Help with electromagnetics: the Poynting Vector and Ampere contradiction

[thatsmessedup]

According to equation (2.33) divergence of the Poynting vector or the outflow of electromagnetic power is equal to the stored magnetic field, stored electric field and ohmic losses.

My contradiction is the following:
Inside a steady state DC current carrying wire, there will presumably be a uniform e-field (E) in the direction of current, and also a steady current density (J) in the direction of current. Therefore equation (2.33) should not be zero because the term J dot E will not be zero.

Now, according to ampere law, in the center of a conductor, the magnetic field should be zero.

Therefore, the Poynting vector, equation (2.36), at the center of the conductor will be zero and the divergence of the Poynting vector will also be zero. In conclusion, at the center of the wire, using equation (2.33) the left side equals to zero (del(ExH)=0), but the right side equal to non zero (J dot E = non zero). What am I missing? Thank you for your help!

 

[Frabjous]

Why is the divergence zero? E×H points in the radial direction and H is proportional to r.

 

[rude man]

Set up cylindrical coordinates.
$\bf E = ?$
$\bf H = ?$
$\bf P = \bf E \times \bf H = ?$
Then get $\bf P$ and $\nabla \cdot \bf P$

 

[thatsmessedup]

Why is the divergence zero? E×H points in the radial direction and H is proportional to r.

At the center of the wire, the H field is 0 and so ExH is zero, hence the divergence is also zero.

 

[Frabjous]

H field is 0 and so ExH is zero, hence the divergence is also zero.

This is incorrect. The divergence is a measure of spatial change. H is non-zero for 0⁺ and consequently has a non-zero derivative.

 

[phyzguy]

At the center of the wire, the H field is 0 and so ExH is zero, hence the divergence is also zero.

Not true. Just because a vector field is zero at a point does not mean that its divergence is zero at that point. Think of a 1D example like y=x^2 - 4. y is zero at x=2, but dy/dx is not zero at x=2. Similarly for a vector field. In this case you have a vector field in cylindrical coordinates that looks like $\lambda r e_r$, where λ is some constant. If you take the divergence, you will find it is non-zero.

 

[thatsmessedup]

I think I'm rusty with my math skills. It all makes sense that there should be divergence in the center of the conductor as this is where all the Poynting vectors point. I need to whip out my old notes.

I am assuming that J dot E is uniform in the cross sectional area of the wire. So, if I get the math right and find the divergence of ExH at different points on the cross sectional area, should I expect all the values to be the same? I mean, this seems to be the obvious conclusion.

 

[rude man]

I think I'm rusty with my math skills. It all makes sense that there should be divergence in the center of the conductor as this is where all the Poynting vectors point. I need to whip out my old notes.

I am assuming that J dot E is uniform in the cross sectional area of the wire. So, if I get the math right and find the divergence of ExH at different points on the cross sectional area, should I expect all the values to be the same? I mean, this seems to be the obvious conclusion.

$\bf j \cdot \bf E$ is uniform everywhere. Ohm's law applied to electric fields is $\bf j = \sigma \bf E$ and applies at every point inside the wire. $\sigma$ is conductivity.

Yes, $\nabla \cdot \bf P$ is uniform across the cross-sectional surface, in fact everywhere in the wire if it's long. As you can determine and calculate per my post 3.

 

[vanhees71]

The problem is that you didn't solve the magnetostatic equations for the coaxial cable completely. If you do so, you'll find that everything is consistent with energy conservation and Poynting's theorem. For a complete thorough discussion see Sommerfeld, Lectures on Theoretical Physics, vol. 3.

 

[thatsmessedup]

Set up cylindrical coordinates.
$\bf E = ?$
$\bf H = ?$
$\bf P = \bf E \times \bf H = ?$
Then get $\bf P$ and $\nabla \cdot \bf P$

Lol... Its been a few years since I've needed to do cross products on cylindrical coordinates. I wish I could remember how to do them. My EE notes look more like hieroglyphs at this point.

 

[rude man]

Lol... Its been a few years since I've needed to do cross products on cylindrical coordinates. I wish I could remember how to do them. My EE notes look more like hieroglyphs at this point.

Welcome to the club! :-)
Given vectors in cyl. coordinates

$\bf A = a_1 \hat {\bf r } + a_2 \hat {\bf \theta} + a_3 \hat{\bf k}$
and $\bf B = b_1 \hat{\bf r} + b_2 ...$ etc.
form the determinant

1st row : $~~ \hat {\bf r} ~~ \hat {\bf \theta} ~~ \hat {\bf k}$
2nd row: $~~ a_1 ~~ a_2 ~~ a_3$
3rd row: $~~ b_1 ~~ b_2 ~~ b_3$

then determinant = $\bf A \times \bf B$.
(I tried to use the LaTex determinant but failed).

 

[phyzguy]

I think you're overcomplicating it. E points down the wire with constant magnitude, so $E = E_0 e_z$. B wraps around the wire with a magnitude proportional to r, so $B = B_0 r e_{\phi}$. Just apply the right hand rule, and ExB is perpendicular to both of them, and so must be $E \times B = E_0 B_0 r e_r$. Then $\nabla \centerdot (E \times B) = \frac{1}{r} \frac{\partial}{\partial r}(r (E \times B)_r) = 2 E_0 B_0$

 

[rude man]

I think you're overcomplicating it. E points down the wire with constant magnitude, so $E = E_0 e_z$. B wraps around the wire with a magnitude proportional to r, so $B = B_0 r e_{\phi}$. Just apply the right hand rule, and ExB is perpendicular to both of them, and so must be $E \times B = E_0 B_0 r e_r$. Then $\nabla \centerdot (E \times B) = \frac{1}{r} \frac{\partial}{\partial r}(r (E \times B)_r) = 2 E_0 B_0$

Make that ## - 2E_0B_0. and you have a deal. The determinant knows the difference.

 

[thatsmessedup]

I think you're overcomplicating it. E points down the wire with constant magnitude, so $E = E_0 e_z$. B wraps around the wire with a magnitude proportional to r, so $B = B_0 r e_{\phi}$. Just apply the right hand rule, and ExB is perpendicular to both of them, and so must be $E \times B = E_0 B_0 r e_r$. Then $\nabla \centerdot (E \times B) = \frac{1}{r} \frac{\partial}{\partial r}(r (E \times B)_r) = 2 E_0 B_0$

Right on! Thanks for the math help and I guess its not so bad after all.

 

[thatsmessedup]

Make that ## - 2E_0B_0. and you have a deal. The determinant knows the difference.

Yep, the unit vector should be negative since the cross product points inward. It also makes sense because a resistive wire should have an inflow of electromagnetic power and the divergence of a vector represents outflow. Thanks for your help!