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The time integral of torque about a specified origin equals the change in angular momentum about that origin under the usual fixed-origin inertial frame balance. Its units are $\mathrm{N}\mathrm{m}\mathrm{s}$, equivalently $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. For rotation about a fixed principal axis, the magnitude is $I\Delta \omega$.

If two principal moments of inertia coincide, for example $I_1=I_2\ne I_3$, every direction in their two-dimensional eigenspace is a principal axis. The perpendicular axis is determined, but there is no privileged orthogonal pair within that plane. Rotating the pair in the plane leaves the diagonal inertia tensor unchanged. If all three principal moments of inertia coincide, ${\mathbf{I}}_O=I_{0}1$ and every axis through $O$ is principal. This follows from the eigenvalue multiplicity and eigenspace, rather than from a failure of diagonalization. ^[ref]

Equal principal moments of inertia do not prove that a body's outline has continuous rotational symmetry. For example, six equal point masses at $(\pm a,0,0)$, $(0,\pm a,0)$ and $(0,0,\pm a)$ have ${\mathbf{I}}_O=4m{a}^{2}1$. This discrete layout has the same axial moment of inertia in every direction, although its actual mass distribution changes under most rotations. Conversely, even when all three principal moments of inertia are distinct, principal axes are unoriented lines: reversing an eigenvector's sign gives the same axis. Near repeated eigenvalues, small measurement errors can rotate fitted principal axes substantially within the nearly degenerate eigenspace, so quoted directions need an uncertainty assessment as well as quoted principal moments of inertia.

For a rotor spinning about a principal axis with its support at the centre of mass, uniform gravity supplies no torque about that centre. With other external torques negligible, conservation of angular momentum preserves its direction in an inertial frame. A mount that permits free axis motion lets the housing rotate relative to that persistent direction, giving an orientation reference when external landmarks are unavailable. This is the useful limit opposite to the deliberately off-centre support in the spinning-wheel precession experiment. It supplies a direction, not a position or a full three-axis orientation from one axis alone.

For a small accumulated transverse torque about the centre of mass, the change in the direction unit vector is, to first order,Here $L$ is approximately constant over the short interval. Large angular momentum reduces drift for the same integrated transverse torque; parallel torque instead changes its magnitude. For $L=8.0\mathrm{kg}{\mathrm{m}}^2/s$, a constant transverse torque of $0.010\mathrm{N}\mathrm{m}$ over $1.0\mathrm{s}$ produces approximately $0.00125\mathrm{rad}$, or $0.0716^{\circ }$, of direction change. This is another synthetic calculation. The persistent direction belongs to an inertial frame; its components can change in a rotating reference frame even with zero external torque.

For a uniform cube of side $s$ and mass $M$, reflection symmetry makes central off-diagonal inertia components zero. Equal coordinate second moments are $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=s^2/12$, so each diagonal component is $M{s}^2/6$. Thus every central axis is a principal axis. A nonspherical shape can have an isotropic central inertia tensor: that tensor measures only quadratic moments of the mass distribution, not the whole shape.

An exact planar example makes the off-diagonal entries tangible. Join four equal point masses $m$ with negligible-mass rigid connectors at $a(2,1,0)$, $a(-2,-1,0)$, $a(1,-1,0)$ and $a(-1,1,0)$, where $a>0$ is a length. Their centre of mass is $O=(0,0,0)$. Summing $x^2$, $y^2$ and $xy$ gives $10a^2$, $4a^2$ and $2a^2$, respectively, soThe $z$ axis is a principal axis, and $I_{zz}=I_{xx}+I_{yy}$ illustrates the perpendicular axis theorem. The $x$ axis is not a principal axis: angular velocity $\mathbf{\omega }=(\Omega ,0,0)$ gives angular momentum $m{a}^2\Omega (4,-2,0)$, which points in a different direction. A bearing can enforce fixed-axis rotation despite this difference, supplying the necessary transverse torque.

