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Cylinder wrapped with wire on hill with magnetic field

  1. Mar 28, 2016 #1
    1. The problem statement, all variables and given/known data
    The figure shows a cylinder of mass 3.5kg, radius 4.4cm and length 10.2cm with 105turns of wire wrapped around it lengthwise, so that the plane of the wire loop contains the axis of the cylinder. What is the least current which while flowing through the loop will prevent the cylinder from rolling down an inclined plane of inclination 21°, in the presence of a vertical magnetic field of B=0.7T? The plane of the windings is parallel to the inclined plane.

    dfyCbLX.png

    m = 3.5 kg
    r = 4.4cm = 0.044m
    l = 10.2cm = 0.102m
    theta = 21degrees
    B = 0.7T
    n = 105 turns
    I = ?
    2. Relevant equations

    Magnetic torque = (n/l)IABsin(theta)
    Gravitational torque = r x F = mgrsin(theta)
    Torque_CW - Torque_CCW = 0

    3. The attempt at a solution

    Okay so I know that the sum of the torques must be zero, and that there will be a torque from gravity and one from the magnetic field, but I'm confused as to how to actually set up the equations. For gravity I tried doing a free body diagram but I didn't know where to consider gravity to be acting on. And for the magnetic torque I have the equation, but from looking at the picture I would assume that the CW and CCW torques due to magnetism would be equal and impossible to balance with gravity. I think my main problem with this question is coordinates.
     
  2. jcsd
  3. Mar 28, 2016 #2
    suppose there is no field or current flowing -draw a free body diagram of a cylinder which is set to roll down the incline . normally we take the moment of the force from the point of contact of the cylinder with the plane.
    then put on the magnetic fiield and pass a current through the winding which are rectangular in shape with length equal to cylinder length and breadth equal to diameter of the cylinder - calculate the force on four arms and calculate the torque from the same point of contact ......
     
  4. Mar 28, 2016 #3
    Okay so we can use the parallel axis thm to get the moment of inertia at the point of contact with the plane:

    I = 1/2MR2 + MR2 = 3/2MR2

    but then how do you use that to calculate torque?, I know that torque = I*alpha but we can't find alpha because there is no acceleration.
     
  5. Mar 28, 2016 #4
    your body should be at the limit of turning by the torque of the forces i.e. moment of the force should balance rather than accelerated rotation to take place. one does not need alpha.
     
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