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W15D2
Poynting Vector and EM Waves
Radiation Pressure
Final Exam Review
P36 - 1
Announcements
Final Exam Monday Morning May 20
from 9 am-12 noon
Johnson Athletic Center Track 2nd floor
P36 - 2
Before Starting…
All of your grades should now be posted (with
possible exception of last problem set). If this is
not the case contact your grad TA immediately.
We have given you enough information about how
we grade and what the cut lines are that you can
estimate what you need on the final to get the
letter grade of your heart’s desire.
P36 - 3
Poynting Vector and EM Waves
P36 - 4
Energy in EM Waves
Energy densities:
Consider cylinder:
1
1
2
2
u E = e 0 E , uB =
B
2
2 m0
2
æ
1
B ö
2
dU = (uE + uB ) Adz = ç e 0 E + ÷ Acdt
2è
m0 ø
What is the energy flow per unit area?
2
ö cæ
B
EB ö
1 dU c æ
2
S
= ç e 0 E + ÷ = ç e 0 cEB +
A dt 2 è
m0 ø 2 è
cm0 ÷ø
(
)
EB
EB
2
=
e 0 m0 c + 1  
0
2 m0
P36 - 5
Poynting Vector and Intensity
Direction of energy flow = direction of wave propagation
S=
E´B
m0
: Poynting vector
units: Joules per square meter per sec
Intensity I:
2
0
2
0
E0 B0
E
cB
1
2
I S 


 c 0 E0
2 0 2 0 c 2 0 2
P36 - 6
Momentum & Radiation Pressure
EM waves transport energy:
They also transport momentum:
They exert a pressure:
S=
E´B
m0
p =U / c
F 1 dp 1 dU S
P= =
=
=
A A dt cA dt
c
This is only for hitting an absorbing surface. For hitting a
perfectly reflecting surface the values are doubled:
2U
Momentum transfer : p =
;
c
2S
Radiation pressure : P =
c
P36 - 7
Group Problem: Radiation
A light bulb puts out 100 W of electromagnetic radiation.
What is the time-average intensity of radiation from this
light bulb at a distance of one meter from the bulb?
What is the maximum value of the electric field, E , at this
same distance from the bulb in V/m? What is the
pressure this radiation will exert on a very small perfectly
conducting plate at 1 meter. For simplicity, you may
assume the radiation is a plane wave of wavelength λ.
o  1.25664 E  06
 o  8.84194 E  12
c  3 x108 m / s
P36 - 8
Electromagnetism Review
1
òò E ××n̂ da = e òòò r dV
(Gauss's Law)
òò B × ×n̂ da = 0
(Magnetic Gauss's Law)
d
òC E × d s = - dt òòS B × n̂ da
(Faraday's Law)
S
0
V
S
d
ò B × d s = m0 òò J × n̂ da + m0e 0 dt òò E × n̂ da (Maxwell - Ampere's Law)
C
S
S
F = q(E + v ´ B)
òò
closed
surface
(Lorentz Force Law)
d
J × dA = r dV
òòò
dt volume
enclosed
(Charge Conservation)
P36 - 9
Circuits
P36 - 10
Sign Conventions - Battery
Moving from the negative to positive terminal of a
battery increases your potential
DV = Vb - Va = +e
Moving from the positive to negative terminal of a
battery decreases your potential
DV = Vb - Va = -e
P36 - 11
Sign Conventions - Resistor
Moving across a resistor in the direction of current
decreases your potential
DV = Vb - Va = -IR
Moving across a resistor opposite the direction of
current increases your potential
DV = Vb - Va = +IR
P36 - 12
Sign Conventions - Capacitor
Moving across a capacitor from the negatively to
positively charged plate increases the electric
potential
DV = Vb - Va = +Q / C
Moving across a capacitor from the positively to
negatively charged plate decreases the electric
potential
DV = Vb - Va = -Q / C
P36 - 13
(Dis)Charging a Capacitor
1. When the direction of current flow is toward
the positive plate of a capacitor, then
dQ
I=+
dt
2. When the direction of current flow is away from
the positive plate of a capacitor, then
dQ
I=dt
P36 - 14
Sign Conventions - Inductor
Moving across an inductor in
the direction of current
contributes

dI
 L
dt
Moving across an inductor
opposite the direction of
current contributes
e
dI
= +L
dt
P36 - 15
Kirchhoff’s Modified 2nd Rule
dF B
å D Vi = - ò E × d s = + d t
i
dF B
Þ å D Vi =0
dt
i
If all inductance is ‘localized’ in inductors then
our problems go away – we just have:
dI
å D Vi - L d t = 0
i
P36 - 16
Steps For Setting Up Circuit Equations for
Circuit with N loops and M junctions
1.
