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Electromagnetism, Poynting vector

  1. May 10, 2012 #1
    3 (a) ii) A time-dependent electric field in vacuum is given by
    ⃗E= E0(0, 0, sin(ky − ωt))
    where E0 is a constant.
    Derive an expression for the corresponding magnetic field ⃗B. [7]
    Using curl E=-dB/dt
    I end up with B=(E0/c)sin(ky-wt)
    Show that both ⃗E and ⃗B are perpendicular to the wave vector ⃗k = (0, k, 0). [2]
    Using the dot product i found that:
    k.E=(0x0,0xk,0x ⃗E)=0
    k.B=(0x ⃗E/c,0xk,0x0)=0
    What is the Poynting vector for this wave? [4]

    N=(ExB)/μ0μr

    Then we get:

    (1/μ0μr)((E0 2/c)sin(ky-wt)2,0,0)
    Finally giving:
    (-1/μ0.μr.c)(E02Sin(ky-wt)2)

    I think the first two parts are right but have no idea if im doing the right thing on the last part. I have used the cross product between E and B and got my final answer for part 3. Thanks for any help in advance!
     
  2. jcsd
  3. May 13, 2012 #2

    rude man

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    Gold Member

    Basically I see nothing wrong with what you did (except for possibly a sign error.)
    But I'm not sure you took the dot-products right.
    The E field has a z component so the B field has an x component & obviously each of these when dotted into the y direction of propagation gives zero. Is that what you wrote?
     
  4. May 13, 2012 #3
    Erm, i think that's what i did. i knew that the dot products would give 0 as each one is along a different axis. Thanks for the reply btw, its good to know that i'm going along the right line.
     
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