Wire inductance calculator: straight wire, flat strip and parallel pair

The self-inductance of a straight round wire, a flat strip, bar or trace with no return path near it, and a loop of two parallel wires, in millimetres or mils. The formulas are the Bureau of Standards' own, from Rosa and Grover's Scientific Paper 169: the full expression for a cylindrical wire, the geometric-mean-distance method for a rectangular section, and the rectangle formula for a return circuit with its ends, with Kelvin's correction for the field inside the wire at a frequency. The default, 10 mm of 0.5 mm wire, is 7.314 nH; the paper's 10 m of 2 mm wire is 18.31 µH, its 18307.0 cm.

Fig 1 — Round wire: l = 10 mm, d = 0.5 mm. Below: L at low frequency against length with the cross-section held, 7.314 nH at the length entered. Ten times the length gives 118.7 nH, 1.62 times ten times as much: the dashed line is strict proportion. The curve climbs above it because inductance per unit length grows with length.

Conductor

Length unit

Length l (mm)

Wire diameter d (mm)

Relative permeability of the wire

Frequency f (MHz)

Self-inductance L, low frequency: formula (94), uniform current: 7.314 nH
Inductance per unit length: 0.7314 nH/mm
Field outside the wire: the limit at high frequency, (94) less μ·l/2: 6.814 nH
Field inside the wire, μ·l/2: 0.5 nH
Rosa's approximate form (95), 2l[log(2l/ρ) − 3/4]: 7.264 nH
Length over radius, l/ρ: 40

The approximate (95) is 0.7 % below the full formula (94) here: it drops terms in ρ/l, which matter for a wire only 40 radii long. The calculator uses (94).

At high frequency the field inside the wire goes, and L falls towards 6.814 nH, 6.8 % less. Enter a frequency to see how far it has gone.

This is the inductance of a straight piece of a circuit. The current comes back somewhere, and the loop it makes sets the inductance the circuit sees: the parallel-pair geometry is one such loop.

How this is calculated

Standard: Rosa and Grover, Formulas and Tables for the Calculation of Mutual and Self-Inductance, Bulletin of the Bureau of Standards Scientific Paper No. 169 (revised 1916), sections 8–10: formulas (94)–(101), (103), (104), (107), (113), (114), (123)–(126), (144), (154) and (157)–(159), the geometric mean distance table on p. 168 and Table XXIII

