Download UNIT 9: ELECTROMAGNETIC WAVES (E.M.W.)

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UNIT 9: ELECTROMAGNETIC
WAVES (E.M.W.)
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9.1 Electromagnetic Waves (e.m.w.)
{
{
{
Definition – is defined as the transportation of energy because of
the disturbance in electric and magnetic fields.
Electromagnetic waves constitute one of the most common
phenomena in nature. The existence of electromagnetic waves was
predicted theoretically, in the second half of the 19th century by the
English physicist James Clerk Maxwell.
Electromagnetic wave phenomena can be described by using
Maxwell’s equations (in vacuum) :
r r qenc
E
∫ • dA = ε0
r r
B
∫ • dA = 0
(9.1a)
(9.1b)
r r
dΦ
(9.1c)
E
∫ • dl = - dtB
r r
dΦE
B
d
l
µ
I
µ
ε
•
=
+
0
enc
0
0
∫
dt
(Gauss’
(Gauss’s law)
(Gauss’
(Gauss’s law for
magnetism)
(Faraday’
(Faraday’s law of
induction)
(Ampere-Maxwell
(9.1d) (Amperelaw)
They can be summarized in words :
A generalized form of Coulomb’s law relating electric field to its
source, electric charges.
z
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The same for the magnetic field, except that if there are no
magnetic monopoles, magnetic field lines are continuous – they do
not begin or end (likes a closed loop).
z
An electric field is produced by a changing magnetic field.
z
A magnetic field is produced by an electric current or by a changing
electric field.
{
Electromagnetic wave is also known as electromagnetic radiation.
radiation
9.1.1 The properties of the e.m.w.
{
The e.m.w. is produced by the accelerating electric charges.
{
For example, radio waves produced in a radio antenna are a result of
the electrons in the antenna undergoing acceleration when an a.c.
flows in an antenna.
{
The e.m.w. can be represented by using the figure 9.1a.
z
r
B
r
E
r
B
r
E
Fig. 9.1a
{
Properties of electromagnetic waves are :
z
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The electric E and magnetic B fields are perpendicular to each
other and to the direction of motion of the wave.
wave
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Electric E and magnetic B fields are in phase to each other and
vibrates at the same frequency,
frequency the frequency of the e.m.w.
z
All e.m.w. travel at the same speed in vacuum,
vacuum 2.998 x 108 m s-1.
z
E.m.w.s travel in straight lines.
lines
z
Electromagnetic waves are transverse wave only and they are
capable of being polarized.
polarized
z
All electromagnetic waves are experience the phenomena below :
{
Reflection
Refraction
{
{
Interference
{
Diffraction.
Diffraction
9.1.2 Equation of Electromagnetic Waves (Wave functions)
{
Since the electromagnetic wave consists of vibrations of electric field,
E and magnetic field, B, hence the e.m.w. can be represented by
using the wave functions in magnitude form:
form
z
Valid for sinusoidal
e.m.w.
e.m.w. only
E = E0 sin (ωt − kx )
(9.1e)
B = B0 sin (ωt − kx )
(9.1f)
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(ωt − kx )
E.m.w.
E.m.w. propagates in
x-axis
E.m.w.
E.m.w. propagates in positive xx-axis
The meaning of
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{
Wave functions can also being written in vector form:
form
r
E = E0 sin (ωt − kx ) ˆj
Perpendicular to each
r
other.
B = B0 sin (ωt − kx )kˆ
(9.1g)
Electric field E vibrates
along the yy-axis
(9.1h)
Magnetic field B vibrates
along the zz-axis
where ω : angular frequency(velocity)
B0 : amplitude (maximum value) of the magnetic field
E0 : amplitude (maximum value) of the electric field
2π
k : wave number =
λ
Wavelength of e.m.w.
e.m.w.
{
To determine the direction of e.m.w. propagation, the Poynting vector
is used where its formula in a vacuum is given by
(
r 1 r r
S=
E×B
µ0
Poynting vector
Right hand
grip rule
)
(9.1i)
Direction of e.m.w.
e.m.w. propagation (cross
Important
product between E and B)
Thumb – Direction of e.m.w.
