Calculating magnetic flux

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

    A cube of edge length 0.05m is positioned as shown in the figure below. A uniform magnetic field given by B = (5 i + 3 j + 2 k) T exists throughout the region.

    [​IMG]

    a) Calculate the flux through the shaded face.

    2. Relevant equations

    [itex]\phi[/itex] = B.A cos [itex]\theta[/itex]

    3. The attempt at a solution

    The area would simply be 0.0025m^2

    I'm having trouble understanding how to get the angle and also how to interpret the given magnitude of the magnetic field, it's a vector quantity.

    I thought at first the way to get the angle was to assume that the surface of the cube could be considered a vector as well, that way it would only have the j component since it's only got a direction in the y-axis.

    Then using the formula for the angle between two vectors, I got 53.5 degrees, though I'm not too sure how to use the given magnetic field value.
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    Hi NewtonianAlch! :smile:
    Forget angles, forget magnitude of the field …

    just do the inner product! (dot product)​

    the area can be represented by a vector of magnitude A in the normal direction, so just "dot" that with the field, and that's your flux! :wink:

    (or you can "dot" it with the unit normal, and then multiply by the area … same thing)
     
  4. Hi tinytim,

    Do you mean to say:

    (5, 3, 2)[itex]^{T}[/itex].0.0025 which is (5*0.0025 + 3*0.0025 + 2*0.0025)

    B.A
     
  5. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    No, (5,3,2).(the unit normal times 0.0025) :smile:

    (btw, you can't write BT.A …

    it's either BTA or B.A :wink:)
     
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