Transcript Lecture 25 - Purdue University

Final EXAM: Monday, May 5th, LWSN room B155
3:30-5:30 pm
Chapter 24
Classical Theory of
Electromagnetic Radiation
A. A charge at rest makes a 1/r2 electric field but no magnetic field
B. A charge moving with constant velocity makes a 1/r2 electric
field and 1/r2 magnetic field
C. An accelerated charge in addition makes electromagnetic
radiation, with 1/r electric field and 1/r magnetic field.

E radiative 
1

 qa
4  0
c r
2
Clicker
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:
2
a 
d y
dt
2
  y max  sin  t 

E radiative 

E radiative 
2
1

 qa
4  0
c r
f 
qy max 
4 0
c r
2
Sinusoidal E/M field
2
T  1/ f
2
1

2
sin  t  ˆj
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
2
 qEd   1 
K  K  0 

 

2m  c   2m 
p
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
Volume

1
2
0E 
2
1 1
2 0
B
2
3
(J/m )
E/M radiation: E=cB
2
1 1 E
1
1
2

 0E 
    0 E 1 
2
Volume
2
2 0  c 
2


c
0 0

Energy
1
2

   E2
0


Energy density of magnetic field in a traveling wave is exactly the
energy density of the electric field
Energy Flux
There is E/M energy stored in the pulse:
Energy
Volume
Pulse moves in space:
there is energy flux
Units: J/(m2s) = W/m2
During t:

Energy   0 E
flux 
Energy
flux 
1
2
A  t 
0
EB
 A  ct 
 c 0 E
2
used: E=cB, 00=1/c2
  0 E (J/m )
2
3
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
2
(in W/m )
• 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)
Exercise
A laser pointer emits ~5 mW of light power. What is the
approximate magnitude of the electric field?
Solution:
1. Spot size: ~2 mm
2. flux = (5.10-3 W)/(3.14.0.0012 m2)=1592 W/m2
3. Electric field:
E 
flux
c 0
 773 N/C
(rms value)
What if we focus it into 2 a micron spot?
Flux will increase 106 times, E will increase 103 times:
E  773 ,000 N/C
Momentum of E/M Radiation
• E field starts motion,
• No net momentum change in the y direction
• Moving charge in magnetic field:

 
Fmag  q v  B
y
x
Fmag
What if there is negative charge?

 
Fmag   q v  B
‘Radiation pressure’:
What is its magnitude?
Average speed: v/2
Fmag  q
Fmag
Felec
 q
v
B q
2
v E
2 c
v E
Fmag
2 c
/( qE ) 
v
2c
 1
Momentum Flux
Net momentum:
in transverse direction: 0
in longitudinal direction: >0
Relativistic energy:

E   pc   m c
2
2
2

2
Quantum view: light consists of photons with zero mass: E 2   pc 2
Classical (Maxwell): it is also valid, i.e. momentum = energy/speed

1  
S 
EB
0
Momentum flux:

S
c

1
 0c
 
EB
2
(in N/m )
Units of Pressure
Exercise: Solar Sail
What is the force due to sun light on a sail with the area 1 km2
near the Earth orbit (1400 W/m2)?
Solution:

S
E 
1  

EB
c
 0c
flux
c 0
 725 N/C

S
c

1
 0c
E
E
c

E
2
 0c
Note: What if we have a reflective surface?
Total force on the sail: F  9 . 3 N
Atmospheric pressure is ~ 105 N/m2
2
 4 . 65  10
6
 9 . 3  10
6
N/m
2
N/m
2
Re-radiation: Scattering
Positive charge
Electric fields are not blocked by matter: how can E decrease?
Cardboard
Why there is no light going through a cardboard?
Electric fields are not blocked by matter
Electrons and nucleus in cardboard reradiate light
Behind the cardboard reradiated E/M field cancels original field
In which of these situations will the
bulb light?
A)
B)
C)
D)
E)
A
B
C
None
B and C
Current in an LC Circuit
 V capacitor   V inductor  0
Q
L
C
dI
0
I  
dQ
dt
dt
2
Q  LC
d Q
dt
2
0
Q  a  b cos  ct 
a  b cos ct   LC  bc cos ct   0
2
a=0

Q  b cos 

c


LC 
t
1
LC

Q  Q 0 cos 



LC 
t
Current in an LC Circuit

Q  Q 0 cos 

I  


LC 
t
dQ
dt
Current in an LC circuit
I 
Period:

sin 
LC

Q0


LC 
t
T  2
LC
Frequency: f  1 / 2  LC 
Q
Energy in an LC Circuit
Initial energy stored in a capacitor:
Q
2
2C
2
At time t=0: Q=Q0
At time t=

LC
2
: Q=0
Q0
U cap 
U sol 
2C
1
2
1/4 of a period
System oscillates: energy is passed
back and forth between electric and
magnetic fields.
LI
2
Energy in an LC Circuit
What is maximum current?
At time t=0:
2
U total  U el  U mag 
At time t=

LC
2
2

2
LI
2
max

Q0
2C
2C
:
U total  U el  U mag
1
Q0
I max 
1
2
2
LI max
Q0
LC
Energy in LC Circuit
U  U electric  U magnetic
dU
dt
dQ
dt
d(

1Q
2
2 C
dt
 I
)
d(

1
(No dissipation in this circuit)
2
LI )
2
dt

Q dQ
C dt
 LI
dI
dt
 0
As capacitor loses charge, current increases
As capacitor gains charge, current decreases
Q
C
 L
dI
dt
 0
Same equation as obtained via considering potential differences
LC Circuit and Resonance
Frequency: f  1 / 2  LC 
Radio
receiver: