Codex @codex  0

Ferromagnetic interactions enhance the response above the ordering transition. The Curie–Weiss law improves the independent-spin approximation:with positive Weiss temperature $\Theta$ and the same high-temperature coefficient $C$. In a simple mean-field theory, take $\Theta =T_c=70\mathrm{K}$. More generally $\Theta$ is fitted and need not equal $T_c$, and the simple pole does not give exact critical behaviour arbitrarily close to the transition.

At zero temperature, a selected ferromagnetic orientation has all magnetic moments aligned. Its saturation magnetization isEquivalently, there are $28{\mu }_B$ per conventional cell. The moment points along either allowed easy-axis direction at zero magnetic field. This is the local or single-domain spontaneous magnetization; a macroscopic multidomain sample can have smaller, even zero, net magnetization in the absence of a selecting field.

An applied magnetic field changes the alignment of the manganese magnetic moments. The component of their magnetic structure factor perpendicular to the scattering vector contributes to magnetic neutron diffraction. A field-induced uniform magnetization has the same translational periodicity as the nuclear crystal structure, so its scattering amplitude can occur at the existing reciprocal lattice vectors. The diffraction wave intensities can therefore change without moving the peaks. Their angular positions still measure the same interplanar spacings. This mechanism does not require a measurable change of the lattice constant.

In the hydrogenic donor state approximation, the screened Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{ϵ}_{r}r)$ and the kinetic term uses $m^{\ast }=0.067m_e$. Rescaling the hydrogen atom gives the effective Bohr radius of a hydrogenic donor and the positive binding energyThereforeThe radius here is the exponential length of the ground-state wavefunction, not its root-mean-square radius. With $\psi (r)=(\pi a_b^3{)}^{-1/2}{e}^{-r/a_b}$, radial integration gives $⟨r^2⟩=3a_b^2$ and $⟨x^2+y^2⟩=2a_b^2$.

For dilute independent donors of number density $n$, a bound spin one-half electron has magnetic moment magnitude ${\mu }_s=|g|{\mu }_B/2$, with the Bohr magneton defined using the bare electron mass. Its weak-field Curie law response isFor the orbital response, expand the effective mass Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m^{\ast })$ in a uniform magnetic field, using $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. The spherical ground state has zero orbital angular momentum; the quadratic term raises its energy byDifferentiating this energy twice and using $B\simeq {\mu }_{0}H$ givesThe donor-only magnetic susceptibility cancels whenThe negative sign of $g=-0.45$ drops out of the linear magnetic susceptibility. This crossover is far below $E_b/k_B\simeq 63.6\mathrm{K}$, consistent with donor binding. It assumes sufficiently weak magnetic field, negligible exchange between donors and a thermal scale above any inter-donor splitting; sufficiently dilute donors can satisfy those assumptions. The background diamagnetism from part (b) is excluded from this cancellation.

In weak magnetic field and the regime of paramagnetism, localized magnetic moments with total angular momentum $J$ and Landé g-factor $\tilde{g}$ have SI volume magnetic susceptibilitywithin the Curie–Weiss law. Here $n_m$ counts magnetic ions per volume, ${\mu }_0$ is the vacuum permeability, ${\mu }_B$ is the Bohr magneton, and $\Theta$ is the Weiss temperature. The factor $J(J+1)/3$ is the zero-field average of the squared angular projection over $m_J=-J,\dots ,J$.

For one magnetic ion per formula unit, a molar volume $V_m$ contains $N_A$ magnetic ions, where $N_A$ is the Avogadro constant. Consequently $n_m{V}_m=N_A$, and the molar magnetic susceptibility isDefining the dimensionless effective paramagnetic moment $m_{\mathrm{eff}}$ by ${\mu }_{\mathrm{eff}}=m_{\mathrm{eff}}{\mu }_B$, numerical substitution givesThe normalization is per mole of formula units, which here also means per mole of magnetic ions.

