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For an axisymmetric vector field, write $\mathbf{B}=B_R{\mathbf{e}}_R+B_{\varphi }{\mathbf{e}}_{\varphi }+B_z{\mathbf{e}}_z$. Pure differential rotation givesThis vector has no azimuthal component and is independent of $\varphi$, so its curl has zero radial and vertical components. Its azimuthal component isThe final braces are $R\nabla \cdot \mathbf{B}=0$ by Gauss's law for magnetism. The ideal magnetohydrodynamic induction equation therefore yields axisymmetric magnetic winding:Differential rotation creates a toroidal magnetic field from the fixed poloidal magnetic field. A stationary field under the same purely rotational assumptions requires Ferraro's law of isorotation, ${\mathbf{B}}_p\cdot \nabla \Omega =0$.

Let $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ be the material derivative, and write $h$ for specific enthalpy. The ideal magnetohydrodynamic induction equation and Gauss's law for magnetism giveDotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density. Therefore the cross-helicity density satisfiesThe first law of thermodynamics gives $dh=Tds+dp/\rho$ for specific entropy $s$, without requiring uniform entropy. ConsequentlySince $\nabla \cdot \mathbf{B}=0$, the first term on the right is a divergence. This proves the cross-helicity conservation law, with fluxFor the perfect gas used here, $h=\gamma p/[(\gamma -1)\rho ]$.

With no dependence on the azimuthal or axial coordinates, Gauss's law for magnetism in cylindrical coordinates givesRegularity at the axis forces $C=0$, so $B_R=0$. The radial component of the magnetostatic Ampère-Maxwell equation isConsequentlyThis is also the starting point of cylindrical magnetostatic pressure balance.

In the rotating shearing sheet, write the background orbital shear as ${\mathbf{U}}_0=-q\Omega x{\mathbf{e}}_y$. The rotating momentum equation contains Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{u}$ and the shearing-sheet tidal potential; the background shear balances the radial tidal acceleration.

SetGauss's law for magnetism gives ${\partial }_z{B}_z=0$, and the vertical ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_z=0$. Horizontal invariance and $v_z=0$ remove the nonlinear horizontal advection. The remaining background-shear term is $(\mathbf{v}\cdot \nabla ){\mathbf{U}}_0=-q\Omega v_x{\mathbf{e}}_y$. The horizontal magnetic tension is $B_z{\partial }_z{\mathbf{B}}_h/{\mu }_0$, while horizontal pressure gradients vanish. Subtracting background balance yieldsFor ideal magnetohydrodynamics, the ideal magnetohydrodynamic induction equation is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}$. Its horizontal components giveThese horizontally invariant magnetized shearing-sheet equations are exact within the stated local, incompressible ansatz, even for finite horizontal amplitudes. The vertical equation determines the pressure needed to balance vertical gravity and magnetic pressure; it does not add another horizontal evolution equation.

In classical Maxwell's equations, electric charge is a source or sink of the electric field, so in vacuum $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ can be nonzero. There is no magnetic-charge source in the same theory: Gauss's law for magnetism is $\nabla \cdot \mathbf{B}=0$. Magnetic flux entering a closed surface equals the flux leaving it. A magnetic dipole has both polarities; it is not an isolated magnetic monopole.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.

For a steady thin-wire electric current, the Biot-Savart law iswhere the source element follows positive current. For a circular loop, source points are $(R\cos \varphi ,R\sin \varphi ,0)$ and the observer is $(0,0,z)$. Their cross product with $d{\mathbf{\ell }}^{\prime }$ has axial component $R^{2}d\varphi$; transverse components integrate to zero. Thus the on-axis magnetic field of a circular current loop, multiplied by the number of coincident turns, isTake positive current counterclockwise as seen from $+z$. magnetic field lines pass upward through the loop and return around its exterior, closing rather than terminating. They fan outward above its centre and converge below it. The original numerical Biot-Savart law integration below illustrates these directions; the wire singularities are masked.

The axisymmetric magnetic field near a regular axis follows from Gauss's law for magnetism. In cylindrical coordinates, regularity gives $B_{\rho }(0,z)=0$, while symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. Integrate ${\partial }_{\rho }(\rho B_{\rho })=-\rho B_0^{\prime }(z)$ from the axis:A formal $C(z)/\rho$ term is excluded by regularity. For $z>0$ the axial field decreases upward, so the radial component is outward, consistent with the sketch.

