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Chapter 34 Electromagnetic Waves
r r Qenclosed
E
∫ ⋅ dA =
1.
static charge (Ch 21, 22, 23) --> Coulomb and Gauss’s law:
2.
static charge -->
3.
r r
dΦ
E
∫ ⋅ dl = − dtM
r r
current generates field, no magnetic monopole --> ∫ B ⋅ dA = 0
4.
static current (Ch 27) -->
ε0
r r
E
∫ ⋅ dl = 0
flux change and Faraday’s law (Ch 28) -->
r r
B
∫ ⋅ dl = µ0 I enclosed
Lorentz force law (Ch 26): moving charge or current in the magnetic field -->
r r
r
r r
F = qv × B = Il × B
No monopole -->
r r
B
∫ ⋅ dA = 0
Is there any similar formula, like Faraday’s law, for the magnetic field?
r r
r r dΦ
dΦ
∫ E ⋅ dl = − dtM --> ∫ B ⋅ dl ∝ dt
34.1 Displacement Current and the
General Form of Ampere’s Law
r
Ampere’s Law:
r
∫ B ⋅ dl
= µ0 I
C
On loop S1, you enclosed the current I inside, but no
current is enclosed on loop S2. Remember that the loop
S1 can extend the area to the region of zero current.
What happens?
r
The charge ( σ ) or electric displacement ( D ) is
changing with time and generating another kind of current: the Maxwell’s
1
displacement current I d = ε 0
dΦ E
dt
r
The generalized Ampere’s law:
r
∫ B ⋅ dl
= µ 0 I + µ0ε 0
C
dΦ E
dt
Derivation 1:
I=
dQ
dt
r r Q
dQ
d
Φ E = ∫ E ⋅ dA =
-->
= ε0 ΦE ≡ Id
ε0
dt
dt
Current I flows in the capacitor --> change the electric field in the capacitor
The change of the electric field seems to be one kind of current:
r
d
I d = ε 0 ∫ E ⋅ dA
dt
Displacement current density:
r
r
∂E
Jd = ε0
∂t
If the capacitor plates are very close, the electric field
σ
Q
between them is E =
=
. A changing electric field
ε0 ε0 A
dE 1 dQ / dt 1 r
may result in a current
=
= Jd
dt ε 0 A
ε0
r
r
∂E
Jd = ε0
.
∂t
Example: A parallel plate capacitor has closely spaced circular plates of radius R.
Charge is flowing onto the positive plate and off the negative plate at the rate
dQ
= 2.5A . Compute the displacement current through surface S passing
dt
r
between the plates by directly computing the rate of change of the flux of E through
surface S.
I=
Id = ε0
dΦ E
d
d Q
= ε 0 (ES ) = ε 0 
dt
dt
dt  Sε 0
 dQ
S  =
=I
 dt
2
Example: The circulate plates have a radius of R = 3.0 cm. Find the magnetic field
strength B at a point between the plates a distance r = 2.0 cm from the axis of the
plates when the current into the positive plate is 2.5 A.
  2 2 
  2 2  Q
d
4 dQ 4 I


ΦE =   S E =   S 
, Id = ε0 Φ E =
=
 3  
  3   Sε 0
dt
9 dt
9




r r
4
B
∫ ⋅ dl = µ0 I d --> 2π (0.002)B = µ0  9 (2.5 A)
34.2 Maxwell’s Equations and Hertz’s
Discoveries
r r Qenclosed
E
∫ ⋅ dA =
ε0
r
Gauss’s Law for Electric Fields
r
∫ B ⋅ dA = 0
r r
d r r
∫ E ⋅ dl = − dt ∫ B ⋅ dA
r r
d r r
∫ B ⋅ dl = µ0 I + µ0ε 0 dt ∫ E ⋅ dA
Gauss’s Law for Magnetism
+
-
Faraday’s Law
Ampere’s Law with Maxwell’s Displacement
Currents
Displacement Current --> Charge Conservation? Symmetry?
r
r
r r
F = qE + qv × B
Lorentz Force Law
3
34.3 Plane Electromagnetic Waves
1.
The magnetic and electric fields vary with time and displacement.
2.
3.
4.
The electric and magnetic fields are mutually orthogonal.
The magnetic field strength B is equal to E/C.
1
The propagating speed is C =
.
5.
r r
The direction of propagation is E × B
ε 0 µ0
Electromagnetic Waves:
light, infrared waves, radio-frequency waves
microwave oven (2.45 or 2.5 GHz):
Microwaves are absorbed by water, fats and sugars. When they are absorbed they are
converted (through frictional mechanism) into atomic motion - heat. They are not
absorbed by most plastics, glass or ceramics. Metal reflects microwaves, this is why
metal pans do not work well in a microwave oven.
