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Three distinct effects spoil indefinite $T\propto B$ cooling: extra entropy outside the spin subsystem, additional magnetic energy scales, and irreversible heat or work input. They can be distinguished with a thermodynamic derivative. For a smooth equilibrium Helmholtz free energy obeying $dF=-SdT-mdB$, a Maxwell relation givesHere $C_B$ is the total heat capacity participating in the stage's thermal equilibrium. The ideal model has $(\partial m/\partial T{)}_B=-C_{B,\mathrm{mag}}/B$. If its spins share temperature with a field-independent lattice and load of combined heat capacity $C_{\mathrm{other}}>0$, thenThe positive slope still means cooling when $dB<0$, but its smaller magnitude gives less cooling per fractional field reduction. For example, a low-temperature Debye model gives a lattice entropy proportional to $T^3$. As that lattice entropy falls, the spins take up the released entropy; $x$ can no longer remain constant. A weak enough coupling to neglect its equilibrium interaction energy can still be strong enough to exchange heat on the sweep timescale.

Interactions between magnetic moments, including exchange interaction and the fields of neighboring magnetic dipole moments, introduce energy scales that do not vanish with the applied $B$. Once they compete with $k_{B}T$ or $\mu B$, the independent-site partition function fails. Correlations and possible magnetic ordering alter the available magnetic entropy and the cooling curve. Simply setting the external magnetic field to zero does not eliminate those interactions. Different refrigerants therefore need their own measured equilibrium properties; there is no universal final temperature from the applied-field ratio alone.

Finally, a real open heat switch has residual thermal conductance; supports, wires and radiation carry incoming heat. Field changes may also dissipate energy. Locally near thermal equilibrium, with incoming heat rate ${\dot{Q}}_{\mathrm{leak}}$ and internal entropy production rate ${\dot{S}}_{\mathrm{prod}}$, the stage balance becomesThis uses a common stage temperature; a non-equilibrated load requires separate balances. The sweep must allow spin-lattice relaxation while remaining fast enough to limit accumulated parasitic heat. At fixed $B$, the same balance explains warming and finite hold time. Useful refrigeration is the controlled absorption of a load's heat during that finite interval, followed by a renewed isothermal process of magnetization to export its entropy.

A localized steady electric current density satisfies $\nabla \cdot \mathbf{j}=0$. Integrate the divergences of $x_i\mathbf{j}$ and $x_i{x}_k\mathbf{j}$ over a region enclosing its support; all boundary fluxes vanish. The first identity removes the monopole term from the magnetic vector potential expansion. The antisymmetric first moment is equivalent to the magnetic dipole moment $\frac{1}{2}\int \mathbf{x}\times \mathbf{j}dV$.

A localized magnetic source described at leading nonvanishing dipole order by a magnetic dipole moment. A small current loop is one classical model. The source object, its moment vector and the resulting magnetic dipole field are distinct concepts. At large distance from a source with nonzero dipole moment, higher multipoles give smaller corrections; if the dipole moment vanishes, a higher-order term can lead instead.

For a small fixed magnetic dipole moment $\mathbf{m}$ in a slowly varying external magnetic field, $U=-\mathbf{m}\cdot \mathbf{B}$ gives the displayed force. For a compact current loop, integrate $Id\mathbf{\ell }\times \mathbf{B}$ to first order in the field's spatial variation to get the same result. For a coaxial small loop with $m_z=NI\pi a^2$, this becomes $F_z=m_{z}d{B}_z/dz$. A spatially uniform field can exert torque but gives zero net force; a field gradient is essential. The approximation assumes the field does not appreciably vary over the dipole's dimensions.

For a centred magnetic dipole moment pointing toward geographic south, the field at northern latitude $\lambda$ has northward component $B_N=B_{\mathrm{eq}}\cos \lambda$ and downward component $B_D=2B_{\mathrm{eq}}\sin \lambda$. These follow by resolving the magnetic dipole field into local radial and meridional directions. The downward inclination is therefore $\tan \iota =B_D/B_N=2\tan \lambda$, while the magnitude is $B_{\mathrm{eq}}\sqrt{1+3{\sin }^2\lambda }$. This ideal aligned-dipole model has no eastward component and does not capture actual geographic departures of the geomagnetic field.

The orientation-dependent interaction of two distant localized magnetic dipole moments. Its $r^{-3}$ scale makes close approach important in a concentrated magnetic-particle dispersion. Depending on geometry, it can favor chains, collective alignment or frustrated configurations.

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].

In an external field that varies slowly across a localized magnetic dipole,

The magnetic dipole moment determines the coupling to the external magnetic field.

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.

