Direction of Electric Field in an Electromagnetic Wave

Click For Summary

Homework Help Overview

The discussion revolves around the direction of the electric field in an electromagnetic wave, specifically when the magnetic field oscillates parallel to the y-axis. The original poster presents a problem involving the propagation direction of the wave and the orientation of the associated electric field.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to determine the direction of the electric field based on the relationship between the electric and magnetic fields in electromagnetic waves. Some participants question the original poster's understanding of the propagation direction and the orientation of the electric field.

Discussion Status

Participants are actively engaging with the problem, with one suggesting that the electric field should be in the +x-direction based on the cross product of the electric and magnetic fields. There is acknowledgment of the perpendicular relationship between the E and B fields, indicating a productive direction in the discussion.

Contextual Notes

There is an emphasis on understanding the relationship between the electric and magnetic fields in the context of electromagnetic wave propagation, with some participants addressing potential misunderstandings about the directionality involved.

spaghetti3451
Messages
1,311
Reaction score
31

Homework Statement



If the magnetic field of a light wave oscillates parallel to a y axis and is given by ##B_y = B_m\ sin(kz- \omega t)##,

(a) in what direction does the wave travel and

(b) parallel to which axis does the associated electric field oscillate?

Homework Equations



The Attempt at a Solution



(a) The wave is propagating in the ##+ z##-direction because the form of the argument in the sinusoidal function.

(b) I'm having problem here.
 
Physics news on Phys.org
(b) where are you having the problem?
 
I think I figured it out. :D

If the direction of propagation is given by ##\overrightarrow(E) \times \overrightarrow(B)##, then shouldn't the direction of the electric field be in the ##+ x##-direction.
 
That's the one - well done.
Just remember that the E and B fields are perpendicular.

LaTeX note: use curly brackets to enclose what you operate on, but not needed if there is only on letter following the command.
Thus \vec E \times \vec B gets you ##\vec E \times \vec B##
 
Thank you! :)
 

Similar threads

Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K