= Solution
For this <neutral Coulomb pair in a uniform magnetic field>, the reduced <Lagrangian> gives the <canonical momenta>
$$
\mathbf P=M\dot{\mathbf R}-e\mathbf B\times\mathbf r,\qquad
\mathbf p=\mu\dot{\mathbf r}-e^*\mathbf B\times\mathbf r.
$$
Invert these relations and take the <Legendre transform>. The velocity-linear terms cancel against their <momentum> contributions, giving
$$
\boxed{H=\frac{|\mathbf P+e\mathbf B\times\mathbf r|^2}{2M}
+\frac{|\mathbf p+e^*\mathbf B\times\mathbf r|^2}{2\mu}-\frac{e^2}{r}.}
$$
There is no explicit time dependence, so $dH/dt=0$. There is no dependence on $\mathbf R$, so $\dot{\mathbf P}=0$. The distinction between <kinetic momentum and magnetic pseudomomentum> matters: \b[the conserved <vector> is $\mathbf P$, not generally $M\dot{\mathbf R}$]. The mechanical centre-of-mass <momentum> $M\dot{\mathbf R}=\mathbf P+e\mathbf B\times\mathbf r$ generally varies as the separation changes. A neutral pair can therefore have accelerated centre-of-mass motion in a <magnetic field>.
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