Transcript Physics 122B Electromagnetism - Institute for Nuclear Theory

Physics 122B
Electricity and Magnetism
Lecture 25 (Knight: 34.1 to 34.5)
Maxwell’s Equations
Martin Savage
Lecture 25 Announcements
 Midterm 3 is graded and can be picked up at
the end of lecture
The Final Exam is Tuesday June 5 at 2.30 –
4.20 pm
7/20/2015
Physics 122B - Lecture 25
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About the Final Examination
 Final is at 2:30 pM on Tuesday (June 5)
 Total point value of Final = 150 points
 Tutorial multiple-choice question (20 pts)
 Lecture multiple-choice questions (60 pts)
 Lecture long answer questions (25 pts)
 Lab multiple-choice question (20 pts)
 Tutorial long-answer question (25 pts)
You may bring three 8½” x 11” pages on which you may write
anything on both sides. Also be sure to bring a Scantron
sheet (pre-filled-out as much as possible) and a calculator
with a good battery.
 There will be assigned seating. Look up your seat assignment
on Tycho before coming to the Final.

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The Series RLC Circuit
The figure shows a resistor, inductor,
and capacitor connected in series. The
same current i passes through all of the
elements in the loop. From Kirchhoff’s
loop law, E = vR + vL + vC.
Because of the capacitive and inductive
elements in the circuit, the current i will
not in general be in phase with E, so we will
have i = I cos(wt-f) where f is the phase
angle between current i and drive voltage
E. If vL>vC then the current i lags E and
f>0. If vC>vL then i leads E and f<0.
X L  vL / I  wL
X C  vC / I  1/ wC
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E02  vR2  (vL  vC ) 2   R 2  ( X L  X C )2  I 2
I
E0
R  ( X L  XC )
2
2
Physics 122B - Lecture 25

E0
R2  (w L  1/ wC )2
4
Analyzing an LRC Circuit
Draw the current
vector I at some
arbitrary angle.
All elements of
the circuit will
have this current.
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Draw the resistor
voltage VR in phase
with the current.
Draw the inductor
and capacitor
voltages VL and VC
900 before and
behind the current,
respectively.
Draw the emf E0
as the vector sum
of VR and VL-VC.
The angle of this
phasor is wt,
where the timedependent emf is
E0 cos wt.
Physics 122B - Lecture 25
The phasors VR
and VL-VC form
the sides of a
right triangle,
with E0 as the
hypotenuse.
Therefore, E02
= VR2+(VL-VC)2.
5
Impedance and Phase Angle
We can define the impedance
Z of the circuit as:
Z  R2  ( X L  X C )2
 R2  (w L  1/ wC )2
Then I  E / Z
From the phasor diagram ,we
see that the phase angle f of the
current is given by:
VL  VC I  X L  X C 
tan f 

VR
IR
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 X L  XC
R

f  tan 1 
Physics 122B - Lecture 25

1  w L  1/ wC 

tan



R



VR  E0 cos f
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Resonance
I
C
w0 
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1
LC
E0
E

Z
R2  (w L  1/ wC )2
The current I will be a
maximum when wL=1/wC. This
defines the resonant frequency
of the system w0:
I
E0
R   Lw  1  w0 w  


2
2
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2
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Example:
Designing a Radio Receiver
An AM radio antenna picks up a 1000 kHz signal with a peak
voltage of 5.0 mV. The tuning circuit consists of a 60 mH
inductor in series with a variable capacitor. The inductor coil
has a resistance of 0.25 W, and the resistance of the rest of
the circuit is negligible.
(a) To what capacitance should the capacitor be tuned to listen to
this radio station.
(b) What is the peak current through the circuit at resonance?
(c) A stronger station at 1050 kHz produces a 10 mV antenna
signal. What is the current in the radio at this frequency when
the station is tuned to 1000 kHz.
X L  X C so Z  R
w0  1/ LC  1000 kHz = 1 MHz
C
1
1

