The fundamental equation of inductance bridges the physical geometry of a coil with its electrical behavior in a circuit. At the bench, you need two versions of this formula: the physical derivation (to design or wind your own inductor) and the circuit derivation (to size an inductor for a switching power supply or filter). The physical equation for a long solenoid is L = (μ × N² × A) / l, while the circuit equation is L = V / (di/dt).

The Core Equation of Inductance and Symbol Definitions

To design an inductor from scratch, we use the physical geometry equation. This formula calculates the inductance based on the core material, the number of wire turns, and the physical dimensions of the coil. Below is the complete symbol definition table you need for bench calculations.

Table 1: Symbol Definitions for the Physical Inductance Equation
Symbol Parameter Standard SI Unit Typical Bench Unit
L Inductance Henry (H) μH or mH
μ Permeability of the core (μ₀ × μᵣ) Henries per meter (H/m) H/m
N Number of turns (dimensionless) Turns Turns
A Cross-sectional area of the core Square meters (m²) cm² or mm²
l Length of the magnetic path (coil length) Meters (m) cm or mm

For circuit analysis—such as sizing the output inductor for a buck converter—we rely on Faraday’s law of induction, rearranged to solve for inductance:

L = VL / (di / dt)

  • VL: Voltage across the inductor (Volts)
  • di: Change in current (Amperes)
  • dt: Change in time (Seconds)

Rearranged Forms: Solving for Every Variable

When you are reverse-engineering a transformer or figuring out how many turns to add to a toroid to hit a target inductance, you need the physical equation rearranged. Here are the algebraic forms for the solenoid equation L = (μ × N² × A) / l:

  • Solve for Turns (N): N = √((L × l) / (μ × A))
    Use this when you have a specific core and need to know how many wraps of magnet wire to apply.
  • Solve for Permeability (μ): μ = (L × l) / (N² × A)
    Use this to identify an unknown ferrite material by measuring its physical dimensions and inductance.
  • Solve for Area (A): A = (L × l) / (μ × N²)
  • Solve for Length (l): l = (μ × N² × A) / L
Bench Tip: When solving for N, the result is rarely a perfect integer. Always round up to the nearest whole turn, then recalculate your expected L. A fractional turn is physically impossible on a single-layer winding without creating a shorting loop.

Assumptions, Limits, and Unit Traps

When the Formula Applies (and When It Doesn't)

The physical equation L = (μ × N² × A) / l 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, the formula will overestimate your inductance by 10% to 30% due to fringing fields at the ends. For short, stubby coils, you must apply a Nagaoka correction factor (K) to the result.

Furthermore, the formula assumes a linear core material. If you are using powdered iron or ferrite and you push too much DC current through the winding, the core will saturate. When saturation hits, the effective permeability (μ) plummets, and your physical inductance drops regardless of what the equation says.

Unit Mistakes That Break the Math

  1. The μ₀ vs μᵣ Trap: Datasheets for ferrite cores list relative permeability (μᵣ), which is just a multiplier (e.g., 2000). The equation requires absolute permeability (μ). You must multiply the datasheet value by the permeability of free space (μ₀ = 4π × 10⁻⁷ H/m). Forgetting μ₀ is the #1 reason hobbyists calculate 500mH and measure 50μH on their LCR meter.
  2. The Centimeter Squared Trap: Area must be in square meters (m²). If your core cross-section is 1 cm², that is 1 × 10⁻⁴ m², not 0.01 m². Squaring the conversion factor is a frequent point of failure.
  3. Peak-to-Peak vs Instantaneous: In the circuit equation L = V / (di/dt), di is the ripple current (peak-to-peak), but dt is the on-time of the switch, not the full switching period. Using the full period halves your calculated inductance requirement, leading to a saturated inductor and a blown MOSFET.

Worked Examples with Unit Tracking

Problem 1: Physical Design of an Air-Core RF Choke

Scenario: You need to wind an air-core solenoid for an LC tank circuit. You have a 10mm diameter plastic form (non-magnetic, so μᵣ = 1), and you want an inductance of roughly 5 μH. The coil will be wound tightly over a 20mm length. How many turns do you need?

Given:

  • Target L = 5 μH = 5 × 10⁻⁶ H
  • μ = μ₀ × μᵣ = (4π × 10⁻⁷ H/m) × 1 = 1.2566 × 10⁻⁶ H/m
  • Diameter = 10mm = 0.01m → Area (A) = π × r² = π × (0.005m)² = 7.854 × 10⁻⁵ m²
  • Length (l) = 20mm = 0.02 m

Rearranged Formula: N = √((L × l) / (μ × A))

Step-by-Step Calculation:

  1. Numerator: L × l = (5 × 10⁻⁶ H) × (0.02 m) = 1 × 10⁻⁷ H·m
  2. Denominator: μ × A = (1.2566 × 10⁻⁶ H/m) × (7.854 × 10⁻⁵ m²) = 9.869 × 10⁻¹¹ H·m
  3. Divide: (1 × 10⁻⁷) / (9.869 × 10⁻¹¹) = 1013.27 (Unit check: (H·m) / (H·m) = dimensionless)
  4. Square Root: √1013.27 = 31.83 turns

Answer: Wind 32 turns. (Note: Because the length (20mm) is only twice the diameter (10mm), the ideal solenoid assumption is weak here. In practice, you would wind 32 turns, measure it on an LCR meter, and likely find it reads closer to 6.5 μH due to end-fringing. You would then space the turns out slightly to drop the inductance to exactly 5 μH).

