Electromagnetism

Superconductors Explained: How Cold, How Much Field, How Much Current

Definition

A superconductor is a material that, below a critical temperature, carries electric current with no measurable resistance and expels magnetic field from its interior. That state has three limits and crossing any one ends it: the critical temperature, the critical magnetic field, and the critical current that field allows the wire to carry.

Set a current running round a closed loop of superconducting wire, disconnect the battery and walk away. The current keeps going. In a superconducting gravimeter in Belgium one did exactly that for 10467 days — twenty-eight years, seven months and twenty-seven days — with no decay anyone could detect.

Most explanations stop there, at “the resistance falls to zero”. Zero, though, is a limit rather than a measurement, and it leaves the three questions a physicist actually asks unanswered: how cold, how much magnetic field, and how much current. This page answers all three with numbers.

What Is a Superconductor?

A superconductor is a material that, cooled below its critical temperature, carries a current with no measurable resistance and drives magnetic field out of its own interior.

Two properties, not one. Students often learn the first and forget the second, and the second is the one that makes the topic behave strangely. Cooling a perfect conductor would trap whatever field happened to be inside it; a superconductor instead pushes the field out, which is a different and stronger statement.

That expulsion is the Meissner effect, and it is why the topic has a limit at all. Pushing a field out costs energy, and the material only has so much to spend. Turn the applied field up far enough and the superconducting state is simply no longer worth having, so it stops.

The temperature at which it stops with no field applied is the critical temperature, Tc. Lead’s is 7.19 K. That is 7.19 degrees above absolute zero, which is why almost every sentence about this topic quietly assumes a cryostat in the room.

Ordinary metals have no such point. Cool copper and its resistance falls smoothly, flattening out at a residual value set by impurities and never reaching zero. A superconductor does something categorically different at Tc: it changes phase, and Ohm’s law stops having anything to say, because there is no longer a resistance to multiply the current by.

Materials divide into two families. Type-I superconductors, which are mostly pure elements, expel field completely until one critical field destroys the state outright — and everything computed on this page is type-I. Type-II materials, which is what real magnets are wound from, let flux through in quantised threads over a wide band of field and survive far higher fields as a result.

Is a Superconductor’s Resistance Really Zero?

No experiment can measure zero, so the honest answer is an upper bound rather than a value: on the longest measured persistent current, the loop’s resistance is below roughly 1.1 × 10−18 ohms.

Here is where that comes from. A current circulating in a loop of inductance L against a resistance R decays as I(t) = I0e−Rt/L. Rearranging for R gives an inequality, because what an experiment actually establishes is that the current did not change by more than some detectable fraction f.

R ≤ (L/t) × ln(1/(1 − f))
  • R — the upper bound this places on the loop’s resistance, in ohms (Ω).
  • L — the loop’s inductance, in henries (H). A stated assumption, not a measurement.
  • t — how long the current was watched for, in seconds (s). This is the sourced part.
  • f — the largest fractional decay the apparatus could have missed, a pure number. Also a stated assumption.

The 10467 days is the sourced number, from the record of persistent currents in superconductivity. Put it in with a 1 mH coil and a detection floor of one part in a million — both of which are assumptions you are free to change — and the bound lands at 1.10576859 × 10−18 Ω. Problem 8 below does that arithmetic line by line.

Turn it into a resistivity and the scale becomes easier to feel. For a 0.5 mm-radius wire, 100 m of it, that bound is 8.6846862 × 10−27 Ω m. Copper, at the 1.68 × 10−8 Ω m this site publishes in its resistivity calculator, is more resistive than that by a factor of 1.934439 × 1018.

So “zero” is not quite the claim. The claim is bounded, and bounded so far below every other material that the difference has no experimental consequence. Our guide to electrical resistance asks in its own FAQ whether resistance can ever be zero; this is the number that answer was missing.

There is a second reason the decay is not gradual. Flux through a superconducting ring comes in whole multiples of the magnetic flux quantum, and the NIST value of that magnetic flux quantum is 2.067833848 × 10−15 Wb exactly. Since the flux is L times the current, the current in a 1 mH ring can only change in steps of 2.07 pA.

A bigger magnet has a smaller step, because the step is the flux quantum divided by L: for a 100 H coil it is about 20.7 attoamps. There is no smooth way down, which is the real content of “zero resistance”.

