22.1 Physics 6C EM Waves - UCSB Campus Learning Assistance

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Electromagnetic Waves
Physics 6C
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Electromagnetic Waves
Electromagnetic (EM) waves can be produced by atomic transitions
(more on this later), or by an alternating current in a wire. As the
charges in the wire oscillate back and forth, the electric field around
them oscillates as well, in turn producing an oscillating magnetic field.
This magnetic field is always perpendicular to the electric field, and the
EM wave propagates perpendicular to both the E- and B-fields. This
gives us a right-hand-rule relating the directions of these 3 vectors:
1) Point the fingers of your right hand in the direction of the E-field
2) Curl them toward the B-field.
3) Stick out your thumb - it points in the direction of propagation.
Click here for an EM wave animation
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Like any other wave, we know the relationship
between the wavelength and frequency, and
the speed of propagation of the wave:
v wave  f  
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Like any other wave, we know the relationship
between the wavelength and frequency, and
the speed of propagation of the wave:
v wave  f  
In the case of EM waves, it turns out that
the wave speed is the speed of light.
So our formula for EM waves (in vacuum) is:
c  f  ; c 
1
0   0
 3  10
8 m
s
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Like any other wave, we know the relationship
between the wavelength and frequency, and
the speed of propagation of the wave:
v wave  f  
In the case of EM waves, it turns out that
the wave speed is the speed of light.
So our formula for EM waves (in vacuum) is:
c  f  ; c 
1
0   0
 3  10
8 m
s
It turns out that the speed of light is
also related to the strengths of the
Electric and Magnetic fields.
E=cB (in standard metric units)
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Like any other wave, we know the relationship
between the wavelength and frequency, and
the speed of propagation of the wave:
v wave  f  
In the case of EM waves, it turns out that
the wave speed is the speed of light.
So our formula for EM waves (in vacuum) is:
c  f  ; c 
1
0   0
 3  10
8 m
s
It turns out that the speed of light is
also related to the strengths of the
Electric and Magnetic fields.
E=cB (in standard metric units)
The continuum of various
wavelengths and frequencies for
EM waves is called the
Electromagnetic Spectrum
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Examples:
• Find the frequency of blue light with a wavelength of 460 nm.
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Examples:
• Find the frequency of blue light with a wavelength of 460 nm.
c  f   f 
c


3  10
8 m
s
9
460  10
m
 6 . 5  10
14
Hz
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Examples:
• Find the frequency of blue light with a wavelength of 460 nm.
c  f   f 
c


3  10
8 m
s
9
460  10
m
 6 . 5  10
14
Hz
• A cell phone transmits at a frequency of 1.25x108 Hz.
What is the wavelength of this EM wave?
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Examples:
• Find the frequency of blue light with a wavelength of 460 nm.
c  f   f 
c


3  10
8 m
s
9
460  10
m
 6 . 5  10
14
Hz
• A cell phone transmits at a frequency of 1.25x108 Hz.
What is the wavelength of this EM wave?
c  f    
c
f

