Electromagnetic field, finding beta.

[jfierro]

Hi, I am not sure whether this falls into a math category or here, I hope you can help me.
I came across a problem in a book [bare with me, please] (Fundamentals of engineering electromagnetics by Cheng) and asked my electromagnetics profesor, but his response did not help me into getting to the result from the book, I later found that my profesor may have commited a mistake, this is the problem from the book:

6.20.- Given

in the air, determine

[PLAIN]http://latex.codecogs.com/gif.latex?0dpi[/URL] \textbf{H}(x, z; t) \ and \ \beta

The thing is, my electromagnetics profesor insists in saying that for this problem:

[PLAIN]http://latex.codecogs.com/gif.latex?0dpi[/URL] \beta = \omega\sqrt{\mu\epsilon} \approx 62.83

But I think this would apply only if it were only a function of z and not x for this particular wave. The method described in the book we are using consists on applying the following maxwell equations twice and then compare the resulting E field with the original E field so that we can obtain beta:

[PLAIN]http://latex.codecogs.com/gif.latex?0dpi[/URL] \\ \nabla \times \textbf{E} = -j\omega \mu \textbf{H} \\ \nabla \times \textbf{H} = \textbf{J} + j\omega \epsilon \textbf{E}

Following this method I get:

[PLAIN]http://latex.codecogs.com/gif.latex?0dpi[/URL] \beta^2 = \omega^2 \mu\epsilon} - 100\pi^2 \approx 54.41^2

Which is the result from the book.

My profesor goes as far as to say this method is invalid because is as though we were using a single equation to solve a 2 variable system...

Any ideas on how to refute him?

 

[Born2bwire]

I would say your way is correct. We can simply look at the vector wave equation for the electric field in a source free space.

\nabla^2 \mathbf{E} = \mu\epsilon\frac{\partial^2 \mathbf{E}}{\partial t^2}
0.1[-100\pi^2\sin(10\pi x)]\cos(\omega t-\beta z)+0.1\sin(10\pi x)[-\beta^2\cos(\omega t-\beta z)] = \mu_0\epsilon_00.1\sin(10\pi x)[-\omega^2 \cos(\omega t-\beta z)]
100\pi^2+\beta^2 = \mu_0\epsilon_0\omega^2
\beta = \sqrt{\mu_0\epsilon_0\omega^2-100\pi^2}

The way to look at this problem is that you have a standing wave in the x direction times a standing wave in the z direction. In other words, you have plane waves traveling in both the x and z directions at the speed of c. This is why the \beta must be less than the freespace \beta, part of the wave is traveling in the x direction and \beta is only describing the movement of the wave in the z direction.

 

[jfierro]

Thanks for you explanation Born2bwire, again :). This is a more intuitive response which may help me. Another thing I was thinking to do is to plug the \beta my professor says is correct into the resulting H field and then see if it complies with the Maxwell equations, that may disprove it.

Although today he got to a 'the authors of those books are wrong' phase...

 

[jfierro]

Also Born2bwire, what do you mean with freespace when you say

"This is why the \beta must be less than the freespace \beta"

The problem states the wave is traveling in the air and air's \epsilon and \mu are the same as for free space aren't they?

 

[Born2bwire]

I mean that the \beta here is not the actual wave number. The wave number encompasses the phase progression of the wave in space. But in this problem, we have separated the wave equation into different parts, one part for the x direction and another part for the z direction. So the wave number is now shared between these two parts, you should not expect one part to have a "wave number" equal to the total wave number in this medium. Instead, you have the wave number along the x-axis and z-axis, not unlike splitting up a velocity vector into its vector components along desired coordinate axes.

\beta here is an unfortunate choice of variables as k or \beta are typically chosen to denote the total wave number. If we are denoting the component of the wave number along a coordinate axis, we usually write them as k_x or \beta_\rho. That is the main reason why my statement looks weird.