Electromagnetism · 19.1
Magnetic Flux
Induction is about how much field threads a loop — magnetic flux. Φ = BA cosθ counts it: field strength, loop area, and how squarely the loop faces the field.
Build the model
Connect the measurement to the mechanism.
Picture field lines passing through a wire loop. Flux measures that 'threading': maximal when the loop faces the field square-on, zero when it lies edge-on. Everything about induction is about changing this quantity — by field, by area, or by angle.
- Simple definition
- Magnetic flux is the product of field strength and the area it passes through squarely: Φ = BA cosθ.
- Example
- Tilt a hoop in a steady field from face-on to edge-on and the flux through it falls from maximum to zero — no field line threads it edge-on.
Field times area times how squarely they meet — tilt the loop and less field threads through.
θ between B and the loop's normal
One weber is one tesla of field crossing one square metre of face-on loop.
Unit of flux
A coil of N turns links the flux N times over — coils multiply whatever induction the flux change provides.
N turns each catch the flux
Three ways to change flux
Strengthen or weaken B, grow or shrink the threaded area, or rotate the loop. Each is a route to induction in the next lesson.
Face-on vs edge-on
cosθ does the geometry: 0° gives all the flux, 60° gives half, 90° gives none. θ is measured from the loop's normal, not its plane.
Why flux matters
No flux change, no induced e.m.f. — a coil sitting still in a steady field, however strong, generates nothing.
Change one variable at a time
Make the relationship visible.
Tilt the loop toward edge-on and watch cosθ carve the flux to zero, however strong the field.
Flux Φ = BA cosθ50.0 mWb
Fraction of maximum100%
Catch the common trap
Explain before calculating.
A 0.10 m² loop sits with its normal at 60° to a 0.50 T field. The flux through it is…
Choose an answer to test the model.
Worked examples
State the rule, substitute, then check units.
EasyA 0.20 m² loop faces a 0.50 T field square-on. Find the flux through it.
- Φ = BA = 0.50 × 0.20.
- Φ = 0.10 Wb.
Answer0.10 Wb
MediumA 200-turn coil of area 3.0 × 10⁻³ m² faces a 0.40 T field square-on. Find the flux and the flux linkage.
- Φ = BA = 0.40 × 3.0 × 10⁻³ = 1.2 × 10⁻³ Wb.
- Flux linkage = NΦ = 200 × 1.2 × 10⁻³.
- NΦ = 0.24 Wb-turns.
AnswerΦ = 1.2 mWb; NΦ = 0.24 Wb-turns
HardA circular coil of radius 6.0 cm with 250 turns sits with its normal at 40° to a 0.80 T field. Find the flux linkage.
- A = πr² = π × 0.0036 ≈ 0.0113 m².
- Φ = BA cos40° = 0.80 × 0.0113 × 0.766 ≈ 6.9 × 10⁻³ Wb.
- NΦ = 250 × 6.9 × 10⁻³ ≈ 1.7 Wb-turns.
AnswerNΦ ≈ 1.7 Wb-turns
ChallengingA loop of fixed perimeter 1.2 m can be shaped into a square or a circle in a 0.60 T field, face-on. Which shape threads more flux, and by what factor?
- Square: side 0.30 m, A = 0.090 m². Circle: r = 1.2/2π ≈ 0.191 m, A = πr² ≈ 0.1146 m².
- The circle maximises area for a given perimeter.
- Ratio = 0.1146 ÷ 0.090 ≈ 1.27 — 27% more flux through the circle.
AnswerThe circle — 4/π ≈ 1.27× the square's flux