chapter 29: maxwell’s equation and em waves · · 2011-07-14the realm of quantum physics. ......
TRANSCRIPT
Slide 29-1 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 29: Maxwell’s Equation and EM Waves
Slide 29-2 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 29-3 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
©University of Colorado, Boulder (2008)
CT 33.42
According to Biot-Savart, is there a Magnetic Field at the point labeled between the plates?
A: Yes B: No (B=0 there)
Current i Current i
+ + + + +
- - - - -
B=?
Slide 29-4 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Maxwell’s Adjustment to Ampere’s Law • Applying Ampère’s law to a circuit
with a changing current results in an ambiguity. • The result 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 a displacement current density:
• Thus discrepancy goes away if add displacement current to Ampere’s law
�Jd = �0d �E
dt
Q(t) = CVC =�0A
d(Ed) = �0EA = �0ΦE
Id =�
�Jd · d �A = �0dΦE
dt=
dQ
dt= I
��B · d�� = µ0(I + Id)enc = µ0
�I + �0
dΦE
dt
�
enc
Slide 29-5 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 29-6 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 29-7 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
• Field between the plates:
��B · d�� = µ0�0
dΦE
dt
Slide 29-8 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 29-9 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Maxwell’s equations in vacuum • In 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 29-10 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Wave Equation • Maxwell’s equations can be manipulated to derive the
following wave equations:
• 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 = µ0�0∂2 �E
∂t2∇2 �B = µ0�0
∂2 �B
∂t2
∇2 =∂2
∂x2x +
∂2
∂y2y +
∂2
∂z2z
∂2 �E
∂x2= µ0�0
∂2 �E
∂t2∂2 �B
∂x2= µ0�0
∂2 �B
∂t2
Slide 29-11 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y x t x ty x t x t1
2
2 24 2
( , ) sin( )( , ) sin( )
= −
= −
Two traveling waves 1 and 2 are described by the equations.
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.
Clicker Question
Slide 29-12 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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.
• Mathematically
k =2π
λω =
2π
T
( �E × �B gives direction of propagation)
Slide 29-13 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Plane Waves as Solutions
• plane wave expression is a solution if
• This also implies that
��B · d�� = µ0�0
d
dt
�
S
�E · d �A�
�E · d�� = − d
dt
�
S
�B · d �A
= c = 3.0× 108 m/s
=⇒ ∂E
∂x= −∂B
∂t=⇒ ∂B
∂x= −�0µ0
∂E
∂t
fλ = c
kEp = ωBp kBp = �0µ0ωEp
Bp =Ep
c
⇒ ∂
∂x
∂E
∂x= − ∂
∂t(−�0µ0
∂E
∂t)
∂2E
∂x2− �0µ0
∂2E
∂t2= 0
E(x, t) = E0 sin(kx− ωt)
B(x, t) = B0 sin(kx− ωt)
Slide 29-14 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
General Results for EM Radiation • Transverse waves (E and B perpendicular to direction of
propagation): direction of propagation: • 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: • radio waves, light, infrared radiation, X-rays are all the same
phenomena! • In matter, the speed of light is
• No medium required for propagation (no ether)
c = λ f = ω/k
v =1√
�µ=
c
n
�E × �B ∝ k
Slide 29-15 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Clicker question • At a particular point, the electric field of an
electromagnetic wave points in the direction, while the magnetic field points in the direction. Which of the following describes the propagation direction?
A. B. C. either or but you can’t tell which D.
Slide 29-16 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Clicker question • A planar electromagnetic wave is propagating through
space. Its electric field vector is given by
Its magnetic field vector is
�E = Ep cos(kz − ωt)i
1) �B = Bp cos(kz − ωt)j
2) �B = Bp cos(ky − ωt)k
3) �B = Bp cos(ky − ωt)j
4) �B = Bp cos(kz − ωt)k
5) �B = Bp sin(kz − ωt)i
Slide 29-17 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
The Electromagnetic Spectrum
Slide 29-18 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
• 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
Clicker question
Slide 29-19 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 29-20 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 29-21 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 29-22 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Energy in EM waves • Electromagnetic waves transport energy
• The Poynting vector describes the rate of energy flow per unit area (W/m2 in SI):
• For plane waves (traveling in x direction with E oriented in z direction):
• Averaging over the time variations of the oscillating fields gives the average value, also called the average intensity:
• Far from a localized source of radiation, electric field decreases as 1/r. Thus, (as required by conservation of energy)
�S =1µ0
EpBp cos2(kx− ωt)i
S = uc = (uE + uB)c = �E2c
I =< S >=12
1µ0
EpBp =12
E2p
2µ0c=
12�0E
2pc
I ∝ 1r2
Slide 29-23 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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
Clicker Question
Slide 29-24 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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?
Slide 29-25 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
(b) What is the intensity of sunlight incident on Mercury, which is 5.8x1010 m from the Sun?
(c) What is the maximum electric field (if monochromatic light)
Example continued:
Slide 29-26 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Flux and Solar Heating
Slide 29-27 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
How many solar collectors would you need to replace a 4.8 kWatt electric water heater? Assume each collector is 2 meters2 and has an efficiency of 40% for converting light energy to usable energy.
A) 1 panel B) 2 panels C) 4 panels D) 8 panels
Clicker Question
Slide 29-28 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Polarization
Slide 25-24
Slide 29-29 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 25-25
Slide 29-30 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 through both filters?
A: (1/1.4)Io B: (1/2)Io C: (1/4)Io D: (1/8)Io E: None
Io
c
Clicker Question
Slide 29-31 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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 Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley