Lecture 25. Overview
description
Transcript of Lecture 25. Overview
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Final Exam:
This week off. hours: Today: 1:30-2:30pm; Th: 12:30-1:30pm; F: 2:00pm-4:00 pm
Secs. 501, 503, 504, 526, 527: Dec. 10, M., 10:30am - 12:30pmSecs. 516-520: Dec, 12, W., 10:30am - 12:30pm
Lecture 25. Overview
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Evaluation is helpful!• You should all have received an email with a link to
evaluate your PHYS 208 class. • I encourage you all to fill this out• It gives a feedback to me and TAs• This year it’s especially important in view of of coming revolutionary changes in teaching: 1) big
classes (1 lecturer for all),2) flipped lectures (pre-lectures for learning the concepts and no derivations by lecturer on board), 3)multiple choice exams.
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3 Exams (3E) 300
Final Exam (F) 100
Homework (HW) 50
Laboratory (L) 100
Recitation (R) 100
Class Points (CP) not limited
The Course Grade (CG): CG=(3E+2xF+R+L/2+HW+CP)/7.
A: >90, B: 80-90, C: 70-80, D: 55-70, F<55
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Average E1=61+5 (curve)Average E2=56+5 (curve)Average E3=56+5 (curve)Average A3E=(E1+E2+E3+15)/3=63Average lab grade: (AL)=92Average rec. grade (AR)=82For students with A3E=63, AL=92, AR=82, HW=50Before Final Exam:ACG=(3xA3E+R+HW+L/2)/5=73.
For students with A3E=63, L=100, R=100, HW=50Before Final Exam:ACG=(3xA3E+R+HW+L/2)/5=78.
Check your Midterm Exams Grades on elearning!
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The Advanced Course Grade (ACG): ACG=(2E+3xF+R+L/2+HW+CP)/7.
If FE>min {E1,E2,E3} then min{E1,E2,E3} FE Then the course grade is calculated as:
If before curve of FE FE>90 then CG is A;FE>80 and FE>max{CG,ACG} then CG=BFE>70 and FE>max{CG,ACG} then CG=CFE>55 and FE>max{CG,ACG} then CG=D
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Webct homework (HW) is due by Dec.10.
To get full credit (50p) complete HW 1-12Last HW (N13) is not required!
You get 25p if complete >6HW but <12HWIf you complete <6HW you get no credit
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Final Exam: (Ch.21-24, 27-29, 32,33,35,36)
Two Parts:2 problems (50p) old material : Chs. 21-24, 27-30 2 problems (50p) new material : Chs. 32, 33, 35,36
Students may be excused from the first part if1)Each midterm exam is higher or =85 (with a curve)2)An average of 3 exams is higher or = 90 (with a curve)
In this case the Course Grade: CG= (3E+F+R+L/2+HW+CP)/6.
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Preparation to the final exam
• This lecture• Formula sheet• Old Final Exams • Old 1st and 3d exams• Lectures notes (especially, Examples) • Homework• Textbook• Sleep well!
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Ch.21-23 (Lect.1-6)
QQq EqF
1.Electric field, electric force
2. Potential (voltage), potential energy
3. Superposition principle 4. Gauss’s Law 5. Energy conservation law, work
02r
r
kQE
0encl
surface
QAdE
b
a
ab ldEV
V=kq/r VqrU 0)(
Ka+Ua= Kb+Ub Wab =Ua-Ub
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2222
22,)(
xa
kQdq
xa
kdVV
xa
kdq
r
kdqrdV
ringring
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Example. A uniformly charged thin ring has a radius10 cm and a total charge 12 nC. An electron is placed on the ring axis a distance 25 cm from the center of the ring. The electron is then released from rest. Find the speed of the electron when it reaches the center of the ring.
]11
[2
)0(;)(
)]0()([)0()()0(
)0()0()(
22
2
22
xaakeQ
mv
a
kQxV
xa
kQxV
VxVqUxUK
KUxU
smCkg
mmmCCNm
xaam
keQv
/1054.1109
])25.0()1.0(
11.01
[1012106.11092
]11
[2
72231
22
91929
22
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Calculation of E usingthe Gauss’s law
Three types of charge distribution symmetry
Spherical A=4 π r2, V=4 π r3/3
Cylindrical Aside=2 π rL, V= π r2L
L →∞Plane A=L1L2, V=AL3
L1, L2 →∞
0encl
surface
QAdE
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Example. Very long conducting cylinder, L>>R, r
1. r<R: E(r)=0→V(r)=V(R)
2. r>R:
L
For infinitely long charged objects never choose V(r→∞)=0 !!!Choose V(R)=0.
