Deriving Biot-Savart's Law, kinda a math problem not really physics

[flyingpig]

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

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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

$$d\vec{s} = <0,d\vec{y},0>$$

$$\hat{r} = <\sin\theta,\cos\theta,0>$$

$$\begin{vmatrix} i & j & k\\ 0& d\vec{y} &0 \\ \sin\theta & \cos\theta &0 \end{vmatrix} = -\sin\theta d\vec{y} \hat{k}$$

So now my integral is

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

Now here is the problem, I had trouble finding what r² is, I decided to let it be $$x^2 + y^2 = r^2$$

But then my integral becomes$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta d\vec{y} \hat{k}}{x^2 + y^2}$$

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

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

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

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

Now the question is, how do I get rid of the x²?

EDIT: TEX FIXED

 

[flyingpig]

Please help me, it is driving me insane that I can't solve this

 

[flyingpig]

Here is another attempt

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-cos\theta d\vec{y} \hat{k}}{x^2 + y^2}$$

I drew my triangle again and I substitute $$cos\theta = \frac{y}{\sqrt{x^2 + y^2}}$$$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\frac{y}{\sqrt{x^2 + y^2}}d\vec{y} \hat{k}}{x^2 + y^2}$$$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-y d\vec{y} \hat{k}}{(x^2 + y^2)^\frac{3}{2}}$$

So I ran this on Mathematica setting my bounds from -inf to +inf and I got 0...which is bad...very bad

 

[flyingpig]

Please someone, Sammy S, gneil, anyone please lol

 

[flyingpig]

Here is a better picture

http://img39.imageshack.us/img39/6584/pictureson.th.png

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[jhae2.718]

What are you using as your coordinate system? (The magnitude of a x b should be |a||b|sin(theta), and you have a cos(theta)...)

 

[flyingpig]

I know, my x is your standard "y" in this picture and vice-versa

 

[jhae2.718]

Is this the way it is:

Edit: with theta between the x-axis and r...

 

[flyingpig]

No it's like this picture, I should probably switch it...

http://img39.imageshack.us/img39/6584/pictureson.th.png

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[jhae2.718]

I'm not quite sure how you've setup your coordinate system, but it might be easier if you define some angle phi = pi/2 - theta and integrate in terms of phi...

(Of course, if the wire is infinite, the easiest way is to use Ampere's law...)

 

[flyingpig]

I'm not quite sure how you've setup your coordinate system, but it might be easier if you define some angle phi = pi/2 - theta and integrate in terms of phi...

(Of course, if the wire is infinite, the easiest way is to use Ampere's law...)

No I already know how it was derived. I know what you are getting at because that's the one the book used. But I didn't like it because it was too difficult to remember. This one I find much more intuitive. No Ampere's law please

Which part of the coordinate are you confused about?

 

[jhae2.718]

Just to let you know, I haven't forgotten about this, but I'm busy for a little bit...

It would help if you drew a coordinate axes on the drawing; I'm trying to use the same coordinate system as you are.

 

[flyingpig]

No I already drew it, ds = dy in the picture, sorry if it is unclear.

http://img39.imageshack.us/img39/6584/pictureson.th.png

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[SammyS]

If you are deriving the Boit-Savart Law, what is it you are starting with?

 

[flyingpig]

The general equation?

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{\vec{ds} \times \hat{r}}{r^2}$$

 

[SammyS]

The general equation?

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{\vec{ds} \times \hat{r}}{r^2}$$

This is one form of the Biot-Savart Law.

Are you just trying to put it in some form you find easier to work with?

 

[flyingpig]

Yeah basically.

 

[flyingpig]

This is one form of the Biot-Savart Law.

Are you just trying to put it in some form you find easier to work with?

And to derive it, right now that's what I am doing for a rod

 

[jhae2.718]

Are you doing a finite or infinite wire?

 

[flyingpig]

Either, I want to get the indefinite integral solved first

 

[jhae2.718]

From your first post, I believe your r-hat should be $\hat{r} = \langle \sin(\theta), \cos(\theta), 0 \rangle$.

 

[SammyS]

What can you possibly mean by: $$d\vec{y}\hat{k}$$ in your integral?

Mathematica may have interpreted that as a dot product of those two vectors, which does make zero the answer.

 

[flyingpig]

No I left out $$\hat{k}$$

I just typed in int{y/(y^2 + x^2)^(3/2),y}

 

[flyingpig]

From your first post, I believe your r-hat should be $\hat{r} = \langle \sin(\theta), \cos(\theta), 0 \rangle$.

No from the way I defined x and y, it is different

 

[jhae2.718]

Because I get the correct answer if I use that r-hat.

The way you have your angle defined, x = rsin(theta) and y = rcos(theta)...

 

[flyingpig]

But your theta is wrong, at least in my picture it is wrong.

 

[jhae2.718]

x is the green line, right? Then if theta is between r and y (the line ds is along), then x = r*sin(theta)

 

[flyingpig]

yeah, you had $$<sin\theta, cos\theta,0>$$

 

[jhae2.718]

Your vector is in <x,y,z> form, right? Then x = rsin(theta), y = rcos(theta), and z = 0.

 

[flyingpig]

Wait you may be onto something...*goes and fix tex*

 

[flyingpig]

Okay! Tex is fixed on first page, made some other minor substition fixed. BUt my x^2 still remains..

 

[jhae2.718]

Use the approach you did in post #3...

I also TeX-ed up my solution; I'll post it in a little bit...

 

[SammyS]

Okay! Tex is fixed on first page, made some other minor substitution fixed. But my x^2 still remains..

The last I looked, you did have inconsistencies between the way θ is drawn in your figure and the way it's used in some of your equations.$$d\vec{s} = <0,d\vec{y},0>$$  should be  $$d\vec{s} = <0,dy,0>$$

$$\hat{r} = <\cos\theta,\sin\theta,0>$$  should be  $$\hat{r} = <\sin\theta,\cos\theta,0>$$

In general  $$\left|\vec{A}\times\vec{C}\right|=\left|\vec{A}\right|\left|\vec{C}\right|\sin\theta$$ so your cross product should have sin θ, not cos θ.

Jumping down to one of your equations for B:

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-cos\theta d\vec{y} \hat{k}}{x^2 + y^2}$$

After the above corrections and sin θ = x/r, you should have:

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta dy \hat{k}}{x^2 + y^2}$$

$$=-\,\frac{x\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x^2 + y^2)^{3/2}}$$

x comes out of the integral because you're integrating over y, not x.​

 

[jhae2.718]

Here's my solution:
View attachment biot.pdf

 

[flyingpig]

This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?