Magnetic Flux Linkage
- 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:
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)
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.
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)
Examiner Tip
Just like for magnetic flux, the flux linkage through a coil may not be entirely perpendicular.
The magnetic flux linkage through a rectangular coil increases as the angle between the field lines and a normal line to the coil plane decreases
In this case, you can just substitute the equation for B into the equation for φN, such that the flux linkage is calculated by:
As before, you should remember that since cos (0°) = 1, the flux linkage is a maximum when the angle θ is zero. This means the flux and coil face are perpendicular (i.e. the normal line to the coil face and the flux lines are parallel).