For an in-plane unit vector $\mathbf{n}=(\cos \theta ,\sin \theta ,0)$, the planar model hasIts extrema are the principal moments of inertia $(7-\sqrt{13})m{a}^2$ and $(7+\sqrt{13})m{a}^2$; the remaining principal moment of inertia is $14m{a}^2$. These values follow from the two-by-two characteristic polynomial $(4-\lambda )(10-\lambda )-4$, so they can be checked without trusting a numerical eigenvalue routine. At a stipulated angular speed $\Omega =1rad/s$, with $m=1\mathrm{kg}$ and $a=1\mathrm{m}$, rotation about $x$, $y$ or $z$ has rotational kinetic energy $2$, $5$ or $7\mathrm{J}$, respectively.

A truly three-dimensional example adds two more equal point masses at $a(0,0,1)$ and $a(0,0,-1)$. Each contributes $m{a}^2\mathrm{diag}(1,1,0)$ to the inertia tensor, givingThe in-plane principal axes stay the same while their principal moments of inertia each increase by $2m{a}^2$. The $z$ moment of inertia stays unchanged because the added point masses lie on that axis. This comparison changes the total mass from $4m$ to $6m$; it holds the original four point masses fixed rather than redistributing a fixed total mass.

For a rigid body with three distinct positive principal moments of inertia $0<I_1<I_2<I_3$ and zero external torque about its centre of mass, steady rotation about the first or third principal axis is Lyapunov stable in angular velocity, while rotation about the intermediate principal axis is unstable. Here steady rotation means $\mathbf{\omega }=\Omega {\hat{e}}_i$ in the principal body frame, with constant $\Omega \ne 0$. A small sideways perturbation remains small about the extreme principal moments of inertia, but generically grows about the intermediate one. Stable does not mean that the wobble decays: without damping, there is no asymptotic stability. Coincident principal moments of inertia and a stationary body are excluded.^[ref]

No external torque is needed for a body to tumble. Conservation of angular momentum fixes $\mathbf{L}$ in the fixed space frame, but the body can turn relative to $\mathbf{L}$, so its angular velocity components in the principal body frame change. The torque-free rigid-body invariants constrain this motion. A supported gyroscope with gravitational torque has different governing assumptions.

For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

Choose an inertia unit $I_{\ast }$ and an angular speed unit ${\Omega }_{\ast }>0$, so physical principal moments of inertia are $I_{\ast }(1,2,3)$, physical angular velocity is ${\Omega }_{\ast }\mathbf{\omega }$, and normalized time is $t={\Omega }_{\ast }{t}_{\mathrm{physical}}$. The Euler equations for a torque-free rigid body becomeThis is a physically admissible ideal planar mass distribution: $I_3=I_1+I_2$ is the perpendicular axis theorem. A book of finite thickness need not have these exact ratios.

For an original comparison, start with angular velocity $(1,\epsilon ,0)$, $(\epsilon ,1,0)$ or $(\epsilon ,0,1)$, respectively. With $\epsilon =0.01$, the transverse Euclidean norm stays small about axes one and three, but becomes order one about axis two. For the intermediate-axis start, its linearization givesThe hyperbolic functions contain a growing exponential function with rate ${\lambda }_2=1/\sqrt{3}$. Reducing $\epsilon$ tenfold delays reaching the same small transverse amplitude by approximately $\sqrt{3}\log 10\simeq 3.99$ normalized time units, once the growing term dominates. That delays the instability. Once the transverse angular velocity is comparable to the main component, the linearization ceases to describe the motion.

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

First consider rotation predominantly about the third principal axis, writing ${\omega }_3=\Omega +O({\omega }_1^2+{\omega }_2^2)$. To first order the third Euler equation gives $\dot{\Omega }=0$, while the transverse components satisfyThereforeand the same equation holds for ${\omega }_2$.

Order the principal moments as $I_1<I_2<I_3$. The coefficient above is negative, so perturbations about axis three oscillate and remain bounded. Cyclically applying the same linear stability of principal-axis rotation calculation about axis one givesso that rotation is also stable. About axis two, however,which has an exponentially growing solution. Thus the intermediate axis theorem gives

A freely rotating rigid body with no external torque conserves its inertial angular momentum and kinetic energy. In body-fixed principal axes, the Euler equations includeand cyclic permutations. The body can still change its orientation relative to its fixed angular-momentum vector; zero torque does not require every principal axis to remain fixed.

