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?
Best Answer
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.