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

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 = 0.8 Wb turns (2 s.f)

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:

$\Phi N=BAN\cos \theta$

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).