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The tensor relating local current density to electric field. Its inverse, when it exists, is the electrical resistivity tensor; sheet conductivity uses surface current density instead.

Electromagnetic boundary conditions constrain the electric field and magnetic field at the boundary of a region. For a material interface, the interface conditions for electromagnetic fields relate the fields on its two sides. With unit normal $\hat{\mathbf{n}}$ directed from medium 1 to medium 2, free surface charge density ${\sigma }_f$ and free surface current density ${\mathbf{K}}_f$, the usual macroscopic Maxwell equations implyHere $\mathbf{D}$ is the electric displacement field, $\mathbf{B}$ the magnetic flux density, and $\mathbf{H}$ the magnetic field intensity. The relations follow by shrinking a pillbox or rectangular loop across the interface, assuming finite bulk fields and no magnetic surface sources. Material relations such as $\mathbf{D}=\epsilon \mathbf{E}$ then convert them into conditions on the fields used in a particular problem. Even with no free surface sources, all components of both fields need not be continuous when the material properties change.^[ref]

The Lorentz force per unit area on an ideal current sheet is computed using the mean of its limiting magnetic fields. The sheet's own local magnetic field is equal and opposite on its two sides and cancels in this mean. Equivalently, integrate the Lorentz force density through a thin regularized layer: its self-field contribution averages to zero, leaving the external field acting on the surface current density.

In magnetostatics, an ideal infinitely long solenoid is a cylindrical sheet of circumferential surface current density $K=nI$. It has negligible winding advance, no magnetic core, and no imposed external magnetic field. Choose positive electric current to produce magnetic field along $+z$. In this ideal model the interior magnetic field is uniform and axial, $B={\mu }_{0}nI$, and the exterior magnetic field is zero; ${\mu }_0$ is the vacuum permeability. For a finite, closely wound air-core coil, the same result is an approximation well inside the coil, several radii from either end. Air is treated as having the magnetic permeability of vacuum. ^[ref].

For a short proof, the translational and rotational symmetries of the ideal cylindrical current sheet, together with the Biot-Savart law, give an axial magnetic field. Ampère's law on rectangles wholly outside makes that axial exterior value independent of radius; the condition of no imposed magnetic field makes it zero at infinity and hence everywhere outside. Rectangles wholly inside similarly show that the interior value is independent of radius. Now choose an Amperian loop with axial length $s$, one long side inside and the other outside. The transverse sides are perpendicular to the magnetic field, so their contributions to the line integral vanish. The loop crosses $ns$ turns of the winding, givingHere $s$ is the integration-path length, not the whole coil length $\ell$. An Amperian loop surrounding both sides of the winding encloses equal and opposite electric currents; zero net circulation alone would not prove zero magnetic field at every exterior point.

For the $1000$-turn, $0.50\mathrm{m}$ coil with electric current $I=0.50\mathrm{A}$, the long-solenoid approximation givesThis illustrative calculation uses vacuum permeability ${\mu }_0\approx 1.256637\times 10^{-6}H/m$, a rounded value, not an exact constant. Since the 2019 redefinition of SI, vacuum permeability is determined experimentally.^[ref] Doubling $I$ doubles the ideal magnetic field; doubling $N$ at fixed $\ell$ does the same. Increasing $\ell$ while keeping $N$ fixed decreases $n$ and hence the ideal magnetic field.

A surface current density $\mathbf{K}$ concentrated on an interface creates a discontinuity of tangential magnetic field. Let the unit normal $\mathbf{n}$ point from side $-$ to side $+$. In SI units, the microscopic Ampere-Maxwell law givesHere the surface current includes both free current and bound current. An infinitesimal rectangular integration loop straddling the sheet derives the tangential jump. The no-monopole equation separately gives $\mathbf{n}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})=0$.

In a macroscopic material description, the magnetic field intensity instead satisfiesWriting a jump in $\mathbf{B}$ using only free current without accounting for magnetization would mix these two descriptions. In Gaussian units, the total-current magnetic-field jump is $4\pi {\mathbf{K}}_{\mathrm{total}}/c$.

For an isolated uniform sheet in vacuum, reflection symmetry givesFor example, current along $+x$ on $z=0$ gives a field along $-y$ above the sheet and $+y$ below it. The directions agree with the right-hand rule.

In the small-sphere oscillatory-field limit, surface current density excludes the changing magnetic field. Adding a central magnetic dipole to a uniform applied field and imposing zero normal field at the sphere gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_{\mathrm{inc}}$. The minus sign expresses diamagnetism; it does not make this dipole's backward radiation cancel the electric contribution.^[ref]

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.

