Download Electromagnetic Waves

Electromagnetic Waves
March 24, 2014
Chapter 31
1
Electromagnetic Waves
!  We will see that light is an electromagnetic wave
!  Electromagnetic waves have electric and magnetic fields
!  We will see Maxwell’s Equations that describe
electromagnetic phenomena
!  We will see that the speed of light is constant and can be
related to ε0 and μ0
!  We will see that electromagnetic waves can transport energy
and momentum
!  Electromagnetic waves can be polarized
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Chapter 31
2
Induced Magnetic Fields
!  We have seen that a changing magnetic field induces
an electric field
!  Faraday’s Law of Induction tells us
 
dΦ B
∫ Eids = − dt
!  In a similar manner, a changing electric field induces a
magnetic field
!  Maxwell’s Law of Induction tells us
 
dΦ E
∫ Bids = µ0ε 0 dt
!  where B is the magnetic field induced in a closed loop
by a changing electric flux ΦE in that loop
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3
Circular Capacitor
!  To illustrate induced magnetic fields,
consider the example of a circular
parallel plate capacitor
!  We charge the capacitor and disconnect
the battery
!  The charge is constant and the electric
field between the plates is constant
!  There is no magnetic field
!  What if the charge increases with time?
!  A magnetic field is induced
!  The magnitude of the induced magnetic
field is the same along each line and the
direction is tangential to the line
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Changing Magnetic Field
!  Now let’s consider a constant magnetic
field
!  Let’s increase the magnitude of the
magnetic field while keeping it uniform in
space and direction
!  An electric field is induced as shown by
the red loops
!  The magnitude of the electric field is
constant along each loop and the direction
is tangential to each loop
!  Note that the induced electric field points
in the opposite direction from the
magnetic field induced by a changing
electric field
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5
Maxwell-Ampere Law
!  Ampere’s Law relates the integral around a loop of the dot
product of the magnetic field and the integration direction
to the current flowing through the loop
 
∫ Bids = µ0ienc
!  Combining Maxwell’s Law of Induction and Ampere’s Law
describes magnetic fields created by moving charges and by
changing electric fields, the Maxwell-Ampere Law
 
dΦ E
∫ Bids = µ0ε 0 dt + µ0ienc
!  For the case of constant current, such as current flowing in a
conductor, this equation reduces to Ampere’s Law
!  For the case of a changing electric field without current
flowing, such as the electric field between the plates of a
capacitor, this equation reduces to the Maxwell Law of
Induction
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Displacement Current
!  Looking at the Maxwell-Ampere Law
 
dΦ E
∫ Bids = µ0ε 0 dt + µ0ienc
!  The quantity ε0(dΦE/dt) must have the units of current
!  This term is called the displacement current
dΦ E
id = ε 0
dt
!  Note however that no actual current is being displaced
!  We can then rewrite the Maxwell-Ampere Law as
 
∫ Bids = µ0 (id + ienc )
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7
Displacement Current
!  Consider again a parallel plate capacitor with circular plates
!  We place the capacitor in a circuit in
which a current i is flowing while the
capacitor is charging
!  For a parallel plate capacitor with area
A we can relate the charge q to the
electric field E
q = ε 0 AE
!  We can get the current by taking the time derivative of the
charge
i=
dq
dE
= ε0 A
dt
dt
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8
Displacement Current
!  Assuming that the electric field between the plates of the
capacitor is uniform we can obtain an expression for the
displacement current
dE
d ( AE )
dΦ E
= ε0 A
= ε0
id = ε 0
dt
dt
dt
!  The current in the circuit is the same as the displacement
current id
!  Although there is no actual current flowing between the
plates of the capacitor in the sense that no actual charges
flow across the capacitor gap from one plate to the other, we
can use the concept of displacement current to calculate the
induced magnetic field
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9
Displacement Current
!  To calculate the magnetic field between the two plates of the
capacitor, we assume that the volume between the plates can
be replaced with a conductor of radius R carrying current id
!  Thus from Chapter 28 we
know that the magnetic field
at a distance r from the
center of the capacitor is
⎛ µ0id ⎞
B=⎜
r
⎝ 2π R 2 ⎟⎠
!  Outside the capacitor we treat
the system as a current-carrying wire
!  The magnetic field is
µ0id
B=
March 24, 2014
2π r
Chapter 31
10
Maxwell’s Equations
!  The Maxwell-Ampere Law completes the explanation
of the four equations known as Maxwell’s Equations that
describe electromagnetic phenomena
!  We have used these equations to describe electric fields,
magnetic fields, and circuits and now electromagnetic waves
Name
Gauss’ Law for Electric
Fields
Gauss’ Law for Magnetic
Fields
Equation
  qenc

∫∫ EidA = ε
0
 

∫∫ BidA = 0
Faraday’s Law
 
dΦ B
Eid
s
=
−
∫
dt
Ampere-Maxwell Law
 
dΦ E
Bid
s
=
µ
ε
+ µ0 ienc
∫
0 0
dt
March 24, 2014
Chapter 31
Description
Relates the net electric flux
to the net enclosed electric
charge
States that the net magnetic
flux is zero (no magnetic
charge)
Relates the induced electric
field to the changing
magnetic flux
Relates the induced
magnetic field to the
changing electric flux and
to the current
12
Wave Solutions to Maxwell’s
Equations
!  One solution to Maxwell’s Equations is an electromagnetic
wave
!  Here we will assume that the electromagnetic wave
propagates in vacuum (no moving charges or currents)
!  We will make the Ansatz that the electric and magnetic
fields in electromagnetic waves are given by the form
! !
E(r ,t) = Emax sin ( kx − ω t ) ŷ
! !
B(r ,t) = Bmax sin ( kx − ω t ) ẑ
2π
k=
wave number
λ
ω = 2π f angular frequency
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14
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
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15
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. Calculate the frequency of the wave.
Answer:
ω = 1.12 ×108 1s = 2π f is the angular frequency
Thus the frequency is
ω
f=
= 1.78 ×107 Hz
2π
March 24, 2014
Chapter 31
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