Mutual inductance calculator: parallel wires and coaxial loops

The mutual inductance M of two parallel straight wires, or of two circular loops on one axis, with the self-inductance of each conductor and the coupling factor k between them. The formulas are Rosa and Grover's, from the Bureau of Standards paper that collected them: Neumann's exact result for parallel wires and Maxwell's formula in elliptic integrals for coaxial circles, both evaluated without tables. The default pair of wires gives 72.50 nH and k = 0.611; the default loops are the paper's Example 4 in millimetres, and the page's tests hold every formula to the paper's worked examples.

Fig 1 — Two parallel wires 100 mm long, 2 mm apart centre to centre, 0.5 mm in diameter: M = 72.50 nH by Rosa and Grover's (98), against 118.7 nH for each wire alone by (94), so k = 0.6106. The curve is (98) from touching wires (s = one diameter) outwards; it falls slowly, as the logarithm of l/s, while the wires are long compared with their spacing, and as 1/s once they are not. The drawing is not to scale.

Geometry

Length of each wire l (mm)

Separation, centre to centre s (mm)

Wire diameter w (mm)

Mutual inductance M, formula (98): 72.50 nH
Self-inductance of each wire L, formula (94): 118.7 nH
Coupling factor k = M/L: 0.6106
At high frequency, current on the surface: L by (97) · k: 113.7 nH · 0.6377
M per unit length, M/l: 0.725 nH/mm
M by the long-wire approximation (99), and its error: 72.50 nH · +0.0028 %
The pair as a go-and-return circuit, 2L − 2M (formula (100) gives): 92.48 nH (92.38 nH)
Both wires sharing one current, (L + M)/2 (formula (121) gives): 95.62 nH (95.40 nH)
Clear gap between the wires: 1.5 mm

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): formulas (1), (4), (5), (10), (12), (62), (63), (94)–(100), (121), (136), Rosa's formula (p. 39), Table I, and Examples 1–14, 52–55, 70 and 71

$M = 4\pi\sqrt{Aa}\left\{\left(\frac{2}{k} - k\right)F - \frac{2}{k}E\right\}, \qquad k = \frac{2\sqrt{Aa}}{\sqrt{(A + a)^2 + d^2}}$: Maxwell's formula (1), p. 6, for two coaxial circles of radii A and a whose centres are d apart, in the paper's units (lengths and M in centimetres; 1 cm of inductance = 1 nH). F and E are the complete elliptic integrals of the first and second kind to modulus k, evaluated here by the arithmetic-geometric mean. In SI with lengths in metres, 4π becomes μ0.
$M = \frac{\pi^2 k^3}{4}\sqrt{Aa}\left[1 + \frac{3}{4}k^2 + \frac{75}{128}k^4 + \frac{245}{512}k^6 + \cdots\right]$: Series (5), p. 9, for circles far apart, where in (1) the bracket "comes out as the small difference of two large quantities". The calculator sums it, with the general term the paper prints, for k below 0.5. Its first term with k ≈ 2√(Aa)/d is the far-field M ≈ μ0πA²a²/(2d³), derived here.
$M = 4\pi a\left\{\log\frac{8a}{d}\left(1 + \frac{3d^2}{16a^2}\right) - \left(2 + \frac{d^2}{16a^2}\right)\right\}$: Maxwell's formula (12), p. 13, for equal circles near each other; (10) on the same page is the general form for unequal radii, with r = √(c² + d²) and c = A − a. "If d/a is 0.1, the largest term neglected in (12) is less than two parts in a million." Shown in the results as a check on (1).
$M = 2\left[l\log\frac{l + \sqrt{l^2 + d^2}}{d} - \sqrt{l^2 + d^2} + d\right] \approx 2l\left[\log\frac{2l}{d} - 1 + \frac{d}{l}\right]$: Formulas (98) and (99), p. 151: two parallel wires of length l, d apart centre to centre, opposite each other. (98) is "an exact expression when the wires have no appreciable cross section"; (99) holds "when the length l is great in comparison with d".
$L = 2\left[l\log\frac{l + \sqrt{l^2 + \rho^2}}{\rho} - \sqrt{l^2 + \rho^2} + \frac{l}{4} + \rho\right], \qquad L_{hf} = 2l\left[\log\frac{2l}{\rho} - 1\right]$: Formula (94), p. 150, the self-inductance of a straight round wire of radius ρ with the current uniform over its section, and (97), p. 151, the same with the current on the surface, as at high frequency. k = M/L for two equal wires.
$L = 4\pi n^2 a\left\{\left(1 + \frac{\rho^2}{8a^2}\right)\log\frac{8a}{\rho} + \frac{\rho^2}{24a^2} - 1.75\right\}$: Rayleigh and Niven's formula (63), p. 111, "for a circular coil of n turns and of circular section", ρ the radius of the section; with n = 1 a single ring. The surface-current form is (62): 4πa{(1 + 3ρ²/16a²) log(8a/ρ) − ρ²/16a² − 2}. Coupling k = M/√(L1·L2).
$\frac{M}{n_1 n_2} = M_0 + \Delta M$: Rosa's formula, p. 39, for two coaxial coils, "where M0 is the mutual inductance of the central circles" and ΔM "the correction for the section of the coil". The calculator takes M = n1·n2·M0 and leaves out ΔM.
$L_{loop} = 2L - 2M, \qquad L_{parallel} = \frac{L + M}{2}$: Two equal wires used as a go-and-return circuit, and as one conductor sharing a current, from (94) and (98). Derived; with (95) and (99) they become the paper's (100), p. 151, and (121), p. 159, which the calculator shows beside them.

