Electromagnetic Waves
March 25, 2014
Chapter 31
1
Wave Solutions to Maxwell’s
Equations
!  The wave is traveling in the positive x-direction
!  Note that there is no dependence on the y- or z-coordinates,
only on the x-coordinate and time
!  This type of wave is called a plane wave
!  We will show that this general wave Ansatz satisfies
Maxwell’s Equations
March 25, 2014
Chapter 31
2
Wave Solutions to Maxwell’s
Equations
!  The electric field is always perpendicular to the direction the
electromagnetic wave is traveling and is always
perpendicular to the magnetic field
!  The electric and magnetic fields are in phase
!  The wave shown below is a snapshot in time
!  The vectors shown represent the magnitude and direction
for the electric and magnetic fields
!  Now we must show that our proposed solution satisfies
Maxwell’s Equations
March 25, 2014
Chapter 31
4
Consistency of Wave Solutions

E = Emax sin ( kx − ω t ) ŷ

B = Bmax sin ( kx − ω t ) ẑ
!  Ansatz:
!  We need to show that this combination of E- and B-fields
satisfies all four of Maxwell’s equations
!  Let’s do this for vacuum (no free charges and no currents)
 

∫∫ EidA = 0
 

∫∫ BidA = 0
March 25, 2014
 
dΦ B
∫ Eids = − dt
 
dΦ E
∫ Bids = µ0ε 0 dt
Chapter 31
6
Wave Solution: Gauss’s Law, E-Fields
!  First we will show that the plane wave satisfies Gauss’s Law
for electric fields
!  For an electromagnetic wave, there is no enclosed charge
(qenc = 0), so we must show that our solution satisfies
 

∫∫ EidA = 0
!  We can draw a rectangular Gaussian surface
around a snapshot of the wave
!  For the faces in the y-z and x-y planes


 
E ⊥ dA ⇒ E ⋅dA = 0
!  The faces in the x-z planes will contribute +EA and -EA
!  Thus the integral is zero and Gauss’s Law for electric fields is
satisfied
One down, three to go
March 25, 2014
Chapter 31
7
Wave Solution: Gauss’s Law, B-Fields
!  For Gauss’s Law for magnetic fields we must show
 

∫∫ BidA = 0
!  We use the same surface that we used for the electric field
!  For the faces in the y-z and x-z planes


 
B ⊥ dA ⇒ BidA = 0
!  The faces in the x-y planes will contribute
+BA and -BA
!  Thus our integral is zero and Gauss’ Law
for magnetic fields is satisfied
Two down, two to go
March 25, 2014
Chapter 31
8
Wave Solution: Faraday’s Law
!  Next: Faraday’s Law
 
dΦ B
∫ Eids = − dt
!  To evaluate the integral on the left
side, we assume an integration loop
in the x-y plane that has a width dx
and height h as shown by the
gray box in the figure




E(x) ⇒ E(x + dx) = E(x)+ dE
!  The electric and magnetic fields change as we move in the
x-direction
March 25, 2014
Chapter 31
9
Wave Solution: Faraday’s Law
!  The
around the loop is
 integral

∫ Eids
!  We can split this integral over a
closed loop into four pieces
•  a to b, b to c, c to d, d to a
!  The contributions to the integral
parallel to the x-axis, integrating
from b to c and from d to a, are
zero because the electric field is
always perpendicular to the integration direction
!  For the integrations along the y-direction, a to b and c to d,
one has the electric field parallel to the direction of the
integration
!  Because the electric field is independent of the y-coordinate,
it can be taken out of the integral
March 25, 2014
Chapter 31
10
Wave Solution: Faraday’s Law
!  The scalar product simply reduces
to a conventional product
!  The resulting integral is
 
∫ Eids = ( E + dE )h − Eh = dE ⋅h
!  The right hand side is given by
dΦ B
dB
dB
−
= −A = −hdx
dt
dt
dt
!  So we have
dB
dE
dB
hdE = −hdx
⇒
=−
dt
dx
dt
March 25, 2014
Chapter 31
11
Wave Solution: Faraday’s Law
!  The derivatives dE/dx and dB/dt are taken with respect to a
single variable, although both E and B depend on x and t
!  Thus we can more appropriately write
∂E
∂B
=−
∂x
∂t
!  Taking our assumed form for the electric and magnetic
fields we can execute the derivatives
∂E ∂
= ( Emax sin ( kx − ω t )) = kEmax cos ( kx − ω t )
∂x ∂x
∂B ∂
= ( Bmax sin ( kx − ω t )) = −ω Bmax cos ( kx − ω t )
∂t ∂t
March 25, 2014
Chapter 31
12
Wave Solution: Faraday’s Law
!  Which gives us
kEmax cos ( kx − ω t ) = − ( −ω Bmax cos ( kx − ω t ))
!  We can relate the angular frequency ω and the angular wave
number k as
ω 2π f
=
= f λ = c ( c is the speed of light )
2
π
k ⎛ ⎞
⎜⎝ ⎟⎠
λ
!  We can then write
Emax ω
= =c ⇒
Bmax k
E
=c
B
!  Which implies that our assumed solution satisfies Faraday’s
Law as long as E/B = c
Three down, one to go
March 25, 2014
Chapter 31
13
Wave Solution: Ampere-Maxwell Law
!  For EM waves, there is no current
 
