0
$\begingroup$

Say I want to find the total current passing through the following volume, provided by Griffiths' textbook on Electromagnetism. It's not the most useful image, but perhaps imagine the inner curved rectangular prism as a current carrying wire.

enter image description here

If I wanted to find the net current through this volume, I would want to calculate the net flux of the current density through the volume.

$$I_{net} = \oint \mathbf J \cdot d\mathbf a = \int_{S_1} \mathbf J \cdot d\mathbf a + \int_{S_2} \mathbf J \cdot d\mathbf a$$

Where $S_1$ and $S_2$ denote each circular region of this cylindery-shaped volume. The remaining curved surface is perpendicular to $\mathbf J$ constantly, so I needn't consider the flux through it.

Since I am assuming the current is steady, with the area of $S_1 = S_2 = A$, we have

$$\oint \mathbf J \cdot d\mathbf a = -\mathbf JA + \mathbf JA = 0$$

Which implies that the net current is $0$. This seems a bit fishy to me. So, with a steady current, the net current through a volume enclosing a section of the current is $0$?

This would agree with the continuity equation, but it doesn't make physical sense to me. There ought to be current flowing through it, as I don't think I feel safe now sticking my hand inside a cross section of this volume. What's going on here?

$\endgroup$
2
  • $\begingroup$ Can you share some context of the text in the book you cited? Does it really say this is the current "through" the volume? $\endgroup$
    – The Photon
    May 4 '19 at 17:14
  • $\begingroup$ Also, please cite your sources accurately. It doesn't appear that Griffiths has written any book titled Electromagnetism. Are you referring to Introduction to Electrodynamics or some other book? $\endgroup$
    – The Photon
    May 4 '19 at 17:16
2
$\begingroup$

the net current through this volume

Your equation doesn't give the net current through the volume. It gives the net current into the volume.

Since there is equal current flowing in one side of the volume and out the other side, the net inward current is 0.

But the net current through the volume would be given by a different integral (possibly just one of your two surface current terms), and would not be zero.

$\endgroup$
8
  • $\begingroup$ Why is my closed surface integral the net inward current? What would be the current integral? $\endgroup$
    – sangstar
    May 4 '19 at 17:19
  • $\begingroup$ @sangstar, because ${\bf J}\cdot d{\bf a}$ is the element of current inward at each point on the surface of the volume. We actually don't talk about the current through a volume, mostly only about the current through a surface. In this case we might sensibly talk about the current through the volume, but it would be identical to the current through any surface you chose across the path of the current, so why not simplify the problem and talk about the current through one such surface, instead of through the volume, if that's what we wanted to talk about? $\endgroup$
    – The Photon
    May 4 '19 at 17:23
  • $\begingroup$ @sangstar With a closed surface integral, you're effectively subtracting the charge that flows in from the charge that flows out (because the velocity vector is in the same direction as the surface normal vector on one side and in the opposite direction on the other). If charge is not accumulating in the volume, then the amount of charge per second coming into the volume is the same as the amount of charge per second coming out, so the integral (i.e. the difference of these two) is zero. It's only nonzero if charge is accumulating inside the volume, hence, it measures the net inward current. $\endgroup$ May 4 '19 at 17:25
  • $\begingroup$ So essentially, what I'm doing by taking a closed surface integral isn't actually telling me anything particularly useful? $\endgroup$
    – sangstar
    May 4 '19 at 17:29
  • $\begingroup$ @sangstar For an alternative way of seeing this, apply the divergence theorem (aka Gauss's theorem). When you take a closed surface integral of a vector field, that's equivalent to taking the integral of the divergence of that field over the volume. If there are no sources or sinks of charge in the volume, then the divergence should be zero, hence the surface integral is also zero. If, in contrast, charge is accumulating in or disappearing from the volume, then there is a source or a sink of charge in the volume, the current has a nonzero divergence, and the surface integral is nonzero. $\endgroup$ May 4 '19 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.