University Physics IV · Frontiers of Modern Physics · 15.4

Zero Resistance, Meissner & Critical Fields

Two things happen when a metal drops below T_c, and they sound like one fact. The resistance falls below anything an instrument can see, and the magnetic field is pushed out of the interior. Only the second is the definition — and it is the one that hands you a length, a critical field, and a current limit you can actually compute.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The London model is the cheapest theory that produces a Meissner effect, and it is bought at a stated price. Take some fraction of the conduction electrons, density nₛ, to move without scattering; Newton's second law then gives ∂Jₛ/∂t = (nₛ e²/m)E, which is all a perfect conductor needs and which only freezes the flux already inside. The Londons added a second, independent relation — ∇×Jₛ = −(nₛ e²/m)B — not derived from the first but chosen because it is the one constitutive law that makes B = 0 the equilibrium state rather than one remembered state among many.

Feed it into Ampère's law and the field obeys ∇²B = B/λ², so an applied field dies exponentially over λ = √(m/μ₀ nₛ e²), tens of nanometres in a metal. Everything measurable in this lesson follows: the screening currents, the critical field at which expelling the flux costs more than the condensation energy buys, Silsbee's rule tying critical current to that field, and the type II compromise of letting quantised flux in through vortex cores. What the model never supplies is nₛ itself, or T_c, or any reason the electrons should pair.

It is phenomenology with one fitted length.

Simple definition: A superconductor is a material that below its critical temperature carries direct current with no detectable dissipation and actively expels magnetic flux from its bulk, screening an applied field to zero within a penetration depth λ of its surface.
Example: Lead has T_c = 7.20 K and λ ≈ 39 nm, so a 10 mT field applied to a lead block is down to 3.7 mT at 39 nm depth and to 0.1 mT at 180 nm — and the block stays superconducting until the surface field reaches B_c(4.2 K) = 53 mT.

London penetration depthλ = √(m / μ₀ nₛ e²)

nₛ = 4.0×10²⁸ m⁻³ gives λ = 26.6 nm. Working with pair mass 2m, charge 2e and density nₛ/2 returns exactly the same λ.

m and e are the electron mass and charge, nₛ the superconducting carrier density in m⁻³, μ₀ = 4π×10⁻⁷ H m⁻¹; λ in metres

Screened field in a half-space∇²B = B/λ² → B(x) = B₀ e^(−x/λ)

The Meissner effect made quantitative: the flux is not zero at the face, it is gone a few λ in — which is why a film thinner than λ hardly screens at all.

x is depth below the surface in m and B₀ the field just outside in T; valid for a flat face with λ much smaller than the sample

Screening current densityJₛ(x) = (B₀/μ₀λ) e^(−x/λ)K = B₀/μ₀

B₀ = 10 mT with λ = 40 nm demands Jₛ(0) = 2.0×10¹¹ A m⁻², within a factor of a few of the depairing limit.

Jₛ in A m⁻², B₀ in T, λ in m; K is the sheet current in A m⁻¹ found by integrating Jₛ over all depth

Critical field and Silsbee's ruleB_c(T) = B_c(0)[1 − (T/T_c)²]I_c = 2πa B_c/μ₀

Lead at 4.2 K has B_c = 53 mT, so a wire of radius 0.50 mm quenches at 132 A — its critical current is set by its own self-field.

T and T_c in K, B_c in T, a the wire radius in m; the rule is that a wire quenches when its own surface field reaches B_c

Type I or type IIκ = λ/ξtype I if κ < 1/√2μ₀H_c2 = Φ₀/2πξ²

Nb₃Sn has ξ ≈ 3.3 nm, giving μ₀H_c2 ≈ 30 T, some 380 times lead's B_c(0) of 80 mT.

ξ is the coherence length in m and κ the Ginzburg-Landau ratio; 1/√2 is where the interface energy changes sign

Flux quantum and vortex densityΦ₀ = h/2e = 2.068×10⁻¹⁵ Wbnᵥ = B/Φ₀

At 5 T, nᵥ = 2.4×10¹⁵ m⁻² on a triangular lattice of spacing 22 nm — 2.4×10⁹ vortices through every square millimetre.

