# Antenna Theory and Radio Wave Propagation Homework II

#### PROBLEM 2.1
Starting with MAxwell's wquations, derive the Coulomb law of force between two stationary point electric charges $Q_0$ and $Q$:
$$\pmb{\mathscr{F}}=Q_0\pmb{\mathscr{E}}=\frac{Q_0Q}{4\pi\varepsilon_0\varepsilon{R^2}}\mathbf{\hat{R}}$$

Solution:
Assume there is a sphere with radius $R$, equal to the distance between $Q_0$ and $Q$. $Q_0$ locate at the center of the sphere, and $Q$ locate on the surface of the sphere. The direction $\hat{\mathbf{R}}$ is from $Q_0$ to $Q$.Using the integral form of Gauss's Law:
$$\oint_s\pmb{\mathscr{D}\cdot}d\mathbf{s}=\int_v\rho\ d^3\pmb{r}\ \Rightarrow\ \mathscr{D}(4\pi R^2)=\ Q$$
and $\mathscr{D}=\varepsilon_0\varepsilon\mathscr{E}$, thus,
$$\begin{split}
\mathscr{E}&=&\frac{\mathscr{D}}{\varepsilon_0\varepsilon}=\frac{Q}{4\pi\varepsilon_0\varepsilon R^2}\\
\pmb{\mathscr{E}}&=&\frac{Q}{4\pi\varepsilon_0\varepsilon R^2}\hat{\mathbf{R}}\\
\end{split}$$
Therefore,$$
\pmb{\mathscr{F}}=Q_0\pmb{\mathscr{E}}=\frac{Q_0Q}{4\pi\varepsilon_0\varepsilon{R^2}}\mathbf{\hat{R}}$$

#### PROBLEM 2.2
A plane $y=0$ separates the region of free space ($y<0$) from the region occupied by a perfect conductor ($y>0$). A plane wave is traveling in the $+z$ direction with its magnetic field intensity given by
$$\pmb{\mathscr{H}}(z, t)=H_0f\left(t-\frac{z}{v}\right)\mathbf{\hat x}$$
where $f$ is an arbitary function of $t-\frac{z}{v}$ and $H_0$ is a constant.

a. In which region of the space does the wave exist?
Ans: Free Space, but need to proof.
* Hint:
[1] For free space, $\rho=0$ and $\pmb{\mathscr{J}}=0$.
[2] For perfect conductor, $\sigma\rightarrow\infty$ and $\pmb{\mathscr{J}}=\sigma\pmb{\mathscr{E}}$, thus, $\pmb{\mathscr{J}}\rightarrow\infty$.

b. Find the value of $v$.
c. Find the electric field intensity.
d. Find the surface charge and ccurrent densities.
e. Find the Poynting vector.

#### PROBLEM 2.5
When a current-carrying conductor is placed in a uniform magnetic field $\pmb{\mathscr{B}}_0$, an electric field is developed in the direction perpendicular to both the current and the applied magnetic field. This phenomenon is known as the Hall effect.

a. Using an atomic model, justify the folowing "generalized " Ohm's Law:
$$\pmb{\mathscr{J}}+\mu_{H}\pmb{\mathscr{B}_0\times{\mathscr{J}}}=\sigma\pmb{\mathscr{E}}$$ where
$$\mu_H=\frac{q\tau}{m}=\frac{\sigma}{nq}=R_H\sigma$$ is called the Hall mobility. $m$ is the mass of the charge $q$, $n$ is the number density, $\tau$ is the mean collision time and $R_H$ is the Hall coefficient.

