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DOCUMENT RESUME
ED 075 252 24 SE 016 081
TITLE Talking Books (Master Set) .
INSTITUTION Naval Academy, Annapolis, Md.; New York Inst. ofTech., Old Westbury.
SPONS AGENCY Office of Education (DHEW), Washington, D.C. Bureauof Research.
BUREAU NO BR-8-0446PUB LATE 70CONTRACT N00600-68-C-0749NOTE 246p.
EDRS PRICE MF-$0.65 HC-$9.87DESCRIPTORS College Science; *Electricity; Instructional
Materials; *Mechanics (Physics); Physics; ScienceEducation; Supplementary Textbooks; *Talking Books
IDENTIFIERS Self Paced Introduction
ABSTRACTAs one of three audiovisual media in the U. S. Naval
Academy Self-Paced Physics Course, 27 topics relating to mechanics,electricity, and magnetism are presented in this volume for enrichingand supplementary purposes. Each topic is primarily composed ofillustrations and formulas. Terminal behavior objectives anddirections for reaching subsequent study guides are provided at theend of the topic. The material is designed to be used in combinationwith tape recorded lectures. (Related documents are SE 016 065through SE 016 088 and ED 062 123 through ED 062 125.) (CC)
U S Dr:PALITMEIDUCALIONOFFICE OF E
tuts 00CLIMENtOLICI L1 t 'Acrirr AIIIL PERSON ON ONINAIINO 0 POINTSIONS SI Ala, DO
PSESI NI Or I In;OArsiN Potil I ION (1
DEVELOPED AND PRODUCED UNDER THE UNITED STATESNAVY CONTRACT #N00600-68C-0749 AND UNITED STATESOFFICE OF EDUCATION, BUREAU OF RESEARCH, PROJECT#8-0446, FOR THE UNITED STATES NAVAL ACADEMY
NEW YORK INSTITUTE OF TECHNOLOGY, OLD WESTBURY,LONG ISLAND, NEW YORK 11568
TALKING BOOKS
(MASTER SET)
TALKING BOOKS
INDEX
1. PROJECTILE MOTION
2. NEWTON'S FIRST LAW
3. NEWTON'S SECOND LAW
4. NEWTON'S THIRD LAW
5. ATWOOD'S MACHINE
6. CHARACTERISTICS OF CIRCULAR MOTION
7. WORK WHEN FORCE VARIES IN BOTH MAGNITUDE & DIRECTION
8. KINETIC ENERGY
9. 20TENTIAL ENERGY
10. CONSERVATION OF ENERGY
10a. MOVEMENT OF CENTER OF MASS
11. CONSERVATION OF MOMENTUM
12. IMPULSE AND MOMENTUM
13. COLLISIONS
14. GRAVITATION
15. CALCULATION OF E FOR AN INFINITE UNIFORMLY CHARGED. WIRE
16. DEFLECTION OF ELECTRONS IN AN ELECTRIC FIELD
17. FLUX
18. CALCULATION OF t USING GAUSS' LAW
19. CAPACITORS
19a. THE CAPACITOR IN ACTION
20. KIRCHHOFF'S RULES
21. DEFINITION OF "B" FIELD
Continueli
24. FORCE BETWEEN PARALLEL CURRENT-CARRYING CONDUCTORS
23. AMPERE'S LAW APPLIED TO A LONG STRAIGHT CONDUCTOR
25. THE LAW OF BIOT-SAVART
26. FARADAY'S LAW OF INDUCTION
22. MOTION OF. AN ELECTRON IN COMBINED E AND B FIELDS
28. L - R TRANSIENTS
27. R - C TRANSIENTS
iX
PRO IECTILEMOTION
\o
I
Projectilev= Muzzle
Velocity
FIGURE 10Vx
v sin 0
v sin 0 V
FIGUREv cos 0
v = v gtBut At Maximum
Height v = o
vy v sin eso v gt or tg g
FIGURE
time of 2v sinflight
v sin 0
FIGURE
v cos e
t 2 v sin 0flight- g
Range = 2 v2sin cos 0v sin 0
R
FIGURE
2v sin 0g
v cos 0
2v 2 sin 0 cos 0
v2 sin 2 0g
FIGURE
FIGURE
.
I
I
,
1
FIGURE C)
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
mmim"IllIIIIIIIIIIIIIIIIIIIIIII
nommuliIMIMOMMIIIITIENN NM MIME1111111111111111111111111
FIGURE
.1
FIGURE,
5'
.0"
. FIGURE
I
00woe :-:
../
/FIGURE
s'
..0 ..064. *.
A 4IA VN.
vb....v.
///
k
N.111. V.
I"v. v. V.
FIGURE
V.V.
I
I
PROJECTILEMOTION
TERMINAL OBJECTIVES
2/3 B Analyze the trajectory curve of a particle projected
horizontally (no vertical component) from the top of
a structure.
2/3 E Solve position, time velocity, and range problems
involving projectiles with any angle of departure.
Please turn to page 22A of your STUDY GUIDE
to continue with your work.
