Magnitude of flux through wedding ring

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SUMMARY

The discussion centers on calculating the magnetic flux through a wedding ring with a diameter of 2.2 cm in the Earth's magnetic field of 48 μT, directed at 66° below the horizontal. The equation used for this calculation is B*A*cos(theta), where B represents the magnetic field strength, A is the area of the ring, and theta is the angle between the magnetic field and the normal to the surface. The participants clarify that 48 μT refers to micro-Tesla, a unit of magnetic field strength, and correct the misconception regarding the angle, confirming that 66° below the horizontal translates to 24° from the perpendicular.

PREREQUISITES
  • Understanding of magnetic flux and its calculation.
  • Familiarity with the equation B*A*cos(theta).
  • Knowledge of units of magnetic field strength, specifically micro-Tesla (μT).
  • Basic geometry related to angles and their measurements.
NEXT STEPS
  • Calculate the area of a circle using the formula A = π*(d/2)².
  • Learn about the significance of micro-Tesla (μT) in magnetic field measurements.
  • Explore the concept of magnetic flux in different geometrical configurations.
  • Study the relationship between angles in magnetic field orientation and their impact on flux calculations.
USEFUL FOR

Physics students, educators, and professionals in fields related to electromagnetism and magnetic field analysis will benefit from this discussion.

ghostbuster25
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the problem is stated as
The Earth’s magnetic field at a certain location in the UK has a magnitude of 48 μT and is directed at 66° below the horizontal.

a)Determine the magnitude of the flux of the Earth’s magnetic field through a
wedding ring of diameter 2.2 cm when the ring is held in a horizontal plane.

ok so i have used the equation B*A*cos(theta)

the only thing slowing me down is the use of 48 μT . I am not sure how to work this into the equation. I am led to believe that μ is the permuabilty of space so not sure what is meant by μT for the magnetic field of the Earth. I have probably overlooked it in my text and can't find the relevant part now.
Also i am guessing that 66° below the horizontal is equal to 66° from the perpendicular.
 
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ghostbuster25 said:
the only thing slowing me down is the use of 48 μT . I am not sure how to work this into the equation. I am led to believe that μ is the permuabilty of space so not sure what is meant by μT for the magnetic field of the Earth. I have probably overlooked it in my text and can't find the relevant part now.

μT means micro-Tesla. 1 μT is one millionth of a Tesla.
Also i am guessing that 66° below the horizontal is equal to 66° from the perpendicular.

No. Try drawing it out; this is not true.
 
ahhh yes that makes sense now thanks :)

with regards to the 66 below the horizontal...if the horizontal is 0 degrees, and perpendicular is 90 degrees is it simply just 29degrees from the perpendicular?
 

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