Download Notes 26

Physics-272 Lecture
2121
Physics-272
Lecture
Description of Wave Motion
Electromagnetic Waves
Maxwell’s Equations Again!
Maxwell’s Equations
James Clerk Maxwell (1831-1879)
• Generalized Ampere’s Law
• And made equations symmetric:
– a changing magnetic field produces an electric field
– a changing electric field produces a magnetic field
• Showed that Maxwell’s equations predicted
electromagnetic waves and c =1/√ε0µ0
• Unified electricity and magnetism and light.
Maxwell’s Equations (integral form)
Name
Equation
Description
Gauss’ Law for
Electricity
r r Q
∫ E ⋅ dA =
Charge and electric
fields
ε0
Gauss’ Law for
Magnetism
Faraday’s Law
Ampere’s Law
(modified by
Maxwell)
r r
∫ B ⋅ dA = 0
r r
dΦ B
∫ E ⋅ dl = −
dt
r
dΦ E 


∫ B ⋅ dl = µ0  ic + ε 0
dt


Magnetic fields
Electrical effects
from changing B
field
Magnetic effects
from current and
Changing E field
All of electricity and magnetism is summarized by
Maxwell’s Equations.
On to Waves!!
• Note the symmetry now of Maxwell’s Equations in free space,
meaning when no charges or currents are present
r r
∫ E ⋅ dA = 0
r r
∫ B ⋅ dA = 0
r r
dΦ B
∫ E ⋅ dl = −
dt
r
dΦ E
∫ B ⋅ dl = µ 0ε 0 dt
• Combining these equations leads to wave equations for
E and B, e.g.,
∂ 2 Ex
∂ 2 Ex
= µ0ε 0
2
2
∂z
∂t
• Do you remember the wave equation???
∂ 2h 1 ∂ 2h
= 2 2
2
∂x v ∂t
h is the variable that is changing
in space (x) and time (t). v is the
velocity of the wave.
Review of Waves from Physics 170
2
• The one-dimensional wave equation:
has a general solution of the form:
2
∂ h 1∂ h
= 2 2
2
∂x v ∂t
h( x, t ) = h1 ( x − vt ) + h2 ( x + vt )
where h1 represents a wave traveling in the +x direction and
h2 represents a wave traveling in the -x direction.
• A specific solution for harmonic waves traveling in the +x
direction is:
h λ
h x , t = A cos kx − ω t
A
(
)
(
)
2π
ω = 2π f =
k =
λ
T
ω
v = λf =
x
2π
k
A = amplitude
λ = wavelength
f = frequency
v = speed
k = wave number
Clicker problem
y
• Snapshots of a wave with angular
frequency ω are shown at 3 times:
2A – Which of the following expressions
describes this wave?
(a) y = sin(kx-ωt)
(b) y = sin(kx+ω
ωt)
(c) y = cos(kx+ωt)
t=
0
2B • In what direction is this wave traveling?
(a) +x direction
t=0
0
(b) -x direction
t=
0
0
x
π
2ω
π
ω
Clicker problem
y
• Snapshots of a wave with angular
frequency ω are shown at 3 times:
2A – Which of the following expressions
describes this wave?
(a) y = sin(kx-ωt)
(b) y = sin(kx+ω
ωt)
(c) y = cos(kx+ωt)
t=0
0
t=
0
• The t =0 snapshot ⇒ at t =0, y = sinkx
• At t =π
π/2ω and x=0, (a) ⇒ y = sin(-π
π/2) = -1
π/2) = +1
• At t =π
π/2ω and x =0, (b) ⇒ y = sin(+π
t=
0
0
x
π
2ω
π
ω
Clicker problem
y
• Snapshots of a wave with angular
frequency ω are shown at 3 times:
2A – Which of the following expressions
describes this wave?
(a) y = sin(kx-ωt)
(b) y = sin(kx+ω
ωt)
(c) y = cos(kx+ωt)
t=
0
2B • In what direction is this wave traveling?
(a) + x direction
t=0
0
t=
0
(b) -x direction
• We claim this wave moves in the -x direction.
• The orange dot marks a point of constant phase.
• It clearly moves in the -x direction as time increases!!
0
x
π
2ω
π
ω
Velocity of Electromagnetic Waves
• The wave equation for Ex
(Maxwell derived it in ~1865!):
∂ 2Ex
∂ 2Ex
= µ 0ε 0
2
∂ z
∂ t2
• Comparing to the general wave equation: ∂ 2 h
1 ∂ 2h
= 2
2
∂ x
v ∂ t2
we have the velocity of electromagnetic waves in free space:
1
v=
= 3.00 × 10 8 m / s ≡ c
µ 0ε 0
• This value is essentially identical to the speed of light
measured by Foucault in 1860!
– Maxwell identified light as an electromagnetic wave.
