# Homework 6 - Aidan Sharpe
## Exercise 1
Consider a solenoid containing two coaxial magnetic rods of radii $a$ and $b$ and permeabilities $\mu_1 = 2\mu_0$ and $\mu_2 = 3\mu_0$. If the solenoid has $n$ turns every $d$ meters along the axis and carries a steady current, $I$.

### a)
Find $\vec{H}$ inside the first rod, between the first and second rod, and outside the second rod.

For a solenoid regardless of magnetic materials:
$$\vec{H} = {I n \over d}\hat{z}$$
So for all regions:
$$\vec{H} = {I n \over d}\hat{z}$$

### b)
Find $\vec{B}$ inside the three regions.

$\vec{B}$ relies on magnetic material properties:
$$\vec{B} = \mu \vec{H}$$

Therefore:
$$\vec{B} = \begin{cases} 2\mu_0\vec{H} & \text{In region 1}\\ 3\mu_0\vec{H} & \text{In region 2}\\ \mu_0\vec{H} & \text{In region 3} \end{cases}$$

### c)
$\vec{M}$ in each region:

By definition:
$$\vec{M} = \chi_m \vec{H}$$
$$\chi_m = \mu_r - 1$$
Therefore:
$$\vec{M} = \begin{cases} \vec{H} & \text{In region 1} \\ 2\vec{H} & \text{In region 2} \\ 0 & \text{In region 3} \end{cases}$$

### d)
$\vec{J}$ in each region:

By definition:
$$\vec{J}_m = \nabla \times \vec{M}$$

Since $\vec{M}$ is both conservative and solenoidal inside the solenoid, for all three regions:
$$\vec{J}_m = 0$$

## Exercise 2
In a nonmagnetic medium:
$$E = 4\sin(2\pi \times 10^7 t - 0.8x)\hat{z}$$

### a)
Find $\varepsilon_r$ and $\eta$:

The general form:
$$\vec{E}(x,t) = E_0\cos(\omega t -\beta x)\hat{z}$$
$$\beta = \omega \sqrt{\mu \varepsilon} = 0.8$$
$$0.8 = 2\pi\times10^7 \sqrt{\mu_0 \varepsilon_0 \varepsilon_r}$$
$${{\left(0.8 \over 2\pi\times10^7\right)^2} \over \mu_0 \varepsilon_0} = \varepsilon_r$$
$$\boxed{\varepsilon_r = 14.566}$$
$$\eta = \sqrt{\mu \over \varepsilon} = \sqrt{\mu_0 \over \varepsilon_0 \varepsilon_r}$$
$$\boxed{\eta = 98.275}$$

### b)
Find the average poynting vector
$$\vec{P}_\text{avg} = {E_0^2 \over 2 \eta}\hat{x} = {4^2 \over 2(98.275)}$$
$$\vec{P}_\text{avg} = 0.081$$

## Exercise 3
$$E = 3\sin(2\pi\times10^7t - 0.4\pi x)\hat{y} + 4\sin(2\pi\times10^7t - 0.4\pi x)\hat{z}$$

### b)
$$\varepsilon_r = {\left({\beta \over \omega}\right)^2 \over \mu_0 \varepsilon_0} = {\left({0.4\pi\over 2\pi\times10^7}\right)^2 \over \mu_0 \varepsilon_0}$$
$$\boxed{\varepsilon_r = 35.941}$$

### a)
$$\lambda = {v_p \over f}$$
$$v_p = {1 \over \sqrt{\mu \varepsilon}} = {1 \over \sqrt{\mu_0 \varepsilon_0 \varepsilon_r}} = 5\times10^7$$
$$f = {\omega \over 2\pi} = 10^7$$
$$\boxed{\lambda = 5\text{[m]}}$$

### c)
$$H = {E_0 \over \eta}\sin(\omega t - \beta x)\hat{z} + {E_0' \over \eta}\sin(\omega t - \beta x)\hat{y}$$

$$\eta = \sqrt{\mu_0 \over \varepsilon_0 \varepsilon_r} = 62.85$$
$$\boxed{H = 0.0477\sin(2\pi\times10^7t - 0.4\pi x)\hat{z} + 0.0636\sin(2\pi\times10^7t -0.4\pi x)\hat{y}}$$

## Exercise 4
Prove that:
$$\hat{H}_y = - {\hat{E}_x \over \eta_0}$$

Starting with a genering electric plane wave in the $-\hat{z}$ direction:
$$\vec{E}(z, t) = E_0 \cos(\omega t + \beta_0 z)\hat{x}$$

We will use Faraday's Law of Induction to convert a measure in the $\vec{E}$-field to a value for the $\vec{B}$-field:
$$\nabla \times \vec{E} = -{\partial \over \partial t}\vec{B}$$

Converting from $\vec{B}$ to $\vec{H}$:
$$\nabla \times \vec{E} = -\mu{\partial \over \partial t} \vec{H}$$

Assuming free space:
$$\mu = \mu_0$$

This gives the final form for Faraday's Law:
$$\nabla \times \vec{E} = -\mu_0 {\partial \over \partial t} \vec{H}$$

Evaluating the curl of $\vec{E}$:
$$\nabla \times \vec{E} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ \|\vec{E}\| & 0 & 0 \end{vmatrix} = {\partial \over \partial z}\|\vec{E}\|\hat{y} - {\partial \over \partial y}\|\vec{E}\|\hat{z}$$

Since $\vec{E}$ only varies with respect to $z$ and $t$ this can be rewritten as:
$$\nabla \times \vec{E} = {\partial \over \partial{z}}\|\vec{E}\|\hat{y}$$

Evaluate the partial derivative:
$$\nabla \times \vec{E} = -E_0\beta_0\sin(\omega t + \beta_0 z)\hat{y}$$

Back to Faraday's Law:
$$-E_0\beta_0\sin(\omega t + \beta_0 z)\hat{y} = -\mu_0{\partial \over \partial t} \vec{H}$$

Divide out by $-\mu_0$:
$${\partial \over \partial t}\vec{H} = {E_0 \beta_0 \over \mu_0}\sin(\omega t + \beta_0 z)\hat{y}$$

Expand out $\beta_0$:
$$\beta_0 = \omega \sqrt{\mu_0 \varepsilon_0}$$
$${\partial \over \partial t}\vec{H} = {E_0 \omega \sqrt{\mu_0 \varepsilon_0} \over \mu_0}\sin(\omega t + \beta_0 z)dt\hat{y}$$

Integrate both sides with respect to $t$:
$$\vec{H}(z,t) = {E_0 \omega \sqrt{\mu_0 \varepsilon_0} \over \mu_0}\int\sin(\omega t + \beta_0 z)dt \hat{y}$$

Evaluate the integral:
$$\vec{H}(z,t) = -{E_0 \omega \sqrt{\mu_0 \varepsilon_0} \over \mu_0 \omega} \cos(\omega t + \beta_0 z)\hat{y}$$

Simplifying the fraction:
$$\vec{H}(z,t) = -E_0 \sqrt{\varepsilon_0 \over \mu_0} \cos(\omega t + \beta_0 z)\hat{y}$$

Using the definition of $\eta_0$:
$$\eta_0 = \sqrt{\mu_0 \over \varepsilon_0}$$
$$\vec{H}(z,t) = -{E_0 \over \eta_0}\cos(\omega t + \beta z)\hat{y}$$

$$\therefore \vec{H}(z,t) = -{\|\vec{E}(z,t)\| \over \eta_0}\hat{y}$$