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What is the magnitude of the torque acting on the loop? T = IAB

  1. Nov 2, 2005 #1
    A circular wire loop of radius 20.0 cm carries a current of 3.10 A. It is placed so that the normal to its plane makes an angle of 34.0° with a uniform magnetic field of 15.0 T.

    I don't understand why this isnt right, i got the first part right, it wanted me to find the magnetic dipole moment of the loop, which i found to be .3896 Am^2; I got this by using M = IA;
    now it wants me to find the magnitude of the torque acting on the loop. So I used the formula:
    Torque = IAB;
    I = current
    A = area of loop
    B = Magnetic field;
    Torque = (3.10)(PI*.20^2)(15) = 5.84 Nm, which is wrong, any ideas why? :bugeye:
    Thanks.
     
  2. jcsd
  3. Nov 2, 2005 #2

    James R

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    The torque is only IAB if the field lines are perpendicular to the plane of the loop, whioch isn't the case here.

    The complete formula is:

    [tex]\tau = IAB \sin \theta[/tex]

    where [itex]\theta[/itex] is the angle between the normal to the loop and the magnetic field. In your problem, the relevant angle is 34 degrees.

    In vector terms, the torque is:

    [tex]\vec{\tau} = \vec{\mu} \times \vec{B}[/tex]

    where [itex]\vec{\mu}[/itex] is the magnetic moment.
     
  4. Nov 2, 2005 #3
    ohh thanks alot james, worked great!
     
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