Self-Induction
Coils resist changes in their own current.
Lesson at a glanceShow the key points
- Φ = Li (38.5); ℰ = −L di/dt (38.6); henry = Wb/A.
- Long solenoid: L = μ₀n²πr²l (38.7).
- The self-induced emf opposes changes in current.
A coil carrying a current threads its own magnetic flux. Change the current and the coil fights back with an emf of its own: self-induction.
Self-inductance
The field of a current, and so the flux through the loop carrying it, is proportional to the current:
The unit is the weber per ampere, equivalently volt-second per ampere, named the henry (H). For a coil of N turns, add the emf of every turn: ℰ = −N d(∫B⃗·dS⃗)/dt. By Lenz’s law the self-induced emf opposes the change: it pushes against a rising current and tries to keep a falling one going.
An average emf of 0.30 V appears in a coil while its current is reversed from 3.0 A one way to 3.0 A the other way in 0.10 s. Find L.
Inductance of a long solenoid
A solenoid of radius r with n turns per length carries i: B = μ₀ni inside, so each turn links μ₀niπr². A length l has nl turns, giving a total emf −(nl)(μ₀nπr²) di/dt. Comparing with −L di/dt:
Find the self-inductance of a solenoid 30 cm long and 2.0 cm in radius with 2000 turns per metre.
A coil of thick wire has negligible resistance but appreciable inductance: an ideal inductor, drawn as a row of loops.
Inductance resists sudden change
A 12 V cell drives current through an inductor and a rheostat. At one instant the rheostat reads 8 Ω and is being increased. Is the current more or less than 1.5 A at that instant?
- 1Φ = Li (38.5); ℰ = −L di/dt (38.6); unit henry = Wb/A = V s/A.
- 2Long solenoid: L = μ₀n²πr²l (38.7).
- 3Self-induced emf opposes changes in current (Lenz).
Check yourself
Keep exploring
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