For a symmetric top, $I_1=I_2=I_{\perp }$, $I_3=I_{\parallel }$. The axial rate ${\omega }_3$ is constant, and ${\omega }_1+i{\omega }_2$ rotates in body coordinates with signed frequency $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. In space,Thus the symmetry axis precesses about fixed $\mathbf{L}$ at rate $L/I_{\perp }$ and constant inclination. This body-frame and space-frame frequencies of free symmetric-top precession distinction avoids confusing two different frame-dependent rates. A sphere has no dynamically distinguished symmetry axis even though a chosen body marker rotates.

For a general body, rotations about the largest and smallest principal inertias are stable to small perturbations, while rotation about the intermediate inertia is unstable. With $I_1<I_2<I_3$ and rotation ${\omega }_2=\Omega$, linearization gives ${\ddot{\omega }}_1={\Omega }^2(I_3-I_2)(I_2-I_1){\omega }_1/(I_1{I}_3)$, showing exponential growth. This is torque-free motion, distinct from the gravity-driven precession of a supported gyroscope.

The microscopic magnetic response becomes a macroscopic field problem through the magnetization $\mathbf{M}$, the magnetic dipole moment per unit volume. Definein magnetostatics. Bound currents are represented by ${\mathbf{J}}_b=\nabla \times \mathbf{M}$ and surface current ${\mathbf{K}}_b=\mathbf{M}\times \hat{\mathbf{n}}$. Normal $B$ is continuous across an interface, while $\hat{\mathbf{n}}\times ({\mathbf{H}}_2-{\mathbf{H}}_1)={\mathbf{K}}_{\mathrm{free}}$. It is generally incorrect to impose continuity of every component of both fields. A linear isotropic material has $\mathbf{M}=\chi \mathbf{H}$ and ${\mu }_r=1+\chi$, but a ferromagnet usually requires a nonlinear, history-dependent constitutive relation.

Internal field differs from externally applied field. For uniform magnetization along a principal axis of an ellipsoid, the demagnetizing field is $-N_{d}M$ and $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$. A linear specimen therefore has apparent susceptibility $\chi /(1+N_d\chi )$ relative to the applied field. Shape is especially important for large susceptibility. A long closed magnetic circuit reduces stray fields; a bar magnet still has demagnetizing fields despite the absence of magnetic monopoles. Equivalent magnetic-pole descriptions are bookkeeping for magnetization divergence, not free magnetic charge.^[ref]

In paramagnetism, permanent localized moments are preferentially aligned by the field, against thermal disorder. For independent spin-one-half moments with projections $\pm \mu$, the two Zeeman energies givein the dilute weak-field limit with $B\simeq {\mu }_{0}H$. This is a Curie law; the response saturates at $n\mu$ when alignment dominates. More generally, spin $J$ gives $\chi ={\mu }_{0}n(g{\mu }_B{)}^{2}J(J+1)/(3k_{B}T)$. Interactions can produce a Curie–Weiss response $C/(T-\Theta )$ over a suitable range. Conduction-electron Pauli paramagnetism need not show the localized-moment $1/T$ law.

In diamagnetism, the field induces a magnetic moment opposing it, so the linear susceptibility is negative. The microscopic cause is the change in orbital electronic motion, and a closed-shell insulating material has no permanent atomic moment to orient. This weak response is often only mildly temperature dependent away from other transitions. It also exists as a background in materials with a stronger positive response. Perfect flux expulsion in a superconductor is a distinct collective phenomenon and should not be identified with ordinary weak atomic diamagnetism.

