5th semester files
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5th-Semester-Fall-2023/EEMAGS/Engineering_Electromagnetics_8th_Edition.pdf
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5th-Semester-Fall-2023/EEMAGS/Engineering_Electromagnetics_8th_Edition.pdf
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\relax
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\@writefile{toc}{\contentsline {section}{\numberline {I}Constants}{1}{}\protected@file@percent }
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<Transmission_line_element.png, id=1, 1927.2pt x 988.69376pt>
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./Equation-Sheet.tex:114: Missing $ inserted.
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<inserted text>
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$
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l.114
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I've inserted a begin-math/end-math symbol since I think
|
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you left one out. Proceed, with fingers crossed.
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./Equation-Sheet.tex:114: Display math should end with $$.
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<to be read again>
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\tex_par:D
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l.114
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The `$' that I just saw supposedly matches a previous `$$'.
|
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So I shall assume that you typed `$$' both times.
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[2
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<./Transmission_line_element.png>] (./Equation-Sheet.aux) )
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Output written on Equation-Sheet.pdf (2 pages, 137318 bytes).
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@ -0,0 +1,35 @@
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# EEMAGS Equation Sheet
|
||||
|
||||
## Curl and Divergence Identities
|
||||
1. The Laplacian: can operate on a scalar or vector field
|
||||
$$\nabla \cdot (\nabla f) = \nabla^2 f$$
|
||||
$$\frac{\partial^2 f}{\partial x} + \frac{\partial^2 f}{\partial y} + \frac{\partial^2 f}{\partial z}$$
|
||||
2. The curl of a gradient is $0$
|
||||
$$\nabla \times (\nabla f) = 0$$
|
||||
3. The gradient of divergence is a scalar
|
||||
$$\nabla (\nabla \cdot \vec{f})$$
|
||||
4. The divergence of curl is $0$
|
||||
$$\nabla \cdot (\nabla \times \vec{v}) = 0$$
|
||||
5. Curl of curl
|
||||
$$\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}$$
|
||||
|
||||
## Maxwell's Equations
|
||||
Gauss's Law for $\vec{E}$-fields:
|
||||
$$\Phi_E =\int \vec{E} \cdot d\vec{s} = \frac{Q_{enc}}{\varepsilon_0}$$
|
||||
$$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$$
|
||||
|
||||
Gauss's Law for $\vec{B}$-fields:
|
||||
$$\int \vec{B} \cdot d\vec{s} = 0 $$
|
||||
$$ \nabla \cdot \vec{B} = 0$$
|
||||
|
||||
Faraday's Law
|
||||
$$\int \vec{E} \cdot d\vec{l} = -\frac{d}{dt}\vec{B} $$
|
||||
$$ \nabla \times \vec{E} = -\frac{d}{dt}\vec{B}$$
|
||||
|
||||
Ampere's Law
|
||||
$$\int \vec{B} \cdot d\vec{l} = \mu_0\left( \vec{J} + \varepsilon_0 \frac{d\vec{E}}{dt}\right) = \mu_0 I_{enc}$$
|
||||
$$ \nabla \times \vec{B} = \mu_0 J + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}$$
|
||||
|
||||
## Work and Voltage
|
||||
$$\Delta V = -\int \vec{E} \cdot d\vec{l}$$
|
||||
$$-\nabla V = \vec{E}$$
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/EquationSheet/Equation-Sheet.pdf
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5th-Semester-Fall-2023/EEMAGS/EquationSheet/Equation-Sheet.pdf
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143
5th-Semester-Fall-2023/EEMAGS/EquationSheet/Equation-Sheet.tex
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5th-Semester-Fall-2023/EEMAGS/EquationSheet/Equation-Sheet.tex
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@ -0,0 +1,143 @@
|
||||
% DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
|
||||
% Version 2, December 2004
|
||||
%
|
||||
% Copyright (C) 2023 Aidan Sharpe <amsharpe102@pm.me>
|
||||
%
|
||||
% Everyone is permitted to copy and distribute verbatim or modified
|
||||
% copies of this license document, and changing it is allowed as long
|
||||
% as the name is changed.
|
||||
%
|
||||
% DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
|
||||
% TERMS AND CONDITIONS FOR COPYING, DISTRIBUTION AND MODIFICATION
|
||||
%
|
||||
% 0. You just DO WHAT THE FUCK YOU WANT TO.
|
||||
|
||||
|
||||
|
||||
\documentclass[journal]{IEEEtran}
|
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|
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\usepackage{graphicx}
|
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\usepackage{amsmath}
|
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\usepackage{amsfonts}
|
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\usepackage{esint}
|
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\usepackage{physics}
|
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|
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\DeclareMathOperator\arctanh{arctanh}
|
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|
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\setlength{\parindent}{0pt}
|
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|
||||
\title{EEMAGS Equation Sheet}
|
||||
|
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\newcommand{\del}[2]{\frac{\partial #1}{\partial #2}}
|
||||
\newcommand{\veps}{\varepsilon}
|
||||
\newcommand{\vphi}{\varphi}
|
||||
|
||||
\begin{document}
|
||||
\maketitle
|
||||
|
||||
\section{Constants}
|
||||
$$\varepsilon_0 = 8.854 \times 10^{-12}$$
|
||||
$$\mu_0 = 1.257 \times 10^{-6}$$
|
||||
$$c = 3 \times 10^8$$
|
||||
$$\eta_0 = \sqrt{{\mu_0 \over \varepsilon_0}} = 377\Omega \approx 120\pi$$
|
||||
|
||||
\section{Fifth's Bullet Points}
|
||||
Find $\veps_r$ given $\lambda$ and $f$:
|
||||
$$\veps_r = {c^2 \over \mu_r \lambda^2 f^2}$$
|
||||
Find $v_p$ given $\delta$, $\alpha$, and $\beta$ given a good conductor:
|
||||
$$v_p = {\omega \over \beta}$$
|
||||
Find $P_\text{avg}$ given $\vec{H}(t)$ in air:
|
||||
|
||||
Find $\Gamma$ in polar and VWSR given $Z_L$ and $Z_0$:
|
||||
$$\Gamma = {Z_L - Z_0 \over Z_L + Z_0}$$
|
||||
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
|
||||
Find $\vec{E}(z,t)$ given $\vec{H}(z,t)$ in a lossless medium of $\mu_r$ and $\veps_r$:
|
||||
|
||||
Find $\lambda$, $\veps_r$, and $\vec{H}$ given $\vec{E}$:
|
||||
$$\hat{H}_y = {\hat{E}_x \over \eta}$$
|
||||
$$\veps_r = {\beta^2 \over \omega^2 \mu_r \mu_0 \veps_0}$$
|
||||
$$\lambda = {v_p \over f}$$
|
||||
Find $\Gamma^2$ and $\langle P \rangle$ from $\veps$, lossless:
|
||||
|
||||
Find $\beta_\text{air}$, $\beta_\text{material}$, $\hat{\Gamma}$, $\hat{T}$ given $\vec{E}^i$, $\veps_r$, $\mu_r$, $\sigma_r$:
|
||||
|
||||
Find $v_p$, $L$, and $Z_0$ given $\veps_r$ and $C$, lossless:
|
||||
$$v_p = {1 \over \sqrt{\mu_0 \mu_r \veps_0 \veps_r}} = {\omega \over \beta}$$
|
||||
$$Z_0 = \sqrt{L \over C}$$
|
||||
$$L = {\mu_0 \mu_r \veps_0 \veps_r \over C}$$
|
||||
Find $C$ given $Z_0$, $R_L$, $f$, and VWSR:
|
||||
$$C = {L \over Z_0^2} - {G \over j\omega} + {R \over j\omega Z_0^2}$$
|
||||
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
|
||||
$$\Gamma = {R_L - Z_0 \over R_L + Z_0}$$
|
||||
$$\omega = 2\pi f$$
|
||||
Find VWSR, $\Gamma$, $Z_\text{in}$ from $\lambda$, $Z_0$, $Z_L$, length:
|
||||
$$\Gamma = {Z_L - Z_0 \over Z_L + Z_0}$$
|
||||
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
|
||||
$$\max[Z_\text{in}] = Z_0 \cdot \text{VWSR}$$
|
||||
$$\min[Z_\text{in}] = {Z_0 \over \text{VWSR}}$$
|
||||
Find $Z_0$, $\alpha$, $\beta$, $\gamma$, $\lambda$ given $Z_\text{in, sc}$ and $Z_\text{in, oc}$:
|
||||
$$Z_0 = \sqrt{Z_\text{in, sc} Z_\text{in, oc}}$$
|
||||
$$\gamma = {1\over l} \arctanh\left(\sqrt{Z_\text{in, sc} \over Z_\text{in, oc}}\right)$$
|
||||
|
||||
|
||||
\section{Boundary Conditions}
|
||||
Electric field boundary conditions:
|
||||
$$\vec{E}_{T1} = \vec{E}_{T2}$$
|
||||
$$\vec{D}_{N1} = \vec{D}_{N2}$$
|
||||
$$\vec{E} = \vec{E}_N + \vec{E}_T$$
|
||||
$$\vec{D} = \veps \vec{E}$$
|
||||
|
||||
Magnetic field boundary conditions:
|
||||
$$\vec{H}_{T1} = \vec{H}_{T2}$$
|
||||
$$\vec{B}_{N1} = \vec{B}_{N2}$$
|
||||
$$\vec{B} = \vec{B}_N + \vec{B}_T$$
|
||||
$$\vec{B} = \mu \vec{H}$$
|
||||
|
||||
Normal vectors:
|
||||
$$\vec{n} = \vec{E}_1 - \vec{E}_2$$
|
||||
$$\vec{n} = \vec{H}_1 - \vec{H}_2$$
|
||||
$$\hat{n} = {\vec{n} \over \|\vec{n}\|}$$
|
||||
|
||||
|
||||
|
||||
\section{Laplace and Poisson}
|
||||
The Laplacian:
|
||||
$$\vec{\nabla} \cdot (-\vec{\nabla}V) = -\vec{\nabla}^2 V = {\rho \over \veps_0}$$
|
||||
Laplace's Equation (charge free region):
|
||||
$$\vec{\nabla}^2 V = 0$$
|
||||
Poisson's Equation:
|
||||
$$\vec{\nabla}^2 V = -{\rho \over \veps_0}$$
|
||||
|
||||
\section{General Transmission Lines}
|
||||
\begin{center}
|
||||
\includegraphics[width=0.3\textwidth]{Transmission_line_element.png}
|
||||
\end{center}
|
||||
$$\begin{aligned}
|
||||
\gamma &= \alpha + j\beta = \left[(R + j\omega L)(G + j\omega C)\right]^{1/2} \\ & = {1\over l} \arctanh\left( \sqrt{Z_\text{in, sc} \over Z_\text{in, oc}}\right)
|
||||
\end{aligned}$$
|
||||
|
||||
The characteristic impedance:
|
||||
$$Z_0 = \sqrt{R + j\omega L \over G + j\omega C} = \sqrt{Z_\text{in, sc} Z_\text{in, oc}}$$
|
||||
$$Z_\text{in} = Z_0 {Z_L + Z_0 \tanh(\gamma l) \over Z_0 + Z_L \tanh(\gamma l)}$$
|
||||
|
||||
\section{Voltage Standing Wave Ratio}
|
||||
$$\Gamma = {V_\text{reflected} \over V_\text{incident}}\Big|_{z' = 0} = {Z_L - Z_0 \over Z_L + Z_0}$$
|
||||
$$\text{VWSR} = {|V_\text{max}| \over |V_\text{min}|} = {1 + |\Gamma| \over 1 - |\Gamma|}$$
|
||||
|
||||
$$v_p = {\omega \over \beta}$$
|
||||
|
||||
$$P_\text{avg} = {|V_0^+| \over 2 Z_0}(1 - |\Gamma_L|^2)
|
||||
|
||||
|
||||
\section{Transmission Line Design}
|
||||
Lossless lines:
|
||||
$$R = G = 0$$
|
||||
$$\alpha = 0$$
|
||||
$$\beta = \omega \sqrt{LC}$$
|
||||
|
||||
Distortionless lines:
|
||||
$${R \over L} = {G \over C}$$
|
||||
$$\alpha = R \sqrt{C \over L}$$
|
||||
$$\beta = \omega \sqrt{LC}$$
|
||||
|
||||
\end{document}
|
||||
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|
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5th-Semester-Fall-2023/EEMAGS/Exam Notes.png
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|
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21
5th-Semester-Fall-2023/EEMAGS/Homework/H0306a.m
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21
5th-Semester-Fall-2023/EEMAGS/Homework/H0306a.m
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@ -0,0 +1,21 @@
|
||||
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);
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/H0306a.png
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5th-Semester-Fall-2023/EEMAGS/Homework/H0306a.png
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|
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BIN
5th-Semester-Fall-2023/EEMAGS/Homework/H0306b.png
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|
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BIN
5th-Semester-Fall-2023/EEMAGS/Homework/H0306c.png
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5th-Semester-Fall-2023/EEMAGS/Homework/H0306c.png
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|
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36
5th-Semester-Fall-2023/EEMAGS/Homework/H04.m
Normal file
36
5th-Semester-Fall-2023/EEMAGS/Homework/H04.m
Normal file
@ -0,0 +1,36 @@
|
||||
clear;
|
||||
clf;
|
||||
|
||||
N = 50;
|
||||
L = 20.0;
|
||||
h = L/N;
|
||||
c = 1;
|
||||
tau = h/c;
|
||||
nstep = 50;
|
||||
|
||||
coef = c * tau / (2*h);
|
||||
|
||||
sig=0.1; % width of the pulse
|
||||
z = ((1:N) - 1/2)*h - L/2; % value of z at time 0
|
||||
phi = (sech(z / (2 * sig^2))).^2; % wave shape
|
||||
phiset = zeros(nstep, N);
|
||||
|
||||
x = z;
|
||||
t = 0:tau:(nstep - 1) * tau;
|
||||
[xx, tt] = meshgrid(x, t);
|
||||
|
||||
for ist = 1:nstep
|
||||
phiset(ist,:) = phi;
|
||||
|
||||
for k = 1:N
|
||||
if ( k ~= 1 ) && ( k ~= N )
|
||||
phinew(k) = (0.5-coef)*phi(k+1) + (0.5+coef)*phi(k-1);
|
||||
else
|
||||
phinew(1) = (0.5-coef)*phi(2) + (0.5+coef)*phi(N);
|
||||
phinew(N) = (0.5-coef)*phi(1) + (0.5+coef)*phi(N-1);
|
||||
end
|
||||
end
|
||||
phi = phinew;
|
||||
end
|
||||
|
||||
surf(xx, tt, phiset);
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/H0502a.png
Normal file
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/H0502a.png
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|
After Width: | Height: | Size: 45 KiB |
266
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-01.md
Normal file
266
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-01.md
Normal file
@ -0,0 +1,266 @@
|
||||
# Homework 1 - Aidan Sharpe
|
||||
|
||||
## 1.
|
||||
If $\vec{A} = 4\hat{x} + 4\hat{y} - 2\hat{z}$ and $\vec{B} = 3\hat{x} - 1.5\hat{y} + \hat{z}$, find the acute angle between $\vec{A}$ and $\vec{B}$.
|
||||
|
||||
Definition of dot product:
|
||||
$$\vec{A} \cdot \vec{B} = (A_x B_x)+(A_y B_y) + (A_z B_z) =\lVert \vec{A} \rVert \lVert \vec{B} \rVert \cos(\varphi)$$
|
||||
Solve the dot product:
|
||||
$$\vec{A} \cdot \vec{B} = (4 \cdot 3) + (4 \cdot (-1.5)) + ((-2) \cdot 1) = 4$$
|
||||
Magnitudes of the vectors:
|
||||
$$\lVert \vec{A} \rVert = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{36} = 6$$
|
||||
$$\lVert \vec{B} \rVert = \sqrt{3^2 + \left(-\frac{3}{2}\right)^2 + 1^2} = \sqrt{\frac{49}{4}} = \frac{7}{2}$$
|
||||
By the definition of the dot product:
|
||||
$$4 = 6 \cdot \frac{7}{2} \cos(\varphi)$$
|
||||
$$\therefore \cos(\varphi) = \frac{4}{21}$$
|
||||
$$\therefore \varphi = \arccos\left(\frac{4}{21}\right) = 1.379 = 79.02^\circ$$
|
||||
|
||||
## 2
|
||||
If $\vec{A} = \frac{10}{\rho}\hat{\rho} + 5\hat{\varphi} + 2\hat{z}$ and $\vec{B} = 5\hat{\rho} + \cos(\varphi)\hat{\varphi} + \rho\hat{z}$, find (a) $\vec{A} \cdot \vec{B}$ and (b) $\vec{A} \times \vec{B}$ at $x=1$, $y=1$, $z=1$.
|
||||
|
||||
Convert from cartesian to cylidrical:
|
||||
$$\rho = \sqrt{x^2 + y^2} = \sqrt{2}$$
|
||||
$$\varphi = \arctan\left(\frac{y}{x}\right) = \frac{\pi}{4}$$
|
||||
$$z = z = 1$$
|
||||
Find $\vec{A}$ and $\vec{B}$ at $x=1$, $y=1$, $z=1$:
|
||||
$$\vec{A} = \frac{10}{\sqrt{2}}\hat{\rho}+5\hat{\varphi}+2\hat{z} = 5\sqrt{2}\hat{\rho}+5\hat{\varphi}+2\hat{z}$$
|
||||
$$\vec{B} = 5\hat{\rho}+\cos\left(\frac{\pi}{4}\right)\hat{\varphi}+\sqrt{2}\hat{z}=5\hat{\rho}+\frac{\sqrt{2}}{2}\hat{\varphi}+\sqrt{2}\hat{z}$$
|
||||
|
||||
### a) Find $\vec{A} \cdot \vec{B}$
|
||||
$$\vec{A} \cdot \vec{B} = \left(5\sqrt{2} \cdot 5\right) + \left(5 \cdot \frac{\sqrt{2}}{2}\right) + \left(2 \cdot \sqrt{2}\right) = \frac{59\sqrt{2}}{2}$$
|
||||
|
||||
### b) Find $\vec{A} \times \vec{B}$
|
||||
$$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\rho} & \hat{\varphi} & \hat{z} \\ 5\sqrt{2} & 5 & 2 \\ 5 & \frac{\sqrt{2}}{2} & \sqrt{2} \end{vmatrix} = 4\sqrt{2}\hat{\rho} - 20\hat{z}$$
|
||||
|
||||
## 3
|
||||
Two point charges have mass $0.2$g. Two insulating threads of length $1$m are used to suspend the charges from a common point. The gravitational force is $980 \times 10^{-5}\text{N/g}$.
|
||||
|
||||
Define $\varphi$ as the angle between the threads, $\alpha$ as half of that angle, and $\beta$ as $\frac{\pi}{2} - \alpha$. This way, $\alpha$ and $\beta$ make a right triangle with the leg adjacent to $\alpha$ making a perpendicular bisector of the distance between the charges, $r$.
|
||||
|
||||
Find the distance, $r$:
|
||||
$$\frac{r}{2} = (1\text{m})\cos(\beta)$$
|
||||
$$\therefore r = 2\cos(\beta)$$
|
||||
|
||||
By Coulomb's Law, the electric field force on each charge, $\vec{F}_e$, is:
|
||||
$$\vec{F}_e = \frac{q^2}{4\pi\varepsilon_0 r^2}\hat{x}$$
|
||||
Pluggin in for $r$,
|
||||
$$\vec{F}_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}\hat{x}$$
|
||||
|
||||
The force due to gravity on each charge, $\vec{F}_g$, is:
|
||||
$$\vec{F}_g = 0.2 \cdot 980\times 10^{-5} = 0.00196(-\hat{y})\text{N}$$
|
||||
|
||||
Assuming static equilibrium:
|
||||
$$\sum F_x = 0 = F_{e_x} + F_{T_x} + F_{g_x}$$
|
||||
$$\sum F_y = 0 = F_{e_y} + F_{T_y} + F_{g_y}$$
|
||||
Where:
|
||||
|
||||
$F_{T_x}$ is the x-component of the force due to tension in each thread, $F_T$.
|
||||
|
||||
$F_{T_y}$ is the y-component of $F_T$.
|
||||
|
||||
Since $F_g$ only acts in the $\hat{y}$ direction, and $F_e$ only acts in the $\hat{x}$ direction:
|
||||
$$F_{e_x} + F_{T_x} = 0$$
|
||||
$$F_{T_y} + F_{g_y} = 0$$
|
||||
|
||||
Define the components of the tension force $F_T$:
|
||||
$$F_{T_x} = F_T\cos\left(\frac{\pi}{2} + \alpha\right) = -F_T\cos(\beta)$$
|
||||
$$F_{T_y} = F_T\sin\left(\frac{\pi}{2} + \alpha\right) = F_T\sin(\beta)$$
|
||||
|
||||
Solve for $F_T$ using equilibrium in the $\hat{y}$ direction:
|
||||
$$F_T\sin(\beta) - 0.00196 = 0$$
|
||||
$$\therefore F_T = \frac{0.00196}{\sin(\beta)}$$
|
||||
|
||||
Plug in $F_T$ to solve equilibrium in the $\hat{x}$ direction:
|
||||
$$F_e + \left( \frac{0.00196}{\sin(\beta)} \right)(-\cos(\beta)) = 0$$
|
||||
$$\therefore F_e = 0.00196 \cot(\beta)$$
|
||||
|
||||
### a) When $\varphi = 45^\circ$, solve for the charge, $q$:
|
||||
Since $\varphi = \frac{\pi}{4}$, $\alpha = \frac{\pi}{8}$, and $\beta = \frac{3\pi}{8}$.
|
||||
|
||||
Known:
|
||||
$$F_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}$$
|
||||
$$F_e = 0.00196\frac{\cos(\beta)}{\sin(\beta)}$$
|
||||
|
||||
Set equal:
|
||||
$$\frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)} = 0.00196\cot(\beta)$$
|
||||
$$\therefore q^2 = 16\pi\varepsilon_0\frac{\cos^3(\beta)}{\sin(\beta)}$$
|
||||
|
||||
Plug in for $\beta$ and solve for $q$:
|
||||
$$q = \pm \sqrt{16\pi\varepsilon_0(0.00196)\frac{\cos^3\left(\frac{3\pi}{8}\right)}{\sin\left(\frac{3\pi}{8}\right)}} = \pm 2.300\times10^{-7}\text{C}$$
|
||||
|
||||
### b) When $q=0.5\mu\text{C}$, solve for the angle, $\varphi$:
|
||||
Known:
|
||||
$$F_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}$$
|
||||
$$F_e = 0.00196\frac{\cos(\beta)}{\sin(\beta)}$$
|
||||
|
||||
Set equal:
|
||||
$$\frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)} = 0.00196\cot(\beta)$$
|
||||
|
||||
Plug in for $q$:
|
||||
$$\frac{(0.5\times 10^{-6})^2}{16\pi\varepsilon_0(0.00196)} = \frac{\cos^3(\beta)}{\sin(\beta)}$$
|
||||
$$\therefore \frac{\cos^3(\beta)}{\sin(\beta)} = 0.287$$
|
||||
$$\therefore \beta = 0.915$$
|
||||
Using $\beta$ to solve for $\varphi$:
|
||||
$$\alpha = \frac{\pi}{2} - \beta = 0.656$$
|
||||
$$\varphi = 2\alpha = 1.314 = 75.257^\circ$$
|
||||
|
||||
## 4
|
||||
The magnetic field, $\vec{B} = B_0(\hat{x} + 2\hat{y} - 4\hat{z})$ exists at a point. Find the electric field at that point if the force experienced by a test charge with velocity, $\vec{v} = v_0(3\hat{x} - \hat{y} + 2\hat{z})$ is $0$.
|
||||
|
||||
Lorentz Force Law:
|
||||
$$\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}$$
|
||||
|
||||
Since $\vec{F}$ is $0$:
|
||||
$$0 = q\vec{E} + q\vec{v} \times \vec{B}$$
|
||||
$$\therefore q\vec{E} = -q\vec{v} \times \vec{B}$$
|
||||
$$\therefore \vec{E} = -\vec{v} \times \vec{B}$$
|
||||
|
||||
By the definition of the cross product:
|
||||
$$-\vec{v} \times \vec{B} = \vec{B} \times \vec{v}$$
|
||||
In terms of $\vec{E}$:
|
||||
$$\vec{E} = \vec{B} \times \vec{v} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 1 & 2 & -4 \\ 3 & -1 & 2 \end{vmatrix} = -14\hat{y} - 7\hat{z}$$
|
||||
|
||||
## 5
|
||||
A circular loop with radius, $a$, exists in the x-y plane. If the loop is uniformly charged and has total charge, $Q$, determine the $\vec{E}$-field intensity at some point along the axis normal to the loop.
|
||||
|
||||
By Coulomb's Law:
|
||||
$$\vec{E} = \int \frac{dq}{4\pi\varepsilon_0 r^2}\hat{r}$$
|
||||
|
||||
Calling the distance along the normal axis $z$:
|
||||
$$r^2 = a^2 + z^2$$
|
||||
|
||||
Along a uniformly charged line, the charge, $dq$, at any given point is given by the equation:
|
||||
$$dq = \lambda dl$$
|
||||
Where:
|
||||
|
||||
$\lambda$ is the linear charge density
|
||||
|
||||
The linear charge density, $\lambda$ is defined as:
|
||||
$$\lambda = \frac{Q}{L}$$
|
||||
Where:
|
||||
|
||||
$Q$ is the total charge
|
||||
|
||||
$L$ is the total length
|
||||
|
||||
For a circular loop with radius, $a$, and total charge, $Q$:
|
||||
$$\lambda = \frac{Q}{2\pi a}$$
|
||||
$$dl = a d\varphi$$
|
||||
$$\therefore dq = \frac{Qd\varphi}{2\pi}$$
|
||||
|
||||
Plugging into Coulomb's Law:
|
||||
$$\vec{E} = \int\limits_{0}^{2\pi} \frac{Qd\varphi}{8\pi^2\varepsilon_0(a^2 + z^2)} \hat{r}$$
|
||||
$$\therefore \vec{E} = \frac{Q}{8\pi^2\varepsilon_0(a^2 + z^2)} \int\limits_{0}^{2\pi}\hat{r}d\varphi$$
|
||||
|
||||
Find $\hat{r}$ in terms of $\hat{x}$, $\hat{y}$, and $\hat{z}$:
|
||||
$$\hat{r} = \frac{\vec{a} + \vec{z}}{\sqrt{a^2 + z^2}}$$
|
||||
$$\vec{z} = z\hat{z}$$
|
||||
$$\vec{a} = a\cos{\varphi}\hat{x} + a\sin{\varphi}\hat{y}$$
|
||||
|
||||
Back to Coulomb's Law:
|
||||
$$\vec{E} = \frac{Q}{8\pi^2\varepsilon_0(a^2 + z^2)^{(3/2)}} \int\limits_{0}^{2\pi} (a\cos(\varphi)\hat{x} + a\sin(\varphi)\hat{y} + z\hat{z})d\varphi$$
|
||||
|
||||
By components:
|
||||
$$E_x = \frac{aQ}{8\pi^2\varepsilon_0(a^2 + z^2)^{3/2}} \int\limits_{0}^{2\pi} \cos(\varphi)d\varphi = 0$$
|
||||
$$E_y = \frac{aQ}{8\pi^2\varepsilon_0(a^2 + z^2)^{3/2}} \int\limits_{0}^{2\pi} \sin(\varphi)d\varphi = 0$$
|
||||
$$E_z = \frac{zQ}{8\pi^2\varepsilon_0(a^2 + z^2)^{3/2}} \int\limits_{0}^{2\pi} d\varphi = \frac{zQ}{4\pi\varepsilon_0(a^2 + z^2)^{3/2}}$$
|
||||
|
||||
Recombining:
|
||||
$$\vec{E} = \frac{zQ}{4\pi\varepsilon_0(a^2 + z^2)^{3/2}}\hat{z}$$
|
||||
|
||||
## 6
|
||||
Consider a circular ring in the x-y plane with inner radius, $a$, outer radius, $b$, and uniform charge density, $\rho_s$. Find an expression for the $\vec{E}$-field at a point at distance, $z$, along the axis normal to the ring.
|
||||
|
||||
By Coulomb's Law:
|
||||
$$d\vec{E} = \frac{dq}{4\pi\varepsilon_0 r^2}\hat{r}$$
|
||||
|
||||
Calling the radial distance, $\rho$:
|
||||
$$r^2 = \rho^2 + z^2$$
|
||||
|
||||
For surface charge densities:
|
||||
$$dq = \rho_s ds$$
|
||||
|
||||
In cylindrical coordinates:
|
||||
$$ds = \rho d\rho d\varphi$$
|
||||
|
||||
Plugging into Coulomb's Law:
|
||||
$$\vec{E} = \iint \frac{\rho_s \rho}{4\pi\varepsilon_0(\rho^2 + z^2)}\hat{r}d\rho d\varphi$$
|
||||
|
||||
$\hat{r}$ is defined as:
|
||||
$$\hat{r} = \frac{\vec{\rho} + \vec{z}}{\sqrt{\rho^2 + z^2}}$$
|
||||
Where:
|
||||
|
||||
$\vec{z} = z\hat{z}$
|
||||
|
||||
$\vec{\rho} = \rho\cos(\varphi)\hat{x} + \rho\sin(\varphi)\hat{y}$
|
||||
|
||||
Splitting $\hat{r}$ by components:
|
||||
$$\hat{r} = \frac{\rho\cos(\varphi)\hat{x} + \rho\sin(\varphi)\hat{y} + z\hat{z}}{\sqrt{\rho^2 + z^2}}$$
|
||||
|
||||
Back to Coulomb's Law:
|
||||
$$\vec{E} = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho^2\cos(\varphi)\hat{x} + \rho^2\sin(\varphi)\hat{y} + \rho z\hat{z}}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi$$
|
||||
|
||||
By components:
|
||||
$$E_x = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho^2\cos(\varphi)}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi = 0$$
|
||||
$$E_y = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho^2\sin(\varphi)}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi = 0$$
|
||||
$$E_z = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho z}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi = \frac{\rho_s z}{2\varepsilon_0} \left(\frac{1}{\sqrt{z^2 + a^2}} - \frac{1}{\sqrt{z^2 + b^2}}\right)$$
|
||||
|
||||
Recombining:
|
||||
$$\vec{E} = \frac{\rho_s z}{2 \varepsilon_0} \left( \frac{1}{\sqrt{z^2 + a^2}} - \frac{1}{\sqrt{z^2 + b^2}} \right) \hat{z}$$
|
||||
|
||||
## 7
|
||||
Consider two concentric cylindrical surfaces. The inner having radius, $a$, and charge density $\rho_s$, and the outer having radius, $b$, and charge density $-\rho_s$.
|
||||
|
||||
By Gauss's Law:
|
||||
$$\Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \oint_S \vec{E} \cdot d\vec{A}$$
|
||||
|
||||
The $\vec{E}$-field for a cylindrical surface with radius, $\rho$, and length, $l$, is given by the equation:
|
||||
$$2\pi\rho l E = \frac{Q_{enc}}{\varepsilon_0}$$
|
||||
|
||||
### a) For $\rho < a$:
|
||||
$$Q_{enc} = 0$$
|
||||
$$\therefore E = 0$$
|
||||
|
||||
### b) For $a < \rho < b$:
|
||||
$$Q_{enc} = 2\pi a l\rho_s$$
|
||||
Where:
|
||||
|
||||
$l$ is the length of the section of the cylender.
|
||||
|
||||
$$2\pi\rho lE = \frac{2\pi a l \rho_s}{\varepsilon_0}$$
|
||||
$$\therefore E = \frac{a\rho_s}{\rho \varepsilon_0}$$
|
||||
|
||||
### c) For $\rho > b$:
|
||||
$$Q_{enc} = 2\pi l\rho_s(a-b)$$
|
||||
$$2\pi \rho l E = \frac{2\pi l \rho_s(a-b)}{\varepsilon_0}$$
|
||||
$$\therefore E = \frac{\rho_s (a-b)}{\rho \varepsilon_0}$$
|
||||
|
||||
## 8
|
||||
Consider an infinite slab of thickness, $d$, centered on the origin ($x=0$, $y=0$, $z=0$).
|
||||
|
||||
By Gauss's Law:
|
||||
$$\Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \oint_S \vec{E} \cdot d\vec{A}$$
|
||||
|
||||
### a) Find the strength of the electric field inside the slab ($\lvert z \rvert < d/2$):
|
||||
$$Q_{enc} = \rho_v lwh$$
|
||||
$$\oint_S \vec{E} \cdot d\vec{A} = E(2lw + 2lh + 2wh)$$
|
||||
$$E = \frac{\rho_v lwh}{2\varepsilon_0(lw + lh + wh)}$$
|
||||
Where:
|
||||
|
||||
$l$ is the size of the x dimension of a Gaussian rectangular prism centered on the origin,
|
||||
|
||||
$w$ is the size of the y dimension of that rectangular prism,
|
||||
|
||||
$h$ is the size of the z dimension of that rectangular prism
|
||||
|
||||
### b) Find the strength of the electric field inside the slab ($\lvert z \rvert > d/2$):
|
||||
$$Q_{enc} = \frac{\rho_v lwd}{2}$$
|
||||
$$\oint_S \vec{E} \cdot d\vec{A} = E(2lw + 2lh + 2wh)$$
|
||||
$$E = \frac{\rho_v lwd}{4\varepsilon_0 (lw + lh + wh)}$$
|
||||
Where:
|
||||
|
||||
$l$ is the size of the x dimension of a Gaussian rectangular prism centered on the origin,
|
||||
|
||||
$w$ is the size of the y dimension of that rectangular prism,
|
||||
|
||||
$h$ is the size of the z dimension of that rectangular prism
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-01.pdf
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-02.md
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@ -0,0 +1,145 @@
|
||||
# Homework 2 - Aidan Sharpe
|
||||
## 1
|
||||
Consider a long cylindrical wire of radius, $a$, carrying a current $I = I_0 \cos(\omega t)\hat{z}$.
|
||||
|
||||
### a)
|
||||
Write an expression for the magnetic field strength, $B$, outside the wire ($\rho > a$):
|
||||
|
||||
By Ampere's Law:
|
||||
$$\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$
|
||||
|
||||
For a closed loop of radius, $\rho$:
|
||||
$$\int \vec{B} \cdot d\vec{l} = B(2\pi\rho)$$
|
||||
$$\therefore B(2\pi\rho) = \mu_0 I_{enc}$$
|
||||
|
||||
Since ($\rho > a$):
|
||||
$$I_{enc} = I = I_0 \cos(\omega t)$$
|
||||
|
||||
Recombining:
|
||||
$$B(2\pi\rho) = \mu_0 I_0 \cos(\omega t)$$
|
||||
$$\therefore B = {\mu_0 I_0 \cos(\omega t) \over 2\pi\rho} \text{[T]}$$
|
||||
|
||||
### b)
|
||||
Consider a rectangular loop a distance, $d$, from the wire with sidelengths $\alpha$ in the $\hat{x}$ direction, and $\beta$ in the $\hat{z}$ direction.
|
||||
|
||||
#### i)
|
||||
Calculate the magnetic flux, $\Phi_B$, through the loop.
|
||||
|
||||
By Gauss's Law for magnetism:
|
||||
$$\Phi_B = \iint \vec{B} \cdot d\vec{s}$$
|
||||
|
||||
Since the $\beta$ does not vary:
|
||||
$$\Phi_B = \beta \int\limits_{d}^{d+\alpha} {\mu_0 I_0 \cos(\omega t) \over 2\pi\rho} d\rho$$
|
||||
|
||||
Taking out constants:
|
||||
$$\Phi_B = {\beta \mu_0 I_0 \cos(\omega t) \over 2 \pi} \int\limits_d^{d+\alpha} {1 \over \rho}d\rho$$
|
||||
|
||||
Evaluate:
|
||||
$$\Phi_B = {\beta \mu_0 I_0 \cos(\omega t) \over 2\pi} \ln\left( {\lvert d + \alpha \rvert \over \lvert d \rvert} \right) \text{[Vs]}$$
|
||||
|
||||
#### ii)
|
||||
Find the induced EMF, $\cal{E}$:
|
||||
|
||||
$$\mathcal{E} = -{d \over dt}\Phi_B$$
|
||||
|
||||
Plug in:
|
||||
$$\mathcal{E} = -{d \over dt} {\beta \mu_0 I_0 \cos(\omega t) \over 2\pi} \ln\left({\lvert d + \alpha \rvert \over \lvert d \rvert}\right)$$
|
||||
|
||||
Evaluate:
|
||||
$$\mathcal{E} = {\beta \mu_0 I_0 \sin(\omega t) \over 2\pi} \ln\left({\lvert d + \alpha \rvert \over \lvert d \rvert}\right) \text{[V]}$$
|
||||
|
||||
## 2
|
||||
A quarter circle loop of wire with inner radius, $a$, and outer radius, $b$, has current, $I$. Find the magnetic field strength at the center of the circle.
|
||||
|
||||
By superposition:
|
||||
$$B = B_1 + B_2 + B_3 + B_4$$
|
||||
|
||||
By Ampere's law:
|
||||
$$\int \vec{B_1} \cdot d\vec{l} = \mu_0 I_{enc_1}$$
|
||||
$$\int \vec{B_2} \cdot d\vec{l} = \mu_0 I_{enc_2}$$
|
||||
$$\int \vec{B_3} \cdot d\vec{l} = \mu_0 I_{enc_3}$$
|
||||
$$\int \vec{B_4} \cdot d\vec{l} = \mu_0 I_{enc_4}$$
|
||||
|
||||
Since the current in the first and third segments are either parallel or antiparallel:
|
||||
$$I_{enc_1} = 0$$
|
||||
$$I_{enc_3} = 0$$
|
||||
|
||||
Since all of $I$ is enclosed in a loop from either segment two or four:
|
||||
$$I_{enc_2} = I_{enc_4} = I$$
|
||||
|
||||
Back to Ampere's Law:
|
||||
$$\int \vec{B_1} \cdot d\vec{l} = 0 \therefore B_1 = 0$$
|
||||
$$\int \vec{B_2} \cdot d\vec{l} = \mu_0 I$$
|
||||
$$\int \vec{B_3} \cdot d\vec{l} = 0 \therefore B_2 = 0$$
|
||||
$$\int \vec{B_4} \cdot d\vec{l} = \mu_0 I$$
|
||||
|
||||
Determining $d\vec{l}$ for segments two and four:
|
||||
$$\int \vec{B_2} \cdot \vec{a}d\varphi = \mu_0 I$$
|
||||
$$\int \vec{B_4} \cdot \vec{b}d\varphi = \mu_0 I$$
|
||||
|
||||
Determining the bounds:
|
||||
$$\int\limits_0^{\pi \over 2} \vec{B_2} \cdot \vec{a} d\varphi = \mu_0 I$$
|
||||
$$\int\limits^0_{\pi \over 2} \vec{B_4} \cdot \vec{b} d\varphi = \mu_0 I$$
|
||||
|
||||
Evaluate:
|
||||
$$B_2 \left({\pi a \over 2}\right) = \mu_0 I$$
|
||||
$$B_4 \left({-\pi b \over 2}\right) = \mu_0 I$$
|
||||
|
||||
Solve for $B$:
|
||||
$$B_2 = {2 \mu_0 I \over \pi a}$$
|
||||
$$B_4 = -{2 \mu_0 I \over \pi b}$$
|
||||
$$B = {2 \mu_0 I \over \pi a} - {2 \mu_0 I \over \pi b} = {2 \mu_0 I (a-b) \over \pi^2 a b} \text{[T]}$$
|
||||
|
||||
## 3
|
||||
Consider a cylinder of radius, $\rho_0 = 0.5\text{[m]}$, with a current density, $\vec{J} = 4.5e^{-2\rho}\hat{z}[A/m^2]$.
|
||||
|
||||
By Ampere's Law:
|
||||
$$\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$
|
||||
|
||||
For a radial current density:
|
||||
$$I_{enc} = \int\limits_{0}^{2\pi} \int\limits_{0}^{\rho} J sdsd\varphi$$
|
||||
|
||||
Plugging in $J$ with $s$ as the integrating variable:
|
||||
$$I_{enc} = \int\limits_{0}^{2\pi} \int\limits_{0}^{\rho} 4.5e^{-2s} sdsd\varphi$$
|
||||
$$I_{enc} = 4.5\int\limits_{0}^{2\pi} \int\limits_{0}^{\rho} e^{-2s} sdsd\varphi$$
|
||||
$$I_{enc} = 4.5\int\limits_{0}^{2\pi} \left[ \left( {-\rho \over 2} - {1 \over 4}\right)e^{-2\rho} + {1 \over 4}\right]d\varphi$$
|
||||
$$\therefore I_{enc} = {9 \pi e^{-2\rho}(e^{2\rho} -2\rho -1) \over 4}$$
|
||||
|
||||
Back to Ampere's Law:
|
||||
$$\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$$
|
||||
|
||||
For a circular loop:
|
||||
$$\int\limits_{0}^{2\pi} B\rho d\varphi = \mu_0 I_{enc}$$
|
||||
$$B(2\pi\rho) = {9 \pi e^{-2\rho}(e^{2\rho} -2\rho -1) \over 4}$$
|
||||
|
||||
For $\rho \le \rho_0$
|
||||
$$B = {9 e^{-2\rho}(e^{2\rho} -2\rho -1) \over 8\rho}$$
|
||||
|
||||
For $\rho \ge \rho_0$:
|
||||
$$I_{enc} = 4.5\int\limits_{0}^{2\pi} \int\limits_{0}^{\rho_0} e^{-2s} sdsd\varphi = {9 (e-2) \pi \over 4 e}$$
|
||||
|
||||
$$B(2\pi\rho) = {9(e-2)\pi \over 4e}$$
|
||||
$$\therefore B = {9(e-2) \over 8e\rho}\text{[T]}$$
|
||||
|
||||
## 4
|
||||
Consider a solenoidal wire with $n$ coils per unit length. The core becomes magnetized when a current $I=10[A]$ is put into the wire coil, and this causes a bound current to flow around the cylindrical surface of the core as shown in the side view diagram. This bound core surface current density has magnitude, $J = 20n [A/m]$
|
||||
|
||||
$$B = B_s + B_c$$
|
||||
$$B_s = \mu_0 I n = 10\mu_0n$$
|
||||
Pretend the core current is just another solenoid:
|
||||
$$B_c = \mu_0 I n = 20\mu_0n$$
|
||||
$$B = 30\mu_0 n \text{[T]}$$
|
||||
|
||||
## 5
|
||||
A satellite travelling at $5\text{[km/s]}$ enters a current filled curtain. From $t=1\text{[s]}$ to $t=3\text{[s]}$, the satellite's magnetometer increases from $-95\hat{x}\text{[nT]}$ to $95\hat{x}\text{[nT]}$. If the current flows in the $\hat{z}$ direction, find the current density, $J$.
