What is the magnetic flux through the prism's right side?

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SUMMARY

The discussion focuses on calculating the magnetic flux through the hypotenuse of a prism using the formula Flux = BAcos(theta). Given a magnetic field (B) of 3.5 Teslas, a height (y) of 0.640 meters, and a length (z) of 1.18 meters, the hypotenuse was calculated to be 1.22 meters using the Pythagorean theorem. The initial calculation yielded a flux of 2.6432 Weber, but the user questioned the accuracy of this result, suggesting a potential issue with significant figures. Ultimately, a corrected answer of 0.8164 Weber was provided.

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Se Hoon Park
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Homework Statement


e5L8fBt.jpg

Now, ignore the lengths given in the photo.

Height (y) = .640 meters
Length (z) = 1.18 meters
Base (x) = .320 meters
Magnetic Field (B) = 3.5 Teslas in the +X direction

Homework Equations


The problem is to find the magnetic flux through the surface of the prism that is the hypotenuse (side aedf). So, the equation for this should be Flux = BAcos(theta)

The Attempt at a Solution


Well, working out the problem, I figured out the length of the hypotenuse is 1.22m because of pythagorean theorem. So, B*A is simply

(1.22)(.640)(3.5T)=2.7328

Now, theta is the angle that is between the magnetic field and the orthogonal of the surface itself. This angle should be the same as the angle between the base of the triangle and the hypotenuse. So this ratio is (1.18/1.22). So the flux should be

2.7328 * (1.18/1.22) = 2.6432 Wb

But the answer keeps spitting out as wrong! What did I do wrong here?
 
Last edited:
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Perhaps it requires the correct number of significant figures.
good luck
 
I got 0.8164Wb.
If that checks I can give you hints as to how I derived it.
Se Hoon Park said:

Homework Statement


e5L8fBt.jpg

Now, ignore the lengths given in the photo.

Height (y) = .640 meters
Length (z) = 1.18 meters
Base (x) = .320 meters
Magnetic Field (B) = 3.5 Teslas in the +X direction

Homework Equations


The problem is to find the magnetic flux through the surface of the prism that is the hypotenuse (side aedf). So, the equation for this should be Flux = BAcos(theta)

The Attempt at a Solution


Well, working out the problem, I figured out the length of the hypotenuse is 1.22m because of pythagorean theorem. So, B*A is simply

(1.22)(.640)(3.5T)=2.7328

Now, theta is the angle that is between the magnetic field and the orthogonal of the surface itself. This angle should be the same as the angle between the base of the triangle and the hypotenuse. So this ratio is (1.18/1.22). So the flux should be

2.7328 * (1.18/1.22) = 2.6432 Wb

But the answer keeps spitting out as wrong! What did I do wrong here?
Your answer is correct. Guaranteed!
 

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