# Homework 3 - Aidan Sharpe

## 1
An particle with charge $-e$ has velocity $\vec{v} = -v\hat{y}$. An electric acts in the $\hat{x}$ direction. What direction must a $\vec{B}$ field act for the net force on the particle to be 0?

$$\vec{F}_{net} = \vec{F}_E + \vec{F}_B = 0$$
$$\therefore \vec{F}_B = -\vec{F}_E$$
$$\vec{F}_E = q\vec{E} = (-)(\hat{x}) = -\hat{x}$$
$$\vec{F}_B = q\vec{v} \times \vec{B}$$
$$(-)\begin{vmatrix} 
\hat{x} & \hat{y} & \hat{z} \\
0 & (-) & 0 \\
B_x & B_y & B_z \\
\end{vmatrix} \overset{!}{=} \hat{x}$$
$$-[(-B_z)\hat{x} - (0)\hat{y} + (0-(-)B_x)\hat{z})] \overset{!}{=} \hat{x}$$
$$\therefore \hat{a}_B = \hat{z}$$

## 2
Consider a plane wave in free space with electromagnetic field intensity:
$$\hat{E}e^{j\omega t} = 30\pi e^{j(10^8t + \beta z)}\hat{x}$$
$$\hat{H}e^{j\omega t} = H_m e^{j(10^8t + \beta z)}\hat{y}$$

Find the direction of propagation and the values for $H_m$, $\beta$, and the wavelength.

Since $\beta z$ is added to $\omega t$, propagation is in the $-\hat{z}$ direction.

$${E_x \over H_y} = \mu_0 c$$
$$E_x = 30\pi$$
$$H_y = H_m$$
$$\therefore H_m = {30\pi \over \mu_0 c}$$
$$\boxed{H_m = 0.25}$$

$$\beta = \omega \sqrt{\mu \varepsilon}$$
For free space:
$$\beta = \omega \sqrt{\mu_0 \varepsilon_0}$$
$$\omega = 10^8$$
$$\boxed{\beta = 0.334}$$
$$\lambda = {c \over f}$$
$$f = {\omega \over 2\pi} = {5 \over \pi} \times 10^7$$
$$\lambda = {3 \times 10^8 \over {5 \over \pi} \times 10^7} = {30\pi \over 5}$$
$$\boxed{\lambda = 6\pi\text{[m]}}$$

## 3
A uniform electric field has intensity:
$$\vec{E} = 15\cos\left(\pi \times 10^8t +{\pi \over 3}z\right)\hat{y}$$

The $\vec{E}$ field is polarized in the $\hat{y}$ direction.

The wave will propagate in the $-\hat{z}$ direction.

$$f = {\pi \times 10^8 \over 2\pi} = 5 \times 10^7 \text{[s}^{-1}]$$

$$\lambda = {3 \times 10^8 \over 5 \times 10^7} = 6 \text{[m]}$$

$$H_x = {15 \over \mu_0 c} = 0.0398$$

$$\vec{H} = 0.0398\cos\left(\pi \times 10^8 t + {\pi \over 3}z\right)\hat{x}$$

## 4
### a)
The properties of a uniform basic plane wave in free space are:

Polarization, amplitude, angular frequency, and the direction of propagation. All other properties, such as wavelength, and the spatial frequency, $\beta$, can be derrived.

### b)
A uniform plane wave in free space is propagating in the $\hat{z}$ direction. If the wavelength is $\lambda = 3$[cm].

$$f = {c \over \lambda} = {3 \times 10^8 \over 3 \times 10^{-2}} = 10^{10}\text{[s}^{-1}]$$
$$\beta = 2\pi f \sqrt{\mu_0 \varepsilon_0} = 209.613$$

The amplitude of the x-polarized $\vec{E}$-field is:
$$\hat{E}_m = 200e^{j {\pi \over 4}}$$

Real-time $\vec{E}$-field:
$$\vec{E}(z, t) = 200\cos\left(2\pi \times 10^{10}t - 209.613z + {\pi \over 4}\right)\hat{x}$$

Phasor $\vec{H}$-field:
$$\hat{H} = 0.53e^{j({\pi \over 4} - 209.613z)}\hat{y}$$

Real-time $\vec{H}$-field:
$$\vec{H}(z, t) = 0.53\cos\left(2\pi \times 10^{10}t - 209.613z + {\pi \over 4}\right)\hat{y}$$

## 5
A 25[cm] by 25[cm] circuit in the y-z plane grows in the $\hat{y}$ direction by 10[m/s]. The circuit contains a 5-ohm resitor. Find the current through the circuit if it is placed in a uniform $\vec{B}$ field of $-0.5\hat{x}$[T].

By Ohms law:
$$V = IR$$
$$\mathcal{E} = -{d \over dt} \int \vec{B} \cdot d\vec{s}$$
$$\mathcal{E} = -{d B A(t) \over dt}$$
$$B A(t) = -0.5 \times 0.25(0.25 + 10t) = -1.25t - 0.03125$$
$$\mathcal{E} = -{d B A(t) \over dt} = 1.25 \text{[V]}$$
$$I = {V \over R} = 250\text{[mA]}$$

Since the magnetic flux inside the loop is increasing, and the magnetic field induced by the current must counteract the magnetic field inducing the current, the current must flow counter-clockwise around the loop.

## 6
### a)
```matlab
clear all;
k = 9e9;
q1 = 1.0e-6;
q2 = 1.0e-6;
ax = 1.0;
ay = 0;
bx = -1.0;
by = 0;

[X, Y] = meshgrid(-2:0.9:2,-2:0.9:2);

V = 1./sqrt((X - ax).^2 + (Y-ay).^2 ) * k * q1 + 1./sqrt((X-bx).^2 + (Y-by).^2) * k * q2;

surfc(X, Y, V);

[Ex, Ey] = gradient(-V, 0.2, 0.2);
figure
contour(X, Y, V);

hold on;
quiver(X, Y, Ex, Ey);
```

![](H0306a.png)

### b)

![](H0306b.png)

### c)
```matlab
% Removed the negative sign on q2
q1 = 1.0e-6;
q2 = 1.0e-6;
```
![](H0306c.png)