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Homework Help: Current Loop in a nonuniform magnetic field

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data

    A nonuniform magnetic field exerts a net force on a current loop of radius R. The figure shows a magnetic field that is diverging from the end of a bar magnet. The magnetic field B at the position of the current loop makes an angle θ with respect to the vertical, as the same magnitude at each point on the current loop. (I know that I need to solve in terms of R, I, B, and θ
    2013-03-15 19.24.19.jpg
    2. Relevant equations





    3. The attempt at a solution

    I fought the urge to use the force equation after substitution 2∏r for l. Instead I examined the force on one small segment and planned to integrate. The length would be in terms of arc length Δs. The I would be a constant. It seems that the problem indicates that the magnetic field B is constant (surely that is an assumption because the distance from the magnet was not indicated, right?) I also thought that the angle should be the only thing changing (I doubt this is a double integral problem). The subscript i indicates that this is for some segment i.



    The first thing that jumps out at me is that we don't have the necessary Δθ, but we do have Δs. Δs=ΔθR, but this is for a different θ, right? So, here is where I got lost and thought that something was wrong.

    Next I tried relating it with torque.


    But here again we have a different θ value, correct?

    I would very much appreciate some help. Thank you!
  2. jcsd
  3. Mar 15, 2013 #2

    rude man

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    What are you supposed to solve for?
  4. Mar 16, 2013 #3
    Net force
  5. Mar 16, 2013 #4

    rude man

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    Use the differential form of your F equation above, which is
    d F = I d l x B for the force on an element d l of the loop.

    Then integrate around the loop - an easy integration since the force is constant everywhere around the loop.
  6. Mar 16, 2013 #5
    I know that the answer is to be 2πRIBsinθ, but I don't see how or why.

    dF=IdlxB is the same as saying dF=IΔsxB or dF=IBsinθΔs integrating yields

    F=IBssinθ or F=2πRsinθ.

    But doesn't the value of theta change for each segment?

    Oh wait! No, it doesn't. It should remain constant as the ring has rotational symmetry. Am I correct in my reasoning here?
  7. Mar 16, 2013 #6


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    It is important to note that in the formula F = I B Δs sinθ, θ is not the same θ as given in the diagram. Remember, in the formula, sinθ is coming from a cross product of Δs and B.
  8. Mar 16, 2013 #7

    rude man

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    That is correct! And take note of what tsny says about theta. Keep track of angles when you take your cross-product!
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