Be careful with what that does and does not prove. It removes the gradual path, not the possibility — a large enough fluctuation can still move the whole loop by one quantum. That is precisely why every measured lifetime is quoted as a bound and never as an infinity.

The Superconductor Formulas: Critical Field and Critical Current

Two relations carry all the quantitative content, and they are on very different footings. The first is a fit; the second is a derivation.

Bc(T) = Bc0[1 − (T/Tc)²]
  • Bc(T) — the critical magnetic flux density at temperature T, in tesla (T). Above it the metal is an ordinary conductor.
  • Bc0 — the same quantity extrapolated to absolute zero, in tesla (T). Lead’s is 80 mT.
  • T — the operating temperature, in kelvin (K).
  • Tc — the critical temperature, in kelvin (K), where the critical field reaches zero.

That parabola is empirical. It is the standard fit for type-I elements, good to a few per cent, and nothing on this site derives it; two endpoints and a monotonic fall between them are the whole of its claim.

Notice, though, that it meets the temperature axis at a finite angle rather than flattening into it. Superconductivity does not fade away as you warm towards Tc. There is a last field it will hold, right up to the end.

The second relation turns that field into a current, and it is the one exact step in the chain.

Ic = 2πaBc(T)/μ0
  • Ic — the critical current, in amperes (A). Above it the wire goes normal.
  • a — the wire’s radius, in metres (m). Not its diameter, and this is the commonest slip in the whole topic.
  • Bc(T) — the critical field at the working temperature, in tesla (T), from the parabola above.
  • μ0 — the permeability of free space, 1.25663706212 × 10−6 N/A² here, in newtons per ampere squared.

This is Silsbee’s rule, and it is worth saying plainly that it is derived rather than fitted. It inherits the parabola’s few-per-cent error and adds none of its own, which also means a critical current quoted to six figures is precise rather than accurate. Rearranged, the same relation answers a designer’s questions instead of a student’s: how cold, how thin, or what field a target current needs. Our superconductor critical field calculator solves it in any of those directions and prints the critical field, the current density and the penetration depth beside the answer.

One consequence of the formula is easy to miss and worth pinning down now. The wire’s own current puts a magnetic field on its own surface, and the material cannot tell that field apart from one you apply from outside.

Run 100 A through a 0.5 mm-radius lead wire in no applied field, or run no current at all through the same wire sitting in a 40 mT field, and the metal sees the identical surface field and has the identical margin left. Both cases eat the same budget. An applied field is therefore not a separate hazard from a large current — it is the same hazard, arriving from a different direction.

The simulator below is that budget, made adjustable. Set the material’s two numbers, the temperature, the applied field, the transport current and the wire radius, and it plots the operating point against the critical-field curve, draws the field’s decay into the metal, and tells you which side of the boundary you have landed on.

Superconductor Lab

Here is that boundary on its own, drawn once for lead. The simulator’s main panel is this same picture with an operating point loose inside it.

The critical field of lead plotted against temperature. The horizontal axis is temperature from 0 to 8 kelvin; the vertical axis is critical magnetic flux density from 0 to 90 millitesla. A dark curve falls from 80 millitesla at absolute zero to zero at the critical temperature of 7.19 kelvin, meeting the temperature axis at a finite angle rather than flattening into it. The area under the curve is washed in gold and labelled SUPERCONDUCTING, field expelled, no resistance; the area above and to the right of the curve is left plain and labelled NORMAL, an ordinary resistive metal. Wine dashed guides at 4.2 kelvin meet the curve at 52.70 millitesla and that crossing is marked and labelled. Two further marks label Bc0 equals 80 millitesla at 0 kelvin on the vertical axis and Tc equals 7.19 kelvin on the horizontal axis. Notes state that the curve is the empirical parabolic fit, a standard fit for type-I elements good to a few per cent, that it is not derived anywhere on the page, and that lead’s two numbers come from a retrieved table.
One curve, two regions, and every argument on this page is about where the boundary sits.

How Silsbee’s Rule Follows from Ampère’s Law

The derivation is two lines, and following it is the difference between quoting the formula and understanding what it claims.

Take Ampère’s law round a circle drawn on the surface of a round wire of radius a carrying current I. The enclosed current is I and the path length is 2πa, so B × 2πa = μ0I, giving a surface field of B = μ0I/(2πa). No approximation has been made; this is exact for a round wire.

Now add the physics. The material expels field until the field at its own surface reaches Bc(T), and at that point it gives up. Set μ0Ic/(2πa) = Bc(T), rearrange, and Silsbee’s rule drops out with no fitted constant anywhere in it.

