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Magnetic Flux

Key idea: Define magnetic flux Φ and flux linkage NΦ, use Φ = BA cosθ and NΦ = NBA cosθ, and solve induction-style flux questions in exams (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use magnetic flux and flux-linkage relationships.

1. Definitions (Must Know)

A. Magnetic flux, Φ

Magnetic flux, Φ, through a plane surface is:

Φ = BA cos θ

where:

  • B is magnetic flux density (T),
  • A is the area of the surface (m²),
  • θ is the angle between vector B and the normal to the surface.

Unit: weber (Wb), where 1 Wb = 1 T m².

B. Flux linkage, NΦ

For a coil with N turns, the magnetic flux linkage is:

NΦ = NBA cos θ

2. Key Ideas (What Earns Marks)

  • If the field is perpendicular to the surface, θ = 0° and cos θ = 1, so Φ = BA.
  • If the field is parallel to the surface, θ = 90° and Φ = 0.
  • In induction, what matters is usually how fast NΦ changes with time.
Angle trap

θ is measured between vector B and the normal to the surface, not between vector B and the plane.

Magnetic flux and surface orientationThree panels compare a surface normal parallel, angled and perpendicular to a uniform magnetic field, giving maximum, intermediate and zero magnetic flux.θ = 0°Φ = BA (maximum)0° < θ < 90°θΦ = BA cos θθ = 90°Φ = 0
Scroll diagram horizontally to read all labels.
The flux uses the area perpendicular to the field. The angle in Φ = BA cos θ is measured from B to the surface normal, not to the plane.

Magnetic flux vs angle to the normal (scaled)

A cosine curve showing that flux falls from BA to 0 as the angle between the field and the surface normal increases from 0° to 90°.

Scroll across the graph to read all labels.

A cosine curve showing that flux falls from BA to 0 as the angle between the field and the surface normal increases from 0° to 90°.A cosine curve showing that flux falls from BA to 0 as the angle between the field and the surface normal increases from 0° to 90°.
θ is the angle to the normal. If a question gives the angle to the plane (α), then θ = 90° − α so Φ = BA sinα.
Open full-size graph
View figure data
Values for Magnetic flux vs angle to the normal (scaled)
Angle to the normal, θ (°)Φ/(BA) = cosθ
01
150.966
300.866
450.707
600.5
750.259
900

3. Detailed Explanations

A. Why the cosine appears

Only the component of vector B perpendicular to the surface contributes to flux:

B_⊥ = B cos θ ⇒ Φ = B_⊥ A = BA cos θ

B. What changes flux linkage

From NΦ = NBA cos θ, flux linkage changes if any of these change:

  • magnetic flux density B,
  • coil area A,
  • coil orientation θ,
  • number of turns N (usually fixed for a given coil).

4. Common Mistakes

  • Using the wrong angle (using the angle to the plane instead of the normal).
  • Mixing up flux Φ (Wb) with flux density B (T).
  • Forgetting to convert cm² to m².

5. Exam Tips

  • Write down the normal direction as a small arrow on the surface before you decide θ.
  • If the surface is rotated, sketch vector B and the normal at least once to avoid the sin/cos swap.

6. Worked Examples

Modelled example 1

Flux through a surface

Core

Problem

A uniform 0.40 T field is perpendicular to a flat surface of area 2.5 × 10⁻³ m². Find the magnetic flux.
Study the worked solution

  1. Interpret the orientation

    Method

    The field is parallel to the surface normal, so θ = 0°.

    Reason

    The flux angle is measured to the normal, not to the plane.

    Working

    cos 0° = 1.

  2. Calculate flux

    Method

    Φ = 1.0 × 10⁻³ Wb.

    Reason

    Use Φ = BA cos θ = BA.

    Working

    Φ = (0.40)(2.5 × 10⁻³) = 1.0 × 10⁻³ Wb

Guided practice 2

Rotated surface

About 4 min

Problem

A 1.2 × 10⁻² m² surface is in a 0.15 T field at 60° to its normal. Find Φ.

Try this before viewing the solution

Unit: Wb

Hints

Hint 1: project onto the normal
The contributing field component is B cos 60°.
View solution step by step

  1. Use the stated angle

    Method

    Apply Φ = BA cos 60°.

    Reason

    The question already gives the angle between vector B and the normal.

    Working

    Φ = (0.15)(1.2 × 10⁻²) cos 60°

  2. Evaluate

    Method

    Φ = 9.0 × 10⁻⁴ Wb.

    Reason

    cos 60° = 0.50.

    Working

    Φ = 9.0 × 10⁻⁴ Wb

Common misconception 3

Flux linkage of a coil

Find and correct the mistake

Learner claim

A 200-turn coil of area 3.0 × 10⁻⁴ m² is perpendicular to a 0.80 T field. A learner reports 2.4 × 10⁻⁴ Wb as the flux linkage. Diagnose and calculate.

Try this before viewing the solution

Unit: Wb turn

View solution step by step

  1. Distinguish the quantities

    Method

    Flux per turn is Φ = BA = 2.4 × 10⁻⁴ Wb.

    Reason

    The learner stopped before including the number of turns.

    Working

    Φ = (0.80)(3.0 × 10⁻⁴) = 2.4 × 10⁻⁴ Wb

  2. Calculate linkage

    Method

    NΦ = 4.8 × 10⁻² Wb turn.

    Reason

    All 200 turns link the same flux.

    Working

    NΦ = (200)(2.4 × 10⁻⁴) = 4.8 × 10⁻² Wb turn

Examiner practice 4

Angle given to the plane of the coil

3 marks

Examination question

A 5.0 × 10⁻³ m² coil is in a 0.30 T field that makes 20° to the plane. Convert the angle and find the flux. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Convert the angle

    1 mark

    Method

    θ = 70° to the normal.

    Reason

    The normal is 90° to the plane.

    Working

    θ = 90°-20° = 70°

  2. Select the equivalent relation

    1 mark

    Method

    Use Φ = BA cos 70° = BA sin 20°.

    Reason

    Flux uses the perpendicular area component.

    Working

    Φ = BA sin 20°

  3. Evaluate

    1 mark

    Method

    Φ = 5.13 × 10⁻⁴ Wb.

    Reason

    Substitute the field, area and plane-angle sine.

    Working

    Φ = (0.30)(5.0 × 10⁻³) sin 20° = 5.13 × 10⁻⁴ Wb

Challenge 5

Find B from a measured flux

Minimal support

Independent transfer

A square coil of side 8.0 cm has its plane perpendicular to a uniform field. The flux per turn is 1.28 × 10⁻³ Wb. Find B and distinguish why the number of turns is not needed.

Try this before viewing the solution

Hints

Hint 1: find area in square metres
Convert the side to metres and square it; the given quantity is flux, not linkage.
View solution step by step

  1. Find area

    Method

    A = (0.080)² = 6.4 × 10⁻³ m².

    Reason

    Area conversion squares the length conversion.

    Working

    A = 6.4 × 10⁻³ m²

  2. Infer flux density

    Method

    B = 0.20 T.

    Reason

    The plane is perpendicular to vector B, so Φ = BA.

    Working

    B = (1.28 × 10⁻³)/(6.4 × 10⁻³) = 0.20 T

  3. Distinguish turns

    Method

    N is unnecessary because flux per turn, not flux linkage, is given.

    Reason

    Turns enter only through NΦ.

    Working

    Use Φ, not NΦ.

7. Mind Stretchers

Mind stretcher 1: Doubling flux linkageExtension

A coil has fixed N and A. Give two different ways to double NΦ without changing N or A.

Show Answer

From NΦ = NBA cos θ:

  • double B, or
  • change orientation so cos θ doubles (e.g. rotate from θ = 60° to θ = 0°, since cos 60° = 0.5 and cos 0° = 1).

Mind stretcher 2: Can flux be negative?Extension

In calculations, can Φ (or NΦ) be negative? What does a negative sign represent?

Show Answer

Yes. Flux is treated as a signed quantity when you choose a direction for the area normal.

A negative flux means the magnetic field component through the surface is opposite to the chosen normal direction (i.e. it points “into” the surface relative to your chosen positive direction).

8. Optional (Enrichment)

A. Base units of weber

Using 1 Wb = 1 T m² and 1 T = 1 N A⁻¹m⁻¹:

Wb = N m A⁻¹ = J A⁻¹

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027