e.m.w. propagation
Four fingers – pointing along vector E and then
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‘wrapped’
wrapped’ into the vector B.
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{
Note that the direction of poynting vector is in the direction of e.m.w.
propagation.
Example 1 :
A sinusoidal electromagnetic wave of angular frequency ω travels in
vacuum in the +z-direction. The magnetic field B is parallel to the yaxis and has amplitude of B0. The electric field E has amplitude of E0.
a. Write down the wave functions in vector form for that e.m.w.
b. Sketch the e.m.w.
Solution:
a.
Given the e.m.w. propagates in +z+z-direction and vibration of B
parallel to the yy-axis.
axis
Therefore rthe wave functions in vector form are
B = B0 sin (ωt − kz ) ˆj
r
x E = E0 sin (ωt − kz )iˆ
E0
b.
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B0
y
r
E
r
B
c
z
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9.1.3 The Electromagnetic Spectrum
{
Definition – is defined as the range of wavelengths over which
electromagnetic radiation extends.
extends
{
Figure 9.1b shows the electromagnetic spectrum.
Wavelength decreases
Frequency increases
Fig. 9.1b
There is no clear cut boundaries between different types of e.m.
waves. There is overlapping of wavelengths between two neighbouring
types of e.m. waves.
Although they have different value of wavelengths and frequencies but
they travel with the same speed in vacuum, c.
{
{
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{
Table 9.1a shows the e.m. waves and their sources.
Range of
Spectrum
Source
wavelength (m)
Radio waves
Electronic oscillators
105 – 10-3
Microwaves
Magnetron tube
10-3 – 10-6
Infrared (IR)
Hot objects
10-6 – 10-7
Table 9.1a Visible light
{
Incandescent objects,
discharge tubes
Ultraviolet (UV) Sun
10-7 – 10-9
X-rays
X ray tubes
10-9 – 10-11
Gamma rays
Nuclear radioactivity
10-11 – 10-14
Table 9.1b shows the colour of the visible light.
Visible light
Table 9.1b
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4x10-7 – 7x10-7
Range of wavelength (nm)
Red
625– 700
Orange
600 – 625
Yellow
560 – 600
Green
520 – 560
Blue
500 – 520
Indigo
450 – 500
Violet
400 – 450
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9.2 Relation of ε0, µ0 and c
{
By solving the Faraday’s law of induction (eq. 9.1c), we get
E = cB
(9.2a)
or
E0 = cB0
{
(9.2b)
By solving the Ampere-Maxwell law (eq. 9.1d), we get
B = ε0 µ0 cE
{
By substituting eq. (9.2c) into eq. (9.2a), thus the e.m.w. speed in a
vacuum c is given by
c=
where
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{
(9.2c)
1
ε0 µ0
(9.2d)
ε0 : permittivity of free space
= 8.85 × 10 −12 F m −1
µ0 : permeability of free space
= 4 π × 10 −7 H m −1
Substituting the values of ε0 and µ0 into eq. (9.2d),
c=
1
(8.85 × 10 )(4π × 10 )
−12
−7
c = 3.00 × 10 8 m s −1
{
{
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which equal to the speed of
light in the vacuum
The importance of eq.
eq. (9.2d) is that it shows light is a type of
electromagnetic wave.
wave The speed of light in vacuum, 3.00 x 108 m s-1
had been determined before Maxwell made known his equation.
The e.m.w. speed in the vacuum also can be determined by using
equation below :
c = fλ
(9.2e)
9.3 Relation of ε, µ and v
{
{
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Electromagnetic waves may propagate not only in a vacuum but also in
other media.
The properties of electromagnetic waves, especially their propagation
speed are strongly influenced by the type of medium in which the
waves propagate.
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{
The speed v of electromagnetic waves in a medium is given by :
1
εµ
v=
(9.3a)
where ε : permittivity of
{
the medium
µ : permeability of the medium
Since ε = κε0 and µ = µr µ0 then eq. (9.3a) can be written as
v=
where
1
(κε0 )( µr µ0 )
v=
1
ε0 µ0
v=
c
κµr
 1

