5th semester files

5th-Semester-Fall-2023/EEMAGS/Homework/H0306a.m

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+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);

5th-Semester-Fall-2023/EEMAGS/Homework/H04.m

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+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);

5th-Semester-Fall-2023/EEMAGS/Homework/Homework-01.md

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+# 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

5th-Semester-Fall-2023/EEMAGS/Homework/Homework-02.md

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+# 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:
+$$195 \times 10^{-9} = \mu_0 \int\limits_{1}^{3} \int\limits_{0}^{-5000} J dvdt$$
+$$195 \times 10^{-9} = -10000J\mu_0$$
+$$J = {-195 \times 10^{-13} \over \mu_0} \text{[A/m}]$$

5th-Semester-Fall-2023/EEMAGS/Homework/Homework-03.md

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+# Homework 3 - Aidan Sharpe
+
+## 1
+An particle with charge $-e$ has velocity $\vec{v} = -v\hat{y}$. An electric acts in the $\hat{x}$ direction. What direction must a $\vec{B}$ field act for the net force on the particle to be 0?
+
+$$\vec{F}_{net} = \vec{F}_E + \vec{F}_B = 0$$
+$$\therefore \vec{F}_B = -\vec{F}_E$$
+$$\vec{F}_E = q\vec{E} = (-)(\hat{x}) = -\hat{x}$$
+$$\vec{F}_B = q\vec{v} \times \vec{B}$$
+$$(-)\begin{vmatrix} 
+\hat{x} & \hat{y} & \hat{z} \\
+0 & (-) & 0 \\
+B_x & B_y & B_z \\
+\end{vmatrix} \overset{!}{=} \hat{x}$$
+$$-[(-B_z)\hat{x} - (0)\hat{y} + (0-(-)B_x)\hat{z})] \overset{!}{=} \hat{x}$$
+$$\therefore \hat{a}_B = \hat{z}$$
+
+## 2
+Consider a plane wave in free space with electromagnetic field intensity:
+$$\hat{E}e^{j\omega t} = 30\pi e^{j(10^8t + \beta z)}\hat{x}$$
+$$\hat{H}e^{j\omega t} = H_m e^{j(10^8t + \beta z)}\hat{y}$$
+
+Find the direction of propagation and the values for $H_m$, $\beta$, and the wavelength.
+
+Since $\beta z$ is added to $\omega t$, propagation is in the $-\hat{z}$ direction.
+
+$${E_x \over H_y} = \mu_0 c$$
+$$E_x = 30\pi$$
+$$H_y = H_m$$
+$$\therefore H_m = {30\pi \over \mu_0 c}$$
+$$\boxed{H_m = 0.25}$$
+
+$$\beta = \omega \sqrt{\mu \varepsilon}$$
+For free space:
+$$\beta = \omega \sqrt{\mu_0 \varepsilon_0}$$
+$$\omega = 10^8$$
+$$\boxed{\beta = 0.334}$$
+$$\lambda = {c \over f}$$
+$$f = {\omega \over 2\pi} = {5 \over \pi} \times 10^7$$
+$$\lambda = {3 \times 10^8 \over {5 \over \pi} \times 10^7} = {30\pi \over 5}$$
+$$\boxed{\lambda = 6\pi\text{[m]}}$$
+
+## 3
+A uniform electric field has intensity:
+$$\vec{E} = 15\cos\left(\pi \times 10^8t +{\pi \over 3}z\right)\hat{y}$$
+
+The $\vec{E}$ field is polarized in the $\hat{y}$ direction.
+
+The wave will propagate in the $-\hat{z}$ direction.
+
+$$f = {\pi \times 10^8 \over 2\pi} = 5 \times 10^7 \text{[s}^{-1}]$$
+
+$$\lambda = {3 \times 10^8 \over 5 \times 10^7} = 6 \text{[m]}$$
+
+$$H_x = {15 \over \mu_0 c} = 0.0398$$
+
+$$\vec{H} = 0.0398\cos\left(\pi \times 10^8 t + {\pi \over 3}z\right)\hat{x}$$
+
+## 4
+### a)
+The properties of a uniform basic plane wave in free space are:
