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Homework Help: Magnetic field at a pt created from two wires.

  1. Aug 23, 2008 #1
    1. The problem statement, all variables and given/known data

    The two wires carrying the currents are infinite in length. i1 = 5.00 A out
    of the paper and i2 = 7.00 A into the paper. a = 40.0 cm and b = 30.0 cm.
    Calculate the direction of the magnetic field at P, measured counterclockwise
    from the +x axis.

    Answer: 126.4°

    http://answerboard.cramster.com/answer-board/image/5769206ead06ab440f5adef680da9eba.jpg [Broken]

    2. Relevant equations

    http://answerboard.cramster.com/Answer-Board/Image/cramster-equation-20088221440246335501282431425007016.gif [Broken]

    3. The attempt at a solution

    B1 = (μo /2π) (5 A) / (0.30m) = (μo/2π) 16.6667

    B2 = (μo/2π) 14

    Thats it, Im stuck. I tried breaking this into components and finding the angle but it did not work.

    Please Help
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 23, 2008 #2


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    Homework Helper

    All right, you recognize that you need components, which is good. Using the right-hand rule for B1 at point P, what are the components? (This one isn't bad.)

    For dealing with B2 then, we know the components will be

    [tex]B_{2x} = B_2 \cdot cos \theta[/tex] and [tex]B_{2y} = B_2 \cdot sin \theta[/tex].

    So what is the angle? It must point perpendicular to the hypotenuse of the right triangle formed by the two wire cross-sections and P; as you also recognized, this is a 3-4-5 right triangle. So a vector pointing along the hypotenuse from wire 2 to P would have components proportional to
    < -4/5 , 3/5 >. B2 points perpendicular to this vector into the "first quadrant", so the components of B2 are proportional to < 3/5, 4/5 >. Those are the cosine and sine values for our components of B2, so we can get the components themselves from that.

    We're not quite done, since we still have to add the x- and y-components of the two magnetic fields at P to get the components for the total field... then there's getting the direction in which the total field vector points. But I imagine you can take it from there...
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