2.
3.
4.
5.
Simplify resistors in series/parallel
Assign current in every branch
Choose circulation direction for N-1 loops
Assign charges to each side of capacitor.
Determine relation between current and charge
for each branch containing a capacitor
6. Write M-1 current conservation equations for
junctions
7. Write N-1 loop equations.
8. Solve system of equations.
P36 - 17
Displacement Current
P36 - 18
Displacement Current
Q
E=
Þ Q = e 0 EA = e 0 F E
e0 A
dF E
dQ
= e0
º I dis
dt
dt
So we had to modify Ampere’s Law:
d
òC B × d s = m0 òòS J × n̂ da + m0e 0 dt
òò E × n̂ da
S
= m0 (I con + I dis )
P36 - 19
EM Waves
P36 - 20
Traveling Sinusoidal Wave: Summary
y(x,t)  y0 sin(k(x  vt))
Two periodicities:
Spatial period : Wavelength  ; Temporal period T.
Wave Number : k = 2p / l
Dispersion Relation : T = l / v
Direction of Propagation : + x - direction
P36 - 21
Traveling Sinusoidal Wave
Alternative form:
y(x,t)  y0 sin(k(x  vt))  y0 sin(kx   t)
Wave Number : k  2 / 
Angular Frequency :   2 / T
Dispersion Relation :   vT    kv
Frequency :
f  1/ T  v   f
P36 - 22
Plane Electromagnetic Waves
http://youtu.be/3IvZF_LXzcc
P36 - 23
Electromagnetic Waves: Plane
Sinusoidal Waves
Watch 2 Ways:
1) Sine wave
traveling to right
(+x)
2) Collection of out
of phase
oscillators (watch
one position)
Don’t confuse vectors with heights – they are magnitudes of
electric field (gold) and magnetic field (blue)
http://youtu.be/3IvZF_LXzcc
P36 - 24
Traveling Plane Sinusoidal
Electromagnetic Waves
E = E0 sin(kx - w t) ĵ
B = B0 sin(kx - w t) k̂
are special solutions to the 1-dim wave equations
 Ey
1  Ey
 2 2
2
x
c t
2
2
 Bz
1  Bz
 2 2
2
x
c t
2
2
where
k º 2p / l ,
w º 2p / T,
c=l/T
P36 - 25
1 Dim’l Sinusoidal EM Waves
In order for the fields
E = E0 sin(kx - w t) ĵ,
B = B0 sin(kx - w t) k̂
to satisfy either condition below
¶E y
¶Bz
=¶t
¶x
¶Bz
1 ¶E y
=- 2
¶x
c ¶t
then
B0  E0 / c
P36 - 26
Properties of 1 Dim’l EM Waves
1. Travel (through vacuum) with
speed of light
c
1
m
 3.0  10
s
0 0
8
2. At every point in the wave and any instant of time,
electric and magnetic fields are in phase with one another,
amplitudes obey
E0 / B0  c
3. Electric and magnetic fields are perpendicular to one another,
and to the direction of propagation (they are transverse):
4. Direction of propagation = Direction of E ´ B.
P36 - 27
Concept Questions:
EM Waves
P36 - 28
Concept Question: Direction of
Propagation
The figure shows the E
(yellow) and B (blue)
fields of a plane wave.
