Potential energy of magnetic dipole in non-uniform magnetic field

Question

I want to understand why the potential energy of an ideal magnetic dipole with dipole moment $\boldsymbol{m}$ in a non-uniform magnetic field $\boldsymbol{B}$ (neglecting the term to keep the magnitude of the dipole moment fixed) is: \begin{equation} U=-\boldsymbol{m}\cdot\boldsymbol{B} \end{equation}

The force on the dipole is: \begin{equation} \boldsymbol{F}=\nabla(\boldsymbol{m}\cdot\boldsymbol{B}) \end{equation}

The torque is: \begin{equation} \boldsymbol{\tau}=\boldsymbol{m}\times\boldsymbol{B} \end{equation}

The derivations I found either compute (using the gradient theorem)

\begin{equation} U=-\int_{\infty}^\boldsymbol{r}\boldsymbol{F}\cdot d\boldsymbol{l} \end{equation}

with assumptions on the field at infinity, or they use torque:

\begin{equation} U=-\int_{\theta_0}^{\theta}\tau d\theta \end{equation}

that leads to the same result (except and additive constant).

The question is: why are these used separately? In terms of work, we must consider the work to bring the dipole from infinity (integrate the force) and then the work to rotate the dipole (integrate the torque), so what leads to this? Maybe I am considering two different energies? Because if the field is uniform the force is zero (thus zero work moving the dipole), but the torque is not, and we get that same energy from the torque.

Answer 1

If you bring in the dipole from infinity, you can first rotate it while it's at infinity (where, by assumption, the field vanishes, and thus its potential energy is still zero), and then translate it to its final location. In doing so against a force $\nabla (m \cdot B)$, you give it potential energy in the amount $-m \cdot B$.

Alternatively, you can assume that at all times, the dipole is kept perpendicular to the field as it is being brought in from infinity. If that's the case, the force is zero this whole time. Lastly, you can rotate it into its final orientation while keeping its position constant. Now you only have to do work against the torque.

It's the same energy, which can be computed easily using one of the two approaches above. You could also consider a more complicated path where work is being done against both force and torque simultaneously, but that just makes your life harder. But if the dipole does take such a path, the final result must still be the same.