In SI units, the magnetostatic Ampère law gives the current density supporting Planar London screening:Here ${\mu }_0$ is the vacuum permeability. The sign follows from $(\nabla \times \mathbf{B}{)}_y={\partial }_z{B}_x$ in the chosen coordinates. Thus $B_0>0$ requires electric current along $-\hat{\mathbf{y}}$. Its own magnetic field opposes the imposed magnetic field inside the material. The electric current is distributed over a finite layer; its depth integral is the effective surface current densityThe units are $\mathrm{A}{\mathrm{m}}^{-2}$ for $\mathbf{J}$ and $\mathrm{A}{\mathrm{m}}^{-1}$ for $\mathbf{K}$. Shrinking ${\lambda }_L$ at fixed $B_0$ increases the surface current density, while this integrated surface current density stays fixed. The current sheet approximation is the large-length-scale limit of this layer, not a claim that an actual sample carries an infinitely thin electric current. It reproduces the magnetic-field jump across a surface current between the exterior $B_0\hat{\mathbf{x}}$ and a field-free interior, with outward unit normal $-\hat{\mathbf{z}}$.

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.

In vacuum magnetostatics, the two relevant Maxwell equations areOrient the unit normal $\mathbf{n}$ from the minus side to the plus side. Integrating Gauss's law for magnetism over a pillbox straddling the current sheet and shrinking its thickness gives $\mathbf{n}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})=0$.

To obtain the tangential condition, choose an arbitrary tangent unit vector $\mathbf{t}$ and an infinitesimal rectangle with its long sides of length $\ell$ parallel to $\mathbf{t}$ on the two sides of the sheet. Orient its surface normal as $\mathbf{n}\times \mathbf{t}$, so that its plus-side long edge is traversed along $\mathbf{t}$. The line integral in Ampère's law tends to $\ell \mathbf{t}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})$. The enclosed electric current tends to $\ell \mathbf{s}\cdot (\mathbf{n}\times \mathbf{t})$. HenceBecause this holds for every tangent direction and the normal jump is zero,Here $\mathbf{s}\cdot \mathbf{n}=0$, as required for a surface current density.

The force on a current sheet from the mean magnetic field isThe mean removes the sheet's equal and opposite local self-fields, leaving the field that acts on its current through the Lorentz force. Using a one-sided limiting magnetic field would incorrectly include the sheet's own force.

The magnetization $M$ is magnetic dipole moment per unit volume. It produces bound volume and surface current densityStarting from the vector potential of magnetic dipoles and integrating by parts giveswhich has precisely the form generated by those bound currents.

Write the cylinder radius as $R$ and use cylindrical radius $\rho$. Ampère's law for the free axial current gives, inside the material,At $\rho =R$ this is tangential to the cylinder. Since $n=\hat{r}$,andFor $\mu >{\mu }_0$, the bound surface current therefore runs opposite to the free wire current.

Translation symmetry in the plane and reflection symmetry across it force the magnetic field to be uniform on each side, parallel to ${\mathbf{e}}_y$, and opposite on the two sides. Apply Ampère's law to a rectangular loop perpendicular to the sheet whose two long sides have length $L$. The enclosed current is $KL$, soThe right-hand rule fixes the directions:Consequentlywhich is the magnetic jump condition for a surface current density.

In the wing frame the unperturbed plasma moves with velocity $-u_0{\mathbf{e}}_x$. The ideal magnetohydrodynamics condition $\mathbf{E}+\mathbf{u}\times \mathbf{B}/c=0$ therefore produces the motional electric fieldThe wing's surface conductivity then gives the surface current densityIts direction also follows directly from the Lorentz force on the wing's mobile charges.

Let the unit normal $\mathbf{n}$ point from material 1 to material 2, and let their relative magnetic permeabilities be ${\mu }_1,{\mu }_2$, so ${\mathbf{B}}_i={\mu }_0{\mu }_i{\mathbf{H}}_i$ for linear isotropic media. Apply Gauss's law for magnetism to an infinitesimal Gaussian pillbox straddling the interface. The side contribution vanishes as its height tends to zero, leavingThe normal magnetic flux density is continuous.

Next use a narrow rectangular Amperian loop crossing the interface. The two short-side contributions vanish in the zero-height limit. The magnetostatic Ampère law for magnetic field intensity giveswhere ${\mathbf{K}}_f$ is any free surface current density. Thus, if there is no free surface current density on the interface, the tangential magnetic field intensity is continuous. In that case the component relations areThe distinction between free and bound electric currents matters: bound currents due to magnetization are already incorporated in the definition of magnetic field intensity. A change of magnetic permeability alone does not make the tangential magnetic flux density continuous.

