advanced28 min studyLesson 531 in Physics

Magnetic screening requires more than perfect conductivity

Derive the London penetration profile and distinguish equilibrium flux expulsion from frozen magnetic history.

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What you’ll understand

  • Distinguish perfect conductivity from the equilibrium Meissner response.
  • Derive the static London screening equation in SI units.
  • Compute a penetration profile and its screening-current direction.

A conductor with zero resistance can maintain a current indefinitely, but that statement alone does not specify its equilibrium magnetic field. Superconductors additionally exhibit a Meissner state: sufficiently weak applied magnetic fields are expelled from the bulk after equilibration. The London equations encode this extra response. We will derive their exponential screening profile, identify the current that supports it, and separate that equilibrium statement from the magnetic history retained by an ideal perfect conductor.

An acceleration law does not select the magnetic state

For a homogeneous superfluid component, let q=−2e be the pair charge, m* the effective pair mass, and nₚ its effective number density. An electric field accelerates this component and produces the first London equation. Its coefficient contains q squared, so the electrical current accelerates along the field even though the carriers have negative charge. A normal component can also contribute a dissipative current at finite temperature; the London current is only the nondissipative part of that total response.

An ideal conductor with vanishing internal electric field has a time-independent magnetic field by Faraday’s law. That condition freezes an initial field rather than requiring it to vanish. The second London equation imposes an additional relation between current curl and magnetic induction. In a simply connected, vortex-free equilibrium region it selects screening rather than arbitrary frozen flux. Trapped flux in holes, pinning, metastability, and the mixed state of a type-II material require additional information, so flux expulsion should not be asserted for every superconducting configuration.

∂js∂t=npq2m∗E,∇×js=−Bμ0λL2,λL2=m∗μ0npq2\frac{\partial\mathbf j_s}{\partial t}=\frac{n_pq^2}{m^*}\mathbf E,\qquad \nabla\times\mathbf j_s=-\frac{\mathbf B}{\mu_0\lambda_L^2},\qquad \lambda_L^2=\frac{m^*}{\mu_0n_pq^2}
SI conventions: jₛ is current density in A/m², E electric field in V/m, B induction in tesla, λL in metres, and μ₀ the vacuum permeability. The local coefficients are treated as spatially constant; q=−2e and nₚ counts effective pairs.↔ Scroll wide equations horizontally.

The static boundary problem chooses the decaying branch

Neglect displacement current in a static sample and combine Ampère’s law with the second London equation. Taking the curl of Ampère’s law produces the Laplacian of the magnetic field because its divergence is zero. The result is a positive inverse penetration-length squared multiplying the field. This sign gives exponential spatial screening, rather than a sinusoidal bulk field. For a half-space x>0, impose a tangential induction B₀ at x=0 and require a finite field that tends to zero deep inside.

Choose the field along positive z. Its surviving profile is B₀ times exp(−x/λL); the growing exponential is excluded by the bulk boundary condition. Ampère’s law gives a positive y-directed current because the field decreases with x. Integrating that current over depth yields a sheet current B₀/μ₀. Increasing penetration depth spreads the same required sheet current over a thicker layer. Confusing the surface current density with its depth integral would incorrectly predict a different total screening requirement at fixed boundary field.

∇2B=λL−2B,Bz(x)=B0e−x/λL,jy(x)=−1μ0dBzdx=B0μ0λLe−x/λL\nabla^2\mathbf B=\lambda_L^{-2}\mathbf B,\qquad B_z(x)=B_0e^{-x/\lambda_L},\qquad j_y(x)=-\frac1{\mu_0}\frac{dB_z}{dx}=\frac{B_0}{\mu_0\lambda_L}e^{-x/\lambda_L}
The sample occupies x>0, Bz(0)=B₀, and Bz tends to zero at infinity. The displayed current direction follows a right-handed Cartesian system. This solution assumes a weak applied field within the Meissner regime.↔ Scroll wide equations horizontally.

Local screening has a physical range of validity

The London model holds the order-parameter amplitude fixed and treats current response as local. It is useful for long-wavelength, low-frequency electrodynamics away from vortex cores and strong pair-breaking currents. Very clean samples can display nonlocal response when the current at a point depends on fields over a coherence scale. Near a surface or near a critical field, amplitude suppression can matter. These limitations are reasons to move to a richer model, rather than to identify a measured penetration length automatically with a free-electron carrier count.

The function lab illustrates the quadratic magnetic-energy cost, not the screening equation. Reset and set both controls to a=0.5 and x=1, predicting y=0.5. Reset again and set a=0.5 and x=2, predicting y=2. Interpret x as an assigned normalized field and y as its normalized field-energy density. Doubling field costs four times the energy. The curve contains no depth coordinate, current, or superconducting transition, so an exponential London profile must be calculated from the boundary problem rather than read from this parabola.

Resolve a weak-field screening layer

A London half-space has λL=50 nm and B₀=0.20 mT along positive z. Assume these conditions lie safely in its Meissner regime. Find the depth where the field halves, the surface current density, and the current density at that depth. Use μ₀≈4π×10⁻⁷ H/m.

  1. Set exp(−x/λL)=1/2. The required depth is λL ln2≈50×0.693=34.7 nm, and the field there is 0.10 mT.

  2. Convert the inputs: λL=5.0×10⁻⁸ m and B₀=2.0×10⁻⁴ T. The surface current density is B₀/(μ₀λL)≈3.18×10⁹ A/m² along positive y.

  3. The current has the same exponential factor as the field, so at 34.7 nm it is half its surface value: approximately 1.59×10⁹ A/m², still along positive y.

The result

The half-field depth is 34.7 nm; the surface and half-field-depth current densities are 3.18×10⁹ and 1.59×10⁹ A/m² in the positive y direction.

Explore the model

Interactive lab

Interactive lab · function

Test a quadratic field-energy cost

Reset and set BOTH controls to a=0.5 and x=1, predicting y=0.5. Reset and set a=0.5 and x=2, predicting y=2. Assign x a normalized induction; this energy parabola has no London screening dynamics.

Test a quadratic field-energy costThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forField energy grows quadratically, while the screening profile additionally requires currents and spatial boundary conditions.

Check your understanding

What extra statement distinguishes London Meissner screening from perfect conductivity alone?

Try it yourself

At fixed effective mass and pair charge, the effective pair density halves. Find the factor by which λL changes, the factor by which the surface current density changes at fixed B₀, and whether the integrated screening sheet current changes.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use λL proportional to nₚ⁻¹ᐟ² and integrate the exponential current over the half-space.

Compare with the solution

The penetration depth increases by √2. The surface current density B₀/(μ₀λL) decreases by 1/√2. The sheet current remains the integral B₀/(μ₀λL) times λL, namely B₀/μ₀. The same net screening current is distributed over a thicker region, assuming both states remain within the local Meissner regime.

Take it with you

  • Perfect conductivity does not by itself imply equilibrium magnetic expulsion.
  • The London half-space has a finite exponential penetration layer.
  • Local response and a fixed superfluid amplitude limit the model’s domain.

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