Editing Poynting vector (section)

==Example: Power flow in a coaxial cable==
Although problems in electromagnetics with arbitrary geometries are notoriously difficult to solve, we can find a relatively simple solution in the case of power transmission through a section of [[coaxial cable]] analyzed in cylindrical coordinates as depicted in the accompanying diagram. We can take advantage of the model's symmetry: no dependence on θ (circular symmetry) nor on ''Z'' (position along the cable). The model (and solution) can be considered simply as a DC circuit with no time dependence, but the following solution applies equally well to the transmission of radio frequency power, as long as we are considering an instant of time (during which the voltage and current don't change), and over a sufficiently short segment of cable (much smaller than a wavelength, so that these quantities are not dependent on ''Z'').

The coaxial cable is specified as having an inner [[Electrical conductor|conductor]] of radius ''R''<sub>1</sub> and an outer conductor whose inner radius is ''R''<sub>2</sub> (its thickness beyond ''R''<sub>2</sub> doesn't affect the following analysis). In between ''R''<sub>1</sub> and ''R''<sub>2</sub> the cable contains an ideal [[dielectric]] material of [[relative permittivity]] ''ε''<sub>r</sub> and we assume conductors that are non-magnetic (so ''μ'' = [[vacuum permeability|''μ''<sub>0</sub>]]) and lossless (perfect conductors), all of which are good approximations to real-world coaxial cable in typical situations.

[[File:CoaxPoyntingVector.png|center|600px|thumb|Illustration of electromagnetic power flow inside a [[coaxial cable]] according to the <span style="color:green">Poynting vector '''S'''</span>, calculated using the <span style="color:red">electric field '''E'''</span> (due to the voltage ''V'') and the <span style="color:blue">magnetic field '''H'''</span> (due to current I).]]
[[File:Coax-poynting.png|thumb|right|350px|[[Direct current|DC]] power transmission through a [[coaxial cable]] showing relative strength of electric (<math>E_r</math>) and magnetic (<math>H_\theta</math>) fields and resulting Poynting vector (<math>S_z = E_r \cdot H_\theta</math>) at a radius ''r'' from the center of the coaxial cable. The broken magenta line shows the ''cumulative'' power transmission ''within'' radius ''r'', half of which flows inside the [[geometric mean]] of ''R''<sub>1</sub> and ''R''<sub>2</sub>.]]
The center conductor is held at voltage ''V'' and draws a current ''I'' toward the right, so we expect a total power flow of ''P'' = ''V'' · ''I'' according to basic [[Electric power|laws of electricity]]. By evaluating the Poynting vector, however, we are able to identify the profile of power flow in terms of the electric and magnetic fields inside the coaxial cable. The electric fields are of course zero inside of each conductor, but in between the conductors (<math>R_1 < r < R_2</math>) symmetry dictates that they are strictly in the radial direction and it can be shown (using [[Gauss's law]]) that they must obey the following form:
<math display=block>E_r(r) = \frac{W}{r}</math>
''W'' can be evaluated by integrating the electric field from <math>r = R_2</math> to <math>R_1</math> which must be the negative of the voltage ''V'':
<math display=block>-V = \int_{R_2}^{R_1} \frac{W}{r} dr = -W \ln \left(\frac{R_2}{R_1}\right)</math>
so that:
<math display=block>W = \frac{V}{\ln(R_2/R_1)}</math>

The magnetic field, again by symmetry, can only be non-zero in the ''θ'' direction, that is, a vector field looping around the center conductor at every radius between ''R''<sub>1</sub> and ''R''<sub>2</sub>. ''Inside'' the conductors themselves the magnetic field may or may not be zero, but this is of no concern since the Poynting vector in these regions is zero due to the electric field's being zero. Outside the entire coaxial cable, the magnetic field is identically zero since paths in this region enclose a net current of zero (+''I'' in the center conductor and −''I'' in the outer conductor), and again the electric field is zero there anyway. Using [[Ampère's circuital law|Ampère's law]] in the region from ''R''<sub>1</sub> to ''R''<sub>2</sub>, which encloses the current +''I'' in the center conductor but with no contribution from the current in the outer conductor, we find at radius ''r'':
<math display=block>\begin{align}
  I = \oint_C \mathbf{H} \cdot ds &= 2 \pi r H_\theta(r) \\
                      H_\theta(r) &= \frac {I}{2 \pi r}
\end{align}</math>
Now, from an electric field in the radial direction, and a tangential magnetic field, the Poynting vector, given by the cross-product of these, is only non-zero in the ''Z'' direction, along the direction of the coaxial cable itself, as we would expect. Again only a function of ''r'', we can evaluate '''S'''(r):
<math display=block>S_z(r) = E_r(r) H_\theta(r) = \frac{W}{r} \frac {I}{2 \pi r} = \frac{W \, I} {2 \pi r^2}</math>
where ''W'' is given above in terms of the center conductor voltage ''V''. The ''total'' power flowing down the coaxial cable can be computed by integrating over the entire cross section '''A''' of the cable in between the conductors:
<math display=block>\begin{align}
  P_\text{tot}
  &= \iint_\mathbf{A} S_z (r, \theta)\, dA = \int_{R_2}^{R_1} 2 \pi r dr  S_z(r) \\
  &= \int_{R_2}^{R_1} \frac{W\, I}{r} dr = W\, I\, \ln \left(\frac{R_2}{R_1}\right).
\end{align}</math>

Substituting the earlier solution for the constant ''W'' we find:
<math display=block>P_\mathrm{tot} = I \ln \left(\frac{R_2}{R_1}\right) \frac{V}{\ln(R_2/R_1)} = V \, I</math>
that is, the power given by integrating the Poynting vector over a cross section of the coaxial cable is exactly equal to the product of voltage and current as one would have computed for the power delivered using basic laws of electricity.

Other similar examples in which the ''P'' = ''V'' · ''I'' result can be analytically calculated are: the parallel-plate transmission line,<ref name="Morton1979">{{cite journal
| last1 = Morton
| first1 = N.
| title = An Introduction to the Poynting Vector
| journal = Physics Education
| volume = 14
| issue = 5
| year = 1979
| pages = 301–304
| doi = 10.1088/0031-9120/14/5/004| bibcode = 1979PhyEd..14..301M
}}</ref> using [[Cartesian coordinates]], and the two-wire transmission line,<ref name="Boule2024">{{cite journal
| last1 = Boulé
| first1 = Marc
| title = DC Power Transported by Two Infinite Parallel Wires
| journal = American Journal of Physics
| volume = 92
| issue = 1
| year = 2024
| pages = 14–22
| doi = 10.1119/5.0121399
| arxiv = 2305.11827| bibcode = 2024AmJPh..92...14B
}}</ref> using [[bipolar cylindrical coordinates]].