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Magnetic Forces in Moving Reference Frames
Two protons
Electric force:
F21,e
+e
1
r
v
F21,m
B1
+e v
E1
2
F21,e
1 e2
= q2 E1 =
rˆ
2
4pe 0 r
Magnetic field:
m0 q1v1 ´ rˆ
B1 =
4p r 2
Magnetic force:
F21,m = q2v2 ´ B1
F21,m
m0 e 2 v 2
= q2vB1 =
4p r 2
Magnetic Forces in Moving Reference Frames
1 e2 æ v 2 ö
20 ns F = 4pe r 2 çç1 - c 2 ÷÷
0
è
ø
2
1
e
15 ns F =
4pe 0 r 2
Who will see protons hit
floor and ceiling first?
+e
1
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.
Chapter 22
Patterns of Fields in
Space
• Electric flux
• Gauss’s law
• Ampere’s law
• Maxwell equations
Patterns of Fields in Space
What is in the box?
no charges?
vertical charged plate?
Gauss’s law:
If we know the field distribution on closed surface
we can tell what is inside.
Electric Flux: Surface Area
flux through small area:

flux ~ E  nˆA
Definition of electric flux on a surface:

 E  nˆA
surface
Adding up the Flux

 E  nˆA
surface

 E  nˆdA
 
 E  dA

dA
 
electric flux on a closed surface   E  dA
Gauss’s Law

 E  nˆA 
surface

 E  nˆdA 
q
inside
e0
q
inside
e0
Features:
1. Proportionality constant
2. Size and shape independence
3. Independence on number of charges inside
4. Charges outside contribute zero
1. Gauss’s Law: Proportionality Constant

 E  nˆA 
q
surface
inside
e0
1
Q
rˆ  nˆA

2
surface 4e0 r
1
Q
4e0 r 2
 A
surface
1
Q
Q
2
4r 
2
4e0 r
e0
What if charge is negative?
Works at least for one charge and
spherical surface
1
Q
E
4e0 r 2
2. Gauss’s Law: The Size of the Surface

 E  nˆA 
surface
q
inside
e0
1
E~ 2
r
A ~ r2
1
E~ 2
r
universe would be
much different if
exponent was not exactly 2!
1
Q
E
4e0 r 2
3. Gauss’s Law: The Shape of the Surface

 E  nˆA 
surface
q
inside
e0
 E  n̂A   EA

surface
surface
A2   R2   (r2 tan  )2  r22
All elements of the outer surface
can be projected onto corresponding
areas on the inner sphere with the
same flux
A2  / A1  r22 / r12
E2 A2  / E1A1  1
4. Gauss’s Law: Outside Charges

 E  nˆA 
surface

 E  nˆA 
surface
A ~ r 2
1
E~ 2
r
q
inside
e0
 EA

surface
A1 E1  – A2 E2
Outside charges contribute 0 to total flux
5. Gauss’s Law: Superposition

Q1
ˆ
 E1  nA 
surface
e0

Q2
 E2  nˆA 
surface
e0

 E3  nˆA  0
surface

 E  nˆA 
surface
q
inside
e0
Gauss’s Law

 E  nˆA 
q
inside
e0
surface

 E  nˆdA 
q
inside
e0
Is it a law or a theorem?
Can derive one from another
1
Q
E
4e0 r 2
Gauss’s law is more universal:
works at relativistic speeds
Clicker Question
What is the net electric flux on the box?
A. 0 V*m
B. 0.36 V*m
C. 0.84 V*m
D. 8.04 V*m
E. 8.52 V*m
Applications of Gauss’s Law

 E  nˆA 
surface
q
inside
e0
1. Knowing E can conclude what is inside
2. Knowing charges inside can conclude what is E
The Electric Field of a Large Plate
Symmetry:
Field must be perpendicular to surface
Eleft=Eright

 E  nˆA 
q
inside
e0
surface
2EAbox
Q / AAbox


e0
Q / A

E
2e 0
The Electric Field of a Uniform Spherical Shell of
Charge
Symmetry:
1. Field should be radial
2. The same at every location
on spherical surface

