# Homework Help: What is the magnetic flux through the loop

1. Jul 12, 2009

### jlmessick88

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

Φ = AB cos θ
Φ = (pi * r^2)B cos θ

3. 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...

2. Jul 12, 2009

### LowlyPion

Can you elaborate on what you mean by bent?

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

3. Jul 12, 2009

### jlmessick88

here's a picture

#### Attached Files:

• ###### Q8.doc
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4. Jul 12, 2009

### jlmessick88

any ideas? anyone?!?!

5. Jul 12, 2009

### grantrudd

your attachment isnt 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.

6. Jul 12, 2009

### jlmessick88

7. Jul 12, 2009

### 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.

8. Jul 12, 2009

### jlmessick88

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