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A current loop is a closed path carrying electric current. For a thin wire following a closed curve $C$, the volume integral defining its magnetic dipole moment becomes a line integral:The vector area $\mathbf{A}$ is $A\hat{\mathbf{n}}$ for a planar current loop. For a nonplanar current loop, use the vector area, rather than the scalar area of an arbitrarily chosen spanning surface. Reversing the electric current reverses $\mathbf{m}$ without changing the geometry. ^[ref].

For $N$ closely spaced identical turns carrying the same electric current, the magnetic dipole moments add: $\mathbf{m}=NIA\hat{\mathbf{n}}$. Each turn contributes its own vector area; unequal areas or differently oriented turns require a vector sum. As a calculated example, twelve circular turns of radius $R=3.0\mathrm{cm}$ carrying $I=0.40\mathrm{A}$ give $m=NI\pi R^2=1.36\times 10^{-2}\mathrm{A}{\mathrm{m}}^2$. This is an idealized example, not an experimental measurement.

The PHYWE guide Magnetic moment in the magnetic field makes the magnetic dipole moment measurable. A circular test current loop sits between a pair of Helmholtz coils, with separate supplies and meters for the test electric current $I$ and field-producing electric current $I_H$. The test current loop is mounted on a torsion dynamometer. The guide varies $I_H$, the number of turns, the test electric current, diameter, and the angle between the magnetic dipole moment and magnetic field. Its angular settings are spaced by $15°$. The leads hang loosely and are twisted together to suppress extra torque; the instrument zero is checked frequently because moving leads can shift it. Its reported results include approximately linear dependence on each electric current and on winding count, and approximately quadratic dependence on diameter.^[ref]

The readout is a balance of torques. According to the manufacturer's Torsion dynamometer 0.01 N operating instructions, a metallic band supplies the restoring torque. First set the indication to zero with the applied force absent and center the lever using the zero knob. Apply the load, then turn the indicating knob until the lever returns to its original zero position. The scale then reads a force equivalent, not a torque directly. The quoted scale division is $0.1\mathrm{mN}$. For a force applied perpendicular to the calibration lever arm at distance $\ell$, the conversion is $\tau =F\ell$; a current loop applies a couple, so its torque must be compared with the calibrated restoring torque.^[ref]

For a quantitative repeat, independently check that conversion with known forces at a measured perpendicular lever arm, and check the imposed magnetic field at the test current loop. These are proposed calibration checks, not additional measurements reported here. The ideal Helmholtz coil model predictswhere $N_H$ and $R_H$ are the winding count and radius of each field-producing Helmholtz coil, distinct from $N$ and $R$ of the test current loop. Derive $B_0$ by adding the on-axis magnetic field of a circular current loop at distances $R_H/2$ from both coils. Independently calibrating $B_0$ and the restoring torque lets a fitted torque amplitude determine $m$ as amplitude divided by $B_0$, to compare with $NI\pi R^2$.

Useful additional controls are to reverse only $I$, reverse only $I_H$, and reverse both: the signed torque should reverse in either single reversal and be unchanged in the double reversal. Record baselines with each electric current separately zero, and subtract a nearby zero reading to reduce torsion balance zero drift. Repeat angular settings in both directions to reveal lead-dependent offsets. These are suggested controls. Limits include finite magnetic field uniformity, uncertainty in effective winding area and angular zero, meter calibration, Joule heating, and residual lead torque. An apparently sinusoidal trace alone does not establish the absolute magnetic dipole moment. No raw laboratory data or new measured values are claimed here.

Far outside a localized steady source of size $a$, at distance $r\gg a$, its multipole expansion starts with the magnetic dipole moment if that moment is nonzero. In vacuum, using SI units,where ${\mu }_0$ is the vacuum permeability, $\hat{\mathbf{r}}=\mathbf{r}/r$, and the magnetic field is the curl of the magnetic vector potential. The magnetic dipole field falls as $r^{-3}$, and the magnetic vector potential as $r^{-2}$. The localized steady-current moment identities remove the putative $r^{-1}$ term in the magnetic vector potential. No magnetic monopole is needed. If different current loops cancel the total magnetic dipole moment, a higher term of the multipole expansion supplies the leading distant magnetic field. ^[ref].