On the small receiving coil, $d\mathbf{\ell }=ad\varphi \hat{\mathbf{\varphi }}$. The Lorentz force $I_{2}d\mathbf{\ell }\times \mathbf{B}$ has axial component $-I_{2}a{B}_{\rho }(a,d)d\varphi$. Contributions from the axial field cancel radially around the loop. ThereforeWith both currents giving magnetic dipole moments along $+z$ and $d>0$, this is attraction. Reversing one current reverses the force. The small coil has $m_2=N_2{I}_2\pi a^2$, so the result is exactly the leading magnetic dipole force $F_z=m_2{B}_0^{\prime }(d)$. Equivalently $U=-{\mathbf{m}}_2\cdot {\mathbf{B}}_0$. The approximation requires $a\ll R$ and negligible variation over the receiving coil; corrections begin at relative order $a^2/(R^2+d^2)$.

Applying Gauss's law for magnetism to a pillbox and Ampère's law to an interface-crossing rectangle givesHere $\hat{n}$ points from material 1 to 2 and ${\mathbf{K}}_f$ is free surface current. With no such current, $B_n$ and $H_t$ are continuous. For linear isotropic media, $B_{t2}/{\mu }_2=B_{t1}/{\mu }_1$ and ${\mu }_2{H}_{n2}={\mu }_1{H}_{n1}$. The magnetic fields need not have equal magnitudes on either side.

With negligible leakage, Gauss's law for magnetism makes each cross-section of a magnetic flux tube carry the same flux $\mathcal{F}$. High permeability guides it through the core; a short gap permits neglect of fringing. Flux conservation across these particular geometric sections is an approximation, not a consequence of zero divergence without controlling side leakage.

Let $L$ be the constant-radius arc marked in the figure, and ${\mu }_r=\mu /{\mu }_0$. For each straight taper, the radius changes linearly from $R$ to $r$, giving the reluctance of a conical magnetic poleSumming magnetic reluctances to find the gap field of a tapered electromagnet, and using $B_g=\mathcal{F}/(\pi r^2)$ givesIf $L$ is instead defined as the total magnetic-material length, replace the displayed uniform-arc length by $L-2s$. The comparison untapered core has uniform radius $R$ and material length $L+2s$, soThe original PDF specifies $s/L\ll r/R$; the TeX transcription reverses this inequality. In the printed limit the taper reluctances are small relative to that of the broad arc. The gain is approximately $[h+L/{\mu }_r]/[h+L{r}^2/({\mu }_r{R}^2)]$, approaching $(R/r{)}^2$ if the gap reluctance is also negligible. If the gap dominates, both fields instead approach $NI/h$ and the gain approaches one. magnetic saturation near the small poles, gap fringing, leakage and winding heating limit practical enhancement.

Inductance and interface conditions. For a linear coil, self-inductance is the ratio $L=\Lambda /I$ of its own magnetic flux linkage $\Lambda =N\Phi$ to current. Its induced electromotive force is $-LdI/dt$ for fixed $L$. Across an interface with unit normal $\mathbf{n}$ from medium 1 to 2,These electromagnetic boundary conditions follow from Gauss's law for magnetism and Ampère's law. Here ${\mathbf{K}}_f$ is free surface current. At the iron–air faces there is no free surface current, so normal magnetic flux density and tangential magnetic field intensity are continuous. In each isotropic linear medium, $\mathbf{B}={\mu }_0{\mu }_r\mathbf{H}$.

Complete core. A slender toroid has approximately uniform $H$ along mean path length $\ell$. Ampère's law gives $H\ell =NI$, so $B=\mu {\mu }_{0}NI/\ell$. The flux per turn is $\Phi =BA$, yielding the inductance of a toroidal winding,This uses constant relative magnetic permeability $\mu$ and neglects leakage, fringing and saturation.

Why the gap lowers inductance. The gap replaces an easy magnetic path in iron with air. Maintaining the same flux now requires a much larger magnetic field intensity in part of the path and therefore more ampere-turns. Equivalently, magnetic reluctance increases, so the flux linkage for a fixed current and hence self-inductance decrease. A narrow gap can matter greatly because its permeability is smaller by the factor $\mu$.

Oblique-gap calculation. In the original diagram $t$ is the perpendicular separation of the cut faces. Let $\mathbf{s}$ point along the core, and let $\mathbf{n},\mathbf{q}$ be normal and tangential to the gap faces. With the face at angle $\theta$ to the core direction, $\mathbf{s}=\sin \theta \mathbf{n}+\cos \theta \mathbf{q}$. Write the core flux density as $B\mathbf{s}$. The boundary conditions givesoThe gap area normal to its nearly normal field is $A/\sin \theta$, not $A$. Its leading reluctance is consequently $t\sin \theta /({\mu }_{0}A)$. Adding the core reluctance gives the inductance of a toroid with an oblique narrow air gap,Thus at $\theta =45^{\circ }$,The approximation retains the possibly order-one ratio $\mu t/\ell$ while neglecting $t/\ell$, fringing and the small tangential gap contribution. More precisely, a local uniform-field path along the core, with missing iron length $t/\sin \theta$ subtracted, gives denominator $\ell /\mu +t\sin \theta (1-1/\mu )$. Its correction is beyond the stipulated large-$\mu$, narrow-gap accuracy; simply neglecting $\mu t$ would not be justified.

