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## Main Question or Discussion Point

This start out as an example in the book:

Find Poynting Vector of a long conducting wire with radius = b and conductivity [itex] \sigma [/itex] with DC current through the wire. Assume wire in z direction and use cylindrical coordinates.

Since DC current, current distribute evenly inside the wire.

[tex] \vec I \;=\; \vec J \pi b^2 \;=\; \sigma \vec E \pi b^2 \;\Rightarrow\; \vec E \;=\; \hat z \;\frac {I}{\sigma \pi b^2} [/tex]

[tex] \vec H \;=\; \hat {\phi} \;\frac {I}{2\pi b} \;\hbox { on the surface of the wire.}[/tex]

[tex] \hbox {Poynting vector = }\; \vec E X \vec H \;=\; \hat z \;X\; \hat{\phi} \;\frac {I^2}{2\sigma \pi^2 b^3}[/tex]

This is all nice and good. MY question is: The book consider

Find Poynting Vector of a long conducting wire with radius = b and conductivity [itex] \sigma [/itex] with DC current through the wire. Assume wire in z direction and use cylindrical coordinates.

Since DC current, current distribute evenly inside the wire.

[tex] \vec I \;=\; \vec J \pi b^2 \;=\; \sigma \vec E \pi b^2 \;\Rightarrow\; \vec E \;=\; \hat z \;\frac {I}{\sigma \pi b^2} [/tex]

[tex] \vec H \;=\; \hat {\phi} \;\frac {I}{2\pi b} \;\hbox { on the surface of the wire.}[/tex]

[tex] \hbox {Poynting vector = }\; \vec E X \vec H \;=\; \hat z \;X\; \hat{\phi} \;\frac {I^2}{2\sigma \pi^2 b^3}[/tex]

This is all nice and good. MY question is: The book consider

**E**is on the surface of the conductor. From the calculation about, I see that the**E**is evenly inside the wire also since it is DC current. Why the book calculate as if the**E**is on the surface or in the air right above the surface of the wire?