Magnetic Flux through a bent loop.

AI Thread Summary
To calculate the magnetic flux through a bent loop in a uniform magnetic field, the area must be treated differently than for a flat surface. The bent loop can be analyzed as two flat surfaces, allowing for the calculation of flux through each section separately. Alternatively, one can consider the effective area outlined by the two sides of the loop, noting that the angle of the magnetic field relative to the surface also changes. The key is to apply the magnetic flux formula correctly, taking into account both the area and the angle. Understanding how the bend affects the area and angle is crucial for accurate calculations.
Jnumen
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Homework Statement


A 10cm x 10cm square is bent at a 90deg angle as shown in the figure.

A uniform 0.050T magnetic field points downdard at 45deg.

What is the magnetic flux through the loop.


Homework Equations



Flux= [(Aeff)*(B)].
Flux= [(Aeff)*(B)]*cos(theta).

The Attempt at a Solution



First I solved for the side due to the bend by using d= SQRT of [(a)^2*(c)^2].

Then, I calculated the Area using A= [(d)*(b)].

Then, I tried calculating with or without the 45deg angle using the following (which is not giving the correct answer):

Flux= [(Aeff)*(B)].
Flux= [(Aeff)*(B)]*cos(theta).
 

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Does anyone know how I am supposed to treat the bend?

I have done problems when the area was flat, but I am not sure what I need to do differently when the area is bent. Isn't it going to decrease the area?
 
Jnumen said:
Does anyone know how I am supposed to treat the bend?

I have done problems when the area was flat, but I am not sure what I need to do differently when the area is bent. Isn't it going to decrease the area?
Two ways to go about it:

1) Treat it as two flat surfaces of area b*c and b*a. Find the flux through each of those surfaces and add them up.

2) Find the surface outlined by the by the two b sides. The angled piece doesn't matter, since the field is parallel to the bend. Sure the area changes, but so does the angle that the field makes with that surface.
 
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