I'm confused about the use of the Ampere's Law and the Biot-Savart Law due the inconvenience of each law.

enter image description here

I want to calculate the magnetic field due to current carrying a circular loop over itself, i.e. not the magnetic field outside the loop but $B$ over the loop. For this, I use the two laws:

1. Ampere's Law

It states that: $$\oint B\cdot dl = \mu_0 I$$

The problem with the Ampere's Law is that $B$ is inside the integral, so in order to solve $B$ I need to use a closed line $L$, such that $B$ that does not depend of $dL$. In that case:

$$\oint B\cdot dl = \mu_0 I$$ $$B \oint dl = \mu_0 I$$ $$B = \frac{\mu_0 I}{L}$$

But, what type of trajectory $L$ should I choose?

2. Biot Savart Law

Let the trajectory: $$c(\theta) = R(\cos\theta\hat i + \sin\theta\hat j)$$ $$dc(\theta) = R(-\sin\theta\hat i + \cos\theta\hat j)d\theta$$

The magnetic field at point $c(t)$ is:

$$ dB = \frac{\mu_0}{4\pi}\frac{Idc\times r}{|r|^3}$$ $$ B(t) = \int_0^{2\pi}\frac{\mu_0}{4\pi}\frac{Idc\times (c(t)-c(\theta))}{|c(t)-c(\theta)|^3}$$ $$ B(t) = \int_0^{2\pi}\frac{\mu_0}{4\pi}\frac{IR(-\sin\theta\hat i + \cos\theta\hat j)d\theta\times R((\cos t-\cos\theta)\hat i+(\sin t-\sin\theta)\hat j)}{|R((\cos t-\cos\theta)\hat i+(\sin t-\sin\theta)\hat j)|^3}$$ $$ B(t) = \int_0^{2\pi}\frac{\mu_0I}{4\pi R}\frac{(-\sin\theta \sin t+\sin^2\theta - \cos\theta \cos t +\cos^2\theta)\hat k}{\sqrt{\cos^2 t-2\cos t\cos \theta+\cos^2\theta+\sin^2 t-2\sin t\sin\theta+\sin^2\theta}^3}d\theta$$ $$ B(t) = \int_0^{2\pi}\frac{\mu_0I}{4\pi R}\frac{1-\cos(t-\theta)}{(2(1-\cos(t-\theta)))^{3/2}}d\theta\hat k$$ $$ B(t) = \frac{\mu_0I}{8\sqrt{2}\pi R}\int_0^{2\pi}\frac{d\theta}{\sqrt{1-\cos(t-\theta)}}\hat k$$

This integral tends to infinity, because in some point $t$ (that is $c(t)$ is one point in the circular loop) tends to $\theta$ and the denominator becomes 0. So, is impossible to calculate the magnetic field over the own spiral.

And I think that the principal reason of this is that in Biot-Savart law the $r$ is in the denominator, so when I try to calculate the magnetic field very close to the current, this $r$ tends to zero and the magnetic field tends to infinity.

If I try this calculation with the formula for volumes ($ B = \int_V \frac{\mu_0}{4\pi}\frac{Jdv\times r}{|r|^3}$) the problem persists due the $r$ is in the denominator and the magnetic field near some point $dv$ will tend to infinity because $r$ tends to zero.

What is the way to do this calculation?

  • $\begingroup$ See formula $(28)$ page $8$ in this paper $\endgroup$
    – Trimok
    Commented Nov 14, 2013 at 19:56
  • $\begingroup$ If you see the formula (14) in my case $k=1$ because $a=r$ and $z=h$. So, if $k=1$ the formula (28) tends to infinity $\endgroup$
    – Julian
    Commented Nov 14, 2013 at 20:31
  • $\begingroup$ In fact, for $z=h$, you have only the second term in $(28)$, and the coefficient of $E$ has both a numerator ($k^2(r+a) - 2a$) and a denominator ($2r(1-k^2)$), which are zero (for $r=a, k=1$), so you need a more detailed analysis of the limit. $\endgroup$
    – Trimok
    Commented Nov 15, 2013 at 10:53
  • $\begingroup$ For information, write a $@user$ if you want that your message goes directly in the user mail (this is not necessary if you answer to a question, and in some other cases), in fact, just type $@$ at the beginning of the comment, and if it is necessary, you will get the list of possible users. $\endgroup$
    – Trimok
    Commented Nov 15, 2013 at 10:58
  • $\begingroup$ @Trimok : the coefficient in the second term in (28) tends to $-1$, because the numerator tends to $k^2(2r)-2r=-2r(1-k^2)$ which is the same denominator $2r(1-k^2)$. So it tends to $-1$. But the first term is also an undefined value 0/0 due this $(z-h)/(1-k^2)$. Resolving this: $(z-h)/(1-\frac{4r^2}{4r^2+(z-h)^2})$$=\frac{4r^2+(z-h)^2}{z-h}$$=\frac{4r^2}{z-h}+(z-h)$... and that tends to $\infty$ for $z=h$. $\endgroup$
    – Julian
    Commented Nov 15, 2013 at 12:10

2 Answers 2


Using Biot Savart or Ampere's Law you will come to the same problem $B$ is not defined on the ring.

This is the same problem that trying to find the Electric field $E$ of a puntual charge just in the point where the charge is placed $1/r²$ becomes $\infty$...

