Download Lecture 29 - Purdue Physics

Physics Education Research:
Submit your on-line course evaluation (you should have
information in your e-mail, deadline this week).
I will add extra point (Exam 2) for these who completed course
evaluation. When completed evaluation e-mail me with
“I have completed Phys272H course evaluation” and your name to
claim your extra point.
EXAM 2 free response: These who have 0-3 points on free response
problem can earn extra 2 points by solving problem from scratch in
my office. Offer valid till May 2nd.
Final EXAM: May 3rd, Friday 3:30-5:30 pm CL50 (room 224)
you are responsible for all material covered in lecture slides.
Practice problems are available on course web page.
“Extra” problems (not related to final) are also available on course
web page for these who want to try out their skills.
Physics Education Research: Clicker 1
Physics Education Research: Clicker 2
Physics Education Research: Clicker 3
Maxwell’s Equations
Four equations (integral form) :
q
Gauss’s law

 E  nˆdA 
Gauss’s law for magnetism
—
 B  n̂dA  0
Faraday’s law
Ampere-Maxwell law
+ Lorentz force
inside
0


 
d 
 E  dl   dt  B  nˆdA
 
d elec 

 B  dl  0  Iinside_ path   0 dt 


 
F  qE  qv  B
A Pulse: Speed of Propagation
Ampere’s law
B  0 0vE
E=Bv
Faraday’s law
B  0 0vBv
1  0 0v 2
v
1
0 0
 3  108 m/s
E=cB
Based on Maxwell’s equations, pulse must propagate at speed of light
 
Direction of speed is given by vector product: E  B
Maxwell’s Theory of Electromagnetism
• Light is electromagnetic wave!
• Challenge: Design an electric
device which emits and detects
electromagnetic waves
(1831-1879)
Accelerated Charges
Electromagnetic pulse can propagate in space
How can we initiate such a pulse?
Short pulse of transverse
electric field
Accelerated Charges
1. Transverse pulse
propagates at speed of
light
2. Since E(t) there must
be B
3. Direction
 of v is given
by: E  B
E
v
B
Accelerated Charges: 3D
Magnitude of the Transverse Electric Field
We can qualitatively predict the direction.
What is the magnitude?
Magnitude can be derived from
Gauss’s law
Vectors a, r and E always in one plane
Field ~ -qa

Eradiative 

1  qa
40 c 2 r
1. The direction of the field is opposite to qa
2. The electric field falls off at a rate 1/r
Exercise
a
An electron is briefly accelerated in the
direction shown. Draw the electric and
magnetic vectors of radiative field.
E
B
1. The direction of the field is opposite to qa
 
2. The direction of propagation is given by E  B
Exercise
An electric field of 106 N/C acts on an electron
for a short time. What is the magnitude of
electric field observed 2 cm away?
1. Acceleration a=F/m=qE/m=1.78.1017 m/s2
E=106 N/C
B
2 cm
a
2. The direction of the field is opposite to qa

3. The magnitude: 
1  qa
.10-7 N/C
E=1.44
Eradiative 
40 c 2 r
 
4. The direction of propagation is given by E  B
Erad
What is the magnitude of the Coulomb field at the same location?
q
6
E 

3
.
6

10
N/C
2
40 r
1
Question
A proton is briefly accelerated as shown below. What
is the direction of the radiative electric field that will
be detected at location A?
B
A
A
+
D
C
Sinusoidal Electromagnetic Radiation
Acceleration:
d2y
a  2   ymax  2 sin t 
dt


1  qa
Eradiative 
40 c 2 r

f 
2
T  1/ f

1 qymax  2
Eradiative 
sin t  ĵ
2
40 c r
Sinusoidal E/M field
Energy and Momentum of E/M Radiation
According to particle theory of light:
photons have energy and momentum
Classical E/M model of light:
E/M radiation must carry energy and momentum
Energy of E/M Radiation
A particle will experience electric
force during a short time d/c:
Felec  qE
 d
p  p  0  Felec t  qE  
 c
What will happen to the ball?
It will oscillate
Energy was transferred from E/M field to the ball
2
p 2  qEd   1 
K  K  0 

 

2m  c   2m 
Amount of energy in
the pulse is ~ E2
Energy of E/M Radiation
Ball gained energy:
2
 qEd   1 
K  
 

 c   2m 
Pulse energy must decrease
Energy 1
1 1 2
2
 0E 
B (J/m 3 )
Volume 2
2 0
E/M radiation: E=cB
Energy 1
1 1 E
1
1 
2
2
2

 0E 
    0 E 1 


E
2 
0
Volume 2
2 0  c 
2


c
0 0


Energy density of magnetic field in a traveling wave is exactly the
energy density of the electric field
2
Energy Flux
There is E/M energy stored in the pulse:
Energy
  0 E 2 (J/m 3 )
Volume
Pulse moves in space:
there is energy flux
Units: J/(m2s) = W/m2
During t:


Energy   0 E 2 A  ct 
Energy
flux 
 c 0 E 2
 A  t 
flux 
1
0
EB
used: E=cB, 00=1/c2
Energy Flux: The Poynting Vector
flux 
1
0
EB
 
The direction of the E/M radiation was given by E  B
Energy flux, the “Poynting vector”:
 1  
S
EB
0
(in W/m 2 )
• S is the rate of energy flux in E/M radiation
• It points in the direction of the E/M radiation
John Henry
Poynting
(1852-1914)