Magnetic flux and flux linkage
Key idea: H2 Physics lessons on magnetic flux, induction laws, applications and ideal transformers.
Continue where you stopped
The core idea
Build the idea
Learn the idea
Big question: What does it mean for a magnetic field to link a coil?
Magnetic flux through area A is Φ = BA cosθ for a uniform field, where θ is measured from the field to the area normal. A coil of N turns has flux linkage NΦ. Changing B, area or orientation can therefore change flux linkage even when the other quantities stay fixed.
Distinguish flux from flux linkage
Magnetic flux through a flat loop in a uniform field is Φ = BA cosθ, where θ is between B and the area normal. A field parallel to the plane gives zero flux; a field perpendicular to the plane gives maximum magnitude.
For N turns experiencing the same flux, flux linkage is NΦ. Its unit is weber-turn, conventionally written Wb turn. Flux can change through B, area, orientation or how much of the loop lies in the field.
Check your understanding: A loop's plane is parallel to B. What is its flux?
Zero, because its area normal is perpendicular to B, so cos 90° = 0.
Use motional e.m.f. as flux change
A conductor of length l moving at speed v across a uniform field experiences magnetic separation of charges, producing e.m.f. Blv when motion, field and conductor are mutually perpendicular.
This is consistent with Faraday's law: the swept area grows at rate lv, so flux changes at rate Blv. If any of the three directions is not perpendicular, use only the relevant component.
Check your understanding: Why is no e.m.f. induced when a rod moves along magnetic field lines?
Magnetic force qv × B is zero and the rod sweeps no magnetic flux.
Key ideas to keep
- Use the angle to the normal, not the plane.
- Flux is measured in webers; flux linkage in weber-turns.
- A field parallel to the plane gives zero flux through it.
See the reasoning
Worked example
Move from general orientation to perpendicular area
Question: A loop of area A has its normal at angle θ to B. Connect the general orientation model to the syllabus form Φ = BA.
Step 1: Mark the area normal
Why: The angle in flux is measured to the normal, not the coil plane.
Working: Φ = BA cosθ when θ is between B and the normal.
Step 2: Express perpendicular area
Why: The syllabus form can use the projected area normal to B.
Working: A⊥ = A cosθ, so Φ = BA⊥.
Step 3: Check the maximum orientation
Why: When the plane is perpendicular to B, its normal is parallel to B.
Working: θ = 0, so Φ = BA and linkage is NBA.
Answer: General flux is Φ = BA cos θ, where θ is between B and the area normal. The syllabus wording uses cross-sectional area perpendicular to B, so its effective perpendicular area is A⊥ = A cos θ and Φ = BA⊥. When the loop plane is perpendicular to B, θ = 0 and Φ = BA.
Check: A plane parallel to B has its normal at 90° and therefore zero flux.
Use a hint if needed
Practise with support
Try this
Area doubles and B halves while orientation and N stay fixed. State the linkage factor.
Hint: Track N, B and perpendicular area separately.
Check your answer
NΦ = NBA is unchanged because the factors 2 and 1/2 cancel.
Now work without the hint
Practise independently
Your turn
Define magnetic flux and flux linkage with units, orientation and the meanings of N and area.
Check your answer
Magnetic flux through a surface is flux density times cross-sectional area perpendicular to B: Φ = BA⊥, unit weber. Flux linkage NΦ is the sum of flux linked by N turns, unit Wb turn. N is total linked turns; area is per turn.
Avoid these traps
Common mistakes
Common mistake
Flux is BA regardless of orientation.
What is wrong with this reasoning?
Show better thinking
Use the area perpendicular to B, or equivalently BA cos θ where θ is between B and the area normal.
Common mistake
Flux and flux linkage are identical for every coil.
What is wrong with this reasoning?
Show better thinking
Flux is per surface/turn; linkage is NΦ for N turns linking the same flux.
Write for the examiner
Exam guidance
Draw the area normal and mark θ before substituting into BA cosθ.
Exam-style practice [5 marks]
A 120-turn coil of area 5.0 cm² has its plane perpendicular to 0.80 T. Find Φ and NΦ.
Plan before you answer
- Convert cm² to m².
- Find flux per turn.
- Multiply by turn count.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
A = 5.0 × 10⁻⁴ m². Φ = BA = 4.0 × 10⁻⁴ Wb and NΦ = 4.8 × 10⁻² Wb turn.
Come back in three days
Check what stayed with you
Recall question
A loop plane turns from perpendicular to parallel with B. State the flux change.
Check the answer
Its area normal turns from parallel to perpendicular to B, so flux falls from BA to zero.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Flux uses area perpendicular to B; flux linkage is NΦ for N linked turns. Faraday's law uses the rate of change of linkage and Lenz's law fixes polarity from the change being opposed. The simple Blv motional-e.m.f. form requires mutually perpendicular conductor length, velocity and uniform field. Ideal transformer ratios assume common linked flux, alternating operation and no winding or core losses. More advanced induction applications are not required here.
- GCE A-Level H2 PhysicsTopic 18(a) / Topic 18(b) / Topic 18(c) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027