Newton'sst Lom
p
AIRRESISTANCE
FIGURE
rAirmift'Ur
AIRRESISTANCE
FIGURE
FIGURE C)
NEWTON'S FIRST LAWOF MOTION
A BODY. REMAINS AT REST OR INMOTION WITH UNIFORM VELOCITYUNLESS ACTED UPON BY ANEXTERNAL, UNBALANCED FORCE
.s\
ONCE A BODY HAS BEEN SET IN MOTIONTT ME:NO LONGER NECESSARY TO EXERTA IM EON IT TO KEEP IT MOVING.
FIGURE C)
FIGURE C)
INE_MOTION OF AN OBJECT CANNOTSE 'SPECIFIED UNLESS THIS MOTIONCAN BE REFERRED TO SOME OTHER
BODY.
FIGURE C)
ANCIF IE TEAT WHICH CHANGES THE°' '4,' itiE.MOTION OF A BODY.
FIGURE C)
0,
Newton's1 st Law
, *r.,:>70'';?4
-
. TERMINAL OBJECTIVES
3/2 A Analyze and interpret a vammr.mtr.of natural phenomena
relevant to Newton's Firs: of Motion in terms of the
First Law.
Please turn now to pagpLISA of your STUDY GUIDE
to continue with your7mork.
E4 t!C
'OM
,1 $.74 ',:! -,,z.. ar i.- i, 1' :11:-:":P. $4. , .':.;..ar ;.. 1 ..-mok9a ..Tgi .'1513;:f;i7 -,!.cf,--tr.; t......
Or`
L 1 N j At_ r1 [ A\,'A
r
FIGURE
FIGURE
FIGURE
a,0-1
FIGURE
r
I
a -1.5 in sec
F=0.15ntal kg
FIGURE
S..'
Newton's2nd Law
ow
TERMINAL OBJECTIVES
3/2 B Analyze and interpret a variety of natural
phenomena relevant to Newton's Second Law
in terms of the Second Law.
Please turn to page 23A of your STUDY GUIDE
to continue with your work.
Newton's3rd Law
IF tODY A EXERTS A FORCEON PODY 8, THEN BODY BEXiMITS A FORCE OF EQUALNMOINITUDE QPPOSITELYD I MKT ED, ON BODY A
1000"-
FIGURE
IVY WT WS 1RD LAWOF MOTION
F 11;BRE
I
FIGURE
f LGURE
FIGu
0
N tows3 rd Law
P.
INAL CEBJECLIVES
C Analyze aiad inasarpmett a vamiety of natural
phenomena relamant to Newton s Third di Law
of Motion in to (of the Third Law.
:'Please, -turn to page 3 ,of your STUDY ,GNIJDE
to orrartinue with your waork.
I
-of
D'S
111NE
ATWOOD'SORIGINAL MACHINE
FIGURE
1: URE
TENSION
WEIGHT
FIGURE
m1
mig
mig --T n
FIGURE
,
;Pi-, g, = ;Pi,
nil M2T2 *igmi÷m2
FIGURE
mi m2T 2 *gini÷ M2
1000x10002
1000 +1000
= 1000F IGURE
ml m2mi+ m2 fg
1400 x 10002g1400 +1000
1170
FIGURE
2
,40.111111
ATWOOD'SMACHINE
TERMINAL OBJECTIVES
3/3 D Apply the "free body" approach to
problem solutions.
Please turn to page 13A of your STUDY GUIDE
to continue with your work.
. Yr n - :. 1., s.,.., .+'n.
- -
. . rernmcwosimmIsarrpruvresori.
FIGURE
CENTRIPETAL & aNTILIFUGAL FORCESIN C1 RC f4 R MOTION
FIGURE
Centripetal Acceleration
Substituted into the Equation of M-atiox
F =nw
Yields an Equation for Gocular Motion
Fc
FIGURE
;
I
FIGURE
r
FIGUR7 .
FT GIME ,r;
1
Li
FIGURE t
F ICUPE
s-
T (-. ,4,
F I G Ull E''.; u
)
FIGURE
1
.1'
TERMINAL OBJNCTIVES
5/1 B Gnlculate work associated with veriable
forces.
Please turn to page 26A of your STUDY GUIDE
to continue with your work.
KINETICENERGY
FIGURE
WORK DONE ON BODY
= 1-0 ds
F-*
resultant forceon body
FIGURE
I
KINETICENERGY
TERMINAL OBJECTIVES
5/2 D Answer qualitative questions about Kinetic
Energy.
Please turn to page 38A of your STUDY GUIDE
to continue with your work.
POTENTIALENERGY
FIGURE
POTENTIAL ENERGY
Energy ofPosition
or state.
