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Electromagnetism Question

  1. Feb 14, 2009 #1
    1. "A wire segment carries a current of 1.8A from a to b. There is a uniform magnetic field of 1.2T acting in the z-direction. Find the total force on the wire and show that the total force is the same as if the wire were a straight segment from a to b".

    I've attached the diagram showing what the segment looks like and I've called the point between a and b, c.

    2. Relevant equations: F=BIL

    3. Would this be right?:
    Fac = ILBcos90 = 0N (90° since the x-direction is orthogonal to z)
    Fcb = ILBsin90 = ILB = 1.8 x 0.04 x 1.2 = 0.0864N (90° since the y-direction is orthogonal to z)
    => Facb = 0.0864N

    Fab = ILBsinθ = ILBsin(arctan(4/3)) = 0.8ILB = 0.8 x 1.8 x 0.05 x 1.2 = 0.0864N

    Therefore, Facb = Fab

    Or is it just a coincidence that the two values come out to be the same? Because I don't really understand why it would be cos90 from a to c and sin90 from c to b =s

    Attached Files:

  2. jcsd
  3. Feb 14, 2009 #2


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    Welcome to PF :smile:

    There are some problems with your derivation. For one, all currents (ac, cb, and ab) are perpendicular to the magnetic field. So use F=BIL to find each force, with no cos or sin factors. None of the three forces will be zero.

    You'll also need to use the right-hand rule to figure out the direction of each force, and combine the ac & cb forces.
  4. Feb 14, 2009 #3
    Thanks for replying,
    So, would this be right:
    Fac = ILB = 0.0648N in the -ve y-direction so -0.0648j
    Fcb = ILB = 0.0864N in the +ve x-direction so 0.0864i
    => Facb = 0.0864i - 0.0648j

    Fab = ILB = 0.108N
    The direction is sin(arctan(4/3))Fi - cos(arctan(4/3))Fj
    = 0.0864i - 0.0648j

    Not sure about the signs though because the right-hand rule really confuses me =s
    Last edited: Feb 14, 2009
  5. Feb 14, 2009 #4


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    Looks good! The signs are correct too.
  6. Feb 14, 2009 #5
    Oh good :smile:
    Thanks a lot.
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