Take dimensionless spin operators and $H=D{S}_z^2+g{\mu }_{B}B{S}_z$, with $g>0$. The four wavefunctions are $|m⟩$, $m=-3/2,-1/2,1/2,3/2$, withAt zero field they form doublets at $D/4$ and $9D/4$. A transverse microwave magnetic field drives $\Delta m=\pm 1$ transitions. When $h\nu >2D$, their positive resonance fields in ascending order areThey connect $(1/2,3/2)$, $(-1/2,1/2)$, and $(-3/2,-1/2)$, respectively. ThusHere $D/h=4.25\mathrm{GHz}$ and the zero-field doublet gap is $2D$.

At lower frequency it is essential to use absolute energy gaps, not discard a transition after its levels cross. For $0<h{\nu }^{\prime }<2D$, the distinct candidate positive fields areThe first two coincide whenThis axial spin-three-halves resonance coincidence leaves two distinct resonance positions, at $0.1275$ and $0.3825\mathrm{T}$, with two transitions superposed at the lower position. At ${\nu }^{\prime }=2D/h$ there are also only two strictly positive-field lines, but an additional zero-field resonance remains; this is a different counting convention. Observable intensities also require thermal occupation of the relevant lower levels.

For competing exchanges on a body-centred cubic antiferromagnet, use the Heisenberg model convention $H={\sum }_{⟨ij⟩}{J}_{ij}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, with each bond counted once, dimensionless spins, and positive antiferromagnetic $J$. There are eight nearest and six next-nearest neighbours. Bond counting in the two classical collinear structures givesFor type II the eight $J_1$ contributions cancel, while all six $J_2$ bonds are antiparallel. Their crossing yields $c=2/3$. One can also exclude a lower spiral within the classical bilinear model: writing $u_i=\cos (q_{i}a/2)$ givesFor a favourable sign of the product, $u_x^2+u_y^2+u_z^2\ge 3|u_x{u}_y{u}_z{|}^{2/3}\ge 3|u_x{u}_y{u}_z|$. Hence when $J_2\ge 2J_1/3$ its minimum is at $u_x=u_y=u_z=0$, the type-II ordering vectors. When $J_2\le 2J_1/3$, compare with the type-I value and use $3-\sum u_i^2\le 3(1-|u_x{u}_y{u}_z|)$ to obtain the complementary bound. ThusThis is a classical/mean-field energy comparison; an exact quantum ground state is not generally a collinear product state.

For type I a magnetic cell is the simple cube of side $a$, with opposite moments at $0$ and $a(1,1,1)/2$. Its magnetic structure factor is proportional to $1-e^{i\pi (h+k+l)}$. Allowed magnetic reflections have integer $h,k,l$ with $h+k+l$ odd. The first two families are $(100)$ and $(111)$, with symmetry-equivalent members understood.

For type II choose magnetic primitive translations $a(1,1,0)$, $a(1,0,1)$ and $a(0,1,1)$, enclosing four spins in volume $2a^3$. Equivalently, a side-$2a$ cube is a convenient nonprimitive magnetic cell. On each original simple-cubic sublattice the moment at $a\mathbf{n}$ is proportional to $(-1{)}^{n_x+n_y+n_z}$. Accordingly,The translated body-centre basis gives $1+e^{i\pi (h+k+l)}=1\pm i$, so there is no further extinction. The first two families are $(1/2,1/2,1/2)$ and $(3/2,1/2,1/2)$. The neutron polarization factor may additionally suppress a reflection for a particular moment direction; the stated rules are the positional selection rules.

Each sublattice has number density $n_A=1/a^3$ and saturation magnetization $M_s=g{\mu }_{B}S/a^3$. Replacing neighbouring spins by their averages makes their exchange energy equivalent to a molecular magnetic field. Its minus sign expresses antiferromagnetic exchange, not an externally applied field. Since $⟨S_B⟩=M_B/(n_{A}g{\mu }_B)$,For $M_B=-M_A$, $B_{\mathrm{mf}}=({\lambda }_{AB}-{\lambda }_{AA})M_A$. Linearizing the Brillouin magnetization formula givesThe g-factor cancels because exchange is specified in energy units. A convention with $2J{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ doubles these exchange energies and $T_N$.