F I CURE
rn
F IGURE
F
kV
F d.
x2f F dx
X2
kx dxx,
FIGURE
FIGURE
x2
F dxx,
FIGURE
X2
F dxx,
X2
kx dxxi
k,. 2
2 6'2
FIGURE
1
2 x,2
I pRin+:52
FIGURE
CHANGE INPOTENTIAL ENERGY
f gecti'
f112G Mdh
h, (R4-h)2
FIGURE
(9)
Tmass mmg
FIGURE
h
FIGURE
Work Done
mghChange inPotentialEnergy
FIGURE
POTENTIAL
TERMINAL OBJECTIVES
5/3 A Use. the concept of potential energy ±or
objects near the surface of the Earth and
for springs.
Please turn to page 33A of your STUDY GUIDE
to continue with your work.
/I/
CON VATION.OF
EN Y
F dfdv 4 4.m dt
mv
FIGURE
WORK DONECHANGE IN
KINETIC ENERGY(free body)
FIGURE
0 APPLIED FORCE
Illimm" RESTOR/NG FORCEOF SPRING
FIGURE
WORK DONE INCOMPRESSING
SPRING
pdx fxkxdx0
1 kx22
FIGURE
POTENTIALENERGY
21 kx2
FIGURE
Fa
FIGURE
Fa
FIGURE
1
RESU LTANT Fa -kxFIGURE
Fa 14.1cr
Free Body DiagramFIGURE
WORK DONE,f) F dx
RE-7/kx dx
0F IGURE
WORK DON EfF dx
A RE. =f kx dx0
K.E. (Fa kx) dxFIGURE
EXTERNAL WORK DONE
HANGE in PE.
CHANGE IN K.E.
TOTAL ENERGY
F IGURE
NO EXTERNAL WORK DONE
TOTAL ENERGYFIGURE
Negative
CHANGE IN RE.
FIGURE
PositiveCHANGE IN K.E.
CONS NATIONOF
ENERGY
TERMINAL OBJECTIVES
5/2 C Answer questions pertaining to the statement
of conservation of energy.
5/3 B Apply conservation of energy to a simple pendulum.
5/3 C Demonstrate a knowledge of specifies required for
the application of the Conservation of Energy
theorem.
Please turn to page 34A of yOur STUDY GUIDE-
to continue with your work.
CENTER OF MASS
(a) for a solid ball
(b) for a hollow ball
EQUAL MASS CARS
CENTER OF MASS0
' ;VV.) ..X:ttN; .11,411,..S.T-ruCr=tVA.'t; ,41 ;7", C.717.a..511..113-° Jr".... tr. tc
CENTER Of MASS00
th4tn.r 11.-N104 - ..tar.....:AM?"M7W47....,, .1 .a.5AfAkn 7'..:31.*.77.2.1.W:.s. 14 Of1io7-.---01.x. X
CENTER OF MASS4.30
4,..-et.riv.;177.1,71.C579.77;:r..,IM pi ;vLInwor.me
X
.1
FIGURE
UNEQUAL MASS CARS
CENTEROF
MASS
74\27. N Lsnef.7.T.TC'egrif: t/-774:75.' i'..74a5:0!...^..-uZe A .Gr
CENTEROF MASS
0
rx 71 .0 . 1 AiI"' A.':
2x
. .
-CENTER OF MASS.
0
".-40 hn: "t?..4.7c- wytf2. c476-7- .;77:177.
2x' x'
, 1"t \I 1
/ \/.>
FIGURE
//i/ ,/ " 7 ".\Ir /// / .'/ // r/ ////7/ /,
,.- .......- ..- ..................-.....-...-..
......... .e. 1 1 I...... r
....I ii
...... ro
... t HI 0 i
CONSERVATION
OF
MOMENTUM
mA
A
VA
mB
L.B
VB
FIGURE 0
mA mA
A
40EXIIIMIZEMS*
VA
mB
B
IIMMir!1111010v
B
F I GURE
A B
mA
A
mB
B
FIGURE
mA
A
mB
B
VB
B
B
FBA
Impulse
f dt=Area
FIGURE ()
t
1GURE
Impulse Impulseapplied to applied to B
FIGURE
&Ake - Impulseapplied to A"-- l applied to B
momentum change of Amomentum change of B
FIGURE
Impulse Impulseapplied to A----"applied to B
momentum change of Amomentum change of B
No change in momentum
FIGURE
mA
A
-7A
mB
IIMIsmimsr.r.=111111.
vA
mB
vB
A B
mA
_11111110.
vA
mB
g=m1111111.
vB.
F IGURE
mA
A
all110.vA
m B
vB
vA mB v ;;B A' 4- 61B .1-m 13
MOMENTUM MOMENTUMBEFORE A FTER
COLLISION COLLISION
CONSERVATION
OF
MOMENTUM
TERMINAL OBJECTIVES
6/2 B Solve momentum problem$ involving bodies with
variable macs.
6/2 C Analyze situations and phenomena in which momentum
is a significant factor.
Please turn to Page 21A of your STUDY GUIDE
to continue with your work.