The Wave Equation for Electromagnetic Waves
How can we get a differential equation of electromagnetic waves from the four
fundamental equations for electricity and magnetism?
A propagated plane wave function can be e i ( kx −ωt ) or sin (kx − ωt ) . Since
d2
d2
2
sin
(
kx
−
ω
t
)
=
−
k
sin
(
kx
−
ω
t
)
and
sin (kx − ωt ) = −ω 2 sin (kx − ωt ) , we may
dx 2
dt 2
 d2 
 d2 
suggest a differential equation of  2 Ψ ω 2 =  2 Ψ k 2 . --> A differential
 dx

 dx

4
d2
1 d2
=
ψ
ψ.
dx 2
c 2 dt 2
A changing electric field induced a magnetic field and, at the same time, the varying
magnetic field induced an electric field. The induction process maintain the
propagation of electromagnetic waves.
What is a plane wave?
equation maybe written as
What is a spherical wave?
Derivation of the Wave Equation
When electromagnetic waves propagate through vacuum
space, the current and charge are zero (I = 0 and q = 0).
The four fundamental equations (Maxwell’s equations)
are
r r
E
∫ ⋅ dA = 0 --- (1)
r r
B
∫ ⋅ dA = 0
--- (2)
r r
dΦ B
E
∫ ⋅ dl = − dt --- (3)
r r
dΦ E
B
∫ ⋅ dl = µ 0ε 0 dt --- (4)
from eq. 3 -> ( E + dE ) ⋅ l − E ⋅ l = −
dE =
d
(B ⋅ l ⋅ dx )
dt
∂E
∂E
dB
dx ,
dx ⋅ l = −
l ⋅ dx
∂x
∂x
dt
∂E
∂B
=−
∂x
∂t
-- (important) --- (5)
from eq. 4 ->
− ( B + dB ) ⋅ l + B ⋅ l = µε
−
d
(E ⋅ l ⋅ dx ) -> − dB ⋅ l = µε d (E ⋅ l ⋅ dx ) & dB = ∂B dx
dt
∂x
dt
∂B
dE
= µε
-- (important) --- (6)
∂x
dt
∂2
∂ ∂
∂2E
Eq. (5) & Eq. (6)-->
E=−
B = µε 2
∂t ∂x
∂x 2
∂t
2
∂ ∂E
∂ B
∂  ∂B 
= − µε 2
−
 = µε
∂x  ∂x 
∂x ∂t
∂t
r
r
E = Emax cos(kx − ωt )iˆ & B = Bmax cos(kx − ωt ) ˆj
5
Transverse waves
∂
(E = E max cos(kx − ωt ))− > ∂E = −kE max sin (kx − ωt )
∂x
∂x
∂
(B = Bmax cos(kx − ωt ))− > ∂B = ωBmax sin (kx − ωt ), ∂B = − ∂E
∂t
∂t
∂t
∂x
kE max
E
E
= 1 , max = c ->
=c
ωBmax
Bmax
B
Example: The electric field of an electromagnetic wave is given by
r
E ( x, t ) = E0 cos(kx − ωt )kˆ . (a) What is the direction of propagation of the wave? (b)
What is the direction of the magnetic field in the x = 0 plane at time t = 0? (c) Find the
r r
magnetic field of the same wave. (d) Compute E × B .
(a) iˆ
( )
(b) kˆ × ˆj = −iˆ --> kˆ × − ˆj = iˆ
( )
r
E
(c) B ( x, t ) = 0 cos(kx − ωt ) − ˆj
c
r r E2
(d) E × B = 0 cos 2 (kx − ωt )iˆ
c
Example: The electric field of an electromagnetic wave is given by
r
E ( x, t ) = ˆjE0 sin (kx − ωt ) + kˆE0 cos(kx − ωt ) . (a) Find the magnetic field of the same
r r
r r
wave. (b) Compute E ⋅ B and E × B .
(a) The propagation direction is in the x-direction iˆ .
( )
ˆj × kˆ = iˆ & kˆ × − ˆj = iˆ
( )
r
E
E
B( x, t ) = kˆ 0 sin (kx − ωt ) + − ˆj 0 cos(kx − ωt )
c
c
r r
r r
E2
(b) E ⋅ B = 0 , E × B = iˆ 0
c
6
34.4 Energy Carried by Electromagnetic
Waves
Intensity:
Pav U av / ∆t u avV / ∆t
L
=
=
= uav
= uav c
A
A
A
∆t
I=
uav = ue + um
1
B2
E2
ε0E2
B 2 EB
2
ε 0 E 2 , um =
=
=
-->
u
=
u
+
u
=
E
=
=
ε
av
e
m
0
2
2µ0 2µ0c 2
2
µ0 µ 0c
r
E B
EB
I = uav c = rms rms = 0 0 = S
av
µ0
2µ0
ue =
Poynting Vector:
r r
r E×B
S=
a direction in which the energy is transmitted?