R
rk
r
Rk
r
drkrV
R
r
ln2ln22)( r
V(r)
L
Q
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Ch.27-28 (Lect.12-15)BvqF
Motion of the charged particles in B; crossed B and E
Force on a segment of a current lBIF
Magnetic force on a moving charge
Amper’s law IldB
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Example1. At t=0 proton with v=(1.5x105 i+2x105 j)m/s enters at the origin of the coordinate system the region with B=0.5iT. Describe and plot the pass.
Find the coordinates of the proton at t=T/2 where T is the period of the circle.
F
y
z
y
z B
Vy
F
R
R
mvBqvmaF y
y
2
BvqF
2/)2/(
,2)2/(,0)2/(
22,
TvTtx
RTtzTty
qB
m
v
RT
qB
mvR
x
y
y
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Motion of q in B┴E.Velocity selector
B
EvqEqvB Independent on the mass and q
Particles with this velocity will be undeflected. Particles with larger velocity will be deflected by B. Particles with smaller velocity will be deflected by E.
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Amper’s law: Conducting cylinder
r
IB
IrB
RrR
IBRr
R
IrB
R
IJrJrB
Rr
2
2
.22
:
2
,2
.1
0
0
0
20
22
0
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Lect.16,17,20 (Ch.29,32)Law of EM Induction (Faraday’s law) dt
d B
Motional emf a
b
ldBv
)(
vBE
EBu
BE
S
TA
PdtUAdSP ,
c
IPrad
c
IPrad
2Absorbing plane
Major characteristics of e.m. waves
Nonabsorbing plane
c
PAPF rad
2cosinout II Malus’s law 2
cos2 ininout
III Random
polarization
KKn mn
cv
n0
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Lenz’s lawMagnetic field produced by induced current opposes change of magnetic flux
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Slide-wire generator
Origin of this emf is in separation of charges in a rod caused by its motion in B due to magnetic force
vBl
lBIFm
'
lBIFF mext
'
R
vBlBlv
RIBlv
dt
dxF
dt
dWP
R
vBl
RRIP
extmech
el
2
222
)(
)(
The secondary magnetic force
BvqF
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Example. Find motional emf in the rod.
IL
d
Example. Find induced current in the loop with resistance R.
I=0I
V
V
d
ldIv
r
drIv
r
IB
ldBv
Ld
d
L
ln22
2
)(
00
0
0
BvqF
+ -
+ -
+ -
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Induced nonelectrostatic electric field when B(t)Find E(r).
dt
dldE B
r
R
dt
dinE
dt
dBRrE
Rr
rdt
dinE
KniBdt
dBrrE
Rr
m
2
2
0
2
2
2
)22
,
2
.1
R
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)(dt
dildB E
General form of Amper’s law (displacement current)
Let’s find B between the plates.
r
iB
irB
RrR
riB
R
i
A
ij
R
rirjrB
Rr
c
d
c
cdd
cd
2
2
.22
2
.1
0
0
20
2
2
2
02
0
B
rR
μ= Kmμ0, ε=Kε0,
In free space K=1, Km =1
X X X X X
X E
B
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E and B in e.m. wave
)cos(
)cos(
0
0
kxtBB
kxtEE
z
y
)cos(
)cos(
0
0
kxtBkB
kxtEjE
or
)cos(
)cos(
0
0
kxtBB
kxtEE
z
y
)cos(
)cos(
0
0
kxtBkB
kxtEjE
or
This is y-polarized wave. The direction of E oscillations determines polarization of the wave..
