Download Chapter 22 – Electromagnetic Waves

College Physics 150 Chapter 22 – Electromagnetic Waves
• Production of EM waves
• Maxwell’s Equations
• Antennae
• The EM Spectrum
• Speed of EM Waves
• Energy Transport
• Polarization
• Doppler Effect
Maxwell’s Equations and EM Waves
A stationary charge produces an electric field.
A charge moving at constant speed produces electric and magnetic fields. A charge that is accelerated will produce variable electric and magnetic fields. These are electromagnetic (EM) waves. If the charge oscillates with a frequency f, then the resulting EM wave will have a frequency f. If the charge ceases to oscillate, then the EM wave is a pulse (a finite-­‐‑sized wave).
Maxwell’s Equations
Gauss’s Law:Electric fields (not induced) must begin on + charges and end on -­‐‑ charges.
Gauss’s Law for magnetism:There are no magnetic monopoles (a magnet must have at least one north and one south pole).
Faraday’s Law:A changing magnetic field creates an electric field. Ampère-­‐‑Maxwell Law:A current or a changing electric field creates a magnetic field. Maxwell’s Equations
Gauss’s Law:Electric fields (not induced) must begin on + charges and end on -­‐‑ charges.
Gauss’s Law for magnetism:There are no magnetic monopoles (a magnet must have at least one north and one south pole).
Faraday’s Law:A changing magnetic field creates an electric field. Ampère-­‐‑Maxwell Law:A current or a changing electric field creates a magnetic field. When Maxwell’s equations are combined, the solutions are electric and magnetic fields that vary with position and time. These are EM waves.
An electric field only wave cannot exist, nor can a magnetic field only wave.
Antennae
An electric field parallel to an antenna (electric dipole) will “shake” electrons and produce an AC current.
An EM wave also has a magnetic component. A magnetic dipole antenna can be oriented so that the B-­‐‑field passes into and out of the plane of a loop, inducing a current in the loop. The B-­‐‑field of an EM wave is perpendicular to its E-­‐‑field and also the direction of travel.
Example (text problem 22.5): An electric dipole antenna used to transmit radio waves is oriented vertically. At a point due south of the transmiZer, what is the direction of the wave’s B-­‐‑field?
Looking down from above the Electric Dipole antenna
N
W
E
S
South of the transmiZer, the E-­‐‑field is directed into/out of the page. The B-­‐‑
field is perpendicular to this direction and also to the direction of travel (South). The B-­‐‑field must be east-­‐‑west. The EM Spectrum
EM waves of any frequency can exist.
Energy increases with increasing frequency.
Speed of Light
In 1675 Ole Römer presented a calculation of the speed of light. He used the time between eclipses of Jupiter’s Galilean Satellites to show that the speed of light was finite and that its value was 2.25×108 m/s.
Fizeau’s experiment of 1849 measured the value to be about 3×108 m/s. (done before Maxwell’s work)
Maxwell was able to derive the speed of EM waves in vacuum. EM waves do not need a medium to travel through.
c=
=
1
ε0µ0
1
(8.85 ×10
−12
= 3.00 ×108 m/s
C 2 /Nm 2 ) ( 4π ×10 −7 Tm/A )
When light travels though a material medium, its speed is reduced.
c
v=
n
where v is the speed of light in the medium and n is the refractive index of the medium. When a wave passes from one medium to another the frequency stays the same, but the wavelength is changed.
A dispersive medium is one in which the index of refraction depends on the wavelength of light. Example (text problem 22.22): In order to study the structure of a crystalline solid, you want to illuminate it with EM radiation whose wavelength is the same as the spacing of the atoms in the crystal (0.20 nm).
(a) What is the frequency of the EM radiation?
3.0 ×108 m/s
15
f = =
=
1
.
5
×
10
Hz
−9
λ 0.20 ×10 m
c
(b) In what part of the EM spectrum does it lie?
X-­‐‑ray
Properties of EM Waves
All EM waves in vacuum travel at the “speed of light” c.
Both the electric and magnetic fields have the same oscillation frequency f.
The electric and magnetic fields oscillate in phase.
The fields are related by the relationship
E ( x, y, z, t ) = cB ( x, y, z, t )
EM waves are transverse. The fields oscillate in a direction that is perpendicular to the wave’s direction of travel. The fields are also perpendicular to each other.
E× B.
The direction of propagation is given by The wave carries one-­‐‑half of its energy in its electric field and one-­‐‑half in its magnetic field.
Properties of EM Waves
All EM waves in vacuum travel at the “speed of light” c.
Both the electric and magnetic fields have the same oscillation frequency f.
The electric and magnetic fields oscillate in phase.
The fields are related by the relationship
E ( x, y, z, t ) = cB ( x, y, z, t )
EM waves are transverse. The fields oscillate in a direction that is perpendicular to the wave’s direction of travel. The fields are also perpendicular to each other.
E× B.
The direction of propagation is given by The wave carries one-­‐‑half of its energy in its electric field and one-­‐‑half in its magnetic field.
