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Calculating Electric and Magnetic Field amplitude

  1. Mar 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A circular uniform laser beam has a power of P = 3.2 mW and a diameter of d = 8.5 mm. The beam is a form of electromagnetic raditation with both electric and magnetic field components.
    (a) Calculate the intensity of the beam in units of W/m^2
    (b) Calculate the energy (delta U) delivered in a time of t = .065 s in Joules.
    (c) Use the intensity of the beam (I) to calculate the amplitude of the electric field in V/m
    (d) The amplitudes of the electric and magnetic fields have a fixed relationship. Based on that relationship use the result of part (c) to calculate the amplitude of the magnetic field (B) in Tesla.


    I know that the power is P = 3.2 mW, and the diameter is d = 8.5 mm. From this I calculated intensity of the beam, so part (a) is complete. I can multiply that by time to get energy, so part (b) is complete as well. Where I'm running into a problem is somehow finding the electric field without knowing anything but the intensity and the energy in a small window of time. Typical calculations for electric potential energy haven't worked at all.

    2. Relevant equations
    delta U = Pt
    I = P/A
    A = pi*r^2
    That's about all I know. Our professor left us totally on our own to learn this material.

    3. The attempt at a solution
    All I've managed is to calculate the delivered energy in a small window of time and the intensity of the beam.
     
  2. jcsd
  3. Mar 13, 2016 #2

    blue_leaf77

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    The intensity of a plane monochromatic wave is defined as the time average of Poynting vector. Do you the formula of Poynting vector?
     
  4. Mar 13, 2016 #3
    S = 1/μ0E x B

    However, how is this usable if I don't know both E and B?
     
  5. Mar 13, 2016 #4

    blue_leaf77

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    Good, now realize that ##S=|\mathbf{S}|## is a function of time. While, as I said in my previous post that the intensity is the time average of ##S## over one period of the wave,
    $$
    I = \langle S(t) \rangle_t
    $$
    So, first find out the how ##S## (the magnitude of ##\mathbf{S}##) looks like for a plane wave. (Note: you will also need to employ certain relation between the magnitude of the fields for a plane wave, with which you will solve part d)). Assume the electric field to have the form ##\mathbf{E} = \hat{x}E_0 \sin(\omega t - kz)##, employ Maxwell equations to find the corresponding ##\mathbf{B}##.
     
    Last edited: Mar 13, 2016
  6. Mar 13, 2016 #5

    vela

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    There are no charges involved here, so there's no potential energy you can calculate.

    The magnitudes of the E and B fields are related.
     
  7. Mar 13, 2016 #6
    That's what I keep finding in my text resources, but it's never elaborated, simply stated.

    Would I be able to say that Em/c = Bm?
     
  8. Mar 13, 2016 #7

    blue_leaf77

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    That's correct, and I hope you know how to prove it, for which purpose the last sentence in part #4 was written.
    So, now how does ##S## looks like in terms of ##E_m##?
     
  9. Mar 13, 2016 #8
    I really have no idea how to prove the relationship I stated above, but in terms of Em, S = (1/μ0c) Em2
     
  10. Mar 13, 2016 #9

    blue_leaf77

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    As the first step, follow what I wrote as the last sentence in post #4.
    Not really, you miss the time dependent behavior there. Look, the fields you have are
    $$
    \mathbf{E} = \hat{x}E_m \sin(\omega t -kz) \\
    \mathbf{B} = \hat{y}\frac{E_m}{c} \sin(\omega t -kz)
    $$
    Plug in these equations into ##\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}##.
     
  11. Mar 13, 2016 #10
    S = (1/μ0c)Em2sin2(ωt - kz)
     
  12. Mar 13, 2016 #11

    blue_leaf77

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    Yes, now take the time average of this function over one period ##T=2\pi/\omega##.
     
  13. Mar 13, 2016 #12
    What do I use for ω and z?
     
  14. Mar 13, 2016 #13

    blue_leaf77

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    ##\omega## is used in relation to the period, while ##z## is actually irrelevant for the present calculation but it was introduced in order to make the expression for the field waves make sense, i.e. they represent propagating waves.
     
  15. Mar 13, 2016 #14
    I don't know the period.
     
  16. Mar 13, 2016 #15

    blue_leaf77

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    Look it up in one of the previous posts of mine.
     
  17. Mar 13, 2016 #16
    You've lost me.
     
  18. Mar 13, 2016 #17

    blue_leaf77

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    I denoted the period as ##T##, in particular I also expressed it in terms of ##\omega##.
     
  19. Mar 13, 2016 #18
    So I can set ω = T/2π. Now what?
     
  20. Mar 13, 2016 #19

    blue_leaf77

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    No, the right hand side should be the reciprocal of that one above.
    What did I say in post #11?
     
  21. Mar 13, 2016 #20
    So I'd take the time average from 0 to T = 2π/ω?
     
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