Φ₀ in Wb, B in T, nᵥ in vortices per m² of face perpendicular to B; the 2e is the measured pair charge

Zero resistance is an upper bound, and a very good one

No experiment has ever measured zero; what a measurement gives is an upper limit. Push 100 mA through a strip of cross-section 1.0 mm × 0.10 mm and read the voltage across taps 10 mm apart: a 1 nV noise floor caps the resistance at R = 10⁻⁸ Ω and so the resistivity at ρ = RA/L = 10⁻¹³ Ω m. That is already five orders of magnitude below copper's 1.7×10⁻⁸ Ω m, and it is nowhere near the best bound available. Launch a current round a closed superconducting ring instead and watch its field decay as e^(−Rt/L). File and Mills followed such a current by NMR in 1963 and saw no decay in a run implying a time constant above 10⁵ years; for a ring of inductance 10⁻⁷ H that is R < 3×10⁻²⁰ Ω, and for a loop 0.1 m round of 1 mm² section, ρ below about 10⁻²⁵ Ω m. Seventeen orders below copper is still not zero, and the honest statement is the inequality.

The Meissner effect is an extra fact, not a consequence

Suppose only that the resistance vanishes. Then E = 0 inside, and Faraday's law ∇×E = −∂B/∂t forces ∂B/∂t = 0: whatever field threaded the sample when scattering stopped stays there for good. That is a perfect conductor, and its final state depends on its history — cool it in a 10 mT field, then switch the magnet off, and 10 mT of trapped flux remains, held by a surface current that never decays. Meissner and Ochsenfeld did the experiment on tin and lead in 1933 and found the opposite. Cooling through T_c in a steady field pushes the flux out as the transition passes, and the sample ends field-free whether the field was applied before or after cooling. Zero-field-cooled and field-cooled runs land on the same state, which is the signature of a thermodynamic phase rather than a frozen initial condition. No algebra takes you from R = 0 to that, so it has to be put in — and the second London equation, ∇×Jₛ = −B/μ₀λ², is where it goes in.

Solving the London equation for the penetration depth

Take the London constitutive law ∇×Jₛ = −(1/μ₀λ²)B together with Ampère's law in its static form ∇×B = μ₀Jₛ. Curl the second, use ∇⋅B = 0, and the vector identity leaves ∇²B = B/λ². For a superconductor filling x > 0 with a uniform field B₀ parallel to the face, this reduces to d²B/dx² = B/λ², whose general solution is A e^(−x/λ) + C e^(+x/λ). The growing term is thrown away because B must stay finite deep inside, and matching B(0) = B₀ at the face fixes A, leaving B(x) = B₀ e^(−x/λ). Now read λ off the coefficient: λ = √(m/μ₀ nₛ e²). With nₛ = 4.0×10²⁸ m⁻³ — a few per cent of the conduction electrons — the denominator is μ₀ nₛ e² = 1.29×10⁻¹⁵, the quotient is 7.06×10⁻¹⁶ m², and λ = 26.6 nm. Measured depths run from tens of nanometres in elemental metals to a few hundred in the cuprates. Temperature enters only through nₛ(T), and it is put in by hand: the two-fluid model gives λ(T) = λ(0)/√(1 − (T/T_c)⁴), which diverges at T_c because the screening fluid has run out.

What a field costs: condensation energy and Silsbee's rule

Expelling flux is not free. Holding B = 0 inside against an applied field B₀ costs magnetic energy density B₀²/2μ₀, and the ordered state can only pay for it out of the condensation energy it gained. The two balance at the thermodynamic critical field: B_c²/2μ₀ equals the condensation energy per unit volume. For lead, B_c(0) = 80.3 mT gives 0.0803²/(2 × 1.257×10⁻⁶) = 2.6 kJ m⁻³, and shared among 3.3×10²⁸ atoms per cubic metre that is 0.5 μeV each — which is why T_c is a few kelvin and not a few hundred. The empirical curve B_c(T) = B_c(0)[1 − (T/T_c)²] carries it to zero at T_c. A transport current does the same damage as an external magnet because it makes its own field: a wire of radius a carrying I has surface field μ₀I/2πa, so it quenches when that reaches B_c. That is Silsbee's rule, I_c = 2πa B_c/μ₀, and it says the critical current of a type I wire is a surface property — double the radius and I_c doubles, it does not quadruple.

