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

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.

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

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!

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

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

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.

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!

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

2222

22,)(

xa

kQdq

xa

kdVV

xa

kdq

r

kdqrdV

ringring

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

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

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

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

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

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.

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

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

Lenz’s lawMagnetic field produced by induced current opposes change of magnetic flux

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

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

+ -

+ -

+ -

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

)(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

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..

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)

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

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

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.

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

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

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

0

0

02

0

2:int

0:int

4:int

),cos1(2)2/(cos4

IIerferenceNo

IerefernceeDestructic

IIereferenceveConstructi

III

t

t

t

t

Intensity distribution

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

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

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)

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.

a

1

The position of the first minimum in diffraction pattern

2/sin)2/( a

Diffraction grating

mdm sin2:max

!02

0 ININEE otot

tE0