132 lines
5.0 KiB
Markdown
132 lines
5.0 KiB
Markdown
# Homework 6 - Aidan Sharpe
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## Exercise 1
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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, $I$.
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### a)
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Find $\vec{H}$ inside the first rod, between the first and second rod, and outside the second rod.
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For a solenoid regardless of magnetic materials:
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$$\vec{H} = {I n \over d}\hat{z}$$
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So for all regions:
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$$\vec{H} = {I n \over d}\hat{z}$$
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### b)
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Find $\vec{B}$ inside the three regions.
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$\vec{B}$ relies on magnetic material properties:
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$$\vec{B} = \mu \vec{H}$$
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Therefore:
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$$\vec{B} = \begin{cases} 2\mu_0\vec{H} & \text{In region 1}\\ 3\mu_0\vec{H} & \text{In region 2}\\ \mu_0\vec{H} & \text{In region 3} \end{cases}$$
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### c)
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$\vec{M}$ in each region:
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By definition:
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$$\vec{M} = \chi_m \vec{H}$$
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$$\chi_m = \mu_r - 1$$
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Therefore:
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$$\vec{M} = \begin{cases} \vec{H} & \text{In region 1} \\ 2\vec{H} & \text{In region 2} \\ 0 & \text{In region 3} \end{cases}$$
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### d)
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$\vec{J}$ in each region:
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By definition:
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$$\vec{J}_m = \nabla \times \vec{M}$$
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Since $\vec{M}$ is both conservative and solenoidal inside the solenoid, for all three regions:
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$$\vec{J}_m = 0$$
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## Exercise 2
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In a nonmagnetic medium:
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$$E = 4\sin(2\pi \times 10^7 t - 0.8x)\hat{z}$$
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### a)
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Find $\varepsilon_r$ and $\eta$:
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The general form:
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$$\vec{E}(x,t) = E_0\cos(\omega t -\beta x)\hat{z}$$
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$$\beta = \omega \sqrt{\mu \varepsilon} = 0.8$$
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$$0.8 = 2\pi\times10^7 \sqrt{\mu_0 \varepsilon_0 \varepsilon_r}$$
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$${{\left(0.8 \over 2\pi\times10^7\right)^2} \over \mu_0 \varepsilon_0} = \varepsilon_r$$
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$$\boxed{\varepsilon_r = 14.566}$$
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$$\eta = \sqrt{\mu \over \varepsilon} = \sqrt{\mu_0 \over \varepsilon_0 \varepsilon_r}$$
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$$\boxed{\eta = 98.275}$$
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### b)
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Find the average poynting vector
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$$\vec{P}_\text{avg} = {E_0^2 \over 2 \eta}\hat{x} = {4^2 \over 2(98.275)}$$
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$$\vec{P}_\text{avg} = 0.081$$
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## Exercise 3
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$$E = 3\sin(2\pi\times10^7t - 0.4\pi x)\hat{y} + 4\sin(2\pi\times10^7t - 0.4\pi x)\hat{z}$$
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### b)
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$$\varepsilon_r = {\left({\beta \over \omega}\right)^2 \over \mu_0 \varepsilon_0} = {\left({0.4\pi\over 2\pi\times10^7}\right)^2 \over \mu_0 \varepsilon_0}$$
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$$\boxed{\varepsilon_r = 35.941}$$
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### a)
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$$\lambda = {v_p \over f}$$
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$$v_p = {1 \over \sqrt{\mu \varepsilon}} = {1 \over \sqrt{\mu_0 \varepsilon_0 \varepsilon_r}} = 5\times10^7$$
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$$f = {\omega \over 2\pi} = 10^7$$
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$$\boxed{\lambda = 5\text{[m]}}$$
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### c)
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$$H = {E_0 \over \eta}\sin(\omega t - \beta x)\hat{z} + {E_0' \over \eta}\sin(\omega t - \beta x)\hat{y}$$
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$$\eta = \sqrt{\mu_0 \over \varepsilon_0 \varepsilon_r} = 62.85$$
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$$\boxed{H = 0.0477\sin(2\pi\times10^7t - 0.4\pi x)\hat{z} + 0.0636\sin(2\pi\times10^7t -0.4\pi x)\hat{y}}$$
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## Exercise 4
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Prove that:
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$$\hat{H}_y = - {\hat{E}_x \over \eta_0}$$
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Starting with a genering electric plane wave in the $-\hat{z}$ direction:
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$$\vec{E}(z, t) = E_0 \cos(\omega t + \beta_0 z)\hat{x}$$
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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:
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$$\nabla \times \vec{E} = -{\partial \over \partial t}\vec{B}$$
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Converting from $\vec{B}$ to $\vec{H}$:
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$$\nabla \times \vec{E} = -\mu{\partial \over \partial t} \vec{H}$$
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Assuming free space:
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$$\mu = \mu_0$$
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This gives the final form for Faraday's Law:
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$$\nabla \times \vec{E} = -\mu_0 {\partial \over \partial t} \vec{H}$$
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Evaluating the curl of $\vec{E}$:
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$$\nabla \times \vec{E} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ \|\vec{E}\| & 0 & 0 \end{vmatrix} = {\partial \over \partial z}\|\vec{E}\|\hat{y} - {\partial \over \partial y}\|\vec{E}\|\hat{z}$$
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Since $\vec{E}$ only varies with respect to $z$ and $t$ this can be rewritten as:
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$$\nabla \times \vec{E} = {\partial \over \partial{z}}\|\vec{E}\|\hat{y}$$
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Evaluate the partial derivative:
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$$\nabla \times \vec{E} = -E_0\beta_0\sin(\omega t + \beta_0 z)\hat{y}$$
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Back to Faraday's Law:
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$$-E_0\beta_0\sin(\omega t + \beta_0 z)\hat{y} = -\mu_0{\partial \over \partial t} \vec{H}$$
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Divide out by $-\mu_0$:
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$${\partial \over \partial t}\vec{H} = {E_0 \beta_0 \over \mu_0}\sin(\omega t + \beta_0 z)\hat{y}$$
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Expand out $\beta_0$:
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$$\beta_0 = \omega \sqrt{\mu_0 \varepsilon_0}$$
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$${\partial \over \partial t}\vec{H} = {E_0 \omega \sqrt{\mu_0 \varepsilon_0} \over \mu_0}\sin(\omega t + \beta_0 z)dt\hat{y}$$
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Integrate both sides with respect to $t$:
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$$\vec{H}(z,t) = {E_0 \omega \sqrt{\mu_0 \varepsilon_0} \over \mu_0}\int\sin(\omega t + \beta_0 z)dt \hat{y}$$
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Evaluate the integral:
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$$\vec{H}(z,t) = -{E_0 \omega \sqrt{\mu_0 \varepsilon_0} \over \mu_0 \omega} \cos(\omega t + \beta_0 z)\hat{y}$$
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Simplifying the fraction:
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$$\vec{H}(z,t) = -E_0 \sqrt{\varepsilon_0 \over \mu_0} \cos(\omega t + \beta_0 z)\hat{y}$$
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Using the definition of $\eta_0$:
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$$\eta_0 = \sqrt{\mu_0 \over \varepsilon_0}$$
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$$\vec{H}(z,t) = -{E_0 \over \eta_0}\cos(\omega t + \beta z)\hat{y}$$
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$$\therefore \vec{H}(z,t) = -{\|\vec{E}(z,t)\| \over \eta_0}\hat{y}$$
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