= Solution
Put $\pi=p-eA/c=m\dot q$. The <Hamilton equations> are
$$
\dot q_i=\frac1m\left(p_i-\frac ecA_i\right),\qquad
\dot p_i=\frac e{mc}\sum_j\pi_j\partial_iA_j.
$$
Differentiate the <mechanical momentum> and substitute:
$$
m\ddot q_i=\frac ec\sum_j\dot q_j(\partial_iA_j-\partial_jA_i)-\frac ec\partial_tA_i.
$$
With $B=\nabla\times A$ and $E=-c^{-1}\partial_tA$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot q=eE+(e/c)\dot q\times B}$, the <Lorentz force> in the units of the <Hamiltonian>.
For $A=(-yB_0(z,t),0,0)$, $x$ is cyclic, so $p_x$ is conserved. Choose $p_x=0$, giving $\dot x=\Omega y$, where $\Omega=eB_0/(mc)$. Since $p_y=m\dot y$, its Hamilton equation gives $\dot p_y=-(e/c)B_0\dot x=-m\Omega^2y$, hence
$$
\boxed{\ddot y=-\Omega^2y,\qquad
H=\frac m2\dot z^2+E',\qquad E'=\frac m2(\dot y^2+\Omega^2y^2).}
$$
For nonzero constant $B_0$, the $(y,p_y)$ orbit is an ellipse of semiaxes $\sqrt{2E'/(m\Omega^2)}$ and $\sqrt{2mE'}$. Its enclosed phase-space area is $2\pi E'/|\Omega|$, so the <action variable> is
$$
\boxed{I(E',B_0)=\frac{E'}{|\Omega|}.}
$$
The absolute value keeps the action positive if the signed charge or <magnetic field> is negative.
When the field varies slowly along the trajectory, with $|d\Omega/dt|\ll\Omega^2$ and no zero-frequency crossing, <adiabatic invariance of the action> makes $I$ approximately constant. Thus $E'\simeq I|\Omega(z,t)|$ and the slow motion has effective <Hamiltonian>
$$
\boxed{H_{\rm eff}=\frac{p_z^2}{2m}+I\left|\frac{eB_0(z,t)}{mc}\right|.}
$$
The <effective potential> is the second term. Direct averaging checks its force: the exact equation is $m\ddot z=-m\Omega\partial_z\Omega\,y^2$, while the oscillator has $\langle y^2\rangle=E'/(m\Omega^2)$. Hence $m\ddot z=-I\partial_z|\Omega|$. Spatial field variation along the slow motion must satisfy the same adiabatic condition; slow explicit time variation alone does not control passage through a sharp spatial gradient. A time-dependent <effective potential> need not conserve total energy.
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