A flat, square surface with side length

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SUMMARY

The discussion focuses on calculating the magnetic flux through a flat, square surface with a side length of 3.00 cm in the presence of a magnetic field defined by the vector B = (0.150T)i + (0.350T)j - (0.500T)k. The correct approach involves using the formula for magnetic flux, Φ = BAcos(θ), where θ is the angle between the magnetic field and the normal to the surface. The participants clarify that the area of the square surface must be calculated, and the angle θ should be determined from the magnetic field vector rather than assuming it to be zero degrees.

PREREQUISITES
  • Understanding of magnetic flux and its calculation using the formula Φ = BAcos(θ)
  • Knowledge of vector components in magnetic fields
  • Familiarity with the concept of area for geometric shapes, specifically squares
  • Basic trigonometry to determine angles between vectors
NEXT STEPS
  • Study the calculation of magnetic flux in different orientations of magnetic fields
  • Learn how to derive angles between vectors using dot products
  • Explore the implications of magnetic field direction on flux calculations
  • Review the relationship between Tesla, Weber, and square meters in magnetic field units
USEFUL FOR

Students in physics, particularly those studying electromagnetism, educators teaching magnetic concepts, and anyone involved in solving problems related to magnetic fields and flux calculations.

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Homework Statement



A flat, square surface with side length 3.00cm is in the xy-plane at z=0 .

Calculate the magnitude of the flux through this surface produced by a magnetic field

B=(0.150T)i+(0.350T)j-(0.500T)k

What I'am doing is I know that the magnetic flux= BAcos(theta)

So I multiply my vector B by 0.03m and get (0.0045)i+(0.0105)j, and z=0 so I can ignore that value.

Then they are asking for the magnitude, so √(0.0045)^2+(0.00105)^2= 0.00462Wb , however that is not giving me the correct answer, what am I missing?


Thanks

Homework Equations





The Attempt at a Solution

 
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What angle theta are you using for the magnetic flux?

Also, what's the area of a square? Seeing as a Tesla is 1 Wb/m^2, multiplying by 0.03m should get you an answer of Wb/m, not Wb.
 
Theta is not given so I assumed it was just 0degrees
 
Theta is the angle between the "normal" line of your plane and the magnetic field lines. It would be 0 degrees if the magnetic field lines were parallel with the normal (therefore perpendicular), but this is not the case here. You'll have to find the angle from the vector you're given.
 
Le Answer

Use the BAcos(theta) formula to find the flux. But since the Magnetic field vector is given, keep in mind that flux operates perpendicular to the field, so all you really have to do is calculate (B_k)(A)cos(90). And you'll be squaring that number and then taking the root so because you only have one term in the "magnitude finding" process, just drop the negative sign.
 
Lol I finished that class about 8 months ago. Thank you though
 
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