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