What is the magnetic flux through the loop

[jlmessick88]

Homework Statement

A 10 cm x 10 cm square is bent at a 90° angle . A uniform 0.050 T magnetic field points downward at a 45° angle. What is the magnetic flux through the loop?

Homework Equations

Φ = AB cos θ
Φ = (pi * r^2)B cos θ

The Attempt at a Solution

b = .050 T
cosθ = 45
this is where I'm getting stuck...for A, would i just use length * width (.1m * .1m)
--> .01 * .05 T * cos 45

i'm just unsure of how to approach the problem with the square being bent...

 

[LowlyPion]

Can you elaborate on what you mean by bent?

Bent in the middle? Or at a 90° angle to the horizon?

 

[jlmessick88]

here's a picture

 

[jlmessick88]

any ideas? anyone??

 

[grantrudd]

your attachment isn't working, but i would assume it is bent in the middle forming a triangle with 2 parts being .05m. find the hypotonuse of that triangle, and use that and the part of the square remaining straight to form a new area. the flux will be parallel to the vector that defines the area, so the cosine will drop out. hope this helps.

 

[jlmessick88]

http://i63.photobucket.com/albums/h148/jlmessick88/Image.jpg
here's a pic that should work...
i'm not sure if what you're talking about is exactly what's going on...if it is then i might need you to explain a bit more :)

 

[grantrudd]

i would try using the plane i outlined with red arrows for your area.

Flux= E x dA cos(theta), where cosine is the vector that defines the area of the plane. usually, this vector is normal (perpendicular) to the surface of the plane. when the area vector and the electric field are parallel, the cosine=1 and is the maximum flux that can pass through the plane.

 

[jlmessick88]

so basically that means...
c^2 = a^2 + b^2 = 7
(7*10)(.05)cos(1) = 2.5??