Integrating a vector (Electromagnetism)

[raggle]

Homework Statement

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

Homework Equations

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

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?

 

[HallsofIvy]

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

 

[raggle]

Yeah that's 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?

 

[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!

 

[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.