Electromagnetic Waves
Physics 4
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Maxwell’s Equations
Maxwell’s equations summarize the relationships between electric
and magnetic fields. A major consequence of these equations is
that an accelerating charge will produce electromagnetic radiation.
𝐸 ∙ 𝑑𝐴 =
𝑄𝑒𝑛𝑐𝑙
𝜖0
𝐵 ∙ 𝑑𝐴 = 0
𝑮𝒂𝒖𝒔𝒔′ 𝒔 𝑳𝒂𝒘 𝒇𝒐𝒓 𝑬
𝑮𝒂𝒖𝒔𝒔′ 𝒔 𝑳𝒂𝒘 𝒇𝒐𝒓 𝑩
𝐵 ∙ 𝑑𝑙 = 𝜇0 𝑖𝑐 + 𝜖0
𝑑Φ𝐵
𝐸 ∙ 𝑑𝑙 = −
𝑑𝑡
𝑑Φ𝐸
𝑑𝑡
𝑨𝒎𝒑𝒆𝒓𝒆′ 𝒔 𝑳𝒂𝒘
𝑒𝑛𝑐𝑙
𝑭𝒂𝒓𝒂𝒅𝒂𝒚′𝒔 𝑳𝒂𝒘
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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.
We have a right-hand-rule for plane EM waves:
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:
vwave  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:
vwave  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  108
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:
vwave  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  108
m
s
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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The continuum of various wavelengths and frequencies
for EM waves is called the Electromagnetic Spectrum
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The continuum of various wavelengths and frequencies
for EM waves is called the Electromagnetic Spectrum
• Find the frequency of blue light with a wavelength of 460 nm.
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The continuum of various wavelengths and frequencies
for EM waves is called the Electromagnetic Spectrum
• Find the frequency of blue light with a wavelength of 460 nm.
3  108 m
c
s
c  f   f  
 6.5  1014Hz
 460  109m
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The continuum of various wavelengths and frequencies
for EM waves is called the Electromagnetic Spectrum
• A cell phone transmits at a frequency of 1.25x108 Hz.
What is the wavelength of this EM wave?
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The continuum of various wavelengths and frequencies
for EM waves is called the Electromagnetic Spectrum
• A cell phone transmits at a frequency of 1.25x108 Hz.
What is the wavelength of this EM wave?
8m
3  10
c
s
c  f     
 2.4m
f 1.25  108Hz
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Field of a Sinusoidal Wave
Electromagnetic waves must satisfy the WAVE EQUATION:
In the case of EM waves, both the electric and magnetic fields
need to satisfy this equation. Solving this equation yields
formulas for the E and B fields.
𝜕 2 𝑦(𝑥, 𝑡)
1 𝜕 2 𝑦(𝑥, 𝑡)
= 2
𝜕𝑥 2
𝑣
𝜕𝑡 2
In particular, here are formulas for the E and B fields associated
with a sinusoidal EM plane wave propagating in the +x-direction:
𝐸 𝑥, 𝑡 = 𝐸𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡)𝑗
𝐵 𝑥, 𝑡 = 𝐵𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡)𝑘
2𝜋
𝑘 = 𝑤𝑎𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 =
𝜆
𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 2𝜋𝑓
Notice that these fields are perpendicular to each other, as well as the propagation
direction. A right hand rule comes in handy to remember the directions.
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Field of a Sinusoidal Wave
Electromagnetic waves must satisfy the WAVE EQUATION:
In the case of EM waves, both the electric and magnetic fields
need to satisfy this equation. Solving this equation yields
formulas for the E and B fields.
𝜕 2 𝑦(𝑥, 𝑡)
1 𝜕 2 𝑦(𝑥, 𝑡)
= 2
𝜕𝑥 2
𝑣
𝜕𝑡 2
In particular, here are formulas for the E and B fields associated
with a sinusoidal EM plane wave propagating in the +x-direction:
𝐸 𝑥, 𝑡 = 𝐸𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡)𝑗
𝐵 𝑥, 𝑡 = 𝐵𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡)𝑘
2𝜋
𝑘 = 𝑤𝑎𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 =
𝜆
𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 2𝜋𝑓
Notice that these fields are perpendicular to each other, as well as the propagation
direction. A right hand rule comes in handy to remember the directions.
Example: A sinusoidal EM wave of frequency 6.10x1014Hz travels in vacuum in the
+z-direction. The B-field is parallel to the y-axis and has amplitude 5.80x10-4T.
Write the equations for the E and B fields.
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Field of a Sinusoidal Wave
Electromagnetic waves must satisfy the WAVE EQUATION:
In the case of EM waves, both the electric and magnetic fields
need to satisfy this equation. Solving this equation yields
formulas for the E and B fields.
𝜕 2 𝑦(𝑥, 𝑡)
1 𝜕 2 𝑦(𝑥, 𝑡)
= 2
𝜕𝑥 2
𝑣
𝜕𝑡 2
In particular, here are formulas for the E and B fields associated
with a sinusoidal EM plane wave propagating in the +x-direction:
𝐸 𝑥, 𝑡 = 𝐸𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡)𝑗
𝐵 𝑥, 𝑡 = 𝐵𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡)𝑘
2𝜋
𝑘 = 𝑤𝑎𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 =
𝜆
𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 2𝜋𝑓
Notice that these fields are perpendicular to each other, as well as the propagation
direction. A right hand rule comes in handy to remember the directions.
Example: A sinusoidal EM wave of frequency 6.10x1014Hz travels in vacuum in the
+z-direction. The B-field is parallel to the y-axis and has amplitude 5.80x10-4T.
Write the equations for the E and B fields.
𝐸 𝑧, 𝑡 = 𝐸𝑚𝑎𝑥 cos(𝑘𝑧 − 𝜔𝑡)𝑖
𝜔 = 2𝜋 6.1 ∙ 1014 𝐻𝑧 = 3.83 ∙ 1015 𝑟𝑎𝑑
𝑠
𝐵 𝑧, 𝑡 = 𝐵𝑚𝑎𝑥 cos(𝑘𝑧 − 𝜔𝑡)𝑗
𝜔 3.83 ∙ 1015 𝑟𝑎𝑑
𝑠
7 𝑟𝑎𝑑
𝑘= =
=
1.28
∙
10
𝑚
𝑚
𝑐
3 ∙ 108 𝑠
−4
5𝑉
𝐸𝑚𝑎𝑥 = 𝑐𝐵𝑚𝑎𝑥 = 3 ∙ 108 𝑚
𝑠 5.8 ∙ 10 𝑇 = 1.74 ∙ 10 𝑚
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EM Waves in matter
So far we have assumed that electromagnetic waves propagated through empty space.
If they travel through a transparent material medium (glass, air, water, etc.) the speed
of propagation changes.
𝑐=
𝑣=
1
𝜖0 𝜇0
1
=
𝜖𝜇
This is the speed in vacuum
1
𝐾𝜖0 𝐾𝑚 𝜇0
=
𝑐
𝐾𝐾𝑚
=
𝑐
𝑛
This is the speed in a material
medium with dielectric constant* K
and relative permeability Km
For most materials Km is close to one, so we can effectively ignore it and get
𝑛=
𝐾𝐾𝑚 ≈ 𝐾
n is called the index of
refraction for the medium
Since K>1, the speed of an EM wave in a material medium is always less than c.
*K is not technically a constant – when rapidly oscillating fields are present the value is usually smaller
than with constant fields, so the value of K is dependent on the frequency of the EM 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 = 12𝜖0 𝐸 2 + 2𝜇1 𝐵2 = 𝜖0 𝐸 2
0
This is the energy per unit volume
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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 = 12𝜖0 𝐸 2 + 2𝜇1 𝐵2 = 𝜖0 𝐸 2
0
This is the energy per unit volume
The Poynting vector describes the energy flow rate.