$L = 2\left[\,l\log\frac{l + \sqrt{l^2 + \rho^2}}{\rho} - \sqrt{l^2 + \rho^2} + \frac{l}{4} + \rho\right]$: Formula (94), p. 150: "The self-inductance of a length l of straight cylindrical wire of radius ρ", in the paper's units, lengths and L in centimetres, log natural. The l/4 is the field inside the wire; the calculator writes it μ·l/4 for a wire of relative permeability μ, as (96) does. This is the calculator's round-wire answer.
$L = 2l\left[\log\frac{2l}{\rho} - \frac{3}{4}\right], \qquad L = 2l\left[\log\frac{2l}{\rho} - 1 + \frac{\mu}{4}\right], \qquad L = 2l\left[\log\frac{2l}{\rho} - 1\right]$: Formulas (95), (96) and (97), pp. 150–151: (94) "approximately", the same "Where the permeability of the wire is μ, and that of the medium outside is unity", and the limit "for alternating currents of great frequency, when there is no magnetic field within the wire", "obtained by subtracting from (95) l/2 or from (96) μ l/2, the magnetic flux within the conductor due to unit current." (95) drops terms in ρ/l from (94); the calculator shows it beside (94).
$L' = 2l\left[\log\frac{2l}{\rho} - 1 + \frac{\mu}{4}\left(\frac{4}{x}\frac{Z}{Y}\right)\right], \qquad x = 2\rho\sqrt{\frac{\pi p \mu}{\sigma}} = \frac{\sqrt{2}\,r}{\delta}$: Kelvin's solution, formula (144), p. 174, with p = 2πf, σ the specific resistance and Z, Y the combinations of ber and bei in (144a); x in SI is √2 times the radius over the skin depth. The factor 4Z/(xY) is 1 at low frequency and tends to 2√2/x, (153). The calculator applies it to the internal part of (94), evaluating it as the skin depth calculator does, which is tested against Table XXII.
$L = 2l\left[\log\frac{2l}{R} - 1 + \frac{R}{l}\right], \qquad L = 2l\left[\log\frac{2l}{\alpha + \beta} + \frac{1}{2} + \frac{0.2235(\alpha + \beta)}{l}\right]$: Formulas (103) and (104), pp. 152–153: a straight bar of rectangular section α × β is "the same as the mutual inductance of two parallel straight filaments of the same length separated by a distance equal to the geometrical mean distance of the cross section of the bar." R is that distance. (104) puts R = 0.2235(α + β) and rounds −log 0.2235 − 1 = 0.4983 to 1/2. The calculator uses (103) with R from (124), and shows (104).
$\log R = \log\sqrt{a^2 + b^2} - \frac{1}{6}\frac{a^2}{b^2}\log\sqrt{1 + \frac{b^2}{a^2}} - \frac{1}{6}\frac{b^2}{a^2}\log\sqrt{1 + \frac{a^2}{b^2}} + \frac{2}{3}\frac{a}{b}\tan^{-1}\frac{b}{a} + \frac{2}{3}\frac{b}{a}\tan^{-1}\frac{a}{b} - \frac{25}{12}$: Maxwell's formula (124), p. 167: the geometric mean distance of a rectangular area of sides a and b from itself. For a square it gives R = 0.44705a (125); as one side vanishes, R = 0.22313a, the line (123); for a circle of radius a, (126) gives 0.7788a, which in (103) turns it into (95). The tests check (124) against the table on p. 168.
$L = 2l\left[\log\frac{2l}{b} + \frac{1}{2}\right], \qquad L = 2l\left[\log\frac{2l}{b} - \frac{a}{b} + \frac{1}{2}\right]$: Formulas (113) and (114), p. 156: "a straight, thin tape of length l and breadth b (and of negligible thickness)", and with a thickness a. "A closer approximation to L is given by (104)". The calculator shows (113) for the width entered as the thin-tape bound.
$M = 2\left[\,l\log\frac{l + \sqrt{l^2 + d^2}}{d} - \sqrt{l^2 + d^2} + d\right] \;\approx\; 2l\left[\log\frac{2l}{d} - 1 + \frac{d}{l}\right]$: Formulas (98) and (99), p. 151: the mutual inductance of two parallel wires of length l a distance d apart, "an exact expression when the wires have no appreciable cross section", and its approximation "when the length l is great in comparison with d". (94) less its internal term is (98) with d = ρ.
$L = 4l\left[\log\frac{d}{\rho} + \frac{\mu}{4} - \frac{d}{l}\right], \qquad L = 4l\left[\log\frac{d}{\rho} + \frac{1}{4}\right]$: Formulas (100) and (101), pp. 151–152: a return circuit of two parallel wires "neglecting the effect of the end connections", and "In the usual case of μ = 1 … when d/l is small". On the form d is the spacing s, centre to centre, and ρ half the wire diameter. The per-length row is (101) with the μ/4 term, divided by l.
$L = 4\left[(a + b)\log\frac{2ab}{\rho} - a\log(a + g) - b\log(b + g) - \frac{7}{4}(a + b) + 2(g + \rho)\right], \qquad g = \sqrt{a^2 + b^2}$: Formula (107), p. 155, the rectangle of round wire, with the paper's diagonal d written g here; "If the end effect is large, as when the wires are relatively far apart, use the expression for the self-inductance of a rectangle below (107)" (p. 152). The calculator's loop is (107) with a = l and b = s. Its internal part, a + b, is replaced by μ(a + b)·4Z/(xY) for a magnetic wire or a frequency; that substitution is made here, as (157) makes it for the pair, and is not printed.
$L' = 4l\left[\log\frac{d}{\rho} + \frac{\mu}{4}\left(\frac{4}{x}\frac{Z}{Y}\right)\right], \qquad \frac{\Delta L}{L} = -\frac{1 - \frac{4}{x}\frac{Z}{Y}}{4\log\frac{d}{\rho} + 1}, \qquad \left(\frac{\Delta L}{L}\right)_{x = \infty} = -\frac{1}{4\log\frac{d}{\rho} + 1}$: Formulas (157), (158) and (159), p. 180, for two parallel wires at a frequency; the single wire's limit is (154), −1/(4 log(2l/ρ) − 3). Table XXIII (p. 229) tabulates both limits. Example, p. 180: "With d = 1 cm and ρ = 0.1 cm, and with a frequency of 10⁶, (158) gives ΔL/L = −8.5 per cent."
$1\ \text{H} = 10^9\ \text{cm}, \qquad L_{\text{H}} = \frac{\mu_0}{4\pi}\,L(\text{lengths in m}) = 10^{-7}\,L(\text{lengths in m})$: Every formula above gives L in centimetres from lengths in centimetres; the examples convert at "18307.0 cm = 18.307 microhenrys" (p. 160). With lengths in metres instead the bracket is 100 times larger and 10⁻⁹ × 100 = 10⁻⁷ H/m, which is μ0/4π. So 2l[…] is (μ0/2π)·l[…] and 4l[…] is (μ0/π)·l[…] in henries.

