If $\overrightarrow{\mathrm{E}}=\left(\hat{a}_{x}+j \vec{a}_{y}\right) e^{\beta a-k a t}$ and $\overrightarrow{\mathrm{H}}=\left(\frac{k}{\omega \mu}\right)\left(\hat{a}_{y}+j \hat{a}_{x}\right) e^{j u-\mu \mu 1}$, the time averaged Poynting vector is
- null vector
- $\left(\frac{k}{\omega \mu}\right) \hat{a}_{z}$
- $\left(\frac{2 k}{\omega \mu}\right) \hat{a}_{x}$
- $\left(\frac{k}{2 \omega \mu}\right) \hat{a}_{z}$