Physics 3rd Secondary Chapters 1-7 revision مراجعة فيزياء 3 ثانوى من 1-7 مستر أحمد هيكل

MR.AHMED HEKAL
PHYSICS 3RD SECONDARY
CHAPTERS (1-7)

Mr. AhmedHekal
1
Chapter 1
A. Definitions
1- Electrical current:
- Electronsmovementthroughaconductorfromnegative pole topositive pole inthe presence of electric
source
2- Traditional Electrical current:
- Movementof positive chargesthroughaconductorfrompositive pole tonegative pole
3- Currentintensity:
- Quantityof electrical chargesthroughasection of a conductor in1 second
4- Ampere:
- Currentintensityresultsinpassingquantityof chargesof 1 Coulombthroughaconductor in1 second
5- Coulomb
- The quantityof electrical chargesthatwhenpassesthrough a conductorintime of 1 secondit generates
currentof 1 Ampere
6- Potential difference:
- The work done injoule usedtotransferquantityof chargesof 1 C between2points
7- Volt:
- The potential difference bet.2pointswhenworkdone of 1 J usedto transferquantityof chargesof 1 C
8- Electromotive Force:
- The whole workdone outside andinside the batterytotransfer 1coulombof charges inthe electrical circuit
- Potential difference betweenthe polesof the batteryincase of opencircuit
9- Electrical Resistance:
- The opposition ‫الممانعة‬ tothe flowof the current
- The ratio betweenpotentialdifference involtacrossthe terminalsof conductorandcurrentintensitythat
flowsthroughthisconductorinampere
10- Ohm’sLaw:
- At constanttemperature,currentintensityisdirectlyproportionalwithpotential difference
11- Ohm:
- The resistance of conductorthat allowspassage currentof 1 Ampere whenpotentialdifference across
terminalsof thisconductoris1 volt
12- Resistivity:
- It’sthe resistance of conductorof length1 m and crosssectional areaof 1 m2
at constanttemperature
13- Conductivity:
- The reciprocal of resistivity
- Reciprocal of resistance of conductorof length1 m and cross sectional areaof 1 m2
at constant temperature
14- Kirchhoff’sfirstlow:
- In a closedcircuit,summationof currentsenteringapointisequal to summationof currentsexitingthis
point
- The algebraicsummationof currentsina specificpointinclosedcircuitequalszero
15- Kirchhoff’ssecondlaw:
- Algebraicsummationof e.m.f.inaclosedcircuitequalsthe algebraicsummationof potential differencesin
thiscircuit
- Algebraicsummationof potential differencesinaclosedbranchequal zero

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B. What’s meant by?
1- Current intensitypassesthrough a conductor is5 A
- The quantity of electrical chargesthat passesthrough a sectionof this conductor in 1second equals5
coulombs
2- Potential difference betweenterminalsofa conductor is 20 V
- The work done to transfer 1 C ofcharges bet. terminalsof this conductor is 20 Joule
3- Electromotive force of a battery = 1.5 V
- The whole workdone outside andinside the batterytotransfer1 coulombof chargesinthe circuit= 1.5
joule
4- Electrical Resistance of aconductor100 Ohm
- The ratio betweenpotentialdifference acrossthe terminalsof conductor andcurrentintensitythatflows
throughthisconductor is100 V/A
5- Resistivityof aconductor = 6 *10-6
ohm.m
- Resistance of aconductor of thismaterial of length1 m and cross sectional areaof 1 m2
ata constanttemp.
is 6 *10-6
ohm
6- Conductivityof material is5.6*107
ohm-1
.m-1
- Reciprocal of resistance of conductorof length1 m and cross sectional areaof 1 m2
at constant temperature
Is 5.6 *107
ohm-1
C. Deductions
1- Resultant resistance ofgroup ofresistances connectedin series
- Current intensityisconstantthrough all resistances
- Potential difference isdividedbetweenthem
V = V1 + V2 + V3
V = I R
I R` = I R1 + I R2 + I R3
Then
2- Resultant resistance ofgroup ofresistancesconnectedin parallel
- Potential difference isconstantacrossall resistances
- Current intensityisdividedbetweenresistances
I = I1 + I2 + I3
I = V / R
V / R` = V / R1 + V / R2 + V / R3
Then
R` = R1 + R2 + R3
1 / R` = 1 / R1 + 1/ R2 + 1 / R3

