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Torque from magnetic field/current

  1. Oct 20, 2004 #1
    The figure shows a cylinder of mass 5 kg, radius 3.6 cm and length 9.2 cm with 145turns of wire wrapped around it lengthwise, so that the plane of the wire loop contains the axis of the cylinder. Assuming that the plane of the windings is parallel to the incline, what is the current (in Amps) which when flowing through the loop will make the cylinder stationary on an inclined plane of inclination 41° in the presence of a vertical magnetic field of B=0.1 T?

    Ok, so I'm equating the the torque generated by gravity to that of the current. So for gravity I have:

    T = (5 kg)(9.8sin(41)) X (.092m) = 2.96 Nm

    for the current I have:

    T = (145 turns)(i)(pi*(.036m)^2) X (.1T)

    When I equate that to 2.96 and solve for i, I get i = 50.14 A, which isn't the right answer.

    The only part that confused me is that the wire is wrapped lengthwise, which would mean it goes 145 times around the ends of the cylinder? I have no idea how to solve the problem if that is the case.
     
  2. jcsd
  3. Oct 21, 2004 #2

    Galileo

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    Maybe the plane of the windings should be parallel to the surface normal.
     
  4. Oct 21, 2004 #3
    Your first torque is right almost but the radius from the pivot point is the radius from the cylinder not its length. Your second torque should read something like 145*i*0.092*(2*.036)*sin(41)*0.1, that is Area of one turn*number of turns*current*the field*sin(theta). Set the two equal since gravity is clockwise torque and the current would make counterclockwise.
     
    Last edited by a moderator: Oct 21, 2004
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