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