Magnetic field ($\mathbf{B}$)

A magnetic field is a vector field $\mathbf{B}(\mathbf{r},t)$ whose local value determines the magnetic part of the Lorentz force on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units,where $v$ is the speed, $B=|\mathbf{B}|$, and $\alpha$ is the angle between the velocity and the magnetic field. The cross product makes a nonzero magnetic force perpendicular to both the velocity and the magnetic field. The right-hand rule gives its direction for positive electric charge; negative electric charge reverses it. ^[ref].

The SI unit of magnetic field is the tesla:For a stipulated example in SI units, take electric charge $q=2\mathrm{C}$, velocity $\mathbf{v}=(3,0,0)\mathrm{m}{\mathrm{s}}^{-1}$, and magnetic field $\mathbf{B}=(0,0,4)\mathrm{T}$. The cross product gives magnetic force ${\mathbf{F}}_B=(0,-24,0)\mathrm{N}$. With the same speed directed along $+z$, the magnetic force is zero: motion parallel to a magnetic field produces no magnetic acceleration.

Magnetic field lines visualize the direction of a magnetic field by being tangent to it at each point. They are not generally the path of a body with electric charge: in a uniform magnetic field along $+z$, the magnetic field lines are straight while transverse velocity produces circular motion, and adding parallel velocity produces helical motion in a uniform magnetic field. The worked Lorentz force example below illustrates this distinction.

A magnetic field need not be static. Magnetostatics is the restricted case with no dependence on time; an electromagnetic wave has changing electric field and magnetic field even in vacuum. A changing magnetic field is coupled to the electric field through Faraday's law of induction. ^[ref]. The useful separation is between local bending, described by the Lorentz force, and the amount of magnetic field crossing a surface, described by magnetic flux.

In vacuum or an approximately nonmagnetizable plasma, the field energy per volume is $B^2/(2{\mu }_0)$. magnetic pressure has the same scale, but a magnetic field also exerts directional tension, so its stress is not universally an isotropic pressure. Material magnetization can require a different constitutive treatment.

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 tesla is the SI unit of magnetic field. From the magnetic Lorentz force, $1\mathrm{T}=1\mathrm{N}\mathrm{s}{\mathrm{C}}^{-1}{\mathrm{m}}^{-1}$. ^[ref].

A solenoid is a coil of electrical conductor, usually wire wound into a helix around a circular cylinder. The same electric current $I$ flows through its successive turns. Each turn contributes a magnetic field; the superposition principle makes their axial contributions reinforce inside the coil. Curl the fingers of your right hand in the direction of the electric current around the turns: the thumb gives the interior magnetic field direction by the right-hand rule. Throughout this branch, $N$ is the total number of turns, $\ell$ is the coil length, and $n=N/\ell$ is the solenoid turn density. ^[ref].

The attraction of a long solenoid is a controllable, nearly uniform magnetic field over an interior working region: change the electric current to tune its strength, or reverse the electric current to reverse its direction. This makes it a useful setting for studying the Lorentz force, measuring magnetic flux, and constructing an inductor. Its geometry turns the many contributions from individual turns into the simple long-solenoid approximation. A real finite coil has end fields and leakage; its exterior magnetic field is not zero. The on-axis magnetic field of a finite solenoid below quantifies that difference.

Historically, the coil helped connect magnetic fields to electric currents. In his memoir on the mathematical theory of electrodynamic phenomena, Ampère modeled the interactions of magnets using the ends of solenoids in place of microscopic magnetic constituents. This was a route toward explaining magnetism by electric currents, rather than a modern microscopic account of magnetization. ^[ref]. The historical setting is documented by the ^[ref].

