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March 25, 2014 Chapter 31 2
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 4
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 6
! 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)
Consistency of Wave Solutions
E = Emax sin kx −ωt( ) yB = Bmax sin kx −ωt( ) z
EidA∫∫ = 0
BidA∫∫ = 0
Eids∫ = − dΦB
dt
Bids∫ = µ0ε0
dΦE
dt
March 25, 2014 Chapter 31 7
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
! We can draw a rectangular Gaussian surface around a snapshot of the wave
! For the faces in the y-z and x-y planes
! 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
EidA∫∫ = 0
E ⊥ d
A ⇒
E ⋅dA = 0
One down, three to go
March 25, 2014 Chapter 31 8
Wave Solution: Gauss’s Law, B-Fields ! For Gauss’s Law for magnetic fields we must show
! We use the same surface that we used for the electric field ! For the faces in the y-z and x-z planes
! 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
BidA∫∫ = 0
B⊥ d
A ⇒
BidA = 0
Two down, two to go
March 25, 2014 Chapter 31 9
Wave Solution: Faraday’s Law ! Next: Faraday’s Law
! 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
! The electric and magnetic fields change as we move in the x-direction
Eids∫ = − dΦB
dt
E(x) ⇒
E(x +dx)=
E(x)+d
E
March 25, 2014 Chapter 31 10
Wave Solution: Faraday’s Law ! The integral around the loop is
! 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
Eids∫
March 25, 2014 Chapter 31 11
Wave Solution: Faraday’s Law ! The scalar product simply reduces
to a conventional product ! The resulting integral is
! The right hand side is given by
! So we have − dΦB
dt= −A dB
dt= −hdx dB
dt
hdE = −hdx dB
dt ⇒ dE
dx= − dB
dt
Eids∫ = E +dE( )h− Eh = dE ⋅h
March 25, 2014 Chapter 31 12
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
! Taking our assumed form for the electric and magnetic fields we can execute the derivatives
∂E∂x
= − ∂B∂t
∂E∂x
= ∂∂x
Emax sin kx −ωt( )( ) = kEmax cos kx −ωt( )
∂B∂t
= ∂∂t
Bmax sin kx −ωt( )( ) = −ωBmax cos kx −ωt( )
March 25, 2014 Chapter 31 13
Wave Solution: Faraday’s Law ! Which gives us ! We can relate the angular frequency ω and the angular wave
number k as
! We can then write
! Which implies that our assumed solution satisfies Faraday’s Law as long as E/B = c
kEmax cos kx −ωt( ) = − −ωBmax cos kx −ωt( )( )
ωk= 2π f
2πλ
⎛⎝⎜
⎞⎠⎟= f λ = c c is the speed of light( )
Emax
Bmax= ω
k= c ⇒ E
B= c
Three down, one to go
March 25, 2014 Chapter 31 14
! For EM waves, there is no current
! 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
Wave Solution: Ampere-Maxwell Law
Bids∫ = µ0ε0
dΦE
dt
Bids∫ = − B+dB( )h+ Bh = − dB( ) h( )
March 25, 2014 Chapter 31 15
! The right hand side can be written as
! Substituting back into the Ampere-Maxwell relation we get
! Simplifying and expressing this equation in terms of partial derivatives as we did before we get
! Putting in our assumed solutions gives us
Ampere-Maxwell Law
µ0ε0
dΦE
dt= µ0ε0
dE ⋅Adt
= µ0ε0dE ⋅h ⋅dx
dt
−dB ⋅h = µ0ε0
dE ⋅h ⋅dxdt
− ∂B∂x
= µ0ε0∂E∂t
− kBmax cos kx −ωt( )( ) = −µ0ε0ωEmax cos kx −ωt( )
March 25, 2014 Chapter 31 16
Ampere-Maxwell Law ! We can rewrite the previous equation as
! This relationship also holds for the electric and magnetic field at any time
! So our assumed solutions satisfy the Ampere Maxwell law if the ratio of E/B is
Emax
Bmax= kµ0ε0ω
= 1µ0ε0c
EB= 1µ0ε0c
1µ0ε0c Four down!
March 25, 2014 Chapter 31 17
The Speed of Light
! Our solutions for Maxwell’s Equations are correct if
! We can see that
! 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
EB= c and E
B= 1µ0ε0c
c = 1
µ0ε0c ⇒ c = 1
µ0ε0
c = 1
1.26 ⋅10−6 H/m( ) 8.85 ⋅10−12 F/m( )= 2.99 ⋅108 m/s
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
Thus the wavelength is
March 25, 2014 Chapter 31 18
λ = 2π
k= 2πω / c
= 2πcω
=16.8m
ωk= c and k = 2π
λ where λ is the wavelength
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 Thus the amplitude of the magnetic field is
March 25, 2014 Chapter 31 19
B =1.44×10−8T
EB= c ⇒ B = E
c
March 25, 2014 Chapter 31 20
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 21
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
! Examples of electromagnetic waves include light, radio waves, microwaves (radar), and X-rays
! This diverse spectrum is illustrated on the next slide
c = λ f