_j]=\frac ec(\partial_iA_j-\partial_jA_i).}
$$
For $\boldsymbol A=(0,0,F(r))$ with $r=\sqrt{x_1^2+x_2^2}$, the Hamiltonian is independent of $x_3$, so $\dot p_3=0$. The radial force in the horizontal plane is $\dot p_r=e(p_3-eF/c)F'/(mc)$. For circular motion the radial acceleration is $-r\Omega^2$, and $p_r$ is the mechanical radial momentum because the horizontal <vector potential> vanishes. Thus
$$
\boxed{\Omega^2=-\left(p_3-\frac{eF}{c}\right)\frac e{m^2cr}\frac{dF}{dr}.}
$$
A circular orbit at that radius is possible when the right side is nonnegative; positive values give either sense of horizontal rotation, and the longitudinal velocity is the constant $(p_3-eF(r)/c)/m$ along that orbit.
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