Chapter
34
Ampere’s Law is Wrong!
•Maxwell realized Ampere’s Law is not self-consistent
•This isn’t an experimental argument, but a theoretical one
•Consider a parallel plate capacitor getting charged by a wire
•Consider an Ampere surface between the plates B  ds  0 I1  0
•Consider an Ampere surface in front of plates
B  ds   0 I 2   0 I
•But they must give the same answer!


•There must be something else that creates
B-fields
B  ds  0  I  ? 
I
•Note that for the first surface, there is also
an electric field accumulating in capacitor
•Maybe electric fields?
•Take the time derivative of this formula
•Speculate : This replaces I for first surface

Q
E
0 A
 0 E  Q
I
d  E dQ
I
0

dt
dt
Ampere’s Law (New Recipe)
dE 

 B  ds  0  I   0 dt 
•Is this self-consistent?
•Consider two surfaces
with the same boundary
d  E1 
dE2 


 B  ds  0  I1   0 dt   0  I 2   0 dt 
d
I1  I 2    0  E 2   0  E1 
dt
d
I net    0  E 
dt
I1
I2
•Gauss’s Law for electric fields
 0 E  qin
•This makes sense!
I net
dqin

dt
dE
 B  ds  0 I   0 0 dt
E1
B
E2
CT – 1 For a charging capacitor, the total displacement current between the plates is equal to the total
conduction current I in the wires. The capacitors in the diagram have circular plates of radius R. In (a), points A
and B are each a distance d > R away from the line through the centers of the plates; in this case the magnetic
field at A due to the conduction current is the same as that at B due to the displacement current. In (b), points P
and Q are each a distance r < R away from the center line.
Compared with the magnetic field at P, that at Q is
A.
B.
C.
D.
bigger.
smaller.
the same.
need more information.
Maxwell’s Equations
We now have four formulas that describe how to get electric and
magnetic fields from charges and currents
E  nˆ dA  qin  0
•Gauss’s Law
S
•Gauss’s Law for Magnetism
B  nˆ dA  0
•Ampere’s Law (final version)
S
•Faraday’s Law
dE
•Collectively, these are called
B  ds  0 I in  0 0
Maxwell’s Equations
dt



There is also a formula for forces on charges
•Called Lorentz Force
F  q E  v  B
dB
 E  ds   dt
All of electricity and magnetism
is somewhere on this page
•Inside materials, these formulas are sometimes modified to include the
effects of the reaction of the materials (like dielectric constants)
•These formulas commonly rewritten in derivative form
Wave solutions
•We can solve Maxwell’s Equations
(take my word for it) and come up with two
“simple” differential equations.
E  x, y, z , t   E0 sin  kx  t 
B  x, y , z , t   B 0 sin  kx  t 
2E
2E
 0 0 2
2
x
t
2B
2B
 0 0 2
2
x
t
with 0 0  1 2
c
•I could have used cosine instead, it makes no difference
•I chose arbitrarily to make it move in the x-direction
•These are waves where we have .   2 f   f  c
k
2 
Show that
E  x, y, z , t   E0 sin  kx  t 
B  x, y , z , t   B 0 sin  kx  t 
satisfies
2E
2E
 0 0 2
2
x
t
2B
2B
 0 0 2
2
x
t
with 0 0  1 2
c
On board
Wave Equations summarized:
E  x, y, z , t   E0 sin  kx  t 
•Waves look like:
B  x, y , z , t   B 0 sin  kx  t 
  ck
•Related by:
•Two independent solutions to these equations:
E0  cB0
B0
E0
E0
B0
•Note E  B is in direction of motion
E y 0  cBz 0
or
Ez 0  cE y 0
•Note that E, B, and
direction of travel are all
mutually perpendicular
•The two solutions are
called polarizations
•We describe polarization by telling which
way E-field points
Understanding Directions for Waves
E0  cB0
•The wave can go in any direction you want
•The electric field must be perpendicular to the wave direction
•The magnetic field is perpendicular to both of them
•Recall: E  B is in direction of motion
A wave has an electric field given by E = j E0 sin(kz – t).
What does the magnetic field look like?
A) B = i (E0/c) sin(kz - t) B) B = k (E0/c) sin(kz - t)
C) B = - i (E0/c) sin(kz - t) D) B = - k (E0/c) sin(kz - t)
•The magnitude of the wave is B0 = E0 / c
•The wave is traveling in the z-direction, because of sin(kz - t).
•The wave must be perpendicular to the E-field, so perpendicular to j
•The wave must be perpendicular to direction of motion, to k
•It must be in either +i direction or –i direction
•If in +i direction, then E  B would be in direction j  i = - k, wrong
•So it had better be in the –i direction
CT – 2 A planar electromagnetic wave is propagating through space. Its electric
field vector is given by E = Eo cos(kz – wt) î . Its magnetic field vector is
A. B = Bo cos(kz – wt) ˆj
B. B = Bo cos(ky – wt) k̂
C. B = Bo cos(ky – wt) î
D. B = Bo cos(kz – wt) k̂
E. None of the above
The meaning of c:
  ck
E  x, y, z , t   E0 sin  kx  t 
•Waves traveling at constant speed B  x, y , z , t   B 0 sin  kx  t 
•Keep track of where they vanish
kx  t  0
x

k
t  ct
•c is the velocity of these waves
c
1
 0 0
 2.99792458 108 m/s
c  3.00 108 m/s
•This is the speed of light
•Light is electromagnetic waves!
•But there are also many other types of EM waves
•The constant c is one of the most important fundamental constants of
the universe
Wavelength and wave number
•The quantity k is called the
wave number
•The wave repeats in time
•It also repeats in space f  1 T
k  2
E  E0 sin  kx  t 
B  B 0 sin  kx  t 
  2 f