For a near-axis beam of equal incoming speed $v$ through a uniform solenoid of length $l$, define $\omega =|qB|/(2\gamma m)$. Neglecting the small axial-speed changes, the internal radius is $\rho ={\rho }_0\cos (\omega z/v)$. The exit fringe removes the azimuthal velocity for zero incident canonical angular momentum. The field-free radial trajectory then reaches the axis a distancebeyond the exit, when $0<\omega l/v<\pi /2$. In the weak-lens limit, $f\simeq 4(\gamma mv{)}^2/(q^2{B}^{2}l)$. Without the paraxial approximation, the radius-dependent axial speed produces aberration; the ideal common focus is then only approximate.

Let a particle enter a uniform solenoid at transverse position $u_0=x_0+i{y}_0$ with zero radial velocity after an ideal fringe. Define signed $\Omega =qB/(2\gamma m)$ and initial field-free speed $v$. Its Lorentz force trajectory inside isThe transverse motion is a circle about the guiding centre $u_0/2$ at cyclotron angular frequency $2|\Omega |$. Cylindrical radius is $|u_0\cos \Omega t|$; its apparent azimuthal jumps when the orbit crosses the axis are coordinate singularities. Forward entry requires $v>|\Omega u_0|$ in this ideal model.

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.

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

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.

Model a finite solenoid as a continuum of coaxial turns shaped as circles of radius $a>0$, uniformly distributed over length $\ell$ with solenoid turn density $n=N/\ell$. Put its centre at $z=0$, so its ends are $z=\pm \ell /2$. With the same electric current orientation as in the long-solenoid approximation, its axial magnetic field isThis formula applies only on the axis, where symmetry cancels the transverse magnetic field. It does not describe points off the axis. It is exact for the continuum circular-turn model, which neglects winding thickness, discrete wire spacing, and helical advance. ^[ref].

The derivation is an especially direct use of the superposition principle. The on-axis magnetic field of a circular current loop at source position $z^{\prime }$ is ${\mu }_{0}I{a}^2/[2(a^2+(z-z^{\prime }{)}^2{)}^{3/2}]$. There are $nd{z}^{\prime }$ turns in an axial slice, soSubstitute $u=z-z^{\prime }$ and use the antiderivative $u/[a^2\sqrt{a^2+u^2}]$ to obtain the displayed expression. If the centre is instead at $z=z_0$, replace every $z$ in that expression by $z-z_0$.

At the centre and at either end, respectively,For aspect ratios $\ell /a=2,6,20$, the centre ratios are approximately $0.7071,0.9487,0.9950$. The end ratio tends to $1/2$ as $\ell /a$ becomes large, even though the centre ratio tends to $1$. Thus a long coil can have an excellent central magnetic field while the ends still differ strongly from that value. Convergence to the long-solenoid approximation is not uniform across the end region.

For a concrete geometry, give the earlier coil radius $a=0.025\mathrm{m}$, so $\ell /a=20$. Its central magnetic field is approximately $1.2504\mathrm{mT}$, its end value is $0.6275\mathrm{mT}$, and at $z=\ell$ its axial magnetic field is still $0.00277\mathrm{mT}$. The last point is outside the coil: finite length does not permit the zero-exterior conclusion of the infinite model. Off the axis, the magnetic field lines bend around to form the return field. A central axial value alone also does not determine the total magnetic flux through every turn of a short coil.

For a long vacuum-core or air-core solenoid with fixed geometry and linear magnetic response, neglecting end effects gives self-inductanceHere $N$ is the total number of turns, $n=N/\ell$ is the solenoid turn density, and $A$ is the interior cross-sectional area. The reason for the square of $N$ is physical: more turns produce more magnetic field, and more turns link the resulting magnetic flux. Approximately equal magnetic flux $\Phi \approx BA$ through each turn gives magnetic flux linkage $\Lambda \approx N\Phi \approx {\mu }_0{N}^{2}AI/\ell$; division by the electric current yields $L$. ^[ref].

