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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.
- Field inside (Ampere's law): B = μ0 n i
- Flux through each turn: Φ = BA = μ0 n i A
- Flux linkage: NΦ = (nl)(μ0 n i A) = μ0 n² A l i
- 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
- Amperian loop of radius r inside the core: B (2πr) = μ0 N i
- B = μ0 N i / (2πr)
- Flux per turn: Φ = BA = μ0 N i A / (2πr)
- 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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