Problem 2: Sizing a Buck Converter Inductor

Scenario: You are designing a 12V-to-5V buck converter switching at 500 kHz. The maximum load current is 3A, and you want to limit the inductor ripple current (di) to 30% of the max load (0.9A). What inductance value is required?

Given:

  • VL during the switch ON time = Vin - Vout = 12V - 5V = 7V
  • di (ripple current) = 0.9A
  • dt (switch ON time) = Duty Cycle / Frequency. Duty Cycle ≈ Vout/Vin = 5/12 = 0.416.
    dt = 0.416 / 500,000 Hz = 8.33 × 10⁻⁷ seconds (833 ns)

Formula: L = VL / (di / dt) which rearranges to L = (VL × dt) / di

Step-by-Step Calculation:

  1. Numerator: VL × dt = 7V × (8.33 × 10⁻⁷ s) = 5.831 × 10⁻⁶ V·s
  2. Divide by di: (5.831 × 10⁻⁶ V·s) / 0.9A = 6.47 × 10⁻⁶ H
  3. Unit check: (V·s) / A = Henry (H)

Answer: You need a 6.47 μH inductor. You would select a standard off-the-shelf value like 6.8 μH with a saturation current rating well above the 3A peak load (e.g., a 5A+ Isat part like the Würth Elektronik 7447742068).

Realistic Magnitudes: What Should Your Answer Look Like?

If your calculator spits out '4500', you need to know if that means 4500 Henrys, milliHenrys, or microHenrys. Context is everything. Use this benchmark table to sanity-check your results against real-world applications.

Table 2: Realistic Inductance Magnitudes by Application
Application Typical Magnitude Real-World Example
RF Matching / GHz Filters 0.5 nH to 50 nH 0402 SMD ceramic chip inductor
High-Freq Switching (GaN/SiC) 100 nH to 2 μH Metal alloy powder core SMD inductors
Standard DC-DC Buck/Boost 4.7 μH to 47 μH Shielded ferrite drum core (e.g., 10 μH)
Audio Crossovers / EMI Chokes 1 mH to 10 mH Thick wire wound on toroidal iron powder
Line Frequency Transformers 1 H to 100+ H Primary winding of a 60Hz mains transformer
Sanity Check Warning: If you are designing a PCB-level switching regulator and your equation yields an inductance of 2.5 H, you have dropped a metric prefix (likely milli or micro) in your time or voltage conversion. A 2.5 Henry inductor capable of handling 1A of DC current would be the size of a brick and cost over $50.

Frequently Asked Questions

How does the equation of inductance change for a toroidal core?

For a toroid, the magnetic path length (l) is not the physical height or width of the core, but the mean circumference of the magnetic path. You calculate this as l = 2 × π × r_mean, where r_mean is the distance from the center of the toroid hole to the center of the core's cross-section. Additionally, because the inner turns are closer together than the outer turns, the field isn't perfectly uniform. For standard hobbyist toroids, the basic solenoid equation using the mean path length is accurate to within 5%, which is well within the manufacturing tolerance of the core material itself.

Why does the equation of inductance use N squared instead of just N?

This is a classic point of confusion. The first 'N' accounts for the fact that N turns of wire create a magnetic field (H) that is N times stronger than a single turn. The second 'N' accounts for Faraday's law of induction: the induced voltage is proportional to the number of turns the changing magnetic flux passes through. Because the flux generated by the coil is passing back through the exact same coil, you multiply the generation factor (N) by the interception factor (N), resulting in N². Doubling your turns quadruples your inductance.

Can I use the equation of inductance for AC circuits with high frequencies?

The physical equation L = (μ × N² × A) / l calculates the theoretical, low-frequency inductance. At high frequencies (typically above 1 MHz for standard ferrites, or lower for iron powder), two things break the model. First, the core material's effective permeability drops due to magnetic lag (eddy currents and hysteresis). Second, parasitic capacitance between the wire windings creates a self-resonant frequency (SRF). Above the SRF, the component stops acting like an inductor and acts like a capacitor. For high-frequency RF design, you must rely on the manufacturer's S-parameter tables rather than the raw geometric equation.

What happens to the inductance equation when the core saturates?

When the DC bias current exceeds the core's saturation limit, the magnetic domains in the ferrite or iron align completely. At this point, the core can no longer amplify the magnetic field, and the relative permeability (μᵣ) crashes from a value like 2000 down to 1 (the equivalent of air). In the equation L = (μ × N² × A) / l, the μ term plummets, causing the inductance to drop drastically. This is why power inductor datasheets specify an Isat (saturation current) rating alongside an Irms (thermal heating) rating. Always ensure your peak current stays below Isat to maintain the inductance your circuit equations rely on.