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Magnitude of torque on a current loop

  1. Apr 30, 2006 #1
    I've given this one a good effort and I cannot seem to solve it, been trying for a couple days on my own now...Anyone able to lend a hand?

    I was using t=uxB(u cross B) to find the torque and then take the negative of it to find the torque that is to hold it?

    vector u = I(vector A), and for this I am getting vector(A)=.0692i+.08j+.0346k
    When I do the matrices for the cross(x) I get 0i+.321088j+.7424k

    The only part of this that is right is the i component which is zero...

    I'll be in your debt forever if you could help with a) or c)

    The rectangular loop in Fig is pivoted about the y-axis and carries a current of 16.0 A in the direction indicated.((It's 16A, don't mind the picture saying 15A, this is the diagram below, click link))

    a) If the loop is in a uniform magnetic field with magnitude 0.580 T in the +x-direction, find the magnitude and direction of the torque required to hold the loop in the position shown.

    b) Repeat part (a) for the case in which the field is in the z-direction.

    c) For each of the above magnetic fields, what torque would be required if the loop were pivoted about an axis through its center, parallel to the y-axis?
    a =
    b =
  2. jcsd
  3. Apr 30, 2006 #2
    [tex] \tau = NIAB sin \Theta [/tex]

    N = number of loops
    I = current
    A = area of loop
    B = magnetic field
    theta = angle between loop and field
  4. Apr 30, 2006 #3
    I love you! why doesn't my book have this?? :*(.

    Thanks alot!
  5. Oct 1, 2007 #4
    well it is sinQ if we take the angle of rotatation in the geomtrical center of the loop......

    just for information..heh
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