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Magnetic Field from Two Wires

  1. Oct 27, 2007 #1
    Hi, can someone help me out? I've tried 5 different solution that I thought were definitely right and they all give me the same response: "Your answer is off by a multiplicative factor."

    1. The problem statement, all variables and given/known data
    [​IMG]

    Find the magnitude of the net magnetic field B_L created at point L by both wires.
    Express your answer in terms of I, d, and appropriate constants.

    2. Relevant equations
    [tex]
    B_{\rm wire}=\frac{\mu_0 I}{2\pi d} ,
    [/tex]

    3. The attempt at a solution
    The question before this: there was a question:

    Point L is located a distance d\sqrt 2 from the midpoint between the two wires. Find the magnitude of the magnetic field B_1L created at point L by wire 1.


    I put in the answer and got it right.
    [tex]
    B_1L =\left({\mu}_{0}I\right)/\left[2{\pi}\sqrt{\left(d^{2}+\left(d\sqrt{2}\right)^{2}\right)}\right]
    [/tex]

    So for the question at hand, I put in the same equation and multiplied by 2, figuring the net force would be the 2 separate forces added, which are equal since they both give the same direction of magnetic force and to do the same degree.

    It didn't work.

    So maybe it works like a loop since the top current is coming out at distance d and the bottom current is going in at distance d. d is now a radius.

    so I tried
    [​IMG]

    Also wrong. Tried to input it in various ways, ^(3/2), cancel out the pi's. Still wrong, "my answer is off by a multiplicative factor."

    Help would be awesome!
     
    Last edited: Oct 27, 2007
  2. jcsd
  3. Oct 27, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    While the magnetic field from each wire does have the same magnitude, they point in different directions. So you can't just double the field of one to get the field of both. Add them like vectors (which they are).

    While you have the correct formula for the magnitude of the field from a current carrying wire, what's the direction of that field? (Hint: Use the right hand rule.)
     
  4. Oct 27, 2007 #3
    I think I get what your saying.

    Point 1 @ L will have a up-right vector for B. Point 2 @ L will have a down-right vector for B. Both will be a hypetnuse of equal magninute. The y's will cancel. But I don't see how I determine the x-components that add up. I need angle an angle or at least another side to do some trig and I'm not sure how to go about it. Is it 180 degrees minus the angle created by the d and d(sqr-root(2))?
     
  5. Oct 28, 2007 #4

    Doc Al

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    Staff: Mentor

    The direction of the field from wire 1 at L is perpendicular to the line from point 1 to L. The angle it makes with the horizontal (x-axis) is the same angle as you'll find in the upper corner of the right triangle formed by 1-K-L.
     
  6. Feb 3, 2008 #5
    I still didnt get it... can you explain more?
     
  7. Feb 3, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

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