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Homework Help: Force on triangular current loop: line integrals

  1. Aug 20, 2012 #1
    1. The problem statement, all variables and given/known data

    http://pokit.org/get/img/65e8ba92c1d00bf7fc8be2b178757ed8.jpg [Broken]

    If a=5b, and I1 and I2 are known, find the force on the triangular loop.
    2. Relevant equations
    3. The attempt at a solution

    For start, the field from the infinitely long wire is :

    [itex] \vec B=\large -\frac{\mu _{0} I_{1}}{2 \pi x}\vec a_{z}[/itex]

    This is understood and very trivial.

    To find the total force on the current loop we will divide the problem into 3 parts, and solve forces on the (1) (2) (3) separately, and then add them all up.

    For the force we will use the formula:

    [itex] \vec F = I_{2} \int{\vec {dl} × \vec B} [/itex]

    From this current view I want first question answered:

    Will the only [itex] \vec a_{x} [/itex] component of the force be present?

    I got that from the right hand rule, part of the y component vector-crossed with -z will give a positive x component.

    The part with the x component crossed with the -z component will give both positive and negative contribution, leading to net 0 y component of the force.

    Is this correct and if it is, i have further questions regarding the calculations and solving the line integrals, because I am getting weird answers.

    Namely in case (1) and (2).

    Is [itex] \vec{dl} =dx \vec a_{x} + dy \vec a_{y} [/itex] in both cases the same or is in second case [itex] \vec{dl} =-dx \vec a_{x} + dy \vec a_{y} [/itex].

    If I use the fact that dl is in both cases the same, I get right results.

    But keep in mind that when substituting dy in first case i used:

    [itex] \vec{dy}=\frac {c}{2b} dx [/itex]

    and in second case:

    [itex] \vec{dy}=-\frac {c}{2b} dx [/itex]

    I should also add that case 3 is not a problem because I got that right.

    If i use [itex] \vec{dl} =-dx \vec a_{x} + dy \vec a_{y} [/itex] I get a net y component of the force which is not right in my intuition.

    How do you solve these line integrals the right way?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 21, 2012 #2


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    Yes, you're right. By symmetry, the vertical component of the forces will cancel, so any net force has to act in the horizontal direction.

    By definition, ##d\vec{l} = dx\,\vec{a}_x + dy\,\vec{a}_y + dz\,\vec{a}_z##. The relative sign dependence of the first two pieces will come in through the relationship between x and y, e.g. y=(2b/c)x or y=-(2b/c)x, and dz=0 for these contours.
    Last edited by a moderator: May 6, 2017
  4. Aug 22, 2012 #3
    oh so I dont have to account for it in the differential of l?

    It will account for it self when I change the dy into dx through that relationship?

    Makes sense. Thank you for the answer. I actually did it this way, but this was purely shot in the dark because I knew the final answer.

    Thank you for Your help, You have been very helpful.
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