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.(adsbygoogle = window.adsbygoogle || []).push({});

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]

and

[tex]r=\sqrt{x^2+y^2}[/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(dLxB)

[tex]

=I Rd\theta (-cos\theta(-\vec{\i})-sin\theta(-\vec{\j})) \times \frac {a}{r^3} (-\vec{k} )

[/tex]

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!

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

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