 Solution

Put $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c=m\dot{\mathbf{q}}$. From Hamilton's equations,For the stated time-independent vector potential, differentiating ${\pi }_i$ gives $m{\ddot{q}}_i=(e/c){\dot{q}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, orUse the Poisson bracket convention $[F,G]={\sum }_i(F_{q_i}{G}_{p_i}-F_{p_i}{G}_{q_i})$. Then direct differentiation givesFor $\mathbf{A}=(0,0,F(r))$ 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 ${\dot{p}}_r=e(p_3-eF/c)F^{\prime }/(mc)$. For circular motion the radial acceleration is $-r{\Omega }^2$, and $p_r$ is the mechanical radial momentum because the horizontal vector potential vanishes. ThusA 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.