I'm trying to solve the following question, but to no avail.

Let $B=\left(0,B_0\left[1+\sin(kx+\omega t)\right],0\right)$ be the magnetic field of some electromagnetic plane wave.

  1. Find the wave direction vector.
  2. Find the electric field.
  3. Compute the Poynting vector and the energy density.

Part $1$:

We have $$B_y=B_0+B_0\sin(kx+\omega t)=B_0+B_0\cos\left(\frac{\pi}{2}-kx-\omega t\right)=B_0+B_0\mathbf{Re}\left[e^{j\left(\frac{\pi}{2}-kx-\omega t\right)}\right]$$ hence, $\vec{k}=\left(k,0,0\right)\Rightarrow \hat{k}=(1,0,0)$.

Now, my problem is with part $2$. I tried to use Maxwell equations $\displaystyle \nabla\times B=\mu_0 J+\frac{1}{c^2}\frac{\partial E}{\partial t}$ and $\displaystyle \nabla\times E=-\frac{\partial B}{\partial t}$. In the first equation, I don't $J$ and in the second one, I don't know how to solve for $E$.

Part $3$ is simple after solving part $2$.

How should one solve part $2$? any help would be appreciated.



I'll just give you hints on each question.

  1. I assume you are using 3D cartesian coordinates. Two components appear to be zero. What can you deduce on the vector's direction ? Is this related to the direction of propagation ?

  2. You indeed have two Maxwell equations that involve both electric and magnetic field. You thus have to choose one of them. Derivate something with respect to time is a simple operation, while taking the rotational is a much more complicated one. You already know the expression of the magnetic field, so I would apply the complicated operation to this field rather than to the unknown electric field.

In what medium are you calculating the field ? Is there conducting charges to support an electric current $J$ ?

  • $\begingroup$ Thanks! for $1$ I believe you mean that the vector direction has to be in the direction of one axis only (and it only depends on $x$, thus it have to be $(1,0,0)$, is this correct? The medium is not given, may I assume it is vacuum? $\endgroup$
    – Galc127
    Mar 18 '16 at 9:10
  • $\begingroup$ If the medium is not given, I would assume that this is vacuum. For question 1 there is a mistake in your reasoning. The norm of your field indeed only depends on $x$, but this has nothing to do with the field's direction. Write your field in cartesian coordinates over the basis vectors $\vec{u}_x$, $\vec{u}_y$ and $\vec{u}_z$ ; what is the only component that is non-zero ? $\endgroup$
    – Dimitri
    Mar 18 '16 at 9:23
  • $\begingroup$ Unfortunately I don't understand. Could you elaborate please? $\endgroup$
    – Galc127
    Mar 18 '16 at 9:29
  • $\begingroup$ You see that the only non-zero coordinate is the $\vec{u}_y$ one. This means that the magnetic field vector is directed along $\vec{u}_y$, which isn't incompatible with the fact that the field can still vary in the $x$ direction (its norm at least) while always being directed along $\vec{u}_y$. $\endgroup$
    – Dimitri
    Mar 18 '16 at 10:29
  • 1
    $\begingroup$ Sorry, my mistake. You are right, the question is indeed about the direction of propagation (which is the direction of the wave vector), and this direction is along $\vec{u}_x$ because the norm of the field reads $\sin(kx)$. The fact that the only nonzero component is along $\vec{u}_y$ means that the wave is polarized along $\vec{u}_y$. $\endgroup$
    – Dimitri
    Mar 18 '16 at 10:56

It sometimes helps to sketch the waveform of the magnetic field at a given time.

Perhaps $t=0$ in this example?

enter image description here

The direction of the wave that you are given can readily be identified from what is in the bracket.
In this example both terms have a plus sign in front of them $(+kx+\omega t)$ so it is a wave travelling in the negative $x$ direction which would also be the case if it was $(-kx-\omega t)$.
If the signs were different $(+kx-\omega t)$ or $(-kx+\omega t)$ then it would be a wave travelling in the positive $x$ direction.

A way of deciding is to see that when $x=0$ and $t=0$ then $B=B_o$

Now if the time advances a little for what value of $x$ will the magnetic field be $B_o$?

Looking at the equation $B_y=B_0 ( 1+\sin(kx+\omega t))$ we need $kx + \omega t$ to be zero.

$\omega t$ is positive so $kx$ must be negative which means that $x$ is negative.

The wave is travelling in the negative $x$ direction and that must be the direction of the Poynting vector $\vec S$.

This will immediately enable you to find the direction of the electric field at any point as $\vec S = \dfrac {\vec E \times \vec B}{\mu_o}$

To find the electric field it is probably best to use $\nabla \times \vec B = \dfrac {1}{\mu_o \epsilon_o } \dfrac {\partial \vec E}{\partial t}$ with there being no $\vec J $ term as there are no charges moving around.

From the diagram you can see that the only no zero term on the left hand side is $\dfrac{\partial B_y}{\partial x} \hat Z$.


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