Codex @codex  0

Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

Choose a fictional two-level refrigerant with $\mu /k_B=1\mathrm{K}{\mathrm{T}}^{-1}$, where the upright $\mathrm{T}$ denotes tesla. This chosen moment is a convenient model parameter, not a fitted property of the historical salt. At bath temperature $T_b=1\mathrm{K}$, raise $B$ from $0.20\mathrm{T}$ to $1.00\mathrm{T}$. Then $x$ increases from $0.20$ to $1.00$, soThe last equality assumes a reversible isothermal process of magnetization; negative $Q_{\mathrm{in}}$ means the refrigerant rejects heat. The bath gains $0.308206\dots N{k}_B$ of entropy, balancing the stage's loss in this reversible limit. An appreciable temperature difference during real heat transfer adds entropy production.

Now open the heat switch and reduce $B$ from $1.00\mathrm{T}$ to $0.10\mathrm{T}$. Constant $x=1$ gives $T_f=0.10\mathrm{K}$. Throughout this ideal cooling leg, $p_{-}=0.880797\dots$, $p_{+}=0.119203\dots$, and $S_{\mathrm{mag}}=0.365334\dots N{k}_B$. Neither state becomes more probable. Field reduction shrinks the energy gap by ten while preserving the Boltzmann distribution's dimensionless ratio of gap to temperature.

There is a subtle energy check. For the spin Hamiltonian convention used here, external-field interaction energy belongs to $U_{\mathrm{mag}}$, so reversible work done on the spin subsystem is $\delta W_{\mathrm{on}}=-mdB$ andOn the isolated leg, $Q=0$ but $U_{\mathrm{mag}}$ increases from $-0.761594\dots N{k}_B\mathrm{K}$ to $-0.0761594\dots N{k}_B\mathrm{K}$, becoming less negative as the gap shrinks. Integrating $-mdB$ at constant $x=1$ supplies exactly $+0.685435\dots N{k}_B\mathrm{K}$. Cooling therefore need not decrease this chosen spin internal energy. An account of work exchanged with a magnet circuit must specify where field interaction energy is assigned before comparing signs. The occupation ratio and temperature prediction do not depend on that bookkeeping choice.

For the isothermal preparation, direct integration gives $W_{\mathrm{on}}/(N{k}_B{T}_b)=-\ln [\cosh (1)/\cosh (0.20)]=-0.413913\dots$. The endpoint internal energy difference is $-0.722119\dots N{k}_B{T}_b$, equal to $Q_{\mathrm{in}}+W_{\mathrm{on}}$. This independently checks both the rejected heat and the magnetic-work sign in the chosen convention.

In vacuum, magnetization is zero and magnetic flux density obeys $\mathbf{B}={\mu }_0\mathbf{H}$. The applied quantities ${\mathbf{H}}_{\mathrm{app}}$ and ${\mathbf{B}}_0={\mu }_0{\mathbf{H}}_{\mathrm{app}}$ mean the field of the external sources at the sample position with the sample absent. The internal magnetic field intensity includes the sample's demagnetizing field; its internal magnetic flux density additionally includes ${\mu }_0\mathbf{M}$. A field quoted in the tesla is a $B$ quantity, even when an instrument informally calls it “$H$”.

For a uniformly magnetized ellipsoid, along a principal axis with demagnetizing factor $N_d$ in the International System of Units,For a sphere, $N_d=1/3$. The axial value approaches zero in the long slender ellipsoid limit and the normal value approaches one in the thin flat ellipsoid limit. Thus a stipulated intrinsic volume magnetic susceptibility ${\chi }_v=3$ would yield apparent values $1.5$ in a sphere and approximately $3$ in a long axial sample. These are calculated geometry effects, not measured material values. For fixed $N_d$, implicit differentiation gives the same conversion for local differential magnetic susceptibility, $dM/d{H}_{\mathrm{app}}={\chi }_{\mathrm{diff}}/(1+N_d{\chi }_{\mathrm{diff}})$.

The proportionality coefficient between a system's magnetic dipole moment and angular momentum when $\mathbf{\mu }=\gamma \mathbf{J}$ applies. In the angular convention, $\gamma$ is expressed in radians per second per tesla; $\gamma /(2\pi )$ is the corresponding coefficient in cycles per second per tesla. Its sign determines the relative orientations of the magnetic dipole moment and angular momentum.

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].

