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Torque on a current-carrying loop

  1. Nov 9, 2003 #1
    When a current carrying loop is placed in a uniform magnetic field and the normal to the plane of the loop is ⊥ to the direction of the field (iow the plane of the loop is parallel to the direction of the field), at that moment the torque (about any axis) exerted on the loop is supposed to be
    T = IAB
    where I is the current, A is the area of the loop, and B is the field.

    So, I want to try this out on an equilateral triangular loop with base = 2x, height = x√3, and hypotenuse = 2x.
    I'll put my axis of rotation along a vertical line from the apex to the center of the base. This way, the torque on half of the base is clockwise & on the other half counterclockwise, so the base can be ignored in the calculations. By symmetry, the torque contributions of the other two legs are equal, so we can just calculate one of them & then double it. So, now I'm looking at half of the original equilateral triangle: a 30-60-90 triangle, with the axis of rotation along the vertical (long) leg. The measurement along this axis is h = x√3. The measurement along the base is just x.

    The magnetic force is exerted along the hypotenuse, L = 2x.
    The force exerted on each little piece dL is:
    F = BIdL
    but
    L = (2/√3)h; therefore
    dL = (2/√3)dh; so
    F = (2/√3)BIdh
    The distance from each dL to the axis:
    d = (1/√3)h
    Therefore the torque contribution from each dL should be
    dT = Fd
    dT = (2/√3)BIdh*(1/√3)h
    dT = (2/3)BIhdh
    So the total torque on this half of the triangle is
    T = (2/3)BI∫0x√3 hdh
    T = (2/3)BI[h2/2]|0x√3
    T = (2/3)BI*(3x2/2)
    T = BIx2

    The torque on the entire equilateral triangle should be double this, or
    T = 2BIx2

    But this is wrong. The formula says that the torque should be
    T = IAB
    For an equilateral triangle with side = 2x, A = x2√3
    T = IBx2√3

    Where did I go wrong?
     
  2. jcsd
  3. Nov 9, 2003 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The force is given by F = Idl X B, where "dl X B" is the cross product of the vectors dl and B. Since they are not perpendicular, you must include a factor of sin 60. Include that missing factor and your answer is perfect.
     
  4. Nov 9, 2003 #3
    Thanks, Doc. I forgot that now I'm dealing with the L vector instead of just the orthogonal to the plane of the loop.
     
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