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Maxwell's equations

1. 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


2. Homework Equations

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

3. 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
 
Last edited:
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!
 
Last edited:
383
0
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 spacial 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].
 
383
0
Now that I've had a good night's rest, you can actually ignore the fact that I said the real components of the spacial 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!
 

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