IMPULSE
ANDMOMENTUM
KE
X
FIGURE
FIGURE 1CD
F madi5
dtF IGURE
in
dt
dij m di3
FIGURE C4 )
F ma
.11
dt
Fdt m dv
F dt m dv
(vector equation)
in di)
Fdt = mdv
ftFdt dc,t,2
V2
v,
I GURE
Fdtrt2
F dtt2
F dt
ft,
FIGURE
Fdtt2
F dt2t
F dt
Impulse
FIGURE
m df,
m d,
m dv
m dfV/
M i32 f,
change inmomentum
V 30 miihr44 ft/sec
FIGURE
.91644°11312LIV°410041.Pg
FIGURE
Impulse
Mass mwalmilat
11111MME
v2
F IGURE
change inmomentum
5 slug32
44 .ft /sec
0 fitilsec
change in x 44M 0 inentum 32
FIGURE
change in 5 x 44momentum
impulse = Ft
1
100FIGURE
slug ftu sec
slug ftu sec
averageforce
F x 1100 -N4f4t
(100 x 352 x 44; lhs
687 Ms or 1 ton iAPPROX3
FIGURE
IMPULSE
ANDMOMENTUM
TERMINAL OBJECTIVES
6/2 A Solve momentum problems involving bodies with
constant, mass .
6/3 A Analyze situations which involve net impulsive
fortes acting on bodies of constant mass.
Please turn to page 31A of your STUDY GUIDE
to continue with your work.
COLLISION-S
FIGURE
m = m = mass of each glider
ul = velocity of glider 1 BEFORE collision
u2 = velocity of glider 2 BEFORE colliion = 0
vl = velocity of glider 1 AFTER collision
v2 = velocity of glider 2 AFTER collision
FIGURE
Before collision the system momentum = mu1
After the collision, the system momentum = mv1+ mv
2
FIGURE
mu1
= mv1+ mv
2and dividing by m
u1
=, v1+ v
2
FIGURE
1 2 1 2 1 2
mul2 "1 2
my2
2 2 2u1= v
1+ v
2
FIGURE C)
ul vl v2
2 2u1
= v1
v2
2
FIGURE C)
Subtracting
u12 2 2= vi + 211,112 + v2
2=
2 2u1
vl v2
(linear equation squared)
0 = 2v1v2
so v1
= 0 or v2
= 0 or both v1
and v2
= 0
FIGURE
1F v1 = 01 then
u1
= 0 + v2
u1
= v2
FIGURE
1F v2 = 0, then
°
ul
FIGURE
COLLISIONS
TERMINAL OBJECTIVES
7/1 A Analyze a two-body collision problem in terms
of the impulse mentum theorem.
7/1 C Apply the principle of conservation of momentum
to the solution of problems involving inelastic
collision.
cr'
"Every object in the universeattracts every other object witha force that is directly propor-tional to the product of theirmasses and inversely proportionalto the square of the distancebetween their centers"
F = G M1 M2
FIGURE
r2
Ca) F ck m^
(b) F = kma
(c) F = ma since k==1 if m is
in kilograms, a is
in m/sec and F is
in newtons.
FIGURE
G= m m1 2
Fr2
FIGURE
IFIRST
POSITION
QUARTZTHREAD
MIRROR
SECONDPOSITION
/ - Jm,
FIRSTm POSITION
S E COMBPOSITrellit
\ / /
LIGHT:SOURCE
FIGURE
SCALE
where :
k e r2
m1 m2 L
= torsional constant ofsuspension thread
e = angle of twist
r = d i stance from centerof mi to center of m2
L = length of hor izontal bar
G
FIGURE
!K.:9 r
mi m2 L
kg m2radians m
sect
kg . kg . m
nt 1112
kg2
FIGURE
2
VII TION
TERMINAL OBJTELIIVES
8/1 A Analyze gravitational force actionshetween,two particles in terms ofthe gravitational field.
; (., . 4^,
F I GURE
1 X dyrA
4reo r2
FIGURE
FIGURE
_These fit components areequal and 'opposite
.% They will cancel
-.These_ _L components:are inn the same
:direction, Theywill_ add.
(2) (11E
-.111MMI-^almom
1 Xdy A
4ire0 142 r
1 Adyos
dy
FIGURE:
acos 0
r der dt9cos 0
1 &cos-0- -F-(10dEx 4 ire° r-2-
1 X cos 0reo a
Exe2co dO
4 7reo a
1 A4 7re a
sin Oy sin 0;1o
11111111111
1111111110
For a infinitely long wire:
Ex 1
2ireo a
FIGURE
CALCULATION OF E
FOR AN INFINITEUNIFORMLYCHARGED WIRE
I?A.RV OVareke,V11*-..7:
TERMINAL OBJECTIVES
10/2 B Answer questions and solve problems relating to atomicmodels based on sperically symmetric charge distributions.
11/1 A Solve problems and answer questions on the relationshipbetween potential and field intensity.
