General form of Faraday’s Law
Ub  Ua
Vba  Vb  Va 
   E ds
q
a
b
So the electromotive force around a closed path is:
   E ds
And Faraday’s Law becomes:
d B
   E ds  
dt
A changing magnetic flux produces an electric field.
This electric field is necessarily non-conservative.
Induced emf and Electric Fields
• An electric field is created in the conductor as
a result of the changing magnetic flux
• Even in the absence of a conducting loop, a
changing magnetic field will generate an
electric field in empty space
• This induced electric field is nonconservative
– Unlike the electric field produced by stationary charges
• The emf for any closed path can be expressed as
the line integral of E.ds over the path
E produced by changing B
d B
 E d 
dt
E 2r  r
d
B
2
dt
rdB
E 
2 dt
How about outside ro ?
Problems with Ampere’s Law

C
B  d  o Iencl  ?
B 2r  o I
o I
B
2r
But what if…..

C
B  d  o Iencl  ?
B 2r   o  0 
B 0
?????
Maxwell’s correction to Ampere’s Law
Q  CV
o A
C
d
V  Ed
 o A 
Q
  Ed    o AE   o  E
 d 
dQ
d E
Called “displacement current”, Id
I
 o
dt
dt
Maxwell’s Equations
q
S E  dA  εo
 B  dA  0
Gauss's law  electric 
Gauss's law in magnetism
S
dB
 E  ds   dt
Faraday's law
dE
 B  ds  μo I  εo μo dt
Ampere-Maxwell law
•The two Gauss’s laws are symmetrical, apart from the absence of the term for
magnetic monopoles in Gauss’s law for magnetism
•Faraday’s law and the Ampere-Maxwell law are symmetrical in that the line
integrals of E and B around a closed path are related to the rate of change of
the respective fluxes
•
•
•
•
•
•
•
•
Gauss’s law (electrical):
The total electric flux through any
closed surface equals the net charge
inside that surface divided by o
This relates an electric field to the
charge distribution that creates it
Gauss’s law (magnetism):
The total magnetic flux through
any closed surface is zero
This says the number of field lines
that enter a closed volume must
equal the number that leave that
volume
This implies the magnetic field
lines cannot begin or end at any
point
Isolated magnetic monopoles have
not been observed in nature
q
S E  dA  εo
 B  dA  0
S
•
•
•
•
Faraday’s law of Induction:
This describes the creation of an electric field by a
changing magnetic flux
The law states that the emf, which is the line
integral of the electric field around any closed
path, equals the rate of change of the magnetic flux
through any surface bounded by that path
One consequence is the current induced in a
conducting loop placed in a time-varying B
•
The Ampere-Maxwell law is a generalization of
Ampere’s law
•
It describes the creation of a magnetic field by an
electric field and electric currents
The line integral of the magnetic field around any
closed path is the given sum
•
dB
 E  ds   dt
dE
 B  ds  μo I  εo μo dt
The Lorentz Force Law
• Once the electric and magnetic fields are known at
some point in space, the force acting on a particle
of charge q can be calculated
• F = qE + qv x B
• This relationship is called the Lorentz force law
• Maxwell’s equations, together with this force law,
completely describe all classical electromagnetic
interactions
Maxwell’s Equation’s in integral form
Q 1
 A E  dA  o  o

A
B  dA  0

V
dV
Gauss’s Law
Gauss’s Law for Magnetism
d B
d
 C E  d   dt   dt A B  dA
Faraday’s Law

d E
dE 
C B  d  o Iencl  oo dt  o A  J  o dt   dA
Ampere’s Law
Maxwell’s Equation’s in free space
(no charge or current)


A
A
E  dA  0
Gauss’s Law
B  dA  0
Gauss’s Law for Magnetism
d B
d
 C E  d   dt   dt A B  dA
d E
d
 C B  d  oo dt  oo dt A E  dA
Faraday’s Law
Ampere’s Law
Vector Calculus Theorems


A
C
F  dA    FdV
V

Gauss’ Divergence Theorem

F  dl    F  dA
A
Stokes Theorem
And an Important Identity

 
  
