Inductance equals permeability times turns squared times area divided by coil length

Solution

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How It Works

The inductance of a solenoid (cylindrical coil) is set by four factors: the core material's permeability, the number of wire turns, the cross-sectional area, and the coil length. Inductance rises with the square of the turns count, so doubling the turns quadruples inductance. A ferromagnetic core can boost permeability by thousands of times compared to air.

Example Problem

Design a 1 mH air-core inductor with 200 turns and a 0.001 m² cross-section. What coil length is needed?

  1. μ for air ≈ 4π × 10−7 = 1.2566 × 10−6 H/m
  2. l = μN²A / L = (1.2566e−6 × 200² × 0.001) / 0.001
  3. l = (1.2566e−6 × 40,000 × 0.001) / 0.001 = 0.0503 m (5 cm)

Frequently Asked Questions

Why does inductance scale with turns squared?

Each turn of wire both creates and links magnetic flux. Doubling the turns doubles the flux created and doubles the flux linked, giving a quadratic (N²) relationship.

What is the permeability of free space?

μ0 = 4π × 10−7 H/m ≈ 1.2566 × 10−6 H/m. Iron cores have relative permeabilities of 1,000–10,000, dramatically increasing inductance.

When is this solenoid formula not accurate?

It assumes a long, tightly wound single-layer coil. For short coils (length < 2× diameter), toroidal shapes, or multi-layer windings, use Wheeler's approximation or Nagaoka's correction factor.

What are common applications of inductors?

Inductors are used in power supply filters, RF circuits, transformers, impedance matching networks, and energy storage. Switching regulators depend on inductors to smooth output current.

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