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Strength of a magnetic field around a wire with current

  1. Aug 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Three parallel wires are arranged in a partial square configuration with point P as the fourth corner. In two of the wires at opposite corners of the square, a 5.00×10^-4 A current flows toward you, and in the third wire opposite the point P, a 7.50×10−4 A current flows away from you. The sides of the square are 5.36cm. What is the strength of the magnetic field at the point P, at the "missing corner" of the square? (T)

    [​IMG]

    2. Relevant equations
    I believe I should be using:
    (Magnetic field)=[(Permeability of free space)*(current)]/[2*pi*(distance from wire)]
    The permeability of free space = (4*pi*10^-7)


    3. The attempt at a solution
    B1+B2-B3=BP
    [(5E-4)((4*pi)E-7)]/(2*pi*.0536) = B1 = B2 = 1.9788436E-9 T

    distance from wire3 and point P = r = sqrt(.0536^2+.0536^2) = .0758018469 m
    [(7.5E-4)((4*pi)E-7)]/(2*pi*r) = B3 = 1.97884361E-9 T

    So, B1+B2-B3 = 1.752E-9 T
    which is not correct unfortunately. Do the parallel wires do something weird I am unaware of? I thought only the distance from the wire and the current determined the magnitude of the magnetic field, and the opposing currents caused opposing magnetic fields. Help!
     

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    Last edited: Aug 16, 2008
  2. jcsd
  3. Aug 16, 2008 #2

    Redbelly98

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    It might help to indicate the B-field vectors at P, due to each individual wire. Be sure to draw each vector in the correct direction.
     
  4. Aug 17, 2008 #3
    I am still having trouble solving this one. Anyone have any advice?
     
  5. Aug 17, 2008 #4

    Redbelly98

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    It's vector addition, where you must account for the angles of the vectors (B1, B2, and B3) you are adding up.

    Simple addition & subtraction (i.e. "B1+B2-B3") does not work here.
     
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