5th semester files

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