For the same $N=1000$, $\ell =0.50\mathrm{m}$, and $a=0.025\mathrm{m}$, the area is $A=\pi (0.025{)}^2{\mathrm{m}}^2\approx 1.9635\times 10^{-3}{\mathrm{m}}^2$. The long-coil estimate isAt $I=0.50\mathrm{A}$, each turn links approximately $2.4674\times 10^{-6}\mathrm{Wb}$ and the total magnetic flux linkage is approximately $2.4674\times 10^{-3}\mathrm{Wb}$. Their ratio to $I$ is the stated self-inductance. A real finite coil need not give precisely these values: the axial centre correction above is not by itself a correction formula for inductance, which depends on magnetic flux across all the turns.

At fixed $A$ and $\ell$, doubling $N$ quadruples $L$. At fixed solenoid turn density $n$ and area $A$, doubling $\ell$ doubles both $N$ and $L$. These are different changes to the geometry, so there is no contradiction between $L\propto 1/\ell$ at fixed $N$ and $L\propto \ell$ at fixed $n$. A magnetic core requires its own magnetic permeability model; nonlinear magnetization can make a constant-$L$ description inadequate.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.

A magnetic compass uses the orientation of a magnetic needle to indicate the direction of a magnetic field.

The magnetic field has zero divergence. The ideal magnetohydrodynamic induction equation preserves this constraint from initial data, since the divergence of its curl is zero.

A magnetic field line is an integral curve tangent to the magnetic field. In coordinates $x^i$, it obeys $d{x}^i/B^i=d{x}^j/B^j$ wherever the relevant components are nonzero.

A magnetic arcade is a family of arch-shaped magnetic field lines joining two regions of a boundary. The boundary intersections are its footpoints. In a simple line-tied family, paired footpoints can be labelled by Euler potentials. A decaying linear force-free magnetic arcade is one explicitly solvable planar example; a general arcade need not be linear or force-free.

The curvature vector of a magnetic field line parametrized by arc length. For circular lines, it points towards their center and has magnitude $1/R$. The curvature contribution to magnetic tension is $B^2\mathbf{\kappa }/(4\pi )$ in Gaussian units.

A flux tube is a bundle of magnetic field lines bounded by a surface everywhere tangent to the magnetic field. Gauss's law for magnetism makes the magnetic flux through any transverse cross-section constant along the tube. For a narrow tube with nearly uniform field magnitude $B$ and transverse area $A$, this gives $BA=\mathrm{constant}$.

The cross-sectional area of a narrow flux tube, measured normal to its magnetic field. Constancy of the magnetic flux gives $A|\mathbf{B}|=\mathrm{constant}$ along the tube. This area controls the mass transport when flow follows the field.

The magnetic-field-line equation states that the tangent to a field line is parallel to $\mathbf{B}$. In spherical coordinates for an axisymmetric poloidal field, it can be written $dr/(rd\theta )=B_r/B_{\theta }$.

Outside a localized magnetic dipole $\mathbf{\mu }$, the vacuum field in Gaussian conventions isFor a dipole along the polar axis, $B_r=2\mu \cos \theta /r^3$ and $B_{\theta }=\mu \sin \theta /r^3$.

For coaxial circular loops with radii $a_1,a_2\ll d$, the first loop's on-axis magnetic field of a circular current loop has far value $B_0\simeq {\mu }_0{I}_1{a}_1^2/(2d^3)$. Its near-axis radial magnetic field at the second loop is $B_{\rho }\simeq 3{\mu }_0{I}_1{a}_1^2{a}_2/(4d^4)$. Integrating the axial Lorentz force from $I_2{a}_{2}d\varphi \hat{\mathbf{\varphi }}\times B_{\rho }\hat{\mathbf{\rho }}$ gives the displayed magnitude. Equal current senses attract; opposite senses repel. This is the leading dipole limit and is not an exact close-coil force law.

