EM radiation is produced by accelerating
charges. example Radio waves
+
+
EM waves
3.2
-
• Producing EM waves
• Electromagnetic Spectrum
• Energy in EM waves
E field lines
B field lines
at t=0
Time->
Evolution of the E and B fields
Dipole Antenna
+q
+
-q
-
Dipole antenna
d(t) Oscillating Dipole
λ/4
Electric Field
lines
λ/2
radiation ⊥ to the
dipole axis
Half-wave antenna
Quarter wave antenna
Electric field lines
No radiation || to the
dipole axis (Why?)
Question
A cell phone uses a frequency of 2.0 GHz.
Find the length of the quarter wave
antenna.
1) 3.8mm
2) 0.38 cm
3) 3.8 cm
4) 38 cm
Interactions of EM radiation with an
Antenna.
Oscillating Electric field drives the movement of charges in
a conductor
+
E
• An optimal antenna to capture energy from an EM
wave has a size close λ. (λ/2 for a dipole antenna)
1
Passage of EM radiation through
holes in a conductor
reflected
Spectrum of EM waves
transmitted
• EM waves pass easily through holes in a conductor that
are larger than λ but are blocked by holes smaller than λ.
• When the size of the hole is close to the λ, interference
and diffraction effects are observed (discussed later).
Radio Waves
Microwaves
Fm radio, TV
f~ 100 MHz
λ~1m
f~ 1-10 GHz
λ~ 1-10 cm
dipole antenna
length =λ/4
Visible Light
Silver nano-particles
f~1015 Hz
λ~ 400-700 nm
stained glass, gold particles
with diameter ~100 nm
scatter specific colors of light
Electron micrograph
light micrograph
2
X-rays
X-rays
f ~ 1016-1020 Hz
X-rays penetrate
soft tissue but
are absorbed by
heavy atoms,
such as Calcium
in bones.
λ~ 10-8 -10-12 M
Atomic dimensions
x-ray diffraction
pattern
3-dimensional
structure
Energy carried by an EM wave
Energy density of E and B fields (from capacitor and inductor)
1
2
Intensity
Average power
P=
µE = ε o E 2
µB =
Joules/m3
2
B
2 µo
Average energy density (average of
Eo and Bo are the peak fields
µE =
1
1
1 Bo2
ε o Eo 2 = ε o c 2Bo2 =
4
4
4 µo
µB =
Use
E o2 = c 2Bo2
Bo2
4 µo
Use
=>
1
ε o E o2c
2
1 Bo2
S=
c
2 µo
=>
S=
1
µE = ε o Eo 2 =>
4
µB =
Bo2
4 µo
Eo = cBo
In general expression for the intensity of EM radiation
uv uv
uv ExB
S=
µo
The average
intensity
Saverage =
Erms xBrms
µo
uuv uuv
uv
E xB
S average = o o
2 µo
Units W/m2
λ
c=λ/T
S=
The average energy density for E and B are the same
Poynting Vector
T
P
S = = µtotal c = 2µE c = 2µB c
A
Average Intensity
sin2θ=1/2)
µtotal Aλ
V=Aλ
A
1 Bo Eo
2 µo
34.4 Light intensity
The light intensity at noon on a sunny day is about 1kW/m2.
What is the peak E field due to the light? What is peak B
field?
Electric Field
S=
Eo =
1
ε o Eo2c
2
2S
2(1000)
=
= 8.7x102 V / m
εo c
8.85x10−12 (3x108 )
Magnetic Field
Bo =
Eo 8.7x102
=
= 3x10−6 T
c
3x108
In terms of normal laboratory fields the E field is more
significant than the B field. (i.e. for interactions with matter.
3
Cell phone
Microwave ovens and cell phones.
If a cell phone antenna can detect a peak electric field of
10-3 V/m how much power must a transmitter that is 10 km away
have if it transmits waves in all directions.
Frequency =2.45GHz
Power ~ 1kW
compare to cell phone
power ~1W.
• Microwaves are reflected from the walls
of the cooking chamber and the energy is
confined to cook the food.
• Microwaves form standing waves in the
chamber.
Microwaves are absorbed by water
molecules reoriented by the E field
E field
O
H
H
electric dipole
orients with the
fluctuating field
4