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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.

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 $-\hat{\mathbf{z}}$.

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.

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 $B={\mu }_{0}nI$.