µ0
Example: Fields due to a point source
A point source of em radiation has an average power output of 800 W.
Calculate the
maximum values of the electric and magnetic fields at a point 3.5 m from the source.
I=
P
800
=
= 5.2W / m 2
2
2
4πr
4π ⋅ 3.5
E max = 2 µ 0 cI = 62.6V / m ,
34.5 Momentum and Radiation Pressure
Charged particle experience electric fields of the EM Waves:
qE
q2E 2 2
t --> K =
t
m
2m
 qE 
FB = qvB = q
t B
 m 
v y = at =
the transferred momentum in time t1: p = ∫ Fdt =
q 2 EB t12
m 2
7
--> p =
1 q2E 2 2 K
t =
c 2m
c
momentum change when adsorbing em waves: p =
U
c
if the intensity is I and the pressure is P
P=
1 U av U av l 1
1 I
F 1 p
=
=
=
= uav ⋅ c ⋅ =
A A t
At c
Al t c
c c
for a perfect reflector: P = 2
I
c
Laser Tweezer:
Example: Solar energy
The sun delivers 1000 W/m2 of energy to the Earth’s surface.
(a) Calculate the total
power incident on a roof of dimension 8 m x 20 m.
Power: P = 1000 × 8 × 20 = 1.6 × 10 5 W
Example: Pressure from a laser pointer
If a 3mW pointer creates a spot with a diameter of 2 mm.
Determine the radiation
pressure on a screen that reflects 70% of the light striking it.
3 × 10 −3
Intensity: I =
π (10
Pressure: P =
I
(1 + 0.7 ) = 5.4 × 10 −6 N / m 2
c
)
−3 2
= 9.6 × 10 2 W / m 2
Space Sailing by Using the Radiation Pressure?
Example: You are stranded in space a distance of s from your spaceship. You carry a
laser with power Pav. If your total mass, including your space suit and laser, is m, how
long will it take you to reach the spaceship if you point the laser directly away from
it?
Pav =
dU
dp d  U
, F=
= 
dt
dt dt  c
--> a =
Pav
& t=
mc
2s
=
a
 1 dU 1
= Pav = ma
=
c
 c dt
2smc
Pav
8
34.6 Production of Electromagnetic
Waves by an Antena
−
Q
dI
−L
=0
C
dt
d 2Q
d 2Q Q
Q + LC 2 = 0 ,
+
=0
LC
dt
dt 2
k
d 2x
F = m 2 = − kx , x = x0 cos(ϖt ) , ω =
m
dt
2
d Q Q
1
+
= 0 , Q = Q0 cos(ωt ) , ω =
2
LC
dt
LC
Generating EM Waves:
r
a≠0
1.
accelerate or decelerate free charges
2.
light wave: when outer electrons bounded to atoms make transitions
3.
macroscopic electric currents oscillating in radio transmission antennas
4.
vibration --> infrared EM waves
deceleration of electrons when crashing into a metal target --> bremsstrahlung
Ref: http://www.ghettodriveby.com/bremsstrahlung/
9
X-Ray:
a
v
Synchrotron Radiation:
Electric Dipole Radiation
Generator:
Electric Dipole Radiation
I ∝ sin 2 θ
θ
Detector:
10
Example: A loop antenna consisting of a single 10-cm radius loop of wire is used to
detect electromagnetic waves for which Erms = 0.15 V/m. Find the rms emf induced
in the loop if the wave frequency is (a) 600 kHz and (b) 60 MHz.
Brms =
( )
( )
Erms
d
E
, ε = − (BA) --> ε rms = ωBrms πr 2 = ω rms πr 2
c
dt
c
34.7 The Spectrum of Electromagnetic
Waves
Radio waves, Microwaves, Infrared waves, Visible light, Ultraviolet waves, X-rays,
Gamma rays
11
3 × 10 8
= 7.5 × 1014 Hz -->
−9
400 × 10
/ 1.602 × 10 −19 = 3.1 eV
Visible Light: λ = 400 nm --> f =
(
)(
E = hf = 6.626 × 10 −34 7.5 × 1014
)(
)
3 × 108
= 3 × 1018 Hz -->
0.1× 10 −9
3 × 1018 / 1.602 × 10 −19 = 12.4 keV
X-Ray: λ = 0.1 nm --> f =
(
E = hf = 6.626 × 10 −34
)(
)(
)
12