;,./ ri . : ,
FOCUSSINGCYLINDER
DEFLECTIONPLATES 1i.5 cm
ANODECATHODE
ANDHEATER
41:)15cm 1=2cm
FIGURE (2)
FIGURE
CATHODEOV
Loss in PotentialEnergy
FIGURE
CATHODEOv
Gain in KineticEnergy
Loss in PotentialEnergy
11111111111
CATHODEOv
Gain in KineticEnergy
1eV2 "
vh/2eV
V m
I
FIGURE
CATHODEOV
1
0.02m
0.015m
FIGURE
+NM IN MIN
Vh
Vh/sr+101....
a.a.a time of
flight
.4k 4P-.4"
FIGURE
Vh
vhhv - ..............-4.. 1sammisom
time of 1flight vh
force on electrons . EeFIGURE
111111 =milmvh..............
.11. of ara
time of 1flight viz
j ice on tOttctrons = Ee
acceleration
FIGURE
F Eem m.
1
time offlight vh
force on electrons = Ee
acceleration Ee
deflection plate region
F IGURE
I
vh .............time offlight vh
force on electrons Ee
acceleration Ee
deflection in plate region
deflec*" acceleration
FIGURE
Ee
deflection in
defl:,(4, on
lif
.... * 14,
time offlight
on electrons = Ee
acceleration
plate region
BeaccelerationMt
frial deflected velocity .FIGURE
Eem
1 Eevh
i.
1
vh........... 1 IP. .4.
II 4r411.."; Ilf
yI time of
flight yhforce on electrons = Ee
acceleration Ee
1 Edeflection in plate region =2
ate 2 me
deflectioucceleration Ee
final deflect& velocity Ee
al deflects* = drft time
FIGURE
vh
2
deflection a Awe region =
additional *election
.11101.,AMMON.
food deflection
FIGURE
1 ar2..= 1 Ee (I2 2 ra vh
dr* time xL vvh d
sumo c these.* 1Vh 2 (F2
= t2 .x 10-2m
DERECTIMELECTRONS AN
ELECTRICYAW ,
tT4,-NA.;
TERMINAL OBJECTIVES
10/3 B Answer questionsand solve ptlobLems relating
to potential and field strength.
T
NO.
VI
FIGURE
FIGURE
IF ECARIKE
FIGURE
AFIGURE
FIGURE
dii
FIGURE
(i) d A
F IGURE FIGURE
0
t
FIGURE
z =w cos 30°E =az
=aw cos 30°
FIGURE
d+=t dAEL dwcos36awLcosi2 30°dw
z =w cos30°E =eiz
=aw cos 30°
d awL cos' 30° dw
(1) = awL cost 30° dw0
FIGURE
d awL cos2 30° dw
01)
rawL cos' 30 °dw0
2 a W2L cos2 300
FIGURE
TERMINAL OBJECTIVES
10/1 A Answer questions and solve problems
concerning electric field flux.
STR' Y
(1) Draw a closed symmetricalsurface around the charge
(2) GAUSS' LAW(1) qlEo
F IGURE
STRATEGY
(i) Draw a closed symmetricalsurface around the charge
(2) GAUSS' LAW(I) 91E0
(3) 04) =f E dA
FIGURE
1 dA - EAFIGURE
CHARGE DENSITY X
E DENSITY= X
A.LIE0
CHARGE NY- Xq =XL(0= XL/E0
(I) EA
CHARGE
FIGURE 6141
S1TY X
E=2-471-A;r-EA
(11 43= 2 71 E04)=. q 1E0
q=XL(2) 4)= AL/E0
(3) 2777LE =XLIE0
E=211Vor
CHARGE DENSITY
CHARGE DENSITY-- G,
FIGURE
CHARGE IIIENSITY--= G
FIGURE
q CIA
4)=GA
(4)
( 5 ) (1) Eqo
(6) 04)=
F I GURE
ITY
q(to OA
E0
(4) q=c5A
(5) itsEt)
(6) (0 ?A
(7) ocp= 2EA
2EAcAE0
E= °
TERMINAL OBJECTIVES
10/2 A Answer questions and solve problems using Gauss's
Law for cases of spherically symmetric charge
distributions.
10/2 E Apply Gauss' Law to charged bodies.
7
FIGURE
FIGURE
FIGURE
WITH PRESSURE CONSTANT
QA 9B
OF Q
SO C =
Q,
Cl
F I CURE
Q 2
so c
C2
FIGURE
V
MO.
FIGURE
FIGURE
Ams
FIGURE
+0-2E0
Po'
1
FIGURE
E 217-E0
Imr
E
FIGURE
Eo
FIGURE
Eo
V fdE di0
=Ed
6dEo
dE0
6aFIGURE
FIGURE
V
gA0-cl/E0
E0Ad
14
CAPACITORS
TERMINAL OBJECTIVES
11/3 A Answer questions and solve numerical problems involving
the physical significance and units (basic and
submultiples) of capacitance, C.
11/3 D 'Solve problems involving various conductor-pair
geometries' and the corresponding capacitances.
12/1 A Solve discriptive and numerical problems involving
capacitors in series and parallel combinations.
(Note: All interconnecting wires are resistanceless).
12/1 D Predict the effect of adding a dielectric of known
dimensions and material to a vacuum capacitor in both
descriptive and quantitative situations.
