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The magnitude of the magnetic torque on the loop

  1. Feb 12, 2009 #1
    A plastic circular loop of radius R and a positive charge q is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed omega.

    If the loop is in a region where there is a uniform magnetic field B directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.


    well, i did this
    F=qvB
    and know that v = w(omega).R
    it gives me F=qwRB
    and Torque = F. L, then i got T=qwR^2B


    my answer seems not correct.., please advise.
     
  2. jcsd
  3. Feb 14, 2009 #2

    Doc Al

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    Staff: Mentor

    That's the force on a moving point charge (magnitude only) where v and B are perpendicular.
    OK.
    Don't lump all the charge together.
    Note that the force on each element of charge on the loop will give rise to a different torque contribution, since the distance to the axis (a diameter) is different.

    Hint: Review the concept of magnetic dipole moment of a current loop and the torque it experiences in a magnetic field.
     
  4. Mar 27, 2009 #3
    I got close...

    r varies from 0 to R
    F=(q/2R*pi)Bv
    v=omega*r
    dtorque=(q*omega*r^2*B)/(2*pi*R)

    torque= (some constant)(q*omega*B*R^2)/(3*pi)

    I can't figure out the constant though. Have I only found the upper half of the loop?
     
  5. Mar 27, 2009 #4
    wait no, I lost a 2. That still doesn't give me the right answer though.
     
  6. Mar 28, 2009 #5

    Doc Al

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    Staff: Mentor

    To solve it properly using dF = dq v X B requires taking the angle into consideration and integrating. I recommend the hint I gave in my last post.
     
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