The approximation is not valid at the center of a finite current loop. For a circular current loop of radius $R$ and electric current $I$, the exact on-axis magnetic field of a circular current loop and the distant approximation areTheir ratio is $B_z/B_{z,\mathrm{dip}}=(1+R^2/z^2{)}^{-3/2}$: at $|z|=10R$, the approximate magnetic field exceeds the exact magnetic field by about $1.50\%$ relative to the exact value. Near the source, retain its geometry rather than replacing it with a point source specified only by its magnetic dipole moment.

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.

Magnetization is magnetic dipole moment per unit volume, described by a vector field $\mathbf{M}(\mathbf{r})$. Average the microscopic magnetic dipole moments in a small region of volume $\Delta V$ around the point $\mathbf{r}$:The averaging region should contain many microscopic constituents while being small compared with the distance over which the macroscopic magnetization varies. With number density $n$ of identical constituents, this becomes $\mathbf{M}=n⟨\mathbf{m}⟩$. Magnetic dipole moment has SI units $\mathrm{A}{\mathrm{m}}^2$, so magnetization has units $A/m$: amperes per metre. The average is a vector sum, not a sum of magnitudes.^[ref]

Alignment can change magnetization without changing the magnitude of any individual magnetic dipole moment. In a constructed counting example, eight equal magnetic dipole moments of magnitude $m_0$ occupy the same volume $\Delta V$. Six along $+z$ and two along $-z$ give $\mathbf{M}=4m_0\hat{\mathbf{z}}/\Delta V$; turning the two opposing magnetic dipole moments to $+z$ doubles magnetization to $8m_0\hat{\mathbf{z}}/\Delta V$. Both arrangements have the same sum of magnitudes, $8m_0$. Changing the individual magnetic dipole moments provides another possible response; their alignment and their magnitudes are distinct ingredients. Microscopic spin contributes magnetic dipole moment without being a literal rotating charged ball.

Distinguish magnetization $\mathbf{M}$, magnetic field $\mathbf{B}$, and magnetic field intensity $\mathbf{H}$. In SI,Here ${\mu }_0$ is the vacuum permeability, and $\mathrm{T}$ denotes the tesla. Magnetization describes matter's magnetic dipole moment density; magnetic field enters the Lorentz force; magnetic field intensity separates the free electric current contribution in the macroscopic equations. In particular, zero applied magnetic field does not generally imply $\mathbf{H}=0$ inside a finite magnet: its demagnetizing field depends on shape.^[ref]

For stationary electric polarization, magnetization produces bound current in the volume and on its bounding surface:The curl gives a current density in $A/{\mathrm{m}}^2$; ${\mathbf{K}}_{\mathrm{b}}$ is a surface current density in $A/m$. The unit normal $\mathbf{n}$ points outward from the magnetized material into a region with zero magnetization. Across an interface between two magnetized media, the net bound current sheet iswhere the unit normal points from medium 1 to medium 2. Both faces contribute: medium 2 has the opposite outward normal vector. This fixes the cross product sign. If electric polarization changes with time, add its contribution ${\partial }_t\mathbf{P}$ to the bound current.^[ref]

A short proof outline explains why both contributions are needed. Write ${\mathbf{M}}_{\mathrm{ext}}=\mathbf{M}{\chi }_V$, where ${\chi }_V$ is the indicator function of the material's volume. In the sense of distributions, its gradient is $\nabla {\chi }_V=-\mathbf{n}{\delta }_S$, with ${\delta }_S$ the surface delta distribution. The product rule for the curl givesThus a constant interior magnetization can have zero interior curl and still produce a nonzero boundary electric current. Equivalently, neighboring ideal microscopic current loops cancel along shared edges while uncompensated edges remain on the outside. This bound current representation predicts the same macroscopic magnetic field as the magnetization distribution; it does not require a wire or free charge transport around the magnet. Its usefulness is that the familiar Biot-Savart law can now calculate the magnetic field from a specified magnetization.

A time-dependent average magnetization need not vary smoothly under a smooth drive. The Barkhausen effect resolves collective jumps through their induced electromotive force; Barkhausen pulse area explains how that derivative readout relates to the underlying change.

The Biot-Savart law isHere $r^{\prime }$ is the current-element position, $d{\ell }^{\prime }$ follows the electric current, and $r$ is the observation point. For a circular current loop, transverse components cancel on the axis. Each element contributes axial factor $a/\sqrt{a^2+x^2}$ in addition to the inverse-square distance. Integrating around length $2\pi a$ gives the on-axis magnetic field of a circular current loopwith axial direction set by the right-hand rule.

