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

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  • #1
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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 r2 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 x2?

EDIT: TEX FIXED
 
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Answers and Replies

  • #2
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Please help me, it is driving me insane that I can't solve this
 
  • #3
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Here is another attempt

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

I drew my triangle again and I substitute [tex]cos\theta = \frac{y}{\sqrt{x^2 + y^2}}[/tex]


[tex]\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}[/tex]


[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-y d\vec{y} \hat{k}}{(x^2 + y^2)^\frac{3}{2}}[/tex]

So I ran this on Mathematica setting my bounds from -inf to +inf and I got 0...which is bad...very bad
 
Last edited:
  • #4
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Please someone, Sammy S, gneil, anyone please lol
 
  • #6
jhae2.718
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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)...)
 
  • #7
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I know, my x is your standard "y" in this picture and vice-versa
 
  • #8
jhae2.718
Gold Member
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Is this the way it is:
pf4.png

Edit: with theta between the x axis and r...
 
  • #10
jhae2.718
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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...)
 
  • #11
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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?
 
  • #12
jhae2.718
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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.
 
  • #14
SammyS
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If you are deriving the Boit-Savart Law, what is it you are starting with?
 
  • #15
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The general equation?

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{\vec{ds} \times \hat{r}}{r^2}[/tex]
 
  • #16
SammyS
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The general equation?

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{\vec{ds} \times \hat{r}}{r^2}[/tex]

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?
 
  • #17
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Yeah basically.
 
  • #18
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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
 
  • #19
jhae2.718
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Are you doing a finite or infinite wire?
 
  • #20
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Either, I want to get the indefinite integral solved first
 
  • #21
jhae2.718
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From your first post, I believe your r-hat should be [itex]\hat{r} = \langle \sin(\theta), \cos(\theta), 0 \rangle [/itex].
 
  • #22
SammyS
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What can you possibly mean by: [tex]d\vec{y}\hat{k}[/tex] in your integral?

Mathematica may have interpreted that as a dot product of those two vectors, which does make zero the answer.
 
  • #23
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No I left out [tex]\hat{k}[/tex]

I just typed in int{y/(y^2 + x^2)^(3/2),y}
 
  • #24
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From your first post, I believe your r-hat should be [itex]\hat{r} = \langle \sin(\theta), \cos(\theta), 0 \rangle [/itex].

No from the way I defined x and y, it is different
 
  • #25
jhae2.718
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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)...
 

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