In ferromagnetism, exchange interactions stabilize spontaneous magnetic order below the Curie temperature. A specimen can have zero net moment while its magnetic domains remain locally magnetized in different directions. Domain-wall motion and moment rotation produce a large response; anisotropy and pinning lead to remanence, coercivity and hysteresis. Increasing field eventually approaches saturation. Heating above the ordering transition removes spontaneous order, although an applied field still induces a response. Soft magnetic cores favor easy reversal and small cyclic loss; hard permanent-magnet materials favor retained magnetization and resistance to reversal. These properties concern collective order and barriers, not merely independent moments all following a weak field.^[ref]

A practical distinction requires a controlled field sweep and a separate magnetic readout. A toroidal core with drive winding $N_1$, path length $l$ and current $I$ has approximately $H=N_{1}I/l$. A pickup winding measures ${\mathcal{E}}_2=-N_{2}AdB/dt$; integrating after calibration gives $B$, and $M=B/{\mu }_0-H$. Reversing the sweep reveals a loop whose positive loss area per volume is $\oint HdB$. An air-core/empty-holder reference, known geometry and turn count, offset correction, temperature control and slower sweeps test background, heating and rate-dependent eddy-current losses. A single force or susceptibility reading cannot establish spontaneous order or hysteresis. The original curves below are illustrative responses, not measured material data.

For the inertia tensor of a uniform cube, reflection symmetry makes every off-diagonal element of the central inertia tensor vanish. Permutation of the three axes makes its diagonal elements equal. Since $M=8\rho a^3$ and the mean of each squared coordinate in the cube is $a^2/3$,Every central axis is therefore a principal axis, including the body diagonal. The angular momentum is parallel to the angular velocity, and its magnitude is

Choose body-fixed principal axes through a fixed point or the centre of mass, with principal moments $I_1,I_2,I_3$. Angular momentum has components $L_i=I_i{\omega }_i$. The inertial derivative is the body derivative plus $\mathbf{\omega }\times \mathbf{L}$, so Euler's equations readThey avoid differentiating the rotating inertia tensor in laboratory coordinates. Torques must refer to the chosen point, and the displayed fixed moments require a rigid body.

For zero torque, laboratory angular momentum is constant, while its components can change in the body frame. Both $L^2$ and rotational kinetic energy $T=\frac{1}{2}\sum I_i{\omega }_i^2$ are conserved. The intersections of their constant-value surfaces constrain the motion. For distinct $I_1<I_2<I_3$, steady rotation about either extreme principal axis is stable against small perturbations, whereas rotation about the intermediate axis is unstable. For example perturbing rotation $\Omega$ about axis 2 giveswhose coefficient is positive. A symmetric body has additional simplifications and can precess. Conservation of laboratory angular momentum does not imply constant body-frame angular velocity.

For a rigid body about a chosen origin,The inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity for rotation about a principal axis, but need not be parallel in a general direction. Its eigenvalues are the principal moments of inertia.

If the origin is translated from the centre of mass by $\mathbf{a}$, the parallel-axis theorem becomes $I_{ij}=I_{ij}^{\mathrm{CM}}+M(a^2{\delta }_{ij}-a_i{a}_j)$. In body axes, Euler's equation is $\mathbf{\tau }=(d\mathbf{L}/dt{)}_{\mathrm{body}}+\mathbf{\omega }\times \mathbf{L}$. Thus the full tensor is needed whenever the rotation is not confined to one fixed principal axis.

For positions measured from the chosen origin, the inertia tensor isIt relates angular momentum and rotational kinetic energy to angular velocity:The real symmetric tensor has orthogonal principal axes, in which $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$. Its eigenvalues are nonnegative moments of inertia. Rotation about a principal axis has $\mathbf{L}\parallel \mathbf{\omega }$; a general rotation does not. The moment about a unit axis $\mathbf{n}$ is $I_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{In}$.

Changing from the centre of mass to a parallel origin displaced by $\mathbf{a}$ gives the tensor parallel-axis theorem,For example, a uniform solid sphere has $\mathbf{I}=(2M{R}^2/5)1$, while a thin rod along $x$ has principal moments $0,M{l}^2/12,M{l}^2/12$ in the ideal zero-radius limit. The tensor is needed to describe off-axis rotation and torque-free precession; a single scalar inertia only suffices after an axis is specified.^[ref]

For a body-fixed mass element at position $\mathbf{r}$ relative to the fixed tip, $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$. ThereforeThe inertia tensor is real symmetric. Its orthogonal eigenvectors define principal axes and its eigenvalues are principal moments of inertia. angular momentum is parallel to angular velocity when rotation is along a principal axis, but generally not for arbitrary rotation.

