Ch. 34 - Maxwell's equations - UCF Physics

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Nov 16, 2013 (3 years and 8 months ago)

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Chapter 34

(continued)


The Laws of Electromagnetism

Maxwell’s Equations

Displacement Current

Electromagnetic Radiation

The Electromagnetic Spectrum

Maxwell’s Equations of Electromagnetism

in Vacuum (no charges, no masses)

E
dA



0
B
dA



0
E
dl
d
dt
B





B
dl
d
dt
E





0
0

Plane Electromagnetic Waves

x

E
y

B
z

Notes: Waves are in Phase,



but fields oriented at 90
0
.



k=
2p/l.



Speed of wave is c=
w
/欠⠽(
f
l
)







At all times E=cB.

c
m
s



1
3
10
0
0
8
/
/


E
(x, t) = E
P

sin (kx
-
w


B
⡸Ⱐ,⤠㴠B
P

sin (kx
-
w


ˆ

z

ˆ

j

c

Plane Electromagnetic Waves


x

E
y

B
z

Note: sin(
w
t
-
k砩㴠
-
獩渨k
-
w
琩t


湯瑡n楯湳i慲攠楮e敲捨慮来慢汥⸠

獩渨
w
t
-
k砩慮搠獩渨k
-
w
琩r数e敳敮ew慶敳a瑲t癥汩湧n
瑯t慲摳a⭸

w桩h攠獩渨
w
琫t砩瑲t癥汳
瑯t慲摳a
-

.

Energy in Electromagnetic Waves



Electric and magnetic fields contain energy,
potential energy stored in the field: u
E

and u
B

u
E
: ½

0

E
2

electric field energy density

u
B
: (1/

0
) B
2
magnetic field energy density



The energy is put into the oscillating fields by the
sources that generate them.



This energy can then propagate to locations far
away, at the velocity of light.

B

E

Energy in Electromagnetic Waves

area

A

dx

Energy per unit volume is


u = u
E
+ u
B






c

propagation

direction



1
2
1
0
2
0
2
(
)


E
B
B

E

Energy in Electromagnetic Waves

area

A

dx

Energy per unit volume is


u = u
E
+ u
B




Thus the energy, dU, in a box of

area A and length dx is



c

propagation

direction



1
2
1
0
2
0
2
(
)


E
B
dU
E
B
Adx


1
2
1
0
2
0
2
(
)


B

E

Energy in Electromagnetic Waves

area

A

dx

Energy per unit volume is


u = u
E
+ u
B




Thus the energy, dU, in a box of

area A and length dx is



Let the length dx equal cdt. Then all of this energy leaves

the box in time dt. Thus energy flows at the rate

c

propagation

direction



1
2
1
0
2
0
2
(
)


E
B
dU
E
B
Adx


1
2
1
0
2
0
2
(
)


dU
dt
E
B
Ac


1
2
1
0
2
0
2
(
)


Energy Flow in Electromagnetic Waves

area

A

dx

c

propagation

direction

B

E

dU
dt
c
E
B
A


2
1
0
2
0
2
(
)


Rate of energy flow:

Energy Flow in Electromagnetic Waves

area

A

dx

c

propagation

direction

We define the
intensity

S
, as the rate

of energy flow per unit area:

S
c
E
B


2
1
0
2
0
2
(
)


B

E

dU
dt
c
E
B
A


2
1
0
2
0
2
(
)


Rate of energy flow:

Energy Flow in Electromagnetic Waves

area

A

dx

c

propagation

direction

We define the
intensity

S
, as the rate

of energy flow per unit area:

S
c
E
B


2
1
0
2
0
2
(
)


Rearranging by substituting E=cB and B=E/c, we get,

S
c
cEB
c
EB
c
EB
EB





2
1
1
2
1
0
0
0
0
0
2
0
(
)
(
)






B

E

dU
dt
c
E
B
A


2
1
0
2
0
2
(
)


Rate of energy flow:

The Poynting Vector

area

A

dx

B

E

propagation

direction

In general, we find:



S

=
(1/

0
)

E

x
B



S

is a vector that points in the

direction of propagation of the

wave and represents the rate of

energy flow per unit area.