|
||||
|
||||
By Ampere's Law:
|
||||
$$\int \vec{B} \cdot d\vec{l} = \mu_0 \iint \vec{J} \cdot d\vec{s}$$
|
||||
|
||||
The total increase in $B$ is $195 \text{[nT]}$, so:
|
||||
$$195 \times 10^{-9} = \mu_0 \iint \vec{J} \cdot d\vec{s}$$
|
||||
|
||||
Plug in bounds to get the total distance through the curtain:
|
||||
$$195 \times 10^{-9} = \mu_0 \int\limits_{1}^{3} \int\limits_{0}^{-5000} J dvdt$$
|
||||
$$195 \times 10^{-9} = -10000J\mu_0$$
|
||||
$$J = {-195 \times 10^{-13} \over \mu_0} \text{[A/m}]$$
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-02.pdf
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BIN
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-02.pdf
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134
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-03.md
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-03.md
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@ -0,0 +1,134 @@
|
||||
# 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);
|
||||
```
|
||||
|
||||
|
||||
|
||||
### b)
|
||||
|
||||
|
||||
|
||||
### c)
|
||||
```matlab
|
||||
% Removed the negative sign on q2
|
||||
q1 = 1.0e-6;
|
||||
q2 = 1.0e-6;
|
||||
```
|
||||
|
||||
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-05.md
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|
||||
# Homework 5 - Aidan Sharpe
|
||||
|
||||
## 1
|
||||
A nonuniform time-varying electric field given in the cylindrical coordinates by:
|
||||
$$\vec{E} = \left(3\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho}\hat{\varphi}\right) \sin(3 \times 10^8t) [\text{V/m}]$$
|
||||
|
||||
The field is applied to the following homogeneous, isotropic dielectric materials:
|
||||
1. Teflon
|
||||
- $\mu_r = 1$
|
||||
- $\varepsilon_r = 2.1$
|
||||
- $\sigma = 0$
|
||||
1. Glass
|
||||
- $\mu_r = 1$
|
||||
- $\varepsilon_r = 6.3$
|
||||
- $\sigma = 0$
|
||||
1. Sea Water
|
||||
- $\mu_r = 1$
|
||||
- $\varepsilon_r = 81$
|
||||
- $\sigma = 4$[S/m]
|
||||
|
||||
### a)
|
||||
Find the polarization vector, the polarization current density, and the polarization charge density for each material.
|
||||
|
||||
$$\vec{P} = \varepsilon_0 \chi_e \vec{E}$$
|
||||
$$\chi_e = \varepsilon_r - 1$$
|
||||
$$\vec{J}_p = {\partial \over \partial t} \vec{P}$$
|
||||
$$\rho_p = - \nabla \cdot \vec{P}$$
|
||||
|
||||
In cylindrical coordinates:
|
||||
$$\nabla = {1\over\rho}{\partial (\rho A_\rho) \over \partial \rho} + {1\over\rho}{\partial A_\varphi \over \partial \varphi} + {\partial A_z \over \partial z}$$
|
||||
|
||||
For teflon:
|
||||
$$\vec{P} = \varepsilon_0 (2.1 - 1) \left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \sin(3 \times 10^8t)$$
|
||||
$$\vec{J}_p = (3\times10^8)\varepsilon_0 (2.1 -1)\left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \cos(3 \times 10^8t)$$
|
||||
$$\nabla \cdot \vec{P} = 1.1 \varepsilon_0 \sin(3\times10^8t)\left[{\partial \over \partial \rho} 9\rho^2\cot(\varphi) + {\partial \over \partial \varphi} {1\over\rho^2}\cos(\varphi)\right]$$
|
||||
$$\rho_p = 1.1 \varepsilon_0 \sin(3 \times 10^8) \left[-18 \rho \cot(\varphi) + {1\over\rho^2} \sin(\varphi)\right]$$
|
||||
|
||||
|
||||
For glass:
|
||||
$$\vec{P} = \varepsilon_0(6.3 - 1) \left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \sin(3 \times 10^8t)$$
|
||||
$$\vec{J}_p = (3\times10^8)\varepsilon_0 (6.3 -1)\left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \cos(3 \times 10^8t)$$
|
||||
$$\nabla \cdot \vec{P} = 5.3 \varepsilon_0 \sin(9\times10^8t)\left[{\partial \over \partial \rho} 9\rho^2\cot(\varphi) + {\partial \over \partial \varphi} {1\over\rho^2}\cos(\varphi)\right]$$
|
||||
$$\rho_p = 5.3 \varepsilon_0 \sin(9 \times 10^8) \left[-18 \rho \cot(\varphi) + {1\over\rho^2} \sin(\varphi)\right]$$
|
||||
|
||||
For sea water:
|
||||
$$\vec{P} = \varepsilon_0(81 - 1) \left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \sin(3 \times 10^8t)$$
|
||||
$$\vec{J}_p = (3\times10^8)\varepsilon_0 (81 -1)\left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \cos(3 \times 10^8t)$$
|
||||
$$\nabla \cdot \vec{P} = 80 \varepsilon_0 \sin(3\times10^8t)\left[{\partial \over \partial \rho} 9\rho^2\cot(\varphi) + {\partial \over \partial \varphi} {1\over\rho^2}\cos(\varphi)\right]$$
|
||||
$$\rho_p = 80 \varepsilon_0 \sin(3 \times 10^8) \left[-18 \rho \cot(\varphi) + {1\over\rho^2} \sin(\varphi)\right]$$
|
||||
|
||||
### b)
|
||||
Find the ratio of the conduction to the displacement currents in the sea water.
|
||||
|
||||
$${4 \over J_p} = {1 \over (6\times10^9) \varepsilon_0\left(9\rho^2 \cot(\varphi) \hat{\rho} + {\cos(\varphi) \over \rho^2}\hat{\varphi}\right) \cos(3 \times 10^8t)}$$
|
||||
|
||||
**NOTE**: I am unsure why the book seems to be taking a slightly different approach to finding $\rho_p$. In the example we did in class, I was also off by a factor of three, so I am likely missing something here. The example we did in class was:
|
||||
$$\vec{P} = k\vec{r} = kr\hat{r}$$ Taking the negative of the dot product with $\nabla = {\partial \over \partial r} \hat{r}$ gives me $-k$ and not $-3k$, as you got. ## 2
|
||||
Three coaxial cylinders separated by two different dielectric are charged as follows:
|
||||
- The inner cylinder of radius $a$ has a positive linear charge density $\rho_{l1}$[C/m].
|
||||
- The middle cylinder of radius $b$ is grounded
|
||||
- The outer cylinder of radius $c$ has negative linear charge density $-\rho_{l2}$[C/m]
|
||||
|
||||
### a)
|
||||
Determine and draw sketches showing the variation of electric flux density and the electric field intensity between the cylinders and outside them.
|
||||
|
||||
For the inner cylinder, the electric field inside is 0, and the flux through a cylindrical surface with length, $z$, and radius, $\rho$, enclosing the cylinder is given by Gauss's law:
|
||||
$$\Phi_E = {Q_\text{enc} \over \varepsilon_0 \varepsilon_r} = {\rho_{l_1} z \over \varepsilon_0 \varepsilon_1} = \int \vec{E} \cdot d\vec{s}$$
|
||||
|
||||
By inspection, the electric field strength through the label of the cylinder is uniform, so the surface integral of the electric field evaluates to:
|
||||
$$E 2\pi \rho z$$
|
||||
|
||||
Solving for the $E$-field strength at the surface yields
|
||||
$$E = {\rho_{l_1} \over 2 \pi \rho \varepsilon_0 \varepsilon_1} : \{a < \rho < b\}$$
|
||||
|
||||
Since the middle cylinder is grounded the electric field is defined to be zero at and around its extent.
|
||||
$$E = 0 : \{b \le \rho \le c\}$$
|
||||
|
||||
Outside the large cylinder, more charge is added, this time $-\rho_{l_2}$. The total electric field here is:
|
||||
$$E = {- \rho_{l_2} \over 2 \pi \rho \varepsilon_0} : \{\rho > c\}$$
|
||||
|
||||
The total electric field strength at a distance, $\rho$ is given by the piecewise function:
|
||||
$$E(\rho) = \begin{cases} {\rho_{l1} \over 2 \pi \rho \varepsilon_0 \varepsilon_1} & a < \rho < b \\ 0 & b \le \rho < c \\ {- \rho_{l_2} \over 2 \pi \rho \varepsilon_0} & \rho > c \\ \end{cases}$$
|
||||
|
||||
Using some dummy values to plot the field strength. $\rho_{l1} = 1.5$, $\rho_{l2} = 1.3$, $\varepsilon_0 =1$, $a=1$, $b=2$, $c=3$, $\varepsilon_1 = 1.2$, $\varepsilon_2 = 1.4$.
|
||||
$$E(\rho) = \begin{cases} {\rho_{l1} \over 2 \pi \rho \varepsilon_0 \varepsilon_1} & a < \rho < b \\ 0 & b \le \rho < c \\ {-\rho_{l_2} \over 2 \pi \rho \varepsilon_0} & \rho > c \\ \end{cases}$$
|
||||
|
||||
### b)
|
||||
Determine the induced surface charge on the middle conductor.
|
||||
$$\sigma(2 \pi b z) = -\rho_{l_1} z$$
|
||||
$$\therefore \sigma = -{\rho_{l_1} \over 2\pi b}$$
|
||||
|
||||
## 3
|
||||
An $N$ turn toroid of rectangular cross section consists of 3 regions. Region 1 with relative permeability $\mu_{r_1} = 3000$, region 2 with $\mu_{r_2} = 1 + 2/\rho$, region 3 is air.
|
||||
|
||||
### a)
|
||||
Find the magnetic field intensity, $\vec{H}$, the magnetic flux density, $\vec{B}$, and the magnetization, $\vec{M}$ in each of the three regions.
|
||||
|
||||
Region 1:
|
||||
$$M_1 = (3000 -1) {N I \over 2\pi \rho} = {2998 N I \over 2\pi \rho}\hat{\varphi}$$
|
||||
$$H_1 = {N_I \over 2\pi \rho}$$
|
||||
$$B_1 = \mu_0{3000 N I \over 2\pi \rho} = {1500 N I \mu_0 \over \rho}\hat{\varphi}$$
|
||||
|
||||
Region 2:
|
||||
$$M_2 = {2\over \rho}{N I \over 2\pi \rho} = {NI \over 2\pi \rho^2}\hat{\varphi}$$
|
||||
$$H_2 = {N_I \over 2\pi \rho}$$
|
||||
$$B_2 = \mu_0\left(1 + {2\over\rho}\right){N I \over 2\pi \rho}\hat{\varphi}$$
|
||||
|
||||
Region 3:
|
||||
$$M_3 = 0\hat{\varphi}$$
|
||||
$$H_3 = {N_I \over 2\pi \rho}$$
|
||||
$$B_3 = \mu_0{N I \over 2\pi \rho}\hat{\varphi}$$
|
||||
|
||||
### b)
|
||||
Find the magnetization current density in region 2:
|
||||
$$J_m = \nabla \times \vec{H} = {1\over\rho}{\partial \vec{H} \over \partial \rho} = -{NI \over \pi\rho^3}$$
|
||||
|
||||
## 4
|
||||
A uniform plane wave is travelling in a medium in the $\hat{x}$ direction. $\lambda = 25$[cm], $v_p= 2 \times 10^8$[m/s], and $\hat{E}$ is polarized in the $\hat{z}$ direction.
|
||||
|
||||
### a)
|
||||
Find the frequency of the wave and $\varepsilon_r$:
|
||||
$$\lambda = {v_p \over f} \therefore f = {v_p \over \lambda} = 8 \times 10^9[\text{s}^{-1}]$$
|
||||
|
||||
$$v_p = {1 \over \sqrt{\mu \varepsilon}}$$
|
||||
$$\mu = \mu_0$$
|
||||
$$\varepsilon = \varepsilon_0 \varepsilon_r$$
|
||||
$$\therefore \varepsilon_r = {1 \over \mu_0 \varepsilon_0 v_p^2} = 2.246$$
|
||||
|
||||
### b)
|
||||
$$\lambda = {2\pi \over \beta} \therefore \beta = 8\pi$$
|
||||
$$\hat{E} = 50e^{j(16\pi \times 10^9-8\pi x)}\hat{z}$$
|
||||
$$\hat{H} = -0.1989e^{j(16\pi \times 10^9 - 8\pi x)} \hat{y}$$
|
||||
|
||||
## 5
|
||||
An electromagnetic plane wave of frequency $f$ is given by:
|
||||
$$\omega = 2\pi f$$
|
||||
$$\hat{E} = E_0e^{j\omega t - \beta z)}\hat{x}$$
|
||||
$$\hat{H} = {E_0 \over \eta} \cos(\omega t - \beta z)\hat{y}$$
|
||||
|
||||
$$\mathcal{E} = -{d\over dt}\mu \int_s \vec{H} \cdot d\vec{s}$$
|
||||
$$\mathcal{E} = \mu {E_0 \over \eta} 2\pi\omega \sin(\omega t - \beta z) = \mu {E_0 \over \eta} 2\pi\omega e^{j(\omega t - \beta z - {\pi \over 2})}$$
|
||||
|
||||
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-05.pdf
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-06.md
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|
||||
# 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 $\eta_0$:
|
||||
$$\eta_0 = \sqrt{\mu_0 \over \varepsilon_0}$$
|
||||
$$\vec{H}(z,t) = -{E_0 \over \eta_0}\cos(\omega t + \beta z)\hat{y}$$
|
||||
|
||||
$$\therefore \vec{H}(z,t) = -{\|\vec{E}(z,t)\| \over \eta_0}\hat{y}$$
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-06.pdf
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-06.pdf
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121
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-08.md
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5th-Semester-Fall-2023/EEMAGS/Homework/Homework-08.md
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@ -0,0 +1,121 @@
|
||||
# Homework 8 - Aidan Sharpe
|
||||
|
||||
## 1
|
||||
A transmission line is terminated with a matched 50$\Omega$ load. The transmitter puts out 100W of power, and the transmission line is 100ft long. What for what value of $\alpha$ will the power loss be 10W over the length of the line?
|
||||
|
||||
$$\alpha = {\ln\left({P_\text{in} \over P_\text{out}}\right) \over l} = {\ln\left({100 \over 90}\right) \over 30.48[\text{m}]} = 0.00345671\left[{\text{np} \over \text{m}}\right]$$
|
||||
|
||||
## 2
|
||||
Evaluate the phase velocity and attenuation constant for a distorionless line, and compare it to a lossless line.
|
||||
|
||||
$$\gamma = \alpha + j\beta = \sqrt{(R + j\omega L)(G + j\omega C)}$$
|
||||
$$\gamma ^2 = (R + j\omega L)(G + j\omega C)$$
|
||||
$$\gamma^2 = \omega^2 LC(j + {R \over \omega L})(j + {G \over \omega C})$$
|
||||
|
||||
For a distorionless line:
|
||||
$${R \over L} = {G \over C}$$
|
||||
$$(j + {R \over \omega L}) = (j + {G \over \omega C})$$
|
||||
So $\gamma^2$ becomes:
|
||||
$$\gamma^2 = \omega^2 L C (j + {R \over \omega L})^2$$
|
||||
Solving for $\gamma$
|
||||
$$\gamma = \omega\sqrt{L C}(j + {R \over \omega L})$$
|
||||
Separating the real and imaginary components:
|
||||
$$\alpha = R \sqrt{C \over L}$$
|
||||
$$\beta = j\omega\sqrt{L C}$$
|
||||
|
||||
For a lossless line, there is no attenuation (by definition). Therefore:
|
||||
$$\alpha = 0$$
|
||||
To actually be able to build this lossless line, $R=0$ and $G=0$.
