Lecture 25 (Knight: 34.1 to 34.5)

 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
2 4/14/2020 Physics 122B - Lecture 25

 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.
4/14/2020 Physics 122B - Lecture 25 3

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 = v
R
+ v
L
+ v
C
.
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( w tf ) where f is the phase angle between current i and drive voltage
E . If v
L
>v f
>0
. If v
C
C
>v then the current i lags E and
L then i leads E and f
<0.
X
L
 v
L
/ I
 w
L
E
0
2  v
2
( v v
R

L

C
)
2 

2 
(

)
2
R X X I
L C
2
X
C
 v
C
/ I

1/ w
C
I

E
0
R
2 
( X
L

X
C
)
2

R
2
E
0

( w
L

1/ w
C )
2
4/14/2020 Physics 122B - Lecture 25 4

Draw the current vector
I at some arbitrary angle.
All elements of the circuit will have this current.
Draw the resistor voltage
V
R in phase with the current.
Draw the inductor and capacitor voltages
V
L and
V
C
90 0 before and behind the current, respectively.
Draw the emf E
0 as the vector sum of
V
R and
V
L
The angle of this phasor is w
-V
C
t, where the time-
. dependent emf is
E
0 cos w t.
4/14/2020 Physics 122B - Lecture 25
The phasors
V
R and
V
L
-
V
C form the sides of a right triangle, with
=
V
E
0 as the hypotenuse.
Therefore,
R
2 +
(V
L
E
-
V
0
2
C
)
2 .
5

We can define the impedance
Z of the circuit as:
Z

R
2 
( X
L

X
C
)
2

R
2 
( w
L

1/ w
C )
2
Then I
 E / Z
From the phasor diagram ,we see that the phase angle f of the current is given by: tan f 
V
L

V
C
V
R


L

X
C

IR f  tan

1
X
L

X
C
R
 tan

1 w
L

1/ w
C
R
V
R
 E
0 cos f
4/14/2020 Physics 122B - Lecture 25 6

C w 
0
1
LC
I

E

Z
R
2
E
0

( w
L

1/ w
C )
2
The current
I will be a maximum when w L=1/ w C. This defines the resonant frequency of the system w
0
:
I

E
0
R
2   
2
 1
  w w
0

2
2
4/14/2020 Physics 122B - Lecture 25 7

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 m H 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 w
0

1/ LC

1000 kHz = 1 MHz I
1
 E
0
/ R
  
3 W 
0.020 A

20 mA
C

1
L w
0
2
4/14/2020

1
 -6  6
(60 10 H)(6.28 10 rad/s)
2
  -10
F

423 pF
X
L
'
 w
'
I
2

L

396
W
X
C
 w
' C

358
E
0
'
R
2 
( X
L
'

X
C
')
2

0.26 mA
W
Physics 122B - Lecture 25 8

4/14/2020
E

1
4

0 q r 2 r
ˆ

1
4

0 q r 2
, away from q


B
 m
0
4
 qv r
ˆ
2


 r m
0
4
 qv sin
 r
2
,

to and by RHR


Physics 122B - Lecture 25 9

Field lines start and stop on charges (if any).
Q -Q
Field lines never cross.
Field line spacing indicates field strength.
strong weak
Field lines form closed loops only when there is a current or a flux change in the other field (i.e., energy flow).
4/14/2020 Physics 122B - Lecture 25 10

4/14/2020
  e

  
Q in

0
  m

  
0
(magnetic monopoles go here)
Physics 122B - Lecture 25 11

F
E
 qE
Coulomb’s electric force law
F
B
 
Magnetic force on a moving charge
F

(
 
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.
4/14/2020 Physics 122B - Lecture 25 12

v
^
A proton is launched with velocity
j into a region of space where an electric field E
0
B
0
0
^
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 E
0
= 10 kV/m, B
0
= 0.1 T, v
1.0x10^5 m/s, and L = 10 cm.
0
= a x
 eE
0 m p
L
 1
2 a t x
2  eE t
0
2
2 m p f cyc
 eB
0
2
 m p
4/14/2020 t