Assumptions

Straight conductors in air, with no magnetic material or other conductor near them. A bend, a nearby plane or a nearby trace changes the answer.
Uniform current over the section at low frequency: "Excepting in a very few specified cases, the formulas of the preceding sections apply only to conductors carrying direct current or alternating currents of frequencies so low that the error, due to the assumption that the current is uniformly distributed over the cross section of the wire, is negligible" (p. 172). For a round wire and a pair, a frequency entered applies Kelvin's correction to the internal field; for a strip the paper gives none, and none is applied.
A single wire or strip is an open conductor: its self-inductance is a share of the inductance of whatever loop it is part of, not the inductance of a circuit.
The parallel pair is two equal round wires joined at both ends into a rectangle; proximity effect between the wires is not included, which Rosa and Grover found negligible at d/ρ = 10 and 1 MHz.
A magnetic wire has a constant permeability.
The defaults are illustrative, not from the paper: 10 mm of 0.5 mm wire, a 10 mm trace 0.25 mm wide in 1 oz copper, and a pair of the same wire 2 mm apart.

What sets the inductance of a straight wire

A current in a wire sets up a magnetic field around it, in rings, and inside it. The self-inductance is the flux that field links per ampere. For a straight round wire of length l and radius ρ, Rosa and Grover's Bureau of Standards paper gives it as formula (94) (p. 150), "The self-inductance of a length l of straight cylindrical wire of radius ρ", and, for a wire long against its radius, the shorter form (95), 2l[log(2l/ρ) − 3/4], with every length in centimetres and L in centimetres too, the old electromagnetic unit: 1 µH is 1000 cm. The calculator works in SI, which multiplies the same bracket by 10⁻⁷ H/m; the conversion is the last entry under How this is calculated.

Two things follow from the shape of the formula. First, L depends on the radius only through a logarithm, so the wire's thickness matters much less than its length: halving the diameter of the default 10 mm of 0.5 mm wire takes it from 7.31 nH to 8.68 nH, while doubling the length takes it to 17.35 nH. Second, L is not proportional to length. The bracket grows with log l, so a longer wire has more inductance per millimetre: 0.313 nH/mm for 1 mm of the 0.5 mm wire, 0.731 nH/mm for 10 mm, 1.187 nH/mm for 100 mm. A figure in nH per millimetre belongs to one length and one diameter and is wrong for any other; the chart under the calculator shows how far the curve climbs above strict proportion over two decades of length.

The formula has two parts. Most of it is the field outside the wire. The rest, l/2 in the paper's units, is the field inside the conductor, which (97) removes: "This is obtained by subtracting from (95) l/2 or from (96) μ l/2, the magnetic flux within the conductor due to unit current" (p. 151). That internal part is the only part a wire's permeability touches, which is why (96) carries μ/4 and why an iron wire's inductance is dominated by it. For the default wire it is 0.500 nH of 7.314 nH,6.8 %. It is also the part that skin effect takes away at high frequency, below.

A single straight wire is not a circuit. Its self-inductance is a share of the inductance of whatever loop the current makes, with mutual inductances to every other piece of that loop. The paper makes the point with its Example 73 (pp. 161–162): a 1 cm wire of 1 mm radius has4.4915 cm of self-inductance, "which is a little more than one-half of the mutual inductance of AB and BC, BC being one thousand times the length of AB." The parallel-pair geometry is the simplest whole loop.

Wire inductance by length and diameter

The calculator's formula, (94), for copper at low frequency. Cells where the length is under four diameters are left out: a conductor that short is a stub, not a wire.