Choose $+x$ along the electric current, $+z$ upward normal to the sheet, and $+y$ to complete a right-handed frame. In the interior region far from its edges, treat the sheet as infinite in its plane. Its surface current density isReflection and translational symmetry make the magnetic field parallel to the sheet, perpendicular to its current, and equal in magnitude with opposite direction above and below. Choose an Ampère's circuital law rectangle with length $s$ along $y$, straddling the full thickness. The two long sides each contribute $Bs$ with the same sign; on the vertical sides $\mathbf{B}\cdot d\mathbf{l}=0$ since $\mathbf{B}$ is horizontal. The enclosed electric current is $Ks=Is/w$. ThusThe right-hand rule fixes the directions:The current density inside the finite-thickness conductor is $\mathbf{J}=I/(wt)\hat{\mathbf{x}}$, so integrating through thickness $t$ gives the same $\mathbf{K}$. These are exterior fields in the broad-sheet approximation, not exact finite-width edge fields.

Choose conventional surface current density along $+x$, magnetic field along $+z$, and negative electron charge. An isotropic in-plane response has the following sheet conductivity and sheet resistivity tensors for $s\ge 0$, $k>0$:These are matrix inverses, rather than componentwise reciprocals. The sheet conductivity has units siemens, the sheet resistivity ohms. With $K_y=0$, the sign is $E_y/K_x<0$, agreeing with the negative Hall coefficient of a single electron species. Contact polarity can reverse the measured voltage sign, so quote the Hall magnitude. For a homogeneous strip of width $W$ and longitudinal probe separation $L$, $I=W{K}_x$ givesAt $s=0$, $k=N$, these reduce to $R_K/N$ and zero. A nonzero transverse sheet resistivity creates no local Joule heating: the antisymmetric contribution cancels in $\mathbf{E}\cdot \mathbf{K}=(e^2/h)s|\mathbf{E}{|}^2$.

For an original nonideal example, choose $k=2$, $s=0.01$ and $L/W=2$. Then $V_{xx}/I\simeq 129.061\Omega$, while the Hall magnitude is below $R_K/2$ by the relative amount $s^2/(k^2+s^2)\simeq 25.0$ parts per million. An unwanted admixture $0.02V_{xx}$ in the Hall readout would instead introduce approximately $2.58\Omega$. The chosen calculation shows why a small intrinsic tensor correction does not by itself guarantee an accurate contact measurement.

The sheet resistivity tensor relates tangential electric field to surface current density: ${\mathbf{E}}_{\parallel }={\mathbf{\rho }}_s\mathbf{K}$. It is the matrix inverse of the sheet conductivity tensor, when that inverse exists. Its SI unit is the ohm, rather than the ohm times metre used for bulk electrical resistivity. With a transverse response, individual tensor components are not generally reciprocals of the corresponding sheet conductivity components.

The solenoid turn density $n$ counts turns per unit axial length in a uniformly wound solenoid. The total turn count $N$ is dimensionless, whereas $n$ has units ${\mathrm{m}}^{-1}$. A $1000$-turn coil of length $0.50\mathrm{m}$ has $n=2000{\mathrm{m}}^{-1}$; an axial interval of length $s=0.010\mathrm{m}$ then contains $ns=20$ turns. The effective circumferential surface current density in the closely spaced circular-turn model is $K=nI$, with units $A/m$. ^[ref].

For a single-layer helix, the axial advance per turn is approximately $1/n$. The circular-turn model neglects this advance and the associated axial component of electric current, as well as the detailed magnetic field close to individual wires. The number of turns around the whole coil is always $N=n\ell$ here; it must not be substituted for $n$ in $B={\mu }_{0}nI$.

For a thin layer with essentially uniform tangential electric field, integrating electrical conductivity through its thickness gives the sheet conductivity tensor ${\mathbf{\sigma }}_s$. The two-dimensional form of Ohm's law is $\mathbf{K}={\mathbf{\sigma }}_s{\mathbf{E}}_{\parallel }$, with surface current density $\mathbf{K}$. Its SI unit is $\mathrm{S}=A/V$: amperes per metre divided by volts per metre. When response is the same in every in-plane direction, the sheet conductivity reduces to a scalar $\Sigma$; for uniform bulk electrical conductivity $\sigma$ and thickness $t$, $\Sigma =\sigma t$.

A surface current density is electric current per unit transverse length confined to a surface. Integrating a thin layer's current density through its thickness gives $\mathbf{K}$.