Magnetic Flux Linkage (AQA A Level Physics): Revision Note

Exam code: 7408

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

Magnetic Flux Linkage

  • More coils in a wire mean a larger e.m.f is induced

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

  • The flux linkage is defined as:

The product of the magnetic flux and the number of turns of the coil

  • It is calculated using the equation:

Magnetic flux linkage = ΦN = BAN

  • 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 ΦN has the units of Weber turns (Wb turns)

Flux Linkage in a Rotating Rectangular Coil

  • When the magnet field lines are not completely perpendicular to the area A, then the perpendicular component of magnetic flux density B is taken

  • The equation then becomes:

Φ = BA cos(θ)

  • Where:

    • Φ = magnetic flux (Wb)

    • B = magnetic flux density (T)

    • A = cross-sectional area (m2)

    • θ = angle between magnetic field lines and the line perpendicular to the plane of the area (often called the normal line) (degrees)

  • This means the magnetic flux is:

    • Maximum = BA

      • when cos(θ) =1 therefore θ = 0o

      • The magnetic field lines are perpendicular to the plane of the area

    • Minimum = 0

      • when cos(θ) = 0 therefore θ = 90o

      • The magnetic fields lines are parallel to the plane of the area

  • An e.m.f is induced in a circuit when the magnetic flux linkage changes with respect to time

  • This means an e.m.f is induced when there is:

    • A changing magnetic flux density B

    • A changing cross-sectional area A

    • A change in angle θ

Flux Linkage, downloadable AS & A Level Physics revision notes

The magnetic flux through a rectangular coil decreases as the angle between the field lines and plane decrease

  • Magnetic field lines may not be completely perpendicular to the plane of the area that they pass through

  • Therefore, the component of the flux density which is perpendicular is equal to:

ΦN = BAN cos(θ)

  • Where:

    • N = number of turns of the coil

Worked Example

A solenoid of circular cross-sectional radius 0.40 m and 300 turns is placed 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

    • Number of turns of the coil, N = 300 turns

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

ΦN = BAN

Step 3: Substitute in values and calculate

ΦN = (5.1 × 10-3) × 0.503 × 300 = 0.7691 = 0.77 Wb turns (2 s.f)

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.