Diameter	2 mm	5 mm	10 mm	20 mm	50 mm	100 mm	1 m
1 mil, 0.0254 mm	2.004 nH	5.921 nH	13.23 nH	29.22 nH	82.22 nH	178.3 nH	2.243 µH
0.1 mm	1.463 nH	4.558 nH	10.49 nH	23.75 nH	68.52 nH	150.9 nH	1.969 µH
0.25 mm	1.111 nH	3.657 nH	8.675 nH	20.1 nH	59.37 nH	132.6 nH	1.786 µH
0.5 mm	0.8575 nH	2.988 nH	7.314 nH	17.35 nH	52.46 nH	118.7 nH	1.647 µH
1 mm	—	2.343 nH	5.977 nH	14.63 nH	45.58 nH	104.9 nH	1.509 µH
2 mm	—	—	4.686 nH	11.95 nH	38.75 nH	91.17 nH	1.37 µH

Read down any column and the diameter's effect is small; read across a row and L grows faster than the length. Every tenfold step in length in the table multiplies L by between 12.6 and 20.2, never by ten.

Worked example: Rosa and Grover's Example 70

The paper's first example of the straight-wire formulas (pp. 159–160) is "A straight copper wire 100 cm long and 0.2 cm diameter", worked in its own units, followed by the same wire twice as long, in iron, at high frequency, and ten metres long:

(95)       l = 100 cm, ρ = 0.1 cm: 200(log 2000 − 0.75) = 1370.18 cm
           twice as long: 400(log 4000 − 0.75)        = 3017.62 cm
(96)       iron, μ = 1000: 200(log 2000 − 1 + 250)    = 51320 cm
(97)       no field inside: 200(log 2000 − 1)         = 1320.18 cm
(95)       l = 1000 cm: 2000(log 20000 − 0.75)        = 18307.0 cm
           in henries, × 10⁻⁹                         = 18.307 µH
(94)       l = 100 cm, in full                        = 1370.38 cm

The paper prints 1370.18, 3017.62, 51320, 1320.18 and "18307.0 cm = 18.307 microhenrys", and every line is reproduced by the page's tests. Enter 10 m and 2 mm in the calculator to get the last one in SI. The paper adds that "The more exact formula (94) gives practically the same result where ρ is so small compared with l", and so it does: 1370.38 cm against 1370.18 cm. The difference is almost exactly 2ρ, the last term of (94), which (95) drops.

That term is not always negligible. For the short wire of Example 73, 1 cm long and 1 mm in radius, (95) gives the paper's 4.4915 cm, and (94) gives 4.6865 cm, 4.3 % more. At l/ρ = 10 the approximation is off by more than most components' tolerance, which is why the calculator uses (94) and shows (95) beside it.

Flat strips, bars and PCB traces: the geometric mean distance

For a conductor of rectangular section the paper uses Maxwell's geometric mean distance. The self-inductance of a straight bar "is, to within the accuracy of the approximate formula (99), the same as the mutual inductance of two parallel straight filaments of the same length separated by a distance equal to the geometrical mean distance of the cross section of the bar" (p. 152), formula (103). The geometric mean distance R of an area from itself is the distance whose logarithm is the average of log r over every pair of points in the section. Maxwell's formula (124) gives it for a rectangle exactly, and the calculator evaluates (124) directly; for a circle it is 0.7788ρ (126), which turns (103) back into the round-wire formula (95). One method covers both.

The paper's table on p. 168 shows that R is very nearly 0.2235 times the sum of the sides for any rectangle, which is its formula (128) and the origin of the 0.2235 in (104). Formula (124) as the calculator evaluates it, beside the printed table:

Sides a : b	R ÷ a, printed	R ÷ a, (124)	R ÷ (a + b), printed	R ÷ (a + b), (124)
1 : 1	0.44705	0.44705	0.22353	0.22352
1.25 : 1	0.40235	0.40236	0.22353	0.22353
1.5 : 1	0.37258	0.37259	0.22355	0.22355
2 : 1	0.33540	0.33540	0.22360	0.22360
4 : 1	0.27961	0.27961	0.22369	0.22369
10 : 1	0.24596	0.24596	0.22360	0.22360
20 : 1	0.23463	0.23463	0.22346	0.22345
1 : 0	0.22313	0.22313	0.22313	0.22313

Every cell agrees to within one unit in the fifth place. Where it is one unit off (1.25 : 1, 1.5 : 1 and 20 : 1) the printed value is what the rounded ratio gives when worked back, so the table was evidently built from its last column. The paper's convenient formula (104) uses the 0.2235 and rounds −log 0.2235 − 1 = 0.4983 to 1/2; it runs a fraction of a per cent above (103) with the exact R, and the calculator shows both.

