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## Homework Statement

Basically I am deriving Biot-Savart's Law using the vector method. The one my book gives me is impossible to remember

## The Attempt at a Solution

http://img39.imageshack.us/img39/6584/pictureson.th.png [Broken]

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So I am actually just going to do the indefinite integral first, but I got so many differentials, it's ridiculous

[tex]d\vec{s} = <0,d\vec{y},0>[/tex]

[tex]\hat{r} = <\sin\theta,\cos\theta,0>[/tex]

[tex]\begin{vmatrix}

i & j & k\\

0& d\vec{y} &0 \\

\sin\theta & \cos\theta &0

\end{vmatrix} = -\sin\theta d\vec{y} \hat{k}[/tex]

So now my integral is

[tex]\vec{B} = \frac{\mu_0 I}{4\pi}\int \frac{-\sin\theta d\vec{y} \hat{k}}{r^2}[/tex]

Now here is the problem, I had trouble finding what r

^{2}is, I decided to let it be [tex]x^2 + y^2 = r^2[/tex]

But then my integral becomes

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta d\vec{y} \hat{k}}{x^2 + y^2}[/tex]

Now the problem is, I could do a trig substitution and let y = xtanθ

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta d\vec{y} \hat{k}}{x^2 + x^2 tan\theta ^2}[/tex]

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta d\vec{y} \hat{k}}{x^2 sec\theta ^2}[/tex]

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta \cos^2\theta d\vec{y} \hat{k}}{x^2 }[/tex]

Now the question is, how do I get rid of the x

^{2}?

EDIT: TEX FIXED

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