Field in the slot. The component expression above givesFor the $45^{\circ }$ gap and $\mu \gg 1$,Its units are $\mathrm{A}{\mathrm{m}}^{-1}$. The field is almost perpendicular to the cut faces, rather than following the iron direction straight across the cut. Its small angle away from the face normal satisfies $\tan \alpha =1/\mu$ at $45^{\circ }$; the sense reverses if the current reverses.

Field law and centre value. The Biot-Savart law for a steady thin wire isHere $d{\mathbf{\ell }}^{\prime }$ follows the conventional electric current and the displacement points from source to observation point. For a circular loop of radius $a$, every element at the centre has distance $a$ and contributes in the same axial direction. Choose $+z$ by the right-hand rule. Then

On-axis integration. At axial distance $z$, every wire element has distance $\sqrt{a^2+z^2}$. Azimuthal symmetry cancels transverse components. Parameterizing ${\mathbf{r}}^{\prime }=(a\cos \phi ,a\sin \phi ,0)$ gives axial cross-product element $a^{2}d\phi$, so the on-axis magnetic field of a circular current loop isIt points the same way on the two halves of the axis; its strength is even in $z$.

Directions near the axial point. At a point on the positive axis the field is axial, but nearby lines spread outward because $B_0$ decreases as $z$ increases. On the negative-axis side they converge toward the axis while still progressing in the positive axial direction. There is no azimuthal field. The near-axis sketch shows this local spreading; the complete magnetic field lines return around the outside of the loop and form closed curves.

Radial component from zero divergence. Smooth axial symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$ and $B_{\rho }=O(\rho )$. By Gauss's law for magnetism,Integrate radially from zero, excluding a singular $1/\rho$ term, to obtain the near-axis radial magnetic field, $B_{\rho }=-\rho B_0^{\prime }(z)/2+O({\rho }^3)$. Differentiating the axial field therefore givesFor $z>0$ it is outward; for $z<0$ it is inward. This supplies the direction reasoning used in the sketch.

Force on the second loop. Put the second loop at $z=d$ and radius $\rho =a$. For $d\gg a$, the radial field due to the first loop is $B_{\rho }\simeq 3{\mu }_{0}I{a}^3/(4d^4)$. The element $d\mathbf{\ell }=ad\phi \hat{\mathbf{\phi }}$ experiences Lorentz forceThe radial contributions cancel as vectors around the loop. The axial contributions add, giving the force between distant coaxial current loops,The sign follows $\hat{\mathbf{\phi }}\times \hat{\mathbf{\rho }}=-\hat{\mathbf{z}}$. Both the small radial coordinate relative to $d$ and the dipole approximation are justified by $a/d\ll 1$.

Force direction. The second loop is pulled back toward the first: the force is attractive for the same current sense. Reversing one current reverses the force. The much larger axial field chiefly produces radial element forces; retaining the smaller radial field is essential for the net axial attraction.

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.

The starting point is Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$: a flux tube carries the same magnetic flux through successive transverse sections. To identify that tube approximately with the physical core, one must also make a negligible-leakage approximation. The large core magnetic permeability makes the core route much less reluctant than competing paths through the surrounding air; the small gap further reduces the incentive for a long external return path. The gentle taper, $s\gg R-r$, lets the field follow the changing section with only small transverse corrections.

For two cuts and the intervening lateral surface, the divergence theorem gives ${\Phi }_2-{\Phi }_1=-{\Phi }_{\mathrm{side}}$. Neglecting lateral leakage givesThe normal magnetic flux density is continuous at each core–air interface by part (a), so this same magnetic flux crosses the gap. To use the gap's geometrical area $\pi r^2$ in the later formulas, the magnetic fringing must also be negligible, normally requiring $h\ll r$. This is the small-gap interpretation of the paper's approximation. Large magnetic permeability alone does not eliminate leakage or magnetic fringing for arbitrary gap geometry.

In SI units, the microscopic Maxwell's equations for total charge density $\rho$ and current density $\mathbf{J}$ areThe first is Gauss's law: electric charge sources electric flux. The second is Gauss's law for magnetism, expressing the absence of magnetic monopole sources in this theory. Faraday's law relates changing magnetic flux to circulating electric field. The Ampere-Maxwell law relates magnetic circulation to both material current and changing electric flux.

