Jordan's lemma   
  Delta function  
  Young's Modulus  
  Fourier Transforms  
  ask us  


A sigle wave

Superposition of waves

© The scientific sentence. 2010

Electromagnetic waves

1. The relationship E= cB

We have set that Maxwell's equations allow to set that light is an electromagnetic wave. This wave is composed by an electric field and magnetic field propagating together at the speed of light. These two fields are transverse ans perpendicular to each other. We have found the two wave equations for the fields E and B. This allow us to write their expression as:
Ey = Eo sin (kEx - ωEt)
Bz = Bo sin (kBx - ωBt + φ)

We have seen the following equation:
∂Ey/∂x = - ∂Bz/∂t

By substituting in the expressions of E and B, we have:
Eo kE cos (kEx - ωEt) = ωB Bo cos (kBx - ωBt + φ)
This equation requires:
Eo kE = ωB Bo ,
kE = kB ,
ωE = ωB , and
φ = 0

We have finally:
Eo k cos (kx - ωt) = ω Bo cos (kx - ωt)
With ω = c k, we have:

Eo = c Bo
The expressions of E and B become:
Ey = Eo sin (kx - ωt)
Bz = Bo sin (kx - ωt)

2. Electromagnetic wave energy

An electromagnetic wave transports energy by the means of the electric and magnetic fields. We know that the energy density associated with the electric field is ue = (1/2) εo E2, With magnetic field the density energy is um = (1/2μo) B2.
As the speed of light c = 1/[εoμo]1/2, and E = cB, then εo = 1/μoc2 = B2oE2.
Thus: ue = (1/2) εo E2 = (1/2) B2oE2 E2 = B2/2μo = um.
The electromagnetic energy density is the sum : uem = ue + um = 2 ue = εo E2
uem = εo E2

If the electromagnetic wave passes through a slab of volume Δx L2, the energy density in this volume is U = uem Δx L2. The wave travels at speed "c"; then the rate at which energy passes through the surface L2 of the slab perpendicular to the wave is:
U/Δt = U/(Δx/c) = Uc/Δx = c uem L2

The wave intensity is then: S = (U/Δt ) (1/L2) = c uem

We introduce the Poynting vector S = S i that points in the propagation direction (over "x" axis) with the magnitude is equal to the wave intensity:
S = (1/μo) E x B
(x is the cross product).

We use the average value of S, calculated as:
Sav = Average(S) = Average [(1/μo) EoBo sin2(kx - ωt)] = (1/μo) EoBo (1/2) = EoBo /2μo
Sav = EoBo /2μo

©: The 2007.


chimie labs
scientific sentence
green cat

© Scientificsentence 2010. All rights reserved.