Magnetic Flux
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Magnetic flux measures how much field passes through a surface: ΦB = ∫B · dA. For a flat surface in a uniform field it’s BA cos θ, where θ is between B and the surface’s normal. The unit is the weber (1 Wb = 1 T·m²).
1. Counting field through a loop
2. When B isn’t uniform
Slice the loop into thin strips where B is nearly constant, write dΦ = B dA for each strip, and integrate.
For a loop beside a long wire, B = μ0I/(2πr) changes with r, so use strips parallel to the wire: dA = (height) dr.
Closed surfaces: field lines always form loops, so every line that enters a closed surface leaves it. The net flux is zero: there are no magnetic “charges.”
Only the area where the field actually is counts: a big loop around a solenoid has flux B(πrsolenoid²).
3. Example, then try it
A 0.10 m × 0.30 m loop sits in a uniform 0.80 T field that makes 60° with the loop’s normal. Find the flux. What angle gives the most flux?
Φ = BA cos θ = (0.80)(0.030)(cos 60°) = 0.012 Wb.
The most flux is at θ = 0: field perpendicular to the loop’s surface, giving 0.024 Wb.
A square loop of side 0.20 m lies in the xy-plane with corners at x = 0 and x = 0.20 m. The field is B = (2.0 T/m)x, pointing along +z, so it grows across the loop. Find the flux.
Slice: use strips of width dx and height a = 0.20 m. Each strip has dΦ = B a dx = (2.0x)(0.20) dx.
Integrate: Φ = ∫00.20 0.40x dx = 0.20x² from 0 to 0.20.
Result: Φ = 0.20(0.040) = 0.0080 Wb.
Check: B grows linearly, so its average is 0.20 T; 0.20 T × 0.040 m² = 0.0080 Wb.
Answer: 8.0 × 10⁻³ Wb.
4. Common mistakes
A loop’s plane is parallel to a uniform magnetic field. What is the flux through it?
The normal is perpendicular to B, so θ = 90° and BA cos 90° = 0. No field lines pass through.
Ready to practice?
5 questions on Magnetic Flux, each with a hint and an explanation for every wrong answer.