Self-induction, eddy currents and the AC generator
Week 9 · Chapter 3 (B): self-induction (emf = −L ΔI/Δt, the henry), what the self-inductance depends on, eddy currents and induction furnaces, the emf in a straight wire moving across a field (emf = Bℓv sin θ), and the construction and emf of the AC generator · 3 lessons
1
Self-induction and its coefficient L
Physics Sec 3 Term 1, Chapter 3: pp. 67–72
GoalExplain why a coil opposes changes of its own current and calculate emf = −L ΔI ÷ Δt.
New words
self-induction · the induction of an emf in a coil by the changing current in the same coilcoefficient of self-induction (L) · the self-induced emf for each unit rate of change of the current: L = |emf| ÷ (ΔI ÷ Δt)self-induced emf · the emf that appears in the coil itself and opposes the change of its current
Explanation
The current in a coil makes a magnetic flux through the coil itself. If the current changes, so does this flux, and by Faraday’s law an emf is induced in the same coil. By Lenz’s rule it opposes the change. When the current is switched on, the self-induced emf opposes the rise, so the current climbs gradually instead of jumping to its final value. When the current is cut, the collapsing flux induces an emf that opposes the fall: it tries to keep the current flowing in its old direction, and it can be much larger than the battery’s emf because the fall is so quick. The size is emf = −L ΔI ÷ Δt, with the positive direction taken along the current. The classic demonstration with a large electromagnet, a spark at the switch and a neon lamp is never to be tried: the idea is enough, and sparks and high voltages are not for home or paper tasks.
Worked examples
L = 0.6 H and the current rises from 0 to 2.0 A in 0.040 s (positive direction: along the current). emf = −L ΔI ÷ Δt = −0.6 × 2.0 ÷ 0.040 = −30 V. It has size 30 V and opposes the rise.
L = 0.30 H and the current falls from 5.0 A to 1.0 A in 0.020 s (positive direction: along the old current). ΔI = −4.0 A, so emf = −0.30 × (−4.0) ÷ 0.020 = +60 V. The sign is positive: the emf pushes the current the same way as before, trying to keep it.
A coil shows a self-induced emf of size 27 V while its current changes by 1.8 A in 0.05 s. L = |emf| × Δt ÷ |ΔI| = 27 × 0.05 ÷ 1.8 = 0.75 H.
Class activity
Paper practice, no switch, spark or lamp. Draw two graphs of current against time for a coil on a cell: one for switching on (a gradual rise) and one for switching off (a fall that takes a short time). On each, write the direction of the self-induced emf: against the current or along the current?
Practice
1
A coil of L = 0.8 H has its current rise from 0 to 2.5 A in 0.10 s. Find the size of the self-induced emf.
|emf| = 0.8 × 2.5 ÷ 0.10 = 20 V.
2
A coil of L = 1.5 H has its current fall from 4.0 A to 0 in 0.25 s. Find the size of the self-induced emf and say whether it acts along the old current or against it.
|emf| = 1.5 × 4.0 ÷ 0.25 = 24 V. The current is falling, so the emf tries to keep it: it acts along the old current.
3
A coil of L = 0.4 H has a self-induced emf of size 12 V. At what rate is its current changing?
ΔI ÷ Δt = |emf| ÷ L = 12 ÷ 0.4 = 30 A/s.
4
A lamp is in series with a coil of large L and a cell. Why does the lamp come on slowly when the circuit is closed?
The current is rising, so the self-induced emf opposes the rise and the current grows gradually.
