Electric field due to a current carrying wire

AI Thread Summary
The discussion revolves around calculating the radiated electric field from a circular loop carrying a time-harmonic current, positioned above a perfectly conducting half-space. Participants emphasize the importance of determining the vector magnetic potential to derive the electric field intensity, while also grappling with the coordinate system, particularly the placement of the z-axis in relation to the x-y plane. The loop is identified as a small vertical antenna, and its behavior in the far field is likened to that of a small dipole, with considerations regarding the effects of the conducting surface on the electric field. Participants encourage experimentation with different approaches to solve the problem, highlighting the learning process involved. Overall, the conversation reflects a collaborative effort to understand complex electromagnetic concepts related to antennas and fields.
AgerPl
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Homework Statement


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Circular loop of diameter d located in the vertical plane (x-y) at a distance h from the perfectly conducting half-space. The current source creates a time-harmonic current i(t)=Acos (ωt).
  • Find the radiated electric field vector in the plane (x-y) in the far zone of the loop. Assuming that R>>λ
  • Find E at the axis z, for any value of this coordinate -∞<z<+∞
Arguments of E are radius vector r and time t.
d<<λ and h<<λ

Homework Equations



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The Attempt at a Solution


Should I find the vector magnetic potential and then the magnetic flux density? Like a magnetic dipole and then try to find the electric field?

I know that nabla X B due to a loop far away is = 0.

And regarding the z how I know to where is the axis if he isn't point out in the problem.AgerPl
 
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You find the dircetion of the z axis using your understanding of the cartesian coordinate system and the fact that x and y axes are shown.
If you are unsure how to proceded with the problem - pick a method and give it a try: the point of these problems is that you learn from trying things out. The most important thing to learn is to try things out instead of relying on someone to tell you the best approach.
 
Simon, first of all thank you for your answer.

I researched an read some literature and I think I found some nice insights to resolve the problem.
I have to calculate the vector magnetic potential, using that I can infere the magnetic flux density and after I will be capable of getting the electric field intensity. The thing is the fact that I will have to use cartesian or cylindrical coordinates instead of polar ones are really messing with my mind.

I don't know where I will locate the z-axis. Because the professor says that we have an x-y plane and then he asks for the z axis, really ninja stuff that I don't have no insights how to solve.

Can you help me with at least the coordinate system approach?Ager
 
After some research and thought I have some conclusions:

  • This is a small loop vertical antenna above the surface of the Earth (i.e.)
  • The plane x-z is parallel to the surface of the Earth.

Still didn't managed to get more things done. Some help over here?

BR
 
If you don't know how to find the z axis given the x and y axes you have a serious problem.
Go online and look up "cartesian axes" and look at some pictures.
 
I already wrote that I found that x-z axis is parallel to the surface of the PEC conductor.

Still trying to find the influence of the PEC in the small antenna.Regards
 
Our posts crossed.
 
So what does the magnetic field do to the loop?
(Is this a section of your course about antennas or just about magnetic fields?)
 
Antennas.
More research give me more insights but still can't do the exercise. Really need some help here.

  • A small current loop in the far field has the same behavior (in electric field terms) of one small dipole.
  • The dipole is horizontally polarized and is from me to the screen. And in his image below the conductor has his polarity reversed, since the tangential component of electric field has to vanish on the conducting plane. The fields are zero in the ground plane.
  • I0=Acos(ωt)=e-jωt

Thanks for your time.
 
  • #10
Still in need for help
 
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