|
||||
$$\gamma^2 = (R + j\omega L)(G + j\omega C)$$
|
||||
Since $R$ and $G$ are both 0:
|
||||
$$\gamma^2 = (j\omega L)(j \omega C)$$
|
||||
$$\gamma^2 = -\omega^2 LC$$
|
||||
$$\gamma = \sqrt{-\omega^2 LC} = j\omega\sqrt{LC}$$
|
||||
Separating out real and imaginary:
|
||||
$$\alpha = 0$$
|
||||
$$\beta = j\omega\sqrt{LC}$$
|
||||
|
||||
For lossless and distortionless lines, the attenuation constant differs, but the phase constant does not. Since the phase velocity only depends on $\omega$ and $\beta$, $v_p$ is the same for both lossless and distortionless lines.
|
||||
|
||||
$$v_p = {\omega \over \beta} = {-j \over \sqrt{LC}}$$
|
||||
|
||||
## 3
|
||||
$$\hat{Z}_L = 75 + j150 \Omega$$
|
||||
$$f = 2[\text{MHz}]$$
|
||||
$$\omega = 2\pi f = 4\pi \times 10^6$$
|
||||
$$r = 150\left[{\Omega \over \text{km}}\right]$$
|
||||
$$l = 1.4\left[{\text{mH} \over \text{km}}\right]$$
|
||||
$$c = 88\left[{\text{nF} \over \text{km}}\right]$$
|
||||
$$g = 0.8\left[{\mu\text{s} \over \text{km}}\right]$$
|
||||
$$\hat{V}_G = 100 e^{j0^\circ}$$
|
||||
$$z = 100[\text{m}]$$
|
||||
|
||||
$$R = 15[\Omega]$$
|
||||
$$L = 140[\mu\text{H}]$$
|
||||
$$C = 8.8[\text{nF}]$$
|
||||
$$G = 80[n\Omega]$$
|
||||
|
||||
### a)
|
||||
Find $\hat{Z}_0$:
|
||||
$$Z_0 = \sqrt{R + j\omega L \over G + j\omega C} = \sqrt{15 + j(4\pi\times10^6)(140\times10^{-6}) \over 80\times10^{-9} + j(4\pi \times 10^6)(8.8\times10^{-9})}$$
|
||||
$$Z_0 = 126.1324 - j0.5377$$
|
||||
$$\hat{\gamma} = \alpha + j\beta = \sqrt{(R + j\omega L)(G + j\omega C)}$$
|
||||
$$\hat{\gamma} = 0.0595 + j13.9482$$
|
||||
|
||||
### b)
|
||||
Find the input imedance
|
||||
$$Z_\text{in} = Z_0{Z_L + Z_0 \tanh(\gamma z) \over Z_0 + Z_L \tanh(\gamma z)}$$
|
||||
$$Z_\text{in} = -126.1324 + j0.5377$$
|
||||
|
||||
### c)
|
||||
Find the average power delivered
|
||||
$$\alpha = {\ln\left({P_\text{in} \over P_\text{out}}\right) \over z}$$
|
||||
$$\alpha = 0.0595$$
|
||||
$$P_\text{in} = {V_G^2 \over Z_\text{in}} = 79.2802 + j0.3373$$
|
||||
$$P_\text{out} = P_\text{in} e^{-\alpha z} = 0.2073 + j0.0009$$
|
||||
|
||||
## 4
|
||||
$$\Gamma = {Z_L - Z_0 \over Z_L + Z_0}$$
|
||||
|
||||
### a)
|
||||
$$Z_L = 3 Z_0$$
|
||||
$$\Gamma = 0.5$$
|
||||
|
||||
### b)
|
||||
$$Z_L = (2 - j2)Z_0$$
|
||||
$$\Gamma = 0.5385 - j0.3077$$
|
||||
|
||||
### c)
|
||||
$$Z_L = -j2Z_0$$
|
||||
$$\Gamma = 0.6 - j0.8$$
|
||||
|
||||
### d)
|
||||
$$Z_L = 0$$
|
||||
$$\Gamma = -1$$
|
||||
|
||||
## 5
|
||||
|
||||
### a)
|
||||
$$\Gamma = 0.06+j0.24$$
|
||||
|
||||
### b)
|
||||
$$\text{VWSR} = {1 + |\Gamma| \over 1 - |\Gamma|} = 1.657$$
|
||||
|
||||
### c)
|
||||
$$Z_\text{in} = Z_0 {Z_L + Z_0\tanh(\gamma l) \over Z_0 + Z_L\tanh(\gamma l)} 30.50 - j1.09$$
|
||||
|
||||
### d)
|
||||
$$Y_\text{in} = {1 \over Z_in} = 0.03274 + j0.0012$$
|
||||
|
||||
### e)
|
||||
$$0.106\lambda$$
|
||||
|
||||
### f)
|
||||
$$z = -0.106\lambda$$
|
||||
|
||||
## 6
|
||||
$$L = {3\lambda \over 8}$$
|
||||
$$Z_\text{in} = -j2.5$$
|
||||
$$Z_L = {-j2.5 \over 100} = -j0.025$$
|
||||
At ${3\lambda \over 8}$:
|
||||
$$Z_L = j95$$
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Homework/Homework-08.pdf
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23
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.aux
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23
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.aux
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@ -0,0 +1,23 @@
|
||||
\relax
|
||||
\providecommand\hyper@newdestlabel[2]{}
|
||||
\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
|
||||
\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
|
||||
\global\let\oldcontentsline\contentsline
|
||||
\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}}
|
||||
\global\let\oldnewlabel\newlabel
|
||||
\gdef\newlabel#1#2{\newlabelxx{#1}#2}
|
||||
\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
|
||||
\AtEndDocument{\ifx\hyper@anchor\@undefined
|
||||
\let\contentsline\oldcontentsline
|
||||
\let\newlabel\oldnewlabel
|
||||
\fi}
|
||||
\fi}
|
||||
\global\let\hyper@last\relax
|
||||
\gdef\HyperFirstAtBeginDocument#1{#1}
|
||||
\providecommand\HyField@AuxAddToFields[1]{}
|
||||
\providecommand\HyField@AuxAddToCoFields[2]{}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {I}Introduction \& Methods}{2}{section.1}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {II}Results \& Analysis}{2}{section.2}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {III}Conclusions}{2}{section.3}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {IV}Further Comments}{2}{section.4}\protected@file@percent }
|
||||
\gdef \@abspage@last{1}
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.dvi
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.dvi
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.log
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.log
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4
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.out
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4
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.out
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@ -0,0 +1,4 @@
|
||||
\BOOKMARK [1][-]{section.1}{\376\377\000I\000n\000t\000r\000o\000d\000u\000c\000t\000i\000o\000n\000\040\000\046\000\040\000M\000e\000t\000h\000o\000d\000s}{}% 1
|
||||
\BOOKMARK [1][-]{section.2}{\376\377\000R\000e\000s\000u\000l\000t\000s\000\040\000\046\000\040\000A\000n\000a\000l\000y\000s\000i\000s}{}% 2
|
||||
\BOOKMARK [1][-]{section.3}{\376\377\000C\000o\000n\000c\000l\000u\000s\000i\000o\000n\000s}{}% 3
|
||||
\BOOKMARK [1][-]{section.4}{\376\377\000F\000u\000r\000t\000h\000e\000r\000\040\000C\000o\000m\000m\000e\000n\000t\000s}{}% 4
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.pdf
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BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.pdf
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66
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.tex
Normal file
66
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/Lab01.tex
Normal file
@ -0,0 +1,66 @@
|
||||
\documentclass[conference]{IEEEtran}
|
||||
\usepackage[siunitx]{circuitikz}
|
||||
\usepackage{graphicx}
|
||||
%\usepackage{lipsum}
|
||||
\usepackage{color}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{epsfig}
|
||||
\usepackage{mathptmx}
|
||||
\usepackage{times}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{float}
|
||||
\usepackage{hyperref}
|
||||
%\usepackage{setspace}
|
||||
%\usepackage{tikz}
|
||||
\usepackage{circuitikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage{textcomp}
|
||||
%\usepgfplotslibrary{external}
|
||||
%\tikzexternalize
|
||||
%\usepackage[utf8]{inputenc}
|
||||
%\usepackage[english]{babel}
|
||||
%\usepackage{pgf}
|
||||
\usepackage{pgfpages}
|
||||
\usepackage[margin=0.4in]{geometry}
|
||||
\usepackage{lmodern}
|
||||
\usepackage{datetime}
|
||||
\usepackage{ragged2e}
|
||||
\input{border.tex}
|
||||
\pgfpagesuselayout{boxed}
|
||||
|
||||
\hyphenation{op-tical net-works semi-conduc-tor}% correct bad hyphenation here
|
||||
|
||||
\font\myfont=cmr12 at 15pt
|
||||
\title{\myfont Getting Situated with Ansys Electronics Desktop for EEMAGS}
|
||||
\author{Aidan Sharpe 916373346}
|
||||
|
||||
|
||||
\providecommand{\keywords}[1]{\textbf{\textit{Keywords---}} #1}
|
||||
\providecommand{\e}[1]{\ensuremath{\times 10^{#1}}}
|
||||
\setlength{\columnsep}{7mm}
|
||||
\pgfplotsset{compat=1.15}
|
||||
\begin{document}
|
||||
|
||||
\makeatletter
|
||||
\@twocolumnfalse
|
||||
\maketitle
|
||||
\@twocolumntrue
|
||||
\makeatother
|
||||
|
||||
\section{Introduction \& Methods}
|
||||
The primary goals of this lab were to get situated with the \textit{Ansys Electronics Desktop} software and the Rowan Virtual Desktop environment as a whole. Within the \textit{Ansys Electronics Desktop} software, the primary goal was to learn the placements of specific tool bars and widgets by constructing objects in the primary view port and analyzing the electric field around some primitive objects.
|
||||
|
||||
The lab consisted of two parts, with the first acting as a guide to using the basic features of the software, and the second allowing for further exploration into different object primitives and the electrical interactions between them.
|
||||
|
||||
\section{Results \& Analysis}
|
||||
In part one, a perfectly conducting cylinder was charged and centered in a square-based rectangular prism with the same height. The electric field inside the rectangular prism was then analyzed. Since there were no reference voltages, the field lines were un-ordered and fairly chaotic. To resolve this, one of the vertical faces of the prism was defined as $0$V. Then, electric field intensity and direction were analyzed inside the rectangular prism. The resulting electric field, in accordance with predictions, extended from the cylinder to the grounded side of the rectangular prism. On the side of the rectangular prism opposing the grounded side, however, the electric field lines still exhibited some chaotic behavior, similar to the analysis before grounding. While quite simple in theory, this part of the lab still proved to be moderately challenging since some parts of the software were quite unintuitive. After completing part one, however, creating and analyzing objects should be much more straightforward in the future.
|
||||
|
||||
While part one was a guided exercise, part two was much more open-ended. A similar process was to be followed with some other shape. For this we created a sheath-ground coaxial cable. By making a hollow copper cylinder defined as $0$V, and a much skinnier concentric solid copper core with charge $1000$C, the electric field in and around the coaxial cable could be measured.
|
||||
|
||||
\section{Conclusions}
|
||||
While parts of this lab did not directly align with the course material, it definitely served its purpose as a crash course in the basic features of \textit{Ansys Electronics Desktop}. Going forward, using the software will require much less direction, and hopefully less fiddling around with. We found the software to be quite unintuitive to use at times, but we could not find any real suitable alternatives online. While it can be difficult to use at times, using visualization software at all is better than relying on calculations and analogy alone.
|
||||
|
||||
\section{Further Comments}
|
||||
Unfortunately, I was unable to access Ansys to take screenshots for this lab. I have an IRT ticket and they are working it out. Thank you for understanding.
|
||||
\end{document}
|
||||
18
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/border.tex
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18
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-01/border.tex
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@ -0,0 +1,18 @@
|
||||
\pgfpagesdeclarelayout{boxed}
|
||||
{
|
||||
\edef\pgfpageoptionborder{0pt}
|
||||
}
|
||||
{
|
||||
\pgfpagesphysicalpageoptions
|
||||
{%
|
||||
logical pages=1,%
|
||||
}
|
||||
\pgfpageslogicalpageoptions{1}
|
||||
{
|
||||
border code=\pgfsetlinewidth{2pt}\pgfstroke,%
|
||||
border shrink=\pgfpageoptionborder,%
|
||||
resized width=.95\pgfphysicalwidth,%
|
||||
resized height=.95\pgfphysicalheight,%
|
||||
center=\pgfpoint{.5\pgfphysicalwidth}{.5\pgfphysicalheight}%
|
||||
}%
|
||||
}
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/BackAir.png
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BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/BackAir.png
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|
After Width: | Height: | Size: 25 KiB |
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/BackDiamond.png
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BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/BackDiamond.png
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|
After Width: | Height: | Size: 24 KiB |
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/CenterAir.png
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BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/CenterAir.png
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|
After Width: | Height: | Size: 27 KiB |
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/Coax.png
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BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/Coax.png
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|
After Width: | Height: | Size: 23 KiB |
32
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/Lab-02.aux
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32
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/Lab-02.aux
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@ -0,0 +1,32 @@
|
||||
\relax
|
||||
\providecommand\hyper@newdestlabel[2]{}
|
||||
\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
|
||||
\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
|
||||
\global\let\oldnewlabel\newlabel
|
||||
\gdef\newlabel#1#2{\newlabelxx{#1}#2}
|
||||
\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
|
||||
\AtEndDocument{\ifx\hyper@anchor\@undefined
|
||||
\let\newlabel\oldnewlabel
|
||||
\fi}
|
||||
\fi}
|
||||
\global\let\hyper@last\relax
|
||||
\gdef\HyperFirstAtBeginDocument#1{#1}
|
||||
\providecommand\HyField@AuxAddToFields[1]{}
|
||||
\providecommand\HyField@AuxAddToCoFields[2]{}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {I}Introduction \& Methods}{2}{section.1}\protected@file@percent }
|
||||
\newlabel{eqn:capacitance-parallel-plate}{{1}{2}{Introduction \& Methods}{equation.1.1}{}}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {II}Results \& Analysis}{2}{section.2}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox {II-A}}Air as a Dielectric}{2}{subsection.2.1}\protected@file@percent }
|
||||
\newlabel{eqn:efield-line-voltage}{{2}{2}{Air as a Dielectric}{equation.2.2}{}}
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
\@writefile{toc}{\contentsline {subsection}{\numberline {\mbox {II-C}}Changing the Form of the Conductors}{2}{subsection.2.3}\protected@file@percent }
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces Using diamond as a dielectric}}{3}{figure.3}\protected@file@percent }
|
||||
\newlabel{fig:diamond-dielectric}{{3}{3}{Using diamond as a dielectric}{figure.3}{}}
|
||||
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-02/Lab-02.pdf
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||||
|
||||
\font\myfont=cmr12 at 15pt
|
||||
\title{\myfont Simulating Simple Electrostatic Capacitors}
|
||||
\author{Aidan Sharpe 916373346}
|
||||
|
||||
|
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|
||||
\begin{abstract}
|
||||
The primary goal of this lab was to understand the electric fields between differently charged conductors. Since the definition of a capacitor is two separated conductors of differing charge, capacitance was also predicted and measured. Along with a simple parallel plate model, coaxial conductors were also analyzed and found to change the characteristics of the electric field.
