Planes waves produced from a current sheet

[time of 5422]

Homework Statement

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.

Homework Equations

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

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.

 

[mark726]

a) The wave generated by the given current source is a propagating electromagnetic wave with a polarization in the y-direction. It is propagating in both the z and y directions, with a phase shift in the y-direction due to the exponential term in the current source.

b) The electric field is in the x-direction and the magnetic field is in the y-direction.

c) The electric and magnetic fields outside of the current sheet are given by:
\begin{gather*}E_+ = \frac{\eta}{2}J_0e^{-j(k_1 y + k z)} + E_{x0}\\ H_+ = \frac{k}{\omega \mu} \frac{\eta}{2}J_0e^{-j(k_1 y + k z)}\\ E_- = \frac{\eta}{2}J_0e^{+j(k_1 y + k z)} + E_{x0}\\ H_- = -\frac{k}{\omega \mu} \frac{\eta}{2}J_0e^{+j(k_1 y + k z)}\end{gather*}
where \eta = \sqrt{\frac{\mu}{\epsilon}} is the intrinsic impedance of free space.

These equations describe an electromagnetic wave traveling in the z-direction with a phase shift in the y-direction due to the current source. The amplitude of the wave is proportional to the current density and the intrinsic impedance of free space. The electric and magnetic fields are in phase and perpendicular to each other, as expected for an electromagnetic wave.