Write $\mathbf{n}={\hat{\mathbf{e}}}_3$ and $s=I_3{\omega }_3$. Axial symmetry gives $\mathbf{L}=I_1\mathbf{\omega }+(I_3-I_1){\omega }_3\mathbf{n}$. Since $\dot{\mathbf{n}}=\mathbf{\omega }\times \mathbf{n}$, the vector triple-product identity gives $\mathbf{n}\times \dot{\mathbf{n}}=\mathbf{\omega }-{\omega }_3\mathbf{n}$. HenceDifferentiate in the inertial frame, using constant $s$ and $\dot{\mathbf{n}}\times \dot{\mathbf{n}}=0$. The gravitational torque about the fixed tip is $(\ell \mathbf{n})\times (-mg\hat{\mathbf{z}})$, yielding the symmetry-axis equation of a heavy symmetric top:The tip reaction exerts no moment about that tip.

For steady precession, $\mathbf{n}=(\sin \theta \cos {\Omega }_{P}t,\sin \theta \sin {\Omega }_{P}t,\cos \theta )$. At azimuth zero, put $s_{\theta }=\sin \theta$ and $c_{\theta }=\cos \theta$. Then $\dot{\mathbf{n}}=(0,{\Omega }_P{s}_{\theta },0)$, $\ddot{\mathbf{n}}=(-{\Omega }_P^2{s}_{\theta },0,0)$, and the $y$ component gives $s{\Omega }_P{s}_{\theta }-I_1{\Omega }_P^2{s}_{\theta }{c}_{\theta }=mg\ell s_{\theta }$. For a nonvertical axis this becomesReal steady branches require a nonnegative discriminant. In particular, for a top pointing above the horizontal there is a spin threshold for these fixed-angle solutions. At $\cos \theta =0$ use the quadratic's linear limit ${\Omega }_P=mg\ell /s$, rather than dividing by zero. For a vertical axis, azimuth does not describe a physical conical motion and the division by $\sin \theta$ is inappropriate.

For rapid positive spin $s$ with $s^2\gg 4mg\ell I_1\cos \theta$, the slow and fast steady precession roots areThe approximation assumes a nonsingular fixed inclination. For $g=0$, the quadratic instead gives ${\Omega }_P=0$ or ${\Omega }_P=s/(I_1\cos \theta )$. The zero branch has a stationary symmetry axis and pure axial spin. On the other branch angular momentum is constant and parallel or antiparallel to the chosen $z$ axis; $L_z=I_1{\Omega }_P$ and $s=L_z\cos \theta$. This is torque-free precession in space, not the different rate at which transverse components rotate in body coordinates.

Take axes fixed in a rigid body, chosen along its principal axes, with moments of inertia $I_1,I_2,I_3$. The angular momentum components are $H_i=I_i{\omega }_i$. Newton's rotational equation is $\mathbf{\tau }=(d\mathbf{H}/dt{)}_{\mathrm{inertial}}$, while the rotating-frame transport rule givesThus Euler's equations areThe nonlinear cross terms arise because the principal axes rotate. Even with no external torque, the body-frame components of angular momentum can vary while the angular momentum vector stays fixed in inertial space.

With zero torque, the equations conserveUniform rotation about any principal axis is a torque-free solution. If $I_1<I_2<I_3$, linearization about the intermediate-axis solution ${\omega }_2=\Omega$ givesThe positive coefficient gives exponential growth: intermediate-axis rotation is unstable, whereas rotation about the largest or smallest moment is linearly stable. For a symmetric body, Euler's equations also describe the precession of the transverse angular velocity around its symmetry axis. Torques from gravity or an applied couple enter through the right-hand sides rather than being absorbed into the inertial cross terms.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

Magnetization and field response. A magnetic material contains microscopic magnetic moments associated with electron spin and orbital motion. The magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI the macroscopic fields satisfyHere $\mathbf{H}$ separates free-current excitation from the material response. A uniform isotropic linear material has $\mathbf{M}={\chi }_m\mathbf{H}$, givingThis constant scalar magnetic permeability model is useful only within its linear range. Crystals can have a susceptibility tensor; strongly magnetic materials depend on field and magnetic history. For a finite specimen the internal $\mathbf{H}$ also includes its demagnetizing field. In a uniformly magnetized ellipsoid, $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$ along a principal axis, so sample shape affects the apparent susceptibility.

