W15D1:
Poynting Vector and Energy
Flow
Today’s Readings:
Course Notes: Sections 13.6, 13.12.3-13.12.4
1
Announcements
Final Math Review
Week 15 Tues from 9-11 pm in 32-082
Final Exam Monday Morning
May 20 from 9 am-12 noon
Johnson Athletic Center Track 2nd floor
2
Outline
Poynting Vector and Energy Flow
Examples
3
Maxwell’s Equations
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
4
Electromagnetism Review
Conservation of charge:
òò
closed
surface
d
J × dA = r dV
òòò
dt volume
enclosed
E and B fields exert forces on (moving) electric
charges:
Fq = q(E + v ´ B)
Energy stored in electric and magnetic fields
e0 2
U E = òòò uE dV = òòò
E dV
2
all space
all space
1 2
U B = òòò uB dV = òòò
B dV
2 m0
all space
all space
5
Energy Flow
6
Poynting Vector
Power per unit area:
Poynting vector
S=
E´B
m0
Power through a surface
P=
òò
open
surface
S × n̂ da
7
Group Problem: Resistor Power
a
I
L
Consider the above cylindrical resistor with resistance R
and current I flowing as shown.
a) What are the electric and magnetic fields on the surface
of the resistor?
b) Calculate the flux of the Poynting vector through the
surface of the resistor (power) in terms of the electric and
magnetic fields.
c) Express your answer to part b) in terms of the current I
and resistance R. Does your answer make sense?
8
Energy Flow: Resistor
S=
E´B
m0
On surface of resistor direction is INWARD
9
Energy Flow: Capacitor
S=
E´B
m0
What is the magnetic field?
10
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 )
11
Sign Conventions: Right Hand Rule
d
òC B × d s = m0 òòS J × n̂da + m0e 0 dt
òò E × n̂da
S
Integration direction counter clockwise for line
integral requires that unit normal points out
page for surface integral.
Current positive out of page. Negative into
page.
Electric flux positive out of page, negative into
page.
B field clockwise, line integral negative.
B field counterclockwise, line integral positive.
12
Concept Questions:
Poynting Vector
13
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.
14
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.
15
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.
16
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.
17
Group Problem: Capacitor
A circular capacitor of spacing
d and radius R is in a circuit
carrying the steady current i
shown. Ignore edge effects. At
time t = 0 it is uncharged
1.
2.
3.
4.
Find the electric field E(t) at A (mag. & dir.)
Find the magnetic field B(t) at A (mag. & dir.)
Find the Poynting vector S(t) at A (mag. & dir.)
What is the flux of the Poynting vector into/out of the
capacitor?
5. How does this compare to the time derivative of the energy
stored in the electric field?
6. Does this make sense?
18
Another look at Inductance
19
Faraday & Inductors
L
LI = FSelf
I
e
dF B
dI
== -L
dt
dt
20
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.
21
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.
22
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.
23
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.
24
Group Problem: Inductor
A solenoid of radius a and length h
has an increasing current I(t) as
pictured.
Consider a point D inside the
solenoid at radius r (r < a). Ignore
edge effects.
1.
2.
3.
4.
5.
Find the magnetic field B(t) at D (dir. and mag.)
Find the electric field E(t) at D (dir. and mag.)
Find the Poynting vector S(t) at D (dir. and mag.)
What is the flux of the Poynting vector into/out of the inductor?
How does this compare to the time derivative of the energy
stored in the magnetic field?
6. Does this make sense to you?
25
Energy Flow: Inductor
S=
E´B
m0
Direction on surface of inductor with
increasing current is INWARD
26
Energy Flow: Inductor
S=
E´B
m0
Direction on surface of inductor with
decreasing current is OUTWARD
27
Power & Energy in Circuit Elements
P=
òò
S× dA
Surface
uE = e 0 E
1
2
uB =
1
2 m0
B
2
2
Dissipates
Power
Store
Energy
POWER
When
(dis)charging
28
Poynting Vector and EM Waves
29
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
30
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:
E0 B0
E
cB
I º<S >=
=
=
2 m0
2 m0 c 2 m0
2
0
2
0
31
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
32
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 are the maximum values of electric and magnetic
fields, E and B, at this same distance from the bulb?
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.
33