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Homework Help: HELP - Magnetism Physics Problem

  1. Dec 4, 2004 #1
    HELP - Magnetism Physics Problem!!

    The problem says:

    A long piece of wire with a mass of .100 kg and a total length of 4.00m is used to make a square coil with a side of 0.100 m. The coil is hinged along a horizontal side, carries a 3.40-A current, and is placed in a vertical magnetic field with a magnitude of 0.0100 T. (a) Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. (b) Find the torque acting on the coil due to the magnetic force at equilibrium.

    I am having trouble solving for the angle that the coil makes with the vertical...

    I know that the equation T = BINAcos(theta) has to do with it and i can solve the torque to be .0034 Nm without the cos(theta) from the angle that the plane makes with the vertical but I have absolutely no idea how to solve for theta from the information that is given in the problem.

    I also tried to solve for theta using F = BILsin(theta) using the mass to calculate the force by setting it equal to the weight which is mg then plugging in B, I and L but when I tried to use arcsin to solve for theta the number that I got was out of the domain of sin.

    What I had was

    wt = mg
    wt = (.100kg)(9.81 m/s/s)
    wt = .981 N
    F = .981 N
    F = BILsin(theta)
    .981 N = (.0100T)(3.40A)(4.00m)sin(theta)
    then I got:
    sin(theta) = 7.213 and obviously this number is out of the domain of sin so I can't take the arcsin of it..

    can someone please tell me what I did wrong or point me in a more correct direction because i am 100% stumped..

    (This is for my high school physics class but it might as well be college level)
  2. jcsd
  3. Dec 4, 2004 #2


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    The magnetic field is not supporting the entire weight of the coil. The hinge also provides support. What you need to do is balance the torques produced by both the magnetic field and the hinge. Also, the current gets mulitplied by the number of windings in the coil.
  4. Dec 4, 2004 #3
    I don't understand how I balance the two torques, how do I find the torque that is produced by the hinge?
  5. Dec 5, 2004 #4


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    OK. Since the magnetic field is uniform then it exerts no net force on the loop - only torque and it is given by [itex]NIAB \sin \theta[/itex] where A is the area of the loop. The hinge supports the enitire weight of the loop applying an upward force of mg at a distance L/2 (0.05 m) from the center of the loop and at an angle of [itex]\pi/2 - \theta[/itex] with respect to a line joining the hinge and the center of the loop. Therefore, the hinge produces torque [itex]mg (L/2) \cos \theta[/itex] in the opposite direction.
  6. Dec 5, 2004 #5
    Ahhh, I still don't know what I'm doing... The solutions im getting are out of the domain of sin and I have no idea what to do... can someone please walk me through this?
    Last edited: Dec 5, 2004
  7. Dec 5, 2004 #6
    I either get sin(theta) = 28.85 or another number out of its domain...
  8. Dec 5, 2004 #7


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    You should be getting [itex]\tan \theta[/itex] equals something.
  9. Dec 5, 2004 #8
    ah hah! I figured it out at last, thank you very much for the help.
  10. May 13, 2005 #9
    I'm having trouble with the same exercise, but I still can't figure it out. What I did was try to calculate the magnetic force acting at each side of the coil. The magnetic forces at the right and left sides of the coil are balanced, and the magnetic force at the lower side doesn't matter because this side is hinged. Therefore, in order to have a net force equal to zero, I need to equal the normal component of the weight (perpendicular to the coil) to the normal component of the only magnetic force that matters (the one that acts at the upper side). Thus:

    F.cos(theta) = W.sin(theta)
    F/W = tan(theta)
    [N.i.l.B.sin(90º)]/(mg) = tan(theta)
    tan(theta) = 0.0347 => theta = 1.987º


    N = 10
    i = 3.40 A
    l = 0.100 m
    B = 0.010 T
    m = 0.100 kg
    g = 9.8 m/s^2

    According to the answer, theta = 3.97º (two times my answer!).

    Could anyone help me please?!!

    Last edited: May 13, 2005
  11. May 13, 2005 #10

    Doc Al

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    What you need to equate is not the forces, but the torques they generate. (Note that the forces do not act at the same point; the weight acts at the center of mass.)
  12. May 13, 2005 #11
    I understand that the forces don't act at the same point. But I thought that, in order to have equilibrium, two things were necessary:
    1) Net force = 0
    2) Net torque = 0

    That's why I tried to equal the forces. What I don't get is that question "b" asks you to find the torque. If there is a torque, the object will rotate; is this equilibrium?! I'm confused... :confused:

    Thanks a lot for the help
  13. May 13, 2005 #12

    Doc Al

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    Staff: Mentor

    Yes, it's true that for equilibrium both conditions apply. But to apply the first you must consider all the forces acting on the coil. In addition to the weight and magnetic force there is the force provided by the hinge.

    All you need to solve this problem is the second condition. By choosing the hinged side as your pivot you can say that the torques produced by the magnetic force and the weight must be equal and opposite.

    Question "b" asks for the torque due to the magnetic force, not the net torque on the coil. You are correct that the net torque will be zero.
  14. May 13, 2005 #13
    OK. I think I got it now. At least my final answer is correct. What I understand is that the only movement the coil can have is a rotation, since it's hinged at one side (there's no way it could have a displacement in translation). Therefore, there's no point in finding the net force, because it should be zero already (?!). Equating the torques, I got:

    Torque = F.cos(theta).l - W.sin(theta).(l/2)
    Torque = 0 => F.cos(theta).l = W.sin(theta).(l/2)
    2F.cos(theta) = P.cos(theta)
    2.F/P = tan(theta)
    tan(theta) = [2NilBsin(90º)]/mg

    And the torque produced by the magnetic force (this part puzzled me!) is:

    Torque = [NilBsin(90º)].l.cos(theta)
    Last edited: May 13, 2005
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