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D. Factors that depends on
Physical quantity Factors that this quantity dependson
Resistance of a conductor
R = Ꝭe L/A
1- Lengthof conductor(directlyproportional)
2- Its cross sectional areaA (inversely
Proportional)
3- Material type of thisconductor
4- Temperature of thisconductor
Resistivityofconductor 1- Material type of thisconductor
Conductivityof a conductor 1- Material type of thisconductor
F. Comparisons
resistancesconnectedin series resistancesconnectedin parallel
Shape of
connection
Target To obtaina large resultantresistance
froma groupof small resistances
To obtaina small resultantresistance
froma groupof large resistances
Current intensity Equal or constant inall resistances The whole currentequalsthe summation
of all currentsinall resistances
Potential difference The whole P.D.equalsthe summationof
all P.D.sin all resistances
V = V1 + V2 + V3
Equal or constant across all resistances
Law of resultant
If all resistancesare
equal
R` = N R
Where Nis number of them
R` = R / N
R` = R1 + R2 + R3 1 / R` = 1 / R1 + 1/ R2 + 1 / R3

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E. What Happens in these cases
1- Increasingpotential difference todouble value for current intensityand power consumed?
- Currentintensitywill increase todouble asI= V / R
- Powerwill increase 4timesas Pw = V2
/ R
2- Current intensityincreasesto double forthe resistance value
- Resistance remainsconstantasitdoesn’tdependsoncurrentitdependson
- Lengthof conductor(directlyproportional)
- Crosssectional areaA (inverselyProportional)
- Material type
- Temperature
3- Increasingcross sectional area of a conductor to double and decreasingits lengthto half value for the
resistance
- L1 = 2 L , L2 = L , A1= A , A2 = 2 A
- R1 / R2 = L1 A2 / L2 A1 = 2 L * 2 A / L * A
- R2 = ¼ R1
4- Connectingtwo resistance in parallel on of themhas a value of1 ohm to the resultant resistance
- Resultantresistance will be lessthan1 ohm
5- Current doesn’tflowfrom an electricsource to potential difference betweenthe terminalsofthis electric
source
- Potential difference betweenthe terminalsof thiselectricsource will be equal toelectromotive force of the
electricsource accordingtothis relation
(V = VB – I r) and I = 0 then V = VB
F. Give reason
1- Work shouldbe done to transfer charges from point to point
- To get ridof resistance bet.the twopointsandcurrentcan flow
2- Some materials can conduct electricitybutothers cannot
- As some materialshave aplentyof free electronssoitallowsflow of currentwhile othermaterials don’t
have free electronsortheirelectronsare stronglycorrelatedtotheiratoms
3- Increasingradius of a wire of copper leadsto decreasingits resistance to quarter value
- Accordingto thisrelation
R = Ꝭe l/r2
Resistance isinverselyproportional withsquare of radius
4- Whena conductor isshaped to be parallelogramits ribs resistancesare differentwhile ifshapedas a
cube itsribs resistancesare equal
- As ribsof parallelogramare differentinlengthsotheirresistance differsaccordingtorelation
R = Ꝭe l/A but cube ribs are equal in length and equal in resistance
5- Resistance increase whenincreasingtemperature
- Whenraisingtemperature thisincreasethe speedof vibratingitsmoleculesandincrease the rate of
collisions betweenelectronsof currentandconductormoleculessothe opposition ‫الممانعة‬ of current
increases

