1
$\begingroup$

I'm stuck on one really simple example, I can't figure out what's happening to energy here...

(This is not homework)

Let's consider an uncharged electric cable, we'll model it by an infinite cylinder on the axis $(Oz)$ with radius $a$ and conductivity $\gamma$ with uniform, constant current $I$, and we'll obviously use cylindrical coordinates $(\vec{e_r},\vec{e_\theta},\vec{e_z})$.

If I haven't made any mistake, we should have the electric and magnetic fields $\vec{E},\vec{B}$ as follows :

$\vec{E}=\begin{cases}\frac{1}{\gamma}\frac{I}{\pi a^2}\vec{e_z}&r<a\\\vec{0}& r>a\end{cases}$, $\vec{B}=\begin{cases}\frac{\mu_0I}{2\pi}\frac{r}{a^2}\vec{e_\theta}&r\le a\\\frac{\mu_0I}{2\pi}\frac{1}{r}\vec{e_\theta}&r\ge a\end{cases}$

Thus the poynting vector $\vec{\Pi}=\begin{cases}\frac{-r}{2\pi^2\gamma a^4}I^2\vec{e_r}&r< a\\\vec{0}&r> a\end{cases}$

Hence, if we consider an lateral surface $(\mathcal{S})$ orientated towards the inside and a height $h$ of cable, $$\iint_{(\mathcal{S})}\vec{\Pi}\cdot\vec{dS}=\begin{cases}\frac{h}{\pi\gamma a^2}I^2&r< a\\\vec{0}&r> a\end{cases}$$.

That expression is most puzzling. The energy comes from the side, however there is no energy outside.... where does the energy come from ? Something weird is going on here.

At first I thought it was because we had made the hypothesis of an infinite cable, but it doesn't seem to be related at all.

To allow the energy flux, there would need to be charges inside the cable, which is initially not charged. Thus I thought there might be some effect like the Hall effect with local charges appearing, but once again I do not see why that would be.

$\endgroup$
  • $\begingroup$ I believe you have some typos. Is $\vec{E}$ in the $r$ direction or the $z$ direction. The Poynting vector must be perpendicular to $\vec{E}$ and $\vec{B}$. $\endgroup$ – Bill N Feb 5 '15 at 19:15
2
$\begingroup$

There is a mistake in your working. The surface integral of the Poynting vector $$ \int \vec{\Pi} \cdot d\vec{S} = \Pi\ 2\pi a h = \frac{a I^2}{2\pi^2 \gamma a^4} 2\pi a h = \frac{I^2 h}{\pi \gamma a^2}$$

As pwf correctly points out, this power matches the power dissipated as Ohmic heating, $VI$, where $$V I = E h I = \frac{I^2 h}{\pi \gamma a^2}$$

This power does not appear to me to come into the wire from outside The Poynting vector immediately outside the wire is zero, as is the flux into the wire from outside.

You also say there is "no energy outside... [the wire]". This is not relevant because there is no energy flux into the wire. It is also not true, because the energy density outside the wire is given by $B^2/2\mu_0$.

I agree that it requires some thought to work out physically why the Poynting vector must be pointed inwards, but there is no conservation of energy problem. The energy dissipated per unit volume is a constant; as we consider cylindrical shells moving out from the centre, each shell has a volume proportional to the radius $dV = 2\pi rh\ dr$. Thus as we move into the wire from the outside, there needs to be a gradient in the flux deposited by the Poynting vector, such that more energy needs to be deposited in the outer shells. This tells you that the radial component of the Poynting vector must depend linearly on $r$. It is inwards because the Poynting theorem tells you that the divergence of the Poynting vector in cases where the energy density of the fields is constant (true for static fields) will be given by minus $\vec{J} \cdot \vec{E}$, so will always be negative for static fields in a standard conductor.

$\endgroup$
1
$\begingroup$

This is a great question! I'm not sure where the power is coming from, but I do know that it corresponds to the power flowing out due to Ohmic heating:

Note that $R = \frac{h}{\gamma \pi a^2}$ and $V = IR$, so $E_z = \frac{V}{h}$ (not surprising).

Then $\Pi_r|_a = \frac{V I}{2\pi a h}$, and $\oint\Pi\cdot dS = \Pi_r|_a 2\pi a h = VI$, the ohmic dissipated power. (I think the 2 in your last denominator is wrong.)

So the power influx you're calculating is exactly the electromagnetic power being delivered to the wire that turns up as Joule heating. As for why it appears to be coming in from the sides, I'm not sure.

The fact that there is an inward stress may be a manifestation of the "pinch effect" - the current will self-attract, causing it to want to compress towards the axis.

$\endgroup$
0
$\begingroup$

Firstly: when the current flows in the positive z-direction and the magnetic field obeys the usual right hand rule then the Poynting vector points into the wire, so there is a minus sign missing in your expression:

$$\vec{\Pi}=\begin{cases}\frac{r}{2\pi^2\gamma a^4}I^2(-\vec{e_r})&r< a\\\vec{0}&r> a\end{cases}$$

Secondly: There is actually a paragraph about this in the feynman-lectures: search for figure 27–5. To make this some kind of answer I will paraphrase his argument:

The Poynting vector tells us the flow of energy of the electromagnetic field. Now, when you push electrons down some wire you obviously need energy from other charges moving somewhere. Since this "somewhere" is outside of the wire the Poynting vector is pointing from outside into the wire, this way the wire obtains the energy needed from the outside.

This also makes sense with your result you got from the flux from a surface oriented inwards: even with the minus in the Poynting vector ($\propto -\vec e_r$) and the minus sign from the surface oriented inwards (also $\propto -\vec e_r$) you get a positive result. Positive in this context means that there is a flux into the surface.

$\endgroup$
  • $\begingroup$ Sorry, I forgot the "-" sign indeed $\endgroup$ – Hippalectryon Feb 5 '15 at 20:23
  • $\begingroup$ I understand all you say there, but this isn't my issue here. My problem is : how can there be a flux from the outside to the inside, if there is no energy ($\iint_{(\mathcal{S})}\vec{\Pi}\cdot\vec{dS}$) outside ? $\endgroup$ – Hippalectryon Feb 5 '15 at 20:26
  • $\begingroup$ Yeah this is a good question, I obviously didn't read the question carefully ^^ ... but after thinking a bit, this equations says that there is flux into the wire and none into the "outside". Although there is no flux into the "outside", this doesn't mean that there is no energy out there. It just says that the system "wire" consumes energy, so the flux through the surface is not zero while no energy goes into the "outside", so the second term ($r>a$) is zero. That's the best I can come up with right now. $\endgroup$ – user42076 Feb 5 '15 at 20:40
-2
$\begingroup$

The Surface integral goes over the surface element of cylinder's lateral Surface, i.e. $d \vec S = r d \phi dz \vec e_r$. Then, you evaluate the integral at $r = a = const.$; you set $r = a$ for the Integrand. Integration over $z \in [0,h]$ and $\phi \in [0,\pi]$ gives the result.

$\endgroup$
  • $\begingroup$ I already integrated over the lateral surface. That's exacty what I do above. $\iint_{(\mathcal{S})}\vec{\Pi}\cdot\vec{dS}=\Pi dS=\Pi\pi a^2h$ $\endgroup$ – Hippalectryon Feb 5 '15 at 19:36
  • $\begingroup$ $\pi a^2 h$ is volume. $\endgroup$ – Rob Jeffries Feb 5 '15 at 22:59

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.