The arithmetic hands you a shortcut worth keeping. Because μ0/(2π) is 2.000000001089 × 10−7, the surface field of a round wire in millitesla is very nearly 0.2 × I(A) / a(mm) — so 100 A in a 0.5 mm-radius wire is 40 mT, done in your head.

Do not call that an identity. It was one before 2019, when μ0 was defined as 4π × 10−7; the SI redefinition made the permeability a measured quantity, and NIST’s current vacuum magnetic permeability agrees with the value used here to nine significant figures. Excellent approximation, not an equality.

Look at what the rule says about size and something counter-intuitive falls out. The critical current is proportional to the radius, not to the area, because the quantity that kills the wire lives on the surface. Four times as much metal buys exactly twice the current.

Two stacked line charts of the critical current and the critical current density of a lead wire at 4.2 kelvin against its radius. Both panels share one horizontal axis, wire radius from 0 to 2.1 millimetres, and each has its own vertical scale in its own units. Panel A, critical current in amps from 0 to 600, holds a dark straight line rising in proportion to the radius, starting not at the origin but at the left-hand end of the plotted range, a equals 0.050 millimetres. Two marked points are keyed in a legend as a equals 0.50 millimetres, Ic equals 131.8 amps, and a equals 1.00 millimetres, Ic equals 263.5 amps, twice the current for four times the metal. Panel B, critical current density in amps per square millimetre from 0 to 1800, holds a gold curve falling steeply as the radius grows; its legend reads a equals 0.50 millimetres, Jc equals 167.8 amps per square millimetre, and a equals 1.00 millimetres, Jc equals 83.9 amps per square millimetre, half of it, with a note that a equals 0.050 millimetres is the curve’s left-hand end, the highest density the panel shows, at 1677.6 amps per square millimetre. Notes state that doubling the radius doubles the current and halves the density to machine precision because the quantity that kills the wire lives on its surface, that the panels share the radius axis and nothing else so no vertical distance in one may be compared with a vertical distance in the other, and that both curves are Silsbee’s rule for lead at 4.2 kelvin where the critical field is 52.70 millitesla.
Double the radius and the current doubles while the current density halves. Both lines are the same rule, read two ways.

The copper comparison has to name which copper rule it is contrasting, because there are two. At a fixed current density, copper’s ampacity goes as the area and therefore as the square of the radius, and that is the rule of thumb being overturned here. At a fixed watts-per-metre limit, copper scales with the radius too, so the contrast would be no contrast at all.

The same result says something blunter about the middle of a superconducting wire: it does nothing. Silsbee’s rule sees only the outer radius, so a tube of the same outer radius carries exactly the same critical current as a solid rod, on a small fraction of the metal.

Problem 7 works the figure out. A 0.5 mm-radius tube with a 1 µm wall matches the solid rod on 0.3996 per cent of the lead, a saving of over 250 times, which is why real superconducting cable is thousands of fine filaments in a normal-metal matrix rather than one thick rod. The wall cannot shrink for ever, and the next section says exactly where it has to stop.

Critical Temperature and Critical Field of the Type-I Elements

Eleven elements carry both numbers in the table retrieved for this cluster, and they are the complete legal material space for every calculation on this page.

Element Critical temperature Tc (K) Critical field at 0 K, Bc0 (mT)
Lead (Pb)7.1980
Tantalum (Ta)4.4890
Mercury (α-Hg)4.1540
Tin (Sn)3.7230
Indium (In)3.430
Thallium (Tl)2.3920
Aluminium (Al)1.2010
Gallium (Ga)1.0835.8
Zinc (Zn)0.8555
Cadmium (Cd)0.522.8
Titanium (Ti)0.3910

Read the two columns against each other and the table says something a single ranking would hide. Tantalum has a lower critical temperature than lead, 4.48 K against 7.19 K, and yet a higher critical field, 90 mT against 80 mT.

Titanium makes the same point harder. It tolerates the same 10 mT as aluminium while transitioning at 0.39 K, barely a third of aluminium’s 1.20 K. Ranking materials by critical temperature alone is therefore wrong, and these numbers say so rather than an opinion saying so.

Real magnets are not wound from any of them. They use type-II materials, whose field column is a genuinely different quantity: the upper critical field Bc2, above which superconductivity finally ends, with a broad band below it in which flux threads through the material and the state survives anyway. Neither the parabola nor Silsbee’s rule applies to the rows below, and this page never applies them there.

