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Self-inductance of a toroid

Short answer

Electromagnetic induction is emf from any changing flux; self-induction is emf from a coil’s own changing current. For a toroid, Ampere’s law gives B = μ₀Ni/(2πr), so flux linkage NΦ = μ₀N²Ai/(2πr), yielding self-inductance L = μ₀N²A/(2πr).

Differentiate between electromagnetic induction and self-induction. Develop an expression for self-inductance of a toroid having N turns, radius r and carrying current i.

Electromagnetic induction versus self-induction

Electromagnetic inductionSelf-induction
An emf is induced in a circuit when the flux linked with it changes, from any causeAn emf is induced in a coil due to the change of current in that same coil
The changing flux may come from another circuit, a moving magnet or a change in area or orientationThe changing flux is produced by the coil's own current
Mutual induction involves two coilsInvolves one coil
Law: ε = −dΦ/dtε = −L di/dt
Depends on the rate of flux changeDepends on the inductance L and di/dt
Example: generator, transformerExample: choke coil, solenoid

Self-induction is a special case of electromagnetic induction.

Self-inductance of a toroid

Toroid with N turns, mean radius r, core area A, carrying current i.

       .------------.
     /   N turns, i    \
    |  .------------.   |   mean radius r
     \ '------------' /
       '------------'
  1. Take a circular Amperian loop of radius r inside the core: B (2πr) = μ0 N i
  2. B = μ0 N i / (2πr)
  3. Flux through one turn: Φ = B A = μ0 N i A / (2πr)
  4. Total flux linkage: NΦ = μ0 N² A i / (2πr)
  5. Inductance L = NΦ/i:

L = μ0 N² A / (2πr)

It depends only on the geometry and core material, increasing with N² and A and decreasing with r. With an iron core of relative permeability μr, L is multiplied by μr.

✓ Answer checked by Claude Opus 5.5 (high effort)

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