Circular polarization and finding B(x,t) from Faraday's law

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Homework Statement



Draw the state of polarization of the electromagnetic (EM) wave defined by

*****PLEASE NOTE EQUATION SHOWN IN NEXT POST****** (For some reason can't change it in this post...

with Eo real. Use a sentence to describe in words the state of polarization of this EM wave.

Use the differential form of Faraday's law to obtain B(x,t) for the same EM wave


Homework Equations



*****PLEASE NOTE EQUATION SHOWN IN NEXT POST****** (For some reason can't change it in this post...

The Attempt at a Solution



I have done the first part and found the EM wave to be circularly polarized (anticlockwise) with an amplitude of E0

i am unsure how to use faradays law. I tried breaking the LHS of faraday into its respective partial differential vector form, which i am currently working my way through but i am fairly sure it is wrong!

Any ideas on how to apply to get the magnetic field from the electric using faraday-maxwell's law would be greatly appreciated!

Many Thanks
 
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Sorry i copied and pasted the equation and it has come out wrong...

It should read

E(x,t) = Eo( y + e^(i3[tex]\Pi[/tex]/2) z )e^(i(kx-[tex]\omega[/tex]t)) ,


Bold Letters denote unit vectors

Relevant Equations;

[tex]\nabla[/tex] x E = - [tex]\partial[/tex]B / [tex]\partial[/tex] t (Faraday-Maxwell Equation)

Excuse my mistake!
 
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Your electric field is given by

[tex] \mathbf{E}=E(x,y,z,t)=E_0\exp[i(kx-\omega t)]\hat{\mathbf{y}}+E_0\exp[i(kx-\omega t)]\exp\left[i\frac{3\pi}{2}\right]\hat{\mathbf{z}}[/tex]

correct? But we also know [itex]\mathbf{E}=Re(\mathbf{E}'\exp[i\omega t])[/itex] where [itex]\mathbf{E}'[/itex] is the spatial component of the electric field. Then through Faraday's law,

[tex] \mathbf{B}'=-\frac{1}{i\omega}\nabla\times\mathbf{E}'[/tex]

You can then solve [itex]\mathbf{B}=Re(\mathbf{B}'\exp[i\omega t])[/itex].
 
Now that I've had a good night's rest, you can actually ignore the fact that I said the real components of the spatial electric and magnetic components. This would only be true if your original electric field were given by cosine.
 
Thank you, that really helps! I appreciate it!