Deriving the equation for magnetic induction

[raul_l]

Homework Statement

We have a straight wire with radius R. What is the magnetic induction through point A at a distance of r?
In other words, derive the equation B=\frac{\mu_{0}i}{2\pi r} by using d\vec{B}=\frac{\mu_{0}i}{4\pi r^2}(d\vec{l} \times \vec{u_{r}})

Homework Equations

d\vec{B}=\frac{\mu_{0}i}{4\pi r^2}(d\vec{l} \times \vec{u_{r}})

The Attempt at a Solution

Now, I know there are a couple of methods for doing this, some of them being quite simple. But I'm specifically interested in solving the problem by using just one equation (the Biot-Savart law) and integrating it over the entire volume of the wire (assuming we have an infinitely long straight wire with constant radius and density). I guess this means bringing in a triple integral. As far as I know dB=\frac{\mu_{0}i}{4\pi s^2}\cos{\alpha}dl where \cos{\alpha}=\frac{r}{s} (see drawing1) in which case we are dealing with an infinitely thin wire and it would be not too difficult to solve B=\int^{\infty}_{-\infty} \frac{\mu_{0}i}{4\pi s^2}\cos{\alpha}dl. However, assuming that we have a real wire with a radius R, we need one more angle to describe the position of dl with respect to point A. (see drawing2). If any of this makes any sense at all, I should get dB=\frac{\mu_{0}i}{4\pi x^2}\cos{\alpha}\cos{\phi}dl where \cos{\alpha}=\frac{a}{s}, \cos{\phi}=\frac{s}{x} and r is the distance between point A and the center of the wire. I haven't learned yet how to create triple integrals (only double integrals so far) so this is why it's a little bit beyond my level. Any thoughts on how to continue? Am I even moving in the right direction here?

 

[jhicks]

Be careful in choosing your differential elements. As it stands right now, yours doesn't make too much sense because you ultimately wish to integrate over a volume V containing a constant current density \vec{J}.

Use the form d\vec{B} = \frac{\mu_0}{4 \pi r^2} (\vec{J}dV \times \vec{r}) to start with. By forgoing the use of vector calculus, however, setting up the integrals properly may be pretty difficult. For example, the bounds of \phi over which to integrate are not nearly as simple as \alpha. As to the solution, you may not be able to find an exact result I'm afraid. I have done a few similar calculations and I ended up computing them in MATLAB.

 

[raul_l]

Interesting. I guess that makes sense. I was wondering about this however. Is it right for me to assume that the bounds of \phi are functions of \alpha and a?

 

[raul_l]

I finally figured it out. I'm going to divide the wire into infinitely thin planes, each plane perpendicular to r (the distance between point A and the center of the wire). Each plane I will divide into infinitely thin straight lines, so as if the whole wire represents a bundle of infinitely thin straight "wires" (see wire.jpg, where each line represents a single point on the cross-section of the wire). Each of those lines has a current di running through them inducing a magnetic field vector d\vec{B} at point A. All I have to do is find a way to express the sum of all the modules of d\vec{B} created by each line to get the answer I'm looking for.

symbol definitions:
r - distance between point A and the center of the wire
h - distance between the observable plane and the center of the wire (r+h is the distance between point A and the observable plane)
R - radius of the wire
R' - half the width of the observable plane. It depends on the value of h, therefore R'(h).
x - distance between line B (in the drawing it appears as point B) and the central line of the observable plane
s - distance between line B and point A
phi - angle between r and s (see drawing)

First, I will observe the magnetic induction created at point A by the central line of the wire. It's easy to show that that would be dB=\frac{\mu_{0}di}{2\pi r}. For any other line I would have to substitute r with s (distance between the observable line and point A). For example for line B I get dB=\frac{\mu_{0}di}{2\pi s}\cos{\phi}=\frac{\mu_{0}jdS}{2\pi s}\frac{r+h}{s}=\frac{\mu_{0}j(r+h)dxdh}{2\pi s^2}=\frac{\mu_{0}j(r+h)dxdh}{2\pi ((r+h)^2+x^2)}
By integrating it over x from -R' to R' I get the magnetic induction at point A induced by each plane. To find the sum of all planes I have to integrate it over h from -R to R. B=\int^{R}_{-R}{\int^{R'(h)}_{-R(h)}{\frac{\mu_{0}j(r+h)}{2\pi ((r+h)^2+x^2)}dx}dh}=\int^{R}_{-R}{2\int^{R'(h)}_{0}{\frac{\mu_{0}j(r+h)}{2\pi ((r+h)^2+x^2)}dx}dh}=\int^{R}_{-R}{\int^{R'(h)}_{0}{\frac{\mu_{0}j(r+h)}{\pi ((r+h)^2+x^2)}dx}dh}

h^2+R'^2=R^2 \Rightarrow R'(h)=\sqrt{R^2-h^2}
Also, I want to keep the current constant, independent of the radius of the wire, therefore I will substitute j=\frac{i}{4\pi R^2} to get B(r,R,i)=\int^{R}_{-R}{\int^{R'(h)}_{0}{\frac{\mu_{0}\frac{i}{4\pi R^2}(r+h)}{\pi ((r+h)^2+x^2)}dx}dh}=\int^{R}_{-R}{\int^{R'(h)}_{0}{\frac{\mu_{0}i(r+h)}{4R^2 ((r+h)^2+x^2)}dx}dh} which gives me the magnetic induction of an infinitely long straight wire with the radius R at point A. So far I haven't found a computer program that could solve this kind of integral but by using different values of r, R and i to numerically find B(r,R,i) I believe that I can safely conclude that
B(r,R,i)=\int^{R}_{-R}{\int^{R&#039;(h)}_{0}{\frac{\mu_{0}i(r+h)}{4R^2 ((r+h)^2+x^2)}dx}dh}=\left\{\begin{array}{cc}\frac{\mu_{0}ir}{2\pi R^2},&amp;\mbox{ if }<br /> r&lt;R \\\frac{\mu_{0}i}{2\pi r}, &amp; \mbox{ if } r \geq R\end{array}\right.