+
+Polarization, amplitude, angular frequency, and the direction of propagation. All other properties, such as wavelength, and the spatial frequency, $\beta$, can be derrived.
+
+### b)
+A uniform plane wave in free space is propagating in the $\hat{z}$ direction. If the wavelength is $\lambda = 3$[cm].
+
+$$f = {c \over \lambda} = {3 \times 10^8 \over 3 \times 10^{-2}} = 10^{10}\text{[s}^{-1}]$$
+$$\beta = 2\pi f \sqrt{\mu_0 \varepsilon_0} = 209.613$$
+
+The amplitude of the x-polarized $\vec{E}$-field is:
+$$\hat{E}_m = 200e^{j {\pi \over 4}}$$
+
+Real-time $\vec{E}$-field:
+$$\vec{E}(z, t) = 200\cos\left(2\pi \times 10^{10}t - 209.613z + {\pi \over 4}\right)\hat{x}$$
+
+Phasor $\vec{H}$-field:
+$$\hat{H} = 0.53e^{j({\pi \over 4} - 209.613z)}\hat{y}$$
+
+Real-time $\vec{H}$-field:
+$$\vec{H}(z, t) = 0.53\cos\left(2\pi \times 10^{10}t - 209.613z + {\pi \over 4}\right)\hat{y}$$
+
+## 5
+A 25[cm] by 25[cm] circuit in the y-z plane grows in the $\hat{y}$ direction by 10[m/s]. The circuit contains a 5-ohm resitor. Find the current through the circuit if it is placed in a uniform $\vec{B}$ field of $-0.5\hat{x}$[T].
+
+By Ohms law:
+$$V = IR$$
+$$\mathcal{E} = -{d \over dt} \int \vec{B} \cdot d\vec{s}$$
+$$\mathcal{E} = -{d B A(t) \over dt}$$
+$$B A(t) = -0.5 \times 0.25(0.25 + 10t) = -1.25t - 0.03125$$
+$$\mathcal{E} = -{d B A(t) \over dt} = 1.25 \text{[V]}$$
+$$I = {V \over R} = 250\text{[mA]}$$
+
+Since the magnetic flux inside the loop is increasing, and the magnetic field induced by the current must counteract the magnetic field inducing the current, the current must flow counter-clockwise around the loop.
+
+## 6
+### a)
+```matlab
+clear all;
+k = 9e9;
+q1 = 1.0e-6;
+q2 = 1.0e-6;
+ax = 1.0;
+ay = 0;
+bx = -1.0;
+by = 0;
+
+[X, Y] = meshgrid(-2:0.9:2,-2:0.9:2);
+
+V = 1./sqrt((X - ax).^2 + (Y-ay).^2 ) * k * q1 + 1./sqrt((X-bx).^2 + (Y-by).^2) * k * q2;
+
+surfc(X, Y, V);
+
+[Ex, Ey] = gradient(-V, 0.2, 0.2);
+figure
+contour(X, Y, V);
+
+hold on;
+quiver(X, Y, Ex, Ey);
+```
+
+![](H0306a.png)
+
+### b)
+
+![](H0306b.png)
+
+### c)
+```matlab
+% Removed the negative sign on q2
+q1 = 1.0e-6;
+q2 = 1.0e-6;
+```
+![](H0306c.png)

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:
+$$\omega = 2\pi f$$
+$$\hat{E} = E_0e^{j\omega t - \beta z)}\hat{x}$$
+$$\hat{H} = {E_0 \over \eta} \cos(\omega t - \beta z)\hat{y}$$
+
+$$\mathcal{E} = -{d\over dt}\mu \int_s \vec{H} \cdot d\vec{s}$$
+$$\mathcal{E} = \mu {E_0 \over \eta} 2\pi\omega \sin(\omega t - \beta z) = \mu {E_0 \over \eta} 2\pi\omega e^{j(\omega t - \beta z - {\pi \over 2})}$$
+

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 $\eta_0$:
+$$\eta_0 = \sqrt{\mu_0 \over \varepsilon_0}$$
+$$\vec{H}(z,t) = -{E_0 \over \eta_0}\cos(\omega t + \beta z)\hat{y}$$
+
+$$\therefore \vec{H}(z,t) = -{\|\vec{E}(z,t)\| \over \eta_0}\hat{y}$$

5th-Semester-Fall-2023/EEMAGS/Homework/Homework-08.md

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+# 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 ${3\lambda \over 8}$:
+$$Z_L = j95$$