THE CAPACITORIN ACTION
THREAD
COATED5TYROi::0AM
BALL--
UNCHARGEDMETALSTAN
CHARGEDMETAL
STA N
I N L ATED BASE
FIGURE
Aluminum-)-4' Disc
PlasticInsulator
F IGURE
NEGATIVELYCHARGEDBY CONTACT
NEGATIVE ROD CLOSE BUT_a) NOT
---°*-1TOUCHING
FIGURE
FIGURE
PCWITIVi ROD CLOSE BUT
FIGURE
LARGE
contact:
C = I C EA
°
(2) V= C)
PARALLEL METALPLATES (CAPACITOR)
SUPPORTED ONINSULATED STAND
(NOT SHOWN )
(V is measured by 0
F I GURE
The Changes
(1) C = KE041
(2) V=
hence 0 OF
M=WIr/P
FIGURE
V
= KE
WINNEIMMO
The Changes
(1) C =KE0
(2) V=
= as shown
140---d constant
FIGURE
The Changes
(1) C= KE0Ad
(2)
disysmaiierthan
A CONSTANT
d CONSTANT
PLASTICSHEET II
7 -r
,A4FIGURE
(1) C = KE0
(2)
4--d constant
V=
0 = as shown
The Changes
(2)
C A=KE0 d
V
Of is smallerthan 0
(19)A
THE CAPACITORIN ACTION
TERMINAL OBJECTIVES:
11-1.080-00
Solve descriptive and numerical problems involvingcapacitors in series and parallel combinations.(Note: All interconnecting wires are resistanceless)
11-1.083-00
Predict the effect of adding a dielectric of knowndimensions and material to a vacuum capacitor inboth descriptive.and quantitative situations.
...
11
BRANCHPOINT
(NODE)
FIGURE
KIRCYR0FF4' RULES
41. AT ANY JUNCTION,THE ALGEBRAIC SUMOF THE CURRENTSMUST BE ZERO.
FIGURElih 7../n
KIRCHHOFF'S RULES
2. THE SUM OF THECHANGES IN POTENTIAL
ENCOUNTERED IN MAKINGA COMPLETE LOOP IS ZERO.
--te, 24v 4=12v
FIGURE
TE;.--24v
FIGURE
= (811 i2 6.n.
24v= i, 14n
F I GURE
ma-12v = iz 10.n. 6n
FIGURE
24v= i, 1461 6.n.
-12v = on - 6i1FIGURE
= 1. 62 amps
= - .25 amps
FIGURE
FIGURE
schunai7
O = SJUIV9?'+sduiv9L.
0 " `1.
U9
I I NI
a/1110I a
zi + GDU91.1
U9 UOL =An
U9 zi ugh = AVZ
8n
El 24v ta2=12v
FIGURE
FIGURE
+ i, 011
i2 8n
811+ 411)
KIRCHHOFF'SRULES
TERMINAL OBJECTIVES
13/1 B Answer questions relative to the methods of
application of Kirchhoff's Current Law to
electrical networks.
13/1 D Apply Kirchhoff's Laws to the solution of
numerical problems ranging from simple to
more complex multiloop networks.
r
FIGURE 0-
FIGURE
FIGURE
FIGURE
UNITS of "B"(a vector quantity)
nt ntcoul (m/ s) amp m
weberm2 tesla
FIGURE
DEFINITIONOF "B" FIELD
TERMINAL OBJECTIVES
14/1 B Answer qualitative questions relating to the
magnetic induction vector B.
FORCE BETWEENPARALLELCURRENT-CARkYINGCONDUCTORS
r
If two wires are freely suspended very close to one another, and if
a current is then p,ssed through each of the wires, a force of at-
traction or repulsion can be detected between them. The direction
of the force is a function of the relative current directions; if
the current directions are the setae in each wire, the conductors
will attract one another but if the direction of the current in one
of the wires is reversed, the force changes to repulsion. Please
refer to Figure 1.
Analysis of the electromagnetic fields that surround each conductor
indicates that both the magnitude and the direction of the force
can he theoretically predicted. Let us assume that the wires
shown in Figure 2 are connected directly to a source of emf, in
series with one another, so that the currents are opposite in
direction but equal in magnitude.
The current in wire a is directed downward while that in wire b is
upward. In order to make the analysis easier to perform in two
dimensions, imagine that both wires have been rotated about a
horizontal axis so that they present the picture shown in Figure 3.
The wires now appear in cross-section as small discs; wire a
carries a dot to indicate that the current is directed toward the
observer and wire b contains a cross to show that the current in
this wire is directed into the plane of the paper, away from the
observer. Considering wire a alone for the moment, as in. Figure 4,
the B-lines surrounding it may be drawn as concentric circles to
conform with experimental facts obtained from Oersted's Experiment.
#24 - 1
a
FlexibleSuspension
IGURE
FIGURE
WIRE
FIGURE
FIGURE
Applying the right-hand rule for wires (Oersted's Rule),
the thumb of the right hand is pointed in the directioa
of the conventional current so that the fingers then
encircle the wire in the direction of the magnetic field.