Type-II material, for context only Critical temperature Tc (K) Upper critical field Bc2 (T)
Niobium (Nb)9.260.82
Niobium–titanium (NbTi)1015
Niobium–tin (Nb3Sn)18.330
Magnesium diboride (MgB2)3974
YBCO cuprate95120 to 250 (a range)

YBCO gets a range and not a number because the source gives a range, and inventing a midpoint would be the kind of tidy-looking fiction this page exists to avoid.

The last column of that table is also the whole reason the cuprates caused a fuss. Liquid nitrogen boils at 77.36 K and liquid helium at 4.22 K, two figures our article on absolute zero and the temperature scale already publishes. On the sourced numbers exactly one row here — the 95 K cuprate — can be cooled with nitrogen. Everything else, magnesium diboride included, needs something colder than nitrogen.

Where This Superconductor Model Stops Working

Three things break this model, and each of them has a measurable boundary rather than a vague warning. A fourth heading below simply lists what was never sourced at all.

The parabola is a fit, and so is the penetration depth

Every critical field on this page rests on an empirical parabola that nothing here derives, and the same goes for the temperature dependence of the penetration depth, λ(T) = λ0/sqrt(1 − (T/Tc)4). Both are standard fits for type-I elements. Silsbee’s rule sitting on top of them is exact, so the answers carry the fits’ error and no more.

Why a thin enough wire breaks it

Share the critical current out over the whole cross-section and you have a density, Jc = 2Bc/(μ0a). Shrink a towards nothing and that expression climbs without bound, promising any current density you care to name from a fine enough filament. No metal can honour that, so somewhere the arithmetic has run past the physics.

Where it runs past it is fixed by the only length a superconductor owns, the penetration depth. Screening the critical field across a depth λ needs a density of about Bc/(μ0λ), and that bare field scale carries no fitted coefficient of any kind.

Set the two equal and the crossing falls at a = 2λ exactly, independently of material, field and temperature. Taking 100 nm as the order of magnitude for λ, a 1 mm-diameter wire sits 2500 times below that scale, which is comfortable. The 5 µm filaments inside real cable sit only about 12.5 times below it, which is precisely where a page should stop quoting the rule as if it were the answer.

A line chart of magnetic flux density against depth below the surface of a lead wire at 4.2 kelvin. The horizontal axis is depth from 0 to 560 nanometres; the vertical axis is flux density from 0 to 56 millitesla. A grey strip to the left of the axis is labelled vacuum, and a thick dark line marks the metal surface. A dark curve starts at 52.70 millitesla at the surface and decays exponentially towards zero. Three marked points are keyed in a legend: the surface value 52.70 millitesla at zero depth; one penetration depth at 106.4 nanometres, where the field is 19.39 millitesla, which is one over e of the surface field and is picked out by the left dashed guide; and the depth at which only 1 per cent is left, 489.9 nanometres, picked out by the right dashed guide. Notes state that the wire is 1.00 mm across so the frame is its outermost 560 nanometres, that the drawing is not to the wire’s scale because on a shared scale the decay would be thinner than the axis line, that the penetration depth is the sourced order of magnitude of 100 nanometres warmed to 4.2 kelvin by the empirical two-fluid form, and that no material here is given a penetration depth of its own.
The field does not stop dead at the metal’s face. It fades over a few hundred nanometres, and that depth is the only length this model owns.

The penetration depth runs away at the critical temperature

Silsbee’s rule quietly assumes the current hugs a surface, and λ(T) diverges at Tc, so the rule must die once the screening depth reaches the wire’s radius. Solving for that temperature gives a window, and for a wire you can hold the window is absurdly thin: for a 0.5 mm-radius lead wire it is 71.9 nanokelvin wide.

The divergence is real and, at that width, irrelevant. For a 2.5 µm-radius filament the same window opens to 2.88 mK, about forty thousand times wider, and that is the direction that matters.

What this cluster did not source, and therefore never states

No material on this page is given a penetration depth of its own: 100 nm is the published order of magnitude for most superconductors, and it is used as an order and nothing finer. No coherence length, carrier density, molar mass or low-temperature heat capacity appears for any material either, because none was sourced.

Nor is anything claimed here about superconductivity at or near room temperature. Nothing was sourced for it, so nothing is said.

