Problem 5.6.2: From a phasor back to the time domain

$\omega =2\pi f=2\pi \times 10^{7}rad/s$. The phasor $20e^{-jx}$ has the amplitude $20A/m$ and the phase $-\varphi (x)=-x$, so $\beta =1rad/m$. The factor $e^{-jx}$ marks a wave towards $+x$.

$\varphi (x)=x$, so $\nabla \varphi =a_x$ and $|\nabla \varphi |=1$:

Since the medium is lossless and nonmagnetic,

In free space a $10\mathrm{MHz}$ wave is ${\lambda }_0=c/f=30\mathrm{m}$ long. Here it is $2\pi \mathrm{m}$ long, shorter by $30/2\pi =4.77=\sqrt{{\epsilon }_r}$. The wave slows down and its wavelength shrinks by the same factor, while the frequency stays the same.