0
$\begingroup$

So, most of the documents on the internet use Biot Savart law and integrate over length of the wire (from minus infinity to infinity) for an infinite wire magnetic field, but I have tried to integrate over the angle.

Here is how I do it. The point P is placed at radius R away from the wire. The wire fragment $dl$ is at distance $r$ from this given point. The angle between $r$ and the wire itself is $\theta$.

I have decided to use this form of Biot Savart law: $$ dB = \frac{\mu_0 I}{4 \pi} \frac{\vec{dl} \times \vec{r}}{|\vec{r}|^3} $$

So, I have started derivng all required variables.

First of all, for a change in angle $ d \theta$, the $dl$ is equal $dl = r(\theta) d \theta$, where $r(\theta)$ is equal to $\frac{R}{sin(\theta)}$, so finally: $$ dl = \frac{R d\theta}{sin(\theta)} $$

Now, the vector component of $\vec{r}$ which is perpendicular to the wire is always equal to $R$.

This means, that the cross product $\vec{dl} \times \vec{r}$ is simply equal to: $$\vec{dl} \times \vec{r} = \frac{R^2 d\theta}{sin(\theta)}$$

Now I just need the ${|\vec{r}|^3} $ part which is simply equal to: $${|\vec{r}|^3}= \frac{R^3}{sin^3(\theta)}$$

Plugging all those values to my equation i get: $$dB = \frac{\mu_0 I}{4 \pi} \frac{sin^2(\theta)}{R} d\theta$$

Which I then tried to integrate over $\theta$ in range $[0, \pi]$, however the result was wrong. I think I'm somehow getting a $sin^2(\theta)$ where I just should have gotten $sin(\theta)$

Now I know that similar questions have been posted here before, but I know how I can do this so that I get the right result, but the right result isn't really important - I just want to know, why my approach is wrong.

I have also tried to sketch a picture of my "setup", here it is:

enter image description here

$\endgroup$
6
  • $\begingroup$ This might help you: physics.stackexchange.com/questions/172894/… $\endgroup$ May 28, 2020 at 11:40
  • $\begingroup$ This post is helpful, but sadly the poster has found dl in a slightly different way - he has actually differentiated l over theta, while I try to calculate dl using trigonometry $\endgroup$ May 28, 2020 at 12:26
  • $\begingroup$ Why do you assume $dl =r d\theta$? If you want to integrate over $\theta = \pi - \alpha -\phi$, that should be $dl = r d\theta/\sin\theta$. $\endgroup$ May 28, 2020 at 13:37
  • $\begingroup$ @JánLalinský I have just noticed, that I have used phi instead of theta in my image. I assume, that $rd\theta$ because alpha is equal to $d\theta$ so for small angle $dl$ is equal to $rd\theta$. I basically have a triangle of sides $r$, $r$, and $dl$ (the other leg is also $r$ because it's length isn't affected much by $d\theta$. $\endgroup$ May 28, 2020 at 13:46
  • $\begingroup$ It is true that $d\alpha= - d\theta$, but $dl$ isn't $rd\theta$. Expression $rd\theta$ gives length of circle section at distance $r$. This section is a projection of $dl$ in the direction of $r$. $\endgroup$ May 28, 2020 at 13:50

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.