Real-World Examples of Superconductors

The state turns up in three places most readers have already met: a lecture bench, a hospital corridor, and a very long spool of wire.

A magnet hovering above a chilled superconductor, the demonstration most people have seen of the effect
The magnet is held by flux pinned inside the cooled disc, which is a type-II mechanism and outside the model on this page.

The floating magnet on the lecture bench

A small magnet hanging in mid-air above a chilled disc is the image nearly everyone has of this topic, and it is a real effect. It is not, however, the effect this page computes. The magnet is held by flux pinned inside a type-II material, which is a different mechanism from the complete field expulsion modelled here, so no number on this page describes that demonstration.

Scanner magnets, and why they are not made of lead

A medical scanner’s bore is a very large superconducting solenoid carrying an enormous current with no resistance, and our article on the electromagnet already notes that tens of tesla are a superconducting speciality. Those coils are type-II, for a reason the energy makes obvious.

The condensation energy density of lead, the free-energy advantage that makes the superconducting state worth having, is 2546.5 J/m³ at absolute zero. A 1 T field carries 397887.4 J/m³, which is 156.25 times as much, and that ratio is exactly (1 T / 80 mT) squared. A lead coil would destroy itself long before it became a useful magnet.

That advantage is also startlingly small in thermal terms. A cubic centimetre of lead holds 2.546479 mJ of it and weighs 11.34 g; at lead’s room-temperature specific heat that is the heat needed to warm it by 1.754 mK. Heat capacity falls steeply on cooling, so the true figure at 4 K is larger than that — the honest statement is a lower bound, because no low-temperature heat capacity was sourced here.

Cable, and the price of resistance in watts

Here is what the whole business buys, in units an electrician would recognise. A 1 mm-diameter lead wire at 4.2 K carries 131.8 A with nothing dissipated at all. The same wire in copper, carrying the same current, burns 371.3 W for every metre of it — roughly a hot plate per metre of cable.

Double the radius and something neat happens. The critical current doubles, and copper’s loss at that new current is unchanged at 371.3 W/m, because the current squared quadruples exactly as the resistance per metre quarters. Silsbee’s rule and Joule’s law move in lockstep there.

Common Misconceptions About Superconductors

“A superconductor has exactly zero resistance”

It has no measurable resistance, which is a different and weaker claim, and the difference is the whole of section two. The strongest statement any experiment supports is a bound — about 1.1 × 10−18 Ω on the stated assumptions — not a zero.

“The critical temperature is the number that matters”

Tantalum quenches at a lower temperature than lead and survives a stronger field. Titanium matches aluminium’s field tolerance from a third of the temperature. A material has at least two numbers, and a design that ignores the field column will fail in a magnet while looking fine on paper.

“A thicker superconducting wire carries proportionally more current”

Four times the metal buys twice the current, because the critical current follows the circumference rather than the area. The rule of thumb being overturned is copper’s fixed-current-density one, not copper’s thermal limit, which also scales with the radius.

“A modest applied field costs a modest slice of the current”

It does not. Because the critical current is linear in the headroom left over, the fraction of the rating a field costs is exactly the fraction of the critical field it takes. Problem 4 puts a 30 mT field on lead at 4.2 K and watches 56.9 per cent of the current rating disappear.

“A persistent current runs for ever”

It runs until something moves the trapped flux by one quantum, and nothing on this page rules that out. Flux quantisation removes the smooth path downward; it does not remove the possibility. Twenty-eight years of no detectable change is a magnificent bound and still a bound.

