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For a planar magnetic field $\mathbf{B}=({\psi }_y,-{\psi }_x,0)$, ideal induction advects the Cartesian magnetic flux function: $D_t\psi =0$. Incompressible flow also preserves the area inside each material contour. During relaxation, the arms of a magnetic separatrix can approach each other while their field-line connectivity remains fixed. A transition of width $ϵ$ with a finite change of tangential field has electric current density of order ${ϵ}^{-1}$ and can approach a current sheet. This is a possible limiting mechanism, not a claim that every planar equilibrium is singular.

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.

Magnetic reconnection changes connectivity between magnetic field lines. It requires departure from smooth magnetic flux freezing, for example through finite resistivity or other nonideal terms concentrated in a current sheet. Ideal viscous magnetic relaxation dissipates mechanical energy while retaining frozen connectivity, so viscosity alone does not provide reconnection.

Magnetic relaxation converts magnetic energy into material motion through the Lorentz force density, and dissipates the resulting kinetic energy through viscosity. In a perfect conductor, magnetic flux freezing preserves field-line connectivity throughout every smooth stage. A regular limiting state has zero velocity and is a magnetostatic equilibrium, but its magnetic field can contain current sheets. Conservation of one global magnetic helicity is weaker than preservation of the complete frozen field-line structure.

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

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

The relevant competition is between energy reduction and magnetic flux freezing. Ideal evolution transports the entire connectivity of the magnetic field lines, including the separation and linkage of flux regions. Viscous dissipation can remove kinetic energy but cannot permit magnetic reconnection between such regions. As each region relaxes toward lower magnetic energy, neighboring flux regions can approach with different limiting tangential fields. A smooth transition can become increasingly thin; its field remains bounded while its curl, and therefore its electric current density, becomes large.

The limiting object is a current sheet. If $\mathbf{n}$ points from its minus to its plus side, its surface electric current density isFor a tangential discontinuity, ${\mathbf{B}}_{\pm }\cdot \mathbf{n}=0$. The normal balance in magnetostatic equilibrium requiresThus a finite jump of tangential field can coexist with equilibrium: the jump of magnetic pressure is balanced by the ordinary pressure jump.

This need not happen for every initial field; a field already in smooth magnetostatic equilibrium is an immediate exception. The expectation concerns topology-constrained relaxation for which a smooth limiting arrangement is obstructed. Current-sheet formation is consistent with the relaxation framework developed in Moffatt's primary analysis. Retaining complete frozen connectivity is essential; minimizing energy subject only to the value of total magnetic helicity would generally permit field rearrangements forbidden by this evolution.

In a planar cross-section write the magnetic field asThe Cartesian magnetic flux function $\psi$ labels the magnetic field lines. Ideal induction, with a suitable additive gauge, givesThus each flux contour is material. Incompressibility also preserves the area inside any material contour. Neither a magnetic island's flux distribution nor the connections of a separatrix can be freely altered during relaxation.

A configuration with two or more magnetic islands separated by an X-type saddle illustrates the mechanism. As the magnetic islands reshape to lower magnetic energy, the arms of the separatrix can be pressed together. Magnetic reconnection would change the topology, so it is excluded in a perfect conductor. The X-type structure can instead flatten into an extended interface, with oppositely directed tangential fields on its two sides. The transition shrinks and the electric current density concentrates into a current sheet. Such collapse in two-dimensional relaxation is documented in section 4 of Moffatt's primary discussion. It is a possible asymptotic outcome, not a finite-time singularity assertion or a property of every planar field.

The local equations show precisely how the interface can support equilibrium. For the planar field,On a regular connected flux region, magnetostatic equilibrium therefore implies $p=P(\psi )$ andDifferent disconnected flux regions can have different functions $P$. Their limiting fields need not join with continuous first derivatives of $\psi$ across a common separatrix; continuity of total pressure allows the tangential-field jump.

For an explicit local sheet, takeThese satisfy magnetostatic equilibrium exactly, withThe finite limiting surface current agrees with the field-jump formula. This is a local equilibrium demonstration of the sheet balance, not a claim that this particular family is the full ideal relaxation trajectory from the given initial data.

Strictly planar fields of the form above have zero magnetic helicity: choose $\mathbf{A}=\psi {\mathbf{e}}_z$, giving $\mathbf{A}\cdot \mathbf{B}=0$. Thus this strictly planar example interprets the last part as a separate illustration of topology-constrained relaxation. It still has an energy obstruction: if $\psi =0$ on the cross-sectional boundary, the advected integral $\int {\psi }^{2}dxdy$ is constant and the Poincaré inequality gives $M\ge {\lambda }_1\int {\psi }^{2}dxdy/(2{\mu }_0)$ per unit length. If the earlier nonzero-helicity condition is retained instead, a possible extension is a field depending on only two coordinates but permitting an axial component, with periodic boundary conditions in the invariant direction. WriteThe axial force balance makes the axial component a function of $\psi$ on each regular connected region; the remaining Cartesian magnetostatic flux-function equilibrium isIn that periodic extension, with consistent flux and gauge conventions, and $\psi =0$ on the cross-sectional boundary, integration by parts gives helicity per unit length $H_M/L_z=2\int \psi B_z,dxdy$. It can be nonzero. The same separatrix-collapse mechanism still concentrates current in a sheet. Three-dimensional linkage is therefore unnecessary for sheet formation; planar frozen flux-contour topology already supplies the constraint.

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.

For the field generated by the infinitely long, axially symmetric shell currents, axisymmetry and the right-hand rule give $\mathbf{B}=B_{\varphi }(r){\mathbf{e}}_{\varphi }$. A circular loop has $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi r{B}_{\varphi }(r)$. Its enclosed electric current is zero inside the inner shell, $I$ between the shells and $2I$ outside both. Ampère's law therefore givesThe positive ${\mathbf{e}}_{\varphi }$ direction is counterclockwise viewed from positive $z$. The ideal current sheets make the tangential magnetic field discontinuous at each shell; these formulas specify the three open regions, with no additional imposed background field.

Only the signed enclosed electric current changes. Between the shells it remains $I$, but outside both it is $I-I=0$. The same circular-loop application of Ampère's law givesOpposite currents confine the generated magnetic field to the annulus, as in an ideal infinitely long coaxial cable. The two shell current sheets account for the field jumps.

Integrating the $y$ component of Ampère's law,through the current sheet gives the magnetic-field jump across a surface currentReflection in the wing plane reverses the tangential perturbation $\delta B_x$. Ifthe upper and lower boundary values on the wing are consequentlyOutside the wing, the corresponding boundary value is zero.