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

1. Nov 24, 2008

### Thomassino

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.

$$B=\frac{a}{r^3} (-\vec{k} )$$
and
$$r=\sqrt{x^2+y^2}$$

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

which results in

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

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

I then integrate $$2\pi IRa \int_{-R}^{R} \frac{dx}{x^3}$$

but my final result ends up being

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

Also, I believe the correct answer is
$$\frac{2Ia}{R^2}(-\vec{j})$$ 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. Nov 24, 2008

### Thomassino

here are some pics

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

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