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I have to calculate the magnetic field along the axis of a ring of radius $R$ on which circulates a current $I$ using the Biot-Savart law. The Biot-Savart law as given in my (really bad) course states

$$\mathbf{B}=\frac{\mu_0}{4\pi}\oint_{C}\frac{\mathbf{I}\times \mathbf{r}}{|\mathbf{r}|^3}\mathrm{d}l $$

where in this case

$$C=\{(\rho, \phi, z): \rho=R, \, z=0,\,\phi\in[0,2\pi[ \}$$

and $\mathbf{I}=I\hat{e}_{\phi}$

we have $\mathbf{r}=\rho\hat{e}_{\rho}+z\hat{e}_z$, thus

$$\hat{e}_\phi\times \mathbf{r}=\hat{e}_\phi\times(\rho\hat{e}_{\rho}+z\hat{e}_z)=-\rho\hat{e}_z+z\hat{e}_\rho$$ and thus

$$\mathbf{B}=\frac{\mu_0}{4\pi}\int_{0}^{2\pi}\left[\frac{-\rho\hat{e}_z+z\hat{e}_\rho}{(\rho^2+z^2)^\frac{3}{2}}\right]_{\rho=R, \,z=0}\mathrm{d}\phi$$

which of course gives something which is constant and wrong. I really don't understand how this formula could give anything which makes sense, since every spatial variable will disappear with the integration. I found other versions of the law around which include notations such as $\vec{dl}\times\vec{r}$ which I just don't understand, I don't know what it means to take the cross product of a differential with something. Do I have a wrong Biot-Savart law? If not, what am I doing wrong? Thank you.

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  • $\begingroup$ I have not looked at your work in detail, but if your problem is that $z$ disappears because $z=0$ on the ring, watch out. There are really two $z$'s, one for the coordinate of the region of integration (the ring), often called $z'$, the other is the coordinate of the observation point. $z'$ gets integrated away, but $z$ does not. In other words, do not set $z=0$. If that's not the issue you are having, please explain what you mean when you say "every spacial variable will disappear". $\endgroup$ – garyp Jun 5 '17 at 14:37
  • $\begingroup$ Also, you will have to be careful with the integral because $\hat{e}_\rho$ depends on $\phi$. $\endgroup$ – garyp Jun 5 '17 at 14:44
  • $\begingroup$ That is my problem, and I see what you're saying, the problem is that then if I don't set $z=0$ the $z\hat{e}_\rho$ part doesn't disappear, then I have a radial component that should not be there. How do I distinguish $z$ and $z'$? And how does $\hat{e}_\rho$ depend on $\phi$? $\endgroup$ – user2723984 Jun 5 '17 at 14:48
  • $\begingroup$ $\hat{e}_\rho$ depends on $\phi$ so you will have to account for that explicitly. Write $\hat{e}_\rho$ explicitly in terms of vectors that do not change with $\phi$, and, of course, $\phi$. The simplest vectors that I can think of are $\hat{x}$ and $\hat{y}$. That is, find $\hat{e}_\rho(\hat{x},\hat{y},\phi)$ $\endgroup$ – garyp Jun 5 '17 at 14:55
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    $\begingroup$ @Chris : I don't think that the limit you are using is right. They should be left just like that. The actual law is something like this : $$\int \frac {\mathbf I \times \mathbf {(r-r')} }{\mathbf{|r-r'|^3}} dl$$ Where $r$ is the point where field is estimated and $r'$ is the curve $\endgroup$ – Icchyamoy Jul 18 '17 at 13:01
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First of all, let's say the current flow in the direction of increasing $\phi$. Now, for the ring, let's write Biot-Savart law in the form

$$ \textbf{B}(\textbf{x}) = \frac{\mu_o i}{4\pi} \oint_C \frac{d\textbf{l}'\,\times\,\textbf{r}}{|\textbf{r}|^3},$$

where $d\textbf{l}'$ is the infinitesimal displacement vector and $\textbf{r} = \textbf{x} - \textbf{x}'$ is the separation vector from the field source (current along the wire's segment $dl'$) at $\textbf{x'}$ to the point $\textbf{x}$ where you want to measure the field. Note I have used $\textbf{i}\,dl'=i\,d\textbf{l}'$ (personally, I prefer to use $i\,d\textbf{l}'$).