|
||||
\end{abstract}
|
||||
|
||||
\section{Introduction \& Methods}
|
||||
The idea of a charged object in a vacuum can be quite helpful when developing simple models and fundamental understandings. Unfortunately, this approach can only be taken so far before a system becomes \textit{oversimplified}. To better understand these complex interactions, more complex models must be made.
|
||||
|
||||
Parallel plate capacitor with square plates of side length 0.5[mm], and spacing 0.1[mm]. The bottom plate was defined as 0[V], and the top plate was given a charge, $5 \times 10^{-6}$[C]. Since the measurements are being taken with air between the plates, and air has a relative permittivity of 1.0006, for air, $\varepsilon = 8.859 \times 10^{-12}$. The capacitance was calculated by plugging the known values for area and distance into equation \ref{eqn:capacitance-parallel-plate}. Doing so yields the theoretical capacitance of $2.215\times 10^{-14}$ [F].
|
||||
|
||||
\begin{equation}
|
||||
C = {\varepsilon A \over d}
|
||||
\label{eqn:capacitance-parallel-plate}
|
||||
\end{equation}
|
||||
Where:\\
|
||||
$C$ is capacitance\\
|
||||
$A$ is the inside area of the plates\\
|
||||
$d$ is the distance between the plates\\
|
||||
$\varepsilon$ is the permittivity of the material between the plates.
|
||||
|
||||
\section{Results \& Analysis}
|
||||
\subsection{Air as a Dielectric}
|
||||
After setting up the capacitor, analysis began. First, as seen in figure \ref{fig:field-air-edge} and figure \ref{fig:field-air-center}, the electric field between the plates was analyzed at the edge of the plates and through the center of the area between the plates. It is immediately apparent that changing the location of measurement inside the region between the plates does not affect the results of the measurement. For this reason, we can determine that the electric field strength between the plates is constant. Additionally, the strength of the electric field outside the region between the plates drops off quickly and is very small above and below the plates.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{BackAir.png}
|
||||
\caption{Measuring electric field strength at the edge}
|
||||
\label{fig:field-air-edge}
|
||||
\end{figure}
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{CenterAir.png}
|
||||
\caption{Measuring electric field strength in the center of the capacitor}
|
||||
\label{fig:field-air-center}
|
||||
\end{figure}
|
||||
|
||||
Since one of the plates was charged and the other was set as a reference ground, there exists some voltage between the two plates. This voltage can be found using equation \ref{eqn:efield-line-voltage}. To calculate this voltage, the electric field strength was measured along a straight line from the center of one plate to the center of the other. The resulting voltage was 196[MV].
|
||||
|
||||
\begin{equation}
|
||||
V = \int\vec{E} \cdot d\vec{l}
|
||||
\label{eqn:efield-line-voltage}
|
||||
\end{equation}
|
||||
|
||||
To find the experimental capacitance, equation \ref{eqn:cap-charge-potential}. Using the predetermined charge of $5 \times 10^{-6}$[C] and the tool evaluated line integral for voltage, the experimental capacitance was determined to be $2.549 \times 10^{-14}$[F]. The assumption of a uniform electric field between the plates is likely the source of the error. Since the plates have thickness, modelling them as plane charges will cause some discrepancy.
|
||||
|
||||
\subsection{Changing the Dielectric}
|
||||
The region surrounding the plates was air in this context, however, changing the material between the plates will change the capacitance. However, changing this material into a conductor will short the plates and effectively ruin the capacitance. For this reason, the material was chosen to be diamond for its insulating and dielectric strengths. Diamond has a relative permittivity of 16.5. Therefore, it should be expected that the capacitance be multiplied by the same factor. Additionally, since capacitance is increasing and charge is remaining constant, the voltage between the plates should reduce by the same factor as well. Sure enough, the evaluated voltage between the plates is 11.86[MV], a factor of 16.53 smaller.
|
||||
|
||||
Similarly, since voltage is related to electric field strength by distance, and the distance is constant, the electric field strength should also decrease. This is verified by the test results shown in figure \ref{fig:diamond-dielectric}. With diamond as a dielectric, the maximum field strength was $1.192 \times 10^{11}$[V/m] compared to $2.004 \times 18^{11}$[V/m] with air as a dielectric.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{BackDiamond.png}
|
||||
\caption{Using diamond as a dielectric}
|
||||
\label{fig:diamond-dielectric}
|
||||
\end{figure}
|
||||
|
||||
\subsection{Changing the Form of the Conductors}
|
||||
Finally, the dielectric was reset to air, and for this test, the parallel plates were replaced with concentric conductors in the shape of a coaxial cable. Since the surface area of the inner conductor is smaller than the surface area of the outer conductor, the electric field cannot be uniform between them. As seen in figure \ref{fig:coax-efield}, the electric field is stronger around the inner conductor, and drops off until the surface of the outer conductor. It is also apparent that there is no electric field beyond the outer conductor, so it has some shielding effects. Additionally, equation \ref{eqn:capacitance-parallel-plate} only applies to parallel plate capacitors, the capacitance of the coaxial conductors was evaluated experimentally.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{Coax.png}
|
||||
\caption{Electric field between coaxial conductors}
|
||||
\label{fig:coax-efield}
|
||||
\end{figure}
|
||||
|
||||
The charge of the inner conductor was set to the same $5 \times 10^{-6}$[C] as before, and the outer conductor was defined as a reference ground. The radius of the inner conductor was 0.05[mm], the inner radius of the outer conductor was 0.1[mm], and the outer radius of the outer conductor was 0.15[mm], leaving a 0.05[mm] gap between the conductors. The line integral evaluated voltage was 248[MV] between the conductors. Again using equation \ref{eqn:efield-line-voltage}, the capacitance was calculated to be $2.016 \times 10^{-14}$[F].
|
||||
|
||||
\section{Conclusions}
|
||||
Using a parallel plate capacitor is quite limiting. Since capacitance is larger with high surface area, getting larger capacitance requires very large plates. Beyond the range of picofarads, other capacitor form factors are preferred for their compactness.
|
||||
|
||||
Additionally, parallel plate capacitors, are superb for creating a uniform electric field. However, they are also create electromagnetic disturbances in the surrounding region. This can be reduced with shielding, but adding this shielding will affect the quality and uniformity of the internal electric field. This behavior is exemplified by the coaxial cable. There are no disturbances surrounding the central conductor, but this comes at the cost of lost uniformity.
|
||||
|
||||
Overall, the primary takeaway is that the form factor makes a large contribution to a capacitors behavior. There are trade-offs to picking one design over another, so application must be considered.
|
||||
\end{document}
|
||||
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\hyphenation{op-tical net-works semi-conduc-tor}% correct bad hyphenation here
|
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||||
\font\myfont=cmr12 at 15pt
|
||||
\title{\myfont Simulating Simple Electrostatic Capacitors}
|
||||
\author{Aidan Sharpe 916373346}
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|
||||
\begin{abstract}
|
||||
This lab focuses on measuring the inductance of a rectangular coil of wire. The coil is modeled to be representative of a standard circuit board coil. The inductance was measured parametrically to understand the impact of trace width.
|
||||
\end{abstract}
|
||||
|
||||
\section{Introduction \& Methods}
|
||||
Lab setup was probably the most time-consuming portion of the process. Before taking any measurements, the model circuit board coil had to be created. This was done by drawing a path for the trace to follow, and then dimension was provided to create a trace with a rectangular cross section.
|
||||
|
||||
Due to a simpler geometry, measuring a coil following a rectangular path is less computationally intensive than measuring one with a circular path. To work with the computational restrictions imposed by the virtual machine's limited resources, the rectangular path was the obvious choice. It also happens that most PCB design packages default to eight trace directions, so rectangular path coils are somewhat common in the real world.
|
||||
|
||||
Every dimension of the coil plays some role in its inductance---its reaction to a changing current. The coil, as seen in figure \ref{fig:coil} has the dimensions shown in table \ref{tbl:coil-dimensions}. Width and height are the width and height of the rectangular section of the coil, terminal is the length of the terminals, via is the vertical via distance for bringing the second terminal back outside the coil. Trace is the width of the trace, thickness is the thickness of the trace, dist is the distance between parallel traces, and pitch is the distance between the centers of parallel traces.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{coil.png}
|
||||
\caption{Final coil design}
|
||||
\label{fig:coil}
|
||||
\end{figure}
|
||||
|
||||
\begin{table}[h!]
|
||||
\caption{Initial coil dimensions}
|
||||
\begin{center}
|
||||
\begin{tabular}{c c c}
|
||||
Dimension & Magnitude & Units \\
|
||||
\hline
|
||||
Width & 4 & mm \\
|
||||
Height & 4 & mm \\
|
||||
Terminal & 700 & $\mu$m \\
|
||||
Via & 200 & $\mu$m \\
|
||||
Trace & 200 & $\mu$m \\
|
||||
Thickness & 50 & $\mu$m \\
|
||||
Dist & 100 & $\mu$m \\
|
||||
Pitch & trace + dist & -- \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\label{tbl:coil-dimensions}
|
||||
\end{table}
|
||||
|
||||
After constructing the coil, a box of air, just larger than the coil, was created. Getting the box to fit properly around the coil took some messing with, but the final positioning and dimensions cab be seen in table \ref{tbl:box-dimensions}. This box of air defined the region over which the magnetic field was analyzed.
|
||||
|
||||
|
||||
\begin{table}[h!]
|
||||
\caption{Air box dimensions}
|
||||
\begin{center}
|
||||
\begin{tabular}{c c c}
|
||||
Dimension & Magnitude & Units \\
|
||||
\hline
|
||||
X Position & -width/2 - trace & -- \\
|
||||
Y Position & -1.3 & mm \\
|
||||
Z Position & -1 & mm \\
|
||||
X Size & 2 $\times$ trace + width & -- \\
|
||||
Y Size & length + trace + terminal & -- \\
|
||||
Z Size & 2 & mm \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\label{tbl:box-dimensions}
|
||||
\end{table}
|
||||
|
||||
\section{Results \& Analysis}
|
||||
A 1[A] current was passed through the coil, and the magnetic field was analyzed inside the box of air. In doing so, the inductance of the coil could be found. Using the coil dimensions in table \ref{tbl:coil-dimensions}, the inductance was found to be 22.822[nH]. Given that it is a small coil with only two windings, a small, but non-zero inductance is expected.
|
||||
|
||||
Analyzing equation \ref{eqn:inductance}, the inductance of the coil is directly proportional to magnetic flux. Therefore, inductance can be increased by either making the loop area larger or by increasing the number of turns of the coil.
|
||||
|
||||
\begin{equation}
|
||||
L = {\Phi_M(i) \over i}
|
||||
\label{eqn:inductance}
|
||||
\end{equation}
|
||||
|
||||
The goal of this lab was to determine the effect that trace width had on inductance. Originally, the trace width was increased from 100[$\mu$m] to 300[$\mu$m] at 100[$\mu$m] intervals to decrease analysis time, but the result of this simulation was too low resolution to make out any distinct patterns. To mitigate this, the range was increased to extend up to 400[$\mu$m] with a finer step value of 10[$\mu$m]. This drastically increased the resolution, and the results can be seen in figure \ref{fig:inductance}.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{inductance.png}
|
||||
\caption{Inductance as it relates to trace width}
|
||||
\label{fig:inductance}
|
||||
\end{figure}
|
||||
|
||||
The magnetic field produced by the coil will add around parallel wires and cancel out between the wires. This means that as the coils get wider, the magnetic field has a smaller area to travel through. The net effects of the current in the coil can be seen in figure \ref{fig:magnetic-field}.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{magnetic-field.png}
|
||||
\caption{Magnetic field around the coil}
|
||||
\label{fig:magnetic-field}
|
||||
\end{figure}
|
||||
|
||||
\section{Conclusions}
|
||||
Looking closer at figure \ref{fig:inductance}, increasing the trace width had a positive effect up until about 230[$\mu$m] before levelling off and trending downwards. Since the coil is made of copper and not a perfect conductor, it has some resistance. Therefore, increasing the trace width will increase cross-sectional area and thereby decrease resistance. This possibly explains why the inductance increases as resistance goes down.
|
||||
|
||||
However, as the traces get wider without increasing the size of the coil, the area of magnetic flux must decrease. This is because the area that contributes the most to the magnetic flux is inside the coil. This combination of behavior could explain the increase and decrease of inductance as a function of trace width.
|
||||
|
||||
Making an inductor on a circuit board therefore poses a few challenges. Firstly, the current must be limited as the trace width impacts inductance. Second, the number of windings is limited by the area given to the coil. Typically, inductors are mounted components that use all three dimensions allotted. This is because they are defined by their geometry and can often be quite difficult to make compact.
|
||||
|
||||
The PCB inductor has its benefits, however. If the required inductance is small, it can be quite easy to put together a PCB inductor to use as a small antenna. Unfortunately, this design takes up a lot of board space and makes the center more or less unusable. For this reason, a better, more compact design should probably be used for most applications.