E & B in Electromagnetic Wave
• Plane Harmonic Wave:
E x = E 0 sin( kz − ω t )
B y = B0 sin( kz − ω t )
where
ω = kc
x
z
y
By is in phase with Ex
B0 = E0 / c
The direction of propagation ŝis given by the cross product
( )
sˆ = eˆ × bˆ
where eˆ , bˆ are the unit vectors in the (E,B) directions.
Nothing special about (Ex, By); e.g., could have (Ey, -Bx)
Clicker Problem
• Suppose the electric field in an e-m wave is given by:
r
E = − yˆ E0 cos( − kz + ω t )
3A In what direction is this wave traveling ?
(a) + z direction
(b) - z direction
3B Which of the following expressions describes the magnetic field
associated with this wave?
(a) Bx = -(Eo/c) cos(kz +ω t)
(b) Bx = +(Eo/c) cos(kz -ω t)
(c) Bx = +(Eo/c) sin(kz -ω t)
Clicker Problem
• Suppose the electric field in an e-m wave is given by:
r
E = − yˆ E0 cos( − kz + ω t )
3A – In what direction is this wave traveling ?
(a) + z direction
(b) - z direction
• To determine the direction, set phase = 0: − kz + ωt = 0
• Therefore wave moves in + z direction!
• Another way: Relative signs opposite means + direction
z = +
ω
k
t
Clicker Problem
• Suppose the electric field in an e-m wave is given by:
r
E = − yˆ E0 cos( − kz + ω t )
3A – In what direction is this wave traveling ?
(a) + z direction
(b) - z direction
3B • Which of the following expressions describes the magnetic field
associated with this wave?
(a) Bx = -(Eo/c) cos(kz +ω t)
(b) Bx = +(Eo/c) cos(kz -ω t) )
(c) Bx = +(Eo/c) sin(kz -ω t)
• B is in phase with E and has direction determined from: bˆ = sˆ ×eˆ
• At t=0, z=0, Ey = -Eo
• Therefore at t=0, z=0, bˆ = sˆ ×eˆ = kˆ ×(− ˆj) = iˆ
r
ˆ E0
B = +i
c
c o s (k z − ω t )
10) An electromagnetic wave is travelling along the x-axis,
with its electric field oscillating along the y-axis. In what
direction does the magnetic field oscillate?
a) along the x-axis
b) along the z-axis
c) along the y-axis
Note: the direction of propagation ŝ is given by the cross product
( )
sˆ = eˆ × bˆ
where eˆ , bˆ are the unit vectors in the (E,B) directions.
In this case, direction of
ŝ
is x and direction of
ê is y
xˆ = yˆ × zˆ
The fields must be perpendicular to each other
and to the direction of propagation.
Leads to the possibility of “polarization” of light to
be discussed in next chapter.
Electromagnetic (EM) Waves in free space
Maxwell’s eqns predict electric & magnetic fields can
propagate in vacuum.
Examples of EM waves include; radio/TV waves, light,
x-rays, and microwaves.
2
2
∂ Ex
∂ Ex
= µ0ε 0
2
2
∂z
∂t
• A specific solution for harmonic waves traveling in the +x
direction is:
h λ
A
h x , t = A cos kx − ω t
(
)
(
)
2π
ω = 2π f =
k =
λ
T
ω
v = λf =
x
2π
k
A = amplitude
λ = wavelength
f = frequency
v = speed
k = wave number
No – moving in the minus z direction
No – has Ey rather than Ex
Most of the EM spectrum
is invisible to our eyes
X-rays are deeply
penetrating because of their
short wavelengths
The earth’s atmosphere
blocks out much of the EM
spectrum (e.g. much of UV
and X-ray band)
Students must become familiar with the wavelengths
of common types of EM radiation
Wavelength is equal to the speed
of light divided by the frequency.
c 300, 000, 000 1
λ= =
=
f
900, 000, 000 3
y
An electromagnetic wave is described by:
where ĵ is the unit vector in the +y direction.
x
r
E = ˆjE0 cos(kz − ωt )
z
Which of the following graphs represents the z-dependence of Bx at t = 0?
X
(A)
X
(B)
(C)
(D)
E and B are “in phase” (or 180o out of phase)
r
E = ˆjE0 cos(kz − ωt )
Wave moves in +z direction
y
r r
E
E × B points in direction of propagation
x
r
ˆ 0 cos(kz − ωt )
B = −iB
B
z
y
An electromagnetic wave is described by:
r iˆ + ˆj
E=
E0 cos(kz + ωt )
2
What is the form of B for this wave?
A)
r iˆ + ˆj
B=
( E0 / c)cos(kz + ωt )
2
C)
r iˆ − ˆj
B=
( E0 / c)cos(kz + ωt )
2
r iˆ + ˆj
E=
E0 cos(kz + ωt )
2
B)
r −iˆ − ˆj
B=
( E0 / c)cos(kz + ωt )
2
D)
r −iˆ − ˆj
B=
( E0 / c)cos(kz + ωt )
2
Wave moves in –z direction
y
+z points out of screen
-z points into screen
E
x
B
r r
E × B points in negative z-direction
B must be perp to E
x
z
Plane Waves
• For any given value of z, the magnitude of
the electric field is uniform everywhere in
the x-y plane with that z value.
x
z
y
Shown is an EM wave at an instant
in time. Points A, B, and C lie in
the same x-y plane.
