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Homework Help: Torque from a magnetic field

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A rectangular loop of wire of 150 turns, 34 by 23 cm, carries a current of 2.5 A and is hinged at one side. What is the magnitude of the torque that acts on the loop, if it is mounted with its plane at an angle of 28 degrees to the direction of a 1 T uniform magnetic field in the positive x direction?

    2. Relevant equations
    torque = u x B, where u is the magnetic dipole moment and B is the magnetic field vector

    u = I*A (current * area)

    u x B = u*B*Sin(theta)

    3. The attempt at a solution

    Basically, I tried plugging this into the formulas above:
    Torque = (.34m)(.23m)(150 turns)(2.5 Amps)(1 Tesla)(Sin (28))
    Torque ~ 13.767 N*m

    However, this was flagged as wrong. I also tried doing this without the (150 turns) term, and still was flagged as wrong. I'm stumped. Does anyone have any thoughts?

  2. jcsd
  3. Mar 31, 2009 #2


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    Hi maxsthekat,

    In your formulas, you have: u x B = u*B*Sin(theta). What exactly does the angle theta represent? If you look in your book at the definition of this angle I think you'll see where you are going wrong.
  4. Mar 31, 2009 #3
    As I understand it, theta represents the angle between the magnetic dipole moment and the B-field. As such, shouldn't it be the 28 degrees given in the problem? If not, why not?
  5. Mar 31, 2009 #4


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    Exactly right; that angle is what is needed in the formula.

    In a problem they can give you any angle whatsoever, and in this problem the 28 degrees is between the plane of the loop and the B-field. But you need the angle between the magnetic moment and the B-field. How is the direction of the magnetic dipole moment of a current loop related to the plane of that loop?
  6. Mar 31, 2009 #5
    Interesting... By looking at the drawing of the problem (sorry I can't post it online-- no scanner), I see the current goes up the "hinge" on the positive z-axis. So, by applying the right hand rule, I get the magnetic moment pointing back towards the y-axis.

    Since the magnetic moment is perpendicular to the "area" of the loop, I take 90 degrees, and add that to the angle formed from the x-axis (90 degrees - 28 degrees = 62 degrees) for a total of 152 degrees. Does that sound right?

    Thanks for your help, by the way :)
  7. Mar 31, 2009 #6
    Bah. Just noticed that Sin(152) = Sin(28).

    My only other conjecture is that it might be just 62 degrees (angle between the plane and the field as measured from the x-axis)?

    Also, am I using the 150 turns correctly? I believe I tried punching in 62 degrees in the equation, but got it kicked back to me.
  8. Mar 31, 2009 #7


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    I do wish I could see the picture, but yes, it should be the 62 degrees. (But I'm not sure why you call that the angle between the plane and the field, because that has already been said to be 28 degrees.)

    Imagine the coil is laying flat on a table, so the plane of the coil is horizontal. Then according to the problem the B-field is 28 degrees above or below the horizontal (either one would be 28 degrees to the plane of the loop). And since the loop is horizontal, then the magnetic moment u is either vertically upwards or downwards. Based on that, the only angles possible between B and u are 62 degrees and 118 degrees, so your angle of 62 degrees would be right.

    Yes, the factor N does have to be included in the calculation.
  9. Mar 31, 2009 #8

    Found out how to get the link for that image :)

    The 28 degrees is measured from the y-axis to the plane, and the 62 is from the x-axis to the plane, with the B-field going in the +x direction... So, if I understand what you're saying, since the B-field is in the direction of the x-axis, we have to use the angle between the x-axis and the plane (62 degrees).

    I'll give it a shot :)
  10. Mar 31, 2009 #9
    Worked! Thanks so much for your help!

    When I tried using 62 degrees earlier, it kicked it back. Now I know why; the grading program is very particular about the precision, and my calculator was rounding :P
  11. Mar 31, 2009 #10


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    Yes, it works out that way here, but remember that there is nothing in the direction of the x-axis.

    So you might also want to think of it this way: in the following picture, I am looking straight down the z-axis of your picture. Then all you see is the top side of the loop, and you can get the directions a bit easier (I think):

    http://img208.imageshack.us/img208/7809/loopu.jpg [Broken]

    In this picture, the B-field is in the y-direction, which is what you are saying in your post with the figure (when you say the 28 degrees is from the loop to the y-axis), but that's different from what it in your original post, so I'm a bit confused. But anyways, if we assume the B-field is in the y-direction, then my diagram has the correct directions, and you can see that the angle between B and u is 62 degrees.
    Last edited by a moderator: May 4, 2017
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