Assumptions

Filaments along the conductor centres: M of two round wires is taken as that of their axes, spaced centre to centre. The paper calls (98) "not appreciably in error even when the section is large and d is small if l is great compared with d" (p. 151).
Parallel wires of equal length lying opposite each other; offset or unequal wires are not covered.
Coaxial loops: both circles on one axis, their planes parallel. Tilted or offset loops are not covered.
Air, with nothing magnetic or conducting nearby. No core, shield or ground plane.
M does not depend on frequency; each self-inductance is given with the current uniform over the wire (low frequency) and on its surface (high frequency), the two limits the paper gives. No capacitance between the conductors is modelled.
For coils of several turns, the winding section is small compared with the radius and the spacing, so Rosa's ΔM can be neglected; the section diameter entered is used for both loops.
The default wires are illustrative, not from the paper. The default loops are the paper's Example 4, A = 25, a = 20, d = 10 cm, in millimetres, which scales M by one tenth: 248.7874 cm there, 24.88 nH here.

What mutual inductance is, and what sets it

Two conductors near each other share flux. A current in the first sets up a magnetic field, some of that field threads the second, and a change in the current induces a voltage there in proportion: v₂ = M·di₁/dt. M is the mutual inductance. Rosa and Grover, in the Bureau of Standards paper this calculator follows, define it for two parallel wires as "the number of lines of force, due to unit current in one, which cut the other when the current disappears" (p. 151). It is the same in both directions, and in air it depends on nothing but the geometry: the lengths, the radii and the spacing. There is no material property in any of the formulas.

That makes mutual inductance one of the few quantities in a circuit you can calculate from a ruler. What the geometry does not fix is how large M is compared with the inductance each conductor has alone. That comparison is the coupling factor, k = M/√(L₁L₂), which runs from 0 for conductors that share no flux to 1 for two windings that share all of it. To compute k you need the self-inductances too, and those do depend on the wire diameter, so the calculator asks for it. The same paper gives both: formula (94) for a straight round wire and Rayleigh and Niven's formula (63) for a circular ring, which the paper reports experimental work at the Bureau had shown to be consistent with (94) and (98): "formula (100), and therefore (94) and (98) are consistent with the formula (63) for the inductance of a circular ring" (p. 152).

Every formula in the paper is written in centimetres with inductance "in centimeters", a unit equal to a nanohenry. Lengths in metres turn the same expressions into henries once multiplied by μ₀/4π = 10⁻⁷ H/m. The calculator works in millimetres; entering the paper's centimetre figures as millimetres gives exactly one tenth of the paper's number, in nanohenries, which is how its default loops reproduce the paper's Example 4.

