CIE A Level Physics复习笔记20.2.2 Magnetic Flux Linkage

Magnetic Flux Linkage

  • The magnetic flux linkage is a quantity commonly used for solenoids which are made of N turns of wire
  • Magnetic flux linkage is defined as:

The product of the magnetic flux and the number of turns

  • It is calculated using the equation:

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)
  • 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)

Worked Example

A solenoid of circular cross-sectional radius of 0.40 m2 and 300 turns is facing perpendicular to a magnetic field with 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)

 

 

 

 

 

转载自savemyexams

更多Alevel课程
翰林国际教育资讯二维码