Let $m=M_A/M_s$, $A=(S+1)/(3S)$ and let $b_B>0$ denote the cubic coefficient of the Brillouin magnetization formula. The argument is $y=(T_N/T)m/A$. Thus $m=Ay-b_B{y}^3$ impliesConsequently $\boxed{\beta =1/2}$ in mean-field theory. The positive branch falls from saturation at zero temperature to zero at $T_N$ and remains zero above it. An elastic magnetic neutron-reflection intensity scales as $I\propto M_A^2$; a log–log fit against $T_N-T$ has slope $2\beta$. Background subtraction and a suitable critical interval are needed. Fluctuations can change the measured exponent from its mean-field value; uniform bulk magnetization vanishes for a compensated antiferromagnet and is not itself this order parameter.

In weak-coupling BCS theory, a useful scale isHere the phonon spectrum sets the attractive-interaction cutoff, $N(0)$ is the consistently defined Fermi-level density of states and $V_{\mathrm{eff}}$ the effective pairing attraction after repulsion is accounted for. Light hydrogen supports high vibrational frequencies; strong electron–phonon coupling and favourable metallic density of states can raise $T_c$. Pressure changes bonding, phase stability, electronic structure and phonon frequencies. It is not simply an independent multiplier in this expression. In a real strongly coupled hydride, Eliashberg theory uses the full spectral function and a screened Coulomb pseudopotential; the elementary exponential is a mechanism guide rather than a controlled quantitative fit to $203\mathrm{K}$.

Measure a four-terminal resistance that falls below the instrumental bound, and test that a magnetic field suppresses the transition. Independently detect diamagnetic screening and, where feasible, field-cooled flux expulsion or an AC susceptibility signal with controlled cell background. A resistance drop alone can reflect contact changes or a filamentary path; a consistent magnetic response is important evidence for superconductivity. A reproducible isotope shift upon replacing hydrogen by deuterium tests the involvement of lattice motion, although it is not by itself proof of superconductivity.

Extreme-pressure samples are microscopic and enclosed in a diamond-anvil cell. Reliable separate electrical contacts, homogeneous pressure, pressure calibration during cooling, chemical stability and the tiny magnetic signal relative to the anvils and gasket are practical difficulties. Structural identification matters because a sulfur-rich hydrogen compound formed under pressure need not retain the starting stoichiometry.

Historical correction: the original primary report associates the $203\mathrm{K}$ transition with pressures near $155\mathrm{GPa}$, while roughly $90\mathrm{GPa}$ is the metallization scale, not that same transition point. The printed pairing of $203\mathrm{K}$ with $90\mathrm{GPa}$ should not be treated as an established measured pair. The statement about the highest conventional transition temperature is also a historical statement in this 2018 examination.^[ref]

The displayed horizontal variable is ${\mu }_{0}H$ in tesla. The source formulas written with a flux quantum actually have the dimensions of magnetic flux density, so define $B_{c1}={\mu }_0{H}_{c1}$ and $B_{c2}={\mu }_0{H}_{c2}$ when applying them:From the onset of curvature away from the low-field line in panel A, $B_{c1}$ is roughly $0.06\mathrm{T}$ with uncertainty of order $0.01\mathrm{T}$. Panel B's branches approach the normal background over approximately $2$–$3\mathrm{T}$; use $B_{c2}\simeq 2.5\mathrm{T}$ as a coarse central estimate. Resolving irreversible loop closure from the true thermodynamic upper field, and correcting demagnetization, would require better data and geometry. These readings support scales, not precise material constants.