 E  nˆA 
q
inside
e0
surface
A. Outer sphere:
Q
2
E 4r 
e0
B. Inner sphere:
0
2
E 4r 
e0
1
Q
E
4e0 r 2
E 0
The Electric Field of a Uniform Cube

 E  nˆA 
surface
q
inside
e0
Is Gauss’s law still valid?
Can we find E using Gauss’s law?
Clicker Question
What is the electric flux through the area A?
cos(30o)  0.866
sin(30o)  0.5
E = 100 V/m
q  30o
A = 2 m2
A)
B)
C)
D)
E)
0
100 V*m
173 V*m
50 V*m
87 V*m
Gauss’s Law: Properties of Metal
Can we have excess charge inside a metal that is in
static equilibrium?
Proof by contradiction:

 E  nˆA 
q
surface
=0
q
inside
e0
0
inside
e0
Gauss’s Law: Hole in a Metal

 E  nˆA 
surface
=0
q
inside
e0
q
inside
e0
0
What is electric field inside the hole?
VACB =
0
VADB
 
   E  dl  0
1. No charges on the surface of an
empty hole
2. E is zero inside a hole
Gauss’s Law: Screening

 E  nˆA 
surface
q
inside
e0
Gauss’s Law: Charges Inside a Hole

 E  nˆA 
q
surface
=0
inside
e0
q
inside
e0
+5nC
q
+ qinside 0
q
 5 nC
surface
surface
0
Gauss’s Law: Circuits
Can we have excess charge inside in steady state?

 E  nˆA 
surface

 E  nˆA  
left _ surface

 E  nˆA
right_ surface
q
inside
e0
q
inside
e0
0
Gauss’s Law: Junction Between two Wires
n2<n1
u2<u1
i1=i2
n1Au1E1 = n2Au2E2
n1u1
E2 
E1  E1
n2u2

 E  nˆA 
surface
q
inside
q
inside
e0
 e 0 (E1 A  E2 A)  0
There is negative charge along the interface!
Patterns of Magnetic Field in Space
Is there current passing
through these regions?
There must be a relationship between the
measurements of the magnetic field along a closed
path and current flowing through the enclosed area.
Ampere’s law
Quantifying the Magnetic Field Pattern
m0 2 I
Bwire 
4 r
 
Curly character – introduce:  B  dl
  m0 2 I
m0 2 I
 B  dl  4 r  dl  4 r 2r
 
 B  dl  m0 I
Similar to Gauss’s law (Q/e0)
Will it work for any circular path of radius r ?
A Noncircular Path (home study)
 
 B  dl  m0 I
 
 
Need to compare B1  dl1 and B2  dl2
 
m0 2 I
B1 
B  dl  Bdl||
4 r1
dl1 dl2||

r1
r2
m0 2 I r1
B2 
 B1
4 r2 r2
 
 r1  r2 
B2  dl2  B2dl2||   B1  dl1 
 r2  r1 
   
B2  dl2  B1  dl1
Currents Outside the Path (home study)
 
 B  dl  m0 I
 
 
Need to compare B1  dl1 and B2  dl2
dl1 dl2||

r1
r2
r1
B2  B1
r2
 
 
B2  dl2   B1dl1
 
 B  dl  0
for currents outside the path
Three Current-Carrying Wires (home study)
 
 B1  dl  m0 I1
 
 B2  dl  m0 I 2
 
 B3  dl  0
r
r
r
—
 B1 + B2 + B3  dl  m0 I1  I2 


Ampere’s law
 
 B  dl  m0  Iinside_ path
Ampère’s Law
 
 B  dl  m0  Iinside_ path
All the currents in the universe contribute to B
but only ones inside the path result in nonzero path integral
Ampere’s law is almost equivalent to the Biot-Savart law:
but Ampere’s law is relativistically correct
Ampere, 1826: Memoir on the Mathematical Theory of Electrodynamic
Phenomena, Uniquely Deduced from Experience
Maxwell: We can scarcely believe that Ampère really discovered the law of action by means of the experiments
which he describes. We are led to suspect, what, indeed, he tells us himself, that he discovered the law by some
process which he has not shown us, and that when he had afterwards built up a perfect demonstration he
removed all traces of the scaffolding by which he had raised it.