Riemann surface
matlab

Exercise 1.3 The electric field of a traveling electromagnetic wave is given by
$$E(z, t)=10 \cos \left(\pi \times 10^7 t+\pi z / 15+\pi / 6\right) \quad(\mathrm{V} / \mathrm{m})$$
Determine (a) the direction of wave propagation, (b) the wave frequency $f$, (c) its wavelength $\lambda$, and (d) its phase velocity $u_{\mathrm{p}}$.

Solution:
(a) $-z$-direction because the signs of the coefficients of $t$ and $z$ are both positive.
(b) From the given expression,
$$\omega=\pi \times 10^7(\mathrm{rad} / \mathrm{s})$$
Hence,
$$f=\frac{\omega}{2 \pi}=\frac{\pi \times 10^7}{2 \pi}=5 \times 10^6 \mathrm{~Hz}=5 \mathrm{MHz} .$$
(c) From the given expression,
$$\frac{2 \pi}{\lambda}=\frac{\pi}{15}$$
Hence $\lambda=30 \mathrm{~m}$
(d) $u_{\mathrm{p}}=f \lambda=5 \times 10^6 \times 30=1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$.

Fig. 1.1
(b) The total charge is
\begin{aligned} Q & =\int_{\text {all space }} \rho d V \ & =-\int_0^{\infty} \frac{\varepsilon_0 A b e^{-b r}}{r^2} \cdot 4 \pi r^2 d r+\int_{\text {all space }} 4 \pi \varepsilon_0 A \hat{i}(\mathbf{r}) d V^{\prime} \ & =4 \pi \varepsilon_0 A\left[e^{-b r}\right]_0^{\infty}+4 \pi \varepsilon_0 A \ & =-4 \pi \varepsilon_0 A+4 \pi \varepsilon_0 A=0 . \end{aligned}
It can also be obtained from Gauss’ flux theorem:

\begin{aligned} Q & =\lim {r \rightarrow \infty} \oint_S \varepsilon_0 \mathbf{E} \cdot d \mathbf{S} \ & =\lim {r \rightarrow \infty} \frac{\varepsilon_0 A e^{-b r}}{r^2} \cdot 4 \pi r^2 \ & =\lim _{r \rightarrow \infty} 4 \pi \varepsilon_0 A e^{-b r}=0, \end{aligned}
in agreement with the above.

Exercise 1.6 An electromagnetic wave is propagating in the $z$-direction in a lossy medium with attenuation constant $\alpha=0.5 \mathrm{~Np} / \mathrm{m}$. If the wave’s electric-field amplitude is $100 \mathrm{~V} / \mathrm{m}$ at $z=0$, how far can the wave travel before its amplitude will have been reduced to (a) $10 \mathrm{~V} / \mathrm{m}$, (b) $1 \mathrm{~V} / \mathrm{m}$, (c) $1 \mu \mathrm{V} / \mathrm{m}$ ?

Solution:
(a)
\begin{aligned} 100 e^{-\alpha z} & =10 \ 100 e^{-0.5 z} & =10 \ e^{-0.5 z} & =0.1 \ -0.5 z & =\ln 0.1=-2.3 \ z & =4.6 \mathrm{~m} . \end{aligned}
(b)
$$\begin{gathered} 100 e^{-0.5 z}=1 \ z=\frac{\ln 0.01}{-0.5}=9.2 \mathrm{~m} . \end{gathered}$$
(c)
\begin{aligned} & 100 e^{-0.5 z}=10^{-6} \ & z=\frac{\ln 10^{-8}}{-0.5}=37 \mathrm{~m} \end{aligned}

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