2 m L p eE
0
N orbits
 f t cyc
 eB
0
2

2 m L p m eE p 0

B
0
2

Physics 122B - Lecture 25
2 eL m E p 0

15.6
13

In what direction is the net force on the moving charge?
(a) Left; (b) Right; (c) Into page;
(d) Up and left at 45 0 ; (e) Down and left at 45 0
4/14/2020 Physics 122B - Lecture 25 14

 r r
Ampere’s Law m
0
I
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.
4/14/2020 Physics 122B - Lecture 25 15

 r r m
0
I
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 r r
B.dl = m
0
( I thro
+ I disp
)
4/14/2020 Physics 122B - Lecture 25 16

   e
EA
Q

0
A
A

Q

0 d
 e dt

1

0 dQ dt


I
0
I disp
 
0 d
 e dt


B ds
 m
0

I through

I disp

 m
0
I through
 
0 d
 e dt
  m
I
0 through
 m 
0 0 d
 e dt
Physics 122B - Lecture 25 4/14/2020
 
17

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.
4/14/2020 Physics 122B - Lecture 25 18

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?
 
EA
 
2 r E
  r
2
Q
 
0
R
2
  2 r Q

0


B ds
 m 
0 0 d
 e dt
 m 
0 0
 
 
2 d dt



Q
0

  m
0
B

2
1
 m
0 r R
2
I
 m
0
2
 r
R
2
I
 
5.0 10 T
2
I
4/14/2020 Physics 122B - Lecture 25 19

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.
4/14/2020 Physics 122B - Lecture 25 20

    in
/

0
Gauss’s Law
 
B dA

0
(magnetic monopole charge goes here)
Gauss’s Law for magnetism
 
E ds
  d
 m dt
Faraday’s Law
(magnetic monopole current goes here)
 
B ds
 m
I
0 through
 m 
0 0 d
 e dt
Ampère-Maxwell Law
F

(
 
B )
Lorentz Force Law
Physics 122B - Lecture 25 4/14/2020 21

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:
2 d E dx 2
 m 
0 0
2 d E dt 2
This can be recognized as describing an electromagnetic wave traveling through space with a velocity of: v
EM wave

1 m 
0 0

   9 2 2
(4 9.0 10 Nm /C )
  
7 2
(4 10 N/A )
4/14/2020
  8
3.0 10 m/s
Physics 122B - Lecture 25 22

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.
4/14/2020
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?
Physics 122B - Lecture 25 23

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 '
 
or 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’.
4/14/2020 Physics 122B - Lecture 25 24

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
E '
 
B
F '
 qE '
More generally, if an electric field E is present in S, then in S’:
E ' E V B
4/14/2020 Physics 122B - Lecture 25 25

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 =
1.0x10
5 m/s in the +y direction.
E ' E V B (10 i
ˆ  k
ˆ
E '
 
4/14/2020 Physics 122B - Lecture 25 26

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 field E:
E
1 q r
ˆ 
4
 r
2
0
From Sharon’s perspective there is the same electric field E’, since q and r are the same as in
Bill’s frame:
E '
 
1
4
 q
2 r
ˆ
0 r
However, Sharon also sees a magnetic field
B’ produced by the charge moving at -V:
B '
  m
0
4
 q r
2
V
    m
0 0
V



1
4

0 q r
2 r
ˆ


   m
0 0
V

E
B '
   m
0 0
4/14/2020

V

E

B
1 c
2

V

E

Physics 122B - Lecture 25 27

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 '
    
B
B '
 
E '

BVk c
ˆ
1
2


V

E


B
(1.0 T)(1000 m/s) k
ˆ  k
ˆ
B '
  .
j
4/14/2020 Physics 122B - Lecture 25 28

Reference frame S observes
E and B fields as shown.
Which diagram shows the fields in reference frame S’?
4/14/2020 Physics 122B - Lecture 25 29

 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
30 4/14/2020 Physics 122B - Lecture 25