Lw02 (60 10-6 H)(6.28 106 rad/s)2
 4.23 10
-10
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F  423 pF
I1  E0 / R  (5.0 103 V) /(0.25 W)  0.020 A  20 mA
X L '  w ' L  396 W X C '  1/ w ' C  358 W
I2 
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E0 '
R  ( X L ' X C ')
2
2
 0.26 mA
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Electromagnetic Fields and
Forces
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E
 1 q

q
ˆ
r

,
away
from
q


2
4 0 r 2
4

r
0


B
m0 qv  rˆ  m0 qv sin 


,

to
v
and
r
by
RHR


2
4 r 2
 4 r

1
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Field Lines
Field lines start and stop
on charges (if any).
Q
-Q
Field lines never cross.
Field line spacing indicates
field strength.
weak
strong
Field lines form closed loops
only when there is a current
or a flux change in the other
field (i.e., energy flow).
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Gauss’s Law Revisited
e 

 E  dA 
Qin
0
m 
 B  dA  0
(magnetic monopoles go here)
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The Lorentz Force
FE  qE
FB  qv  B
Coulomb’s electric force law
Magnetic force on a moving charge
F  q(E  v  B)
Lorentz Force Law
The most general statement of electromagnetic forces on a charge.
E and B may be frame-dependent (see the later part of this lecture),
but the Lorentz Force does not change with frame.
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Example:
The Motion of a Proton
A proton is launched with velocity
v0^
j into a region of space where an
electric field E0^i and a magnetic field
B0^i are parallel.
How many cyclotron orbits will the
proton make while traveling a distance
L along the x axis? Find an algebraic
expression and evaluate your answer
for E0 = 10 kV/m, B0 = 0.1 T, v0 =
1.0x10^5 m/s, and L = 10 cm.
eE0
ax 
mp
f cyc
eB0

2 m p
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eE0t 2
L  ax t 
2mp
1
2
N orbits
2
eB0
 f cyct 
2 m p
t
2m p L
2m p L
eE0
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eE0

B0
2
2eL
 15.6
m p E0
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Question
In what direction is the net force on the moving charge?
(a) Left;
(b) Right;
(d) Up and left at 450;
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(c) Into page;
(e) Down and left at 450
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The Amperian Surface

r r
B.dl = m0 I
Ampere’s Law
Question: What restricts the shape
and extent of the surface bounded
by the integration path?
Answer: The shape of the
surface does not matter.
Any surface should be valid.
If the surface intersects no
current, the line integral is
zero. Otherwise, it has a
non-zero value.
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
r r
B.dl = m0 I
Something is
Missing !!!!
Maxwell’s Paradox: Consider a capacitor that is
being charged by a battery, with a current flow
to the positive plate and from the negative plate.
If the Ampere’s Law surface goes through the
wire, a current passes through it.
If the Ampere’s Law surface goes through the
capacitor gap, no current passes through it.
Thus there is a paradox. The line integral of
Ampere’s Law appears to depend on which
surface is used, bringing its validity into question.
Maxwell’s Solution: Add a “displacement current”
term that depends on the changing electric field in
the gap.
r r
 B.dl = m0 (Ithro + Idisp)
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Displacement Current
Q
Q
e  EA 
A
0 A
0
d  e 1 dQ I


dt
 0 dt  0
I disp   0
d e
dt

d e 

 B  ds  m0  I through  Idisp   m0  I through   0 dt 

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d e
B  ds  m0 I through  m0 0
dt
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Induced Magnetic Field
Thus, the situation is
symmetric: a changing
magnetic field induces an
electric field, and a changing
electric field induces a
magnetic field.
In both cases, the induced
field lines are in closed loops,
and represent potential sources
of energy.
Note, however, that there
is a sign difference. The loops
are in opposite directions.
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Example:
Fields in a Charging Capacitor
A 2.0 cm diameter parallel plate
capacitor with a 1.0 mm gap is being
charged at the rate of 0.50 C/s.
What is the magnetic field strength
in the gap at a radius of 0.5 cm?
2
Q
r Q
2
2
 e  EA   r E   r
 
2
 0 R  R   0

d e
r  d Q
r
B  ds  m0 0
 m 0 0  

m
 
0
 I
dt
 R  dt   0 
R
2
2
m0 r
r
6
B
m0   I 
I

5.0

10
T
2
2 r  R 
2 R
1
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A Prelude to
Maxwell’s Equations
Suppose you come across a vector
field that looks something like this.
What are the identifiable
structures in this field?
1. An “outflow” structure:
2. An “inflow” structure:
3. An “clockwise circulation”
structure:
4. An “counterclockwise
circulation” structure:
Maxwell’s Equations will tell us that the “flow” structures are charges
(+ and -) and the “circulation” structures are energy flows in the field.
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Maxwell’s Equations

