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Math Help - magnetic flux

  1. #1
    Senior Member
    Joined
    Jan 2007
    Posts
    477

    magnetic flux

    A house has a floor area of 104 m^2 and an outside wall that has an area of 27 m^2. The earth's magnetic field here has a horizontal component of 3.00 x 10^-5 T that points due north and a vertical component of 3.40 x 10^-5 T that points straight down, toward the earth.

    (a) Determine the magnetic flux through the wall if the wall faces north.
    in Wb (weber)

    (b) Determine the magnetic flux through the wall if the wall faces east.
    in Wb (weber)

    (c) Calculate the magnetic flux that passes through the floor.
    in Wb (weber)

    ====================

    i know that flux = B*A*cos(angle) will give the answer in Weber.

    however, i'm not sure what angles to use? Also, if i have to use components to get the angles.
    please show the example.

    thankyou.
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  2. #2
    JessicaLondon
    Guest

    This was a homework problem for me too!

    You don't need an angle to solve this one!! But if you wanted to, the angle would be 90 degrees. Here's the way I would solve it....

    Formula : Φ=BA
    Area of the Floor: 104m2
    Area of the Wall: 27m2
    Magnetic Field (horizontal): 3.0 x 10-5T
    Magnetic Field (vertical): 3.4 x 10-5T

    a) The magnetic flux through the wall if the wall faces north
    Φ = BA
    = (3.0 e-5 T) * (27m2)
    = _____ Wb
    b) The magnetic flux through the wall if the wall faces east
    Φ = 0 Wb
    c) The magnetic flux that passes through the floor
    Φ = BA
    = (3.4 x e-5T) * (104m2)
    = ______ Wb
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  3. #3
    Senior Member
    Joined
    Jan 2007
    Posts
    477
    awesome. Thanks, i was not sure how to set it up, much easier than i anticipated for this one.

    i got a.) 8.1 x 10^-4 Wb
    b.) 0 Wb

    and c.) .003536 Wb

    Thankyou, great help!
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