A PCB trace is a very flat rectangle. For 10 mm of 1 oz copper, taken as0.0350 mm thick as on the trace width calculator, with no return path near it:

Width	L, (103)	Per mm	GMD R	Round wire of the same GMD
0.1 mm	11 nH	1.100 nH/mm	0.0302 mm	0.000 mm diameter
0.25 mm	9.51 nH	0.951 nH/mm	0.0637 mm	0.000 mm diameter
0.5 mm	8.263 nH	0.826 nH/mm	0.1196 mm	0.000 mm diameter
1 mm	6.967 nH	0.697 nH/mm	0.2312 mm	0.001 mm diameter
2 mm	5.66 nH	0.566 nH/mm	0.4544 mm	0.001 mm diameter

The last column is the diameter of round wire with the same geometric mean distance, 2R/0.7788, which has the same inductance to within the R/l term of (103). A 0.25 mm trace behaves like a 0.000 mm wire: a flat conductor has less inductance than a round one of the same cross-sectional area, because its current is spread further apart. Widening the trace is the strip's equivalent of using thicker wire, and with the same logarithm it buys as little.

A trace over a ground plane is a transmission line

The isolated-strip formula applies to a trace with nothing nearby: a jumper strap, a bus bar, a trace on a two-layer board with the return current far away. Put a solid plane under the trace and the return current flows in the plane directly beneath it; the field is confined to the thin dielectric between them and the inductance per length drops a long way. That case is a microstrip, and its inductance follows from the line's impedance and delay: for a lossless line Z0 = √(L′/C′) and the delay per length is √(L′C′), so L′ = Z0 × delay (derived here, not from Rosa and Grover).

Take the 0.25 mm trace in 1 oz copper over 0.2 mm of dielectric with εr = 4.3, an illustrative two-layer stack. The microstrip calculator's IPC-2141 formulas give Z0 = 59.2 Ω and 140 ps per inch, so L′ = 0.327 nH/mm, against 0.951 nH/mm for the same 10 mm of trace with no plane. Use themicrostrip impedance calculatorfor a trace over a plane, and this page for one that has no return path close by. A short trace between a capacitor and its via is somewhere between the two; thebypass placement calculator works that case from measured mounting inductances.

Worked example: the tape of Example 79 and the bar of Example 74

Example 79 (pp. 164–165): "Let the tape of thin copper be 10 meters long and 1 cm wide." The paper works the thin tape by (113) and the same tape 2 mm thick by (114):

(113)      l = 1000 cm, b = 1 cm: 2000(log 2000 + 0.5) = 16202 cm
(114)      a = 0.2 cm thick: 2000(log 2000 − 0.2 + 0.5) = 15802 cm
(124)      GMD of 1 cm × 0.2 cm                       = 0.26842 cm
(103)      with that GMD                              = 15833 cm
(104)      R = 0.2235 × 1.2 cm                        = 15838 cm

The paper prints "= 2000 × 8.1009 = 16202 cm = 16.202 microhenrys" and "L = 2000 × 7.9009 cm = 15.802 microhenrys". (114) is the first-order form of (104) for a thin tape, and the paper itself says "A closer approximation to L is given by (104)", which here is 15.838 µH; the exact GMD gives 15.833 µH, the calculator's figure. Example 74 (p. 162), a square bar 1000 cm long and 1 cm on a side, gives by (104) without its last term "2000 (6.908 + 0.5) = 14816 cm"; unrounded it is 14815.5 cm, and with the exact square GMD, 0.44705 cm, (103) gives 14812.9 cm. Enter 10 m, 10 mm and 10 mm to check.