Diamagnetic and paramagnetic response. Diamagnetism is a weak negative susceptibility caused by induced orbital response opposing the applied field. It occurs in all matter but can be masked by larger positive contributions. Closed-shell materials such as bismuth or many organic compounds show it clearly. Ordinary diamagnetism is often only weakly temperature dependent; superconducting flux exclusion is a separate collective phenomenon and is not described by an arbitrarily small negative susceptibility.

Paramagnetism arises when available magnetic moments partially align with a field. Thermal disorder prevents complete alignment at weak fields. For independent localized moments of magnitude $m$, the classical weak-field result iswhere the second expression uses $B\simeq {\mu }_{0}H$ in a weakly magnetic sample. Thus localized moments obey a Curie law in this regime. Quantum angular momentum changes the coefficient to the appropriate $J(J+1)$ expression and gives saturation at strong field. Conduction electrons instead exhibit Pauli paramagnetism: only states near the Fermi surface can change spin populations, so a degenerate gas has a nearly temperature-independent response. Its orbital diamagnetism must also be included for a full metallic susceptibility.

Collective magnetic order. Exchange interactions can favor parallel alignment, producing ferromagnetism below the Curie temperature. Iron, cobalt and nickel are familiar examples. A material may contain regions of spontaneous magnetization, or magnetic domains, while having little net external moment because the domains point in different directions. Applied fields move domain walls and rotate moments; at large fields the magnetization approaches saturation. Pinning and energy barriers make this process hysteretic. Remanence is the magnetization left after removing the driving field, and coercivity measures the reverse field needed to bring the chosen response back to zero.

The area of a closed $B$–$H$ loop is the energy dissipated per unit volume per quasistatic cycle, $\oint HdB$, with positive dissipated magnitude. Consequently the magnetic permeability is neither one universal constant nor enough information to describe a ferromagnet: differential magnetic permeability $dB/dH$, saturation and hysteresis all matter. Above the ordering temperature the spontaneous moment disappears, and a Curie–Weiss law is often an appropriate approximate paramagnetic description away from the critical region.

In antiferromagnetism, ordered opposing sublattices have equal moments and cancel macroscopically in the ideal compensated case. The order disappears above the Néel temperature. In ferrimagnetism, opposing sublattices have unequal moments and leave a net magnetization. Ferrites combine ferrimagnetic response with relatively high electrical resistivity, making them valuable when eddy-current losses would be substantial in a metallic core.

Electromagnets and engineering choice. A coil supplies free-current magnetomotive force $NI$. For a simple uniform magnetic path, Ampère's law gives $H\ell \simeq NI$ and flux $\Phi \simeq \mu ANI/\ell$. A high-magnetic permeability soft magnetic core concentrates flux and reduces the current required for a given field. With an air gap $g$,Even a short gap can dominate the magnetic reluctance. The approximate attractive force across one uniform gap is $F\simeq B^{2}A/(2{\mu }_0)$, obtained from the field energy, with fringing neglected.

Soft magnetic materials have small coercivity and narrow hysteresis loops; they suit relays, electrical transformer cores and controllable electromagnets. Hard magnetic materials retain their magnetization and suit permanent magnets. Laminating metallic cores breaks large eddy-current loops; ferrites are useful at higher frequencies because of their resistivity. Increasing current cannot increase $B$ indefinitely through the linear formula: core saturation, coil heating and leakage limit performance. A good design selects material for the field amplitude, switching frequency and acceptable loss, rather than simply choosing the largest quoted magnetic permeability.