For this case, the B-lines are counterclockwise in
direction as indicated in Firl.re 5. At a point P near
the current-carrying wire, the line of magnetic in-
duction is tangent to the circle of the B-line surround-
ing the wire.
The magnitude of the field at point P is given by
Ampere's Law and may be written as indicated in Figure 6,
in which B is the magnitude of the field, uo
is the
permeability constant, is is the current in wire a, and
r is the distance between the center of wire a and point P.
#24 - 2
Ito
a \ \lp)
FIGURE
A -\ Bp
FIGURE
To review another concept briefly, please refer to Figure 7.
In this diagram, a wire is immersed in a magnetic field; the
wire carries a current into the plane of the diagram. The
source of the magnetic field is not indicated, nor is this
information seeded to analyze the problem. The B-lines
from this unknown source are directed upward in the plane of
the paper as indicated. Applying the Palm Rule to determine
the direction of the force acting on the current-carrying
conductor immersed in the given field, the fingers of the
right hand are placed so that they point in the direction of
the B-lines while the extended thumb points in the direction
of the current. The direction of the force on the wire is
then given by the direction in which the palm would exert a
thrust if the hand were used in the-normal manner. In the
example given in Figure 5, the direction of the force would
be that shown in Figure 8, namely to the right as viewed by
the observer.
The Palm Rule may always be used in this way and will be
found to be a great help in analyzing this kind of situation
and others similar to it.
#24 - 3
F IGURE
FIGURE
The magnitude of the force on the current-carrying wire is given
by the relation shown in Figure 9. Thus, both the magnitude of the
force and its direction are determinable for the example given.
Please refer to Figure 10; this is reiteration for review. Also
refer to Figure 11.
These ideas may now be combined to determine the nature of the force
in a specific case; that is, to determine whether to expect attraction
or repulsion when the current directions are known. Working with
conductors carrying oppositely directed currents as in Figure 12, it
can be readily shown that the force is one of repulsion in the follow-
ing manner.
The line of magnetic induction at wire b due to the current in wire
a is labeled Ba
. Applying the Palm Rule to wire b, it is seen that
the force on this wire is directed to the right away from wire a as
illustrated in Figure 13. The magnitude of the force is given in
the same Figure. In this relationship, Fb is the force acting on
wire b, ib is the current in wire b, lb is the length of wire b, and
Ba
is the magnetic induction due to the current in wire a.
FIGURE
FIGURE
it B.L.
A
la
FIGURE
FIGURE
a
la
FIGURE
Bp
Exactly the same process may be followed to find the force acting
on wire a due to the current in wire a and the magnetic induction
produced by the current in wire b. The right-hand rule is first
applied to wire b; this demonstrates that the B-line at wire a is
directed upward. Then the Palm Rule is applied to wire a, showing
that the force on this wire acts to the left away from wire b.
The direction and magnitude of this force is diagrammed in Figure
14. The student should confirm this for himself.
Thus, the wires repel each other. From Third Law considerations
alone, one may conclude that the force on wire a must equal the
force on wire b since they form an action-reaction pair. The fact
that the forces are equal may also be shown directly as in Figure
15. In the first step, the magnitude of Fb is given in equation
form. In the second step, Ba has been replaced by its equivalent,
i.e.uoia/27rr. BOth sides are then divided by the wire length
to yield the force per unit length in the third step. The
remainder is self-explanatory.
#24 - 5
FIGURE
Fb
lb
ito2 71r
and assuming equallengths and currents
2AO i for either wire2 71-r
F I CURE
In summary, as presented in Figure 16, the force between
current-carrying wires is one of REPULSION if the currents
are OPPOSITELY DIRECTED; the force is ATTRACTION if the
currents have the SAME DIRECTION. The force per unit.
length on either wire for equal currents and equal lengths
is given by
F/1 =)uoi2/27Tr
Summary
REPULSION, if currents areoppositely directed;
ATTRACTION, if currents havesame direction.
F i 2
2 71r
for either wire if currents are equal.
FIGURE
-
4,
FORCE BETWEENPARALLEL
CURRENT-CARRYINGCONDUCTORS
TERMINAL OBJECTIVES
14/3 A Describe the magnetic field around a straight-current-
carrying conductor.
14/3 D Prove that the force between wires a and b in
the diagram is an attractive force, the magnitude
of the force on either wire being given by (equation).
AMPERE'S LAW APPLIEDTO A LONG
STRAIGHT CONDUCTOR
AMPERE'S LAW
0
0
00
FIGURE
B dl =,uoiFIGURE 0
I
0
7 0/0
0
0
\P
dl0
O /I/
FIGURE
ri di yoiwhere )So = 47T (10-7)
web Iampinand
di B di cos 0
FIGURE
I
AMPERE'S LAW
F IGURE
FIGURE
F IGURE
(9a)
(9b)
(9c)
(9d)
g (17= B dl cos 0
Bdl cos 0= B dl
Bdl= BPI
B (27Tr)=,u0 i or B=27n,
(10) BLweb)2i (amp) -7 web
amp. mm2/ x 10(in)
AMPERE'S LAW APPLIEDTO A LONG
STRAIGHT CONDUCTOR
TERMINAL OBJECTIVES
14/3 A Describe the magnetic field around a straight-
current- carrying conductor.