For a centred magnetic dipole moment pointing toward geographic south, the field at northern latitude $\lambda$ has northward component $B_N=B_{\mathrm{eq}}\cos \lambda$ and downward component $B_D=2B_{\mathrm{eq}}\sin \lambda$. These follow by resolving the magnetic dipole field into local radial and meridional directions. The downward inclination is therefore $\tan \iota =B_D/B_N=2\tan \lambda$, while the magnitude is $B_{\mathrm{eq}}\sqrt{1+3{\sin }^2\lambda }$. This ideal aligned-dipole model has no eastward component and does not capture actual geographic departures of the geomagnetic field.

An axisymmetric dipole field line satisfies $r=C{\sin }^2\theta$, where $C$ is its equatorial crossing radius. This follows by integrating $dr/(rd\theta )=B_r/B_{\theta }=2\cot \theta$.

Near the polar axis of a magnetic dipole field, $B\propto r^{-3}$. Conservation of magnetic flux in a flux tube consequently gives $A(r)\propto r^3$. A dipole magnetic-field line intersecting a sphere of radius $R_{\star }$ at small polar angle ${\theta }_{\star }$ obeysThis area is for one polar tube. The approximation requires ${\theta }_{\star }^{2}r/R_{\star }\ll 1$; the full magnetic dipole field does not remain a narrow polar tube at arbitrarily large radius.

A magnetostatic equilibrium is a time-independent configuration in which magnetic, pressure, gravitational, and any other body-force densities balance.

For an axisymmetric magnetic field $\mathbf{B}=\psi {\mathbf{e}}_{\varphi }/(r\sin \theta )+\nabla \times [\chi {\mathbf{e}}_{\varphi }/(r\sin \theta )]$, the poloidal magnetic flux function gives $B_r={\chi }_{\theta }/(r^2\sin \theta )$ and $B_{\theta }=-{\chi }_r/(r\sin \theta )$. Force balance implies locally $p=p(\chi )$ and $\psi =\psi (\chi )$ on regular connected flux surfaces. Direct calculation gives $(\nabla \times \mathbf{B})\times \mathbf{B}=-({\Delta }_{\ast }\chi +\psi {\psi }^{\prime })\nabla \chi /(r^2{\sin }^2\theta )$, with ${\Delta }_{\ast }\chi ={\chi }_{rr}+\sin \theta {\partial }_{\theta }({\chi }_{\theta }/\sin \theta )/r^2$. Equating this to ${\mu }_0\nabla p$ proves the displayed magnetostatic equilibrium equation, including its sign and dimensions.

Set $\psi =\alpha \chi$, $p=p_0+P\chi$ and $\chi =F(r){\sin }^2\theta$. A regular family is $F=A[\cos (\alpha r)-\sin (\alpha r)/(\alpha r)]+c{r}^2$, where $P=-{\alpha }^{2}c/{\mu }_0$. A zero magnetic field at radius a requires $F(a)=F^{\prime }(a)=0$. For nonzero A this yields the displayed condition with $x=\alpha a$ and $c=-A[\cos x-\sin x/x]/a^2$. The boundary pressure is $p_0$, independent of angle. The trigonometric pair is tied by regularity and the radial differential equation; its coefficients are not independent.

A plane-parallel magnetostatic equilibrium with constant downward gravitational acceleration $g$, horizontal magnetic field, constant isothermal sound speed $c_s$, and constant Alfvén speed $v_a$ hasIndeed the magnetohydrodynamic total pressure is $(c_s^2+v_a^2/2)\rho$, and its vertical derivative is $-g\rho$. The magnetic pressure therefore increases the scale height above that of an unmagnetized atmosphere.

A cylindrical magnetostatic equilibrium is an axisymmetric vector field and pressure configuration independent of the axial coordinate. Its radial Lorentz force density balances the pressure gradient, as expressed by cylindrical magnetostatic pressure balance.

In a Z-pinch, an axial electric current produces a toroidal magnetic field whose inward magnetic tension confines a plasma. A cylindrical magnetostatic equilibrium balances the resulting Lorentz force density against a pressure gradient.