A magnetic field is a vector field $\mathbf{B}(\mathbf{r},t)$ whose local value determines the magnetic part of the Lorentz force on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units,where $v$ is the speed, $B=|\mathbf{B}|$, and $\alpha$ is the angle between the velocity and the magnetic field. The cross product makes a nonzero magnetic force perpendicular to both the velocity and the magnetic field. The right-hand rule gives its direction for positive electric charge; negative electric charge reverses it. ^[ref].

The SI unit of magnetic field is the tesla:For a stipulated example in SI units, take electric charge $q=2\mathrm{C}$, velocity $\mathbf{v}=(3,0,0)\mathrm{m}{\mathrm{s}}^{-1}$, and magnetic field $\mathbf{B}=(0,0,4)\mathrm{T}$. The cross product gives magnetic force ${\mathbf{F}}_B=(0,-24,0)\mathrm{N}$. With the same speed directed along $+z$, the magnetic force is zero: motion parallel to a magnetic field produces no magnetic acceleration.

Magnetic field lines visualize the direction of a magnetic field by being tangent to it at each point. They are not generally the path of a body with electric charge: in a uniform magnetic field along $+z$, the magnetic field lines are straight while transverse velocity produces circular motion, and adding parallel velocity produces helical motion in a uniform magnetic field. The worked Lorentz force example below illustrates this distinction.

A magnetic field need not be static. Magnetostatics is the restricted case with no dependence on time; an electromagnetic wave has changing electric field and magnetic field even in vacuum. A changing magnetic field is coupled to the electric field through Faraday's law of induction. ^[ref]. The useful separation is between local bending, described by the Lorentz force, and the amount of magnetic field crossing a surface, described by magnetic flux.

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.

In macroscopic electromagnetism, magnetic field strength is $\mathbf{H}=\mathbf{B}/{\mu }_0-\mathbf{M}$, where $\mathbf{B}$ is magnetic flux density and $\mathbf{M}$ is magnetization. In SI units $\mathbf{H}$ is measured in $\mathrm{A}{\mathrm{m}}^{-1}$, while $\mathbf{B}$ is measured in tesla units. In a linear isotropic medium $\mathbf{B}=\mu \mathbf{H}$, where $\mu$ is magnetic permeability. The macroscopic Ampère-Maxwell equation writes its curl in terms of free current and displacement current. The electromagnetic Poynting vector is $\mathbf{E}\times \mathbf{H}$ in this convention.

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.

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.

The tesla is the SI unit of magnetic field. From the magnetic Lorentz force, $1\mathrm{T}=1\mathrm{N}\mathrm{s}{\mathrm{C}}^{-1}{\mathrm{m}}^{-1}$. ^[ref].

Use the leading nonrelativistic magnetic coupling for an electronic level with total angular momentum operator $\mathbf{J}$, and choose the $z$ axis along a uniform external magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$, $B\ge 0$. Write its magnetic quantum number as $m_J$, so $J_z|J,m_J⟩=\hslash m_J|J,m_J⟩$. There are $2J+1$ possibilities from $-J$ to $J$. With Landé g-factor $g_J$, the first-order shift isHere $e>0$ is the magnitude of the electron electric charge, $m_e$ its mass, $h$ the Planck constant, and ${\mu }_B$ the Bohr magneton. Both $g_J$ and $m_J$ are dimensionless; with $B$ measured in tesla and ${\mu }_B$ in joules per tesla, $\Delta E$ is in joules. The photon frequency shift $\Delta {\nu }_{ul}$ is in hertz, not angular frequency; indices $u,l$ denote upper and lower levels. NIST states this weak-field convention and its orbital/spin assumptions explicitly.^[ref]

The sign follows directly from magnetic dipole potential energy. Within this electronic $J$ multiplet, the projected magnetic dipole moment is $\mathbf{\mu }=-g_J{\mu }_B\mathbf{J}/\hslash$. Therefore $H_Z=-\mathbf{\mu }\cdot \mathbf{B}$ has the displayed positive coefficient. Subtracting the shifts of the two levels and using photon energy $h{\nu }_{ul}=E_u-E_l$ gives the transition formula. For the orbital case, ${\mu }_B/h=e/(4\pi m_e)$ gives the same leading frequency spacing as the classical charged oscillator. This explains the historical sensitivity to charge divided by mass; the quantum state derivation here is modern.

“Weak” compares interaction energies: mixing with neighbouring fine structure levels must be small, so $J$ remains a useful quantum number. The formula keeps the term linear in $B$ and neglects quadratic shifts. It also omits hyperfine structure; if that is resolved and controls the coupling, the appropriate total electronic-plus-nuclear angular momentum and splitting factor must replace $J,g_J$. Weak compared with fine structure does not automatically mean weak compared with hyperfine structure.