dΦ E
∫ Bids = µ0ε 0 dt
!  To evaluate the integral, we
assume an integration loop with
a width dx and height h
!  The plane is in the x-z plane, such
that the field is either parallel or
orthogonal to the integration direction
!  The parts of the loop parallel to the axis do not contribute
!  The integral around the loop in a counter-clockwise
direction (a to b to c to d to a) is given by
 
∫ Bids = − ( B + dB)h + Bh = − (dB)(h )
March 25, 2014
Chapter 31
14
Ampere-Maxwell Law
!  The right hand side can be written as
dΦ E
dE ⋅ A
dE ⋅h⋅dx
µ0ε 0
= µ0ε 0
= µ0ε 0
dt
dt
dt
!  Substituting back into the Ampere-Maxwell relation we get
dE ⋅h⋅dx
−dB⋅h = µ0ε 0
dt
!  Simplifying and expressing this equation in terms of partial
derivatives as we did before we get
∂B
∂E
−
= µ0ε 0
∂x
∂t
!  Putting in our assumed solutions gives us
− ( kBmax cos ( kx − ω t )) = − µ0ε 0ω Emax cos ( kx − ω t )
March 25, 2014
Chapter 31
15
Ampere-Maxwell Law
!  We can rewrite the previous equation as
Emax
k
1
=
=
Bmax µ0ε 0ω µ0ε 0c
!  This relationship also holds for the electric and magnetic
field at any time
E
1
=
B µ0ε 0c
!  So our assumed solutions satisfy the Ampere Maxwell law if
the ratio of E/B is
1
µ0ε 0c
March 25, 2014
Four down!
Chapter 31
16
The Speed of Light
!  Our solutions for Maxwell’s Equations are correct if
E
E
1
= c and
=
B
B µ0ε 0c
!  We can see that
c =
1
µ0ε 0c
1
µ0ε 0
⇒ c=
!  Thus the speed of an electromagnetic wave can be expressed
in terms of two fundamental constants related to electric
fields and magnetic fields, the magnetic permeability and
the electric permittivity of the vacuum
!  If we put in the values of these constants we get
c=
March 25, 2014
(1.26⋅10
1
−6
H/m )( 8.85⋅10−12 F/m )
Chapter 31
= 2.99⋅108 m/s
17
Example problem: Plane wave
!  The components of the electric field in an electromagnetic
wave traveling in vacuum are described by Ex = 0, Ey = 0,
and Ez = 4.31 sin(kx - 1.12E8 t) V/m, where x is measured
in meters and t in seconds.
1. Determine the wavelength of the wave.
Answer: Wave number and angular frequency are related by
ω
2π
= c and k =
where λ is the wavelength
k
λ
Thus the wavelength is
2π 2π 2π c
λ=
=
=
= 16.8m
k ω /c ω
March 25, 2014
Chapter 31
18
Example problem: Plane wave
!  The components of the electric field in an electromagnetic
wave traveling in vacuum are described by Ex = 0, Ey = 0,
and Ez = 4.31 sin(kx - 1.12E8 t) V/m, where x is measured
in meters and t in seconds.
2. Calculate the amplitude of the magnetic field of the wave.
Answer: Electric field and magnetic field in the plane wave are
related by E
E
=c ⇒B=
B
c
Thus the amplitude of the magnetic field is
−8
B = 1.44 ×10 T
March 25, 2014
Chapter 31
19
The Speed of Light
!  This implies that all electromagnetic waves travel at the
speed of light (… in vacuum …)
!  And furthermore that light itself is an electromagnetic wave
!  We will see later that the speed of light plays an important
role in how we observe reality:
•  The speed of light is always the same in any reference frame
•  Thus if you send an electromagnetic wave out in a specific
direction, any observer, regardless of whether that observer is
moving toward you or away from you, will see that wave moving at
the speed of light
•  This amazing result leads to the theory of relativity
!  The speed of light can be measured extremely precisely
!  So now the speed of light is defined as precisely
c = 299,792,458 m/s
March 25, 2014
Chapter 31
20
The Electromagnetic Spectrum
!  All electromagnetic waves travel at the speed of light
!  However, the wavelength and frequency of electromagnetic
waves can vary dramatically
!  The speed of light c, the wavelength λ, and the frequency f
are related by
c=λf
!  Examples of electromagnetic waves include light, radio
waves, microwaves (radar), and X-rays
!  This diverse spectrum is illustrated on the next slide
March 25, 2014
Chapter 31
21
The Electromagnetic Spectrum
March 25, 2014
Chapter 31
22