𝑆 = 𝜇1 𝐸 × 𝐵
0
This vector usually oscillates rapidly, so it makes sense to talk about the average value,
which turns out to be the INTENSITY of the radiation, with units W/m2.
For a sinusoidal wave in vacuum we can write this in several forms:
𝐼 = 𝑆𝑎𝑣
𝐸𝑚𝑎𝑥 𝐵𝑚𝑎𝑥 𝐸𝑚𝑎𝑥 2 1
=
=
= 2𝜖0 𝑐𝐸𝑚𝑎𝑥 2
2𝜇0
2𝜇0 𝑐
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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 a 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 a 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  1012 sJ )  (4.0  109 s)  8  103 J  8000J
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 a 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  r2    (2.5  10 6 m)2  2.0  10 11 m2
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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 a 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  r2    (2.5  10 6 m)2  2.0  10 11 m2
Now divide to get intensity:
Power
2.0  1012 W
21 W
Intensity 


1
.
0

10
m2
100  r2 2.0  109 m2
This is the total area
of all 100 cells.
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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 a 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 intensity formula:
𝐼 = 12𝜖0 𝑐𝐸𝑚𝑎𝑥 2
𝐸𝑚𝑎𝑥 =
𝐵𝑚𝑎𝑥
2𝐼
=
𝜖0 𝑐
2 ∙ 1021
11 𝑉
=
8.68
∙
10
𝑚
(8.85 ∙ 10−12 )(3 ∙ 108 )
𝐸𝑚𝑎𝑥
=
= 2.89 ∙ 103 𝑇
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.
𝑆𝑎𝑣
𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑐
This is the same as the total energy absorbed by the surface.
If the energy is reflected, the pressure is doubled.
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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
intensity of sunlight is 1410 W/m2.
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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
intensity of sunlight is 1410 W/m2.
𝑊
𝑆𝑎𝑣 1410 𝑚2
−6
𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
=
=
4.7
∙
10
𝑃𝑎
𝑚
𝑐
3 ∙ 108 𝑠
Recall that Pressure = Force/Area. We can use this and F=ma to get our formula:
F
F PA
A
F
F  ma  a 
m
P
PA
a
m
Since the sunlight reflects from our solar sail we should double the given pressure.
a
2(4.7  106
N )  2.59  106 m2
m2
4
2.5  10 kg
 9.72  104
m
s2
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Standing EM Waves
When EM waves are reflected we can have a superposition of waves traveling in
opposite directions, forming a STANDING WAVE. After combining the formulas for
the opposite-directed waves, and applying a bit of trigonometry, we arrive at
formulas for the E and B fields of a standing EM wave.
𝐸𝑦 𝑥, 𝑡 = −2𝐸𝑚𝑎𝑥 sin 𝑘𝑥 𝑠𝑖𝑛(𝜔𝑡)
𝐵𝑧 𝑥, 𝑡 = −2𝐵𝑚𝑎𝑥 cos 𝑘𝑥 𝑐𝑜𝑠(𝜔𝑡)
We can find the positions where these
fields go to zero (at all times t). These
are called the NODAL PLANES:
For the E-field we need sin(kx)=0, which
leads to the following locations:
𝑥 = 0, 𝜆2, 𝜆, 3𝜆
2 , 2𝜆, …
Similarly for the B-field we need
cos(kx)=0, which gives:
5𝜆
𝑥 = 𝜆4, 3𝜆
,
4 4 ,…
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Standing EM Waves
If we have 2 reflecting surfaces parallel to each other we can “trap” a standing EM wave in a
box, just like having a standing wave on a stretched string. The formulas are even the same:
2𝐿
(𝑛 = 1,2,3, . . )
𝑛
𝑐
𝑓𝑛 = 𝑛
(𝑛 = 1,2,3, . . )
2𝐿
𝜆𝑛 =
These formulas give the wavelengths
and frequencies for standing waves that
will “fit” in a box of length L
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