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Magnetic Force parabolic wire

  1. Nov 24, 2003 #1
    I'm stuck on this problem. I think it's the 3D geometry that's giving me the biggest problem.

    The current i exists in the direction shown (from left to right) in a parabolic wire segment whose equation is y = (x^2)/b in the XY plane between x = -a and x = +a. Magnetic flux density, given by B = py^2 and the B field lines are everywhere out of the paper making an angle 60 degrees to the +X axis and 60 degrees to the +Y axis.

    a) Show that B can be written as [(py^2}/2)i + [(py^2}/2]j + [(py^2)/(2^(1/2))]

    b) Find X, Y, and Z components of the total force on the wire.

    Part a) was easy the projections in the x and y directions are just Bcos60 and z^2 = B^2 - x^2 - y^2.

    Part b) is really giving me fits.

    I know that dF(magnetic) = I(dl)Bsin(theta) and when B is perpendicular to dl life is great because dF = IdlB and the dFx and dFy are dFsintheta and dFcostheta respectively.

    But I'm totally thrown by B not being perpendicular and having trouble with the 3-dimensionality of the problem.

    Can someone please help me see.....

  2. jcsd
  3. Nov 24, 2003 #2

    Doc Al

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    Staff: Mentor

    Consider it this way: dF = I dl X B (where X is the vector cross product)

    The length of wire, dl, has components in the x and y directions.

    Also, take advantage of symmetry about the y-axis. The current though the y component of a wire segment (dl) on one side of the y-axis is exactly opposite the current through the y component of the segment on the opposite side of the y-axis. What does that tell you about its contribution to the total force on the wire?
  4. Nov 24, 2003 #3
    Thanks for your help, but the real problem I'm having is with the angle between B and dl, since the magnitude of dF is IdlBsin(theta).

    Symmetry will make Fx = 0. So now I only have the Y and Z components to figure out.

    Thanks again.
  5. Nov 24, 2003 #4
    Do I need to use the direction cosines in my equation? The angle between dl and B is changing depending on where you are in the wire right?
  6. Nov 24, 2003 #5

    Doc Al

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    Symmetry will make the contribution to the force due to the y-component of dl equal to zero. So all you care about is the x-component of dl.
  7. Nov 24, 2003 #6

    Doc Al

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    Staff: Mentor

    Yes, the angle keeps changing. But one way to attack this problem is as I suggested above. Rather than try to visualize the angle, use components and take the cross product.
  8. Nov 24, 2003 #7
    I'm still stumped. I understand what you're saying about dy not contributing anything to the total force on the wire, but there's still a y component in the total force.

    If I integrate the x component, I'm going to integrage for -a to +a and Fx = 0,

    I have the y component of B from part a) of the problem. The y component of dl is just dy right?

    So is dFy = Idly X By or |dFy| = IdlyBysin(angle between the dly and By) = IdlyBySin60.

    That can't be right because Fy = [2^(1/2)Ipa^5]/[5b^2] and
    Fz = [Ipa^5]/[5b^2]
  9. Nov 24, 2003 #8

    Doc Al

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    Staff: Mentor

    Well, yes, but: I thought we agreed that the y-component of dl does not contribute to the force? Forget about it!
    You are correct... That can't be right!

    What you have is this:

    I dx X (Py2/2 [j] + Py2/√2 [k]) ...Right? (Don't go beyond this point until you get this!)

    Now do the cross product --- like I've been telling you! (i x j = k, etc.)

    Then integrate from x= -a to +a. (Don't forget: y = x2/b) And you will see the light!:wink:
  10. Nov 24, 2003 #9

    I see the light now. It's not a super bright light, but it is bright enough for me to think clearly about what's going on in the problem.

    I took the cross product and integrated and, low and behold, I got the correct answer.

    I was thinking, incorrectly, that I should isolate both the dl and B into its components and insisting on finding the magnitude of the cross product instead of taking the actual cross product which in this case makes life much easier.

    I need a lot more practice, but this is a good start.

    Your help is very much appreciated. Thanks.
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