 Solution

For $\psi =\mu /r$, the radial gradient givesThe force $-\nabla \varphi$ is radial and has no torque about the origin. For the remaining force, use the vector triple product:Meanwhile the radial unit vector obeysSince $\psi \mathbf{r}=\mu \hat{\mathbf{r}}$, these derivatives cancel. Thus the generalized angular momentum in a radial monopole field isThe argument applies away from $r=0$, where the field is defined, for any radial $\varphi$. In contrast to ordinary angular momentum, this conserved vector has $\mathbf{L}\cdot \hat{\mathbf{r}}=-\mu$; it fixes a cone rather than generally fixing a plane.