Inductance per unit length is a fundamental parameter for high-frequency transmission lines and coax cables. This page explains how to estimate L’ with a simple calculator and the classic formula. By entering the medium’s permeability and the conductor radii, you’ll see how spacing and material choice influence energy storage along the line. The result aids impedance planning, signal integrity, and safer, more accurate designs.
Inductance per unit length calculator
Introduction
When designing high-frequency systems, the magnetic energy stored per meter of a conductor affects how a line behaves under load, how it interacts with other components, and how much power is dissipated as heat. The inductance per unit length, L’, is a convenient single parameter that captures those effects for a uniform cross-section. For many transmission lines, including coaxial cables, L’ depends mainly on geometry and the magnetic properties of the material between the conductors. In most practical cases, this medium is air or a dielectric with a known relative permeability close to that of air, so L’ can be estimated with a clean, closed-form expression.
The calculator on this page uses the classic coaxial formula L’ = μ/(2π)·ln(b/a), where μ is the permeability of the medium between the conductors, a is the inner conductor radius, and b is the outer conductor inner radius. The logarithm is natural (base e). This simple relationship makes it easy to compare different cable designs or verify estimates from more complex simulations.
How to use the calculator above
Here is a quick, practical guide to get a reliable L’ estimate:
- Identify the medium between the conductors. Use μ = μ0 × μr, where μ0 is 4π×10^-7 H/m and μr is the relative permeability of the dielectric. For air or most dielectrics, μr is very close to 1, so μ ≈ μ0.
- Measure or choose the inner radius a of the central conductor in meters. This is the radius of the wire or rod inside the coax assembly.
- Measure or choose the outer radius b of the inner surface of the outer conductor in meters. For a coaxial cable, this is the distance from the axis to the inner surface of the outer sheath.
- Enter μ, a, and b into the calculator. The output will be the inductance per unit length L’ in henries per meter (H/m). Remember that many engineers convert to nanohenries per meter (nH/m) by multiplying by 1e9.
- Interpret the result in the context of your design. A larger radius ratio (b/a) increases L’, while a higher μ (more magnetic material) also increases L’.
Worked example with specific numbers
Suppose we have air between the conductors (μ ≈ μ0 = 4π×10^-7 H/m ≈ 1.256637061e-6 H/m). Let the inner conductor radius be 1.5 mm (0.0015 m) and the outer conductor inner radius be 6 mm (0.006 m). The ratio b/a is 0.006/0.0015 = 4, so the natural logarithm is ln(4) ≈ 1.3862943611.
Applying the formula L’ = μ/(2π)·ln(b/a):
- μ/(2π) = 1.256637061e-6 / (2 × π) = 1.256637061e-6 / 6.283185307 ≈ 2.000000000e-7 H/m
- L’ ≈ 2.0e-7 × 1.3862943611 ≈ 2.772588722e-7 H/m
So, the inductance per unit length is approximately 2.77×10^-7 H/m, or about 277 nH/m. This aligns with what one would expect for a coaxial line with modest radii separation in air. If you used a different dielectric with μr > 1, L’ would scale proportionally with μ = μ0×μr.
Why the formula works and how material choices matter
The coaxial formula emerges from solving Maxwell’s equations for a cylindrical geometry with a conductive inner rod and an outer cylindrical conductor. The magnetic flux between the conductors forms a coaxial field that decays with radius. Because the energy stored in the magnetic field is proportional to L’ and to the integral of the magnetic field squared, the logarithmic dependence on the ratio b/a reflects how more space between conductors increases the field region and thus the energy stored per ampere of current.
Material choice matters mainly through μ. If you replace air with a dielectric of relative permeability μr > 1—such as a ferromagnetic core or a magnetizable surrounding material—L’ increases as μ = μ0×μr. In practice, most RF and microwave coax uses non-magnetic dielectrics with μr ≈ 1, so the standard air or dielectric estimates are quite accurate. For specialized devices, engineers may tailor μr by design, then re-evaluate L’ to gauge its impact on impedance and bandwidth.
Practical considerations and related concepts
Relation to impedance and capacitance
In a uniform transmission line, the characteristic impedance Z0 depends on both inductance and capacitance per unit length (L’ and C’). For a coax with a dielectric, Z0 ≈ sqrt(L’/C’). Estimating C’ typically requires the dielectric constant εr, and the geometry; together, these values determine how the line will terminate and how reflections behave. The calculator focuses on L’, but understanding Z0 requires both L’ and C’.
