Magnetic flux and Faraday's laws of induction
Magnetic flux Φ = BA cos θ measures field lines through an area. The induced emf equals the rate of change of flux: ε = −N dΦ/dt.
Magnetic flux
Magnetic flux through a surface tells us how much magnetic field passes through it. Think of it as the number of field lines crossing the surface. For a plane area A in a uniform field B, Φ = BA cos θ. Here θ is the angle between B and the area vector, an arrow drawn normal (perpendicular) to the surface. Flux is greatest, BA, when the field is perpendicular to the surface (θ = 0°). It is zero when the field runs along the surface (θ = 90°). Flux is a scalar. Its SI unit is the weber (Wb), and 1 Wb = 1 T m². For a coil of N turns, the flux linkage is NΦ.
Faraday's laws of electromagnetic induction
- First law: whenever the magnetic flux linked with a circuit changes, an emf is induced in it. The emf lasts only as long as the flux keeps changing.
- Second law: the magnitude of the induced emf equals the rate of change of magnetic flux linked with the circuit.
- In symbols, for one turn: ε = −dΦ/dt. For a closely wound coil of N turns: ε = −N dΦ/dt.
- The minus sign shows the direction of the emf. It comes from Lenz's law.
- For a change ΔΦ in a time Δt, the average emf is ε = N ΔΦ/Δt in magnitude.
Three ways to change the flux Φ = BA cos θ
| What changes | How | Example |
|---|---|---|
| B, the field | Move a magnet, or change the current in a nearby coil | Faraday's magnet and coil experiment |
| A, the area inside the field | Slide a rod on rails, or squeeze or stretch the loop | Motional emf |
| θ, the angle | Rotate the coil in the field | AC generator |
Worked example: a changing field
A square coil of side 10 cm has 500 turns. Its plane is perpendicular to a uniform magnetic field of 0.20 T. The field is reduced steadily to zero in 0.10 s. Find the emf induced in the coil.
- Given: N = 500, side = 10 cm = 0.10 m, so A = 0.10 m × 0.10 m = 1.0 × 10⁻² m².
- The field is perpendicular to the plane, so θ = 0° and cos θ = 1.
- Initial flux per turn Φ₁ = BA = 0.20 T × 1.0 × 10⁻² m² = 2.0 × 10⁻³ Wb. Final flux Φ₂ = 0.
- Change in flux per turn ΔΦ = 2.0 × 10⁻³ Wb.
- Formula: ε = N ΔΦ/Δt.
- ε = 500 × 2.0 × 10⁻³ Wb ÷ 0.10 s = 1.0 ÷ 0.10 = 10 V.
Induced emf = 10 V
Worked example: turning a coil
A coil of 200 turns and area 0.05 m² lies with its plane perpendicular to a uniform field of 0.4 T. It is turned in 0.2 s until its plane is parallel to the field. Find the average emf induced.
- Given: N = 200, A = 0.05 m², B = 0.4 T, Δt = 0.2 s.
- At the start, the area vector is along B, so θ = 0°: Φ₁ = BA cos 0° = 0.4 × 0.05 = 0.02 Wb.
- At the end, the plane is parallel to B, so θ = 90°: Φ₂ = BA cos 90° = 0.
- ΔΦ = 0.02 Wb per turn.
- Average emf ε = N ΔΦ/Δt = 200 × 0.02 Wb ÷ 0.2 s = 4 ÷ 0.2 = 20 V.
Average induced emf = 20 V
When a debit card's magnetic stripe is swiped, its tiny magnetised patches move past a coil in the reader. The changing flux induces small emfs, which the machine reads as data.
1 marks: “Write the SI unit of magnetic flux and express it in terms of the tesla.” Weber (Wb); 1 Wb = 1 T m².
2 marks: “State Faraday's laws of electromagnetic induction.” First: an emf is induced whenever the flux linked with a circuit changes, and it lasts only while the flux changes. Second: its magnitude equals the rate of change of flux linkage, ε = −N dΦ/dt.
3 marks: “A coil of 100 turns and area 0.04 m² is held perpendicular to a field that rises from 0.1 T to 0.6 T in 0.5 s. Find the induced emf. What would it be if the rise took 0.25 s?” ΔΦ per turn = 0.5 T × 0.04 m² = 0.02 Wb; ε = 100 × 0.02 Wb ÷ 0.5 s = 4 V. In 0.25 s the rate doubles, so ε = 8 V.
Flux needs B, A and cos θ: change any one and an emf appears. Faraday counts how fast it changes.