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A magnetic dipole moment is a vector that summarizes the strength and orientation of a localized magnetic source. For a planar current loop, it is the electric current times the vector area: $\mathbf{m}=IA\hat{\mathbf{n}}$. The right-hand rule fixes the unit normal vector $\hat{\mathbf{n}}$: curl the fingers of the right hand along the conventional current, and the thumb points along $\mathbf{m}$. This single vector predicts both the distant magnetic dipole field and the turning effect of an imposed magnetic field.^[ref]

The SI unit is ampere times square metre, $\mathrm{A}{\mathrm{m}}^2$, equivalently joule per tesla. A magnetic dipole moment is neither a magnetic field nor a magnetic monopole: a closed current loop can have a nonzero magnetic dipole moment while its net outward magnetic flux through any closed surface remains zero, as required by Gauss's law for magnetism.

For a localized, steady electric current density $\mathbf{J}(\mathbf{r})$, the more general definition isHere the cross product is integrated over the complete source, including return paths. Steadiness requires $\nabla \cdot \mathbf{J}=0$, and localization requires sufficiently rapid decay for the boundary terms to vanish. Under these assumptions the localized steady-current moment identities give $\int \mathbf{J}{d}^{3}r=0$: shifting the origin by $\mathbf{a}$ changes the proposed magnetic dipole moment by $-\frac{1}{2}\mathbf{a}\times \int \mathbf{J}{d}^{3}r=0$. An isolated open wire segment does not satisfy these assumptions. ^[ref].

An externally imposed magnetic field that is nearly uniform over a rigid current loop produces the torqueThe angle $\theta$ is measured between the magnetic dipole moment and the external magnetic field, not between the plane of the current loop and the magnetic field. The magnetic dipole potential energy uses a dot product, whereas the torque uses a cross product. The torque is largest when the magnetic dipole moment is perpendicular to the magnetic field; parallel alignment minimizes the potential energy.^[ref]

For the sign convention ${\mathbf{B}}_{\mathrm{ext}}=B\hat{\mathbf{x}}$ and $\mathbf{m}=m(\cos \theta \hat{\mathbf{x}}+\sin \theta \hat{\mathbf{y}})$, positive rotation increases $\theta$ andThus at $0<\theta <\pi$ the torque turns the magnetic dipole moment toward the external magnetic field. Both parallel and antiparallel alignment have zero torque, but they are respectively a stable equilibrium and an unstable equilibrium. With no damping, a freely released current loop can undergo oscillation around the stable equilibrium; settling requires damping.

In the twelve-turn example, take $B=0.020\mathrm{T}$ and $\theta =60°$. The predicted torque is ${\tau }_z=-2.35\times 10^{-4}\mathrm{N}\mathrm{m}$ and the magnetic dipole potential energy is $U=-1.36\times 10^{-4}\mathrm{J}$. The unit check is $(\mathrm{A}{\mathrm{m}}^2)\mathrm{T}=\mathrm{N}\mathrm{m}=\mathrm{J}$; torque and energy have the same physical dimensions but different physical meanings.

A short derivation starts with the magnetic force on a current-carrying wire, $d\mathbf{F}=Id\mathbf{r}\times {\mathbf{B}}_{\mathrm{ext}}$. In an exactly uniform magnetic field, the net force is $I({\oint }_{C}d\mathbf{r})\times {\mathbf{B}}_{\mathrm{ext}}=0$, but its torque need not vanish: opposite sides produce a turning couple. Integrating $\mathbf{r}\times d\mathbf{F}$ gives $\mathbf{m}\times {\mathbf{B}}_{\mathrm{ext}}$. Integrating $dU=-{\tau }_{z}d\theta$ at fixed $m$ and $B$ gives the orientation potential energy. This is the mechanical orientation potential energy for a maintained electric current, with the external source held fixed; the energy accounting for field and power supply is a separate problem. A varying external magnetic field requires corrections across a finite current loop, and can also give a net force.