The mutual inductance of two parallel wires

For two straight wires of the same length l, side by side and opposite each other, a distance d apart, the paper gives Neumann's result as formula (98), p. 151: M = 2[l log((l + √(l² + d²))/d) − √(l² + d²) + d], in its centimetre units. When the wires are long compared with their spacing it reduces to (99), M = 2l[log(2l/d) − 1 + d/l], which is the form most often quoted as the mutual inductance formula for parallel wires. The calculator uses (98) and shows (99) with its error beside it.

The shape of (99) says most of what matters. While the wires are long against their spacing, M falls only with the logarithm of the spacing: every doubling of d takes off the same amount, 2l·log 2, which for a 100 mm pair is 13.86 nH. Once d approaches l the logarithm gives way and M falls as 1/d. The table is an illustrative pair, two 0.5 mm wires 100 mm long, the calculator's default, with d from touching to twice the length.

Spacing d	M (98)	Error of (99)	k = M/L	2L − 2M
0.5 mm	99.93 nH	< 0.001 %	0.842	37.63 nH
1 mm	86.17 nH	< 0.001 %	0.726	65.15 nH
2 mm	72.50 nH	+0.0028 %	0.611	92.48 nH
5 mm	54.77 nH	+0.023 %	0.461	128.0 nH
10 mm	41.86 nH	+0.12 %	0.353	153.8 nH
20 mm	29.85 nH	+0.67 %	0.251	177.8 nH
50 mm	16.51 nH	+7.4 %	0.139	204.5 nH
100 mm	9.343 nH	+48.4 %	0.079	218.8 nH
200 mm	4.903 nH	4.08× too high	0.041	227.7 nH

Halving the spacing from 2 mm to 1 mm adds 13.66 nH, and the step from 20 mm to 200 mm takes M from 29.85 nH to 4.903 nH. Touching, the two wires couple with k = 0.842, the most this pair can reach; by (95) and (99) k rises towards 1 only as the wires grow far longer than they are thick. At d = l/10, 10 mm here, the long-wire form (99) is +0.12 % out, and the paper restricts it in Example 71: "The approximate formula (99) is only applicable when the length of the conductors is great compared with their distance apart" (p. 160).

The last column is the pair used as a go-and-return circuit, the current flowing out along one wire and back along the other. Its inductance is the two self-inductances less twice the mutual, which is exactly how the paper's return-circuit formula (100) is built: twice (95) less twice (99) gives (100) term for term, and the tests check it. This is why close spacing lowers the inductance of a supply and return pair: M is subtracted, and the closer the wires, the more of each wire's own inductance it cancels. For two wires carrying one current between them, in parallel, M is added instead, and the calculator gives that case too, beside the paper's formula (121) for it.

The mutual inductance of two coaxial loops

For two circular loops on a common axis, the paper opens with the formula it calls "the first and most important of the formulas for the mutual inductance of coaxial circles", Maxwell's formula in elliptic integrals (p. 6): M = 4π√(Aa)·{(2/k − k)F − (2/k)E}, with k = 2√(Aa)/√((A + a)² + d²). A and a are the radii, d "the distance between their centers", and F and E the complete elliptic integrals of the first and second kind to modulus k. "Formula (1) is an absolute one, giving the mutual inductance of two coaxial circles of any size at any distance apart" (p. 7). This is the mutual inductance formula between two coils of one turn each, and the calculator evaluates F and E by the arithmetic-geometric mean, the same routine the air-core inductor calculator uses for Lorenz's formula, so no table of elliptic integrals is needed.

There is one numerical trap, which the paper names. When the loops are far apart the bracket "comes out as the small difference of two large quantities" (p. 9), and for that case it expands (1) as series (5), M = (π²k³/4)√(Aa)[1 + (3/4)k² + (75/128)k⁴ + …], with the general term printed. The calculator sums (5) whenever k is below 0.5 and uses (1) above, and its tests hold the two together to 1 part in 10¹¹ at the switch. The first term of (5), with k ≈ 2√(Aa)/d, is the far-field limit M ≈ μ₀πA²a²/(2d³), derived here rather than printed, which the calculator shows with its error.

