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Self induction; solenoid and toroid inductance

Short answer

Self-induction is the property where a changing current induces an opposing emf in the same coil, ε = -L di/dt. For a solenoid, L = μ0 N^2 A / l. For a toroid, L = μ0 N^2 A / (2πr). Both depend on geometry and core material.

Explain the meaning of self induction. Calculate the inductance of a solenoid and a toroid.

Self-induction is the property of a coil by which a changing current in it induces an emf in the same coil, opposing the change (Lenz's law). The induced emf is ε = −L di/dt, where the self-inductance L = NΦ/i has unit henry (H).

Inductance of a solenoid

Length l, cross-sectional area A, N turns, n = N/l turns per metre, current i.

  1. Field inside (Ampere's law): B = μ0 n i
  2. Flux through each turn: Φ = BA = μ0 n i A
  3. Flux linkage: NΦ = (nl)(μ0 n i A) = μ0 n² A l i
  4. L = NΦ/i

L = μ0 n² A l = μ0 N² A / l

With a core of relative permeability μr, multiply by μr.

Inductance of a toroid

Toroid of N turns, mean radius r, cross-section area A (small compared with r), current i.

       .------------.
     /    ( toroid )   \
    |     (  core  )    |
     \    (        )   /
       '------------'
         mean radius r
  1. Amperian loop of radius r inside the core: B (2πr) = μ0 N i
  2. B = μ0 N i / (2πr)
  3. Flux per turn: Φ = BA = μ0 N i A / (2πr)
  4. L = NΦ/i

L = μ0 N² A / (2πr)

Both depend only on geometry (and core material), not on the current.

✓ Answer checked by Claude Opus 5.5 (high effort)

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