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Torque on Circular Current

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Note getting the right answer.

    a). What is the magnitude of the torque on the circular current loop in the figure?
    b). What is the loop's equilibrium position.

    2. Relevant equations
    Torque = [(I*A)*B*(sin(theta))]

    L= 2.0cm
    a= 2.0mm
    Iwire = 2.0A
    Iloop = 0.20A

    3. The attempt at a solution

    First, I tried to find theta using tan^-1(x/y).
    Second, I tried to find B, using B= [(1.257E-6T*(m/A)*(I)] /(2*pi*r).
    Finally, I tried Torque = [(I*A)*B*(sin(theta))].
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Feb 16, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Not sure what you did here. Theta is the angle between the field from the wire and the magnetic moment of the loop (which is perpendicular to the loop).
    Looks OK.
  4. Feb 16, 2009 #3
    So my angle is 90deg, and I didn't need to solve for it?
  5. Feb 16, 2009 #4
    What did you use for r? It's not L, you have to use the pythagorean theorem with L and a as your legs.
  6. Feb 16, 2009 #5
    For r, I did the square root of [(L)^2+(a/2)^2].
  7. Feb 16, 2009 #6
    OK good, you've got that. Maybe:

    1) Did you account for both torques, one for each wire in the loop?
    2) Did you use the correct area for the loop?
    3) Are the directions correct? I think the force on the bottom piece would be up and to the left, and the force on the top would be up and to the right..

    If not that I can't see what else might be wrong.
  8. Feb 16, 2009 #7
    1). Aren't both torques the same, so I would double it?
    2). Isn't the area A= [(pi)*(R)^2]?
    3). I don't think direction is important, because it just wants the magnitude of torque.

    Let me know I made any bad assumptions on these 3.

    **Also, can someone confirm that the angle in my calculation for Torque = [(I*A)*B*(sin(theta))] is 90deg because of the figure being perpendicular.

    Thanks so far
  9. Feb 17, 2009 #8
    Still could use a reply, to my last post (especially about the 90deg).

  10. Feb 18, 2009 #9

    Doc Al

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    Staff: Mentor

    I wouldn't bother with that, since the distance from wire to loop segment varies along the loop. Instead I would make the approximation that the loop is small enough that the field from the wire can be considered uniform across the loop. Use the field at a distance L from the wire.

    The loop is circular, not rectangular.

    Yes. In the orientation shown in the diagram, the angle between the loop magnetic moment (perpendicular to the loop) and the magnetic field is 90 degrees.
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