The table is an illustrative pair of equal loops, 25 mm radius in 0.5 mm wire, against the distance between them. The (12) column is Maxwell's simpler series for equal circles near each other, and the far-field column the cube law; both are errors against (1). Where (12) comes out below zero the table says so: its logarithm has run out, which is how far from its range it has been taken.

Distance d	d/a	M (1)	k	Error of (12)	Error of far field
0.5 mm	0.02	125.4 nH	0.809	< 0.001 %	49,187× too high
1 mm	0.04	103.7 nH	0.669	< 0.001 %	7,438× too high
2.5 mm	0.10	75.07 nH	0.484	< 0.001 %	657× too high
5 mm	0.20	53.85 nH	0.347	+0.0036 %	115× too high
10 mm	0.40	33.77 nH	0.218	+0.067 %	22.8× too high
25 mm	1.00	12.35 nH	0.0797	+3.5 %	4.00× too high
50 mm	2.00	3.546 nH	0.0229	+55.9 %	+73.9 %
100 mm	4.00	650.0 pH	0.00419	negative	+18.6 %
250 mm	10.00	47.91 pH	0.000309	negative	+3.0 %

Close in, M falls with the logarithm of the distance, as Maxwell's (12) shows with its log(8a/d), much as the parallel wires do; far out it falls as the cube of the distance, so that going from 100 mm to 250 mm divides it by 13.6 against the 2.5³ = 15.6 of the pure cube law. At d equal to the radius, M is 12.35 nH and k = 0.080. Maxwell's (12) is excellent close in and +3.5 % out by then, as the paper warns: "if d/a is 0.1, the largest term neglected in (12) is less than two parts in a million. If, however, d = a, this term will be more than one per cent, and the formula will be quite inexact" (p. 13).

Worked example: two parallel wires, Example 71

The paper's Example 71 (p. 160) works (98) and (99) for three pairs. "Two parallel copper wires of length 100 cm and distance apart 200 cm will have a mutual inductance of" 49.02 cm, it prints; with the lengths reversed, 200 cm long and 100 cm apart, 330.24 cm; and by (99), "two conductors 10 meters long are 10 cm apart", 8616.6 cm, "= 8.6166 microhenrys". Entered in the calculator as millimetres the first pair gives one tenth of the paper's figure in nanohenries.

l, d     100 cm, 200 cm, by (98)              = 49.029 cm    paper: 49.02 cm
         as the paper rounds it               = 49.022 cm   
l, d     200 cm, 100 cm, by (98)              = 330.24 cm    paper: 330.24 cm
l, d     10 m, 10 cm, by (99)                 = 8616.6 cm    paper: 8616.6 cm
         the same by (98)                     = 8616.6 cm

The second and third agree with the paper to the figures it prints. The first is 49.029 cm unrounded: the paper writes the bracket as log 1.61803 − 0.2361, rounding √5 − 2 = 0.23607 up to four places, and that alone takes it to 49.022 cm. For the 10 m pair the approximate (99) and the exact (98) differ by 5.8 parts per million, which is the point of the example: at a spacing of 1 % of the length, the long-wire form is as good as exact.

To get a coupling factor the wires need a diameter. Example 70, on the page before, takes "A straight copper wire 100 cm long and 0.2 cm diameter" and gives its self-inductance by (95) as 1370.18 cm; the calculator's (95) gives 1370.18 cm, and the fuller (94), which the paper says "gives practically the same result", 1370.38 cm. Taking that wire for the conductors of the first pair, a combination the paper does not make, k = 49.03/1370.38 = 0.0358 for two wires a metre long and two metres apart.

Worked examples: coaxial circles, Examples 1, 4 and 6

Example 1 (p. 20) is two equal circles: "Let a = A = 25 cm", d = 20 cm. The paper reads log F and log E from Legendre's tables and gets M = 167.08562 cm, then repeats the calculation from Table I, Maxwell's table of the bracket against γ = sin⁻¹ k, and gets 167.08546 cm, "agreeing almost exactly with the above value" (p. 21).

k        50/√(2500 + 400)                     = 0.92847669   paper: 0.9284766
bracket  (2/k − k)F − (2/k)E                  = 0.5318506    paper: 0.5318500
M        4π × 25 × bracket                    = 167.08578 cm paper: 167.08562 cm
         by Table I, interpolated             =              paper: 167.08546 cm

The calculator's 167.08578 cm sits 1.0 parts per million above the paper's elliptic-integral result and matches the 167.08577 cm the paper gets by Nagaoka's formula in Example 9; the paper's bracket, from seven-place logarithms, is the source of the difference. The k it prints, 0.9284766, is 50/√2900 = 0.92847669 cut off rather than rounded.