Take a wire segment on the $y$ axis, with source positions ${\mathbf{r}}^{\prime }=(0,s,0)$ for $s_1\le s\le s_2$, and positive electric current along $+y$. Observe at $\mathbf{r}=(a,0,0)$, where $a>0$ is the perpendicular distance to the wire's supporting line. Choose $x$ rightward and $y$ upward, so $+z$ points out of the page. ThenThus the magnetic field points into the page by the right-hand rule. The Biot-Savart law and an elementary antiderivative giveFor an original worked example, choose $s_1=-0.12\mathrm{m}$, $s_2=0.12\mathrm{m}$, $a=0.05\mathrm{m}$, and $I=3\mathrm{A}$. Use ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$ for these rounded calculations. Each endpoint contributes $0.12/0.13=12/13$, soThese are calculated inputs and output, not laboratory data. The two endpoint angles measured from the perpendicular each have sine $12/13$; this recovers the equivalent endpoint-angle formula without depending on a diagram's angle-sign convention.

For a segment centered on the perpendicular foot, $s_1=-L$, $s_2=L$. Its magnetic field magnitude isThe limit $L/a\to \infty$ gives $B_{\infty }={\mu }_{0}I/(2\pi a)$, consistent with Ampère's circuital law for an infinite straight wire. In the example, $B_{\infty }=12\mu \mathrm{T}$ and $B/B_{\infty }=12/13$: the infinite-wire value overestimates the segment value by $1/12$, or about $8.33\%$ of the segment value. For $a\gg L$, the isolated segment contribution instead behaves as $B\sim {\mu }_{0}I(2L)/(4\pi a^2)$. The complete localized electrical circuit has a different far-field behavior: return-path contributions cancel this leading segment term, leaving a magnetic dipole field when its magnetic dipole moment is nonzero. Segment and whole-circuit limits must be distinguished.

The SI units check directly: ${\mu }_{0}I/a$ has units $(\mathrm{T}m/A)\mathrm{A}/\mathrm{m}=\mathrm{T}$, the tesla. The apparent divergence as $a\to 0$ is a failure of the thin-wire model near the conductor, not a prediction of infinite magnetic field inside a real wire. This calculation gives only this segment's contribution; include the return wires before comparing with an apparatus.

A uniformly charged solid sphere of total charge $Q$, radius $R$ and angular velocity $\mathbf{\omega }$ has magnetic dipole momentFor a uniformly charged thin spherical shell the coefficient is $1/3$ instead of $1/5$. At large distances the magnetic field is a magnetic dipole field, $\mathbf{B}={\mu }_0[3(\mathbf{m}\cdot \hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}]/(4\pi r^3)$. Its angularly averaged magnitude and root mean square magnitude both scale as $r^{-3}$; its angularly averaged vector is zero. Steady rotation of this axisymmetric charge distribution produces magnetostatics, not radiation.

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.

Let $m_z={\int }_V{M}_{z}dV$ be the axial magnetic dipole moment, where $M_z$ is the axial magnetization. Write $J_z$ for the internal angular momentum associated with that magnetic dipole moment, and $I\dot{\theta }$ for the body's rotational angular momentum, with moment of inertia $I$. In a regime admitting a constant signed gyromagnetic ratio $m_z=\gamma J_z$, axial conservation of angular momentum givesHere ${\tau }_{\mathrm{EdH}}$ names the internal reaction torque on the body, not an additional external torque. The external term includes the suspension. The model assumes that the changing internal angular momentum is transferred to the measured mechanical motion, with other storage or escape channels negligible on the observation timescale. A changing or non-collinear mixture of orbital angular momentum and spin angular momentum need not admit a single constant $\gamma$.

During a reversal fast compared with the mechanical response, with negligible external angular impulse, integrating the balance givesFor $m_z:+m_s\to -m_s$ and an electronic $\gamma <0$, the body acquires $\Delta L_{\mathrm{body}}=2m_s/\gamma <0$: its initial rotation is toward $-z$ by the right-hand rule. Reversing the reversal reverses the impulse. Over a complete periodic cycle, $\Delta m_z=0$, so the net reaction angular impulse vanishes; a resonant alternating response does not imply steadily accumulating angular momentum.

The orbital angular momentum prediction can be derived without a material model. A charge $q$ traveling around a circle of radius $r$ at angular frequency $\omega$ has loop electric current $q\omega /(2\pi )$ and magnetic dipole moment $m=q\omega r^2/2$. Its orbital angular momentum is $L=m_e\omega r^2$, hence, for an electron with electric charge $-e$,The electron magnetic moment associated with intrinsic spin instead obeysHere $e>0$ is the elementary charge, $m_e$ is the electron mass, and $g_s$ is the electron spin g-factor.^[ref] Thus the same change of magnetic dipole moment corresponds to about half as much transferred angular momentum in the pure-spin model as in the orbital angular momentum model. This factor of two is a relation between magnetic dipole moment and angular momentum, not evidence that an electron is literally a rotating charged sphere. In a material with both contributions, the inferred coefficient is a material-dependent magnetomechanical gyromagnetic ratio.

A magnetometer measures a magnetic field or a specimen's magnetic dipole moment. A vibrating-sample magnetometer determines the latter from the electromotive force induced in pickup coils by specimen motion.