For a finite straight wire parametrized by $z\in [-b,b]$, at perpendicular distance $a$, the Biot-Savart law givesThe direction is azimuthal about the current axis by the right-hand rule. In the hairpin, the semicircle contributes ${\mu }_{0}I/(4a)$ and each semi-infinite straight segment contributes ${\mu }_{0}I/(4\pi a)$. For the indicated current, all three contributions point out of the page. Thus the magnetic field at the center of a hairpin wire is

For a circular current loop, the Biot-Savart law iswhere $\mathbf{s}$ points from a current element to the field point. Choose the loop in the $yz$ plane, with observation point $x\hat{\mathbf{x}}$. A parametrization consistent with positive axial magnetic field isThen $s=\sqrt{a^2+x^2}$ andThe transverse contributions cancel around the circle, leaving the on-axis magnetic field of a circular current loopThe right-hand rule fixes its direction; reversing the electric current reverses the magnetic field. At the centre this becomes ${\mu }_{0}I/(2a)$, and far away it decays as $|x{|}^{-3}$, as expected for a magnetic dipole.

An electric dipole consists of opposite charges separated by a small displacement, with electric dipole moment $\mathbf{p}=q\mathbf{d}$ directed from negative to positive. At distances large compared with its size,The net charge term vanishes; the field decays faster than a monopole field. In an imposed slowly varying electric field,for a permanent fixed dipole. A uniform field produces a torque but no net force. For a linear induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the interaction energy is instead $-\alpha E^2/2$, because building the dipole also costs energy.

A small planar current loop has magnetic dipole moment $\mathbf{\mu }=IA\hat{\mathbf{n}}$, with the normal fixed by the right-hand rule. Outside the source,Magnetic field lines form closed loops; a current-loop dipole does not require magnetic monopoles. Dipole moments describe the leading directional field of a neutral localized source, and their torques explain alignment in external fields.

Prescribe constant magnetization $\mathbf{M}=M\hat{\mathbf{z}}$, with $M>0$, inside a circular cylinder of radius $R$ and length $L$, surrounded by zero magnetization. In cylindrical coordinates, the outward unit normal on the curved surface is $\hat{\mathbf{r}}$, soThe endcaps have outward unit normals $+\hat{\mathbf{z}}$ and $-\hat{\mathbf{z}}$. Their cross products with magnetization vanish, so ${\mathbf{K}}_{\mathrm{b}}=0$ on each endcap. The side bound current circulates counterclockwise when viewed from $+z$, by the right-hand rule. Zero endcap surface current density does not imply zero endcap magnetic field.

For an original example, let $M=800A/m$, $R=0.020\mathrm{m}$, and $L=0.10\mathrm{m}$. The volume and total magnetic dipole moment areCheck the result independently using the equivalent bound current: a side strip of axial length $dz$ forms a current loop with $dI=Mdz$ and magnetic dipole moment $d\mathbf{m}=\pi R^{2}Mdz\hat{\mathbf{z}}$. Its integral over $L$ gives the same $M\pi R^{2}L\hat{\mathbf{z}}$. In particular, $ML=80\mathrm{A}$ is the sum of the equivalent loop electric currents across a full axial cut; it is not a surface current density.

The side surface current density has exactly the geometry of a continuum solenoid with $nI=M$, where $n$ is solenoid turn density. In the infinite-cylinder idealization with no applied magnetic field, the long-solenoid approximation therefore becomesThe limit is useful well away from the ends of a sufficiently long circular cylinder. A finite circular cylinder instead has fringing magnetic field and a demagnetizing field. For this prescribed uniform magnetization, integrate the on-axis magnetic field of a circular current loop, or substitute $nI=M$ in the on-axis magnetic field of a finite solenoid, to obtain the central magnetic field:Our example has $L/(2R)=2.5$ and $B_z(0)/({\mu }_{0}M)=5/\sqrt{29}\approx 0.92848$, rather than unity. With the rounded vacuum permeability ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$, this is approximately $0.9334\mathrm{mT}$, compared with the infinite-cylinder value $1.0053\mathrm{mT}$. These are model calculations for a prescribed magnetization, not measured values or a solution for the self-consistent magnetization of an arbitrary material.