|
||||
\end{document}
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18
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5653
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-04/Lab-04-Coil-Square.step
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5653
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-04/Lab-04-Coil-Square.step
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-04/Lab-04.FCStd
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-04/Lab-04.FCStd
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789
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-04/Magnet.step
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789
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-04/Magnet.step
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|
||||
\hyphenation{op-tical net-works semi-conduc-tor}% correct bad hyphenation here
|
||||
|
||||
\font\myfont=cmr12 at 15pt
|
||||
\title{\myfont Simulating Simple Electrostatic Capacitors}
|
||||
\author{Aidan Sharpe 916373346}
|
||||
|
||||
|
||||
\providecommand{\keywords}[1]{\textbf{\textit{Keywords---}} #1}
|
||||
\providecommand{\e}[1]{\ensuremath{\times 10^{#1}}}
|
||||
\setlength{\columnsep}{7mm}
|
||||
\pgfplotsset{compat=1.15}
|
||||
\begin{document}
|
||||
\makeatletter
|
||||
\@twocolumnfalse
|
||||
\maketitle
|
||||
\@twocolumntrue
|
||||
\makeatother
|
||||
\begin{abstract}
|
||||
This lab applies Faraday's Law of Induction, which states that a changing magnetic field induces an EMF voltage in a conductor. In this specific case, the changing magnetic field was supplied by a permanent magnet with a constant, non-zero velocity moving through the center of a copper coil. The ends of the coil were connected through a resistor, which allowed the induced voltage to create a current by Ohm's law.
|
||||
\end{abstract}
|
||||
|
||||
\section{Introduction \& Setup}
|
||||
This lab started as far from the simulation software as possible, locally in FreeCAD. While it is possible (and encouraged for most) to model a system directly in the simulation software, it is finicky at best. Furthermore, the documentation for it is almost as proprietary as the software itself. For this reason, it made sense to use an external tool for design purposes. The technical drawings are included after the appendix.
|
||||
|
||||
All components were exported from FreeCAD into the simulation software, positioned properly and the simulation was set up. Currents were measured at the ends of the coil, and the magnet was moved along a path starting below the coil and ending above it. A 100$\Omega$ resistor was placed in series with the coil, so current could be measured through the coil.
|
||||
|
||||
The magnet started 2[mm] below the coil and traveled upwards at 500[mm/s] for 10[ms]. The speed was initially 10[mm/s], but this needed to be increased as the magnet only travelled 0.1[mm]. This was so slow that the magnet did not even reach the bottom of the coil before the simulation ended. 500[mm/s] seemed like the most reasonable speed since it would travel a full 5[mm] over the course of the simulation. At this speed, the magnet travels from -2[mm] to +3[mm], which is above the top of the coil.
|
||||
|
||||
The importance of it travelling through the coil is that the magnetic field is strongest close to the magnet, so it is critical to get the magnet as close as possible to the coil to see the best induction effects. The induced currents are in accordance with Faraday's Law of Induction seen in equation \ref{eqn:faraday-law}. For a coil of wire, the effects of Faraday's law are multiplied scaled up with each winding. The voltage across a coil with $N$ windings and magnetic flux, $\Phi_B$, through the center is given by equation \ref{eqn:solenoid-emf}. Therefore, by Ohm's Law, the current through the inductor is just the voltage given by equation \ref{eqn:solenoid-emf} divided by 100$\Omega$.
|
||||
|
||||
\begin{equation}
|
||||
\nabla \times \vec{E} = -{\partial \vec{B} \over \partial t}
|
||||
\label{eqn:faraday-law}
|
||||
\end{equation}
|
||||
|
||||
\begin{equation}
|
||||
\mathcal{E} = -N {d\Phi_B \over dt}
|
||||
\label{eqn:solenoid-emf}
|
||||
\end{equation}
|
||||
|
||||
\section{Results \& Analysis}
|
||||
Passing the magnet through the coil creates a changing magnetic field from the coil's perspective. From the magnet's perspective field is static, and the coil is moving towards it. Therefore, when the animation is played, the magnetic field appears to move upwards at the same rate as the magnet. The shape of the magnetic field around the magnet, passing through the coils is seen in figure \ref{fig:magnetic-field}. The magnetic field is originating from the magnet, curling back on itself, and then terminating back at the magnet. This makes sense, since all magnets have a north and south pole, with the magnetic field lines travelling between them.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{Magnetic-Field.png}
|
||||
\caption{Magnetic field vectors in the region}
|
||||
\label{fig:magnetic-field}
|
||||
\end{figure}
|
||||
|
||||
As seen in figure \ref{fig:coil-current}, there are two spikes in current. As the magnet approaches the coil, it makes sense that the magnetic field increases rapidly. It also makes sense, that as the magnet leaves the coil, around 9[ms] or so, the current will decrease. However the dropout in the middle, may be unexpected. Faraday's Law of Induction relates voltage to the time rate of change of the magnetic field, and not directly to its strength. Following this line of thought, it must be that the most change in magnetic field strength happens when the magnet is either entering or leaving the coil.
|
||||
|
||||
\begin{figure}[h]
|
||||
\includegraphics[width=0.48\textwidth]{CurrentPlot.png}
|
||||
\caption{Current through the coil over time}
|
||||
\label{fig:coil-current}
|
||||
\end{figure}
|
||||
|
||||
\section{Conclusion}
|
||||
The drop in current while the magnet is in the coil was quite unexpected at first, but after recalling Faraday's Law of Induction, the results made more sense. Since the change in magnetic field is highest when the magnet is approaching or receding from the coil, it makes sense that spikes in current occur here. However, when the magnet is centered in the coil, the magnetic field is symmetric with respect to the coil. For this reason, magnetic field changes very little at this point, and in a perfectly symmetric system, the current would probably go to zero. Thinking back to one dimensional kinematics, at the apex of a projectile being thrown straight into the air, the instantaneous velocity goes to zero.
|
||||
|
||||
Measuring the time rate of change of the magnetic flux is analogous to projectile motion in this way. Think about it as the magnetic field strength reaching a maximum in the coil before decreasing and going to zero at $t=\infty$. Since the strength reaches a maximum, the rate of change of strength is by definition, zero.
|
||||
|
||||
The voltage across the coil can be recovered since the resistor has a known value of 100$\Omega$. Simply by scaling the current results by a factor of 100 will yield the EMF voltage of the coil.
|
||||
Finally, by working backwards from Faraday's Law of Induction, since the number of coils is known to be $N=5$, by taking the negative of the integral of the EMF voltage with respect to time, the magnetic flux through the coil can be extrapolated.
|
||||
|
||||
By knowing, the current, voltage, and magnetic flux of the coil with respect to time, much is learned about the system as a whole, and even more information, such as inductance characteristics, can be derived.
|
||||
\end{document}
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/Lab 6 - Numerical.pdf
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|
||||
# Created by Octave 8.4.0, Mon Nov 27 11:05:07 2023 EST <sharpe@dhcp-150-250-217-15>
|
||||
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/Lab_6_EEMags.pdf
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|
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/Part3-Potential.png
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|
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/Part4-Potential.png
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|
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/Part5-EField.png
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|
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5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/Part5-Potential.png
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|
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129
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/cemLaplace01.m
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129
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/cemLaplace01.m
Normal file
@ -0,0 +1,129 @@
|
||||
% cemLaplace01.m
|
||||
% Solution of Laplace's equation
|
||||
|
||||
clear all
|
||||
close all
|
||||
clc
|
||||
tic
|
||||
|
||||
% INPUTS ================================================================
|
||||
% Number of XY grid points (integer)
|
||||
Nx = 10; Ny = 10;
|
||||
% Range for X and Y values [minX maxX minY maxY minZ maxZ]
|
||||
XY = [0 2 0 1];
|
||||
% tolerance for ending iterations
|
||||
tol = 0.1;
|
||||
% number of iteriations
|
||||
maxN = 50;
|
||||
% limits for electric field plot
|
||||
minXL = 0 ; maxXL = 2;
|
||||
minYL = 0; maxYL = 1;
|
||||
|
||||
% Boundary values for the potential
|
||||
V0 = 10;
|
||||
V = zeros(Ny,Nx);
|
||||
V(:,1) = 20;
|
||||
V(:,end) = 10;
|
||||
V(1,:) = -5;
|
||||
V(end,:) = -10;
|
||||
|
||||
% SETUP ==================================================================
|
||||
|
||||
minX = XY(1); maxX = XY(2);
|
||||
minY = XY(3); maxY = XY(4);
|
||||
x = linspace(minX, maxX,Nx);
|
||||
y = linspace(minY, maxY,Ny);
|
||||
[xx, yy] = meshgrid(x,y);
|
||||
hx = x(2) - x(1); hy = y(2) - y(1);
|
||||
Kx = hx^2/(2*(hx^2+hy^2)); Ky = hy^2/(2*(hx^2+hy^2));
|
||||
|
||||
|
||||
|
||||
% CALCULATIONS ===========================================================
|
||||
|
||||
% dSum difference in sum of squares / n number of iterations
|
||||
dSum = 1; n = 0; cc = 0;
|
||||
|
||||
while n < maxN
|
||||
%while dSum > tol
|
||||
sum1 = sum(sum(V.^2));
|
||||
|
||||
for ny = 2: Ny-1
|
||||
|
||||
for nx = 2: Nx-1
|
||||
V(ny,nx) = Ky * (V(ny,nx+1) + V(ny,nx-1)) + Kx * (V(ny+1,nx) + V(ny-1,nx));
|
||||
% end
|
||||
|
||||
% V(ny,nx) = Ky * (V(ny,nx+1) + V(ny,nx-1)) + Kx * (V(ny+1,nx) + V(ny-1,nx));
|
||||
end
|
||||
% V(ny,Nx) = Ky * (V(ny,Nx) + V(ny,Nx-2)) + Kx * (V(ny+1,Nx) + V(ny-1,Nx));
|
||||
end
|
||||
|
||||
sum2 = sum(sum(V.^2));
|
||||
dSum = abs(sum2 - sum1);
|
||||
n = n + 1; cc = cc+1;
|
||||
end
|
||||
% electric field
|
||||
[Exx, Eyy] = gradient(V,hx,hy);
|
||||
Exx = -Exx; Eyy = -Eyy;
|
||||
E = sqrt(Exx.^2 + Eyy.^2);
|
||||
|
||||
% GRAPHICS ===============================================================
|
||||
|
||||
figure(1) % VECTOR FIELD V -----------------------------------------
|
||||
set(gcf,'units','normalized','position',[0.05 0.5 0.4 0.35]);
|
||||
surf(xx(1:5:end,1:5:end),yy(1:5:end,1:5:end),V(1:5:end,1:5:end));
|
||||
xlabel('x [m]'); ylabel('y [m]'); zlabel('V [ V ]');
|
||||
set(gca,'fontsize',14)
|
||||
rotate3d
|
||||
box on
|
||||
axis tight
|
||||
colorbar
|
||||
view(43,54);
|
||||
|
||||
figure(2)
|
||||
set(gcf,'units','normalized','position',[0.5 0.5 0.4 0.35]);
|
||||
c = 1; yStep = zeros(1,8);
|
||||
for n = 1 : round(Ny/15): round(Ny/2)
|
||||
xP = xx(n,:); yP = V(n,:);
|
||||
plot(xP,yP,'b','linewidth',1.2);
|
||||
hold on
|
||||
c = c+1;
|
||||
yStep(c) = yy(c,1);
|
||||
|
||||
end
|
||||
tt = num2str(yStep,2);
|
||||
text(0.2,95,'y =');
|
||||
text(0.4,95,tt,'fontsize',13)
|
||||
xlabel('x [m]'); ylabel('V [ V ]');
|
||||
set(gca,'fontsize',16)
|
||||
|
||||
figure(3)
|
||||
set(gcf,'units','normalized','position',[0.05 0.05 0.4 0.35]);
|
||||
index1 = 1 : Nx; index2 = 1 : Ny;
|
||||
p1 = xx(index1, index2); p2 = yy(index1, index2);
|
||||
p3 = Exx(index1, index2); p4 = Eyy(index1, index2);
|
||||
h = quiver(p1,p2,p3,p4);
|
||||
set(h,'color',[0 0 1],'linewidth',2)
|
||||
xlabel('x [m]'); ylabel('y [m]');
|
||||
set(gca,'xLim',[minXL, maxXL]);
|
||||
set(gca,'yLim',[minYL, maxYL]);
|
||||
|
||||
figure(4)
|
||||
set(gcf,'units','normalized','position',[0.5 0.05 0.4 0.35]);
|
||||
surf(xx,yy,E);
|
||||
shading interp
|
||||
xlabel('x [m]'); ylabel('y [m]'); zlabel('|E| [ V/m ]');
|
||||
set(gca,'fontsize',14)
|
||||
set(gca,'zTick',[0 1000 2000]);
|
||||
rotate3d
|
||||
box on
|
||||
axis tight
|
||||
colorbar
|
||||
view(43,54);
|
||||
|
||||
|
||||
cc
|
||||
V
|
||||
|
||||
toc
|
||||
204
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/cemLaplace02.m
Normal file
204
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/cemLaplace02.m
Normal file
@ -0,0 +1,204 @@
|
||||
% cemLaplace01.m
|
||||
|
||||
% Solution of Laplace's equation
|
||||
clear
|
||||
clc
|
||||
tic
|
||||
|
||||
% INPUTS ================================================================
