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?


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.

  • $\begingroup$ Thank you Philip; the problem does not go away by just assuming J instead of I, as, even there, at some point you will divide by a zero distance between the observation and source points. You're right that there are quite a few approximations out there; however, they never satisfied me... Unfortunately, I think there is really no other answer than that in this case, the loop indeed has infinite inductance, and therefore infinite magnetic flux or the other way around. $\endgroup$ – HD Lang Aug 5 '17 at 22:03
  • $\begingroup$ If you insist on the wire having zero thickness, the difficulty does not go away by using $J$ instead of $I$, but for a wire of finite thickness it surely does. After all, the measured inductance of a coil of wire is finite! But I like your use of the semicolon. $\endgroup$ – Philip Wood Aug 6 '17 at 13:59

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.

  • $\begingroup$ I do calculate B at every point; that's exactly why there's the third integral, to go $dB_z\rightarrow B_z$ in the same step, with the $dB_z$ as given above, along the current, with $d\phi'$. I added two steps, to clarify. $\endgroup$ – HD Lang Aug 4 '17 at 18:43

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