The upper field givesThe leading-log critical-field inversion starts by dividing the two field formulas to eliminate $\xi$:The large-$\kappa$ solution is $\kappa \simeq 6.15$, henceThe penetration-depth range is an indicative graph/model scale, not a propagated confidence interval. There is a mathematical ambiguity worth making explicit: $f(\kappa )=\ln \kappa /(2{\kappa }^2)$ rises from zero at one, peaks at $\sqrt{e}$, then falls. The same central ratio also gives $\kappa \simeq 1.055$ and $\lambda \simeq 12.1\mathrm{nm}$, so the printed restriction $\kappa >1$ alone does not select a unique solution. The leading-log lower-field formula is intended for appreciably large $\kappa$, which motivates the $\kappa \simeq 6$ branch; near one it needs the omitted core-energy correction. Even the larger-branch estimate has systematic uncertainty from that approximation.

$\lambda$ is the London penetration depth, the magnetic screening scale, while $\xi$ is the superconducting coherence length, the recovery scale of the superconducting order parameter after spatial suppression. In the mixed state the complex order parameter vanishes in vortex cores of radius of order $\xi$, recovers between separated cores and winds in phase by $2\pi$ around each singly quantized vortex. Currents and magnetic field spread over scale $\lambda$; vortex density grows with the mean internal field according to $n_v{\Phi }_0=\overline{B}$. Near $H_{c2}$ cores overlap and the order-parameter amplitude becomes small throughout the sample. A schematic core profile appears in solution.

The hysteresis is not the reversible equilibrium magnetization: defects pin vortices, so entry and exit leave different spatial flux distributions. Macroscopic currents oppose changes in flux, trapped vortices give remanence, and depinning and creep depend on history and temperature. Surface barriers can also contribute. The loop width probes an irreversible current scale, but its conversion to current density additionally needs sample geometry and an appropriate critical-state model.

A sufficiently clean two-dimensional electron gas at low temperature in a perpendicular magnetic field has quantized orbital Landau levels. In the integer quantum Hall regime, the longitudinal resistivity is nearly zero while the Hall resistance has plateauxThe Hall sign depends on carrier charge and lead convention. Fractional plateaux can occur in strongly interacting states, but the following single-particle counting and calibration use the integer effect.

One resolved spin branch of a Landau level has one state per flux quantum $h/e$: in area $A$, its guiding-centre degeneracy is $Ae{B}_z/h$ for positive $B_z$. Defining the nominal filling factor as the number of electrons per available orbital state givesThe same count follows in Landau gauge: a $2\pi /L_y$ increment in $k_y$ moves the orbit centre by $2\pi {\ell }_B^2/L_y$, giving $A/(2\pi {\ell }_B^2)=Ae{B}_z/h$ states. Include each spin branch separately; an unresolved spin-degenerate orbital level contains twice this count, and common plateau indices are then even.

Plateau qualification: $\nu$ as defined by density is continuous when $B_z$ changes at fixed $N_S$. The integer plateau index $p$ can remain unchanged over a field interval because localized states accommodate the changing occupation. Therefore $N_S=pe{B}_z/h$ is not an exact identity across an entire plateau. It is the ideal filled-level relation, often used approximately at a suitable plateau centre; confusing the nominal filling with the plateau index would contradict fixed density.

The plateau resistance realizes the von Klitzing constant $R_K=h/e^2\simeq 25812.8\Omega$, divided by a known integer. A four-terminal Hall measurement with negligible longitudinal voltage and contacts in the quantized regime supplies a reproducible resistance reference, which can be compared to other resistors using precision bridges. Its robustness comes from the quantized transport state, not from measuring the Hall bar's width accurately.

The Hall-plateau voltmeter calibration is obtained as follows. Let $D_p$ be a plateau height in displayed units after any voltage offset is removed, with identified integer plateau index $p$. The true Hall voltage is $I_x{R}_K/p$, so the calibration factor would beThe nominal current is $15\mathrm{nA}$. If several plateau heights are available, their inverse-integer ratios help identify indices and a fit of $D_p$ against $1/p$ can separate a constant offset from the gain. Reversing magnetic field and antisymmetrizing the voltage also removes an even offset. A unique absolute index should not be asserted solely from an unspecified sequence of steps.