For an axisymmetric vector field independent of the axial coordinate and regular on the axis, Gauss's law for magnetism gives $B_R=0$. The magnetostatic Ampère-Maxwell equation then gives $j_R=0$, $j_{\varphi }=-c{B}_z^{\prime }/(4\pi )$, and $j_z=c(R{B}_{\varphi }{)}^{\prime }/(4\pi R)$ in Gaussian units. Radial magnetostatic equilibrium isFor enclosed axial electric current $I(R)$, the magnetostatic Ampère-Maxwell equation gives $B_{\varphi }=2I/(cR)$, henceThe first term contains pressure and axial magnetic pressure; the second combines toroidal magnetic pressure with the hoop force from magnetic tension.

For a regular cylindrical flux tube with $\mathbf{B}=(0,B_{\varphi }(s),B_z(s))$, vanishing field beyond radius $s_0$, and exterior pressure $p_e$, define $⟨h⟩=2s_0^{-2}{\int }_0^{s_0}sh(s)ds$ and $B_p^2=2{\mu }_0{p}_e$. Integrating cylindrical magnetostatic pressure balance against $s^2$ gives the displayed identity. The toroidal magnetic pressure and hoop magnetic tension cancel in this cross-sectional axial-field average. Nonnegative gas pressure gives $⟨B_z^2⟩\le B_p^2$, with strict inequality precisely when $⟨p⟩>0$.

For a flux tube with $B_{\varphi }=\gamma (s/s_0)B_z$, regularity at the axis and cylindrical magnetostatic pressure balance give $B_z(0{)}^2=B_p^2-2{\mu }_{0}p(0)+{\gamma }^2⟨B_z^2⟩$. The flux-tube axial field virial identity and $p\ge 0$ give the bound. A twisted toroidal magnetic field can support a central axial field above the untwisted magnetic pressure limit even though its cross-sectional mean obeys that limit.

For constant axial magnetic field and $p=p_0(1+R^2/a^2{)}^{-k}$ with $p_0>0$ and $k>0$, cylindrical magnetostatic pressure balance gives, with $x=R^2/a^2$,The nontrivial electric current is finite at infinity exactly when $k>1$, in which case $I^2(\infty )=2\pi c^2{p}_0{a}^2/(k-1)$. Either sign of $I$ is possible because the pressure balance fixes its square.

With no axial electric current, cylindrical magnetostatic pressure balance reduces to constant $p+B_z^2/(8\pi )$. For $p=p_0{e}^{-R^2/a^2}$ and $B_z(0)=0$, its radial solutions areAs $R\downarrow 0$, $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ and $j_{\varphi }\to -\sigma c\sqrt{8\pi p_0}/(4\pi a)$. These finite scalar limits do not give a smooth function of Cartesian position: $R=\sqrt{x^2+y^2}$ has a cusp at the axis, and the azimuthal unit vector has no unique limit there. Thus these data describe radial equilibrium away from the axis, but cannot satisfy smooth vector-field regularity on the axis without modifying the data.

A conductor or plasma moving with nonrelativistic velocity $\mathbf{u}$ through a magnetic field experiences the motional electric field $\mathbf{E}=-\mathbf{u}\times \mathbf{B}/c$ in Gaussian units when the electric field vanishes in its rest frame.

The London equations relate the superconducting electric current to the electromagnetic field: ${\partial }_t\mathbf{J}=\mathbf{E}/({\mu }_0{\lambda }_L^2)$ and $\nabla \times \mathbf{J}=-\mathbf{B}/({\mu }_0{\lambda }_L^2)$. In a gauge with constant condensate phase the second follows from $\mathbf{J}=-\mathbf{A}/({\mu }_0{\lambda }_L^2)$; it implies exponential magnetic field screening.

Combining the static London equations with the magnetostatic Ampère law and $\nabla \cdot \mathbf{B}=0$ gives the displayed bulk equation for a uniform superconducting carrier density. A planar boundary gives exponentially decaying field on the London penetration depth scale; boundary geometry and vortices require their corresponding field solutions.

