= Solution
For $\psi=\mu/r$, the radial <gradient> gives
$$
\mathbf B=-\nabla\psi=\frac{\mu}{r^3}\mathbf r.
$$
The force $-\nabla\phi$ is radial and has no <torque> about the origin. For the remaining force, use the <vector triple product>:
$$
\frac d{dt}(m\mathbf r\times\dot{\mathbf r})
=\mathbf r\times\mathbf F
=\frac\mu{r^3}\mathbf r\times(\dot{\mathbf r}\times\mathbf r)
=\frac\mu r(\dot{\mathbf r}-\dot r\,\widehat{\mathbf r}).
$$
Meanwhile the radial unit vector obeys
$$
\frac d{dt}\widehat{\mathbf r}
=\frac{\dot{\mathbf r}-\dot r\,\widehat{\mathbf r}}r.
$$
Since $\psi\mathbf r=\mu\widehat{\mathbf r}$, these derivatives cancel. Thus the <generalized angular momentum in a radial monopole field> is
$$
\boxed{\mathbf L=m\mathbf r\times\dot{\mathbf r}-\mu\widehat{\mathbf r},\qquad\dot{\mathbf L}=0.}
$$
The argument applies away from $r=0$, where the field is defined, for any radial $\phi$. In contrast to ordinary <angular momentum>, this conserved vector has $\mathbf L\cdot\widehat{\mathbf r}=-\mu$; it fixes a cone rather than generally fixing a plane.
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