Calculate Power Transported in Example 7.13 Cables w/V & I

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SUMMARY

The discussion focuses on calculating the power transported in a long coaxial cable as described in Example 7.13, using Poynting's Theorem. The key variables include the potential difference V and the current I flowing through the conductors. The Poynting vector S is determined to be radially inward with a magnitude of \(\frac{VI}{2\pi sl}\) at the outer surface of the cable. The integral evaluation confirms that the power transported is equal to VI, validating the solution presented by the participant.

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Homework Statement


Calculate the power transported down the cables of Example 7.13 assuming the two conductors are held at potential difference V and carry current I.

The cable in example 7.13 is "a long coaxial cable" with inner radius a and outer radius b.


Homework Equations



Poynting's Theorem: [tex]\frac{dW}{dt} = -\frac{dU\sub{em}}{dt} - \int{\vec{S}\vec{da}}[/tex], where S is the Poynting vector and the integral is over a closed surface.

The Attempt at a Solution


Not confident my solution is correct. Seems somehow too easy. The first term of the LHS disappears because there is no time dependence. E is parallel to the z axis and has magnitude V/L at the surface of the outer wire, while B is circumferential and has magnitude [tex]\frac{\mu I}{2\pi s}[/tex]. Thus S points radially inward and has magnitude [tex]\frac{VI}{2\pi sl}[/tex]. The integral should be evaluated at the outer surface (s = b) over a cylindrical section of length L, yielding:

[tex]\int{\vec{S}\vec{da}} = S2\pi bL = VI[/tex]

Like I said, I'm not confident my solution is correct (the book is very sparse on examples), and I'd like someone to confirm whether it is correct or incorrect.
 
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You solution is indeed correct :approve:. You should have more confidence in yourself!
 

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