* See also:
[1] [Drude model](https://en.wikipedia.org/wiki/Drude_model)
[2] [10202林秀豪教授普通物理二_第9B講 Hall effect](https://youtu.be/IhfkjkRcXHU)

b. Show that the result of a. may be expressed as
$$\pmb{\mathscr{J}}=\pmb{\bar{\sigma}\cdot\mathscr{E}}$$ where the conductivity tensor take the form
$$\pmb{\bar\sigma}=\sigma_\perp\left(\bar{\mathbf{I}}-\hat{\mathbf{b}}\hat{\mathbf{b}}\right)-\sigma_H\left(\hat{\mathbf{b}}\pmb{\times}\bar{\mathbf{I}}\right)+\sigma_\parallel\hat{\mathbf{b}}\hat{\mathbf{b}}$$$\hat{\mathbf{b}}$ is the unit vector in the direction of $\pmb{\mathscr{B}}_0$ and
$$
\begin{split}
\sigma_\parallel&=&\ \sigma=\mathrm{longitudinal\ conductivity}\\
\sigma_\perp&=&\ \frac{\sigma}{1+\mu_H^2\mathscr{B}_0^2}=\mathrm{transverse\ conductivity} \\
\sigma_H&=&\ \frac{\mu_H\mathscr{B}_0\sigma}{1+\mu_H^2\mathscr{B}_0^2}=\mathrm{Hall\ conductivity}
\end{split}
$$

Solution：
$$
\begin{split}
\pmb{\mathscr{J}}+\mu_{H}\pmb{\mathscr{B}_0\times{\mathscr{J}}}&=&\sigma\pmb{\mathscr{E}}\\
\mathbf{\bar{I}}\cdot\pmb{\mathscr{J}}+\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)\cdot\pmb{\mathscr{J}}&=&\sigma\pmb{\mathscr{E}}\\
\left(\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)\cdot\pmb{\mathscr{J}}&=&\sigma\pmb{\mathscr{E}}\\
\pmb{\mathscr{J}}&=&\sigma\left(\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)^{-1}\cdot\pmb{\mathscr{E}}
\end{split}
$$

Let $\mathbf{\bar{A}}=\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}$, and $\mathbf{\bar{A}}^{-1}={adj(\mathbf{\bar{A}})}/{det(\mathbf{\bar{A}})}$, where
$$
\begin{split}
adj(\mathbf{\bar{A}})&=&\ adj(\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})\\\\
&=&\ adj(\mathbf{\bar{I}})+adj(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})+\mathbf{\bar{I}}\cdot(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})+(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})\cdot\mathbf{\bar{I}}\\
&&-\mathbf{\bar{I}}_t(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})-(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})_t\mathbf{\bar{I}}+\mathbf{\bar{I}}_t(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})_t\mathbf{\bar{I}}-(\mathbf{\bar{I}}\cdot(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}))_t\mathbf{\bar{I}}\\\\
&=&\ \mathbf{\bar{I}}+\left(adj\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)_t\mathbf{\bar{I}}-\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)_t\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)+\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)^2\right)\\
&&+\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)+\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)-3\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)-0+0-0\\\\
&=&\ \mathbf{\bar{I}}+\left(-\mu_H^2\mathscr{B}_0^2\mathbf{\bar{I}}-0+\left(\mu_{H}\pmb{\mathscr{B}_0}\right)\left(\mu_{H}\pmb{\mathscr{B}_0}\right)+\left(\mu_{H}\pmb{\mathscr{B}_0}\right)\cdot\left(\mu_{H}\pmb{\mathscr{B}_0}\right)\mathbf{\bar{I}}\right)-\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)\\\\
&=&\ \mathbf{\bar{I}}+\left(-\mu_H^2\mathscr{B}_0^2\mathbf{\bar{I}}+\mu_H^2\mathscr{B}_0^2\hat{\mathbf{b}}\hat{\mathbf{b}}+\mu_H^2\mathscr{B}_0^2\mathbf{\bar{I}}\right)-\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)\\\\
&=&\ \mathbf{\bar{I}}-\mu_{H}\mathscr{B}_0\left(\mathbf{\hat{b}}\pmb{\times}\mathbf{\bar{I}}\right)+\mu_H^2\mathscr{B}_0^2\hat{\mathbf{b}}\hat{\mathbf{b}}
\end{split}
$$and
$$
\begin{split}
det\left(\mathbf{\bar{A}}\right)&=&\ det(\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}})\\\\
&=&\ det\left(\mathbf{\bar{I}}\right)+det\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)+\left(adj\left(\mathbf{\bar{I}}\right)\cdot\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)\right)_t+\left(\mathbf{\bar{I}}\cdot adj\left(\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)\right)_t\\\\
&=&\ 1+0+0+\mu_H^2\mathscr{B}_0^2\ =\ 1+\mu_H^2\mathscr{B}_0^2
\end{split}
$$thus,
$$
\left(\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)^{-1}=\mathbf{\bar{A}}^{-1}=\frac{adj(\mathbf{\bar{A}})}{det(\mathbf{\bar{A}})}=\ \frac{\mathbf{\bar{I}}-\mu_{H}\mathscr{B}_0\left(\mathbf{\hat{b}}\pmb{\times}\mathbf{\bar{I}}\right)+\mu_H^2\mathscr{B}_0^2\hat{\mathbf{b}}\hat{\mathbf{b}}}{1+\mu_H^2\mathscr{B}_0^2}
$$therefore,
$$
\begin{split}
\pmb{\mathscr{J}}&=&\sigma\left(\mathbf{\bar{I}}+\mu_{H}\pmb{\mathscr{B}_0\times}\mathbf{\bar{I}}\right)^{-1}\cdot\pmb{\mathscr{E}}\\\\
&=&\ \sigma\left(\frac{\mathbf{\bar{I}}-\mu_{H}\mathscr{B}_0\left(\mathbf{\hat{b}}\pmb{\times}\mathbf{\bar{I}}\right)+\mu_H^2\mathscr{B}_0^2\hat{\mathbf{b}}\hat{\mathbf{b}}}{1+\mu_H^2\mathscr{B}_0^2}\right)\cdot\pmb{\mathscr{E}}\\\\
&=&\ \sigma\left(\frac{\left(1+\mu_H^2\mathscr{B}_0^2\right)\hat{\mathbf{b}}\hat{\mathbf{b}}}{1+\mu_H^2\mathscr{B}_0^2}+\frac{\mathbf{\bar{I}}-\hat{\mathbf{b}}\hat{\mathbf{b}}}{1+\mu_H^2\mathscr{B}_0^2}-\frac{\mu_{H}\mathscr{B}_0\left(\mathbf{\hat{b}}\pmb{\times}\mathbf{\bar{I}}\right)}{1+\mu_H^2\mathscr{B}_0^2}\right)\cdot\pmb{\mathscr{E}}\\\\
&=&\ \left(\sigma_\parallel\hat{\mathbf{b}}\hat{\mathbf{b}}+\sigma_\perp\left(\bar{\mathbf{I}}-\hat{\mathbf{b}}\hat{\mathbf{b}}\right)-\sigma_H\left(\hat{\mathbf{b}}\pmb{\times}\bar{\mathbf{I}}\right)\right)\pmb{\cdot{\mathscr{E}}}\\\\
&=&\ \pmb{\bar{\sigma}\cdot{\mathscr{E}}}
\end{split}
$$

#### PROBLEM 2.6 Show that for a time-varying field, the boundary condition (2.26) implies (2.23) and the coundary condition (2.27) implies (2.24).

(2.23)
$$\left(\mathscr{\pmb{B}_2-\pmb{B}_1}\right)\pmb\cdot\hat{\mathbf{q}}=0$$
(2.24)
$$\left(\mathscr{\pmb{D}_2-\pmb{D}_1}\right)\pmb\cdot\hat{\mathbf{q}}=\lim_{\Delta h\rightarrow0}{\rho\ \Delta h}=\rho_s$$
(2.26)
$$\hat{\mathbf{q}}\pmb\times\left(\mathscr{\pmb{E}_2-\pmb{E}_1}\right)=\pmb 0$$
(2.27)
$$\hat{\mathbf{q}}\pmb\times\left(\mathscr{\pmb{H}_2-\pmb{H}_1}\right)=\lim_{\Delta h\rightarrow0}{\pmb{\mathscr{J}}\ \Delta h}=\pmb{\mathscr{J}}_s$$