Example 4 (pp. 22–23), the calculator's default loops, is unequal: "A = 25, a = 20, d = 10 cm". The paper sums F and E from series and uses (1):

k²       4 × 20 × 25 / (45² + 10²)            = 16/17       
F, E     arithmetic-geometric mean            = 2.8302431, 1.0688879 paper: 2.8302430, 1.0688878
bracket  (2/k − k)F − (2/k)E                  = 0.8853876    paper: 0.885388
M        4π√500 × bracket                     = 248.7874 cm  paper: 248.7875 cm
in mm    A = 25, a = 20, d = 10 mm            = 24.88 nH

Weinstein's formula in Example 8 and Havelock's in Example 14 give 248.7873 cm for the same circles; the calculator's 248.7874 cm agrees with all three to 0.6 parts per million or better. In millimetres the loops are 50 and 40 mm across, 10 mm apart, about the size of a pair of wireless-charging coils, and M is 24.88 nH turn on turn, with k = 0.184 in 0.5 mm wire.

Example 6 (p. 24) is the far-apart case, two 10 cm circles a metre apart, worked by series (5). The paper sums the bracket to 1.02974111 and gets M = 0.19164962 cm. The full series sums to 1.02974106; the difference is the k⁴ term, printed 0.00086684 where (75/128)k⁴ with k² = 1/26 is 0.00086677. The calculator's M is 0.19164953 cm, nearer the 0.19164958 cm the paper finds on the same page by formula (6), which it reports "differs by only one part in five million from the value given by (5)". The far-field first term alone gives 0.197392 cm, +3.00 % out at this spacing of five diameters.

Close together, Examples 2 and 3 (pp. 21–22) take the same 25 cm circles 1 cm apart and get 1036.6664 cm by Maxwell's formula (2) and 1036.6652 cm by (1) with F and E from series; the calculator gives 1036.6648 cm. Example 13 (p. 28) gives 1036.663 cm by the simpler (12), and the calculator's (12) is 1036.6649 cm.

Maxwell's table against the exact formula

Table I (pp. 190–192) lists log₁₀[M/(4π√(Aa))] for γ from 60° to 89°54′ in steps of 6′, and closes with the claim that "The values given are sufficiently accurate to give M within one part in a million." The page's tests check 32 of its rows, every whole degree plus 68°6′ and the last, against (1). All 32 agree within 1.1 × 10⁻⁶ in the logarithm; 25 hold the claimed part in a million and 7 do not, the worst being 63°, printed 1̄.582 5962, which is 2.6 parts per million out in M. For any design that is irrelevant; it matters only if the table is used to check another program, which is better done against the worked examples above.

The coupling factor, and the self-inductance of each loop

The self-inductance of a single round-wire ring is Rayleigh and Niven's formula (63), p. 111, "for a circular coil of n turns and of circular section": L = 4πn²a{(1 + ρ²/8a²) log(8a/ρ) + ρ²/24a² − 1.75}, ρ the radius of the section. "When n = 1, this will be the self-inductance of a single circular ring." The paper's Examples 52 to 54 (p. 114) compare it with three older formulas; the calculator's (63) against the printed values, in the paper's π cm:

Ring	(63) printed	(63) here	(62), surface current
a = 25 cm, ρ = 0.05 cm	654.40548π	654.40540π	629.40556π
a = 25 cm, ρ = 0.5 cm	424.1781π	424.17808π	399.18889π
a = 10 cm, ρ = 1 cm	105.517π	105.51683π	95.58472π

The (62) column is the same ring with its current on the surface; the paper's Example 55 (p. 115) prints 629.40556π, 399.1889π and 95.585π for it, and remarks that "the error increases rapidly as the ratio ρ/a is increased" for Maxwell's expression against its own refinement (65).

