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Homework Help: Magnetic Field at a point due to two parallel wires

  1. Oct 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Two long parallel wires are a center-to-center distance of 4.90 cm apart and carry equal anti-parallel currents of 3.70 A. Find the magnetic field intensity at the point P which is equidistant from the wires. (R = 10.00 cm).

    http://imageshack.us/a/img11/3812/twoparallelwires.jpg [Broken]

    2. Relevant equations

    dB = μo*I*(dL[itex]\vec{}[/itex]xdR[itex]\vec{}[/itex])/(4*pi*r^3)

    3. The attempt at a solution

    Well first I tried multiplying the equation for magnetic field of an infinite line (μo*I/(2*pi*r) by two since there are two wires. then I realized that since the current is flowing in opposite directions, the y-component of the field, so to speak, would cancel out, and now I'm stuck.

    Here is the solution, but I want to know how to arrive here (and more specifically I'd like to know why we're multiplying the infinite line equation by the y-component): d*I*μo/(2*pi*(R^2+ (d/2)^2)
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 17, 2012 #2


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    hi smithisize! :smile:
    no, if you draw a vector diagram (with arrows), you'll find the x components cancel out :wink:
  4. Oct 17, 2012 #3
    Well, the vector equations for dr are R[itex]\hat{i}[/itex] -(d/2)[itex]\hat{j}[/itex] and R[itex]\hat{i}[/itex]+(d/2)[itex]\hat{j}[/itex] therefore the vector sum states that the vertical component cancels out.

    But, I ended up figuring it out. If you draw the b-field for the top wire, and manipulate theta (of the upper left corner of a triangle drawn on the top wire) a bit, you can end up with the proper configuration and see that where it all comes from.
  5. Oct 18, 2012 #4


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    no, if you draw the arrows, you'll see that one goes to the left, and the other to the right,

    so it's -R[itex]\hat{i}[/itex] +(d/2)[itex]\hat{j}[/itex] and R[itex]\hat{i}[/itex]+(d/2)[itex]\hat{j}[/itex] :wink:
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