How to use Wave equations for uniform plane waves

1. Oct 2, 2014

ycgoat

1. The problem statement, all variables and given/known data
Home work 3 Q1

Study the E
field in free space and a source-free region, E= (a + b)exp(-jkx), where a and b are nonzero real constants, and in the x,y plane respectively.
Does it satisfy Maxwell’s equations? If so, find the k and H fields . If not, explain why not.

2. Relevant equations
wave equation - dell^2 E + omega^2 mu epsilon E =0
Faraday's law - curl E = -j omega mu H
dispersion relation - k^2 = omega^2 mu epsilon

3. The attempt at a solution

Please excuse my lack of understanding I am a bit old for trying to get this degree.

curl E = -bjkexp(-jkx) in the z direction,
= -j omega mu H
so H = [(b/(omega mu))exp(-jkx)]z = [(b/omega mu)cos(omeg t - kx)]z

for finding k using the wave formula I think I am to curl E twice and get
[-bk^2exp(-jkx)]z + k^2 ([a exp(-jkx)]x + [b exp(-jkx)]y) = 0
= exp(-jkz) ([k^2 a ]x + [k^2 b ]y + [-bk]z) = 0
then ([k^2 a ]x + [k^2 b ]y + [-bk]z) = 0
I feel like I am beating a dead horse here.

2. Oct 2, 2014

Simon Bridge

Your equation is $$\vec E = (a\hat\imath + b\hat\jmath)e^{-jkx}$$ ... is that correct?

The first step is to see if the equation satisfies Maxwell's equations.
So you start by listing Maxwell's equations - the differential form is probably easiest.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/maxeq.html

3. Oct 3, 2014

ycgoat

Thanks, yes that is the equation. I will work on those after work. So is $k^2=μ\varepsilon\omega^2$ always the solution for k or does it vary with E and H?

4. Oct 3, 2014

Simon Bridge

What k turns out to be will depend on the situation.

5. Oct 4, 2014

ycgoat

So back to the original problem $\underline{E}=(\hat{x}a+\hat{y}b)exp(-jkx)$

to start solving Maxwells equations
$\nabla x \underline{E} = j \omega\mu\underline{H}$
I get $\nabla x \underline{E} = [-jbk exp(-jkx)]\hat{z}$
So $[-jbk exp(-jkx)\hat{z} = j\omega\mu\underline{H}$

solving for H gives me $\underline{H} = [-1/(\omega\mu)]kbexp(-jkx)]\hat{z}$

assuming this is correct so far I now solve $\nabla x \underline{H}$
and I get $[(1/(\omega\mu))]k^2 bexp(-jkx)]\hat{y}$
which should $= j\omega\varepsilon\underline{E}$
but solving for $\underline {E}$ I get $\underline{E}= [(1/(\omega^2 \mu\varepsilon))k^2bexp(-jkx)]\hat{y}$
If $k^2 =\omega^2\mu\varepsilon$ then $\underline{E}=[bexp(-jkx)]\hat{y}$
so for this to be true a would have to equal 0, a=0, but the original problem states that a and b are non zero real constants so this can not exist.

does this appear correct?

Last edited: Oct 4, 2014
6. Oct 4, 2014

Simon Bridge

Just some notes on typography first:

Are you using an underline to indicate a vector?
Try using "\vec" instead, or "\mathbf" ...

the exponential function \exp[-jkx] gets $\exp [-jkx]$ ... in general, putting a backslash in front of a function name will (probably) typeset the function properly.

The cross product sign is \times so $$\vec\nabla\times\vec E = \frac{\partial}{\partial t}\vec B$$

aside: the tilde under a letter as in $\underset{\sim}{E}$ is a typographical notation that tells the typesetter to "make this character boldface". So it would get printed as $\mathbf{E}$. The more modern text for a vector is $\vec E$

Last edited: Oct 4, 2014
7. Oct 4, 2014

Simon Bridge

... that's what I am thinking also.

I got there from $\vec\nabla\cdot\vec E = 0$