Worked Problems

Problem 1
Lead has a critical temperature of 7.19 K and a critical field of 80 mT extrapolated to absolute zero. What field will it still expel at 4.2 K?
Show Solution
Solution: Step 1: Use the empirical parabolic law, Bc(T) = Bc0[1 − (T/Tc)²], with T = 4.2 K, Tc = 7.19 K and Bc0 = 80 mT. Step 2: Form the reduced temperature squared as a ratio of squares, which avoids rounding the ratio itself: (T/Tc)² = 4.2² / 7.19² = 17.64 / 51.6961 = 0.341224967. Step 3: Subtract from one and scale: 1 − 0.341224967 = 0.658775033, so Bc = 80 mT × 0.658775033 = 52.702003 mT. Answer: Bc(4.2 K) = 52.702003 mT A tool rounding fields to two decimal places shows that as 52.70 mT. Keep the long form for any further arithmetic; problem 4 explains why.
Problem 2
A lead wire 1 mm in diameter carries 50 A. What magnetic field does that current put on the wire's own surface?
Show Solution
Solution: Step 1: Ampère’s law round the wire surface gives B = μ0I/(2πa). The radius is half the diameter, so a = 0.500 mm = 5.00 × 10−4 m. Step 2: Top: μ0I = 1.25663706212 × 10−6 × 50 = 6.2831853106 × 10−5. Bottom: 2πa = 2π × 5.00 × 10−4 = 3.14159265359 × 10−3. Step 3: Divide: 6.2831853106 × 10−5 / 3.14159265359 × 10−3 = 2.00000000109 × 10−2 T = 20.0000000 mT. Answer: 20.0000000 mT, which a two-decimal display shows as 20.00 mT The mental shortcut agrees: 0.2 × 50 / 0.5 = 20. It is very nearly exact, and the tail of digits above is the gap between the measured permeability and the pre-2019 defined one.
Problem 3
Same wire, same temperature, no applied field. How much current can it carry before it goes normal?
Show Solution
Solution: Step 1: Apply Silsbee’s rule, Ic = 2πaBc(T)/μ0, with Bc(4.2 K) = 52.702003 mT = 0.052702003 T from problem 1. Step 2: Top: 2πa × Bc = 3.14159265359 × 10−3 × 0.052702003 = 1.6556822545 × 10−4. Step 3: Divide by the permeability: 1.6556822545 × 10−4 / 1.25663706212 × 10−6 = 131.755007 A. Answer: Ic = 131.755007 A, which a one-decimal display shows as 131.8 A Sanity check on the magnitude: that is more current than any domestic circuit carries, down a wire a millimetre across, with nothing at all to dissipate.
Problem 4
The same wire now sits in an applied field of 30 mT. What is left of its current rating, and what fraction has the field cost?
Show Solution
Solution: Step 1: The applied field and the wire’s self-field draw on one budget, so the headroom left for the self-field is Bc(T) minus the applied field: 52.702003 − 30 = 22.702003 mT. Step 2: Put that headroom through Silsbee’s rule. Top: 3.14159265359 × 10−3 × 0.022702003 = 7.1320445847 × 10−5. Divide: 7.1320445847 × 10−5 / 1.25663706212 × 10−6 = 56.755007 A. Step 3: The fraction lost is the field fraction, because the current is linear in the headroom: 30 / 52.702003 = 0.56923833, which is 56.923833 per cent. Answer: 56.755007 A remains, and the 30 mT field has cost 56.923833 per cent of the field-free rating One caution, and it is the reason this page prints long figures. Do not re-derive any of this from the rounded 52.70 mT: that route gives 131.7 A in problem 3 where the exact chain gives 131.8 A. Round once, at the end, and never from a printed figure.
Problem 5
Compare a lead wire 1 mm in diameter with one 0.1 mm in diameter, both at 4.2 K. What happens to the critical current and to the critical current density?
Show Solution
Solution: Step 1: The critical current is proportional to the radius, so a tenth of the radius carries a tenth of the current: 131.755007 / 10 = 13.1755007 A. Step 2: Current density is the current divided by the full cross-section. For the thin wire, a = 0.0500 mm, so πa² = π × 0.0500² = 7.85398163 × 10−3 mm². Step 3: Divide: 13.1755007 / 7.85398163 × 10−3 = 1677.55685 A/mm². The thick wire, on π × 0.500² = 0.785398163 mm², gives 131.755007 / 0.785398163 = 167.755685 A/mm². Answer: 1677.55685 A/mm² in the thin wire against 167.755685 A/mm² in the thick one A tenth of the current at ten times the density, and both factors of ten are exact. This is the trade real cable designers are making.
Problem 6
A copper wire of the same 1 mm diameter carries the 131.755007 A of problem 3. How much power does each metre of it dissipate?
Show Solution
Solution: Step 1: Resistance per metre is R/l = ρ/A with ρ = 1.68 × 10−8 Ω m. The cross-section is A = πa² = π × (5.00 × 10−4)² = 7.853981634 × 10−7 m². Step 2: So R/l = 1.68 × 10−8 / 7.853981634 × 10−7 = 0.0213904244 Ω/m, and the current squared is 131.755007² = 17359.3819. Step 3: Multiply: P/l = 17359.3819 × 0.0213904244 = 371.3245 W/m. Answer: 371.3245 W/m in copper, against 0 W/m in the superconductor That figure is not a saving on a bill. It is the entire price of resistance for this wire, since the lead one dissipates nothing whatsoever at that current.
Problem 7
Silsbee's rule depends only on the outer radius. How much lead does a tube of 0.500 mm outer radius with a 1 micrometre wall save over a solid rod of the same critical current?
Show Solution
Solution: Step 1: The metal fraction of an annulus is 1 − (1 − w/a)², where w is the wall thickness and a the outer radius. Step 2: w/a = 1.00 × 10−6 / 5.00 × 10−4 = 0.002, so (1 − 0.002)² = 0.996004 and the fraction is 1 − 0.996004 = 0.003996, which is 0.3996 per cent. Step 3: The saving is the reciprocal: 1 / 0.003996 = 250.250250. Answer: 0.3996 per cent of the metal, a saving of 250.250250-fold That 1 µm wall is ten penetration depths at the sourced order, which is what makes it defensible at all. Thin it to a single penetration depth and the metal fraction falls to 0.0399960 per cent — but that wall sits under the 2λ limit of the section above, so it is not a design anyone should build out of this arithmetic.
Problem 8
A current in a 1 mH superconducting coil is unchanged to one part in a million after the 10467 days of the Belgian gravimeter record. What upper bound does that put on the coil's resistance?
Show Solution
Solution: Step 1: From I(t) = I0 exp(−Rt/L), the bound is R ≤ (L/t) × ln(1/(1 − f)). The 10467 days are sourced; the 1 mH and the one part in a million are stated assumptions, and they are yours to change. Step 2: Convert and evaluate the pieces. t = 10467 × 86400 = 9.043488 × 108 s, so L/t = 1.00 × 10−3 / 9.043488 × 108 = 1.1057680399 × 10−12, and ln(1/(1 − 1 × 10−6)) = 1.0000005000 × 10−6. Step 3: Multiply: 1.1057680399 × 10−12 × 1.0000005000 × 10−6 = 1.10576859 × 10−18 Ω. Answer: R ≤ 1.10576859 × 10−18 Ω Change either assumption and the bound moves with it, which is the point: a tighter detection floor buys a smaller number, and a bigger coil a larger one. The experiment bounds the resistance; it never measures a zero.

