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Indeed, you can derive also this from a classical picture (a la Ampere). The idea is that an atom ("particle") may be somehow composed of moving charges with some "molecular current". Then the force on this current in a magnetic field is
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{j}(\vec{x}) \times \vec{B}(\vec{x}).$$
Now you can assume that ##\vec{j} \neq 0## only for the very small extent of this "atom". So you can expand the magnetic field around the place of the atom (assumed to be at ##\vec{x}_0=0##). This gives
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{j}(\vec{x}) \times [\vec{B}(0)+\vec{x} \cdot \vec{B}(0)].$$
Now, because of ##\vec{\nabla} \cdot \vec{j}=0## one can derive that
$$\int \mathrm{d}^3 x \vec{j}=0.$$
The 2nd term can be transformed to
$$\vec{F}=\vec{\nabla} (\vec{\mu} \cdot \vec{B})$$
with the magnetic moment
$$\vec{\mu}=\frac{1}{2} \int \mathrm{d}^3 x \vec{x} \times \vec{j}.$$
So indeed the potential of the force is
$$V=-\vec{\mu} \cdot \vec{B}.$$
You can also derive the torque in a similar way
$$\vec{\tau}=\int \mathrm{d}^3 x \vec{x} \times (\vec{x} \times \vec{B})=\cdots=\vec{\mu} \times \vec{B}.$$
From the definition of the magnetic moment you also get the relation to (orbital) angular momentum, rewriting
$$\vec{j}=q n \vec{v}.$$
Here ##\vec{n}## is the particle-number density and ##q## the charge of the particles making up the molecular current. With that we have
$$\vec{\mu}=\int \mathrm{d}^3 x \frac{1}{2} n q \vec{x} \times \vec{v} = \int \mathrm{d}^3 \frac{1}{2} \frac{n q}{m} \vec{x} \times \vec{p}=\frac{q}{2m} \vec{L},$$
where ##\vec{L}## is the angular momentum of the particles making up the molecular current.
Today we know that the magnetic moment is not only made up by such classical currents of moving charges with the corresponding orbital angular momentum, but also by the spin of the particles making up the current, e.g., the electrons around a nucleus in an atom. There is, however, another factor, the Lande or gyromagnetic factor. For an electron this factor is about 2. From the above classical model with the orbital angular momentum of Amperian currents the gyro factor comes out to be 1. Famously Einstein and de Haas thought to have confirmed this in their famous experiment, although de Haas indeed found also values for the gyro factor larger than 1. Not much later other physicists found the gyro factor of ferromagnets to be rather closer to 2, which was of course not understandable in this time (around 1915) . Today we understand it as the empirical evidence that most of the magnetization of a piece of iron is due to electron spins.
For more complicated particles like protons or neutrons and other hadrons the gyro factors' origin are very complicated and even not completely understood today. Here the elementary entities making up the particles are quarks and gluons, and how the spin and magnetic moment is "made up" of these constituents is very complicated and under ungoing research.
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{j}(\vec{x}) \times \vec{B}(\vec{x}).$$
Now you can assume that ##\vec{j} \neq 0## only for the very small extent of this "atom". So you can expand the magnetic field around the place of the atom (assumed to be at ##\vec{x}_0=0##). This gives
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{j}(\vec{x}) \times [\vec{B}(0)+\vec{x} \cdot \vec{B}(0)].$$
Now, because of ##\vec{\nabla} \cdot \vec{j}=0## one can derive that
$$\int \mathrm{d}^3 x \vec{j}=0.$$
The 2nd term can be transformed to
$$\vec{F}=\vec{\nabla} (\vec{\mu} \cdot \vec{B})$$
with the magnetic moment
$$\vec{\mu}=\frac{1}{2} \int \mathrm{d}^3 x \vec{x} \times \vec{j}.$$
So indeed the potential of the force is
$$V=-\vec{\mu} \cdot \vec{B}.$$
You can also derive the torque in a similar way
$$\vec{\tau}=\int \mathrm{d}^3 x \vec{x} \times (\vec{x} \times \vec{B})=\cdots=\vec{\mu} \times \vec{B}.$$
From the definition of the magnetic moment you also get the relation to (orbital) angular momentum, rewriting
$$\vec{j}=q n \vec{v}.$$
Here ##\vec{n}## is the particle-number density and ##q## the charge of the particles making up the molecular current. With that we have
$$\vec{\mu}=\int \mathrm{d}^3 x \frac{1}{2} n q \vec{x} \times \vec{v} = \int \mathrm{d}^3 \frac{1}{2} \frac{n q}{m} \vec{x} \times \vec{p}=\frac{q}{2m} \vec{L},$$
where ##\vec{L}## is the angular momentum of the particles making up the molecular current.
Today we know that the magnetic moment is not only made up by such classical currents of moving charges with the corresponding orbital angular momentum, but also by the spin of the particles making up the current, e.g., the electrons around a nucleus in an atom. There is, however, another factor, the Lande or gyromagnetic factor. For an electron this factor is about 2. From the above classical model with the orbital angular momentum of Amperian currents the gyro factor comes out to be 1. Famously Einstein and de Haas thought to have confirmed this in their famous experiment, although de Haas indeed found also values for the gyro factor larger than 1. Not much later other physicists found the gyro factor of ferromagnets to be rather closer to 2, which was of course not understandable in this time (around 1915) . Today we understand it as the empirical evidence that most of the magnetization of a piece of iron is due to electron spins.
For more complicated particles like protons or neutrons and other hadrons the gyro factors' origin are very complicated and even not completely understood today. Here the elementary entities making up the particles are quarks and gluons, and how the spin and magnetic moment is "made up" of these constituents is very complicated and under ungoing research.