You need to use the formula for volumes but using the superficial current density $J$ and integrating on a torus, then the magnetic field is well defined. Notice that:

$\frac{\mu_0}{4\pi}\int_V \frac{Jdv\times r}{|r|^3}=\frac{\mu_0}{4\pi}\int_V \frac{4\pi r² d\Omega dr J\times r}{|r|^3}=\mu_0\int_V \frac{J\times u_r r³ d\Omega dr }{|r|^3}=\mu_0\int_V J\times u_r d\Omega dr$

Thus even if $r \to 0$ the $\infty$ does not appear.

The problem is that solving volume integrals is more complicated that using a line... but in this case I cannot find a better option.

  • 1
    $\begingroup$ the $r$ that you mention have different meanings. Because the $r$ from ...$Jdv\times r$... is the distance from all points of the torus to one single point in the torus, and the $r$ from $4\pi r^2$ comes from the geometry of the torus, but they are two different $r$'s. $\endgroup$
    – Julian
    Commented Nov 26, 2013 at 15:07

Well... I must be careful with the notation, first start describing the torus that is horizontal and such that the origin lies inside it:

$x_t=\mathrm{sin}\left( \alpha_1\right) \,\left( R+\mathrm{cos}\left( \beta_1\right) \,u\right) $

$y_t= \mathrm{cos}\left( \alpha_1\right) \,\left( R+\mathrm{cos}\left( \beta_1\right) \,u\right)$

$z_t = \mathrm{sin}\left( \beta_1\right) \,u$

For $-\pi/2<\alpha_1\le \pi/2$, $-\pi/2<\beta_1\le \pi/2$ and $0 \le u \le a$ where $R$ is the radius of the torus and $a$ its width. Then a point lie inside the torus if:

$(\sqrt{(R+x)^2+y^2}-R)^2+z^2 \le a^2$

And the current density $||J||=\frac{I}{\pi a^2}$ may point to the direction where $\alpha_1$ grows, that is $u_{\alpha}=[\frac{dx_t}{d \alpha},\frac{dy_t}{d\,\alpha},\frac{dz_t}{d\,\alpha}]=[-y_t,x_t+R,0]$ so finally

$J=||J|| \frac{u_{\alpha}} {||u_{\alpha}||}=\\ =\begin{cases} [-\frac{y\,I}{2\,\pi \,{a}^{2}\,\sqrt{{\left( R+x\right) }^{2}+{y}^{2}}},\frac{I\,\left( R+x\right) }{2\,\pi \,{a}^{2}\,\sqrt{{\left( R+x\right) }^{2}+{y}^{2}}},0] &\mbox{if } \sqrt{(R+x)^2+y^2}-R)^2+z^2 \le a^2 \\ 0 & Otherwise. \end{cases}$

Suppose that you want to compute the magnetic field on the origin, in order to avoid the problem with $1/r^2$ we may use spherical coordinates:

$x=\mathrm{cos}\left( \alpha\right) \,\mathrm{sin}\left( \beta\right) \,r$

$y=\mathrm{cos}\left( \alpha\right) \,\mathrm{cos}\left( beta\right) \,r$

$z=\mathrm{sin}\left( \alpha\right) \,r$

Then you must rewrite the above expressions in terms of the new coordinates and apply Biot Savart law... In addition if you consider the symmetry of the problem you know that $B_x=B_y=0$ and before some lengthy computations the expression for $B_z$ is as follows:

$B_z=\mu_0I\int\limits_0^{2R+a}\int\limits_0^\pi\int\limits_0^{2\pi} {G \frac{\sqrt{R+\left( \mathrm{sin}\left( \beta\right) \,\mathrm{cos}\left( \alpha\right) -\mathrm{cos}\left( \beta\right) \,\mathrm{cos}\left( \alpha\right) \right) \,r}\,\left( \mathrm{sin}\left( \beta\right) \,{\mathrm{cos}\left( \alpha\right) }^{2}\,R+{\mathrm{cos}\left( \alpha\right) }^{3}\,r\right) }{8\,\pi \,{a}^{2}\,\pi\,R+\left( 8\,\pi \,\mathrm{sin}\left( \beta\right) \,{a}^{2}\,\mathrm{cos}\left( \alpha\right) -8\,\pi \,\mathrm{cos}\left( \beta\right) \,{a}^{2}\,\mathrm{cos}\left( \alpha\right) \right) \,\pi\,r} dr d\alpha d\beta}$

Where $G(\alpha,\beta,r)=$

$\begin{cases} 1 &\mbox{if } {\left( \sqrt{{\left( R+\mathrm{cos}\left( \alpha\right) \,\mathrm{sin}\left( \beta\right) \,r\right) }^{2}+{\mathrm{cos}\left( \alpha\right) }^{2}\,{\mathrm{cos}\left( \beta\right) }^{2}\,{r}^{2}}-R\right) }^{2}+{\mathrm{sin}\left( \alpha\right) }^{2}\,{r}^{2}<{a}^{2} \\ 0 & Otherwise. \end{cases}$

Notice that the expression is awful but is well defined, the infinite vanishes (as I told you in the last comment).

As far as I know there is no way to solve this integral analytically, but you can compute a solution numerically for any value of, $R$ and $a$.

  • $\begingroup$ Sorry, but I found a mistake on the computations, the correct expresion is: $\endgroup$
    – Dictino
    Commented Nov 27, 2013 at 17:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.