Uniform plane wave (help with a short walkthrough for a worked problem)

In summary, the instantaneous expression for \vec E for a uniform plane wave in a lossless simple medium propagating in the +z-direction can be written as \vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi), where the phase \psi is introduced to satisfy the initial condition of a maximum value of +10^{-4} at t=0 and z
  • #1
FrogPad
810
0
I don't understand something on this problem:

Q: A uniform plane wave with [itex] \vec E = \hat x E_x [/itex] propagates in a lossless simple medium ([itex] \epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0 [/itex]) in the +z-direction. Assume that [itex] E_x [/itex] is sinusoidal with a frequency of 100 (MHz) and has a maximum value of [itex]+10^{-4} [/itex] (V/m) at [itex] t=0 [/itex] and [itex] z = \frac{1}{8} [/itex] (m)

Write the instantaneous expression for [itex] \vec E [/itex] for any t and z.

A:
Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

[tex] \vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi) [/tex]

Since [itex] E_x [/itex] equals [itex] +10^{-4} [/itex] when the argument of the cosine function equals zero, that is when:

[tex] 2 \pi 10^8 - kz + \psi = 0 [/tex]

we have at [itex] t = 0 [/itex], [itex] z = \frac{1}{8} [/itex]

[tex] \psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6} [/tex]

My question follows from this:
Why do we introduce the phase unknown phase [itex] \psi [/itex] and set [itex] E_x [/itex] according to this. Why couldn't we do this instead,

A:
Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

[tex] \vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz) [/tex]

Set the according t=0, z=1/8 so that,
[tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex] and meet the initial condition of a max of 10^-4 with [itex] E_0 [/itex] instead of setting E_0 originally and changing the phase.
 
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  • #2
I forget -- what is k in terms of lambda? It seems like the phi and the kz can influence the final equation under different constraints -- I could be wrong of course.
 
  • #3
FrogPad said:
I don't understand something on this problem:

Q: A uniform plane wave with [itex] \vec E = \hat x E_x [/itex] propagates in a lossless simple medium ([itex] \epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0 [/itex]) in the +z-direction. Assume that [itex] E_x [/itex] is sinusoidal with a frequency of 100 (MHz) and has a maximum value of [itex]+10^{-4} [/itex] (V/m) at [itex] t=0 [/itex] and [itex] z = \frac{1}{8} [/itex] (m)

Write the instantaneous expression for [itex] \vec E [/itex] for any t and z.

A:
Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

[tex] \vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi) [/tex]

Since [itex] E_x [/itex] equals [itex] +10^{-4} [/itex] when the argument of the cosine function equals zero, that is when:

[tex] 2 \pi 10^8 - kz + \psi = 0 [/tex]

we have at [itex] t = 0 [/itex], [itex] z = \frac{1}{8} [/itex]

[tex] \psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6} [/tex]

My question follows from this:
Why do we introduce the phase unknown phase [itex] \psi [/itex] and set [itex] E_x [/itex] according to this. Why couldn't we do this instead,

A:
Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

[tex] \vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz) [/tex]

Set the according t=0, z=1/8 so that,
[tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex] and meet the initial condition of a max of 10^-4 with [itex] E_0 [/itex] instead of setting E_0 originally and changing the phase.
If you used

[tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex]

the maximum would not be 10^-4 and at t = 0 the maximum would be at z = 0.
 
Last edited:
  • #4
OlderDan said:
If you used

[tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex]

the maximum would not be 10^-4 and at t = 0 the maximum would be at z = 0.

oooooooooooooooooohhhh :)

Thanks man, that definitely makes sense! I really appreciate it.

berkeman:

[tex] k = \omega \sqrt{\mu \epsilon} [/tex]
 

Related to Uniform plane wave (help with a short walkthrough for a worked problem)

1. What is a uniform plane wave?

A uniform plane wave is a type of electromagnetic wave that has a constant amplitude and direction of propagation. It is characterized by having a uniform electric and magnetic field that are perpendicular to each other and to the direction of propagation.

2. How is the direction of propagation determined for a uniform plane wave?

The direction of propagation for a uniform plane wave is determined by the direction of the electric and magnetic fields. The wave travels in the direction of the cross product between these two fields.

3. What is the equation for a uniform plane wave?

The equation for a uniform plane wave is given by E = E0sin(kz - ωt), where E0 is the maximum amplitude of the electric field, k is the wave vector, z is the direction of propagation, and ω is the angular frequency.

4. How can the intensity of a uniform plane wave be calculated?

The intensity of a uniform plane wave can be calculated by taking the square of the electric field amplitude divided by the characteristic impedance of the medium it is traveling through. This can be represented by the equation I = (E0/Z)2, where Z is the characteristic impedance.

5. Can you provide a walkthrough for a worked problem involving a uniform plane wave?

Sure, let's say we have a uniform plane wave with a maximum electric field amplitude of 10 V/m traveling through air with a characteristic impedance of 377 Ω. The wave is propagating in the positive z-direction with a wave vector of 5 rad/m and an angular frequency of 10 rad/s. To find the intensity of the wave, we can use the equation I = (E0/Z)2. Plugging in the values, we get I = (10/377)2 = 0.702 x 10-4 W/m2. This is the intensity of the uniform plane wave in air.

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