A flat, square surface with side length

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Homework Help Overview

The problem involves calculating the magnetic flux through a flat, square surface located in the xy-plane, with a specified magnetic field vector. The original poster attempts to apply the magnetic flux formula but encounters difficulties in arriving at the correct answer.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the angle theta used in the flux calculation, questioning the assumption of it being 0 degrees. They also raise points about the area of the square and the implications of the magnetic field's orientation relative to the surface.

Discussion Status

There is an ongoing exploration of the correct approach to calculating the flux, with participants providing insights about the relationship between the magnetic field and the surface normal. Some guidance has been offered regarding the angle and the area, but no consensus has been reached on the correct method.

Contextual Notes

The original poster's calculations involve a conversion from centimeters to meters and the interpretation of the magnetic field vector's components. The discussion highlights the need for clarity on the angle between the magnetic field and the surface normal, which is not explicitly provided in the problem statement.

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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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