  ck
•EM waves most commonly described
in terms of frequency or wavelength

c
 2 f
k
2

cf
•Some of these equations must be modified when inside a material
Ex- (Serway 34-14) In SI units, the electric field in an electromagnetic wave is
described by Ey = 100 sin(1.00 x 107x -t). (a) Calculate the amplitude of the
corresponding magnetic field. (b) Find the wavelength , (c) Find the frequency f.
Also find an expression for the magnetic field.
Solve on Board
The Electromagnetic Spectrum
•Different types of waves are classified
by their frequency (or wavelength)
 Increasing
f Increasing
cf
Radio Waves
Microwaves
Infrared
Visible
Ultraviolet
X-rays
Gamma Rays
•Boundaries are arbitrary
and overlap
•Visible is 380-740 nm
Red Which of the
Vermillion
following
waves
Orange
hasSaffron
the highest
Yellow speed in vacuum?
Chartreuse
A)
Green Infrared
B)
Orange
Turquoise
Blue C) Green
Indigo
Violet D) Blue
E) It’s a tie
F) Not enough info
Energy and the Poynting Vector
•Let’s find the energy density in the wave
E0  cB0
u E  12  0 E 2  12  0E02 sin 2  kx  t   12  0c2 B02 sin 2  kx  t 
1 0

B02 sin 2  kx  t 
2
B
2  0 0
2
0
u

sin
kx  t 
2
2

B0
B
0
uB 

sin 2  kx  t 
2 0
2 0
1
•Now let’s define the Poynting vector:
S
E  B
1
0
2
S
E0 B0 sin  kx  t   cu
0
•It is energy density times the speed at which the wave is moving
•It points in the direction energy is moving
•It represents the flow of energy in a particular direction,
energy/area/time
J m W
S
uc
•Units:
3
2
m s
m
Intensity and the Poynting vector
•The time-averaged Poynting vector is called the Intensity
•Power per unit area
2
S c u

cB0
0
sin
2
S  cB02 20
 kx  t 
In Richard Williams’ lab, a laser can (briefly) produce
50 GW of power and be focused onto a region 1 m2
in area. How big are the electric and magnetic fields?
P 5.0 1010 W
22
2
S 


5.0

10
W/m
2
6
A
10 m 
7
22
2
2
4


10
T

m/A
5.0

10
W/m



2 0 S
2

B0 
3 108 m/s
c
 4.2  108 T 2
12
E

cB
0
0
E

6.1

10
V/m
B  20, 000 T
0
0
CT -3 At a fixed point, P, the electric and magnetic field vectors in an
electromagnetic wave oscillate at angular frequency w. At what angular frequency
does the Poynting vector oscillate at that point?
A. 2 w
B. w
C. w/2
D. 4 w
Momentum and Pressure
•Light carries energy – can it carry momentum?
p U c
•Yes – but it’s hard to prove
•p is the total momentum of a wave and U the total energy
•Suppose we have a wave, moving into a perfect absorber (black body)
•As they are absorbed, they transfer momentum
•Intensity:
U
S 
A
t
•As waves hit the wall they transfer their momentum
cp cF
 cP
S 

At
A
•Pressure on a perfect absorber: P  S c
•When a wave bounces off a mirror, the momentum is reversed
•The change in momentum is doubled
P2 S c
•The pressure is doubled
Cross-Section:
•To calculate the power falling on an object, all that matters is
the light that hits it
•Example, a rectangle parallel to the light feels no pressure
•Ask yourself: what area does the light see?
•This is called the cross section
P = S 
F = P
P S
c
If light of intensity S hits an absorbing sphere
of radius a, what is the force on that sphere?
A) a2S/c B) 2a2S/c C) 4a2S/c
•As viewed from any side, a
sphere looks like a circle of
radius a
•The cross section for a sphere,
then, is a2
Sample Problem:
A 150 W bulb is burning at 6% efficiency. What is the
force on a mirror square mirror 10 cm on a side 1 m
away from the bulb perpendicular to the light hitting it?
1m
•Light is distributed in all directions
equally over the sphere of radius 1 m
0.06 150 W 
2

0.72
W/m
S 
4 m 2
F  P 
2 S
c

2  0.72 W/m 2 
3 10 m/s
8
F  4.8 1011 N
 0.1 m 
2
Ex- (Serway 34-37) A plane electromagnetic wave of intensity 6.00 W/m2 strikes
a small pocket mirror, of area 40.0 cm2, held perpendicular to the approaching
wave. (a) What momentum does the wave transfer to the mirror each second? (b)
Find the force that the wave exerts on the mirror.
Solve on Board
Sources of EM waves
+
•A charge at rest produces no EM waves
•There’s no magnetic field
•A charge moving at uniform velocity produces no EM waves
•Obvious if you were moving with the charge
•An accelerating charge produces electromagnetic waves
•Consider a current that changes suddenly
•Current stops – magnetic field diminishes
•Changing B-field produces E-field
•Changing E-field produces B-field
•You have a wave
–
Simple Antennas
•To produce long wavelength waves, easiest to use an antenna
+
•AC source plus two metal rods
+
•Some charge accumulates on each rod
+
•This creates an electric field
+
•The charging involves a current
+
•This creates a magnetic field
+
•It constantly reverses, creating a wave
•Works best if each rod is ¼ of a
wavelength long
 –
•The power in any direction is
–
–
–
sin 2 
S
–
2
r
–
Common Sources of EM Radiation
Radio Waves
Microwaves
Infrared
Visible
Ultraviolet
X-rays
Gamma Rays
Antennas
Klystron, Magnetron
Hot objects
Outer electrons in atoms
Inner electrons in atoms
Accelerated electrons
Nuclear reactions