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

AI Thread Summary
The discussion focuses on deriving Biot-Savart's Law using a vector method, with the user struggling to manage multiple differentials and the integral's complexity. They initially set up their equations but encountered confusion regarding the variable r² and the correct coordinate system. Suggestions from other participants include using a different angle for integration and clarifying the definitions of variables to avoid inconsistencies. The user expresses a desire for a more intuitive derivation than the one presented in their textbook, while also seeking assistance with their mathematical approach. The conversation highlights the collaborative effort to resolve the mathematical challenges involved in the derivation.
flyingpig
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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

Uploaded with ImageShack.us

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}<br /> i &amp; j &amp; k\\ <br /> 0&amp; d\vec{y} &amp;0 \\ <br /> \sin\theta &amp; \cos\theta &amp;0 <br /> \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
 
Last edited by a moderator:
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Please help me, it is driving me insane that I can't solve this
 
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
 
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Please someone, Sammy S, gneil, anyone please lol
 
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)...)
 
I know, my x is your standard "y" in this picture and vice-versa
 
Is this the way it is:
pf4.png

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

\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{\vec{ds} \times \hat{r}}{r^2}
 
  • #16
flyingpig said:
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?
 
  • #17
Yeah basically.
 
  • #18
SammyS said:
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
Are you doing a finite or infinite wire?
 
  • #20
Either, I want to get the indefinite integral solved first
 
  • #21
From your first post, I believe your r-hat should be \hat{r} = \langle \sin(\theta), \cos(\theta), 0 \rangle.
 
  • #22
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.
 
  • #23
No I left out \hat{k}

I just typed in int{y/(y^2 + x^2)^(3/2),y}
 
  • #24
jhae2.718 said:
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
 
  • #25
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)...
 
  • #26
But your theta is wrong, at least in my picture it is wrong.
 
  • #27
x is the green line, right? Then if theta is between r and y (the line ds is along), then x = r*sin(theta)
 
  • #28
yeah, you had &lt;sin\theta, cos\theta,0&gt;
 
  • #29
Your vector is in <x,y,z> form, right? Then x = rsin(theta), y = rcos(theta), and z = 0.
 
  • #30
Wait you may be onto something...*goes and fix tex*
 
  • #31
Okay! Tex is fixed on first page, made some other minor substition fixed. BUt my x^2 still remains..
 
  • #32
Use the approach you did in post #3...

I also TeX-ed up my solution; I'll post it in a little bit...
 
Last edited:
  • #33
flyingpig said:
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} = &lt;0,d\vec{y},0&gt;  should be  d\vec{s} = &lt;0,dy,0&gt;

\hat{r} = &lt;\cos\theta,\sin\theta,0&gt;  should be  \hat{r} = &lt;\sin\theta,\cos\theta,0&gt;

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.​
 
Last edited:
  • #35
This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?
 
  • #36
jhae2.718 said:
Here's my solution:
View attachment 33843
I didn't do the integration, but my solution agrees with this.

But, I do think there is a problem with both solutions.

It's late, so I'll look at it tomorrow.

Below added in Edit:

The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.

I intend to redo my solution using (x0, y0) for the coordinates of P.


.
 
Last edited:
  • #37
flyingpig said:
This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?

I used LaTeX :)

MikTeX 2.9 on Windows.

Source:
Code: 
\documentclass[10pt,letterpaper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fullpage}
\usepackage{graphicx}
\usepackage{float}
\usepackage{xfrac}
\newcommand{\custpic}[4]{\begin{figure}[H]
	\begin{center}
		\includegraphics[width=#2\textwidth]{#1}
		\vspace{-15pt}
		\caption{#3}
		\vspace{-20pt}
		\label{#4}
	\end{center}
\end{figure}}

\begin{document}
\custpic{pf5}{.75}{Problem diagram}{pf}

Start with Eq. (\ref{bs}), the Biot-Savart Law:
\begin{equation}\label{bs}
	d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{s} \times \hat{r}}{r^2}
\end{equation} 
From the diagram:
\begin{align*}
	r = \sqrt{x^2 + y^2} \\
	x = r\sin(\theta) \\
	y = r\cos(\theta) \\
	d\vec{s} = d\vec{y} 
\end{align*}
Substitute into Eq. (\ref{bs}):
\begin{equation}
	d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{y} \times \hat{r}}{x^2 + y^2}
\end{equation}
Taking the cross product, $id\vec{y} \times \hat{r} = dy\sin(\theta) (-\hat{k})$. Substitute in:
\begin{equation}
	d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{dy\sin(\theta)}{x^2 + y^2} (-\hat{k})
\end{equation} 
From above, $\sin(\theta) = \dfrac{x}{r}$:
\begin{equation}
	d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation} 
Find $\vec{B}$ by taking the integral over the infinite line:
\begin{equation}
	\vec{B} = \frac{\mu_0 i}{4\pi}\int\limits_{-\infty}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation} 
\begin{equation}
	\vec{B} = 2 \frac{\mu_0 i}{4\pi}\int\limits_{0}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation} 
Integrate (I used Wolfram$|$Alpha) to get:
\begin{equation}
	\vec{B} = 2 \frac{\mu_0 i}{4\pi}\left[\frac{y}{x\sqrt{x^2 + y^2}}\right]_{0}^{\infty} (-\hat{k})
\end{equation} 
The indefinite integral evaluates to:
\begin{equation}
	\vec{B} = \frac{\mu_0 i}{2\pi x} (-\hat{k})
\end{equation}
We can confirm this using Ampere's Law (left as an exercise to the reader...)
\end{document}
 
  • #38
It's getting late here, I will recheck my math tomorrow, thank you to both of you very much. That sentence did not make any sense, but hopefully you understood my appreciation.
 
  • #39
How did you save the file so perfectly? Whenever I saved it as a pdf, and I try reopen it it says it can't read it.
 
  • #40
SammyS said:
...
The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.
...
Simply use x0 for x, and use (y0 - y) for y.

\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}

I think you may find it handy to evaluate the integral from y0 to y .
 
  • #41
flyingpig said:
How did you save the file so perfectly? Whenever I saved it as a pdf, and I try reopen it it says it can't read it.

You need to run latex on it.

SammyS said:
Simply use x0 for x, and use (y0 - y) for y.

\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}

I think you may find it handy to evaluate the integral from y0 to y .

That is an excellent point, and is something I just glossed over.
 
  • #42
WHat dose that mean "run latex" over it...?
 
  • #43
If you have a version of LaTeX installed, open up the command prompt, change directory to the location where you have saved the file, and type:
Code: 
pdflatex filename.tex
replacing filename with the actual filename.
 
  • #44
I am guessing LaTeX is not the same thing as MikTeX lol
 
  • #45
MikTeX is a LaTeX distribution. http://docs.miktex.org/faq/faq.html

Try a LaTeX editor like TeXniCenter, Texmaker, etc to actually write/compile the code if you don't like the CLI.
 

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