Frequently Asked Questions

What is a superconductor in simple terms?
A superconductor is a material that carries electricity with no measurable resistance once it is cold enough, and that pushes magnetic field out of its own interior while it does so. Warm it past its critical temperature, or put it in a strong enough magnetic field, or push too much current through it, and it reverts to being an ordinary resistive metal.
What is the critical temperature of a superconductor?
The critical temperature is the temperature above which a material stops superconducting even with no magnetic field applied. It varies enormously: among the type-I elements measured here, lead is highest at 7.19 K and titanium lowest at 0.39 K. The type-II cuprates reach 95 K, which is the only entry in this page’s tables that liquid nitrogen at 77.36 K can cool.
Does a superconductor really have zero resistance?
It has no resistance anyone has been able to measure, which is not the same as zero. A current in a Belgian gravimeter coil ran for a sourced 10467 days with no detectable decay; assuming a 1 mH coil and a detection floor of one part in a million, that bounds the loop’s resistance below 1.10576859 times ten to the minus eighteen ohms. A bound, not a measurement.
Why does a magnet float above a superconductor?
Because flux threads into a type-II disc and is pinned there, and the pinned flux holds the magnet at the height it was left at. That is the mechanism of the familiar bench demonstration, and it sits outside the type-I model used on this page. A type-I superconductor in the Meissner state expels the field instead and simply repels a magnet, which is a different effect and the one this page models.
How much current can a superconducting wire carry?
As much as its own magnetic field allows. A wire goes normal when the field at its surface reaches the critical field, which gives Ic = 2 pi a Bc(T) / mu0: a lead wire 1 mm in diameter at 4.2 K carries 131.8 A. Note that the current follows the radius, not the area, so four times the metal buys only twice the current.
Why do superconductors need liquid helium or liquid nitrogen?
Because their critical temperatures are far below anything a refrigerator reaches. Liquid helium, boiling at 4.22 K, is the coolant every element in the type-I table needs, and nine of the eleven transition below even that, so they need a pumped or dilution stage under the bath as well. Liquid nitrogen, boiling at 77.36 K, is cheaper and far easier to handle but useless for all eleven; it cools only the 95 K cuprate.
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