Download Lecture 25 - Purdue Physics

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.

Eradiative 

1  qa
40 c 2 r
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:
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
 0E 2 
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)
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
 773 N/C
c 0
(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  qv  B
y
x
Fmag
What if there is negative charge?

 
Fmag   q v  B
‘Radiation pressure’:
What is its magnitude?
Average speed: v/2
v
vE
Fmag  q B  q
2
2 c
Fmag
vE
v
q
/( qE ) 
 1
Felec
2c
2c
Fmag
Momentum Flux
Net momentum:
in transverse direction: 0
in longitudinal direction: >0
Relativistic energy:
 
E   pc  mc
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
1  

E  B (in N/m 2 )
c 0 c
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:
E

S
1  

EB
c 0 c
flux
 725 N/C
c 0

S
1
E
E2
6
2

E 

4
.
65

10
N/m
c 0 c c 0 c 2
Note: What if we have a reflective surface?
Total force on the sail: F  9.3 N
Atmospheric pressure is ~ 105 N/m2
 9.3  106 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
Vcapacitor  Vinductor  0
Q
dI
L
0
C
dt
I 
dQ
dt
d 2Q
Q  LC 2  0
dt
Q  a  b cosct 
a  b cosct   LC  bc2 cosct   0
a=0
 t 
Q  b cos

 LC 
c
1
LC
 t 
Q  Q0 cos

 LC 
Current in an LC Circuit
 t 
Q  Q0 cos

 LC 
I 
dQ
dt
Current in an LC circuit
Q
Q0
 t 
I
sin 

LC  LC 
Period:
T  2 LC

Frequency: f  1 / 2 LC

Energy in an LC Circuit
Q2
Initial energy stored in a capacitor:
2C
At time t=0: Q=Q0
At time t=

2
LC : Q=0
U cap
Q02

2C
U sol
1 2
 LI
2
1/4 of a period
System oscillates: energy is passed
back and forth between electric and
magnetic fields.
Energy in an LC Circuit
What is maximum current?
At time t=0:
Q02
U total  U el  U mag 
2C

LC :
At time t=
2
1 2

LI max
U total  U el  U mag
2
1 2
Q02
LI max 
2
2C
I max
Q0

LC
Energy in LC Circuit
U  Uelectric  Umagnetic
(No dissipation in this circuit)
1 Q2
1 2
d(
)
d(
LI ) Q dQ
dU
dI
 2 C  2

 LI
0
dt
dt
dt
C dt
dt
dQ
 I
dt
As capacitor loses charge, current increases
As capacitor gains charge, current decreases
Q
dI
L 0
C
dt
Same equation as obtained via considering potential differences
LC Circuit and Resonance

Frequency: f  1 / 2 LC
Radio
receiver:
