Rowan-Classes/5th-Semester-Fall-2023/EEMAGS/Homework/Homework-06.md

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}