2
    F    F   F
e j e j
Maxwell’s Equation’s In Differential Form

E 
o
 B  0
Gauss’s Law
Gauss’s Law for Magnetism
B
E  
t
E
  B  o J  oo
t
Faraday’s Law
Ampere’s Law
Hertz’s Experiment
•
•
•
•
•
An induction coil is connected to a
transmitter
The transmitter consists of two spherical
electrodes separated by a narrow gap
The discharge between the electrodes
exhibits an oscillatory behavior at a very
high frequency
Sparks were induced across the gap of the
receiving electrodes when the frequency of
the receiver was adjusted to match that of
the transmitter
In a series of other experiments, Hertz also
showed that the radiation generated by this
equipment exhibited wave properties
–
•
Interference, diffraction, reflection, refraction
and polarization
He also measured the speed of the radiation
Implication
• A magnetic field will be produced in empty space if there
is a changing electric field. (correction to Ampere)
• This magnetic field will be changing. (originally there
was none!)
• The changing magnetic field will produce an electric field.
(Faraday)
• This changes the electric field.
• This produces a new magnetic field.
• This is a change in the magnetic field.
An antenna
Hook up an
AC source
We have changed the magnetic
field near the antenna
An electric field results! This is
the start of a “radiation field.”
Look at the cross section
Accelerating
electric charges
give rise to
electromagnetic
waves
Called:
“Electromagnetic Waves”
E and B are perpendicular (transverse)
We say that the waves are “polarized.”
E and B are in phase (peaks and zeros align)
Angular Dependence of Intensity
• This shows the angular
dependence of the radiation
intensity produced by a dipole
antenna
• The intensity and power
radiated are a maximum in a
plane that is perpendicular to
the antenna and passing
through its midpoint
• The intensity varies as
(sin2 θ / r2
Active Figure 34.3
(SLIDESHOW MODE ONLY)
Harmonic Plane Waves
At t = 0
E
l  spatial period or
wavelength
l
x
phase velocity
l
2 l 
v   fl 

T
T 2 k
At x = 0
E
t
T
T  temporal period
Applying Faraday to radiation
d B
 C E  d   dt

C
E  d   E  dE  y  Ey  dEy
d B dB

dxy
dt
dt
dB
dEy  
dxy
dt
dE
dB

dx
dt
Applying Ampere to radiation
d E
 C B  d  oo dt

C
B  d  Bz   B  dB z  dBz
d E dE

dxz
dt
dt
dE
dBz   o  o
dxz
dt
dB
dE
  o  o
dx
dt
Fields are functions of both
position (x) and time (t)
dE
dB

dx
dt
Partial derivatives
are appropriate
B
E
  o o
x
t
dB
dE
  o  o
dx
dt
E
 B

2
x
x t
2
E
B

x
t
 B
 2E
 o o 2
t x
t
2E
2E
 o o 2
2
x
t
This is a wave
equation!
The Trial Solution
• The simplest solution to the partial differential
equations is a sinusoidal wave:
– E = Emax cos (kx – ωt)
– B = Bmax cos (kx – ωt)
• The angular wave number is k = 2π/λ
– λ is the wavelength
• The angular frequency is ω = 2πƒ
– ƒ is the wave frequency
The trial solution
E  E y  Eo sin  kx  t 
2E
2E
 o o 2
2
x
t
2E
2
 k E o sin  kx  t 
2
x
2E
2
  E o sin  kx  t 
2
t
k 2 Eo sin  kx  t   oo2Eo sin  kx  t 
2
1

2
k
o o
The speed of light
(or any other electromagnetic radiation)
l
2 l 
v   fl 

T
T 2 k

1
vc 
k
o  o
3.
The speed of an electromagnetic wave traveling in a transparent
nonmagnetic substance is , where κ is the dielectric constant of the substance.
Determine the speed of light in water, which has a dielectric constant at optical
frequencies of 1.78.
5.
Figure 34.3 shows a plane electromagnetic sinusoidal wave propagating
in the x direction. Suppose that the wavelength is 50.0 m, and the electric field
vibrates in the xy plane with an amplitude of 22.0 V/m. Calculate (a) the
frequency of the wave and (b) the magnitude and direction of B when the electric
field has its maximum value in the negative y direction. (c) Write an expression
for B with the correct unit vector, with numerical values for Bmax, k, and ω, and
with its magnitude in the form
6.
Write down expressions for the electric and magnetic fields of a
sinusoidal plane electromagnetic wave having a frequency of 3.00 GHz and
traveling in the positive x direction. The amplitude of the electric field is 300
V/m.
The electromagnetic spectrum
l
2 l 
v   fl 

