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Magnetic flux

  1. Feb 20, 2005 #1
    A house has a floor area of 200 m^2 and an outside wall that has an area of 50.0 m^2. The earth's magnetic field here has a horizontal component of 2.58×10^-5 T that points due north and a vertical component of
    4.14×10^-5 T that points straight down, toward the earth. a)Determine the magnetic flux through the wall if the wall faces north. b)calculate the flux that passes through the floor.

    a)
    flux=BAcos(theta)
    =(4.14*10^-5)(50 m^2)(cos 90 deg)
    =0 Wb

    b)
    flux=BAcos(theta)
    =(2.58*10^-5)(200 m^2)(cos 0 deg)
    =5.16*10^-3 Wb

    I'm not sure what I did wrong.
     
  2. jcsd
  3. Feb 20, 2005 #2

    dextercioby

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    Science Advisor
    Homework Helper

    At point a) it's the horizontal component which gives the nonzero flux...

    As for point b),it's viceversa,only the vertical one does.

    Daniel.
     
  4. Feb 20, 2005 #3

    Astronuc

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    Staff: Mentor

    A house has a floor area of 200 m^2 - normal points vertical - a vertical component of 4.14×10^-5 T that points straight down,

    An outside wall has an area of 50.0 m^2. The wall's normal is horizontal, The earth's magnetic field here has a horizontal component of 2.58×10^-5 T that points due north.
     
  5. Feb 21, 2005 #4
    im still in the dark about this vertical horizontal component
    do i need to find some magnitude before i find the flux?
     
  6. Feb 21, 2005 #5

    Astronuc

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    Staff: Mentor

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