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Example. A carbon-dioxide laser emits a sinusoidal e.m. wave that travels in vacuum in the negative x direction. The wavelength is 10.6μm and the wave is z-polarized. Maximum magnitude of E is 1.5MW/m. Write vector equations for E and B as functions of time and position. Plot the wave in a figure.
sradmradsmck
mradm
radk
Tsm
mV
c
EB
kxtBjB
kxtEkE
/1078.1/1093.5)/103(
/1093.5106.10
17.322
105/103
/105.1
)cos(
)cos(
1458
56
38
60
0
0
0
NB: Since B=E/c→B (in T) <<E (in V/m)
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Example. In a microwave oven a wavelength 12.2cm (strongly absorbed by a water) is used. What is the minimum size of the oven? What are the other options? Why in the other options rotation is required?
nfL
vnfn
n
L
L
vvfLL
nn min
maxminmax
max
2...2,1,
2
22
2
If two conductors are placed parallel to each other on the distance L the nodes of E should be on the ends (just as on the string with fixed ends)
...3.18
2.12
1.62min
cmL
middletheinnodeonecmL
cmL
Standing Waves
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Example A radio station on the surface of the earth radiates a sinusoidal wave with an avearge total power 50kW. Assuming that
transmitter radiates equally in all directions, find the amplitudes of E and B detected by a satellite at a distance 100km.
27
210
4
21096.7
1028.6
105
2 m
W
mR
PI
Tc
EB
mVcIE
c
EBEI
1100
200
0
20
0
00
102.8
/105.22
22
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Unpolarized em wave (random polarization)
Ein
Eout
cosinout EE
Malus’s law
2cosinout II
In general case when linear plz wave goes through the filter only its projection on the axis of the filter goes through.
2cos2 in
inout
III
NB: After the filter em wave is always linear polarized along the axis of the filter.
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Lect. 23 (Ch.35)
Conditions for constractive and distractive interference (i) for phase difference (ii) for path difference
Double-slit experiment
Interference in the thin films
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Phase difference and Path difference
S1
S2
r1
r20012
0202
101
)(
)cos(
)cos(
rkrrk
krtEE
krtEE
erferencenoEEIIm
erferenceedestructivEIm
erfernceveconstractiEEIIm
tt
tt
tt
int)2(2)12(2
.3
int)0(0)2
1(2.2
int)2(42.1
000
0
000
Taking into account that
In particular case when 00 2
k
erferencenomr
erferenceedestructivmr
erfernceveconstractimr
int)2
1(
2.3
int)2
1(.2
int.1
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Max and min positions (bright and dark stripes)
)2
1(sin)
2
1(2:min
sin2:max
,sin12
mdm
mdm
rkdrrrdRIf
d
mRy
d
mRy
R
ysmallisIf
m
m
)2/1(:min
:max
tansin
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0
0
02
0
2:int
0:int
4:int
),cos1(2)2/(cos4
IIerferenceNo
IerefernceeDestructic
IIereferenceveConstructi
III
t
t
t
t
Intensity distribution
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Example• A radio station operating at a frequency of 1500kHz has two identical vertical dipole antennas spaced
400m apart, oscillating in phase. At distances much greater then 400m, in what directions is the intensity greatest in the resulting radiation pattern? If intensity produced by each antenna 400km away along y axis is 2mW/m2 what is the total intensity produced by two antennas at this point?
20
6
8
/84.2.1sin2||
902,301,00
2400
)200(sinsin:max
200/1105.1
/103.1
mmWIIimpossiblem
mmm
m
m
mm
d
mmd
ms
sm
f
c
y
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filmfilm
filmfilm
film
n
mmtm
mn
mtm
ktk
0
0
2)2
1(2:min.2
)2
1()
2
1(22:max.1
2,2
Normal coincidence:
Interference in the thin films
n<nfilm
film
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ExampleThe walls of a soap bubble have about the same refractive index as a plain water, n=1.33.In the point where the wall is 120nm thick what colors of incoming white light are the most
strongly reflected?
),(2133
640,1
64033.112044,0
)2
1(2:max
0
0
0
nonvisiblenUVradiationmnm
m
nmnmtnm
mn
t
film
film
(orange)
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Interference at the thin wedge of air
NB: Interference between the light reflected from the upper and lower surfaces of the glass plate is neglected due to the larger thickness of the plate.
0
0
2:min
)2
1(2:max
mt
mt
ExampleA monochromatic light with wavelength in the air 500nm is at normal incidence on the top glass plate (see the figure). What is the spacing of interference fringes?
mmm
mm
h
lxx
h
lmx
mth
tlx
h
l
t
x
mmm 25.1)1002.0(2
1.0)10500(
2,
2
2:min,
3
90
10
0
NB: the fringe at the line of contact is dark, because of phase shift produced by reflection from the second plate.
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a
1
The position of the first minimum in diffraction pattern
2/sin)2/( a
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Diffraction grating
mdm sin2:max
!02
0 ININEE otot
tE0