3  10
8 m
s
8
1 . 25  10 Hz
 2 .4m
You will need to use this formula very often to convert
back and forth between frequency and wavelength.
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Energy and momentum in EM Waves
Electromagnetic waves transport energy. The energy associated with a wave is stored in
the oscillating electric and magnetic fields.
We will find out later that the frequency of the wave determines the amount of energy
that it carries. Since the EM wave is in 3-D, we need to measure the energy density
(energy per unit volume).
u avg 
1
2
2
2
 0 E 0   0 E rms 
1
2 0
2
B0 
1
0
2
B rms
This is the energy per unit volume
Note that the energy can be written in a few equivalent forms. Each can be useful,
depending on the information you know about the wave.
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Energy and momentum in EM Waves
Electromagnetic waves transport energy. The energy associated with a wave is stored in
the oscillating electric and magnetic fields.
We will find out later that the frequency of the wave determines the amount of energy
that it carries. Since the EM wave is in 3-D, we need to measure the energy density
(energy per unit volume).
u avg 
1
2
2
2
 0 E 0   0 E rms 
1
2 0
2
B0 
1
0
2
B rms
This is the energy per unit volume
Note that the energy can be written in a few equivalent forms. Each can be useful,
depending on the information you know about the wave.
We can also talk about the intensity of an EM wave (for light we would think of it as
brightness). Just as for sound, intensity is measured as average power/area.
S avg 
Power
Area
 c  u avg
Just multiply the energy equation above
by the speed of light to get the intensity.
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Example: High-Energy Cancer Treatment
Scientists are working on a technique to kill cancer cells by zapping them with ultrahighenergy pulses of light that last for an extremely short amount of time. These short pulses
scramble the interior of a cell without causing it to explode, as long pulses do.
We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for
4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread
uniformly over the faces of 100 cells for each pulse.
a) How much energy is given to the cell during this pulse?
b) What is the intensity (in W/m2) delivered to the cell?
c) What are the maximum values of the electric and magnetic fields in the pulse?
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Example: High-Energy Cancer Treatment
Scientists are working on a technique to kill cancer cells by zapping them with ultrahighenergy pulses of light that last for an extremely short amount of time. These short pulses
scramble the interior of a cell without causing it to explode, as long pulses do.
We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for
4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread
uniformly over the faces of 100 cells for each pulse.
a) How much energy is given to the cell during this pulse?
b) What is the intensity (in W/m2) delivered to the cell?
c) What are the maximum values of the electric and magnetic fields in the pulse?
Recall that power is energy/time. So 2.0x1012 W is 2.0x1012 Joules/sec.
Energy
 (2 . 0  10
12 J
)
s
 ( 4 . 0  10
9
3
s )  8  10 J  8000 J
This is the total energy, which is spread out over 100 cells, so the
energy for each individual cell is 80 Joules.
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Example: High-Energy Cancer Treatment
Scientists are working on a technique to kill cancer cells by zapping them with ultrahighenergy pulses of light that last for an extremely short amount of time. These short pulses
scramble the interior of a cell without causing it to explode, as long pulses do.
We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for
4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread
uniformly over the faces of 100 cells for each pulse.
a) How much energy is given to the cell during this pulse?
b) What is the intensity (in W/m2) delivered to the cell?
c) What are the maximum values of the electric and magnetic fields in the pulse?
To get intensity, we need to divide power/area. The area for a cell is
just the area of a circle:
Area   r
2
   (2 . 5  10
6
2
m)
 2 . 0  10
 11
m
2
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Example: High-Energy Cancer Treatment
Scientists are working on a technique to kill cancer cells by zapping them with ultrahighenergy pulses of light that last for an extremely short amount of time. These short pulses
scramble the interior of a cell without causing it to explode, as long pulses do.
We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for
4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread
uniformly over the faces of 100 cells for each pulse.
a) How much energy is given to the cell during this pulse?
b) What is the intensity (in W/m2) delivered to the cell?
c) What are the maximum values of the electric and magnetic fields in the pulse?
To get intensity, we need to divide power/area. The area for a cell is
just the area of a circle:
Area   r
2
   (2 . 5  10
6
2
m)
 2 . 0  10
 11
m
2
Now divide to get intensity:
Intensity

Power
100   r
This is the total area
of all 100 cells.
2

2 . 0  10
2 . 0  10
12
9
W
m
2
 1 . 0  10
21
W
m
2
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Example: High-Energy Cancer Treatment
Scientists are working on a technique to kill cancer cells by zapping them with ultrahighenergy pulses of light that last for an extremely short amount of time. These short pulses
scramble the interior of a cell without causing it to explode, as long pulses do.
We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for
4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread
uniformly over the faces of 100 cells for each pulse.
a) How much energy is given to the cell during this pulse?
b) What is the intensity (in W/m2) delivered to the cell?
c) What are the maximum values of the electric and magnetic fields in the pulse?
To get the field strengths, recall our formulas:
u avg 
1
2
2
2
 0 E 0   0 E rms 
1
2 0
2
B0 
1
0
2
B rms
S avg 
Power
Area
 c  u avg
Prepared by Vince Zaccone
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Assistance Services at UCSB
Example: High-Energy Cancer Treatment
Scientists are working on a technique to kill cancer cells by zapping them with ultrahighenergy pulses of light that last for an extremely short amount of time. These short pulses
scramble the interior of a cell without causing it to explode, as long pulses do.
We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for
4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread
uniformly over the faces of 100 cells for each pulse.
a) How much energy is given to the cell during this pulse?
b) What is the intensity (in W/m2) delivered to the cell?
c) What are the maximum values of the electric and magnetic fields in the pulse?
To get the field strengths, recall our formulas:
u avg 
1
2
2
2
 0 E 0   0 E rms 
Since the power was stated as
average power we should
assume that is the rms value. So
our field values should get
multiplied by √2 to find the
maximum.
1
2 0
2
B0 
E0 
B0 
1
0
2
B rms
2S
c  0
 8 . 7  10
20  S
c
S avg 
11
Power
Area
 c  u avg
V
m
3
 2 . 9  10 T
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Photons
The energy carried by an EM wave comes in packets called photons.
The energy of a photon depends on the frequency of the EM wave.
E photon  hf
The constant h is called Planck’s constant.
ℎ ≈ 6.626  10−34  ∙ 
Notice that this is an incredibly small number.
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Photons
The energy carried by an EM wave comes in packets called photons.
The energy of a photon depends on the frequency of the EM wave.
E photon  hf 
hc

