Chapter 13, Problem 13.23P

To determine

(a)

To prove:

The average power density in the $T{E}_{10}$ mode in a rectangular waveguide given by:

   $\langle S\rangle =\frac{{\beta }_{10}}{2\omega \mu }{E}_0^2{\sin }^2\left({\kappa }_{10}x\right)a_z{\text{W/m}}^{\text{2}}$.

To determine

(b)

The integrated value of the result obtained in part (a) and prove that the average powers in watts

transmitted down the guide is given as:

   $P_{avg}=\frac{{\beta }_{10}ab}{4\omega \mu }{E}_0^2=\frac{ab}{4\eta }{E}_0^2{\sin }^2{\theta }_{10}\text{W}$

Where $\eta =\sqrt{\frac{\mu }{\epsilon }}$ and ${\theta }_{10}$ is the wave angle associated with the $T{E}_{10}$ mode