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Homework Help: 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
    vector(B)=.58i+0j+0k
    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)

    Thanks!!!!
    ------------------
    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))
    38333?db=v4net.jpg


    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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