Neither Biot-savart nor Ampere Law can solve this problem?

Question

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

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?

Answer 1

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.

Answer 2

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 $a$.