Maxwell's Equations: Draw State of Polarization & Find B(x,t)

[leoflindall]

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

 

[leoflindall]

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

It should read

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


Bold Letters denote unit vectors

Relevant Equations;

$$\nabla$$ x E = - $$\partial$$B / $$\partial$$ t (Faraday-Maxwell Equation)

Excuse my mistake!

 

[jdwood983]

Your electric field is given by

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

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

$$\mathbf{B}'=-\frac{1}{i\omega}\nabla\times\mathbf{E}'$$

You can then solve $\mathbf{B}=Re(\mathbf{B}'\exp[i\omega t])$.

 

[jdwood983]

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.

 

[leoflindall]

Thank you, that really helps! I appreciate it!