Two parallel wires: the inductance of the loop

Current out along one wire and back along the other is a closed circuit, and its inductance is the self-inductance of the two wires less twice their mutual inductance, because the opposite currents' fields cancel outside the pair. The paper writes it as (100), "neglecting the effect of the end connections" (p. 151), and for non-magnetic wires long against their spacing as (101), 4l[log(d/ρ) + 1/4]: an inductance per unit length that depends only on the ratio of the spacing to the radius. For 1 mm wire:

Spacing s, centre to centre	d/ρ	L per mm of pair, (101)	Most it falls at high frequency, (159)
2 mm	4	0.655 nH/mm	15.3 %
3 mm	6	0.817 nH/mm	12.2 %
5 mm	10	1.021 nH/mm	9.8 %
10 mm	20	1.298 nH/mm	7.7 %
20 mm	40	1.576 nH/mm	6.3 %
50 mm	100	1.942 nH/mm	5.1 %

The ends matter when the wires are far apart against their length: "If the end effect is large, as when the wires are relatively far apart, use the expression for the self-inductance of a rectangle below (107)" (p. 152). The calculator's headline figure for the pair is that rectangle, the two wires and both end connections. For the default, 10 mm of 0.5 mm wire 2 mm apart, the loop is 10.30 nH; the two wires alone are 8.66 nH, and (100) as printed gives 8.52 nH, low because the 2ρ that (95) drops from each wire and the d²/2l that (99) adds to the mutual inductance are no longer small at these proportions.

Worked example: the return circuit of Example 72

Example 72 (p. 161): "Suppose a return circuit of two parallel wires, each 10 meters long and 0.2 cm diameter, distant apart 10 cm, center to center". The paper works it two ways, by (100) and by the wires and their mutual inductance from Examples 70 and 71, then adds the ends:

(100)      4000[log(10/0.1) + 0.25 − 10/1000]         = 19380.7 cm
(95)       2L1, two 10 m wires                        = 36614.0 cm
(99)       2M, 10 cm apart                            = 17233.3 cm
           2L1 − 2M                                   = 19380.7 cm
(95)       the two 10 cm ends                         = 181.9 cm
           with the ends                              = 19.5626 µH
(107)      the rectangle, 1000 cm × 10 cm             = 19.5633 µH

The paper prints "= 4000 × 4.8452 = 19380.8 cm = 19.3808 microhenrys", the bracket rounded to four places; "36.6140 − 17.2332 = 19.3808 microhenrys" by the second route; "2L1 for the two ends by (95) is 181.9 cm"; and "L = 19.5627 microhenrys" with them. The rectangle formula (107), which the calculator uses, gives 19.5633 µH for the same loop, three parts in a hundred thousand more. Enter a pair 10 000 mm long, 2 mm wire, 100 mm apart.

Two of the paper's printed results in this section do not follow exactly from its own formulas, both at the fourth or fifth figure. Example 71 says "The formula (98) gives a value less than two parts in one hundred thousand greater" than (99) for 10 m wires 10 cm apart; evaluated, (98) is 0.050 cm, six parts in a million, smaller, which is d²/2l. And Example 76 says (107) gives "L = 8017.1 cm" for a 2 m by 1 m rectangle of 2 mm wire; (107) as printed gives 8017.86 cm, and 8017.1 is what it gives with 2d in place of its last term 2(d + ρ), the same as the paper's check by the four sides. Neither changes anything at the precision a calculator needs, and the tests record both.

Frequency: skin effect removes the internal field

Everything above assumes the current is spread evenly over the conductor's section. Section 10 of the paper says so plainly: "Excepting in a very few specified cases, the formulas of the preceding sections apply only to conductors carrying direct current or alternating currents of frequencies so low that the error, due to the assumption that the current is uniformly distributed over the cross section of the wire, is negligible" (p. 172). As the frequency rises the current moves to the surface, the field inside the wire shrinks, and "any deviation of the distribution of the current in the wire from uniformity gives rise to a decrease in the inductance." For a round wire the solution is exact, Kelvin's formula (144) (p. 174): the internal term μ/4 is multiplied by a factor 4Z/(xY) that is 1 at low frequency and falls towards zero.

The external part does not change, so the fall is bounded. The default wire, in copper:

Frequency	Kelvin's x	4Z/(xY)	L	Below DC
DC	0	1	7.314 nH	—
100 kHz	1.71	0.9783	7.303 nH	0.15 %
1 MHz	5.42	0.5135	7.071 nH	3.33 %
10 MHz	17.15	0.1647	6.896 nH	5.71 %
100 MHz	54.23	0.0521	6.84 nH	6.48 %
1 GHz	171.49	0.0165	6.822 nH	6.72 %

At 1 MHz the internal part is already down to 51 %; by 100 MHz L has nearly reached the limit (97), 6.814 nH, which is 6.8 % below the DC figure. The limiting change depends only on the proportions of the wire, formula (154): for 2l/ρ = 100 it is 6.48 %, Table XXIII's 0.06485. The same page notes the cost lies elsewhere: "The change of resistance is always relatively much larger than the change in inductance" (p. 173), which is the subject of the skin depth calculator.