For uniform vortex-free superconductivity, the London equation and stationary Maxwell equations give ${\nabla }^2\mathbf{B}=\mathbf{B}/{\lambda }_L^2$. A field tangential to a planar superconducting half-space decays as $\mathbf{B}(z)=\mathbf{B}(0)e^{-z/{\lambda }_L}$. A nonzero normal boundary field is incompatible with this translation-invariant decaying Meissner solution, because divergence-free magnetic flux would require that component to remain constant.

For scalar kinetic term $|D\phi {|}^2$ and condensate magnitude $|\phi |=v$ without a factor $1/\sqrt{2}$, the canonical vector mass is $m_A^2=2e^2{v}^2$. In a constant-phase gauge the current is $\mathbf{j}=-2e^2{v}^2\mathbf{A}$. A static magnetic field obeys ${\nabla }^2\mathbf{B}=m_A^2\mathbf{B}$, so its penetration length is $1/(\sqrt{2}|e|v)$. This normalization differs from defining the vacuum amplitude as $v/\sqrt{2}$.

For a slab of thickness $2a$ with parallel applied field, the local London equation gives $B(x)=B_a\cosh (x/\lambda )/\cosh (a/\lambda )$. Averaging gives $M=-B_a[1-\tanh (a/\lambda )/(a/\lambda )]/{\mu }_0$. Integrating $-Md{B}_a$ and comparing with the bulk condensation energy yields the displayed thin-slab enhancement. It assumes field-independent condensate density and can be limited by pair breaking or nonlocal electrodynamics.

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 ${\lambda }_L=c/{\omega }_p=\sqrt{m/({\mu }_{0}n{q}^2)}$.

In a local London equation description, the measured London penetration depth satisfies ${\lambda }_L^2=m/({\mu }_0{n}_s{e}^2)$ when $n_s$ counts superconducting electrons and $m$ is their effective mass. Thus screening measures $n_s/m$. If instead one counts Cooper pairs, use pair mass $2m$ and pair electric charge $-2e$, giving pair density $n_s/2$ for the same screening length. Identifying this superconducting carrier density with the total conduction-electron density requires the additional assumption that the relevant carriers all participate in screening at the measurement temperature.

The Meissner effect is the equilibrium expulsion of a sufficiently weak, static magnetic field from the bulk of a material exhibiting superconductivity. Expulsion includes a magnetic field that was already present before cooling: zero electrical resistance alone does not predict this. The Field-cooled Meissner experiment tests that distinction. A finite London penetration depth means that the magnetic field and screening electric current occupy a surface layer, rather than disappearing at a mathematical boundary.^[ref]

This makes superconductivity useful for shielding weak magnetic fields, and makes its magnetic response an independent test of the phase. The striking result is that the same screened equilibrium can be reached by two different preparation histories. The assumptions matter: Finite-thickness Meissner screening can leave an appreciable magnetic field throughout a small sample, and the Limits of Meissner screening distinguish quantum vortex penetration and retained magnetic flux from this equilibrium state.

A perpendicular magnetic dipole $m\hat{z}$ a height $h$ above an ideal flux-excluding plane has image $-m\hat{z}$ at $-h$. The exterior field has zero normal component at the plane. Images are a mathematical construction outside the physical solution domain.

For a small fixed-current loop with moment $m=I\pi R^2$ and $R\ll h$, the image field gives upward force $3{\mu }_0{m}^2/(32\pi h^4)$. Differentiating the external field holds the image position fixed; moving both source and image in that derivative would double the force incorrectly.