 κµ
r





1
=c
ε0 µ0
and
(9.3b)
κ@ε r : relative permittivity(dielectric constant)
µr : relative permeability
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{
Rearranging eq. (9.3b), we get
κµr =
n=
{
c
v
and
c
v
κµr = n
(9.3c)
where n : refractive index of the medium
Example 2 :
An electromagnetic wave of wavelength 435 nm is travelling in a
vacuum in the –z direction. The electric field has amplitude of 2.70 x
10-3 V m-1 and is parallel to the x-axis.
a. Calculate i. the frequency ,
ii. the magnetic field amplitude of the e.m.w.
b. Write down the vector equations for E and B.
(Given the speed of e.m.w. in the vacuum, c =3.00 x 108 m s-1)
(Young&Freedman, pg.1243,no.32.4)
λ=435x10-9 m, E0=2.70x10-3 V m-1
a. i. By using the equation relates the c, f and λ in the vacuum, hence
c = λf
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f = 6.90 x10 14 Hz
Solution:
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ii. By using the equation relates the E0, B0 and c in the vacuum,
hence
E0 = cB0
B0 = 9.00 x10 −12 T
b. Given
- the e.m.w. propagates in –z direction
- the vibration (oscillation) of E parallel to the xx-axis.
axis
r
E = E0 sin (ωt + kz )iˆ
r
E
c
x
E0
r
B
− B0
z
r
B = − B0 sin (ωt + kz ) ˆj
B0
y
The angular frequency is ω = 2πf
and the wave number of the e.m.w. is
k=
2π
− E0
ω = 4.34 x10 15 rad s -1
k = 1.44 x107 m -1
λ
Therefore the vector equations for E and B are given by
r
Er = 2.70 x10 −3 sin 4.34 x10 15 t + 1.44 x107 z iˆ
B = −9.00 x10 −12 sin 4.34 x10 15 t + 1.44 x107 z ˆj
(
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{
Example 3 :
An
r electromagnetic wave has an electric field given by
) [(
(
)
(
)
)
13
]
E = − 2.30 x10 5 V m −1 sin 1.45 x10 14 rad s −1 t − kx ˆj
The wave propagates in a vacuum.
a. Determine the direction of the e.m.w. propagation.
b. Calculate the wavelength of the wave.
c. Write down the vector equation for magnetic field.
(Given the speed of e.m.w. in the vacuum, c =3.00 x 108 m s-1)
Solution:
) [(
]
r
5
−1
14
−1
ˆ
E = − 2.30 x10
r V m sin 1.45 x10 rad s t − kx j
ˆ
compare with E = − E0 sin [ω t − kx ] j
14
−1
5
−1
, we get ω = 1.45 × 10 rad s and E0 = 2.30 × 10 V m
(
a. By using the equation given :
) [(
)
]
r
E = − 2.30 x10 5 V m −1 sin 1.45 x10 14 rad s −1 t − kx ˆj
(
)
The e.m.w.
e.m.w. propagates in +x
direction.
b. The frequency of the e.m.w. is
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ω = 2πf
f = 2.31 × 10 13 Hz
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Then the wavelength of the e.m.w. is
c = fλ
λ = 1.30 × 10 −5 m
c.
The magnetic field amplitude is
E0 = cB0
B0 = 7.67 × 10 −4 T
and the wave number is
k=
y
r
B
− B0
2π
λ
k = 4.83 × 10 5 m −1
c
x
z − E0
r
E
Therefore the vector equation for B is
r
B = − B0 sin [ω t − kx ]kˆ
r
B = − 7.67 × 10−4 T sin 1.45 × 1014 rad s −1 t − 4.83 × 105 m−1 x kˆ
) [(
(
)]
) (
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{
Example 4 :
An electromagnetic wave propagates in a dielectric material. At the
frequency of the light, the dielectric constant of the material is 1.83 and
the relative permeability is 1.36. If the magnetic field amplitude is
4.80 x 10-9 T, calculate the electric field amplitude.
(Given ε0=8.85x10-12 F m-1 and µ0=4πx10-7 H m-1)
κ=1.83,µr=1.36, B0=4.80x10-9 T
By using the equation relates the µ, ε and v in the material, thus
Solution
v=
1
µε
v=
1
µ r µ0 κε 0
where µ
= µr µ0
and
ε = κε0
v = 1.90 × 10 8 m s -1
Hence the electric field amplitude is
E0 = vB0
E0 = 0.912V m−1 @ N C −1
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9.4 Intensity of Electromagnetic Wave
{
Definition – is defined as the rate of energy flow across unit area
perpendicular to the direction of the wave propagation
propagation.
Mathematically,
U
At
or
I=
Since P
=
U
t
where
{
I=
(9.4a)
P
A
(9.4b)
I : intensity of the e.m.w.
U : energy of the e.m.w.
A : surface area
P : power of e.m.w. source
t : time
It is scalar quantity and its unit is W m-2.
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{
Since the e.m.w. consists of varying electric and magnetic fields then
the instantaneous energy density (energy per unit volume) stored in
e.m.w. also associated with an electric and magnetic field. It is given by
z
Instantaneous energy stored per unit volume associated with
an electric field,
field U
1
E
(9.4c)
= ε0 E 2
z
V
2
V
2  µ0 
(derivation of eq. 9.4c- refer to Physics for scientists and Engineers
, Serway & Jerway, 6th edition, pg. 807-808)
Instantaneous energy stored per unit volume associated with
an magnetic field,
field U
1  B2 
B
(9.4d)
=  
(derivation of eq. 9.4d- refer to Physics for scientists and Engineers
, Serway & Jerway, 6th edition, pg. 1011-1012)
{
Consider a light beam (e.m.w.) propagates in a vacuum in time t as
shown in figure 9.4a.
ct
Fig. 9.4a
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A
c
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z
From the fig. 9.4a the energy volume of the e.m.w. is given by
z
By rearranging the eq. (9.4c) and eq. (9.4d), we get
V = Act
{
Energy stored in electric field E :
UE =
{
Energy stored in magnetic field B :
UB =
{
1
1
ε0 E 2V = ε0 E 2 ( Act )
2
2
1  B2 
1  B2 
 V =  ( Act )
2  µ0 
2  µ0 
Total energy of e.m.w. is given by
U = UE +UB
1
1  B2 
U = ε0 E 2 ( Act ) +  ( Act )
2
2  µ0 
2

