Planes waves produced from a current sheet

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1. Aug 30, 2016

time of 5422

1. The problem statement, all variables and given/known data
A uniform current sheet in free space is given to be, in phasor form, $$\vec{J}=\vec{x}J_0e^{-jk_1y}$$ located at the $z=0$ plane. $k_1 < k_0$ where $k_0$ is the wavenumber in free space.

a) Describe qualitatively the characteristics of the wave generated by the given current source.
b) Determine the field components involved in this problem.
c) Find the electric and magnetic fields outside of the current sheet.

2. Relevant equations
Maxwell's equations
Boundary conditions
Wavenumber : $k = \frac{\omega}{c}$

3. The attempt at a solution

a) I'm thrown off at this point, I was able to solve part c) but I could not interpret the results. There's a $y$ component in its polarization, when I expected z only. Does this mean that the wave is propagating in both z and y directions?
b) x for electric field, y for magnetic field
c) I managed to solve this

$$(1)\quad \nabla \times \vec{E} = -\hat{x} \frac{\partial E_y}{\partial z} + \hat{y} \frac{\partial E_x}{\partial z} = -j\omega \mu \vec{H}$$
$$(2)\quad \nabla \times \vec{H} = -\hat{x} \frac{\partial H_y}{\partial z} + \hat{y} \frac{\partial H_x}{\partial z} = \vec{J} + j\omega \epsilon \vec{E}$$
Take the derivative of (1) with respect to $\frac{\partial}{\partial z}$ and substitute the results into (2) to give (just the $\hat{x}$ component only):
$$\frac{1}{j\omega \mu} \frac{\partial^2 E_x}{\partial z^2} = J_x + j\omega \epsilon E_x$$
$$\frac{\partial^2 E_x}{\partial z^2} + k^2 E_x= j \omega \mu J_x$$

The case when $z=0$ is:
$$E_{x0}=\frac{j}{\omega \epsilon} J_0 e^{-j k_1 y}$$

The solution of the 1D wave equation is of the form:
$$\begin{gather*}E_+ = Ae^{-j k z} + B e^{+j k z} + E_{x0} \\E_- = Ce^{-j k z} + D e^{+j k z} + E_{x0}\end{gather*}$$
$B = C = 0$ since the field is propagating outwards.

Since $H_y = \frac{j}{\omega \mu} \frac{\partial E_x}{\partial z}$, the magnetic fields are:
$$\begin{gather*}H_+ = \frac{k}{\omega \mu} A e^{-j k z}\\ H_- = -\frac{k}{\omega \mu} D e^{+j k z} \end{gather*}$$
Using the boundary conditions:
$$\begin{gather*}\hat{z} \times (\vec{E_+} - \vec{E_-}) = 0 |_{z = 0}\\ \hat{z} \times (\vec{H_+} - \vec{H_-}) = \vec{J}|_{z=0}\end{gather*}$$
which gives $A=D$ and $A=-\frac{\omega \mu}{2k}J_0 e^{-j k_1 y}$ where we get:
$$\begin{gather*}z>0 \quad E_+ = \frac{\eta}{2}J_0e^{-j(k_1 y + k z)} + E_{x0}\\z=0 \quad E_{x0}=\frac{j}{\omega \epsilon} J_0 e^{-j k_1 y}\\ z <0 \quad E_- = \frac{\eta}{2}J_0e^{+j(k_1 y + k z)} + E_{x0}\end{gather*}$$ and the rest follows.

Last edited: Aug 30, 2016
2. Sep 4, 2016