Download Lecture 23 - Purdue Physics

Physics 42200
Waves & Oscillations
Lecture 23 – Electromagnetism
Spring 2013 Semester
Matthew Jones
Midterm Exam:
Date:
Time:
Room:
Material:
Wednesday, March 6th
8:00 – 10:00 pm
PHYS 203
French, chapters 1-8
Electromagnetism
• Previous examples of wave propagation:
– String with tension and mass per unit length
– Elastic rod with Young’s modulus and density
– Sound propagating in air…
– Waves in water…
– Etc…
• The rest of the course is devoted to the study
of one very important example of wave
phenomena:
Forces on Charges
• Coulomb’s law of electrostatic force:
Charles-Augustin de
Coulomb
(1736 - 1806)
#
#"
• The magnitude of the attractive/repulsive force is
= where
=
and therefore
4
1
= 8.99 × 10 ∙
̂
∙
"∙
= 8.85 × 10 " ∙
(This constant is called the “permittivity of free space”)
Electric Field
• An electric charge changes the properties of the
space around it.
– It is the source of an “electric field”.
– It could be defined as the “force per unit charge”:
"
$=
̂
= 4$
%&'(
– Quantum field theory provides a deeper description…
• Gauss’s Law:
1
)+
* ∙ $,- =
#/01/23
.
Electric “flux” through surface S
Magnetic Field
• Moving charges (ie, electric current) produce a
magnetic field:
6,ℓ × ̂
)
5=
4
A moving charge in a magnetic field
experiences a force:
= 48 × 5
Lorentz force law:
= 4$ + 48 × 5
Gauss’s Law for Magnetism
• Electric charges produce electric fields:
#/01/23
+ ∙ $,- =
)*
.
• But there are no “magnetic charges”:
2
+ ∙ 5,- =
)*
.
0
Magnetic “flux” through surface S
Ampere’s Law
• An electric current produces a magnetic field that
curls around it:
:
:
(Close, but not quite the whole story)
Faraday’s Law of Magnetic Induction
• Magnetic flux: ;< = =. 5 ∙ *> ,-
• Faraday observed that a changing
magnetic flux through a wire loop
induced a current
– It transferred energy to the charge
carriers in the wire
,;<
ℇ = −
,A
,
B $ ∙ ,ℓ = − ) *> ∙ 5,,A .
:
3
The Problem with Ampere’s Law
Current through C" is 6
Current through C is 0!
The current through C is zero,
but the electric flux is not zero.
The electric flux changes as
charge flows onto the capacitor.
Maxwell’s Displacement Current
• We can think of the changing electric flux
through C as if it were a current:
,;3
,
62 =
=
) $ ∙ *> ,,A
,A .
B D ∙ Eℓ = GH I + IE
F
E
+ EQ
= GH J + GH KH ) M ∙ N
EL OP
4
Maxwell’s Equations (1864)
RSNTSEU
+ ∙ M EQ = B N
KH
O
+ ∙ D EQ = H
B N
O
EVW
B M ∙ Eℓ = −
EL
F
EVU
B D ∙ Eℓ = GH I + GH XH
EL
F
1
2
3
4
Maxwell’s Equations in Free Space
In “free space” where there are no electric charges or
sources of current, Maxwell’s equations are quite
symmetric:
+ ∙ M EQ = H
B N
O
+ ∙ D EQ = H
B N
O
EVW
B M ∙ Eℓ = −
EL
F
EVU
B D ∙ Eℓ = GH XH
EL
F
Maxwell’s Equations in Free Space
EVW
B M ∙ Eℓ = −
EL
F
EVU
B D ∙ Eℓ = GH XH
EL
F
A changing magnetic flux induces an electric field.
A changing electric flux induces a magnetic field.
Will this process continue indefinitely?
Light is an Electromagnetic Wave
• Faraday’s Law:
∮: $ ∙ ,ℓ =
2Z[
−
2\
=−
]^_
]\
∆a∆b
B $ ∙ ,ℓ = $c b Δa − $c (b" )Δa
:
≈
x
]hi
∆b∆a
]j
Ex
z1
y
By
Δa
Ex
Δb
z2
z
Light is an Electromagnetic Wave
• Ampere’s law:
∮: 5 ∙ ,ℓ =
2Zk
2\
=
B 5 ∙ ,ℓ = 5m b" Δl − 5m (b )Δl
:
≈−
]^_
]j
x
∆b∆l
]hi
∆l∆b
]\
Ex
∆b
z1 z2
y
By
By
∆l
z
Putting these together…
n5m
n$c
=−
nb
nA
n5m
n$c
−
=
nb
nA
Differentiate the first with respect to b:
n 5m
n $c
=−
nb
nbnA
Differentiate the second with respect to A:
n 5m
n $c
−
=
nbnA
nA
oP M p
oP Mp
= GH KH
P
oq
oLP
Velocity of Electromagnetic Waves
n l
1 n l
=
na
8 nA
oP M p
oP M p
= GH KH
P
oq
oLP
Speed of wave propagation is
1
8=
=
4 × 10
r
/-
1
8.854 × 10
= P. ttu × vHu w/x
(speed of light)
"
/ ∙
Light is an Electromagnetic Wave
Speed of light was measured by Fizeau in 1849:
8 = 315,300km/s
Maxwell wrote:
This velocity is so nearly that of light, that it seems we have
strong reason to conclude that light itself (including radiant
heat, and other radiations if any) is an electromagnetic
disturbance in the form of waves propagated through the
electromagnetic field according to electromagnetic laws.
Electromagnetic Waves
oP Mp
oP Mp
= GH K H
P
oq
oLP
• A solution is $c b, A = $ sin b − €A
where € = • = 2 •/ƒ
• What is the magnetic field?
n5m
n$c
=−
= − $ cos b − €A
nA
nb
5m a, A =
€
$ sin a − €A
Electromagnetic Waves
ax
bz
ly
• $, 5 and 8 are mutually perpendicular.
• In general, the direction is
†̂ = $‡ × 5‡
Energy in Electromagnetic Waves
Energy density of electric and magnetic fields:
1
1
‰3 =
$
‰< =
5
2
2
For an electromagnetic wave, 5 = $ ڥ = $
1
1
‰< =
5 =
$ = ‰3
2
2
The total energy density is
‰ = ‰< + ‰3 =
$
Intensity of Electromagnetic Waves
• Intensity is defined as the average power
transmitted per unit area.
Intensity = Energy density × wave velocity
6=
•$
=
$
•
• = 377Ω ≡ •
(Impedance of free space)
Poynting Vector
• We can construct a vector from the intensity and the
direction †̂ = $‡ × 5‡:
$×5
C=
$
C =
=6
•
• This represents the flow of power in the direction †̂
• Average electric field: $ <1 = $ / 2
$
C =
2•
• Units: Watts/m2
Optics
• If light is an electromagnetic wave, and we just
finished reading that yellow book, why are there still
7 weeks left in the course?
• In practice, the boundary value problem is too
complicated to solve exactly
• We develop approximation techniques instead
• These can be excellent approximations when used
appropriately
• These techniques are so useful for many practical
applications that they deserve specific attention