Because of the geometry of the problem, cylindrical coordinates are more suitable to handle the calculations. For points at the ring axis, $\textbf{x} = z\, \hat{\textbf{e}}_z$. Also, the location of an arbitrary infinitesimal segment of the wire can be represented by $\textbf{x}'=R\, \hat{\textbf{e}}_\rho'$ (remember $\hat{\textbf{e}}_\rho'$ is function of $\phi'$ and $\phi'$ defines where in the wire this segment is). Therefore $\textbf{r}=z\, \hat{\textbf{e}}_z-R\, \hat{\textbf{e}}_\rho'$ and $|\textbf{r}|^3=(z^2+R^2)^{3/2} $. Finally, $d\textbf{l}'=R\, d\phi'\,\hat{\textbf{e}}_\phi'$. I have included all these primes to remember these are variable related to the source of the field and are integration variables in the Biot-Savart law.

The cross-product now reads

$$d\textbf{l}'\,\times\,\textbf{r} = (z\,R\, \hat{\textbf{e}}_\rho'+R^2\,\hat{\textbf{e}}_z')\,d\phi'. $$

Now Biot-Savart law reads

$$ \textbf{B}(\textbf{x}) = \frac{\mu_o i}{4\pi} \int_0^{2\pi} \frac{z\,R\,\hat{\textbf{e}}_\rho'+R^2\,\hat{\textbf{e}}_z'}{(z^2+R^2)^{3/2}}\,d\phi'.$$

The unit vector $\hat{\textbf{e}}_z'$ is constant, and equal to $\hat{\textbf{e}}_z$ so it goes out the integral. On the other hand, $\hat{\textbf{e}}_\rho'=\cos{\phi'}\hat{\textbf{e}}_x+\sin{\phi'}\hat{\textbf{e}}_y$ is a function of $\phi'$ and must be integrated,

$$ \textbf{B}(\textbf{x}) = \frac{\mu_o i}{4\pi}\frac{z\,R} {(z^2+R^2)^{3/2}}\int_0^{2\pi}(\cos{\phi'}\hat{\textbf{e}}_x+\sin{\phi'}\hat{\textbf{e}}_y)\,d\phi'+\frac{\mu_o i}{4\pi}\frac{R^2}{(z^2+R^2)^{3/2}}\hat{\textbf{e}}_z\int_0^{2\pi} d\phi'.$$

The first integral vanishes because $\int_0^{2\pi}\cos\phi'\,d\phi'=\int_0^{2\pi}\sin\phi'\,d\phi'=0$. The second integral is just $2\pi$. Then you get the magnetic field at a distance $z$ along the ring's axis:

$$ \textbf{B}(\textbf{x}) = \frac{\mu_o i}{2}\frac{R^2}{(z^2+R^2)^{3/2}}\hat{\textbf{e}}_z.$$

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    $\begingroup$ The OP should accept this answer and give the writer credit for the work. On the other hand, we generally frown upon complete answers. $\endgroup$ – garyp Mar 21 '18 at 18:59
  • $\begingroup$ Understood. Sorry for that! $\endgroup$ – andrehgomes Mar 21 '18 at 19:36
  • $\begingroup$ That's a very standard textbook example which is not always presented with sufficiently clear mathematical details, so I thought a clear and precise presentation of the solution would help the OP's understanding. $\endgroup$ – andrehgomes Mar 21 '18 at 19:46
  • $\begingroup$ I agree, and it is a nice presentation THAT THE ORIGINAL POSTER SHOULD ACCEPT. (Sorry to shout, but I don't think he or she heard me the first time. $\endgroup$ – garyp Mar 22 '18 at 1:59
  • $\begingroup$ @garyp sorry, I frankly had missed the notification that this old question of mine had been answered, I accepted the answer. A side note: the user who asked the question doesn't receive a notification when someone comments on an answer, moreover, comments written in all capital letters aren't more readable or more likely to be noticed, just more rude. Andrehgomes, thanks for the answer :) $\endgroup$ – user2723984 Jan 19 '19 at 16:54
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I usually approach these problems less mathematical. Griffiths uses the following figure:From Griffiths Introduction to Electrodynamics

The part $dl'$ yields a piece of the magnetic field $dB$. All the horizontal components cancel so we only have to account for the vertical components. The integral becomes:

$$ B(z) = \frac{\mu_0}{4\pi} I \int \frac{d\textbf{l'}\times \textbf{r}}{r^2} = \frac{\mu_0}{4\pi} I \int \frac{dl'}{r^2} \cos\theta $$

The cross product between dl and I is more clearly visible in the figure. You take the cross product of a small part of I with the vector r. The vector r is a unit vector so it's length is 1. The angle between the 2 vectors is 90 degrees so you only have dl left. The cosine picks out the vertical component. Solving this integral is trivial.

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  • $\begingroup$ Thanks, but what I'm looking for is exactly the mathematical way to approach the problem. I don't see why the fact that the horizontal components cancel can't arise mathematically from this law. And besides these things work in a really restricted number of cases, I want to know how to apply the law in general, because at the moment I don't understand how to do it $\endgroup$ – user2723984 Jun 5 '17 at 14:06

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