Power transported in coaxial cable using Poynting vector

[JohnDuck]

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: $$\frac{dW}{dt} = -\frac{dU\sub{em}}{dt} - \int{\vec{S}\vec{da}}$$, 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 $$\frac{\mu I}{2\pi s}$$. Thus S points radially inward and has magnitude $$\frac{VI}{2\pi sl}$$. The integral should be evaluated at the outer surface (s = b) over a cylindrical section of length L, yielding:

$$\int{\vec{S}\vec{da}} = S2\pi bL = VI$$

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.

 

[Hootenanny]

You solution is indeed correct . You should have more confidence in yourself!