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Electric field vector

  1. May 19, 2014 #1
    Q) A harmonic time-dependent electromagnetic plane wave, of angular frequency ω, propagates along the positive z-direction in a source-free medium with σ = 0, ε = 1 and µ = 3. The magnetic field vector for this wave is: H = Hy uy. Use Maxwell’s equations to determine the corresponding electric field vector.


    Ans) I've pretty much forgotten all this stuff from 1st year, so I'm not sure if my answer is correct.

    [tex] \bigtriangledown \times H = \varepsilon_0 \frac{\partial E}{\partial t}​[/tex], [tex]H = (0, Hy, 0)[/tex]

    [tex] \bigtriangledown \times H = \begin{vmatrix} \hat{u}x & \hat{u}y & \hat{u}z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & Hy & 0 \end{vmatrix}[/tex]

    [tex]= \hat{u}x (\frac{-\partial Hy}{\partial z}) - \hat{u}y(0) + \hat{u}z (\frac{\partial Hy}{\partial z}) = (\frac{-\partial Hy}{\partial z}, 0, 0)[/tex]

    and

    [tex]\varepsilon_0 \frac{\partial E}{\partial t}​ = (\varepsilon_0 \frac{\partial E}{\partial t}​, 0, 0)[/tex]

    [tex] \Rightarrow \varepsilon_0 \frac{\partial E}{\partial t} = \frac{-\partial Hy}{\partial z} [/tex]

    At this point I'm somewhat lost as how to find E vector. I know that J = σ E and σ = 0, but how do I get from what I got above to the E vector?
     
  2. jcsd
  3. May 19, 2014 #2

    TSny

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    Looks OK so far. Maybe bringing in another one of Maxwell's equations will help.
     
  4. May 19, 2014 #3
    Any suggestions?
     
  5. May 19, 2014 #4

    rude man

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    Heard of "intrinsic impedance"?

    If you have to derive this I would start with ∇xE = -∂B/∂t.
     
  6. May 19, 2014 #5

    TSny

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    Out of the 3 other equations, which one do you think would be helpful in relating E and H?
     
  7. May 20, 2014 #6
    Hi, thanks for your help.

    Going by what you say, I think then because [tex]\frac{{\partial {E_x}}}{{\partial z}}{u_y} = - \frac{{\partial B}}{{\partial t}} = -\mu \mu_0 \frac{ \partial Hy}{\partial t}[/tex],

    if I take the partial derivatives of both (one with respect to t and the other to z), I get [tex] \varepsilon_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 H_y}{\partial t \partial z}[/tex]

    and

    [tex] \mu \mu_0 \frac{\partial^2 H_y}{\partial t \partial z} = -\frac{\partial^2 E_x}{\partial z^2}[/tex].

    This then give [tex]\mu \mu_0 \varepsilon_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 E_x}{\partial z^2} \Rightarrow \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}[/tex]

    Subbing in the numbers for [tex] \mu \mu_0 \varepsilon_0[/tex]

    [tex]\frac{\partial^2 E_x}{\partial t^2} = 3\times 10^{16} \frac{\partial^2 E_x}{\partial z^2}[/tex]
     
  8. May 20, 2014 #7

    TSny

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    OK. (It might be better to hold off on plugging in numbers.) Use this result with the fact that the time dependence is harmonic with angular frequency ω to deduce how E depends on z.
     
  9. May 20, 2014 #8
    Ok:


    [tex]\frac{\partial^2 E_x}{\partial t^2} = \mu \mu_0 \varepsilon_0 (- k^2 \varepsilon E_x)[/tex]

    [tex] \Rightarrow E_x(z) = \mu \mu_0 \varepsilon_0 A_0 exp[-jkz][/tex]

    [tex]= 3\times 10^{16} A_0 exp[-jkz][/tex]

    Where [tex]k = \frac{\omega}{c} = \omega [\mu_0 \varepsilon_0]^{\frac{1}{2}}[/tex]
     
  10. May 20, 2014 #9

    TSny

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    I'm not quite following your logic. (Maybe you just didn't write out the steps.) You are given that E is harmonic in time with angular frequency ω: Ex ~ exp(jωt).

    So, ∂2Ex/∂t2 = -ω2Ex.

    Using this, you can get a differential equation for Ex as a function of z.
     
  11. May 20, 2014 #10
    But isn't that similar to what I already had where [tex]\frac{\partial^2 E_x}{\partial z^2} = -j\omega^2 Ex[/tex]

    [tex] \Rightarrow \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} A_0 exp[-j \omega z][/tex]
     
  12. May 20, 2014 #11

    TSny

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    This equation isn't correct.

    Go back to what you got from Maxwell's equations:[tex] \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}[/tex]
    From the fact that ##E_x## is given to vary harmonically in time with angular frequency ##\omega##, you can replace the left side of the above equation by ##-\omega^2 E_x##. This will give you the differential equation to solve for finding the z-dependence of ##E_x##.
     
  13. May 20, 2014 #12
    I know the answer is probably really obvious but I'm lost now.

    EDIT: I want to say this has something to do with helmholtz equation?
     
    Last edited: May 20, 2014
  14. May 20, 2014 #13

    TSny

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    You are looking for the electric field ##\vec{E}(z, t)## which you already know only has an x component. So, you are looking for ##E_x(z, t)##.

    You are given the t dependence as harmonic with frequency ##\omega##. So, you can write ##E_x(z, t) = E_x(z)e^{-j\omega t}##.

    Thus, you must now determine the function ##E_x(z)##.

    You have [tex] \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}[/tex]
    What do you get when you substitute ##E_x(z, t) = E_x(z)e^{-j\omega t}## into both sides of this equation and simplify?
     
  15. May 20, 2014 #14

    rude man

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    OP, you're in excellent hands with TSny, but if you want an alternative you could use the equation in post 4 and come up quickly with a 1st-order diff. eq., easily solved if you remember that your B field goes as B0cos(kz - wt) where k = ω/v and v is the phase velocity of the wave in the z direction.

    The catch here is that you need to know v as a function of the basic parameters ε and μ. Which you could easily look up but then you'd lose the insight you gain by pursuing TSny's approach.
     
  16. May 20, 2014 #15

    TSny

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    I might have misinterpreted the problem. I was assuming that the only thing you know is that the time dependence of the wave is harmonic, but that you know nothing of the spatial dependence. Then you can use Maxwell's equations to deduce the spatial dependence and the value of the wavevector k.

    But, if "harmonic wave" is interpreted as already having the form cos(kz-ωt), then you just need to determine k. No need to solve a differential equation.

    Sorry if I have misinterpreted what you are meant to do. Thanks rude man.
     
  17. May 20, 2014 #16
    I appreciate both of your help, but I reckon I really need to study this more. I'm going to try and catch my lecturer tomorrow and ask. Fortunately this isn't homework, the unfortunate thing is this may potentially be a question in an exam I have in a few days.

    I'm an elec eng student and this is from a radio transmissions module tutorial compromising of both the engineering aspects and the physics. Unfortunately I spent too much time studying the engineering part and not enough on the physics.

    The thing is I know that I can do this, it feels like I have it in my head I just can't get at it.

    Anyway thanks. :)
     
  18. May 20, 2014 #17

    rude man

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    Hi T --- if he/she has k you still need to express k (or v = ω/k) in terms of ε and μ the way you indicated, or something similar, right? ε ad μ are introduced via ∇2E = με∂2E/∂t2 far as I know. Seems to me you need to wind up with ε and (especially!) μ to answer the question.
     
  19. May 20, 2014 #18

    TSny

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    Yes, that sounds right. You should be able to show k = (ω/c)√μ.
     
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