# Open current loops in metal conductor

Hello everyone,

I have this question, "can Ac current flow in an open circuit ?", I give the example of a coil carrying an AC current placed above a conducting surface, since we have an AC current, we will induce a current in the metal coductor.

I know there is a current inside the metal, because many experiments proved so, however all what I see is an open current loop in the metal conductor, unless I am wrong, the last time I studied this, current always flows in closed loops, Can someone clarify this issue

• Does the current in the infinite wire in your diagram travel in a loop? – probably_someone Feb 14 '18 at 17:01

Assumptions:

I am assuming that the wire in your question is infinite in length, pointing along the $z$ direction, and that your conductor is finite in the $z$ direction; with boundaries at $z=0$ and $z=a$. (See diagram below).

The fact that current cannot flow in open-circuited conductors is only valid for stationary currents, or cases where the current can be approximated as stationary such as the quasi-static approximation.

First, let's see why current cannot flow in open-circuited conductors for stationary currents.

By the continuity equation (i.e. local conservation of charge), we have: $$\oint_{\partial V} d\boldsymbol{a}.\mathbf J(\mathbf x,t) = - \frac{ d }{d t}\int_V d^3\mathbf x \ \rho(\mathbf x,t)$$ For any (simply connected) region of space $V$. Now for a stationary current , by definition, the right hand side is zero. Which means: $$\oint_{\partial V} d\boldsymbol{a}.\mathbf J(\mathbf x,t) = 0$$ This is nothing but the famous current law of Kirchhoff, which says that the total outgoing current in any region of space is zero (i.e. in going current = out going current). The fact that no current can flow in an open circuit is a direct consequence of this law. Look at the diagram below; because for the open circuit, the out going current from $S_2$ is zero (there is no circuit there), the in going current on the wire ($S_1$) is zero as well.

As you can see, this law is not necessarily true for non-stationary currents. The assumption that the right hand side of the first equation is zero is crucial to this result. But what does this mean physically?
You can easily convince yourself that the right hand side of our first equation is nothing but the rate at which the total charge inside the volume $V$ is decreasing with time. This gives an intuitive picture of what a stationary current (for which this is zero) actually is. A stationary current is a current in which the charge does not experience any compression or expansion; each charge roughly takes the position of the charge before, so that even though there is a current, the charges don't compress or expand. This is not what happens in your particular question. In your question, the charges are being pushed in the $z$ direction, oscillating back and forth as the current of the above wire changes direction. Because the conductor is finite in the $x$ direction, the charges need to compress near the boundaries of the conductor at $z=0$ and $z=a$. This automatically breaks the stationary current assumption; meaning that Kirchhoff's current law does not hold, and there can indeed be a non-zero current in the open-circuited conductor. Notes:

1- This problem is very similar to the problem of current conservation in capacitors (even though a simple capacitor is just to parallel plates with nothing connecting them, how can there be current going through?). The answer to that problem is exactly the same as yours; although it is usually dealt with using displacement currents; which is basically the same approach because a time changing charge density (and hence local compression and expansion of charges) gives rise to a non-zero displacement current. You can see this by taking the divergence of the displacement current and using Gauss's law: $$\boldsymbol {\nabla}.\mathbf J_d(\mathbf x,t) = \boldsymbol {\nabla}.\bigg(\mu_0 \epsilon_0 \frac{\partial {\mathbf E(\mathbf x,t)}}{\partial t} \bigg)=\mu_0 \epsilon_0 \frac{\partial }{\partial t}(\boldsymbol {\nabla}.{\mathbf E(\mathbf x,t)})$$ $$\boldsymbol {\nabla}.\mathbf J_d(\mathbf x,t) =\mu_0 \frac{\partial \rho(\mathbf x,t) }{\partial t}$$ You can see more of this in this link.

2- You can easily argue that the induced electric field due to the time changing magnetic field points along the $z$ direction. We can approximately assume that the magnetic field points in the azimuthal direction (as you have shown in your diagram), using phasor notation, this is $\mathbf B = B_0(\rho,\phi) \boldsymbol {\hat \phi}$. I am assuming that $\mathbf B$ does not depend on $z$ as an approximation by crudely assuming there is still some translational symmetry in the $z$ direction, despite the conductor being finite. By Ampere's law, $\boldsymbol {\nabla} \times \mathbf B = \mu_0 \sigma \mathbf E + i\omega \mu_0 \epsilon_0 \mathbf E$ inside the conductor. By using the expression for the curl in cylindrical coordinates for $\mathbf B = B_0(\rho,\phi) \boldsymbol {\hat \phi}$ you can easily see that the only component left out is the $z$ component; meaning that $\mathbf E$ also points in the $z$ direction by $\mathbf E = \frac 1{\mu_0 \sigma+i\omega \mu_0 \epsilon_0}\boldsymbol {\nabla} \times \mathbf B$.
I'm mentioning this because this rules out the possibility of (stationary) current loops forming in the conductor, as $\mathbf J=\sigma \mathbf E \sim \hat {\mathbf {z}}$. The current roughly points in the same direction as the wire above, and has to be associated with charge compression near the boundaries.

3- The fact that Kirchhoff's current law breaks down for non-stationary currents is very important in high frequency circuits (e.g. microwave circuits). In these devices, the charges get compressed and expanded along the conductors because the fields oscillate so rapidly with time that the charges "can't keep up with the fields". As a result, the current in a high-frequency transmission line can vary as a function of position (contrary to Kirchhoff's law).

• Thank you very much for your detailed answer, I may get back to you on other matters. – Edwardo Newagte Feb 15 '18 at 10:39