 890
 38
 Problem Statement

Problem 8.1 Calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.62, assuming the two conductors are held at potential difference V, and carry current I (down one and back up the other).
Example 7.13. A long coaxial cable carries current I (the current flows down the
surface of the inner cylinder, radius a, and back along the outer cylinder, radius
b) as shown in Fig. 7.40. Find the magnetic energy stored in a section of length l.
 Relevant Equations
 ## \frac {dW}{ dt} =  \frac d {dt} \int_{\upsilon}~~ \frac 1 2~ \left( \epsilon_{\omicron} E^2 + \frac {B^2} \mu_o~\right) d \tau ~~ \frac 1 \mu_o \oint_{S_{in}} \vec E \times \vec B \cdot d\vec a##
Relevant equation:
Since the potential difference V is independent of time and current I is steady, both the electric field and magnetic field are independent of time. Hence, the first term of R.H.S. of (8.9) is 0.
If V >0, then ## V_{in}  V_{out} ## = V,
## \Rightarrow V_{in} \gt V_{out} \Rightarrow ## There is a positive charge distribution ## \sigma ## on the inner cylinder.
Using Gauss's law,
## \vec E_{in} = \frac V {\ln \left (\frac b a \right) } \frac {\hat s } s~~~~~~ ~a\lt s\lt b ## ....(1)
Using Ampere's law,
## \vec B_{in} = \frac {\mu_0 I} {2\pi s} \hat \phi ~~~ ~~~~a\lt s\lt b## .....(2)
Assuming that (1) and (2) is applicable on the surface of the inner cylinder,
on the inner cylinder,
## \frac {dW} {dt} =  \frac 1 {\mu_0} \oint_{S_{in} }\vec E \times \vec B \cdot d \vec a##
## =  VI\frac {aL}{\ln \frac b a } ##
L.H.S. is the power spent on charges due to the electromagnetic force acting on the charges of the inner cylinder, while R.H.S. is the power radiated by the charges of the inner cylinder. In short, the power radiated by the charges is the power given to the charges by the electromagnetic force.
Is this correct?
The question says this is the power transported down the cables. How?