We call this the
“Poynting vector”.


Units of S are Jm
-
2
s
-
1
, or Watts/m
2
.


S
The Inverse
-
Square Dependence of
S

Source

r

A point source of light, or any radiation, spreads

out in all directions:

Source

Power, P, flowing

through sphere

is same for any

radius.

Area
r

2
S
P
r

4
2
p
S
r

1
2
Example:

An observer is 1.8 m from a point light source whose
average power P= 250 W. Calculate the rms fields in
the position of the observer.

Intensity of light at a distance r is S = P / 4
p
r
2


I
P
r
c
E
E
P
c
r
W
H
m
m
s
m
E
V
m
B
E
c
V
m
m
s
T
rms
rms
rms
rms












4
1
4
250
4
10
3
10
4
1
8
48
48
3
10
0.
16
2
0
2
0
2
7
8
2
8
p


p
p
p

(
)(
/
)(
.
/
)
(
.
)
/
/
.
/
Example:

An observer is 1.8 m from a point light source whose
average power P= 250 W. Calculate the rms fields in
the position of the observer.

Intensity of light at a distance r is S = P / 4
p
r
2


I
P
r
c
E
E
P
c
r
W
H
m
m
s
m
E
V
m
B
E
c
V
m
m
s
T
rms
rms
rms
rms












4
1
4
250
4
10
3
10
4
1
8
48
48
3
10
0.
16
2
0
2
0
2
7
8
2
8
p


p
p
p

(
)(
/
)(
.
/
)
(
.
)
/
/
.
/
Example:

An observer is 1.8 m from a point light source whose
average power P= 250 W. Calculate the rms fields in
the position of the observer.

Intensity of light at a distance r is S= P / 4
p
r
2


I
P
r
c
E
E
P
c
r
W
H
m
m
s
m
E
V
m
B
E
c
V
m
m
s
T
rms
rms
rms
rms












4
1
4
250
4
10
3
10
4
1
8
48
48
3
10
0.
16
2
0
2
0
2
7
8
2
8
p


p
p
p

(
)(
/
)(
.
/
)
(
.
)
/
/
.
/
Wave Momentum and Radiation Pressure

Momentum and energy of a wave

are related by,
p = U / c
.


Now,
Force = d p /dt = (dU/dt)/c



pressure (radiation) = Force / unit area


P = (dU/dt) / (A c) = S / c




Radiation Pressure





P
S
c
rad

Example
:
Serious proposals have been made to “sail”
spacecraft to the outer solar system using the pressure of sunlight.
How much sail area must a 1000 kg spacecraft have if its
acceleration is to be 1 m/s
2

at the Earth’s orbit? Make the sail
reflective.

Can ignore gravity.

Need F=ma=(1000kg)(1 m/s
2
)=1000 N

This comes from pressure: F=PA, so A=F/P.

Here P is the radiation pressure of sunlight:

Sun’s power = 4 x 10
26

W, so S=power/(4
p
r
2
) gives


S = (4 x 10
26

W) / (4
p
(1.5x10
11
m)
2
)= 1.4kW/m
2
.

Thus the pressure due to this light, reflected, is:


P = 2S/c = 2(1400W/m
2
) / 3x10
8
m/s = 9.4x10
-
6
N/m
2


Hence A=1000N / 9.4x10
-
6
N/m
2

=1.0x10
8

m
2

= 100 km
2

Polarizatio
n

The direction of polarization of a wave is the direction of

the electric field. Most light is randomly polarized, which





means it contains a mixture of waves






of different polarizations.

x

E
y

B
z

Polarization

direction

Polarizatio
n

A
polarizer

lets through light of only one polarization:

E
0

E

E

E = E
0

cos
q


hence, S = S
0

cos
2
q




-

Malus’s Law

q

Transmitted light

has its E in the

direction of the

polarizer’s

transmission axis.