A LevelPhysicsCIERevision Notes20. Magnetic Fields20.5 Electromagnetic InductionMagnetic Flux Linkage

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First teaching 2023

First exams 2025

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Magnetic Flux Linkage (CIE A Level Physics)

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Magnetic flux linkage

  • The magnetic flux linkage is a quantity commonly used for solenoids which are made of N turns of wire
  • Magnetic flux linkage is defined as:

The product of the magnetic flux and the number of turns

  • It is calculated using the equation:

N capital phi thin space equals space B A N

  • Where:
    • Φ = magnetic flux (Wb)
    • N = number of turns of the coil
    • B = magnetic flux density (T)
    • A = cross-sectional area (m2)
  • The flux linkage NΦ has the units of Weber turns (Wb turns)
  • As with magnetic flux, if the field lines are not completely perpendicular to the plane of the area they are passing through
  • Therefore, the component of the flux density which is perpendicular is equal to:

N capital phi space equals space B A N cos open parentheses theta close parentheses

  • Where:
    • θ = angle between magnetic field lines and the line perpendicular to the plane of the area (degrees)

Worked example

A solenoid of circular cross-sectional radius of 0.40 m and 300 turns is positioned perpendicular to a magnetic field with a magnetic flux density of 5.1 mT.

Determine the magnetic flux linkage for this solenoid.

Answer:

Step 1: Write out the known quantities

  • Cross-sectional area, A = πr2 = π(0.4)2 = 0.503 m2
  • Magnetic flux density, B = 5.1 mT = 5.1 × 10−3 T
  • Number of turns of the coil, N = 300 turns

Step 2: Write down the equation for the magnetic flux linkage

N capital phi space equals space B A N

Step 3: Substitute in values and calculate

N capital phi space equals space open parentheses 5.1 space cross times space 10 to the power of negative 3 end exponent close parentheses space cross times space 0.503 space cross times space 300 space equals space 0.7691

N capital phi space equals space 0.77 Wb turns (2 s.f.)

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