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Expressions for the energy density of electromagnetic waves

  1. Mar 19, 2012 #1
    I hope this is the right forum for a first year undergraduate problem!
    The problem I've been working on is here: http://i.imgur.com/IhTtL.png

    I think that I have the correct answers, but I'm not sure. I think that the energy densities of the E and B field components will be (0.5)(ε0)(E^2) and (B^2)/(2μ0), that the ratio between the two is just 1/(c^2) and that the characteristic impedance of free space will be (Eμ0)/B.

    Can anyone confirm or deny those results? Thank you.
     
  2. jcsd
  3. Mar 19, 2012 #2

    tiny-tim

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    welcome to pf!

    hi 99wattr89! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yes
    which is … ? :wink:
     
  4. Mar 19, 2012 #3
    Thank you! And thanks for the tip!

    I've been playing about with the result, and tried subbing in the energy equations form the first part of the question - that gives me an expression with μ0, ε0, and the E and B energies. I think I see how to finish it now - the E and B field energies at the same, right? Because if they are then the two cancel, and I get √(μ0/ε0), which I found in a couple of books as an expression for vacuum impedance.

    The only trouble is justifying why the E and B waves have equal energy. I thought that maybe it was because in a loss-less system where E is generating B and B is generating E they would have to have the same energy - but in a wave they could both have been created by some unidentified source, which wouldn't necessarily have to give the same energy to both.
     
    Last edited: Mar 19, 2012
  5. Mar 19, 2012 #4

    tiny-tim

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    i suspect the question wants an answer in terms of µo and c :wink:
    no, the electric field and magnetic field (in an EM wave) have equal strength, not energy

    EDIT: actually, that's not correct, i should say that the strengths are related by Emax = cBmax

    (btw, in a more convenient system, both µo and c would be 1, and both the strengths and the energies would be equal :smile:)
     
    Last edited: Mar 19, 2012
  6. Mar 19, 2012 #5
    I see, so the force created by each is the same, but not the energy carried?

    Double checking, I don't actually get this result though, I get μ0 for the impedance. Could my ratio for the two energies be backwards? I was pretty hazy on how to find the ratio for them so I could easily have that wrong. When I did the E energy over the B energy I got μ0ε0E2/B2, so I thought that meant a ratio of μ0ε0 - is that right? What confuses me is that there are E and B terms in the ratio and I don't know how to get rid of them initially I just ignored them, but I don't think that's right.

    If the ratio of E energy over B energy is 1/μ0ε0, then I get a result of μ0c, which I think is correct, but I can't get that ratio.
     
    Last edited: Mar 19, 2012
  7. Mar 19, 2012 #6

    vela

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    I think the problem is looking for a specific value.

    That's backwards. If the waves have the same energy in one set of units, they'll have the same energy in all sets of units. Using SI, the E and B field amplitudes have different units, so you can't compare them directly, but just looking at the numbers and ignoring the units, you'd see "E">>"B". When you use units where c=1, the fields have the same units, and E=B.
     
  8. Mar 19, 2012 #7

    tiny-tim

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    i got that slightly wrong :redface: … see the EDIT above

    using E2 = c2B2 wil give you the correct result …
    in reply to your earlier question …
    yes, they are sort-of generating each other :smile:

    from Maxwell's equations, we get ∇2E = µoεo2E/∂t2 and ∇2B = µoεo2B/∂t2,

    from which we deduce that there's a wave with speed 1/√(µoεo), and Emax = cBmax, see http://en.wikipedia.org/wiki/Electromagnetic_radiation#Derivation
     
  9. Mar 22, 2012 #8
    Thank you both for all your help! I don't understand the equations listed on that wikipedia page, or how E/B=c comes out of them, but maybe that's more second year material?

    I certainly have the answer at least, so thanks again.
     
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