For an established exact integer $p$ and independent current and height errors,The current alone contributes $2\%$ relative uncertainty; it is a common calibration uncertainty and does not average away by fitting more plateaux. Voltage-offset uncertainty and deviations from quantization must also be included where relevant. The $25\pm 2\mu \mathrm{m}$ width cancels because the Hall voltage divided by current is a sheet Hall resistance; it would enter the transverse electric field, not this calibration.

TODO — missing source data: the held PDF's last-page stream refers to an absent /Im6 graphic, and its recovery log reports an unknown XObject. The supplied TeX records the same source gap. No plateau heights, labels or axis values survive, so neither a numerical $D_p$ nor a unique $p$ can be read. The numerical voltmeter calibration and its complete graph-reading error cannot be supplied without that graph; the formula and known $2\%$ current contribution above are determined.

As magnetic field grows from zero, a clean two-dimensional electron gas first has an ordinary finite longitudinal resistivity, then develops Shubnikov–de Haas oscillations as broadened Landau levels pass the chemical potential. Thermal and disorder broadening obscure the low-field oscillations; their visibility grows when the level spacing exceeds those broadening scales. In a well-developed integer quantum Hall regime, ${\rho }_{xx}$ has minima close to zero on Hall plateaux and maxima during transitions between them, when extended states cross the chemical potential. Detailed peak heights and envelopes depend on disorder and scattering and are not fixed by quantization alone.

At fixed density the nominal filling is $\nu (B)=N_{S}h/(eB)$, so level-filling features are approximately periodic in $1/B$, with period $e/(N_{S}h)$ for resolved spin levels or $2e/(N_{S}h)$ for an unresolved spin pair. Increasing $B$ lowers the filling; spin splitting can create additional odd-index features. The original schematic below shows these relations without pretending to reconstruct the missing data.

TODO — missing source graph: the actual $0$–$12\mathrm{T}$ positions and integer plateau labels require $N_S$ or surviving plateau fields from part (b). Neither is provided independently. Thus a uniquely labelled field-axis sketch for this sample cannot be completed. The following DOS diagram distinguishes extended centres and localized tails generically; its chemical-potential markers are illustrative and do not stand for the requested 8 T or 10 T assignments.

Let $\ell =2\pi r-d$ be the mean path length in the core. Neglect leakage and fringing across the narrow air gap, and assume a cross-section small compared with $r$. Continuity of normal magnetic flux density then gives the same tangential magnetic field magnitude $B$ in the core and gap. Ampère's circuital law for the magnetic field strength $H$ isFor a linear core, $H_{\mathrm{core}}=B/({\mu }_0{\mu }_r)$ and $H_{\mathrm{gap}}=B/{\mu }_0$, givingHere $\hat{\mathbf{t}}$ follows the winding's right-hand rule. The gap has much larger $H$ than the core when ${\mu }_r\gg 1$, although $B$ is approximately continuous. Equivalently, the core and gap magnetic reluctances add in series.

On a magnetic hysteresis loop, remanence is the nonzero $B_r$ remaining at $H=0$ after magnetizing the sample. The normal magnetic coercivity $H_{cB}$ is the magnitude of reverse $H$ at which $B=0$ on that branch. The intrinsic magnetic coercivity instead refers to zero magnetization; these two conventions should not be confused. A comparison of coercive $H$ with remanent $B$ requires a unit conversion or a specified constitutive scale, because their SI units differ. A soft ferromagnet is understood here to have a narrow loop and small coercive field.

Use the stated single-valued constitutive equation for the calculation. Inverting it gives $H_{\mathrm{core}}=H_m\mathrm{artanh}(B/B_m)$, soSince $\mathrm{artanh}z=z+O(z^3)$, the small-signal response of a gapped magnetic core isThe small-field core magnetic permeability is $B_m/H_m$. Strictly, $B=B_m\tanh (H/H_m)$ is a reversible approximation: at $H=0$ it has $B=0$, so it cannot itself model nonzero remanence. The displayed loop and this constitutive approximation serve different purposes. Keeping a genuinely remanent branch would generally add a history-dependent offset; it would not imply $B=\kappa I$ through the origin.

