Lecture Notes: Calculating Magnetic Flux

Key Concepts

• Magnetic Flux (Φ): The measure of the quantity of magnetism, considering the strength and extent of a magnetic field.
• Formula: ( \Phi = B \times A \times \cos(\theta) )
  • ( B ) = magnetic field strength (in Tesla)
  • ( A ) = area of the coil (in square meters)
  • ( \theta ) = angle between magnetic field lines and normal to the surface
  • Units: The unit of magnetic flux is the Weber (Wb), where 1 Weber = 1 Tesla ( \times ) 1 square meter.

Problem 1: Circular Coil

• Given:
  • Radius of coil = 25 cm = 0.25 meters
  • Magnetic field (( B )) = 20 Tesla
  • Magnetic field perpendicular to coil
• Solution:
  • Area of circle ( A = \pi r^2 = \pi (0.25)^2 )
  • Magnetic Flux ( \Phi = B \times A = 20 \times \pi (0.25)^2 )
  • Calculation result: ( \Phi \approx 3.93 ) Weber

Problem 2: Square Coil, Parallel Field

• Given:
  • Magnetic field (( B )) = 30 Tesla
  • Field directed parallel to the face of the coil
• Solution:
  • No perpendicular component of magnetic field through the square coil
  • Magnetic Flux ( \Phi = B \times A \times \cos(\theta) )
  • Since ( \theta = 90^\circ ), ( \cos(90^\circ) = 0 )
  • Result: ( \Phi = 0 ) Weber

Problem 3: Calculating Flux for Angled Fields

• Goal: Calculate magnetic flux through squares at various angles.
• Formula: ( \Phi = B \times A \times \cos(\theta) )
  • Angle ( \theta ): Between magnetic field and normal line to the surface
• Example 1:
  • Magnetic field = 10 Tesla
  • Area = 0.5 m ( \times ) 0.5 m = 0.25 m(^2)
  • ( \theta = 40^\circ ), ( \cos(40^\circ) )
  • Calculation: ( \Phi \approx 0.766 ) Weber
• Example 2:
  • Same magnetic field and area
  • ( \theta = 60^\circ ), ( \cos(60^\circ) = 0.5 )
  • Calculation: ( \Phi = 1.25 ) Weber

Summary

• To find magnetic flux through a surface, use ( \Phi = B \times A \times \cos(\theta) )
• Remember: ( \theta ) is the angle between the magnetic field and the normal line of the surface.
• Correct angle selection is crucial for accurate calculation.