Retrieval Practice
Magnetic Flux & Flux Linkage — AQA A-Level Physics
Q1. Define magnetic flux.
- The product of magnetic flux density and the cross-sectional area perpendicular to the field. Φ = BA.
- Unit: weber (Wb).
Q2. State the unit of magnetic flux and express it in terms of base SI units of flux density and area.
The weber (Wb). 1 Wb = 1 T m².
Q3. Write the equation for magnetic flux through a coil at angle θ to the field, and define θ.
Φ = BA cos θ, where θ is the angle between the magnetic field direction and the normal to the plane of the coil.
Q4. When is the flux through a coil (a) maximum and (b) zero?
(a) Maximum when θ = 0° — the field is perpendicular to the coil face (normal parallel to field). Φ = BA.
(b) Zero when θ = 90° — the field is parallel to the coil face (normal perpendicular to field). cos 90° = 0.
Q5. Define flux linkage.
- The product of the number of turns on a coil and the magnetic flux through each turn.
- NΦ = BAN cos θ.
- Unit: Wb turns.
Q6. State four ways to change the flux linkage through a coil.
Change the magnetic flux density B, change the area A of the coil, change the angle θ (rotate the coil), or change the number of turns N.
Q7. A coil rotates at constant angular speed ω in a uniform field. Write the equation for flux linkage as a function of time.
NΦ = BAN cos(ωt), where ω is the angular speed in rad s⁻¹ and t is time in seconds.
Q8. For a rotating coil, when is the induced EMF at its maximum value?
- When the flux linkage is zero (θ = 90° or 270°).
- At these positions, the rate of change of flux linkage is greatest, so the induced EMF is at its peak.
Q9. Explain why the flux linkage and induced EMF for a rotating coil are 90° out of phase.
- EMF equals the negative rate of change of flux linkage.
- NΦ varies as cos(ωt), and its derivative is −ω sin(ωt).
- Cosine and sine are 90° apart.
- When one is at a maximum, the other is zero.
Q10. A coil of 200 turns sits perpendicular to a 0.05 T field. Each turn has area 8.0 × 10⁻⁴ m². What is the flux linkage?
NΦ = BAN = 0.05 × 8.0 × 10⁻⁴ × 200 = 8.0 × 10⁻³ Wb turns = 8.0 mWb turns.