Magnetic Flux (Φ)
0 Wb
Flux Linkage (NΦ)
0 Wb
In mWb
0 mWb
In μWb
0 μWb

What is Magnetic Flux?

Magnetic flux (Φ) is a measure of the total magnetic field (B) passing through a given area (A). Think of it as counting how many magnetic field lines penetrate a surface. The unit is the Weber (Wb), where 1 Wb = 1 T·m².

Magnetic flux is fundamental to understanding electromagnetic induction — the more flux changes through a coil, the greater the induced EMF.

Φ = B × A × cos(θ)

Where:

  • B = magnetic field strength (Tesla, T)
  • A = area of the surface (m²)
  • θ = angle between the field and the normal (perpendicular) to the surface

Understanding the Angle θ

The angle θ is crucial and often misunderstood. It's measured from the normal (perpendicular line) to the surface, not from the surface itself.

Key Cases:
• θ = 0°: Field perpendicular to surface → Maximum flux (Φ = BA)
• θ = 90°: Field parallel to surface → Zero flux (Φ = 0)
• θ = 60°: Φ = BA × cos(60°) = 0.5BA

Flux Linkage

Flux Linkage = N × Φ = N × B × A × cos(θ)

For a coil with N turns, the flux linkage is the total flux through all turns combined. This is the quantity that appears in Faraday's Law: ε = -d(NΦ)/dt

Real-World Applications

  • Transformers: Maximize flux linkage by using iron cores
  • Electric Motors: Rotation changes flux through coils
  • Magnetic Sensors: Detect changes in flux for position sensing
  • MRI Machines: Precise control of magnetic flux for imaging

Worked Examples

✅ Example 1: Basic Flux Calculation

A uniform magnetic field of 0.5 T passes perpendicularly through a rectangular coil of area 0.04 m². Calculate the magnetic flux.

Given: B = 0.5 T, A = 0.04 m², θ = 0° (perpendicular)
Formula: Φ = BA cos θ
Solution: Φ = 0.5 × 0.04 × cos(0°) = 0.5 × 0.04 × 1 = 0.02 Wb = 20 mWb

✅ Example 2: Flux at an Angle

A magnetic field of 0.3 T makes an angle of 60° with the normal to a circular coil of radius 10 cm. Find the flux.

Given: B = 0.3 T, r = 0.1 m, θ = 60°
Step 1: A = πr² = π × 0.1² = 0.0314 m²
Step 2: Φ = BA cos θ = 0.3 × 0.0314 × cos(60°)
Solution: Φ = 0.3 × 0.0314 × 0.5 = 4.71 × 10⁻³ Wb = 4.71 mWb

✅ Example 3: Flux Linkage in a Coil

A solenoid has 500 turns and each turn has an area of 2 cm². If the magnetic field inside is 0.02 T, calculate the flux linkage.

Given: N = 500, A = 2 cm² = 2 × 10⁻⁴ m², B = 0.02 T
Step 1: Φ = BA = 0.02 × 2 × 10⁻⁴ = 4 × 10⁻⁶ Wb
Step 2: Flux Linkage = NΦ = 500 × 4 × 10⁻⁶
Solution: Flux Linkage = 2 × 10⁻³ Wb = 2 mWb

✅ Example 4: Field Required for Given Flux

What magnetic field is required to produce a flux of 5 mWb through a square coil of side 5 cm at 45° to the normal?

Given: Φ = 5 mWb = 5 × 10⁻³ Wb, A = 0.05² = 0.0025 m², θ = 45°
Formula: B = Φ / (A cos θ)
Solution: B = (5 × 10⁻³) / (0.0025 × cos 45°) = (5 × 10⁻³) / (0.0025 × 0.707)
Answer: B = 2.83 T

✅ Example 5: Rotating Coil

A coil of 200 turns and area 0.01 m² rotates in a 0.1 T field. Calculate the flux linkage when the coil is (a) perpendicular to field (b) at 30° to field.

(a) θ = 0°: NΦ = NBA cos(0°) = 200 × 0.1 × 0.01 × 1 = 0.2 Wb
(b) θ = 30°: NΦ = NBA cos(30°) = 200 × 0.1 × 0.01 × 0.866 = 0.173 Wb
💡 Common Mistake: Don't confuse the angle with the surface! The angle θ is always measured from the normal (perpendicular) to the surface, not from the surface itself. If the field is "parallel" to the surface, θ = 90° and flux is zero.