Straight Wire Inductor Calculator
Estimate the inductance of a straight round wire from its length and diameter. Useful for RF circuits, antenna elements, lead wires, and bus bars where even short conductors add reactance.
Inductance Estimator for Straight Round Conductors
Straight Wire Inductance Formula
Where:
- \(L\) — inductance of the wire (µH)
- \(a\) — length of the wire (cm)
- \(d\) — diameter of the wire (cm)
This empirical formula approximates the inductance of a straight round conductor in free air. It is often written in the equivalent form \(L = 0.002\,a\left[\ln\!\left(\frac{4a}{d}\right) - 1\right]\) in µH with \(a\) and \(d\) in cm; both give nearly identical results for thin, straight wires.
Straight Wire Inductance – Calculation Example
Example: a straight copper wire is 10 cm long and has a diameter of 0.1 cm (about 1 mm, roughly AWG 18).
\[ L = 0.00508 \times 10 \times \left( \log_{10}\!\left( \frac{2 \times 10}{0.1} \right) - 0.75 \right) \]
\[ L = 0.0508 \times \left( \log_{10}(200) - 0.75 \right) = 0.0508 \times (2.3010 - 0.75) \]
\[ L \approx 0.0508 \times 1.5510 \approx 0.0788~\mu\mathrm{H} \]
Result: the wire has an inductance of approximately 0.079 µH (79 nH).
How to Use the Calculator
- Enter the Wire Length in centimeters — the end-to-end length of the straight conductor.
- Enter the Wire Diameter in centimeters — the diameter of the round cross-section, not the radius. If you only know the wire gauge, convert it to a diameter first (see the Wire Gauge calculator).
- Read the Inductance result in microhenries (µH). For typical lead wires the value is small, so you may prefer to read it as nanohenries (1 µH = 1000 nH).
The calculator requires \(a > 0\), \(d > 0\), and \(2a > d\). The last condition reflects the fact that the formula is only meaningful for a wire that is long compared to its thickness; a stubby cylinder is not a "straight wire" in the sense of this model.
What Is the Inductance of a Straight Wire?
Any current-carrying conductor is surrounded by a magnetic field, and that field stores energy. A changing current therefore meets an opposing voltage — this is inductance. It is tempting to think of inductance as something that only coils have, but even a perfectly straight piece of wire possesses a small amount of it. The value depends mostly on the wire's length and, more weakly, on its thickness: longer wires have more inductance, and thicker wires have slightly less, because the magnetic flux loops around a fat conductor enclose less length per turn.
The numbers are small but far from negligible in the right context. A 10 cm wire of 1 mm diameter — roughly the length of a component lead or a jumper — contributes about 79 nH, as in the example above. At 100 MHz that translates into a reactance of \(X_L = 2\pi f L \approx 50~\Omega\), which can completely dominate a circuit's intended impedance. This is why RF layouts obsess over short, direct traces, and why a "ground wire" a few centimeters long is no longer a good ground at high frequencies.
Typical figures to keep in mind: a 1 cm wire adds roughly 7–10 nH, and standard through-hole component leads (about 1 cm each, 0.5 mm diameter) contribute several nH per lead. When datasheets for RF components specify layouts with minimal lead length, this parasitic inductance is exactly what they are guarding against.
This calculator is most useful when estimating parasitic inductance of: antenna elements and radials, jumper and lead wires, bus bars in power electronics, and bonding wires in prototypes. If you are designing transmission lines instead, the geometry of both conductors matters — see the Parallel Wire Impedance calculator for the two-wire line case, or the Microstrip PCB Impedance calculator for traces over a ground plane.
Assumptions, Limitations and Practical Tips
- Round conductor in free air. The formula assumes a straight wire of circular cross-section, far from other conductors and from magnetic or conductive materials. Nearby metal (a chassis, a ground plane) or insulation changes the field distribution and the actual inductance.
- Thin-wire approximation. Accuracy is best when \(a \gg d\). The formula becomes inaccurate as the wire approaches its own length in thickness; hence the \(2a > d\) requirement in the calculator.