3) Compare the magnitudes of the electric field at points A and B.
a) Ea < Eb
b) Ea = Eb
c) Ea > Eb
4) Compare the magnitudes of the electric field at points A and C.
a) Ea < Ec
b) Ea = Ec
c) Ea > Ec
Students said …
x
z
Magnitude of field is determined
only by value of z (and t) !!!
y
Right:
• for any given value of z, the magnitude of the electric field is uniform
everywhere in the x-y plane with that z value
Wrong:
• The point A is where the electric field and magnetic field are zero. So the
points c and b have a greater field.
(a common misconception)
EM wave; energy momentum, angular momentum
Previously, we demonstrated the energy density existed in E fields in
capacitors and in B fields in inductors. We can sum these energies,
1
1
2
u = ε0E +
B2
2
2µ 0
Since, E=cB and
c = 1 / µ 0ε 0
then u in terms of E,
2
1
1
E
1


2
2 1
u = ε 0 E + µ0   = ε 0 E + ε 0 E 2 = ε 0 E 2
2
2 c
2
2
=> EM waves have energy and can transport energy at speed c
Almost all energy resources on the earth come directly or indirectly
via sunlight. What doesn’t?
EM wave; energy flow & Poynting Vector
ct
Suppose we have transverse E and B fields moving
at velocity c, to the right. If the energy density is
A
2 and the field is moving through area
u = ε 0 E A at velocity c covers volume Act,
that has energy, uAct. Hence the amount of energy
flow per unit time per unit surface area is,
1 dU
1
(uAct ) = uc = ε 0 cE 2
=
A dt
At
This quantity is called, S. In terms of E&B field magnitudes,
S = ε 0 cE 2 = ε 0 cEcE / c = ε 0 cEcB = EB / µ 0
We also define the energy flow with direction, called the Poynting
vector, r
r r , which is in the direction of propagation
S=
1
µ0
E×B
EM wave; Intensity
The intensity, I, is the time average of the magnitude of the Poynting
vector and represents the average energy flow per unit area.
r
r
1 r
I = S ( x, y , z , t ) =
E ( x , y , z , t ) × B ( x, y , z , t )
µ0
In Y&F, section 32.4, the average S for a sinusoidal plane wave
(squared) is worked out,
r
Emax Bmax
Emax Bmax
1 + cos(2kx − 2ωt ) =
I = S ( x, y , z , t ) =
2µ0
2µ 0
The can be rewritten as,
r
Emax Bmax ε 0 c 2
ε0 2
Emax = Bmax
I = S ( x, y , z , t ) =
=
2µ 0
2
2c
Cell Phone Problem
32.22) A sinusoidal electromagnetic wave emitted by a cell phone
has a wavelength of 35.4 cm and an electric field amplitude of
5.4 x 10-2 V/m at a distance of 250 m from the antenna. (a)
What is the magnetic field amplitude? (b) The intensity? (C)
The total average power?
a.) E = c B;
B= E/c = 5.4 x 10-2 (V/m)/3 x 108 m/s
B = 1.8 x 10-10 T
b.)
1
I = ε 0 cE 2 ;
2
1
I = (8.85 × 10 −12 C 2 / Nm 2 )(3 × 10 8 m / s)(5.4 × 10 − 2 V / m)
2
I = 3.87 × 10 −6 W / m 2
c.) Assume isotropic: I=P/A; P=4πr2 I = 4 π (250m)2 I
P=3W
Waves Carry Energy
Intensity
Intensity = energy delivered per unit time, per unit area
1 dU
(U = energy)
I=
A dt
dU = u ⋅ volume = u Acdt
Length = c dt
Area = A
I =c u
Sunlight on Earth:
I ~ 1000 J/s/m2
~ 1 kW/m2
Comment on the Poynting Vector
Just another way to keep track of all this
- It’s the same thing as I but it has a direction
Light has Momentum!
If it has energy and it’s moving, then it also has momentum:
Analogy from mechanics:
p2
E=
2m
dE 2 p dp
mv dp
=
=
dt 2m dt
m dt
dE
dU
→
= IA
dt
dt
For E-M waves:
= vF
v→c
IA = cF
I
P=
c
Radiation pressure
I F
=
pressure
c A
3 G cell phone networks
in the US use 1710-1756
MHz frequencies.
What is the
corresponding
wavelength range ?
3 ×108 m / s /(1710 ×106 Hz ) = 0.175m
3 ×108 m / s /(1756 ×106 Hz ) = 0.170m