The coupling factor is then M/√(L₁L₂). For coils of several turns the calculator multiplies M by n₁n₂ and each L by n², following Rosa's formula (p. 39), which writes the mutual inductance of two coils as M/(n₁n₂) = M₀ + ΔM, "where M₀ is the mutual inductance of the central circles of the two equal coils of sections b×c, Fig. 5, and ΔM is the correction for the section of the coil". ΔM is left out, so the coils must be small in section compared with their spacing. With the section diameter the same for both windings, the turns cancel out of k: 2 and 3 turns on the default loops give M = 149.3 nH and the same k = 0.184.

The frequency matters for k, though not for M. Section 10 of the paper, on high frequencies, notes: "In the case of standards of mutual inductance the inductance may be regarded as sensibly independent of the frequency, unless the two coils are very close together", while for self-inductance "any deviation of the distribution of the current in the wire from uniformity gives rise to a decrease in the inductance" (p. 172). So the calculator gives each L twice: at low frequency by (63) or (94), and with the current on the surface of the wire by (62) or (97), the limits the paper gives for "a hollow, circular, thin tube" or current "of extremely high frequency" (p. 111). For the default loops k rises from 0.184 to 0.194 between the two.

Where the formulas stop being valid

Filaments, not wires. Formulas (1) and (98) are for lines along the centres of the conductors. The paper is explicit about the parallel-wire case: "Equation (98), which is an exact expression when the wires have no appreciable cross section, 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). For round sections the spacing to use is centre to centre: the geometric mean distance "from one circular area to another is the distance between their centers" (p. 170).
Equal, opposite wires. (98) is for two wires of the same length lying opposite each other. Wires of different lengths, or offset along their length, are outside it, and the calculator has no input for them.
End connections. The go-and-return result is the two straight runs only. The paper's (100) is stated "neglecting the effect of the end connections" (p. 151), and when the wires are far apart it advises adding the self-inductance of the ends or using the formula for a rectangle instead (p. 152).
Thick rings. (63) "neglects higher powers of ρ/a than the second" (p. 111). For the thickest ring in the paper's examples, ρ/a = 1/10, the older formulas already disagree: "for the smallest ring of radius 10 cm and diameter of section 2 cm Maxwell's formula gives a result 1 part in 2500 too small, while the simple approximate formula of Kirchhoff is in error by one in four hundred" (p. 114).
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). M is the exception; the self-inductances are bracketed by the low- and high-frequency forms. When coils are very close, the paper adds, "the capacity between the coils will be a more potent source of error than the departure of the current from a uniform distribution" (p. 172). No capacitance is modelled here.
Air. Every formula assumes non-magnetic surroundings. A ferrite core, a steel chassis or a ground plane under the loops changes M, and none of that is in the paper's formulas.

Common mistakes

Measuring the spacing edge to edge. Two 0.5 mm wires with a 1.5 mm gap are 2 mm apart centre to centre; entering 1.5 mm gives 78.16 nH instead of 72.50 nH, +7.8 %. The calculator shows the clear gap so the two can be told apart.
Using the long-wire formula on short wires. For 20 mm wires 10 mm apart, (99) is +7.4 % out. The paper restricts it to conductors whose length "is great compared with their distance apart" (p. 160).
Entering a diameter where a radius is asked. A and a in (1) are radii to the wire centres. Entering the default loops' diameters, 50 and 40 mm, gives 72.20 nH instead of 24.88 nH.
Reading the paper's centimetres as centimetres. Its inductances are in the cgs unit, 1 cm = 10⁻⁹ H, so Example 71's 8616.6 cm is the 8.6166 µH the paper gives beside it, not a length.
Treating k as a property of the wire alone. M depends only on the geometry of the centre lines, but k also depends on the wire diameter through the self-inductances, and on frequency: the same default loops have k = 0.184 at low frequency and 0.194 with the current on the surface.
Forgetting the turns. For coils, M scales with n₁n₂, not with either alone, and an ideal transformer's turns ratio assumes k = 1, which two separate loops never reach.