Download E - Purdue Physics

Final EXAM: Monday, May 5th, LWSN room B155
3:30-5:30 pm
Magnetic Forces in Moving Reference Frames
1 e2 æ v 2 ö
20 ns F = 4pe r 2 çç1 - c 2 ÷÷
0
è
ø
+e
1
2
1
e
15 ns F =
4pe 0 r 2
Who will see protons hit
floor and ceiling first?
r
v
F21,m
2 B1
+e v
E1
F21,e
Relativistic Field Transformations
Our detailed derivations are not correct for relativistic speeds,
but the ratio Fm/Fe is the same for any speed:
Fm v 2
= 2
Fe c
According to the theory of relativity:
(E
- vBz )
E = Ex
E =
Bx' = Bx
v
æ
ö
B
+
E
ç y
z÷
2
c
ø
By' = è
1 - v 2 / c2
'
x
'
y
y
1- v / c
2
2
E =
'
z
(E
z
+ vB y )
1 - v 2 / c2
v
æ
ö
B
E
ç z
y÷
2
c
ø
Bz' = è
1 - v2 / c2
Magnetic Field of a Moving Particle
q
E=
B=0
2
4pe 0 r
v
æ
ö
v
B
E
ç z
÷
- 2 Ey
y
2
c
ø=
c
Moving: Bz' = è
1 - v2 / c2
1 - v 2 / c2
Still:
1
v 1 q
Slow case: v<<c  B = - 2
c 4pe 0 r 2
1
= c2
m0 qv Field transformation is consistent
'
m0e 0
Bz = with Biot-Savart law
4p r 2
'
z
Electric and magnetic fields are interrelated
Magnetic fields are relativistic consequence of electric fields
Electric Field of a Rapidly Moving Particle
E = Ex
'
x
E =
'
y
E =
'
y
(E
y
- vBz )
1- v / c
2
2
Ey
1- v / c
2
2
E =
'
z
E =
'
z
(E
z
+ vB y )
1 - v 2 / c2
Ez
1 - v 2 / c2
The Principle of Relativity
There may be different mechanisms for different observers
in different reference frames, but all observers can
correctly predict what will happen in their own frames,
using the same relativistically correct physical laws.
Wave Description
 – wavelength: distance between crests (meters)
T – period: the time between crests passing fixed location (seconds)
v – speed: the distance one crest moves in a second (m/s)
f – frequency: the number of crests passing fixed location in one
second (1/s or Hz)
 – angular frequency: 2f: (rad/s)

v
T
1
f 
T
v  f
Wave: Variation in Time and Space
 2 
E  E0 cos
t
 T 
2
 2
E  E0 cos
t

 T
 2 
E  E0 cos
x
  

x

‘-’ sign: the point on wave moves to the right
Wave: Phase Shift
2 
 2
E  E0 cos
t
x
 
 T
But E @ t=0 and x =0, may not equal E0
phase shift, =0…2
2
 2

E  E0 cos
t
x  

 T

 2

E  E0 cos
t     E0 cost   
 T

Two waves are ‘out of phase’
(Shown for x=0)
Wave: Amplitude and Intensity
E  E0 cost   
E0 is a parameter called amplitude (positive). Time dependence
is in cosine function
Often we detect ‘intensity’, or energy flux ~ E2.
Intensity I (W/m2):
I  E02
Works also for other waves,
such as sound or water waves.
Interference
Superposition principle: The net electric field at any location is
vector sum of the electric fields contributed by all sources.
Laser: source of radiation which has
the same frequency (monochromatic)
and phase (coherent) across the beam.
Two slits are sources of two waves
with the same phase and frequency.
What can we expect to see on the screen?
Can particle model explain the pattern?
Interference: Constructive
E1
Two emitters:
E2
Fields in crossing point
E1  E0 cost 
E2  E0 cost 
Superposition: E  E1  E2  2 E0 cost 
Amplitude increases twice: constructive interference
Interference: Energy
E1
Two emitters:
E2
E  E1  E2  2 E0 cost 
What about the intensity (energy flux)?
Energy flux increases 4 times while two emitters produce only
twice more energy
There must be an area in space where intensity is smaller than that
produced by one emitter
Interference: Destructive
E1  E0 cost 
E1
E2  E0 cost   
E2
E  E1  E2  E0 cost   cost     0
 cost 
Two waves are 1800 out of phase: destructive interference
Two-Slit Experiment with Waves
•We measure the Intensity of the wave motion at the detector
(related to the square of the wave height)
Two-Slit Experiment with Bullets
•Bullets arrive in lumps
•We measure the probability of arrival of a lump (bullet)
•P1 = probability bullet went through slit 1 in arriving at x
•P12 = P1 + P2
Differential Form of Gauss' Law (Sec. 22.8)
GAUSS' LAW
Think about a region of space, enclosed by a box.
Divide Gauss' law by the volume of the box:
E || x
Take the limit
of a small box
Work on the left hand side of the equation:
For a general case where E can point in any direction:
GAUSS' LAW
Differential Form
"Parallel Derivative"
where
Differential Form of Ampere's Law (Sec. 22.9)
Ampere's Law
3
Write I in terms of current density J:
2
4
1
Divide Ampere's Law by a very small ΔA:
Current I
out of the board
In our geometry, n = z
Differential Form of Ampere's Law (Sec. 22.9)
Ampere's Law
3
2
4
We divided Ampere's Law by a very small ΔA, and got this:
1
Current I
out of the board
Now work on the left hand side:
Definition of
derivative!
"Crossed derivative"
Differential Form of Ampere's Law (Sec. 22.9)
Ampere's Law
3
2
4
We divided Ampere's Law by a very small ΔA, and got this:
1
Current I
out of the board
For a loop in any direction, this can be re-expressed as:
AMPERE'S LAW
Differential Form
Curl: Here's the Math
copy 1st two colums
+(
-
)
+(
-
)
+(
-
)
set up the answer
Curl: Here's the Math
+(
-
)
+(
-
)
+(
-
)
Maxwell's Equations – The Full Story
Divergence
GAUSS' LAW
Flux
GAUSS' LAW
(Magnetism)
Curl
FARADAY'S LAW
AMPERE'S LAW
Circulation
Maxwell's Equations – No Charges
In the ABSENCE of
"sources" = charges, currents:
GAUSS' LAW
GAUSS' LAW
(Magnetism)
FARADAY'S LAW
AMPERE'S LAW
This says
once a wave
starts,
it keeps going!
Maxwell's Equations – No Charges
What happens if we feed one equation into the other?
Use
This
Maxwell's Equations – No Charges
What happens if we feed one equation into the other?
("Vector identity" -- see Wolfram alpha)
Maxwell's Equations – No Charges
How do you solve a Differential Equation?
Know the answer! (Ask Wolfram Alpha)
 This is a WAVE EQUATION, with speed c
Using similar ideas, you can show that E obeys the same equation: