# Integrating a vector (Electromagnetism)

1. Jan 7, 2014

### raggle

1. The problem statement, all variables and given/known data

Given $\textbf{E}(z,t) = E_{0}cos(kz+ωt)\textbf{i}$
Find B

2. Relevant equations

∇ x E = -$\frac{\partial\textbf{B}}{\partial t}$

3. The attempt at a solution

Taking the curl of $\textbf{E}$ gives $(0, -ksin(kz+\omega t), 0)$
so
$\frac{\partial\textbf{B}}{\partial t} = (0,ksin(kz+\omega t),0)$

I'm not too confident integrating this, I got
$\textbf{B} = (f(z),-\frac{k}{\omega}cos(kz+\omega t), g(z)) + \textbf{c}$
where c is a constant of integration.

Is this right? The next part of the question asks for the poynting vector and it seems like a lot of work calculating $\textbf{E} \times \textbf{B}$ , would i be allowed to set $f = g = 0$?

2. Jan 7, 2014

### HallsofIvy

Staff Emeritus
Where did you get this "f(z)" and "g(z)"? The "constant" of integration is the vector c you have added.

3. Jan 7, 2014

### raggle

Yeah thats what i'm confused about.
My reasoning is that I have $\frac{\partial \textbf{B}}{\partial t}$ in terms of $\textbf{E}$, and since $\textbf{E}$ is a function of z and t I get the functions of z from partially integrating wrt t.
Should they be 0?

4. Jan 7, 2014

### raggle

Oh wait I think i just got it.
I can put those functions of z into the arbitrary constant vector c can't I?

Thanks!

5. Jan 7, 2014

### vanhees71

From the Faraday Law alone, we can in fact only conclude that
$$\vec{B}(t,\vec{r})=-E_0 \frac{k}{\omega} \cos(\omega t+k z)+\vec{B}_0(\vec{r})$$
where $\vec{B}_0$ is an arbitrary static magnetic field.

From Gauss's Law for the magentic field we find
$$\vec{\nabla} \cdot \vec{B}=\vec{\nabla} \cdot \vec{B}_0=0.$$

From the Maxwell-Ampere Law, assuming that there are no currents, we get
$$\partial_t \vec{E}=\vec{\nabla} \times \vec{B} \; \Rightarrow \; \vec{\nabla} \times \vec{B}_0=0.$$
Thus static magnetic field $\vec{B}_0$ is both source and vortex free. Thus if it should vanish at inifinity, it must be 0.

From these additional assumptions we get the usual plane-wave solution.