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A free current is the part of the macroscopic electric current assigned to mobile or externally supplied electric charges, rather than to the bound motion represented by polarization and magnetization. In a conductor, drifting conduction electrons provide a familiar example.

In macroscopic Maxwell's equations, the free-current density appears as $\nabla \times \mathbf{H}={\mathbf{J}}_{\mathrm{f}}+\partial \mathbf{D}/\partial t$. Here $\mathbf{H}$ is the magnetic field strength and $\mathbf{D}$ the electric displacement field. The separation depends on the material model; the total microscopic charge motion remains the source of the electromagnetic field.

The magnetization caused by an applied magnetic field strength. The displayed scalar linear law assumes isotropic susceptibility and neglects magnetic self-interaction; remanent magnetization is an additional component that need not align with the present field.

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.

The magnitude of reverse magnetic field strength at which a previously magnetized specimen's magnetic flux density $B$ vanishes on a specified magnetic hysteresis branch. Intrinsic magnetic coercivity instead refers to zero magnetization. Because $B={\mu }_0(H+M)$, their zero conditions generally occur at different applied fields. The preparation, branch and specimen geometry must be specified when comparing measurements.

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.

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.

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.

For a coaxial cable with inner and outer radii $a,b$, a radial interface $c$, and permittivities $ϵ,{ϵ}^{\prime }$, define $T=\ln (c/a)/ϵ+\ln (b/c)/{ϵ}^{\prime }$. With terminal electric potential difference $V$, the electric displacement field is $D_{\rho }=V/(T\rho )$. The electric field is $E_{\rho }=V/(T{ϵ}_i\rho )$ in each layer $i$. If the outward and return electric currents have magnitudes $I$, the magnetic field strength in the gap is $H_{\phi }=I/(2\pi \rho )$ and the magnetic field is $B_{\phi }={\mu }_{i}I/(2\pi \rho )$.

The proof uses Gauss's law without free interface electric charge, the electric potential difference integral, and the magnetostatic Ampère law. The axial Poynting vector is $VI/(2\pi T{ϵ}_i{\rho }^2)$, so its cross-sectional surface integral is $VI$, independently of the radial material layering and magnetic permeabilities. The empty inner and exterior regions have zero electric field and magnetic field for the regular solution with bounded exterior electric potential and no applied magnetic field.