Magnetic Flux Linkage (AQA A Level Physics)

• 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)

• 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)

• This means the magnetic flux is:
  • Maximum = BA when cos(θ) =1 therefore θ = 0^o. The magnetic field lines are perpendicular to the plane of the area
  • Minimum = 0 when cos(θ) = 0 therefore θ = 90^o. 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

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

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)