|
||||
% Number of XY grid points (integer)
|
||||
Nx = 101; Ny = 101;
|
||||
% Range for X and Y values [minX maxX minY maxY minZ maxZ]
|
||||
XY = [0 4 -1 1];
|
||||
% Boundary potential
|
||||
|
||||
% tolerance for ending iterations
|
||||
tol = 1;
|
||||
% limits for electric field plot
|
||||
minXL = 0; maxXL = XY(2);
|
||||
minYL = 0; maxYL = XY(4);
|
||||
|
||||
% boundary values for the potential
|
||||
V0 = 100; V1 = -5; V2 = -10;
|
||||
V = zeros(Ny,Nx);
|
||||
V(:,1) = V0;
|
||||
% V(:,end) = 0;
|
||||
% V(1,:) = 0;
|
||||
% V(end,:) = 0;
|
||||
|
||||
% linearly increasing potentials on boundaries
|
||||
%enter INPUT details in setup
|
||||
|
||||
|
||||
% SETUP ==================================================================
|
||||
|
||||
minX = XY(1); maxX = XY(2);
|
||||
minY = XY(3); maxY = XY(4);
|
||||
x = linspace(minX, maxX,Nx);
|
||||
y = linspace(minY, maxY,Ny);
|
||||
[xx, yy] = meshgrid(x,y);
|
||||
hx = x(2) - x(1); hy = y(2) - y(1);
|
||||
Kx = hx^2/(2*(hx^2+hy^2)); Ky = hy^2/(2*(hx^2+hy^2));
|
||||
|
||||
% V(:,1) = V0 .* cos(2*pi*y/(4*maxY));
|
||||
|
||||
% linearly increasing potentials on boundaries
|
||||
m1 = 20/maxY; m2 = 20/maxX; m3 = 30/maxY; m4 = 10/maxX;
|
||||
b1 = 0; b2 = 20; b3 = 10; b4 = 0;
|
||||
|
||||
V(:,1) = m1 .* y + b1;
|
||||
V(:,end) = m3 .* y + b3;
|
||||
V(end,:) = m2 .* x + b2;
|
||||
V(1,:) = m4 .* x + b4;
|
||||
|
||||
% CALCULATIONS ===========================================================
|
||||
|
||||
% dSum difference in sum of squares / n number of iterations
|
||||
dSum = 10; n = 0;
|
||||
|
||||
while dSum > tol
|
||||
sum1 = sum(sum(V.^2));
|
||||
|
||||
for ny = 2: Ny-1
|
||||
for nx = 2: Nx-1
|
||||
V(ny,nx) = Ky * (V(ny,nx+1) + V(ny,nx-1)) + Kx * (V(ny+1,nx) + V(ny-1,nx));
|
||||
end
|
||||
% uncomment the next line of code if you need to calculate this column of values
|
||||
% V(ny,Nx) = Ky * (V(ny,Nx) + V(ny,Nx-2)) + Kx * (V(ny+1,Nx) + V(ny-1,Nx));
|
||||
end
|
||||
|
||||
sum2 = sum(sum(V.^2));
|
||||
dSum = abs(sum2 - sum1);
|
||||
n = n+1;
|
||||
dSum
|
||||
end
|
||||
% electric field
|
||||
[Exx, Eyy] = gradient(V,hx,hy);
|
||||
Exx = -Exx; Eyy = -Eyy;
|
||||
E = sqrt(Exx.^2 + Eyy.^2);
|
||||
|
||||
|
||||
% exact solution for boundary conditions
|
||||
% left boundary V0, other boundaries V = 0
|
||||
% Vexact = (2*V0/pi) .* atan(sin(pi*y(51)/maxX)./sinh(pi*x/maxX));
|
||||
% Vexact = (2*V0/pi) .* atan(1./sinh(pi*x/maxX));
|
||||
|
||||
|
||||
% linearly increasing potentials on boundaries
|
||||
m1 = 20/maxY; m2 = 20/maxX; m3 = 30/maxY; m4 = 10/maxX;
|
||||
b1 = 0; b2 = 20; b3 = 10; b4 = 0;
|
||||
|
||||
V(:,1) = m1 .* y + b1;
|
||||
V(:,end) = m3 .* y + b3;
|
||||
V(end,:) = m2 .* x + b2;
|
||||
V(1,:) = m4 .* x + b4;
|
||||
|
||||
% GRAPHICS ===============================================================
|
||||
|
||||
figure(1) % VECTOR FIELD V -----------------------------------------
|
||||
set(gcf,'units','normalized','position',[0.02 0.52 0.3 0.32]);
|
||||
surf(xx(1:5:end,1:5:end),yy(1:5:end,1:5:end),V(1:5:end,1:5:end));
|
||||
xlabel('x [m]'); ylabel('y [m]'); zlabel('V [ V ]');
|
||||
title('potential','fontweight','normal');
|
||||
set(gca,'fontsize',14);
|
||||
rotate3d
|
||||
box on
|
||||
axis tight
|
||||
colorbar
|
||||
view(55,49);
|
||||
% set(gca,'ZTick',[0 5 10]);
|
||||
|
||||
%%
|
||||
figure(2)
|
||||
set(gcf,'units','normalized','position',[0.65 0.52 0.3 0.32]);
|
||||
c = 0; yStep = zeros(1,6);
|
||||
|
||||
for n = 1:10:51
|
||||
xP = xx(n,:); yP = V(n,:);
|
||||
plot(xP,yP,'b','linewidth',1.2);
|
||||
hold on
|
||||
c = c+1;
|
||||
yStep(c) = yy(n,1);
|
||||
|
||||
end
|
||||
tt = num2str(yStep,2);
|
||||
tm1 = 'y = ';
|
||||
tm2 = tt;
|
||||
tm = [tm1 tm2];
|
||||
xlabel('x [m]'); ylabel('V [ V ]');
|
||||
set(gca,'fontsize',16)
|
||||
h_title = title(tm);
|
||||
set(h_title,'fontsize',12,'FontWeight','normal');
|
||||
|
||||
|
||||
% hold on
|
||||
% plot(x,Vexact,'r');
|
||||
|
||||
%%
|
||||
figure(3)
|
||||
set(gcf,'units','normalized','position',[0.65 0.1 0.3 0.32]);
|
||||
|
||||
hold on
|
||||
sx = x(5);
|
||||
for sy = -1 : 0.1 : 1
|
||||
h = streamline(xx,yy,Exx,Eyy,sx,sy);
|
||||
set(h,'linewidth',2,'color',[1 0 1]);
|
||||
end
|
||||
|
||||
|
||||
index1 = 10 : 1: Nx; index2 = 10 : 1 : Ny;
|
||||
p1 = xx(index1, index2); p2 = yy(index1, index2);
|
||||
p3 = Exx(index1, index2); p4 = Eyy(index1, index2);
|
||||
h = quiver(p1,p2,p3,p4);
|
||||
set(h,'color',[0 0 1],'linewidth',2)
|
||||
xlabel('x [m]'); ylabel('y [m]');
|
||||
|
||||
title('electric field','fontweight','normal');
|
||||
hold on
|
||||
axis equal
|
||||
set(gca,'xLim',[minX,maxX]);
|
||||
set(gca,'yLim',[minY,maxY]);
|
||||
set(gca,'xTick',minX:1:maxX);
|
||||
set(gca,'yTick',minY:1:maxY);
|
||||
set(gca,'fontsize',14)
|
||||
box on
|
||||
|
||||
%%
|
||||
figure(4)
|
||||
set(gcf,'units','normalized','position',[0.02 0.1 0.3 0.32]);
|
||||
surf(xx,yy,E);
|
||||
shading interp
|
||||
xlabel('x [m]'); ylabel('y [m]'); zlabel('|E| [ V/m ]');
|
||||
set(gca,'fontsize',14)
|
||||
%set(gca,'zTick',[0 1000 2000]);
|
||||
rotate3d
|
||||
box on
|
||||
axis tight
|
||||
colorbar
|
||||
view(43,54);
|
||||
title('electric field','fontweight','normal');
|
||||
|
||||
figure(5)
|
||||
set(gcf,'units','normalized','position',[0.33 0.52 0.3 0.32]);
|
||||
contourf(xx,yy,V,16);
|
||||
shading interp
|
||||
xlabel('x [m]'); ylabel('y [m]');
|
||||
title('potential','fontweight','normal');
|
||||
set(gca,'fontsize',14)
|
||||
box on
|
||||
colorbar
|
||||
axis equal
|
||||
|
||||
figure(6)
|
||||
set(gcf,'units','normalized','position',[0.33 0.1 0.3 0.32]);
|
||||
contourf(xx,yy,E,32);
|
||||
shading interp
|
||||
xlabel('x [m]'); ylabel('y [m]');
|
||||
title('electric field','fontweight','normal');
|
||||
set(gca,'fontsize',14)
|
||||
box on
|
||||
colorbar
|
||||
axis equal
|
||||
|
||||
|
||||
toc
|
||||
68
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/cemLaplace03.m
Normal file
68
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/cemLaplace03.m
Normal file
@ -0,0 +1,68 @@
|
||||
% cemLaplace01.m
|
||||
|
||||
% Solution of Laplace's equation
|
||||
clear
|
||||
clf
|
||||
tic
|
||||
|
||||
% INPUTS ================================================================
|
||||
|
||||
% Number of XY grid points (integer)
|
||||
Nx = 5;
|
||||
% Range for X and Y values [minX maxX minY maxY minZ maxZ]
|
||||
XY = [0 5];
|
||||
% Boundary potential
|
||||
V0 = 100;
|
||||
% tolerance for ending iterations
|
||||
tol = 0.1;
|
||||
|
||||
% SETUP ==================================================================
|
||||
|
||||
V = zeros(Nx,1);
|
||||
minX = XY(1); maxX = XY(2);
|
||||
x = linspace(minX, maxX,Nx);
|
||||
hx = x(2) - x(1);
|
||||
V(1) = V0; V(Nx) = 0;
|
||||
|
||||
% CALCULATIONS ===========================================================
|
||||
|
||||
% dSum difference in sum of squares / n number of iterations
|
||||
dSum = 10; n = 0;
|
||||
while dSum > tol
|
||||
sum1 = sum(sum(V.^2));
|
||||
|
||||
for nx = 2: Nx-1
|
||||
V(nx) = 0.5*(V(nx+1) + V(nx-1));
|
||||
end
|
||||
sum2 = sum(sum(V.^2));
|
||||
dSum = abs(sum2 - sum1);
|
||||
n = n+1;
|
||||
end
|
||||
|
||||
% exact solution
|
||||
VT = -(V0/maxX)*x + V0;
|
||||
|
||||
% electric field
|
||||
E = -gradient(V,hx);
|
||||
|
||||
% GRAPHICS ===============================================================
|
||||
|
||||
figure(1)
|
||||
set(gcf,'units','normalized','position',[0.05 0.2 0.3 0.4]);
|
||||
subplot(2,1,1)
|
||||
plot(x,V,'b','lineWidth',2);
|
||||
hold on
|
||||
plot(x,VT,'r','lineWidth',2);
|
||||
%set(gca,'xLim',[minX, maxX]);
|
||||
%set(gca,'yLim',[minY, maxY]);
|
||||
xlabel('x [ m ]'); ylabel('V [ V ]');
|
||||
set(gca,'fontsize',14)
|
||||
|
||||
subplot(2,1,2)
|
||||
plot(x,E,'b','lineWidth',2)
|
||||
set(gca,'yLim',[0, 25]);
|
||||
xlabel('x [ m ]'); ylabel('E [ V / m ]');
|
||||
set(gca,'fontsize',14)
|
||||
|
||||
% =======================================================================
|
||||
toc
|
||||
40
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.aux
Normal file
40
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.aux
Normal file
@ -0,0 +1,40 @@
|
||||
\relax
|
||||
\providecommand\hyper@newdestlabel[2]{}
|
||||
\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
|
||||
\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
|
||||
\global\let\oldnewlabel\newlabel
|
||||
\gdef\newlabel#1#2{\newlabelxx{#1}#2}
|
||||
\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
|
||||
\AtEndDocument{\ifx\hyper@anchor\@undefined
|
||||
\let\newlabel\oldnewlabel
|
||||
\fi}
|
||||
\fi}
|
||||
\global\let\hyper@last\relax
|
||||
\gdef\HyperFirstAtBeginDocument#1{#1}
|
||||
\providecommand\HyField@AuxAddToFields[1]{}
|
||||
\providecommand\HyField@AuxAddToCoFields[2]{}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {I}Introduction and Background}{1}{section.1}\protected@file@percent }
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {II}Part 1}{1}{section.2}\protected@file@percent }
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces Approximated and actual electrostatic potential}}{1}{figure.1}\protected@file@percent }
|
||||
\newlabel{fig:part-1}{{1}{1}{Approximated and actual electrostatic potential}{figure.1}{}}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {III}Part 2}{1}{section.3}\protected@file@percent }
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces Electric potential between the plates}}{1}{figure.2}\protected@file@percent }
|
||||
\newlabel{fig:part-2-potential}{{2}{1}{Electric potential between the plates}{figure.2}{}}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {IV}Part 3}{1}{section.4}\protected@file@percent }
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces Electric field strength between the plates}}{2}{figure.3}\protected@file@percent }
|
||||
\newlabel{fig:part-2-efield}{{3}{2}{Electric field strength between the plates}{figure.3}{}}
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Electric field inside the rectangle}}{2}{figure.4}\protected@file@percent }
|
||||
\newlabel{fig:part-3-efield}{{4}{2}{Electric field inside the rectangle}{figure.4}{}}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {V}Part 4}{2}{section.5}\protected@file@percent }
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Potential inside the rectangle}}{2}{figure.5}\protected@file@percent }
|
||||
\newlabel{fig:part-3-potential}{{5}{2}{Potential inside the rectangle}{figure.5}{}}
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {6}{\ignorespaces Electric field inside the new rectangle}}{2}{figure.6}\protected@file@percent }
|
||||
\newlabel{fig:part-4-efield}{{6}{2}{Electric field inside the new rectangle}{figure.6}{}}
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {7}{\ignorespaces Electric potential inside the new rectangle}}{3}{figure.7}\protected@file@percent }
|
||||
\newlabel{fig:part-4-potential}{{7}{3}{Electric potential inside the new rectangle}{figure.7}{}}
|
||||
\@writefile{toc}{\contentsline {section}{\numberline {VI}Part 5}{3}{section.6}\protected@file@percent }
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {8}{\ignorespaces The electric field inside a rectangle with non-constant boundary potential}}{3}{figure.8}\protected@file@percent }
|
||||
\newlabel{fig:part-5-efield}{{8}{3}{The electric field inside a rectangle with non-constant boundary potential}{figure.8}{}}
|
||||
\@writefile{lof}{\contentsline {figure}{\numberline {9}{\ignorespaces The electric potential of a rectangle with non-constant potential}}{3}{figure.9}\protected@file@percent }
|
||||
\newlabel{fig:part-5-potential}{{9}{3}{The electric potential of a rectangle with non-constant potential}{figure.9}{}}
|
||||
\gdef \@abspage@last{3}
|
||||
1402
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.log
Normal file
1402
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.log
Normal file
File diff suppressed because it is too large
Load Diff
6
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.out
Normal file
6
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.out
Normal file
@ -0,0 +1,6 @@
|
||||
\BOOKMARK [1][-]{section.1}{\376\377\000I\000n\000t\000r\000o\000d\000u\000c\000t\000i\000o\000n\000\040\000a\000n\000d\000\040\000B\000a\000c\000k\000g\000r\000o\000u\000n\000d}{}% 1
|
||||
\BOOKMARK [1][-]{section.2}{\376\377\000P\000a\000r\000t\000\040\0001}{}% 2
|
||||
\BOOKMARK [1][-]{section.3}{\376\377\000P\000a\000r\000t\000\040\0002}{}% 3
|
||||
\BOOKMARK [1][-]{section.4}{\376\377\000P\000a\000r\000t\000\040\0003}{}% 4
|
||||
\BOOKMARK [1][-]{section.5}{\376\377\000P\000a\000r\000t\000\040\0004}{}% 5
|
||||
\BOOKMARK [1][-]{section.6}{\376\377\000P\000a\000r\000t\000\040\0005}{}% 6
|
||||
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.pdf
Normal file
BIN
5th-Semester-Fall-2023/EEMAGS/Labs/Lab-06/lab-06.pdf
Normal file
Binary file not shown.
Some files were not shown because too many files have changed in this diff Show More
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Reference in New Issue
Block a user