A Meissner state is an equilibrium state of superconductivity without quantum vortex penetration, with screening electric currents responding to the applied magnetic field. In a solid, simply connected sample much thicker than the London penetration depth, the deep interior magnetic field is approximately zero. This is a statement about a phase and a geometry, not a rule that every sample exhibiting superconductivity always has $\mathbf{B}=0$. In particular, the Meissner state exists for sufficiently weak magnetic fields in both type-I superconductors and type-II superconductors. ^[ref]

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]

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

To isolate finite size from quantum vortex penetration, keep the same weak-field London equations but replace the half-space by an infinite slab $-d/2\le z\le d/2$. Put equal tangential magnetic fields $B_0\hat{\mathbf{x}}$ at its two faces. With $z=0$ now at the slab center, the even solution and its signed current density areThe hyperbolic cosine profile has equal boundary values, while the electric currents on the two sides have opposite signs. In particular,For the stipulated thickness $d=2{\lambda }_L$, the center retains about $0.648B_0$ and the mean is about $0.762B_0$, even though this is a Meissner state. For $d\gg {\lambda }_L$, the center magnetic field is exponentially small and the mean is about $2{\lambda }_L{B}_0/d$, reflecting the two surface layers. For $d\ll {\lambda }_L$, the mean tends to $B_0$: little screening is possible across such a short distance. These are analytic model examples, not material measurements. Real finite objects also need their demagnetizing field and edge geometry accounted for; the slab calculation only isolates the effect of thickness.

Compare two ideal models for a stationary, solid, simply connected sample. In the perfect conductor flux-history model the interior electric field is constrained to zero from the moment perfect conduction starts. Faraday's law of induction then givesConsequently the interior magnetic field retains its value at that moment. Cooling in an already applied magnetic field freezes a nonzero interior value; starting perfect conduction in zero magnetic field and then applying a magnetic field preserves zero. The zero initial interior magnetic field often assumed in perfect conductor wave-reflection problems is one particular preparation, not a general prediction of expulsion. ^[ref]

The Meissner state instead selects the screened equilibrium after either preparation, provided that equilibrium is accessible and the applied magnetic field lies in the Meissner state regime. The first London equation alone illustrates the missing information: at fixed ${\lambda }_L$, combining it with Faraday's law of induction givesThis permits an arbitrary time-independent integration field. The second London equation imposes the additional zero value of that bracket in the vortex-free local model.^[ref] Establishing superconductivity during cooling is not the fixed-parameter, zero-electric field model: transient electric fields accompany changing magnetic flux, and ${\lambda }_L$ itself depends on temperature. The equilibrium London equations predict final spatial screening, not the cooling rate or the time-resolved expulsion pulse.

The 1933 report Ein neuer Effekt bei Eintritt der Supraleitfähigkeit by W. Meissner and R. Ochsenfeld describes cooling tin and lead cylinders in a fixed transverse magnetic field. Two parallel cylinders were about $140\mathrm{mm}$ long and $3\mathrm{mm}$ thick, separated by $1.5\mathrm{mm}$. A roughly $10\mathrm{mm}$ rotating search coil between them drove a ballistic galvanometer. At about $5\mathrm{G}$, the reported below/above-transition induction ratios were $1.70$ for tin single crystals and $1.77$ for polycrystalline lead. The increased gap magnetic field indicates redistribution outside the cylinders, rather than an unchanged frozen field. Coil-winding geometry and noncircular tin specimens limited comparison with the ideal prediction. ^[ref]

Their second arrangement used a lead tube, about $130\mathrm{mm}$ long with $3\mathrm{mm}$ outer and $2\mathrm{mm}$ inner diameters. A rotating search coil could sit inside or alongside it. The bore retained magnetic flux after removing the applied magnetic field; the coil nearly filled the bore, preventing a test of interior field uniformity. A hole containing magnetic field is distinct from magnetic field penetrating the surrounding material exhibiting superconductivity. ^[ref]

The modern Field-cooled Meissner experiment provides a complementary readout. MIT's probe I places a sample in a brass holder, with an $810$-turn test coil of $7.1\mathrm{mm}$ inner diameter, $10.5\mathrm{mm}$ outer diameter and $12.3\mathrm{mm}$ length, inside a longer solenoid. A nearby silicon-diode thermometer tracks cooling by helium gas. With lead still normal, the guide applies about $200\mathrm{mA}$ to the solenoid and connects the test coil to a current-integrator box and oscilloscope. Cooling produces a transient; warming gives the opposite sign. Switching off the solenoid while the sample is normal checks the sign. ^[ref]