- Low frequency / uniform current. At high frequencies the skin effect pushes current toward the surface, slightly reducing internal inductance. The external inductance — which dominates for thin wires — is barely affected, so the formula remains a good first estimate well into the VHF range.
- No proximity or mutual effects. The result applies to a single isolated wire. Two parallel wires couple magnetically; if their fields partially cancel (as in a tightly twisted pair or a coaxial return), the net inductance is much lower than the sum of the individual wires.
- Expect a few percent of uncertainty. This is an empirical engineering formula, not an exact solution. For sign-off on critical RF designs, verify with a field solver or measurement — but for estimating whether a lead is "a few nH or a few dozen nH," it is entirely adequate.
A practical rule: to minimize parasitic inductance, keep current paths short, use the widest conductor practical, and route the return current directly alongside the outgoing current so their magnetic fields cancel. When several inductances combine in a circuit, the Series Inductance and Parallel Inductance calculators can help you total them up.
Frequently Asked Questions
Does the wire material (copper, aluminum, silver) affect the inductance?
No — for a straight wire in air, inductance depends only on geometry (length and diameter), not on conductivity. Copper, aluminum, or silver wires of the same dimensions have virtually the same inductance. Material only affects resistance and skin-effect losses. This is why the formula needs no conductivity input, and why plating or alloying a conductor does not change its parasitic inductance.
How do I calculate straight wire inductance in inches or millimeters?
The formula on this page expects length and diameter in centimeters and returns microhenries. If your dimensions are in inches, multiply them by 2.54 to get centimeters first. Alternatively, use the imperial form \(L = 0.00508\,a\left[\ln\left(\frac{4a}{d}\right) - 1\right]\) with \(a\) and \(d\) in inches, giving \(L\) in µH. Both routes give the same answer.
How do I convert the inductance into a reactance at a given frequency?
Use \(X_L = 2\pi f L\), with \(f\) in hertz and \(L\) in henries. For example, a 79 nH lead at 100 MHz gives \(X_L \approx 2\pi \times 10^8 \times 79 \times 10^{-9} \approx 50~\Omega\). You can also find the frequency where the wire becomes 'significant': reactance reaches just 1 Ω at about \(f = 1/(2\pi L)\) Hz.
What is a typical inductance value for a wire lead or PCB trace?
As practical rules of thumb: a 1 cm length of ordinary wire adds roughly 7–10 nH, and a through-hole component lead (about 1 cm of 0.5 mm wire) contributes several nH per lead. A 10 cm jumper is around 80–100 nH. A PCB trace works out to roughly 0.5–1 nH per millimeter of length. These small values matter mainly above a few tens of MHz.
Why does a thicker wire have less inductance than a thin one of the same length?
Inductance measures how much magnetic flux links the wire per ampere of current. Around a fat conductor, the flux loops 'cut' the wire over a shorter effective length, so fewer flux linkages are produced per amp. The effect is weak — inductance grows only logarithmically as the wire gets thinner — which is why doubling the diameter changes the result by just a few percent, while doubling the length roughly doubles it.
How do I find the wire diameter from an AWG gauge?
Each AWG step changes the diameter by a factor of about \(10^{-1/40} \approx 0.891\), or roughly a factor of 2 every 6 gauges. As quick reference points: AWG 10 ≈ 2.59 mm, AWG 14 ≈ 1.63 mm, AWG 18 ≈ 1.02 mm, AWG 22 ≈ 0.64 mm, and AWG 26 ≈ 0.40 mm. Convert the millimeter value to centimeters (divide by 10) before entering it into the calculator.
Does a bent, L-shaped, or coiled wire follow this formula?
Only approximately, and only for gentle bends. The formula models a perfectly straight isolated wire; sharp corners concentrate the magnetic field and change the inductance by up to a few percent per bend. Once the wire forms even a partial loop, loop area dominates and this model breaks down — use a coil or loop inductance formula instead. For a wire running parallel to a ground plane, the actual inductance is lower than this calculator predicts.
15 thoughts on “Straight Wire Inductor Calculator”