Take $d$ as the centre-to-centre separation and $ϵ={ϵ}_0{ϵ}_r$. Use the ideal lossless two-conductor transmission line model, with nonmagnetic surroundings and negligible internal-wire inductance. Opposite line charges $+{\lambda }_q$ and $-{\lambda }_q$ approximate the cylindrical conductors when $d\gg a$. Integrating their electric fields, or adding their logarithmic electric potentials, gives the potential differenceThis is the two-wire capacitance at large separation. For opposite electric currents $+I$ and $-I$, the magnetic fields between the wires add. The linked magnetic flux per unit length is approximatelyThus the external inductance of a thin two-wire line is $L^{\prime }\simeq {\mu }_0\ln (d/a)/\pi$. The lossless telegrapher's equations give a forward wave with $V/I=\sqrt{L^{\prime }/C^{\prime }}$, so its characteristic impedance isThe product $L^{\prime }{C}^{\prime }={\mu }_0ϵ$ also gives the propagation speed $c/\sqrt{{ϵ}_r}$, an independent consistency check. The exact ideal external-field geometry replaces $\ln (d/a)$ by $\mathrm{arcosh}(d/(2a))$. If the wire separation were specified between surfaces, its centre separation would be $d+2a$, with the same leading large-separation result. Finite electrical conductivity and internal-wire magnetic energy require a more detailed line model.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

Use the constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/\mu$, where $\mathbf{D}$ is the electric displacement field, $\mathbf{H}$ the magnetic field strength, $ϵ$ the permittivity, and $\mu$ the magnetic permeability. Let ${\rho }_f$ be the charge density of free charge and ${\mathbf{J}}_f$ its current density. In this homogeneous medium, Gauss's law and the magnetostatic Ampère law are, respectively,Here $W$ is a volume with closed boundary $\partial W$, and $S$ is an oriented spanning surface whose boundary orientation follows the right-hand rule. Thus the electric field form is $\nabla \cdot \mathbf{E}={\rho }_f/ϵ$ and ${\oint }_{\partial W}\mathbf{E}\cdot \mathbf{n}dS=Q_f/ϵ$; the magnetic field form is $\nabla \times \mathbf{B}=\mu {\mathbf{J}}_f$ and ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{\ell }=\mu I_f$.

The remaining static Maxwell equations areThese express Gauss's law for magnetism and the static limit of Faraday's law. The volume and circulation forms follow from the divergence theorem and Stokes theorem. The electric charge and electric current on the right of the macroscopic laws are free sources; material electric polarization and magnetization are already included in $\mathbf{D}$ and $\mathbf{H}$.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

Work in cylindrical coordinates and use the assumed independence of $\phi$ and $z$. In each material region, the zero curl of the static electric field givesThe tangential electric field is zero at a perfect conductor. The dielectric-interface boundary conditions make its tangential components continuous at each material interface, so $E_z=E_{\phi }=0$ throughout the gap. They also vanish in the empty interior and exterior for the regular solution with no applied external electric field. HenceThis specifies an axis of direction: the sign of $E_{\rho }$ depends on the sign of $V$.

For the static magnetic field, Gauss's law for magnetism impliesThus $\rho B_{\rho }$ is constant in each layer. Continuity of the normal magnetic field across every interface makes that constant global, and regularity at the axis makes it zero. Next use $\nabla \times \mathbf{H}=0$ away from the conducting surfaces. Its $\phi$ component gives $d{H}_z/d\rho =0$. There is no azimuthal surface current density, so $H_z$ is continuous across the conducting surfaces and the material interface. The decay of the exterior magnetic field at infinity therefore forces this global constant to zero. The only remaining component isThe magnetostatic Ampère law fixes its sign: positive inner electric current in the $+z$ direction produces positive $B_{\phi }$ for positive magnetic permeability.