Act
B 
(9.4e)
 ε0 E 2 +

U=
2 
µ0 
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z
Rearrange eq. (9.4e), we get the instantaneous intensity of the
e.m.w. is
2
U c
B
=  ε0 E 2 +
At 2 
µ0
z
Since
U

=I
 and
At

c
B2 
2
(9.4f)


I =  ε0 E +
2
µ0 
1
2
2
, we get B = ε 0 µ0 E
E = cB and c =
ε0 µ 0
then eq. (9.4f) can be written as
c
ε µ E2 
 ε0 E 2 + 0 0 
2
µ0 
2
I = cε0 E
(9.4g)
I=
{
The average intensity of the e.m.w in a vacuum is given by
I av = cε0 E 2
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and
E2 =
1 2
E0
2
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I av =
1
cε0 E02
2
(9.4h)
or
I av =
{
1  cB02 


2  µ0 
The average intensity of the e.m.w in a medium is given by
1
I av = vεE02
2
{
(9.4i)
1  vB02 

I av = 
2  µ 
or
so
Fig. 9.4b
{
(9.4j)
An electromagnetic wave propagates the source in all directions, hence
it is a three-dimensional wave and is said to be a spherical wave as
shown in figure 9.4b.
{
Consider two points at distances r1 and r2
from the source ( power, P is constant),
then
P
I1 =
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1
E0 = cB0 and c =
ε 0 µ0
B02 = ε0 µ0 E02
B02
2
E0 =
ε0 µ0
4πr
2
1
and
I 2 r12
=
I 1 r22
I2 =
P
4πr22
(9.4k)
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Example 5 :
A sinusoidal electromagnetic wave emitted by a cellular phone has a
wavelength of 35.4 cm and an electric field amplitude of 5.40 x 10-2 V m-1
at a distance of 250 m from the antenna. Calculate
a. the frequency of the wave.
b. the magnetic field amplitude.
c. the intensity of the wave.
(Given c=3.00x108 m s-1, ε0=8.85x10-12 F m-1 and µ0=4πx10-7 H m-1)
(Young&Freedman, pg.1243,no.32.18)
λ=35.4x10-2 m, E0=5.40x10-2 V m-1
a. By using the equation relates the c, f and λ , thus
c = fλ
Solution
f = 8.48 x10 8 Hz
b. The magnetic field amplitude is
E0 = cB0
B0 = 1.80 × 10 −10 T
c. The intensity of the wave is
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1
I av = cε0 E02
2
−6
or
I av = 3.87 x10 W m−2
1  cB02 

I av = 
2  µ0 
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{
Example 6 : (exercise)
An electromagnetic wave with frequency of 5.70 x 1014 Hz propagates
with a speed of 2.17 x108 m s-1 in a certain piece of glass. Calculate
a. the wavelength of the wave in the glass.
b. the wavelength of a wave of the same frequency propagating in air.
c. the refractive index n of the glass.
d. the dielectric constant of the glass if the relative permeability is
unity ( ≈1).
(Given c=3.00x108 m s-1, ε0=8.85x10-12 F m-1 and µ0=4πx10-7 H m-1)
(Young&Freedman, pg.1243,no.32.11)
{
{
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Ans. : 3.81 x10-7 m, 5.26 x10-7 m,1.38, 1.90
Example 7 : (exercise)
A monochromatic light source with power output 60.0 W radiates light of
wavelength 700 nm uniformly in all directions. Calculate the maximum
electric field and maximum magnetic field at a distance of 5.00 m from
the source. (Young&Freedman, pg.1243,no.32.19)
(Given c=3.00x108 m s-1, ε0=8.85x10-12 F m-1 and µ0=4πx10-7 H m-1)
Ans. : 12.0 V m-1, 4.00 x 10-8 T
Example 8 : (exercise)
For waves propagating in air, calculate the wavelength in metres of
a. gamma rays of frequency 6.50 x 1021 Hz.
b. visible light of frequency 5.75 x 1014 Hz.
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Ans. :4.61 x 10-14 m, 5.21 x 10-7 m
9.5 Comparison between EM wave and
Mechanical Wave.
E.M. Wave
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Mechanical Wave
{
Produced by the vibration
of E and B.
{
Produced by the vibration
of the medium particles.
{
Transverse wave only
{
Transverse and
longitudinal wave
{
It can propagate in the
vacuum.
{
It needs a medium to
propagate.
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THE END…
Next Unit…
UNIT 10 :
PHOTONS & QUANTIZED ENERGY
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