The constant h is called Planck’s constant.
ℎ ≈ 6.626  10−34  ∙ 
Notice that this is an incredibly small number.
Because photon energies are usually so small, it is often convenient to express their
energy in units of electron-volts (eV) instead of Joules. Recall the conversion factor:
1 eV = 1.6x10-19 J
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Photons
The energy carried by an EM wave comes in packets called photons.
The energy of a photon depends on the frequency of the EM wave.
E photon  hf 
hc

The constant h is called Planck’s constant.
ℎ ≈ 6.626  10−34  ∙ 
Notice that this is an incredibly small number.
Because photon energies are usually so small, it is often convenient to express their
energy in units of electron-volts (eV) instead of Joules. Recall the conversion factor:
1 eV = 1.6x10-19 J
For convenience, Planck’s constant can be converted to eV instead of Joules:
ℎ ≈ 4.14  10−15  ∙ 
This will be useful when dealing with photon energies.
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Photons
Example: A typical x-ray machine scans the body with EM waves of frequency 7x1018Hz.
How much energy is in a typical x-ray photon?
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Photons
Example: A typical x-ray machine scans the body with EM waves of frequency 7x1018Hz.
How much energy is in a typical x-ray photon?
We can give the answer in Joules or eV:
E photon  hf
E photon  ( 6 . 626  10
E photon  ( 4 . 14  10
 34
 15
J  s )( 7  10
18
eV  s )( 7  10
18
Hz )  4 . 64  10
 15
J
Hz )  29 ,000 eV  29 keV
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Energy and momentum in EM Waves
EM waves also carry momentum. This means that a ray of light can actually exert a
force. To get the pressure exerted by a sinusoidal EM wave, just divide the intensity
by the speed of light.
Radiation
Pr essure

S avg
This is the same as the total energy absorbed by the surface.
c
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Energy and momentum in EM Waves
EM waves also carry momentum. This means that a ray of light can actually exert a
force. To get the pressure exerted by a sinusoidal EM wave, just divide the intensity
by the speed of light.
Radiation
Pr essure

S avg
c
Example: Solar Sails
Suppose a spacecraft with a mass of 25,000 kg has a solar sail made of perfectly
reflective aluminized film with an area of 2.59x106 m. If the spacecraft is launched
into earth orbit and then deploys its sail at right angles to the sunlight, what is the
acceleration due to sunlight? Assume that at the earth’s distance from the sun, the
pressure exerted by sunlight on an absorbing surface is 4.70x10-6 Pa.
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Example: Solar Sails
Suppose a spacecraft with a mass of 25,000 kg has a solar sail made of perfectly
reflective aluminized film with an area of 2.59x106 m. If the spacecraft is launched
into earth orbit and then deploys its sail at right angles to the sunlight, what is the
acceleration due to sunlight? Assume that at the earth’s distance from the sun, the
pressure exerted by sunlight on an absorbing surface is 4.70x10-6 Pa.
Recall that Pressure = Force/Area. We can use this and F=ma to get our formula:
P 
F
A
 F  PA
F  m a  a 
F
a 
P A
m
m
Now for the tricky part: When the pressure number was given above, that was for
an absorbing surface. What happens when the sunlight reflects instead?
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Example: Solar Sails
Suppose a spacecraft with a mass of 25,000 kg has a solar sail made of perfectly
reflective aluminized film with an area of 2.59x106 m. If the spacecraft is launched
into earth orbit and then deploys its sail at right angles to the sunlight, what is the
acceleration due to sunlight? Assume that at the earth’s distance from the sun, the
pressure exerted by sunlight on an absorbing surface is 4.70x10-6 Pa.
Recall that Pressure = Force/Area. We can use this and F=ma to get our formula:
P 
F
A
 F  PA
F  m a  a 
a 
F
P A
m
m
Now for the tricky part: When the pressure number was given above, that was for
an absorbing surface. What happens when the sunlight reflects instead?
Twice as much momentum is transferred!
a 
2 ( 4 . 7  10
6
6
N
m
)  2 . 59  10 m
2
4
2 . 5  10 kg
2
 9 . 72  10
4 m
s
2
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