Ez ( y, t ) = Em sin (ky − ωt + φ )
The amplitude
wave number
k=
2π
λ
The wave speed is c = λf
angular frequency
ω = 2πf
=
ω
k
.
phase constant
Example (text problem 22.32): The electric field of an EM wave is given by: π ⎞
⎛
E z ( y, t ) = Em sin ⎜ ky − ωt + ⎟
6 ⎠
⎝
Ex = 0
Ey = 0
(a) In what direction is this wave traveling?
The wave does not depend on the coordinates x or z; it must travel parallel to the y-­‐‑axis. The wave travels in the +y direction. Example (text problem 22.32): The electric field of an EM wave is given by: π ⎞
⎛
E z ( y, t ) = Em sin ⎜ ky − ωt + ⎟
6 ⎠
⎝
Ex = 0
Ey = 0
(b) Write expressions for the magnetic field of this wave.
E× B
must be in the +y-­‐‑direction (E is in the z-­‐‑direction).
Therefore, B must be along the x-­‐‑axis.
Bz = 0, By = 0
π ⎞
⎛
Bx ( y, t ) = Bm sin ⎜ ky − ωt + ⎟
6 ⎠
⎝
Em
with Bm =
c
Energy Transport by EM Waves
The intensity of a wave isI
Pav
=
.
A
This is a measure of how much energy strikes a surface of area A every second for normal incidence.
The rays make a 90° angle with the surface.
Surface
Also,
ΔE uavV uav AΔx
I=
=
=
= uav c
AΔt AΔt
AΔt
where uav is the average energy density (energy per unit volume) contained in the wave.
For EM waves:
u
av
= ε0E
2
rms
=
1
µ0
2
Brms
Example (text problem 22.40): The intensity of the sunlight that reaches Earth’s upper atmosphere is 1400 W/m2.
(a) What is the total average power output of the Sun, assuming it to be an isotropic source?
Pav = IA = I ( 4π R 2 )
= 4π (1400 W/m
2
) (1.50 ×10
11
m)
2
= 4.0 ×10 26 W
(b) What is the intensity of sunlight incident on Mercury, which is 5.8×1010 m from the Sun? Pav
Pav
I=
=
A 4π r 2
4.0 ×10 26 W
=
2
10
4π ( 5.8 ×10 m )
= 9460 W/m 2
What if the EM waves strike at non-­‐‑normal incidence?
Replace A with Acosθ.
θ
Pav = IA cosθ
This is a drawback for solar panels!
Polarization
A wave on a string is linearly polarized. The vibrations occur in the same plane. The orientation of this plane determines the polarization state of a wave.
For an EM wave, the direction of polarization is given by the direction of the E-­‐‑
field.
The EM waves emiZed by an antenna are polarized; the E-­‐‑field is always in the same direction. A source of EM waves is unpolarized if the E-­‐‑fields are in random directions.
A polarizer will transmit linear polarized waves in the same direction independent of the incoming wave.
It is only the component of the wave’s amplitude parallel to the transmission axis that is transmiZed. If unpolarized light is incident on 1 polarizer, the intensity of the light passing through is I = ½ I0.
If the incident wave is already polarized, then the transmiZed intensity is I = I0cos2θ where θ is the angle between the incident wave’s direction of polarization and the transmission axis of the polarizer. (Law of Malus)
Example (text problem 22.45): Unpolarized light passes through two polarizers in turn with axes at 45° to each other. What is the fraction of the incident light intensity that is transmiZed? I0
I1
I2
When the light passes through the first polarizer the intensity goes from I0 to I1. The light then passes through the second polarizer and the intensity goes to I2.
After passing through the first polarizer, the intensity is ½ of its initial value. The wave is now linearly polarized.
Direction of linear polarization
Transmission axis of 2nd polarizer.
45°
1
I1 = I 0
2
1
1
2
I 2 = I1 cos θ = I 0 cos 45° = I 0
2
4
2
The Doppler Effect
For EM waves, the Doppler shift formula is
fo = f s
v
1+
c
v
1−
c
where fs is the frequency emiZed by the source, fo is the frequency received by the observer, v is the relative velocity of the source and the observer, and c is the speed of light.
If the source and observer are approaching each other, then v is positive, and v is negative if they are receding.
When v/c << 1, the previous expression can be approximated as: ⎛ v ⎞
f o ≈ f s ⎜1 + ⎟
⎝ c ⎠
Example (text problem 22.56): Light of wavelength 659.6 nm is emiZed by a star. The wavelength of this light as measured on Earth is 661.1 nm. How fast is the star moving with respect to the Earth? Is it moving toward Earth or away from it?
The wavelength shift is small (Δλ << λ) so v << c.
⎛ v ⎞
f o ≈ f s ⎜1 + ⎟
⎝ c ⎠
c / λo
λs
v fo
= −1 =
− 1 = − 1 = −0.0023
c fs
c / λs
λo
v = −6.8 ×10 5 m/s = −680 km/s
Star is receding.