Find precise inductance, impedance, capacitance, and propagation delay for two parallel wires using the Parallel Wire Inductance Calculator’s exact hyperbolic-cosine formula.
The Exact Two-Wire Inductance Formula (Not the Simplified Approximation)
Two parallel round wires carrying opposite currents form a loop, and that loop has inductance determined entirely by their geometry: how thick they are, how far apart they sit, and how long the run is.
Most references quote a simplified version of this relationship, valid only when the spacing is much larger than the wire diameter. This calculator uses the exact form instead, built on the inverse hyperbolic cosine, so it stays accurate even at tight spacing where the approximation breaks down.
The external inductance per conductor pair is:
$$L = \frac{\mu_0 \mu_r}{\pi} \, l \, \cosh^{-1}\!\left(\frac{D}{d}\right)$$
Here $d$ is conductor diameter, $D$ is center-to-center spacing, $l$ is the length of the run, $\mu_r$ is the relative permeability of the surrounding medium, and $\mu_0 = 4\pi \times 10^{-7}$ H/m is the magnetic constant.
The term $\cosh^{-1}(D/d)$ expands to $\ln\!\left(\frac{D}{d} + \sqrt{(D/d)^2 – 1}\right)$, which is where the “exact” part comes from — it holds for any spacing greater than the wire diameter, not just the widely-spaced case.
That same $\cosh^{-1}(D/d)$ term also drives the transmission-line properties of the pair. Capacitance per pair is $C = \pi \varepsilon_0 l \, / \cosh^{-1}(D/d)$, and characteristic impedance is $Z_0 = \frac{\eta_0 \sqrt{\mu_r}}{\pi}\cosh^{-1}(D/d)$, where $\eta_0 = \sqrt{\mu_0/\varepsilon_0} \approx 376.7\ \Omega$ is the impedance of free space.
One geometric ratio produces inductance, capacitance, and impedance together, because they all come from the same magnetic and electric field distribution around the two conductors.
A 5 cm-Spaced Pair of 2 mm Wires Over a 10-Meter Run
Take a pair of 2 mm diameter wires — roughly 12 AWG — spaced 5 cm apart center-to-center, running 10 meters, surrounded by ordinary air ($\mu_r = 1$). The spacing-to-diameter ratio is $D/d = 50\text{mm}/2\text{mm} = 25$, and $\cosh^{-1}(25) = \ln(25 + \sqrt{624}) \approx 3.91$.
Plugging that into the inductance formula: $$L = \frac{4\pi \times 10^{-7}}{\pi} \times 10 \times 3.91 \approx 15.65\ \mu\text{H}$$. The same $\cosh^{-1}$ value feeds every other output at once.
| Quantity | Result |
|---|---|
| Inductance | 15.65 µH |
| Characteristic impedance | 469.1 Ω |
| Capacitance | 71.10 pF |
| Propagation delay | 33.36 ns (3.34 ns/m) |
| Velocity | 299.79 Mm/s (velocity factor 1.00) |
| Max frequency before radiation loss | 599.58 MHz |
| DC loop resistance | 106.95 mΩ (10.70 mΩ/m) |
The DC resistance comes from a separate calculation entirely — plain $R = \rho l / A$ for annealed copper, applied to the full 20-meter loop (10 m out, 10 m back) — and has nothing to do with the spacing between the wires.
What a 469 Ω Result Actually Tells You
There’s no pass/fail threshold buried in this number the way there is with voltage drop or ampacity. What there is: a useful comparison point.
Commercially sold open-wire and ladder-line feeders typically run in the 300–600 Ω range, with 450 Ω window line being one of the most common products on the market.
A result around 469 Ω sits squarely in that band, which is a reasonable sanity check that the geometry you entered resembles a real feedline rather than something unusual.
The inductance figure by itself is less directly useful. It matters for antenna feedlines, twisted-pair signaling, and loop-antenna design, but it’s rarely the number you act on — impedance and delay usually are, because those are what you match a transmitter, receiver, or driver circuit against.
Why the Wire Doesn’t Radiate Until Almost 600 MHz
The 599.58 MHz figure comes from the common rule that a transmission line stops behaving as a simple guided pair once conductor spacing reaches about a tenth of a wavelength — beyond that point, the line radiates instead of just carrying signal.