Alternative geometries
The same basic idea applies to other geometries, but the formulas change. For two parallel wires of radii a and b separated by distance D, the inductance per unit length is often approximated by L’ = (μ/π) arccosh(D/(2a)). Different geometries yield different logarithmic or hyperbolic expressions, so it’s important to use the right model for the conductor layout you’re working with.
Frequency dependence and losses
In the quasi-TEM regime common to many RF lines, L’ is effectively frequency-independent over a broad range. At very high frequencies, dispersion, skin effect, and magnetic losses can alter the effective permeability and the distribution of currents, causing small changes in L’. For most design tasks at modest frequencies, treating L’ as a constant is a solid approximation.
Units and unit conversions
Always use SI units when entering values into the calculator. If you need L’ in nH/m, multiply the result in H/m by 1e9. Conversely, to express radii in millimeters, convert them to meters first (1 mm = 0.001 m). Keeping units consistent avoids subtle math mistakes and ensures the calculator’s output matches your engineering conventions.
Tips for accurate, practical use
- Check the magnitudes: radii that differ by an order of magnitude produce larger L’ changes than small variations. A small increase in the b/a ratio can noticeably boost L’.
- Document the dielectric, especially μr, if you’re sharing designs. A note on μr helps teammates reproduce your calculations in the future.
- For educational purposes, compare calculator results with a numerical field solver or a published reference for the same geometry to gain intuition about how geometry controls L’.
- When using the calculator for a design, run a quick sweep over different a and b values to see how sensitive L’ is to spacing around your target specs.
Related Calculators
Other calculators that solve closely related problems:
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- Toroid Inductance Calculator
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- Self Inductance Calculator
- Coax Cable Calculator Impedance Inductance Capacitance
Frequently Asked Questions
1. What is inductance per unit length and why is it important?
Inductance per unit length (L’) measures how much magnetic energy is stored per meter of line for a given current. It directly influences the line’s impedance, signal propagation, and how voltage and current phase relate along the cable. Knowing L’ helps engineers design lines that match impedances, minimize reflections, and control loss characteristics.
2. How do I calculate L’ for a coaxial cable?
For a coaxial cable with inner radius a and outer inner radius b, and medium permeability μ, L’ is μ/(2π) multiplied by the natural logarithm of b over a: L’ = μ/(2π)·ln(b/a). Use μ ≈ μ0 × μr, with μ0 ≈ 4π×10^-7 H/m and μr near 1 for non-magnetic dielectrics.
3. What if I have air between the conductors?
If the space is air, μ ≈ μ0 and μr ≈ 1. This is the common case for many RF cables, giving L’ ≈ μ0/(2π)·ln(b/a) and typically a small L’ value in the 10^-7 to 10^-6 H/m range depending on geometry.
4. How can I convert L’ to a more convenient unit like nH/m?
To convert from henries per meter to nanohenries per meter, multiply by 1×10^9. For example, 2.77×10^-7 H/m equals about 277 nH/m.
5. How does the radius ratio (b/a) affect L’?
The logarithmic term ln(b/a) grows as the ratio increases. Greater spacing between the inner conductor and outer shell increases L’ because the magnetic field occupies a larger region, storing more energy for the same current.
6. Can the calculator handle non-coaxial geometries?
The provided formula is tailored for coaxial geometry. For parallel wires or other configurations, you’ll need the corresponding inductance expression, such as L’ = (μ/π)·arccosh(D/(2a)) for two parallel round conductors with center-to-center spacing D.
7. What role does dielectric material play beyond μr?
The dielectric’s relative permittivity εr primarily influences the capacitance per unit length, which in turn affects the characteristic impedance. The permeability governs L’. In non-magnetic dielectrics (μr ≈ 1) the effect on L’ is minimal, but for magnetic dielectrics, L’ can be noticeably larger.
8. Is L’ constant across frequency?
For typical RF ranges and quasi-TEM propagation, L’ is effectively constant with frequency. At very high frequencies and in materials with significant magnetic or dielectric dispersion, L’ can vary slightly due to changes in μ and ε with frequency.
9. How accurate is the coaxial L’ formula?
The formula assumes ideal cylindrical symmetry, uniform dielectric between conductors, and no fringing effects outside the defined radii. In real cables, end effects, imperfections, and connector interfaces introduce small deviations, but the formula provides a solid benchmark for initial design and analysis.
10. Can I verify the calculator results with measurements?
Yes. You can measure impedance Z0 and estimate L’ from Z0 and C’ using Z0 ≈ sqrt(L’/C’). If you know C’ from the dielectric and geometry, you can back-calculate L’ and compare it with the calculator’s output for consistency.