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Magnetic Flux Question 
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#1
Oct913, 07:32 AM

P: 145

1. The problem statement, all variables and given/known data
A horizontal rectangular surface has dimensions 3.10cm by 3.05cm and is in a uniform magnetic field that is directed at an angle of 34.5∘ above the horizontal. What must the magnitude of the magnetic field be in order to produce a flux of 4.5E4 Wb through the surface? 2. Relevant equations Magnetic Flux = BAcos(θ) 3. The attempt at a solution Using the above equation, I solved for B, getting B=Flux/Acos(θ) I then plugged in my numbers: (4.5*10^4)/((.0305)(.031)cos(34.5)) and got that B should equal 0.5775T. However, it says the answer is wrong, any ideas what I did wrong? 


#2
Oct913, 08:28 AM

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P: 11,689

I'd suggest that you make a sketch. You're looking for the flux through a horizontal surface, which means you want the vertical component of the B field...
Alternatively, take a vector equation approach and construct vectors for B and the area normal, then expand ##\Phi = \vec{B}\cdot (A\vec{n})## 


#3
Oct913, 09:11 AM

P: 145

So how do I go about finding that, though? I understand that I need the vertical component, but I have no idea how to set it up if all I know is the angle?



#4
Oct913, 09:29 AM

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P: 11,689

Magnetic Flux Question
You want the component of B that's parallel to the surface normal of your area. 


#5
Oct913, 09:34 AM

P: 145

Okay, so I would say that B_n (normal component of the B field) = Bsin(34.5), then I can do the flux divided by sin(34.5) times the area, so I have B=4.5*10^4/((sin34.5)(.031)(.0305)). That way I get B=0.84?



#6
Oct913, 09:40 AM

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