Download Chapter 29: Maxwell`s Equation and EM Waves

Chapter 29: Maxwell’s Equation
and EM Waves
Slide 29-1
Equations of electromagnetism: a review
•  We’ve now seen the four fundamental equations of electromagnetism,
here listed together for the first time.
•  But one is incomplete: Ampère’s law needs refining
Slide 29-2
Clicker Question
Slide 29-3
Maxwell’s
Adjustment to Ampere’s Law
�
� = µ0 Ienc
� · dr
B
•  The result of Ampere’s law
depends on which surface is
used to determine the encircled
current!
•  Can’t have contradictory results –
either there is a B field or there
isn’t!
•  Notice that electric field is
changing inside conductor.
•  Ampere postulated that a changing
electric flux induces a magnetic
field (similar to how a changing
magnetic flux induces an electric
field)
Slide 29-4
Maxwell’s Adjustment to Ampere’s Law
•  Need to add a term to the right side of
Ampere’s law to account for the changing
electric flux. This is the displacement
current.
•  Notice, for a parallel plate capacitor, the
electric flux is
ΦE = EA =
σ
Q
=
�0
�0
dΦE
I
=
dt
�0
•  So Maxwell added a displacement current
(Ampere-Maxwell
term to Ampere’s law:
•  The the rate of change of the flux is
�
�
dΦE
�
�
B · d� = µ0 (I + Id )enc = µ0 I + �0
dt
•  Corresponding displacement current
density:
�
Law)
enc
�
d
E
J�d = �0
dt
Slide 29-5
Slide 29-6
Slide 29-7
•  Determine the electric
field between the plates:
�
dΦE
�
�
B · d� = µ0 �0
dt
I
2πrB = µ0 �0 πr
�0 πR2
2
r
B = µ0 I
(r < R)
2
2πR
Slide 29-8
Consider a large parallel plate capacitor as shown, charging so
that Q = Q0+βt on the positively charged plate. Assuming the
edges of the capacitor and the wire connections to the plates can be
ignored,
what is the magnitude of the magnetic field B halfway between the
plates, at a radius r?
µ!
A.
s
z
a
r
I
Q
I
-Q
d
B.
0
2"r
µ 0!r
2d 2
C.
µ 0!d
2a 2
D.
µ 0!a
2"r 2
E. None of the aboveSlide 29-9
Maxwell’s equations
•  The four complete laws of electromagnetism are
collectively called Maxwell’s equations.
•  They describe all electromagnetic fields in the universe, outside
the realm of quantum physics.
Slide 29-10
Maxwell’s equations in vacuum
•  In a vacuum there’s no electric charge and therefore also
no electric current.
Maxwell’s equations in vacuum
A changing electric field is a source for a magnetic field, and
a changing magnetic field is a source for an electric field.
These equations infer the possibility of electromagnetic
waves!
Slide 29-11
Wave Equation
•  Maxwell’s equations can be manipulated to derive the
following wave equations:
2�
E
∂
2�
∇ E = µ0 �0 2
∂t
2�
B
∂
2�
∇ B = µ0 �0 2
∂t
2
2
2
∂
∂
∂
∇2 =
x̂ + 2 ŷ + 2 ẑ
2
∂x
∂y
∂z
•  Very similar to wave equation for sound waves! Except
here it is the electric and magnetic field that is
“wiggling.”
•  For plane waves traveling in one direction (say the x
�
�
�
direction): ∂ 2 E�
∂2B
∂2B
∂2E
∂x2
= µ0 �0
∂t2
∂x2
= µ0 �0
∂t2
Slide 29-12
Plane electromagnetic waves
•  A plane electromagnetic wave are waves propagating in
one direction with one wavelength. (E and B do not vary
with respect to the other two dimensions)
•  The fields are perpendicular
to each other and to the
direction of propagation.
� ×B
� gives direction of propagation)
(E
•  Mathematically
r
E ( x, t ) = Ep sin ( kx − ωt ) ˆj
r
B ( x, t ) = Bp sin ( kx − ωt ) kˆ
2π
k=
λ
2π
ω=
T
Slide 29-13
Clicker Question
Two traveling waves 1 and 2 are described by the equations.
y1 ( x, t ) = 2 sin(2 x ! t )
y 2 ( x, t ) = 4 sin( x ! 2 t )
All the numbers are in the appropriate SI (mks) units.
Which wave has the higher speed?
A) Wave 1
B) Wave 2
C) Both have the same speed.
Slide 29-14
Plane
Waves
as
Solutions
�
�
�
�
� · d�� = − d
E
dt
S
� · dA
�
B
∂E
∂B
=⇒
=−
∂x
∂t
� · d�� = µ0 �0 d
B
dt
=⇒
S
� · dA
�
E
∂B
∂E
= −�0 µ0
∂x
∂t
∂ ∂E
∂
∂E
⇒
= − (−�0 µ0
)
∂x ∂x
∂t
∂t
∂2E
∂2E
− � 0 µ0 2 = 0
2
∂x
∂t
E(x, t) = E0 sin(kx − ωt)
B(x, t) = B0 sin(kx − ωt)
kEp = ωBp
kBp = �0 µ0 ωEp
•  plane wave expression is a solution if
ω
k
=
1
ε 0 µ0
fλ = c = c = 3.0 × 108 m/s
•  This also implies that
Ep
Bp =
c
Slide 29-15
General Results for EM Radiation
•  Transverse waves (E and B perpendicular to direction of
propagation): direction of propagation: E
� ×B
� ∝ k̂
•  E and B perpendicular to each other.
