= Solution
The cylindrical entries of $\mathbf A$ are physical <vector> components, so $\mathbf v\cdot\mathbf A=\mu r^2\dot\theta$, not $\mu r\dot\theta$. The <Lagrangian> is
$$
L=\frac m2(\dot r^2+r^2\dot\theta^2+\dot z^2)+e\mu r^2\dot\theta-e\lambda z^2.
$$
Its three equations are
$$
\boxed{m\ddot r=mr\dot\theta^2+2e\mu r\dot\theta,\qquad
\frac d{dt}(mr^2\dot\theta+e\mu r^2)=0,\qquad
m\ddot z=-2e\lambda z.}
$$
For $r>0$, the angular equation can be expanded as $r\ddot\theta+2\dot r\dot\theta+(2e\mu/m)\dot r=0$. Directly computing the fields gives $\mathbf B=2\mu\hat{\mathbf z}$ and $\mathbf E=-2\lambda z\hat{\mathbf z}$, agreeing with the equations above.
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