# Electromagnetism - finding electric field from magnetic field

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.

Thanks!

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 $J$ ?

• 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? – Galc127 Mar 18 '16 at 9:10
• 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 ? – Dimitri Mar 18 '16 at 9:23
• Unfortunately I don't understand. Could you elaborate please? – Galc127 Mar 18 '16 at 9:29
• 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 $\vec{u}_y$. – Dimitri Mar 18 '16 at 10:29
• 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 $\vec{u}_y$. – 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? 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 $\dfrac{\partial B_y}{\partial x} \hat Z$.