Orient the unit normal $\mathbf{n}$ from $R_1$ to $R_2$ and write $\mathbf{H}$ for the magnetic field strength. For static electric fields, $\mathbf{E}=-\nabla \varphi$; the static Ampère's law gives $\nabla \times \mathbf{H}={\mathbf{J}}_f$. The curl product rule therefore gives the local Poynting flux and terminal power identityIntegrate over an oriented piece ${\Sigma }_R$ of the separating surface and apply Stokes theorem:Once all the wires are included, the first surface integral is ${\sum }_k{\varphi }_k{I}_k$: the electric potential is constant on each perfect conductor and the integral of its normal current density is its signed electric current. Equivalently, remove the wire cross-sections and apply Stokes theorem in the surrounding insulator. Each resulting hole has negative boundary orientation, and the magnetostatic Ampère law around it gives the positive contribution ${\varphi }_k{I}_k$.

For a closed separating surface the boundary term is absent. For a noncompact separating surface, take $R\to \infty$ with the usual isolated-circuit condition ${\oint }_{\partial {\Sigma }_R}\varphi \mathbf{H}\cdot d\mathbf{\ell }\to 0$. In either case,For a steady closed circuit the signed electric currents sum to zero, so shifting the reference of electric potential does not change the terminal power.

For a fully general noncompact surface, the hypotheses as printed need an additional condition at infinity. The exact result before imposing it isTime independence alone does not remove this boundary term or ensure a finite Poynting flux; for example, spatially constant crossed electric fields and magnetic fields carry a constant Poynting vector through an infinite plane. The stated terminal-power formula holds with the boundary condition specified above.

With charge density $\rho$, current density $\mathbf{J}$, and speed of light $c=({ϵ}_0{\mu }_0{)}^{-1/2}$, the four Maxwell equations in vacuum areThe electric potential and magnetic vector potential determine the electric field and magnetic field byThese definitions satisfy Gauss's law for magnetism and Faraday's law identically. The Lorenz gauge condition isSubstituting into Gauss's law and using the Lorenz gauge condition givesFor the Ampère-Maxwell equation, the double-curl identity givesThe two gradient terms cancel by the Lorenz gauge condition. Defining the d'Alembert operator with the sign convention $\square ={\nabla }^2-c^{-2}{\partial }_t^2$, the remaining equations areConversely, if the electric potential and magnetic vector potential satisfy these wave equations and the Lorenz gauge condition, the same substitutions in reverse give Gauss's law and the Ampère-Maxwell equation. The other two Maxwell equations follow identically from the definitions, proving equivalence within the Lorenz gauge.

Set $K=1/(4\pi {ϵ}_0)$, $t_0=t-r/c$ and $\mathbf{n}=\mathbf{r}/r$, and evaluate $\mathbf{p}$, ${\mathbf{p}}^{\prime }$, ${\mathbf{p}}^{\prime \prime }$ at $t_0$. The chain rule givesIn particular, for $\mathbf{u}(t_0)$ and any positive integer $m$,Apply this gradient identity to the electric potential $\varphi =K[(\mathbf{n}\cdot \mathbf{p})/r^2+(\mathbf{n}\cdot {\mathbf{p}}^{\prime })/(cr)]$:The magnetic vector potential is $\mathbf{A}=K{\mathbf{p}}^{\prime }/(c^{2}r)$, with time derivative ${\partial }_t\mathbf{A}=K{\mathbf{p}}^{\prime \prime }/(c^{2}r)$. Thus the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ isFor the magnetic field, use the curl product rule and the chain rule:Consequently the fields of a time-dependent electric dipole includeThese formulas hold for $r>0$. The $r^{-3}$ electric field is the retarded electric dipole field, the $r^{-2}$ terms are induction contributions, and the $r^{-1}$ terms describe electric dipole radiation. Retaining the dependence of $t_0$ on position in every spatial derivative is what produces both the induction and radiation contributions.