At 5 cm spacing, that threshold sits well above anything relevant to HF or VHF work, so for most amateur radio and low-frequency signaling applications it’s not a practical concern.
The Spacing-to-Diameter Ratio Drives Everything Else
Because every output depends on $\cosh^{-1}(D/d)$, and that function grows roughly logarithmically, spacing doesn’t move the result the way people expect.
Double the spacing in the example above — from 50 mm to 100 mm — and $D/d$ jumps from 25 to 50, but $\cosh^{-1}(50) \approx 4.60$ versus $\cosh^{-1}(25) \approx 3.91$.
That’s about an 18% increase in inductance and impedance, not the 100% a linear relationship would suggest. Wire diameter has the same diminishing-returns effect in reverse: shrinking $d$ raises $D/d$, but each further reduction buys less.
Permeability is a separate lever, and one that’s easy to misuse. It describes the medium around and between the conductors, not the metal the wire is made from — bare copper or aluminum in air is non-magnetic, so $\mu_r = 1$ is correct for nearly every real case.
Entering a higher value only makes sense if the pair runs through an actual magnetic core or specialized magnetic medium, and doing so scales inductance and impedance up directly while slowing propagation velocity by $1/\sqrt{\mu_r}$.
Two more things worth knowing about what’s under the hood. First, this is external inductance only — the field outside the conductors, which is what dominates at radio frequency once skin effect pushes current toward the surface.
At DC or low frequency, current fills the conductor’s cross-section and adds a small internal-inductance term this formula doesn’t include.
Second, the capacitance and impedance figures assume free-space permittivity between the wires; real insulation raises the effective dielectric constant, which increases capacitance and lowers impedance below what a bare-wire-in-air calculation shows.
Frequently Asked Questions
Does wire insulation change the inductance value?
Barely — insulation has negligible relative permeability, so it doesn’t touch the magnetic-field term. It does affect capacitance and impedance, since those assume an air dielectric, and real insulation typically pulls impedance a bit lower than this calculator shows.
Why does doubling the spacing barely change the inductance?
Because the formula depends on the inverse hyperbolic cosine of the spacing-to-diameter ratio, not the ratio itself. That function grows roughly logarithmically, so a 2x spacing increase typically adds under 20% to the result rather than doubling it.
What’s the difference between this and the “total” inductance of the pair?
This is external inductance only, the part tied to the field outside the conductors and dominant at RF once skin effect takes over. At DC or low frequency, current fills the whole conductor and adds a small internal-inductance term this calculation leaves out.
Can I use this for coaxial cable or twisted-pair instead of open wire?
No. Coax and twisted pair have different conductor geometries — concentric versus side-by-side round wires — and need their own formulas. This one applies specifically to two round, parallel conductors in air, the classic open-wire or ladder-line arrangement.
What permeability value should I use for ordinary copper or aluminum wire?
Use 1. The permeability input describes the medium around the conductors, not the metal itself, and bare wire in air is non-magnetic. Higher values only apply if the pair runs through a magnetic core or medium.
Why does the calculator reject spacing smaller than the wire diameter?
Because the geometry stops making sense at that point — center-to-center spacing smaller than the diameter means the two conductors would physically overlap. The formula is only defined once $D$ exceeds $d$.
Is a higher characteristic impedance better for a feedline?
Neither direction is inherently better — it depends on matching source and load impedance across your whole system. Commercial open-wire feeders commonly land between 300 and 600 Ω because that range balances loss, practical spacing, and compatibility with common antenna and tuner designs.
At what point does spacing cause the line to radiate instead of guide the signal?
Once spacing approaches roughly a tenth of the signal’s wavelength, the line stops acting as a simple guided pair and starts radiating measurably. That’s the max-frequency figure this calculator reports — below it, expect predictable behavior; well above it, expect losses these formulas don’t capture.
This calculates ideal, static-geometry values for a bare, straight, uniformly-spaced pair of conductors in open air — it doesn’t account for insulation raising the dielectric constant, bends or twists changing the effective spacing, proximity effect at very tight D/d ratios, or the rise in copper’s resistivity with temperature that shifts the DC resistance figure over time.