Self-induction, eddy currents and the AC generator
Answers are for parents and teachers.
1
A coil of L = 1.2 H has its current fall from 3.0 A to 0.5 A in 0.05 s. Find the size of the self-induced emf and say whether it acts along the old current or against it.
|emf| = 1.2 × 2.5 ÷ 0.05 = 60 V. The current is falling, so the emf acts along the old current.
2
A self-induced emf of size 40 V appears while the current of a coil changes by 0.8 A in 0.02 s. Find L.
L = 40 × 0.02 ÷ 0.8 = 1 H.
3
A coil of 75 mH has its current changing at 200 A/s. Find the size of the emf. Then say which of a solid iron core and a laminated iron core of the same shape would waste less energy as heat in a changing flux.
75 mH = 0.075 H; |emf| = 0.075 × 200 = 15 V. The laminated core wastes less, because its sheets keep the eddy currents small.
4
A wire 0.9 m long moves at 1.5 m/s through a field of 0.2 T. Find the emf when the velocity is at right angles to the field and when it makes 30° with the field.
At 90°: emf = 0.2 × 0.9 × 1.5 = 0.27 V. At 30°: emf = 0.27 × 0.5 = 0.135 V.
5
A generator coil has N = 300, A = 0.02 m² and B = 0.03 T and turns at 25 Hz (π = 3.14). Find ω, emf(max), and the emf when the normal makes 30° with the field.
ω = 2 × 3.14 × 25 = 157 rad/s. emf(max) = 300 × 0.03 × 0.02 × 157 = 28.3 V. At 30°: 28.3 × 0.5 = 14.1 V.