The inferred change is in linked magnetic flux, not a direct scan of $B(z)$. For a stationary test coil the pulse obeys $\int Vdt=-\Delta \Psi$, where $\Psi$ is its magnetic flux linkage. An integrator or AC-coupled oscilloscope needs its transfer response accounted for before extracting this pulse area. Keep the source electric current fixed during cooling, monitor temperature, and compare warming and cooling slowly enough to reduce sample–thermometer lag. An AC inductance change alone can reflect changed normal-state electrical conductivity; the fixed-field expulsion test distinguishes that alternative. ^[ref] Neither this coil readout nor the historical gap-field measurement resolves the microscopic London penetration depth by itself.

For a bulk type-II superconductor with negligible demagnetizing field, raising the applied magnetic field intensity past the lower critical value $H_{c1}$ favors a superconducting mixed state containing quantum vortex lines; superconductivity survives until the upper critical value $H_{c2}$. The magnetic field extends around each vortex over a penetration scale, not solely inside its core. Thus a nonzero average magnetic flux density can coexist with superconductivity. Below $H_{c1}$, a type-II superconductor also supports a Meissner state; a vortex-free state above the equilibrium entry field may instead be metastable. ^[ref]

Real preparation histories may leave quantum vortex lines held in place by material inhomogeneities, retaining magnetic flux and producing hysteresis. This flux pinning restricts access to equilibrium: it is different from the arbitrary frozen interior magnetic field of the ideal perfect conductor model. Harvard's vortex-imaging work illustrates strong pinning to bulk disorder in a type-II superconductor. ^[ref] Separately, flux conservation in a superconducting loop and fluxoid and flux quantization in a superconductor explain why a hole can retain magnetic flux. Neither bore retention, vortex penetration nor weak screening in a thin sample should be silently identified with the equilibrium Planar London screening calculation.

Magnetic flux measures the signed amount of magnetic field crossing an oriented surface $S$. Choose a unit normal $\hat{\mathbf{n}}$ and use the oriented surface element $d\mathbf{S}=\hat{\mathbf{n}}dA$. The defining surface integral isFor a uniform magnetic field across a flat surface of area $A$, the dot product reduces this towhere $\theta$ is the angle between the magnetic field and the chosen unit normal, rather than the plane of the surface. The SI unit is the weber, $1\mathrm{Wb}=1\mathrm{T}{\mathrm{m}}^2$. ^[ref].

For a stipulated magnetic field $B=2\mathrm{T}$ and area $A=3{\mathrm{m}}^2$, the magnetic flux is $6\mathrm{Wb}$ at $\theta =0$, $3\mathrm{Wb}$ at $\theta =\pi /3$, and zero at $\theta =\pi /2$. In the last case the magnetic field lies in the plane of the surface, so it does not cross it. Reversing the chosen unit normal reverses the sign of magnetic flux without changing the physical magnetic field; at $\theta =\pi$ the same example gives $-6\mathrm{Wb}$.

The geometric idea is projected area: tilting a flat surface reduces the area seen perpendicular to the magnetic field by the factor $\cos \theta$. This explains how rotation alone can change magnetic flux even when the magnetic field is constant; Faraday's law of induction turns such changes into an electromotive force. ^[ref].

The weber is the SI unit of magnetic flux, with $1\mathrm{Wb}=1\mathrm{T}{\mathrm{m}}^2$. A uniform magnetic field of $1\mathrm{T}$ crossing a flat surface of area $1{\mathrm{m}}^2$ along its chosen unit normal gives $1\mathrm{Wb}$. ^[ref].

A device that produces a magnetic field from an electric current.