Find the direction of propagation, mag field, & Poynting vector of an EM wave given its E field

  • #1

Homework Statement


For an electromagnetic wave with an electric field given by
[itex]\mathbf{B}=\mathbf{i} E_o cos (kz-\omega t)+ \mathbf{j} E_o sin (\omega t - kz) [/itex]
where Eo is a constant
Find
(a) its direction of propagation
(b) the magnetic field
(c) Poynting vector
(d) energy density, i.e., the amount of energy that is transmitted by the wave per unit area per unit time.

Homework Equations


(1) [itex]\mathbf{S}=\frac{1}{\mu_o}(\mathbf{E} \times \mathbf{B})[/itex] (Poynting vector)
(2) [itex] u =\frac{1}{2}\big(\epsilon_o E^2 + \frac{1}{\mu_o}B^2 \big) [/itex] (energy density)
(3) [itex]\nabla\times\mathbf{E} = - \frac {\partial \mathbf{B}}{\partial t} [/itex] (magnetic field)
(4) [itex]\mathbf{{\tilde{B}}}(\mathbf{r},t)=\frac{1}{c}\mathbf{\hat{k}}\times\mathbf{\tilde{E}}[/itex] (magnetic field)

The Attempt at a Solution



(a) Because we have the [itex]kz-\omega [/itex] and [itex]\omega t - kz[/itex] terms, the direction of propagation is along the z-axis. The electric and magnetic fields are along the x-y plane. How to I know if the direction of EM wave propagation is + or - z-axis?

(b) Once I get the EM wave propagation direction, then I just use equation Eq. 4. Is that right?

(c) I just use Eq. 1, right?

(d) This is just Eq. 2, right?

Thank you very much.
 

Answers and Replies

  • #2
Wait. I think I have an answer for (a).
Hm, since
[itex]\mathbf{E}=\mathbf{i}E_o cos(kz-\omega t) + \mathbf{j}E_o sin(\omega t-kz)[/itex]
we can have
[itex]\mathbf{E}=\mathbf{i}E_o cos(kz-\omega t) + \mathbf{j}E_o sin(-(kz-\omega t))[/itex]
[itex]\mathbf{E}=\mathbf{i}E_o cos(kz-\omega t) - \mathbf{j}E_o sin(kz-\omega t)[/itex]

After this, can I just assign an arbitrary +z propagation direction so that
[itex]\mathbf{\hat{k}}=+\mathbf{\hat{z}}[/itex]

Then I can easily compute for the magnetic field [itex]\mathbf{B}[/itex] Is this correct? Thanks.
 

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