To calculate the inductance of an ideal solenoid, you use the core geometric formula: L = (μ × N2 × A) / l. This equation links the physical dimensions of your coil and its core material directly to the resulting inductance in Henrys. Whether you are winding a custom choke for a tube amplifier or designing a high-frequency matching network for an RF stage, mastering this formula—and more importantly, mastering the unit conversions—is the difference between a working circuit and a smoked MOSFET.
The Core Inductance Formula and Symbol Definitions
The foundational equation for the inductance of a long, tightly wound solenoid is:
Below is the strict definition of every symbol in this equation, along with the mandatory SI units required for the math to resolve correctly.
| Symbol | Parameter | SI Unit | Notes & Bench Context |
|---|---|---|---|
| L | Inductance | Henry (H) | Often measured in μH or mH on the bench. |
| μ | Core Permeability | Henry/meter (H/m) | μ = μ0 × μr. μ0 is 4π × 10-7 H/m. |
| N | Number of Turns | Dimensionless | Total count of wire loops. Squared in the formula. |
| A | Cross-sectional Area | Square meters (m2) | Area of the core, not the wire. A = π × r2. |
| l | Coil Length | Meters (m) | Physical length of the wound section, not the wire. |
Rearranged Forms: Solving for Any Variable
On the workbench, you rarely solve for L directly. Usually, you have a target inductance and a specific ferrite core, and you need to know how many turns to wind. Here are the algebraically rearranged forms of the core formula:
- Solving for Turns (N): N = √ [ (L × l) / (μ × A) ]
- Solving for Area (A): A = (L × l) / (μ × N2)
- Solving for Length (l): l = (μ × N2 × A) / L
- Solving for Permeability (μ): μ = (L × l) / (N2 × A)
Worked Examples with Strict Unit Tracking
The most common point of failure in inductor design isn't the algebra; it's the unit conversion. Let's walk through two bench-realistic problems, tracking every unit to ensure they cancel correctly.
Example 1: Air-Core RF Choke
Given: You wind N = 500 turns of enameled copper wire on a plastic form. The coil length (l) is 10 cm, and the cross-sectional area (A) is 5 cm2. Calculate L.
• l = 10 cm = 0.1 m
• A = 5 cm2 = 5 × 10-4 m2 (0.0005 m2)
• μ = μ0 (air core) = 4π × 10-7 H/m ≈ 1.2566 × 10-6 H/m
L = [ (1.2566 × 10-6 H/m) × (500)2 × (0.0005 m2) ] / 0.1 m
L = [ (1.2566 × 10-6 H/m) × 250,000 × 0.0005 m2 ] / 0.1 m
L = [ 0.000157075 H·m ] / 0.1 m
L = 0.00157 H (or 1.57 mH)
Example 2: Ferrite-Core Buck Converter Inductor
Given: You need L = 10 mH for a custom power supply. You have a ferrite core with μr = 2000, a cross-sectional area A = 1 cm2, and you want the winding length l = 5 cm. How many turns (N) do you need?
• L = 10 mH = 0.01 H
• A = 1 cm2 = 1 × 10-4 m2
• l = 5 cm = 0.05 m
• μ = μ0 × μr = (4π × 10-7 H/m) × 2000 = 2.513 × 10-3 H/m
N = √ [ (L × l) / (μ × A) ]
N = √ [ (0.01 H × 0.05 m) / (2.513 × 10-3 H/m × 1 × 10-4 m2) ]
N = √ [ 0.0005 H·m / 2.513 × 10-7 H·m ]
N = √ [ 1989.6 ]
N ≈ 44.6 turns (Wind 45 turns in practice).
Assumptions, Realistic Magnitudes, and Unit Pitfalls
When the Formula Applies (and When it Doesn't)
This formula assumes an ideal, infinitely long solenoid where the magnetic field is perfectly uniform inside and zero outside. In reality, if your coil length (l) is not at least 10 times greater than its diameter, edge effects (fringing flux) will cause the actual measured inductance to be slightly lower than the calculated value. For short, stubby coils, you must apply Nagaoka's correction factor or use empirical tools like Wheeler's equations.
What a Realistic Answer Magnitude Looks Like
If your calculator spits out a number, sanity-check it against these typical bench magnitudes:
- Nanohenrys (nH): PCB traces, VHF/UHF RF matching networks, high-speed decoupling.
- Microhenrys (μH): Switch-mode power supplies (buck/boost), EMI filter chokes, Tesla coil primaries.
- Millihenrys (mH): Audio crossover networks, low-frequency line filters, relay coils.
- Henrys (H): Massive utility filter chokes, vintage tube power supply smoothing.
Unit Mistakes That Break the Math
Diameter vs. Radius: The formula requires Area (A = π × r2). If you measure the core's outer diameter with calipers and plug that directly into the radius slot without dividing by 2, your inductance calculation will be off by a factor of 4.
Frequently Asked Questions
How to calculate inductance of a flat spiral coil?
The solenoid formula fails for flat PCB spiral inductors. Instead, use Wheeler's empirical approximation for a flat spiral: L = (r2 × N2) / (8r + 11w), where L is in microhenrys, r is the average radius in inches, N is the number of turns, and w is the winding width in inches. For metric, modern RF designers rely on field solvers or the modified Mohan equation detailed in standard inductance references.
How do you calculate inductance from impedance at a specific frequency?
If you have an LCR meter reading or know the inductive reactance (XL) at a given AC frequency (f), you use the derivative form of Faraday's law: XL = 2πfL. Rearranging to solve for inductance gives L = XL / (2πf). For example, if a coil presents 50Ω of reactance at 100 kHz, L = 50 / (2 × π × 100,000) = 79.5 μH.
How to calculate total inductance in series and parallel?
Assuming the inductors are spaced far enough apart that their magnetic fields do not interact (zero mutual inductance), they combine exactly like resistors. For series: Ltotal = L1 + L2 + ... + Ln. For parallel: 1/Ltotal = 1/L1 + 1/L2 + ... + 1/Ln. However, if you place them close together on the same core, mutual coupling (M) drastically alters the total, requiring the formulas Lseries = L1 + L2 ± 2M depending on the winding phase.