This wave is
propagating in the
1. +x direction
2. –x direction
3. +z direction
4. –z direction
P36 - 29
Concept Question Answer:
Propagation
Answer: 4. The wave is moving in the –z direction
The propagation
direction is given by
the dir E ´ B
(Yellow x Blue)
P36 - 30
Concept Question: Traveling Wave
The B field of a plane EM wave is B( y,t) = B0 sin(ky - w t)k̂
The electric field of this wave is given by
1. E( y,t) = E0 sin(ky - w t) ĵ
2. E( y,t) = E0 sin(ky - w t)(- ĵ)
3. E( y,t) = E0 sin(ky - w t)î
4. E( y,t) = E0 sin(ky - w t)(- î)
P36 - 31
Concept Q. Ans.: Traveling Wave
Answer: 4. E( y,t) = E0 sin(ky - w t)(- î)
From the argument of the sin(ky   t) , we know the
wave propagates in the positive y-direction.
So we have Ê ´ B̂ = ?´ k̂ = ĵ Þ Ê = - î
P36 - 32
Concept Question EM Wave
The electric field of a plane wave is:
E(z,t) = E0 sin(kz + w t) ĵ
The magnetic field of this wave is given by:
1. B(z,t) = B0 sin(kz + w t)î
2. B(z,t) = B0 sin(kz + w t)(- î)
3. B(z,t) = B0 sin(kz + w t)k̂
4. B(z,t) = B0 sin(kz + w t)(-k̂)
P36 - 33
Concept Q. Ans.: EM Wave
Answer: 1. B(z,t) = B0 sin(kz + w t)î
From the argument of the sin(kz   t) , we know
the wave propagates in the negative z-direction.
So we have Ê ´ B̂ = ĵ ´ ? = -k̂ Þ B̂ = î
P36 - 34
Energy Flow
Poynting vector :
Intensity :
Radiation pressure :
S=
E´B
m0
E0 B0
E
cB
I º<S >=
=
=
2 m0
2 m 0 c 2 m0
2
0
2
0
Pabsorb = S / c; Preflect = 2S / c
P36 - 35
Also in Circuit Elements…
S=
E´B
m0
On surface of resistor is INWARD
P36 - 36
Concept Questions:
Poynting Vector
P36 - 37
Concept Question: Capacitor
The figures above show a side and top view of a capacitor with
charge Q and electric and magnetic fields E and B at time t. At
this time the charge Q is:
1.
2.
3.
4.
Increasing in time
Constant in time.
Decreasing in time.
Not enough information given to determine how Q is changing.
P36 - 38
Concept Q. Answer: Capacitor
Answer: 3. The charge Q is decreasing in time
Use the Ampere-Maxwell Law. Choose positive unit normal out
of plane. Because the magnetic field points clockwise line
integral is negative hence positive electric flux (out of the plane
of the figure on the right) must be decreasing. Hence E is
decreasing. Thus Q must be decreasing, since E is proportional
to Q.
P36 - 39
Concept Question: Capacitor
The figures above show a side and top view of a capacitor with
charge Q and electric and magnetic fields E and B at time t. At
this time the energy stored in the electric field is:
1. Increasing in
2. Constant in time.
3. Decreasing in time.
P36 - 40
Concept Q. Answer: Capacitor
Answer: 1. The the energy stored in the electric field is
increasing in time
The direction of the Poynting Flux S (= E x B) inside the
capacitor is inward. Therefore electromagnetic energy is
flowing inward, and the energy in the electric field inside is
increasing.
P36 - 41
Concept Question: Inductor
The figures above show a side and top view of a solenoid
carrying current I with electric and magnetic fields E and B at
time t. The current I is
1. increasing in time.
2. constant in time.
3. decreasing in time.
P36 - 42
Concept Question Answer: Inductor
Answer: 3. The current I is decreasing in time
Use Faraday’s law. Choose positive unit normal out of plane.
Because the electric field points counterclockwise line integral
is positive, therefore the positive magnetic flux must be
decreasing (out of the plane of the figure on the right). Hence B
is decreasing. Thus I must be decreasing, since B is
proportional to I.
P36 - 43
Concept Question: Inductor
The figures above show a side and top view of a solenoid
carrying current I with electric and magnetic fields E and B at
time t. The energy stored in the magnetic field is
1. Increasing in time
2. Constant in time.
3. Decreasing in time.
P36 - 44
Concept Question Answer: Inductor
Answer: 3. The energy stored in the magnetic field is
decreasing in time.
The Poynting Flux S (= E x B) inside the solenoid is directed
outward from the center of the solenoid. Therefore EM energy is
flowing outward, and the energy stored in the magnetic field
inside is decreasing.
P36 - 45