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Homework Help: Can someone help me understand resultant cross product vector on half loop of current

  1. Nov 24, 2008 #1
    I can't seem to integrate this correctly. I only need the half loop integration, as i have the correct integration for the infinite lines.

    1. The problem statement, all variables and given/known data

    We start with the top half of a half-loop of radius R, centered at the origin, with infinite line segments travelling in the +/- x directions.

    [tex]B=\frac{a}{r^3} (-\vec{k} )[/tex]

    2. Relevant equations

    dF=I (dL x B)

    3. The attempt at a solution

    My attempt at a solution ends up with me getting confused over how to determine the resultant vector by cross product.

    =I Rd\theta (-cos\theta(-\vec{\i})-sin\theta(-\vec{\j})) \times \frac {a}{r^3} (-\vec{k} )
    now, [tex]d\theta = \pi [/tex], and
    I integrate [tex]\int_{-\pi /2}^{\pi/2}(-cos\theta(-\vec{i})-sin\theta(-\vec{j}) [/tex]

    which results in

    [tex]\int_{-\pi /2}^{\pi/2}(-cos\theta(-\vec{i})-\int_{-\pi /2}^{\pi/2}(sin\theta(-\vec{j}) [/tex]

    [tex]\left[(-sin\theta(-\vec{i}) \right]_{-\pi /2}^{\pi/2}+\left[(cos\theta(-\vec{j})\right]_{-\pi /2}^{\pi/2} [/tex]
    yielding 2+0=2

    I then integrate [tex]2\pi IRa \int_{-R}^{R} \frac{dx}{x^3}[/tex]

    but my final result ends up being

    [tex]=2IR\pi a [\frac{-1}{2R^2}-\frac{-1}{2R^2}][/tex] which is zero and I need it to be one.

    Also, I believe the correct answer is
    [tex]\frac{2Ia}{R^2}(-\vec{j})[/tex] which does not have the RΠ factor which I end up with...
    Can someone please help me comprehend this better, as it has been many years since I have had calculus and haven't been able to find any examples of this type of integration.

    Thank you very much!
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 24, 2008 #2
    here are some pics

    here are the pictures of my problem and, seemingly, my result.

    Attached Files:

    Last edited: Nov 24, 2008
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