Magnetic flux through current loop

Question

I am trying to calculate the total magnetic flux through the surface of a current loop. I feel that this flux should be finite and nonzero -- so far any attempt in calculating failed. Can this really not be done?

For example: We know \begin{equation}d\textbf{B}=\dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\textbf{l}\times\hat{\textbf{R}}}{|\textbf{R}|^2}\end{equation} where $d\textbf{l}=\hat{\boldsymbol{\phi}}\cdot a\,d\phi'$, for a loop of radius $a$ centered around the origin and lying in the $z=0$ plane in cylindrical coordinates. Hats denote unit vectors. The distance vector is $\textbf{R}=\textbf{r}-\textbf{r}'$ and $|\textbf{R}|=\sqrt{r^2+a^2-2ar\cos(\phi-\phi')}$.

Thus, the magnetic flux density is only $z$-directed: $$dB_z=\dfrac{\mu_0 I}{4\pi}\,\dfrac{a}{|\textbf{R}|^2}\,d\phi'$$

The total flux is simply the surface integral, \begin{multline}\Phi=\int\hspace{-0.5em}\int\textbf{B}\cdot d\textbf{s}=\int\hspace{-0.5em}\int B_z\,r\,d\phi\,dr=\int\hspace{-0.5em}\int\hspace{-0.5em}\int dB_z\,r\,d\phi\,dr\\=\dfrac{\mu_0 Ia}{4\pi}\int_0^a\hspace{-0.5em}\int_0^{2\pi}\hspace{-0.5em}\int_0^{2\pi}\dfrac{r}{r^2+a^2-2ar\cos(\phi-\phi')}\,d\phi'\,d\phi\,dr\end{multline}

However, this does not converge, as far as I can tell. Similar issues arise with formulations via magnetic vector potential or current density $\textbf{J}$ instead of current $I$.

Via inductance, we know that $\Phi=LI$ and that in this theoretical example, the inductance is purely a geometric problem. However... $L\rightarrow\infty$ doesn't make sense to me.

Can anybody help me out and point me into the right direction here?

Answer 1

The trouble arises, I believe, because you're considering the field to be due to a current in a wire of zero thickness, so the flux density approaches infinity as you approach the wire, and this makes the flux integral blow up. If you consider current spread over a finite cross-sectional area of wire this problem goes away. There are other mathematical difficulties, of course, but they can be handled by approximation methods, and you'll find formulae for flux due to a circular loop on the internet.

Answer 2

Your formula is wrong, the flux is not defined like that. The magnetic flux is defined as

$\Phi=\iint \vec{B}\cdot d\vec{S}$

or equivalently $\Phi=\iint \vec{B}\cdot \hat{n}\ dS$

Where $B=B(\vec{r})$ is the magnetic field, and it can depend on the point (and also on the time). By the way, that's the deffinition of "flux" for any vector field.

So what you're doing wrong is saying that $\Phi=\iint dB\cdot dS$. A surface integral cannot have three differentials (as $dS$ are really two differentials, as you wrote above).

You have to calculate $B$ in every point first, and then compute the integral.

The formula of $dB$ is used to calculate the small field created by every current-element. You have to integrate along all the current to get the total $B$ in that point. That's a torture unless you can solve it for a generic point. Then you've got $B(r)$ and you can carry on.

Answer 3

I don't know if you are still interested, but here is your answer. The field near a wire increases with $1/r$. The energy stored per unit volume increases with the square of field, hence $1/r^2$. The volume increases with $r$, so the total energy is proportional to $1/r$. As $r$ approaches zero ... Seems totally counterintuitive, but there are a lot of situations which seem "physical" that are singular. For example, the stress at a point load.

To carry this further, note that flux = inductance times current and look up the formula for the inductance of a round wire. The people who derived these formulas understood this "paradox". You will find that as the diameter tends to zero, the inductance increases without limit (logarithmically), the integral of $1/r$.