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How to use Wave equations for uniform plane waves

  1. Oct 2, 2014 #1
    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. jcsd
  3. Oct 2, 2014 #2

    Simon Bridge

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    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
     
  4. Oct 3, 2014 #3
    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?
     
  5. Oct 3, 2014 #4

    Simon Bridge

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    What k turns out to be will depend on the situation.
     
  6. Oct 4, 2014 #5
    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
  7. Oct 4, 2014 #6

    Simon Bridge

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    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
  8. Oct 4, 2014 #7

    Simon Bridge

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    Back to business:
    ... that's what I am thinking also.

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