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.

The magnetic dipole moment associated with nuclear spin, using dimensionless angular momentum operators and the nuclear magneton ${\mu }_N$. The quoted scalar moment is its expectation in the stretched quantum state, ${\mu }_I=g_I{\mu }_{N}I$. In the same electronic energy level, the leading magnetic-dipole hyperfine structure constant is proportional to ${\mu }_I/I$, apart from isotope-dependent electronic and nuclear-size corrections.

The magnetic dipole moment of an atomic nucleus. Its relation to nuclear spin is expressed by the gyromagnetic ratio; its interaction with an applied magnetic field underlies nuclear magnetic resonance.

A circular loop of radius $R$ carrying electric current $I$ has, on its axis,where the sign of $\hat{z}$ is fixed by the right-hand rule. At its centre, $B(0)={\mu }_{0}I/(2R)$.

Put the loop in the $xy$ plane, centered at the origin, with positive electric current counterclockwise when viewed from $+z$. The right-hand rule then gives the central magnetic field along $+z$. Parameterize source positions by the angle $\varphi$ and observe at $(0,0,z)$:Here $R$ without bold type is the loop radius, whereas $\mathbf{R}$ is the source-to-observer displacement. Their cross product isThe denominator $|\mathbf{R}{|}^3=(R^2+z^2{)}^{3/2}$ is constant around the loop. The integrals of sine and cosine over a full turn vanish, while the axial contribution adds to $2\pi R^2$. Substitution into the Biot-Savart law proves the on-axis formula. Symmetry cancels transverse components; it does not cancel the axial components.

For an original one-turn example in the same vacuum approximation, choose $R=0.04\mathrm{m}$ and $I=2\mathrm{A}$. With ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$,At $z=0.03\mathrm{m}$, the source-to-observer distance is $0.05\mathrm{m}$, so $B(z)/B(0)=64/125=0.512$ and $B(z)\approx 16.08\mu \mathrm{T}$. At $z=-0.03\mathrm{m}$, the magnetic field has the same magnitude and still points along $+z$. The point has crossed the loop plane, but the loop's electric current has not reversed. Reversing the electric current reverses the magnetic field on both sides.

For $|z|\ll R$, a Taylor expansion gives $B(z)/B(0)=1-3z^2/(2R^2)+O(z^4/R^4)$: the center is a smooth maximum, with zero first derivative. For $|z|\gg R$,This is the axial magnetic dipole field of magnetic dipole moment $\mathbf{m}=I\pi R^2\hat{\mathbf{z}}$.^[ref] A compact closed loop has an inverse-cube axial far field, in contrast to the inverse-distance field of the infinite straight wire. The SI units of the exact loop formula are ${\mu }_{0}I{R}^2/(R^2+z^2{)}^{3/2}\sim {\mu }_{0}I/\text{length}$, hence tesla.

For $N$ coincident turns carrying the same electric current, the superposition principle multiplies the ideal result by $N$. A real coil has radial width and axial length: if those are not negligible, integrate or sum over its individual turns. This is why a single measured coil diameter and a turn count may not determine the full spatial magnetic field precisely. A collection of loops becomes a solenoid; integrating their axial contributions gives the on-axis magnetic field of a finite solenoid.

For a light nonmagnetic rod of length $l$ wound uniformly with $N$ thin turns of mean radius $a$, the magnetic dipole moment is $m_B=NI\pi a^2$. Wire of cross-sectional area $A_w$ and density $\rho$ has mass approximately $M=2\pi aN\rho A_w$. The transverse central moment of inertia is $J=M(l^2/12+a^2/2)$ in the thin-winding approximation. Consequently the oscillation period in a transverse magnetic field does not depend on the number of turns: wire inertia and magnetic moment both scale with $N$.

The printed magnetic dipole moment uses dimensionless angular momentum operators: their physical counterparts are $\hslash \mathbf{L}$ and $\hslash \mathbf{S}$. Choose the magnetic field along $z$. Then the Zeeman energy is $H_Z={\mu }_{B}B(L_z+g_s{S}_z)$, whose characteristic size is ${\mu }_{B}B$. In LS coupling, treating this as a time-independent perturbation theory requires it to be much smaller than the relevant fine structure separation, rather than merely smaller than the Coulomb interaction energy.

For a concrete hydrogen atom estimate take the $2p_{1/2},2p_{3/2}$ pair. The hydrogen fine-structure correction iswhere ${\alpha }_{\mathrm{fs}}$ is the fine-structure constant, distinguished from the later potential coefficient. ThusConsequently a suitable weak-magnetic field condition for this pair isThere is no single threshold for every hydrogen energy level: the relevant fine structure separations decrease with $n$. The $1s$ energy level has no same-$n$ spin-orbit interaction partner, so it cannot itself define this comparison. This calculation neglects hyperfine structure; resolving that smaller scale requires an additional weak-magnetic field condition.