The paper's Example 82 (p. 183) works the same thing by hand: 200 cm of copper wire of radius 0.125 cm at 500 000 cycles, "x = 15.146 × 1.25 = 18.932", 4Z/(xY) = 0.14923 from Table XXII, and "the inductance is 85.08 cm or 2.9052 per cent less than the direct current value." The calculator gives x = 18.932, 4Z/(xY) = 0.14923and a fall of 85.08 cm, 2.905 %. Enter 2000 mm, 2.5 mm and 0.5 MHz with Rosa and Grover's copper.

A parallel pair falls further, because its change depends on the sum of the two self-inductances and not the smaller loop inductance. The paper's own case (p. 180): "With d = 1 cm and ρ = 0.1 cm, and with a frequency of 10⁶, (158) gives ΔL/L = −8.5 per cent", and the calculator gives −8.50 % for that pair in the paper's copper. Its limit, (159), is 9.79 % at d/ρ = 10. Table XXIII prints 0.09784 there, and 0.07706 at d/ρ = 20 where (159) gives 0.07702; every other cell of the column checked agrees to the fifth place, and the printed differences match the printed values, so the two cells are most likely arithmetic slips in the original. A third is in Example 84 (p. 186): for iron wires of μ = 100, 1.5 cm apart, the limit is printed "−100/109.94 = −0.9308", but 100/109.94 is 0.9096, and the example's iron results, −0.8940 and −0.1732, carry the 0.9308.

Where the formulas stop being valid

Straight conductors in free space. Every formula on this page is for a straight conductor with nothing magnetic or conducting nearby. A nearby plane, chassis or cable carries induced currents that lower the inductance; that is the microstrip case above.

Low frequency for strips. The geometric-mean-distance method assumes uniform current. The paper's high-frequency section treats straight round wires, two parallel wires and a ring of circular section, and notes that "other cases of linear conductors of circular cross section, may likewise be made to depend on the solution for straight wires" (p. 181); it gives nothing for a rectangular section, and the calculator applies no correction to a strip.

Approximations for long, thin conductors. (95) and (99) hold when the length is large against the radius or spacing: (99) "is only applicable when the length of the conductors is great compared with their distance apart" (p. 160). The calculator uses the full (94) and (98) and the rectangle (107) rather than the short forms. (98) itself "is not an exact expression for the mutual inductance of two parallel cylindrical wires, but is not appreciably in error even when the section is large and d is small if l is great compared with d" (p. 151).

Close spacing at high frequency. Each wire of a pair is treated on its own "Unless the two wires are so near together, relatively to their radius of cross section, that their mutual inductance is appreciably affected by changes in the distribution of the current within the wires" (p. 180). At d/ρ = 10 and 1 MHz the paper found Nicholson's fuller solution changed ΔL/L by "only nine parts in ten thousand"; closer than that, the proximity effect is not modelled.

Magnetic wire. The permeability is a constant, as in the paper's iron example. A real steel wire's permeability depends on the field and the frequency.

Common wire inductance mistakes

Scaling a per-millimetre figure to another length. The default wire has 0.731 nH/mm at 10 mm and 1.187 nH/mm at 100 mm. Compute the length you have.
Treating a straight wire's inductance as the circuit's. The return path is part of the loop; two wires close together have far less inductance than the sum of their self-inductances, and a wire's mutual inductance to its neighbour can exceed its own self-inductance (Example 73).
Using the isolated-strip figure for a trace over a ground plane. With a plane 0.2 mm below it, the 0.25 mm trace above has about a third of the inductance per length.
Using the short formula (95) for a short, thick wire. At l/ρ = 10 it is4.3 % low.
Measuring the pair spacing edge to edge. The formulas take it centre to centre.
Mixing the paper's units. Its L is in centimetres, 1 µH = 1000 cm, and its logarithms are natural except where it says common: its (101) in common logarithms is "0.7411 log₁₀ d/ρ + .0805 in millihenrys per mile of conductor" (p. 152), per mile of each wire, half the figure per mile of line.
Expecting skin effect to change inductance much. It removes at most the internal field, 6.8 % of the default wire; the resistance is what rises without limit.