Poynting vector of current carrying wire

AI Thread Summary
The discussion focuses on calculating the Poynting vector for a long straight wire carrying a constant current. The participants explore the relationship between the electric field E and the magnetic field H at the surface of the wire, noting that the wire is neutral overall but has a non-neutral charge distribution inside. They clarify that the electric field outside the wire is zero, while an electric field exists within due to the flow of electrons. The participants also discuss how to determine the electric field's magnitude using the potential difference across the wire. Overall, the conversation emphasizes understanding the Poynting vector's implications in the context of current-carrying wires.
Physgeek64
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Homework Statement


A long straight wire of radius a and resistance per unit length R carries a constant current I. Find the Poynting vector N = E × H at the surface of the wire and give
a sketch showing the directions of the current, the electric field E, the magnetic field
H, and N. Integrate the Poynting vector over the surface of a length l of the wire and comment on your result

Homework Equations

The Attempt at a Solution


##\integral{H\dot dl}=I_{free}##
##H(2\pi r)=I##
##H|_a=\frac{I}{2\pi a} \hat{\theta}##

E outside the wire will be ##E=-\frac{\lambda}{2\pi\epsilon r } \hat{r}##

I don't know if this is the same as the E field at the surface? Also I don't know how to get ##\lambda## in terms of I and R

then ##S=E\cross H ##

Many thanks
 
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Does a wire piece of length ##l## carry a net charge or is it neutral?
 
kuruman said:
Does a wire piece of length ##l## carry a net charge or is it neutral?
It is neutral. However I know there needs to be a poynting vector so I don't know how this works
 
If it's neutral, there is no E-field outside the wire. How about inside the wire?
 
kuruman said:
If it's neutral, there is no E-field outside the wire. How about inside the wire?
Inside the wire we have a non-neutral charge. Negative electrons flow in one direction, leaving positive ions behind. Hence we have an E-field in the direction of current? Not sure how to calculate its magnitude though
 
Physgeek64 said:
Not sure how to calculate its magnitude though
You know that the potential difference across the two ends of the wire is V. If the distance between the ends is ##l##, what is E?
 
kuruman said:
You know that the potential difference across the two ends of the wire is V. If the distance between the ends is ##l##, what is E?
##\frac{V}{l}##
 
OK, now what? How does that help you answer the question?
 
kuruman said:
OK, now what? How does that help you answer the question?
It does help- thank you very much!
 
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