•  E = cB; E and B oscillate in phase
•  Propagation speed is the speed of light in a vacuum
•  independent of wavelength:
c = λ f = ω/k
•  radio waves, light, infrared radiation, X-rays are all the same
phenomena!
1
c
•  In matter, the speed of light is v = √�µ = n
•  No medium required for propagation (no ether)
Slide 29-16
Clicker question
• 
At a particular point, the electric field of an
electromagnetic wave points in the + y direction, while
the magnetic field points in the − z direction. Which of
the following describes the propagation direction?
A.  + x
B.  − x
C.  either + x or − x but you can’t tell which
D.  − y
Slide 29-17
Clicker question
•  A planar electromagnetic wave is propagating through
space. Its electric field vector is given by
� = Ep cos(kz − ωt)î
E
Its magnetic field vector is
� = Bp cos(kz − ωt)ĵ
1) B
� = Bp cos(ky − ωt)k̂
2) B
� = Bp cos(ky − ωt)ĵ
3) B
� = Bp cos(kz − ωt)k̂
4) B
� = Bp sin(kz − ωt)î
5) B
Slide 29-18
The Electromagnetic Spectrum
Slide 29-19
Clicker question
•  Which type of radiation travels with the highest speed?
1.  visible light
2.  X-rays
3.  Gamma-rays
4.  radio waves
5.  they all have the same speed
Slide 29-20
Slide 29-21
Producing electromagnetic waves
•  Electromagnetic waves are generated ultimately by accelerated
electric charge.
•  Details of emitting systems depend on wavelength, with most efficient
emitters being roughly a wavelength in size.
•  Radio waves are generated by alternating currents in metal antennas.
•  Molecular vibration and rotation produce infrared waves.
•  Visible light arises largely from atomic-scale processes.
•  X rays are produced in the rapid deceleration of electric charge.
•  Gamma rays result from nuclear processes.
A radio transmitter and antenna
Electric fields of an oscillating electric dipole
Slide 29-22
Antennae
An electric field parallel to an antenna
(electric dipole) will “shake” electrons
and produce an AC current.
A magnetic dipole antenna (for AM
radios) should be oriented so that the
B-field passes into and out of the
plane of a loop, inducing a current in
the loop.
Slide 29-23
Energy in EM waves
•  Electromagnetic waves transport energy
•  The Poynting vector describes the rate of energy flow per unit
area (W/m2 in SI):
2
S = uc = (uE + uB )c = �E
�= 1E
� ×B
�
S
µ0
•  For plane waves (traveling in x direction with E oriented in z direction):
c
� = 1 Ep Bp cos2 (kx − ωt)î
S
µ0
•  Averaging over the time variations of the oscillating fields gives the
average value, also called the average intensity:
1 1
1 Ep2
1
I =< S >=
Ep Bp =
= �0 Ep2 c
2 µ0
2 2µ0 c
2
•  Far from a localized source of radiation, electric field decreases as 1/r.
1 (as required by conservation of energy)
Thus,
I∝
r2
Slide 29-24
Clicker Question
Two radio dishes are receiving signals from a radio station which is
sending out radio waves in all directions with power P. Dish 2 is twice
as far away as Dish 1, but has twice the diameter. Which dish receives
more power?
A: Dish 1
B: Dish 2
C: Both receive the same power
Dish 1
Dish 2
Slide 29-25
Example: The intensity of the sunlight that reaches Earth’s upper
atmosphere is 1400 W/m2.
(a) What is the total average power output of the Sun,
assuming it to be an isotropic source?
(
Pav = IA = I 4πR 2
(
)
2
)(
11
= 4π 1400 W/m 1.50 ×10 m
2
)
= 4.0 ×10 26 W
Slide 29-26
Example continued:
(b) What is the intensity of sunlight incident on Mercury, which is
5.8x1010 m from the Sun?
Pav
Pav
I=
=
A 4πr 2
4.0 ×10 26 W
=
4π 5.8 ×1010 m
(
2
)
= 9460 W/m2
(c) What is the maximum electric field (if monochromatic light)
Slide 29-27
Flux and Solar Heating
Slide 29-28
Polarization
Slide
Slide25-24
29-29
Light passed through a polarizing filter has an intensity of 2.0 W/m2.
How should a second polarizing filter be arranged to decrease the
intensity to 1.0 W/m2?
Slide
Slide25-25
29-30
Clicker Question
An unpolarized beam of light passes through 2 Polaroid filters
oriented at 45o with respect to each other. The intensity of the
original beam is Io. What is the intensity of the light coming
I
through both filters?
o
A: (1/1.4)Io
B: (1/2)Io
C: (1/4)Io
D: (1/8)Io
E: None
c
Slide 29-31
Clicker question
• 
Two polarizers are oriented at right angles, so no light
gets through the combination. A third polarizer is
inserted between the two, with its preferred direction at
45° to the others. How will this “sandwich” of
polarizers affect a beam of initially unpolarized light?
A.  All of the initial light will be blocked.
B.  Half of the initial light is blocked.
C.  One-quarter of the initial light is blocked.
D.  None of the initial light will be blocked.
Slide 29-32
Slide 29-33