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Inductance in terms of physical dimensions

Short answer

From Ampere’s law, H = NI/l, B = μH, flux Φ = BA = μ N I A/l. Inductance L = NΦ/I = μ N² A/l, where μ = μ0 μr, N turns, A cross-sectional area, l mean length. L increases with N² and core area, and decreases with length.

Derive the expression for the inductance of inductor in terms of its physical dimensions.

Consider a coil (solenoid or toroid) of N turns wound on a core of mean length l, cross-sectional area A and relative permeability μ_r.

   +-----------------------+
   | ( ( ( ( ( ( ( ( ( ( ( |   N turns, current I
   +-----------------------+
   <---------- l --------->    core area A, permeability μ

Derivation

Step 1: magnetic field strength. By Ampere's circuital law, the MMF of the coil is N I, so

H = N I/l (A/m)

Step 2: flux density. B = μ H = μ₀ μ_r N I/l (tesla)

Step 3: flux. Φ = B A = μ₀ μ_r N I A/l (weber)

Step 4: inductance. Inductance is the flux linkage per ampere:

L = N Φ/I = N × μ₀ μ_r N A/l

L = μ₀ μ_r N² A/l = μ N² A/l henry

where μ₀ = 4π × 10⁻⁷ H/m.

(Equivalent form: L = N²/S, where S = l/(μ A) is the reluctance of the magnetic circuit.)

Conclusions

  • L is proportional to N²: doubling the turns makes the inductance four times larger.
  • L is proportional to the core area A and permeability μ (iron core gives a much larger L than air).
  • L is inversely proportional to the length l.

Example: N = 500, A = 4 cm², l = 20 cm, μ_r = 1000: L = 4π × 10⁻⁷ × 1000 × 500² × 4 × 10⁻⁴/0.2 = 0.628 H.

✓ Answer checked by Claude Opus 5.5 (high effort)

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