☰ Year plan · Physics

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
3

The emf in a moving wire and the AC generator

Physics Sec 3 Term 1, Chapter 3: pp. 67–72
GoalCalculate the emf in a wire moving across a field and in a generator coil, emf = NBAω sin θ.
New words
motional emf · the emf in a wire of length ℓ moving at speed v across a field B: emf = Bℓv sin θ, with θ between v and Bslip rings and brushes · two rings fixed to the ends of the rotating coil and two fixed carbon contacts that touch them and lead the current outangular speed (ω) · the angle turned per second in radians: ω = 2πf, with π = 3.14 here
Explanation

A wire of length ℓ moving at speed v at right angles to a field B sweeps an area ℓvΔt in a time Δt, so the flux through the circuit changes by BℓvΔt and the emf is Bℓv. If the velocity makes an angle θ with the field, only the part v sin θ cuts the field lines: emf = Bℓv sin θ. An AC generator turns a coil of N turns and area A at constant speed in a field B. Its two ends are joined to two slip rings, and two brushes touch the rings and lead the current to the outside. With ω = 2πf, the emf is emf = NBAω sin θ, where θ is the angle that the normal to the coil makes with the field. The emf is zero when the coil’s plane stands at right angles to the field (the normal lies along the field, θ = 0), and it is largest, NBAω, when the plane is parallel to the field (θ = 90°). As the coil turns, θ = 2πft (time counted from the moment the plane is perpendicular to the field) and the emf follows emf = emf(max) sin 2πft.

θemf (V)π/2π3π/22π−30−20−101020300emf = 22.6 sin θ
Worked examples
A wire 0.8 m long moves at 3 m/s at right angles to a field of 0.25 T. emf = Bℓv = 0.25 × 0.8 × 3 = 0.6 V. If the velocity makes 30° with the field, emf = 0.6 × sin 30° = 0.3 V.
A generator coil has N = 120 turns, A = 0.015 m², B = 0.04 T and turns at f = 50 Hz (π = 3.14). ω = 2πf = 2 × 3.14 × 50 = 314 rad/s. emf(max) = NBAω = 120 × 0.04 × 0.015 × 314 = 22.6 V. When the normal makes 30° with the field, emf = 22.6 × sin 30° = 11.3 V.
The same generator is turned twice as fast, f = 100 Hz. ω doubles to 628 rad/s, so emf(max) doubles: 120 × 0.04 × 0.015 × 628 = 45.2 V, and the frequency of the emf is 100 Hz.
Class activity

Use the sine graph above with emf(max) = 22.6 V. Read the emf at θ = π/2, at θ = π, and at θ = 3π/2, and say what position the coil is in at each (plane parallel or perpendicular to the field). Then say what the graph does after θ = 2π.

Practice
1
A wire 0.5 m long moves at 6 m/s at right angles to a field of 0.35 T. Find the emf.
2
The same wire moves at the same speed but its velocity makes 30° with the field. Find the emf.
3
A generator coil has N = 250, A = 0.012 m² and B = 0.06 T and turns at 50 Hz (π = 3.14). Find emf(max) and the emf when the normal makes 30° with the field.
4
In which position of a generator coil is the emf zero, and in which is it a maximum? Describe each position by how the plane of the coil lies against the field.