For fixed loops and surfaces, the integral versions areSurface and loop orientations are related by the right-hand rule. These forms follow from the divergence theorem and Stokes' theorem. In material media, equivalent macroscopic equations can separate free and bound sources using $\mathbf{D}$ and $\mathbf{H}$ together with constitutive relations; the displayed forms use total sources.

The displacement-current term is essential for charge conservation. Taking the divergence of the last differential equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. It also makes magnetic circulation consistent for surfaces spanning the same loop through either a charging capacitor's wire or its dielectric gap.

In source-free vacuum the curl equations combine to give ${\nabla }^2\mathbf{E}={\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$, with the same equation for $\mathbf{B}$. ThusA plane wave is transverse, with $\mathbf{E}$, $\mathbf{B}$ and propagation direction mutually perpendicular and $E=cB$. The fields transport energy with density $u={ϵ}_0{E}^2/2+B^2/(2{\mu }_0)$ and Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Together with the Lorentz force law, the equations connect charges, currents, forces, induction and electromagnetic radiation.

Write $\rho$ for cylindrical radius in this question, not the rod's mass density. A charged relativistic particle in a time-independent magnetic field has the charged-particle electromagnetic Lagrangianwith zero electrostatic potential. The cylindrical speed is ${\mathbf{v}}^2={\dot{\rho }}^2+{\rho }^2{\dot{\theta }}^2+{\dot{z}}^2$. Inside a uniform axial field, choose the symmetric magnetic vector potential $A_{\theta }=B\rho /2$, $A_{\rho }=A_z=0$. Its cylindrical curl gives $B_z={\rho }^{-1}{\partial }_{\rho }(\rho A_{\theta })=B$. HenceIn a field-free region far from either end one can use $A_{\theta }=0$, recovering the requested outside expression. The question's $V=-qB{\rho }^2\dot{\theta }/2$ is a velocity-dependent generalized potential, not electrostatic potential energy.

There is an important end-field qualification. A finite axial field that jumps to zero with no radial component would violate Gauss's law for magnetism. The consistent near-axis idealisation is the solenoid fringe field from an axisymmetric vector potential, $A_{\theta }=\rho B(z)/2$, giving $B_{\rho }=-\rho B^{\prime }(z)/2$ and $B_z=B(z)$. The piecewise Lagrangian is its hard-edge limit and retains fringe impulses through the jump in $A_{\theta }$. Those impulses are essential to the focusing calculation.

Take a homogeneous material exhibiting superconductivity in the half-space $z\ge 0$, bounded by vacuum at $z=0$. Fix the temperature below the superconducting transition temperature, take a constant positive London penetration depth ${\lambda }_L$, and assume a local, linear London equation description in a Meissner state. The applied magnetic field is weak enough that it neither nucleates a quantum vortex nor appreciably suppresses the material's superconductivity. Assume translation invariance in $x,y$, no imposed transport electric current, and a tangential boundary magnetic field $\mathbf{B}(0)=B_0\hat{\mathbf{x}}$. Here $B_0$ is the local surface magnetic flux density, equal to the imposed vacuum magnetic field in this infinite planar geometry; it is not automatically the far-away applied value for a finite object.

For a bounded, decaying interior solution, the London equations, magnetostatic Ampère law and Gauss's law for magnetism giveThe solution isAt one London penetration depth the magnetic field is $e^{-1}\simeq 0.368$ of its surface value; at five it is $e^{-5}\simeq 0.00674$. There is no finite depth where this exponential is exactly zero. The differential equation excludes the growing exponential through the deep-interior boundary condition. A constant normal component cannot decay in this geometry: Gauss's law for magnetism would require ${\partial }_z{B}_z=0$. A perpendicular applied magnetic field on a finite plate therefore requires a different spatial problem.^[ref]

For carrier number density $n$, charge $q$ and effective mass $m$, the plasma frequency is ${\omega }_p^2=n{q}^2/({ϵ}_{0}m)$. Combining the second London equation with the magnetostatic Ampère law and Gauss's law for magnetism gives ${\nabla }^2\mathbf{B}=({\omega }_p^2/c^2)\mathbf{B}$. Therefore the decaying planar solution has length ${\lambda }_L=c/{\omega }_p=\sqrt{m/({\mu }_{0}n{q}^2)}$.

Near the axis of a solenoid, the magnetic vector potential $A_{\theta }=\rho B(z)/2$ givesThe radial fringe component enforces the solenoidal magnetic-field constraint. A sharp axial step in $B$ must retain this fringe as a distributional field; a purely axial field that jumps to zero violates Gauss's law for magnetism. The hard-edge model is a convenient limit for near-axis trajectories, not a literal monopole sheet.