E  dA  Qin /  0
Gauss’s Law
Gauss’s Law for magnetism
dm
 E  ds   dt

Faraday’s Law
(magnetic monopole
current goes here)
 B  dA  0
(magnetic monopole
charge goes here)

d e
B  ds  m0 I through  m0 0
dt
Ampère-Maxwell Law
F  q(E  v  B)
Lorentz Force Law
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A Prelude to Waves
Maxwell’s formulation of electricity and
magnetism has an interesting consequence.
The equations can be manipulated to give a
wave equations for E and B of the form:
d 2E
d 2E
 m0 0 2
2
dx
dt
This can be recognized as describing
an electromagnetic wave traveling
through space with a velocity of:
vEM wave 
1
m 0 0
(4  9.0 109 Nm 2 /C 2 )

(4 107 N/A 2 )
 3.0 108 m/s
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E or B? It’s frame dependent.
Sharon runs past Bill carrying a
positive charge. From Bill’s
perspective the charge is moving,
but from Sharon’s perspective the
charge is at rest.
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Now turn on a magnetic field
into the diagram. From Bill’s
perspective the charge
experiences a upward vxB force.
But from Sharon’s perspective,
the charge is not moving and
should experience no magnetic
force. Do we have a paradox?
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Galilean Relativity
Consider a reference frame S that
is at rest, and another reference
frame S’ that is moving at a constant
velocity V with respect to S.
v '  v  V or v  v ' V
dv ' dv dV dv



dt
dt dt
dt
a' a
ma '  ma
F' F
Therefore, a force F as observed
in S must have the same magnitude
and direction when observed in S’.
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Transformation of E and B
Consequently, in the reference frames of
Bill and Sharon, it wasn’t the force that
changes with the motion.
Therefore, it must have been the fields.
In Sharon’s frame, if there was no magnetic
force, there must have been an electric force.
In other words, in her moving frame there
must have been an induced electric field that
produced a force in the upward direction.
F  qV  B
F '  qE '
E' V B
More generally, if an electric field E is
present in S, then in S’:
E '  E V  B
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Example:
Transforming the Electric Field
In a laboratory at rest there are fields
of E = 10 kV/m and B = 0.10 T , both in
the +x direction in the laboratory frame.
What is the electric field in a
reference frame moving with velocity V =
1.0x105 m/s in the +y direction.
E '  E  V  B  (10iˆ 10kˆ) kV/m
E '  14.1 kV/m, direction 45 below the x axis.
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Producing B from Moving E
Now consider Sharon and Bill again. Now the
charge is at rest in Bill’s reference frame. From
Bill’s perspective B=0, and there is only an electric
1 q
field E:
E
rˆ
4 0 r 2
From Sharon’s perspective there is the same
electric field E’, since q and r are the same as in
1 q
Bill’s frame:
E' E 
rˆ
4 0 r 2
However, Sharon also sees a magnetic field
B’ produced by the charge moving at -V:
B'  
 1 q 
m0 q
ˆ
V

r



m
V

rˆ   0 m0V  E

0 0
2
2 
4 r
 4 0 r 


B '  B   0 m0 V  E  B 
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
1
V E
2
c

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Example:
Two Views of a Magnetic Field
A 1.0 T magnetic field points upward. A
rocket flies by the laboratory, parallel to the
ground, with a velocity of 1000 m/s.
What are the fields between the magnet’s
pole tips, as viewed from a scientist aboard
the rocket?
E '  E V  B  V  B
1
B '  B  2 V  E   B
c
E '  BVkˆ  (1.0 T)(1000 m/s)kˆ  1000kˆ V/m
. ˆj T
B '  B  10
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Question
Reference frame S observes
E and B fields as shown.
Which diagram shows the
fields in reference frame S’?
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Lecture 25 Announcements
 Midterm 3 is graded and can be picked up at
the end of lecture
The Final Exam is Tuesday June 5 at 2.30 –
4.20 pm
7/20/2015
Physics 122B - Lecture 25
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