135 lines
4.1 KiB
Markdown
135 lines
4.1 KiB
Markdown
# Homework 3 - Aidan Sharpe
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## 1
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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?
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$$\vec{F}_{net} = \vec{F}_E + \vec{F}_B = 0$$
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$$\therefore \vec{F}_B = -\vec{F}_E$$
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$$\vec{F}_E = q\vec{E} = (-)(\hat{x}) = -\hat{x}$$
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$$\vec{F}_B = q\vec{v} \times \vec{B}$$
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$$(-)\begin{vmatrix}
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\hat{x} & \hat{y} & \hat{z} \\
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0 & (-) & 0 \\
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B_x & B_y & B_z \\
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\end{vmatrix} \overset{!}{=} \hat{x}$$
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$$-[(-B_z)\hat{x} - (0)\hat{y} + (0-(-)B_x)\hat{z})] \overset{!}{=} \hat{x}$$
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$$\therefore \hat{a}_B = \hat{z}$$
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## 2
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Consider a plane wave in free space with electromagnetic field intensity:
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$$\hat{E}e^{j\omega t} = 30\pi e^{j(10^8t + \beta z)}\hat{x}$$
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$$\hat{H}e^{j\omega t} = H_m e^{j(10^8t + \beta z)}\hat{y}$$
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Find the direction of propagation and the values for $H_m$, $\beta$, and the wavelength.
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Since $\beta z$ is added to $\omega t$, propagation is in the $-\hat{z}$ direction.
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$${E_x \over H_y} = \mu_0 c$$
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$$E_x = 30\pi$$
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$$H_y = H_m$$
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$$\therefore H_m = {30\pi \over \mu_0 c}$$
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$$\boxed{H_m = 0.25}$$
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$$\beta = \omega \sqrt{\mu \varepsilon}$$
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For free space:
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$$\beta = \omega \sqrt{\mu_0 \varepsilon_0}$$
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$$\omega = 10^8$$
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$$\boxed{\beta = 0.334}$$
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$$\lambda = {c \over f}$$
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$$f = {\omega \over 2\pi} = {5 \over \pi} \times 10^7$$
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$$\lambda = {3 \times 10^8 \over {5 \over \pi} \times 10^7} = {30\pi \over 5}$$
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$$\boxed{\lambda = 6\pi\text{[m]}}$$
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## 3
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A uniform electric field has intensity:
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$$\vec{E} = 15\cos\left(\pi \times 10^8t +{\pi \over 3}z\right)\hat{y}$$
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The $\vec{E}$ field is polarized in the $\hat{y}$ direction.
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The wave will propagate in the $-\hat{z}$ direction.
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$$f = {\pi \times 10^8 \over 2\pi} = 5 \times 10^7 \text{[s}^{-1}]$$
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$$\lambda = {3 \times 10^8 \over 5 \times 10^7} = 6 \text{[m]}$$
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$$H_x = {15 \over \mu_0 c} = 0.0398$$
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$$\vec{H} = 0.0398\cos\left(\pi \times 10^8 t + {\pi \over 3}z\right)\hat{x}$$
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## 4
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### a)
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The properties of a uniform basic plane wave in free space are:
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Polarization, amplitude, angular frequency, and the direction of propagation. All other properties, such as wavelength, and the spatial frequency, $\beta$, can be derrived.
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### b)
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A uniform plane wave in free space is propagating in the $\hat{z}$ direction. If the wavelength is $\lambda = 3$[cm].
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$$f = {c \over \lambda} = {3 \times 10^8 \over 3 \times 10^{-2}} = 10^{10}\text{[s}^{-1}]$$
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$$\beta = 2\pi f \sqrt{\mu_0 \varepsilon_0} = 209.613$$
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The amplitude of the x-polarized $\vec{E}$-field is:
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$$\hat{E}_m = 200e^{j {\pi \over 4}}$$
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Real-time $\vec{E}$-field:
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$$\vec{E}(z, t) = 200\cos\left(2\pi \times 10^{10}t - 209.613z + {\pi \over 4}\right)\hat{x}$$
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Phasor $\vec{H}$-field:
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$$\hat{H} = 0.53e^{j({\pi \over 4} - 209.613z)}\hat{y}$$
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Real-time $\vec{H}$-field:
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$$\vec{H}(z, t) = 0.53\cos\left(2\pi \times 10^{10}t - 209.613z + {\pi \over 4}\right)\hat{y}$$
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## 5
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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].
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By Ohms law:
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$$V = IR$$
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$$\mathcal{E} = -{d \over dt} \int \vec{B} \cdot d\vec{s}$$
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$$\mathcal{E} = -{d B A(t) \over dt}$$
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$$B A(t) = -0.5 \times 0.25(0.25 + 10t) = -1.25t - 0.03125$$
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$$\mathcal{E} = -{d B A(t) \over dt} = 1.25 \text{[V]}$$
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$$I = {V \over R} = 250\text{[mA]}$$
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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.
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## 6
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### a)
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```matlab
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clear all;
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k = 9e9;
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q1 = 1.0e-6;
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q2 = 1.0e-6;
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ax = 1.0;
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ay = 0;
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bx = -1.0;
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by = 0;
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[X, Y] = meshgrid(-2:0.9:2,-2:0.9:2);
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V = 1./sqrt((X - ax).^2 + (Y-ay).^2 ) * k * q1 + 1./sqrt((X-bx).^2 + (Y-by).^2) * k * q2;
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surfc(X, Y, V);
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[Ex, Ey] = gradient(-V, 0.2, 0.2);
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figure
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contour(X, Y, V);
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hold on;
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quiver(X, Y, Ex, Ey);
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```
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### b)
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### c)
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```matlab
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% Removed the negative sign on q2
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q1 = 1.0e-6;
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q2 = 1.0e-6;
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```
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