Does infinite magnetic energy $\Leftrightarrow$ non-zero out-of-plane current flux?

Question

Consider a static current density $j_z(x,y)$ flowing out of an infinite 2D plane with $\mu=1$. $j_z(x,y)$ should be physical such as $0<j_z(x,y)<M$ and $\lim_{\sqrt{x^2+y^2} \rightarrow \infty}j_z(x,y) = 0$. The total magnetic energy can thus be calculated.

There are cases of $j_z(x,y)$ where the energy is infinite, such as the single line current $j_z(x,y)=\delta(x,y)$.

However, there are also cases for finite energy, such as the cross-section of a coaxial cable. The finite energy is indicated by the finite effective inductance and finite current, based on transmission-line theory.

Years ago I had the result that non-zero current flux leads to infinite energy by naive Fourier analysis. (See my akward reasoning in Does direct current exist in an infinite straight thin wire? ) The current flux $\int dS~j_z(x,y)$ is the $k=0$ Fourier component of the current density.

But I am not sure if the converse proposition is correct, i.e., \begin{equation} \text{Infinite energy}\Rightarrow \text{non-zero flux}? \end{equation}

If not, what condition should I add to $j_z(x,y)$ to let it satisfied?

Answer 1

Your current density is of the form: $$ j=j_z(x,y)e_z $$ so the magnetic field is of the form (translation along $z$ and reflection perpendicular to $z$): $$ B=B_x(x,y)e_x+B_y(x,y)e_y $$ and your problem is essentially 2D: $$ \begin{align} \partial_xB_x+\partial_yB_y &= 0\\ \partial_xB_y-\partial_yB_x &= \mu_0 j_z \end{align} $$ After a Fourier transform: $$ \begin{align} ik_x\hat B_x+ik_y \hat B_y &= 0\\ ik_x \hat B_y-ik_y \hat B_x &= \mu_0 \hat j_z \end{align} $$ which you can solve: $$ \begin{align} \hat B_x &= \mu_0 \frac{ik_x}{k_x^2+k_y^2}\hat j_z \\ \hat B_y &= \mu_0 \frac{-ik_y}{k_x^2+k_y^2}\hat j_z \end{align} $$ The energy per unit length along $z$ is: $$ \begin{align} E&=\frac{1}{2\mu_0}\int B^2 dxdy \\ &= \frac{1}{2\mu_0}\int |\hat B|^2 \frac{dk_xdk_y}{(2\pi)^2} \\ &= \frac{\mu_0}{2}\int \frac{|\hat j_z|^2}{k_x^2+k_y^2}\frac{dk_xdk_y}{(2\pi)^2} \\ \end{align} $$

Using: $$\hat j_z(0)=\int j_z dxdy$$ I recover your result due to the diverging integral at $0$.

However, the converse is not true. The integral could blowup for different reasons.

For example, you could have a pole of order at least $1$ for $\hat j_z$ at a non zero $k$ (from the reality of $j_z$, you would get another pole at the opposite wave number).

Another problem could be the slow decay of $\hat j_z$ as $k\to\infty$. In real space, this corresponds to an irregular current distribution. A simple example would be two ideal, parallel wires of opposite current $I$ at relative distance $2L$: $$ j_z=I\delta(x-L)\delta(y)-I\delta(x+L)\delta(y) $$ which has a logarithmically diverging energy.

One way to make the converse true would be to impose additional conditions on $j_z$. For example, you can add that it is $L^1$ ($\int dxdy|j_z|<+\infty$) for the poles and that it is $\mathcal C^1$ for the decay at infinity. Note that your original assumptions that $j_z$ is in $L^\infty$ and goes to $0$ at infinity are not necessary.

Hope this helps.

Answer 2

Assuming symmetry of infinite cylinder, if total non-zero current density is localized to insides of some circle, beyond which current density vanishes, then magnetic field outside that circle behaves, at infinity, as $1/r$. This implies magnetic field energy density there is proportional to $1/r^2$ and this means each cylinder of thickness $dr$ contributes amount proportional to $dr/r$. Integral of this function into $r\to \infty$ is positive infinity, so total magnetic energy per unit length is positive infinity.

In practice no wire is infinitely long, so this result is no problem in practice. Magnetic energy of finite-sized circuit is finite. The actual difficulty then is dealing with such finite-sized systems that do not have symmetry of infinite cylinder.