'14/3 F Answer questions and solve problems involving
Ampere's law and its applications.
23
THE LAW OFRIOT-SAVART
AMPERE'S LAW
e56) d /110 i
FIGURE C)
FIGURE I(D
I
,110 i dl sin 0dB 4r2
FIGURE
FIGURE
Bp
r ,a0 i sin 0471 r2
B=fo dl sin 0J 411- r2
}10 i dl coso,r2
FIGURE
or id/
lio dlJ 4ir r 2
R tan
R sec2
ser2 dcz.<_
FIGURE
WIRE
FIGURE
cos
or better
r
(a) di= R sec cc. cos c..,(Ci-c--D)c
r 2
(b) dB Ai R see2 c.c. cos 0<, dcc.47r
(c) dIi47r
R2 sec 2 c
cos 0<, dcR
FIGURE
Aioio417rR
7
(a) it= 'Ara"' f cos_c.0IfirfR
becomes7r
(b) [sin 2
_7r
c)
F IGURE
THE ...LAW OF
BIOT-SAVART
TERMINAL OBJECTIVES
15/1 A Derive the expression for the -nr=etic induction
within an ideal solenoid as (P Hon) is the
actural current in the solenoi&-ulre and n is
the number of turns. (diagram6)
15/1 D Use Fig. 4 as an aid in mathemat±cally deriving
the equation for the magnetic induction at point
P; (equation).
.,
0
'FIGURE
cos 0 dAdA
GALVANOMETER
FIGURE
MAGNET
MAGNET'
GA LVANOMETEIRI
FIGURE
g ot cl(13$
dtFIGURE
FIGURE
. FIELDINCREASED
FIGURE
FIGURE
FIGURE
EMF
SEAMLESSALUMINUM
RING IRONCORE
COIL OF'MANY
TURNS
FIGURE
SEAMLESS:71e---7311,ALUMINUM !
RING >IRONCORE
COIL OFA/MANY
TURNS
FIGURE
FARADAY'S LAW
c14)
dtLENZ'S LAW
The direction of an induced
current is such as to oppose
the change of flux causing it.
FIGURE
t
FARADAY'S LAWOF INDUCTION
TERMINAL OBJECTIVES
15/3 A Trace the development of Faraday's Law of
electromagnetic induction through an analysis
of his basic experiments.
>15/3,D Apply Lenz's Law, L deb :mine the direction of induced
emf's in various induction situations.
MOTION OF ANELECTRON INCOMBINED EAND B FIELDS
%
I
a-.0-
AYMt
iarta:4.1Q`4.A13,1 W->"; -.1 ;":1$'
F IGURE
ectrons4--
5
I
FIGURE
raw
F1 GU
ifI (Aga
I
MOTION OF ANELECTRON INCOMBINED EAND B FIELDS
TERMINAL C4JECTIVES
1
10/3 B Aloft/et questions amid solve problems relating to
potential fiPld strength.
14/1 B Answer rporr41141-ative Glue sittbans 'relating- to tribe magnetic
induction -elector B .
zs
LTRANSIENTS
I
LIGHT
FIGURE
1
FIGURE
Ri+ L didt 0
L didt
Ri
id R dti L
FIGURE
t + ln(constant)
constant)e-RtIL
io e-Rtil,
FIGURE
t
I
-t // RC
RitiLR
time constant
FIGURE
I
Ri L dd L,14
( e
= time constant
F I CURE
t
CURRENT DECAYe-RtIL
CURRENT GROWTH
i 1e-t/4)
For Both
jo
ioo
R
R
FIGURE
TERMINAL OBJECTIVES
15 03 124 00 analyze the general RL current growth equationqualitatively and quantitatively.
15 03 126 00 analyze the general RL current decay equationqualitatively and quantitatively.
F
M
CHAR -DISC HARGE CHKUU
FIGURE
INERE
dqdt ICdqq )
Inqr= +BC (constant)
F WARE
in q = Rtc, + in constant
q = (constant) ctlacgo = constantq= e-tiRC
FIGURE
q0 et/Rc
RC = time constant
FIGURE
t.
FIGURE
C
R dtdqR dt + 0
q (constant
= goc, (1 e-tiRc
FIGURE
DISCHARGE:q
= qo eRC
CHARGING:q go() (i-e-tiRc)
qo goo eC
FIGURE
qor
V
BreakdownVoltage
FIGURE
R CTRANSIENTS
TERMINAL OBJECTIVES
15 02 121 00
15 02 123 00
analyze the general RC circuit charging equationqualitatively and quantitatively.
analyze the general RC circuit discharge equationqualitatively and quantitatively.
