AQA A Level Physics

Revision Notes

7.9.2 Magnetic Flux Linkage

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 component of magnetic flux density B is perpendicular to the area 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)

 

 

 

 

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

  • 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
  • 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:

ΦN = BAN cos(θ)

  • Where:
    • N = number of turns of the coil

Worked Example

A solenoid of circular cross-sectional radius 0.40 m2 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.

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 = 0.8 Wb turns (2 s.f)

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