Integrating a vector (Electromagnetism)

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Homework Statement



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

Homework Equations



∇ x E = -[itex]\frac{\partial\textbf{B}}{\partial t}[/itex]


The Attempt at a Solution



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

I'm not too confident integrating this, I got
[itex]\textbf{B} = (f(z),-\frac{k}{\omega}cos(kz+\omega t), g(z)) + \textbf{c}[/itex]
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 [itex]\textbf{E} \times \textbf{B}[/itex] , would i be allowed to set [itex]f = g = 0[/itex]?
 

Answers and Replies

  • #2
HallsofIvy
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Where did you get this "f(z)" and "g(z)"? The "constant" of integration is the vector c you have added.
 
  • #3
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Yeah thats what i'm confused about.
My reasoning is that I have [itex]\frac{\partial \textbf{B}}{\partial t}[/itex] in terms of [itex]\textbf{E}[/itex], and since [itex]\textbf{E}[/itex] is a function of z and t I get the functions of z from partially integrating wrt t.
Should they be 0?
 
  • #4
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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
vanhees71
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From the Faraday Law alone, we can in fact only conclude that
[tex]\vec{B}(t,\vec{r})=-E_0 \frac{k}{\omega} \cos(\omega t+k z)+\vec{B}_0(\vec{r})[/tex]
where [itex]\vec{B}_0[/itex] is an arbitrary static magnetic field.

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

From the Maxwell-Ampere Law, assuming that there are no currents, we get
[tex]\partial_t \vec{E}=\vec{\nabla} \times \vec{B} \; \Rightarrow \; \vec{\nabla} \times \vec{B}_0=0.[/tex]
Thus static magnetic field [itex]\vec{B}_0[/itex] 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.
 

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