Magnetic Flux Question

  1. 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 4.5E-4 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. jcsd
  3. gneill

    Staff: Mentor

    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})##
     
  4. 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?
     
  5. gneill

    Staff: Mentor

    Make a sketch! Draw a horizontal line to show your area of interest in profile. Draw a vector or two representing the B field. What angle do you need? You can also choose the appropriate trig function and use the angle as given.

    [​IMG]

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

    Attached Files:

  6. 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?
     
  7. gneill

    Staff: Mentor

    Sure. Add the appropriate units and you're good.
     
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