T
T 2 k
Another look
dE
dB

dx
dt
B  Bz  Bo sin  kx  t 
E  E y  Eo sin  kx  t 
d
d
E o sin  kx  t    Bo sin  kx  t 
dx
dt
Eo k cos  kx  t   Bo cos  kx  t 
Eo 
1
 c
Bo k
o  o
Energy in Waves
1
1 2
2
u  0 E 
B
2
2 0
u  0 E 2
Eo 
1
 c
Bo k
o  o
1 2
u B
0
0
u
EB
0
Poynting Vector

1
S
EB
0

EB E 2 c B 2
S


μo μo c
μo
S  cu
• Poynting vector points in the direction the wave moves
• Poynting vector gives the energy passing through a unit
area in 1 sec.
• Units are Watts/m2
Intensity
• The wave intensity, I, is the time average of
S (the Poynting vector) over one or more
cycles
• When the average is taken, the time average
of cos2(kx - ωt) = ½ is involved
2
2
EmaxBmax Emax
c Bmax
I  Sav 


 cuave
2μo
2μo c
2μo
11.
How much electromagnetic energy per cubic meter is
contained in sunlight, if the intensity of sunlight at the Earth’s
surface under a fairly clear sky is 1 000 W/m2?
16.
Assuming that the antenna of a 10.0-kW radio station
radiates spherical electromagnetic waves, compute the
maximum value of the magnetic field 5.00 km from the
antenna, and compare this value with the surface magnetic
field of the Earth.
21.
A lightbulb filament has a resistance of 110 Ω. The
bulb is plugged into a standard 120-V (rms) outlet, and emits
1.00% of the electric power delivered to it by electromagnetic
radiation of frequency f. Assuming that the bulb is covered
with a filter that absorbs all other frequencies, find the
amplitude of the magnetic field 1.00 m from the bulb.
Radiation Pressure
F 1 dp
P 
A A dt
Maxwell showed:
U
p 
c
(Absorption of radiation
by an object)
1 dU Save
P

Ac dt
c
What if the radiation reflects off an object?
Pressure and Momentum
• For a perfectly reflecting surface,
p = 2U/c and P = 2S/c
• For a surface with a reflectivity somewhere
between a perfect reflector and a perfect absorber,
the momentum delivered to the surface will be
somewhere in between U/c and 2U/c
• For direct sunlight, the radiation pressure is about
5 x 10-6 N/m2
26.
A 100-mW laser beam is reflected back upon itself by a
mirror. Calculate the force on the mirror.
27.
A radio wave transmits 25.0 W/m2 of power per unit
area. A flat surface of area A is perpendicular to the direction of
propagation of the wave. Calculate the radiation pressure on it,
assuming the surface is a perfect absorber.
29.
A 15.0-mW helium–neon laser (λ = 632.8 nm) emits a
beam of circular cross section with a diameter of 2.00 mm. (a)
Find the maximum electric field in the beam. (b) What total
energy is contained in a 1.00-m length of the beam? (c) Find the
momentum carried by a 1.00-m length of the beam.
Background for the superior
mathematics student!
Harmonic Plane Waves
In general, We will only be concerned with the real part of
the complex phasor representation of a plane wave.
Using Euler’s formula:
  i bkx t g 
E  Eo e
 Eo cos kx  t  i sin kx  t
b
g b


Re E  Eo cos kx  t
di
k
b
2
g
= propagation number
l
2

= angular frequency

(kx-t) = phase
g
Phase Velocity - Another View
  kx  t
d
dx
k
   kv    0
dt
dt
Since the plane waves remain plane waves,
the phase on a plane does not change with time

v
k