Type II: letting the flux in one quantum at a time

A normal-superconducting boundary carries an energy per unit area set by two lengths: the coherence length ξ, over which the order parameter recovers, and λ, over which the field dies. When ξ > λ the interface costs energy, subdividing is unfavourable, and the sample stays wholly flux-free up to B_c and then goes normal all at once — type I. When λ > ξ, precisely when κ = λ/ξ exceeds 1/√2, the interface energy is negative and the sample gains by making as much boundary as it can. Above B_c1 flux enters as vortices: a normal core of radius about ξ, circulating supercurrent spread over λ, and exactly one flux quantum Φ₀ = h/2e = 2.068×10⁻¹⁵ Wb through each. Their areal density is nᵥ = B/Φ₀, so at 5 T there are 2.4×10¹⁵ per square metre on a triangular lattice of spacing 22 nm. The material stays superconducting between the cores until they overlap at μ₀H_c2 = Φ₀/2πξ², which is why every high-field magnet is type II: NbTi at 4.2 K works to about 11 T and Nb₃Sn beyond 20 T, while lead has given up at 53 mT.

What the London model does not contain

The London equations are a constitutive guess, and it is worth naming what they do not deliver. nₛ is fitted from the measured λ, not computed; there is no prediction of T_c, no reason the electrons should order, and no gap. The theory is also strictly local — Jₛ at a point is taken to respond to B at that point — and that fails whenever the coherence length exceeds λ, as it does in aluminium and lead, where the measured penetration depth comes out larger than λ_L and needs Pippard's non-local kernel or Ginzburg-Landau instead. The vortex picture has its own gap: it says where flux sits, not whether it stays. A vortex in a current-carrying wire feels a force per unit length J × Φ₀, and a vortex that moves dissipates, so the critical current of a real type II conductor is set by how well defects pin the lattice. That is metallurgy, and it is why the same NbTi alloy can differ tenfold in J_c between two heat treatments. Why the electrons pair at all is the next topic.

Change one variable at a time

Make the relationship visible.

Interactive model

Superconducting density nₛ4.0 10²⁸ m⁻³

Slab thickness d200 nm

Applied field B₀20 mT

Start at nₛ = 0.5 and d = 20 nm: the film is a fraction of a penetration depth thick and expels under 1% of its flux. Now take d to 400 nm — the mid-plane goes field-free while the outer λ of each face still carries the whole field and every screening ampere.

PENETRATION DEPTH λ26.6 nm

MID-PLANE FIELD0.927 mT

FLUX EXPELLED73.5 %

SURFACE CURRENT DENSITY5.98 10¹¹ A m⁻²

Live interpretationPENETRATION DEPTH λ: 26.6 nm. MID-PLANE FIELD: 0.927 mT. FLUX EXPELLED: 73.5 %. SURFACE CURRENT DENSITY: 5.98 10¹¹ A m⁻²

Catch the common trap

Explain before calculating.

Misconception check

The convenient shorthand that a superconductor is a perfect diamagnet, χ = −1 with B = 0 inside, is a large-sample, low-field idealisation, and it breaks in two ordinary situations. It is never exactly true at the surface: the field is B₀ at the face and falls as e^(−x/λ), so a film thinner than about λ is barely screened at all — which is precisely why thin films survive fields many times the bulk B_c, since there is almost no flux to expel and almost no energy to be gained by expelling it. And above B_c1 a type II superconductor holds flux in its bulk, one quantum per vortex, while remaining superconducting everywhere outside the cores; its magnetisation curve leaves the M = −H line at B_c1 and creeps to zero only at B_c2. Perfect diamagnetism is the small-field limit of the Meissner effect in a big sample, not the definition of the state.

A tin sphere is cooled through T_c while sitting in a steady 5 mT field, and the magnet is then switched off. Compare what the tin does with what a hypothetical perfect conductor — a metal whose resistivity simply drops to zero at the same temperature — would do.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA superconductor has nₛ = 4.0×10²⁸ superconducting carriers per cubic metre. Find its London penetration depth, then find the depth at which a 12 mT applied field has fallen to 1% of its surface value. Take m = 9.11×10⁻³¹ kg, e = 1.60×10⁻¹⁹ C and μ₀ = 1.257×10⁻⁶ H m⁻¹.

λ = √(m / μ₀ nₛ e²). Build the denominator first: e² = (1.60×10⁻¹⁹)² = 2.56×10⁻³⁸, so μ₀ nₛ e² = 1.257×10⁻⁶ × 4.0×10²⁸ × 2.56×10⁻³⁸ = 1.287×10⁻¹⁵.
Divide, then take the root: m/(μ₀ nₛ e²) = 9.11×10⁻³¹ / 1.287×10⁻¹⁵ = 7.08×10⁻¹⁶ m², so λ = √(7.08×10⁻¹⁶) = 2.66×10⁻⁸ m = 26.6 nm.
The profile is B(x) = B₀ e^(−x/λ). Setting B/B₀ = 0.010 gives x = λ ln(100) = 26.6 nm × 4.605 = 122 nm.
Notice the applied 12 mT never entered either answer. Within the London model the screening length depends on nₛ alone; doubling B₀ doubles the current everywhere and leaves the shape of the profile untouched.

Answerλ = 26.6 nm, and the field is down to 1% of B₀ at x = 122 nm — about 4.6 penetration depths in. The 12 mT figure is not needed.

MediumA lead wire of radius 0.50 mm is cooled to 4.2 K. Lead is type I with T_c = 7.20 K and B_c(0) = 80.3 mT. Find the critical field at 4.2 K and the current at which the wire quenches. Then find the power a copper wire of the same radius would dissipate per metre carrying that same current, taking ρ_Cu = 1.7×10⁻⁸ Ω m.

Critical field at temperature: B_c(T) = B_c(0)[1 − (T/T_c)²]. Here (4.2/7.20)² = 0.5833² = 0.3403, so B_c = 80.3 mT × (1 − 0.3403) = 80.3 × 0.6597 = 53.0 mT.
Silsbee's rule: the wire quenches when its own surface field μ₀I/2πa reaches B_c, so I_c = 2πa B_c/μ₀ = 2π × 5.0×10⁻⁴ m × 0.0530 T ÷ (1.257×10⁻⁶ H m⁻¹).
Numerator: 2π × 5.0×10⁻⁴ = 3.142×10⁻³, times 0.0530 T = 1.665×10⁻⁴. Dividing by 1.257×10⁻⁶ gives I_c = 132 A.
Copper for comparison: A = πa² = π(5.0×10⁻⁴)² = 7.85×10⁻⁷ m², so one metre has R = ρL/A = 1.7×10⁻⁸ ÷ 7.85×10⁻⁷ = 0.0217 Ω.
P = I²R = 132² × 0.0217 = 17400 × 0.0217 = 378 W for every metre — and that heat is why a copper magnet needs a power station while a lead one, kept below B_c, needs none.

AnswerB_c(4.2 K) = 53.0 mT and I_c = 132 A. The same current in copper of the same radius dissipates about 380 W per metre; in the lead, below B_c, the steady-state dissipation is zero.

HardA niobium-titanium magnet wire operates at 4.2 K in a 5.0 T field. NbTi has λ ≈ 240 nm and ξ ≈ 5.0 nm. Show that it is type II, estimate μ₀H_c2, find the areal density and spacing of the vortex lattice at 5.0 T, and say what actually limits the wire's critical current. Take Φ₀ = 2.068×10⁻¹⁵ Wb.

κ = λ/ξ = 240/5.0 = 48. Since 48 is far above 1/√2 = 0.707, the normal-superconducting interface energy is negative and NbTi is strongly type II: it admits flux rather than going normal at B_c.
Upper critical field: μ₀H_c2 = Φ₀/2πξ² = 2.068×10⁻¹⁵ ÷ (2π × (5.0×10⁻⁹)²) = 2.068×10⁻¹⁵ ÷ 1.571×10⁻¹⁶ = 13.2 T, so working at 5.0 T is comfortably inside the mixed state.
Each vortex carries exactly one Φ₀, so the areal density is nᵥ = B/Φ₀ = 5.0 ÷ 2.068×10⁻¹⁵ = 2.42×10¹⁵ vortices per square metre — 2.42×10⁹ through every square millimetre of cross-section.
For a triangular lattice the area per vortex is (√3/2)a², so a = √(2Φ₀ / √3 B) = √(2 × 2.068×10⁻¹⁵ ÷ (1.732 × 5.0)) = √(4.78×10⁻¹⁶) = 2.19×10⁻⁸ m = 22 nm.
Compare the three lengths: cores about 2ξ = 10 nm across are still well separated at 22 nm spacing, so the metal is superconducting between them, while λ = 240 nm is ten times the spacing, so the screening currents overlap heavily and the internal field is nearly uniform.
What limits I_c is not H_c2 but flux motion. The transport current pushes each vortex with a force per unit length J × Φ₀, and a moving vortex dissipates, so the wire carries current losslessly only while pinning centres — α-Ti precipitates, dislocation cell walls — hold the lattice still.

Answerκ = 48, so type II; μ₀H_c2 ≈ 13 T; at 5.0 T there are 2.4×10¹⁵ vortices m⁻² spaced 22 nm apart. The critical current is set by flux pinning on defects, not by H_c2.