The ratio between these two equations shows that the ratio between the fields is equal to the ratio between their amplitudes.
max max
E E B B
Substituting the harmonic solutions to the relationship between the fields E B
x t
w w
w w shows that the ratio of the electric field to the magnetic field is always equal to the speed of light c.
E cB
Example: An electromagnetic wave has a wavelength of 2 10 mu 6 . The amplitude of the electric field is 3 10 N/Cu 3 . Give the harmonic wave solution of this electromagnetic wave as a function of position and time.
Solution:
6 3
2 10 m; Emax 3 10 N/C
O u u
maxcos
E E kxZt
6 6
2 2 10 1/m
k S 2 10S S O u u
Download free eBooks at bookboon.com Click on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more
EXPERIENCE THE POWER OF
6 8 14
10 3 10 3 10 rad/s Z kc uS u u Su
3 103 N/C cos^
103 3 1014`
E u Su x Su t
maxcos
B B kxZt
3 11
3 10 T = 10 T8 max Emax 3 10
B C
u
10 11 T cos^
u 103 3 1011`
B Su x Su t
Energy density of an electromagnetic wave
The energy density of an electromagnetic wave is the sum of the energy densities due to its electric field and magnetic field. As shown in previous chapters, the energy densities due to electric field and magnetic field are respectively given by uE 12H0E2 and
0
1 2 B 2
u B
u . Therefore the electromagnetic energy density u is given by 0
0
2 2
1 1
2 2
E B
u u u H E u B . With B Ec H P0 0E, this simplifies to
0 2 u H E
Also, substituting E H P0 0 B, u can be expressed in terms of B as
0
B2 u P
This is instantaneous energy density. The average of the energy density u for a harmonic wave can be obtained by integrating the energy density with time over one cycle and dividing by its period.
0 0
2 or 2
1 1
2 max 2 max
u H E u P B
Download free eBooks at bookboon.com
Poynting Vector
Poynting Vector SS
is a vector that represents the amount of electromagnetic energy that crosses a unit perpendicular area per a unit time. Its direction is the direction of the propagation of energy which is perpendicular to both the electric and magnetic field. Suppose electromagnetic energy contained in a small cylinder of length dx and cross-sectional area dAA crosses the cross-section of the cylinder in a time interval dt. This amount of energy is equal to udxdAA. Therefore the magnitude of the poynting vector is S udxdAdA dtA udxdtA .
But dx cdt . Therefore the magnitude of the poynting vector is equal to the product of the speed of light and the electromagnetic energy density.
S = cu
Using the expressions for the energy density in terms of the electric field or magnetic field, the magnitude of the poynting vector may also be expressed 2 2
0 0
E C
S B
C
P P . The average of the poynting depends on the amplitudes as 2 2
0 0
1 1
2 max 2 max
E C
S B
cu C
P P ..
Since the poynting vector is perpendicular to both the electric and magnetic fields, a unit vector eˆs in the direction of the poynting vector can be written as eˆs E B
E B u u E B E B
E B. But E B EBu
E B EB because the fields are perpendicular to each other. Therefore S Seˆs cu E B EB u
ˆ c
S Sˆ c E B. Substituting the expressions for the energy density and the magnetic field in terms of the electric field, it can be shown that
0
1 cu EB P .
0
S P11 E Bu S 1 EE BE
Example: The frequency of a certain electromagnetic wave is 4 × 1010 Hz. The amplitude of its electric field is 4 × 10–6 N/m. Assuming the wave is harmonic.
Download free eBooks at bookboon.com
a) Calculate the average energy density for this radiation.
Solution:
10 6
4 10 Hz; max 4 10 N/C; ?
f u E u u
0
2 12 6 2
3 23 3
8.85 10 4 10
J/m J/m
2 7.08 10
max 2
u H E
u
u u
b) Calculate the average of electromagnetic energy that crosses a unit perpendicular are per a unit time.
Solution:
S ?
3 1087.08 10 23 W/m 2 2.1 10 15 W/m2s cu u u u
Download free eBooks at bookboon.com Click on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more