20
March 25, 2014 Chapter 31 1 Electromagnetic Waves

Electromagnetic Waves - web.pa.msu.eduschwier/courses/2014Spring... · For an electromagnetic wave, ... All electromagnetic waves travel at the speed of light ! However, the wavelength

  • Upload
    dinhbao

  • View
    227

  • Download
    0

Embed Size (px)

Citation preview

March 25, 2014 Chapter 31 1

Electromagnetic Waves

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

March 25, 2014 Chapter 31 22

The Electromagnetic Spectrum