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6- Resistivityof aconductor isa physical property
- As itdependsonthe type of the conductor material atconstanttemperature
7- Conductivityof a conductor isa physical property
- Because conductivityisthe reciprocal of resistivitywhichdependsonthe conductormaterial atconstant
temperature
8- Conductivityfactor ofcopper is large
- As resistivityof copperisverysmall incause of plentyof free electronsincopper
9- It’s preferredto use wires ofcopper in the electrical connections
- As resistivityof copperisverysmall soitsresistance islow andthispreventwire fromconsumingelectrical
energy
10- Home devicesare connectedin parallel
- All deviceswill workonthe same potential difference of the source soeachdevice can workalone andif
one of themdamageditdoesn’taffectthe othersand alsoto decrease their total resistance whichdoesn’t
affectthe maincurrent
11- Home devicesaren’tconnectedin series
- Because potential difference will be dividedacrossthemwhichleadstoaprobabilityof insufficient voltage
on a device thatcannot operate and one device cannotworkalone alsotheirtotal resistance will be huge
whichpreventsthe currentfrompassingthroughthe circuit
12- The electrical powerincreasesin case of connectingtwo resistance on parallel
- Whenconnectingresistancesinparallel theirtotal resistancedecreasesandcurrentwill increaseandPower
alsowill increase accordingtothislaw Pw= IV
13- In a circuit ofparallel connectionresistanceswe use thick wiresat the terminalsof battery and thin
wiresat the terminalsof each resistance
- As currentintensityshouldbe maximumatthe inputandoutputof the batteryso we use thickwiresof low
resistance whilecurrentisdividedthrougheachresistance sowe canuse thinwiresinterminalsof each
resistance
14- Potential difference betweenbatterypolesincreaseswhenincreasingresistance of its circuit
- Accordingto thisrelation (V = VB – I r) whenincreasingresistance currentwill decrease andthe internal
potential difference (Ir) will decrease andbecause VB isconstantthenpotential difference acrossthe
batterywill increase
15- E.M.F of a battery is larger than potential difference betweenitsouterterminalswhenclosing the
switch
- Because the internal resistanceof the batteryconsumespowertoallow currenttoflow inside itso
(VB = V + I r) so VB > V

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Laws
To compare betweenPowerconsumedin two
resistances
WhenV isconstant
(Pw)1 / (Pw)2 =R2 / R1
Whencurrentis constant
(Pw) 1 / (Pw)2 = R1 / R2
Resistance
R = V / I
R = Pw / I2
R= V2
/Pw
R = Ꝭe L/A = Ꝭe L/r2
Ꝭ (density) =m (mass) /v (volume)
Ꝭ = m / v
Andv = L * A and L = v / A
Then Ꝭ = m / L A then
A = m / L Ꝭ and
L = m / A Ꝭ
R = Ꝭe L2 Ꝭ / m
R = Ꝭe v (volume) / A2
R = Ꝭe L2/ v (volume)
Compare betweenresistances
R = Ꝭe m / ꝬA2
R1/R2= Ꝭe1 L1 A2 / Ꝭe2 L2 A1
R1/R2 = Ꝭe1 Ꝭ1 L1
2 m2 / Ꝭe2 Ꝭ2 L2
2 m1
R1/R2 = Ꝭe1 L1 r2
2 / Ꝭe2 L2 r1
2
Conductivity
σ = 1 / Ꝭe= L / R A
Laws
Quantity of charges
Q = I t
Q = n qe
Q = W / V
Q: total quantityof charges (electrons)
n: numberof electrons
qe: charge of one electron
W: work done
V: potential difference
I: currentintensity
t: time
Potential difference V
V = W / Q
V = W / nqe
V = I R
V = Pw / I
Current Intensity
I = Q / t
I = n qe / t
I = V / R
I = Pw / V
Electrical Power
Pw = W / t
Pw = V I
Pw = I2
R
Pw = V2
/ R

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To Solve Kirchhoff’sProblemsfollowthe following:
1- Divide the circuittono. of loops(2 or 3 loops)
2- Finda point(junction) inwhichall currentsare enteringorleavingit
3- Write the firstequation usingKirchhoff’sfirstlaw (sumof currentsenteringapointequalstosumo
currentsleavingthispoint
4- Specifymydirectionasfollow
A. If there is one batteryinthe loop,specifythe directiontobe from +ve pole to –ve pole ofthis
battery
B. If there are twobatteriesinthe loop,specifythe directiontobe from +ve pole to –ve pole of
the largestbattery
5- Write the 2nd
and 3rd
equationsusingKirchhoff’ssecondlaw (sumof potential difference inside a
loopequalszero),we have twochoices:
A. If there is one batteryinthe loop,thenwrite itsvalue direct V=IR + IR
B. If there are twobatteriesinthe loop,thenwe have two choices
 If theyare connectedparallel ( -ve connectedto –ve and +ve connectedto+ve) then
subtract theirvaluesV1-V2=IR +IR
 If theyare connectedinseries( -ve connectedto+ve and+ve connectedtonegative)
Thenadd theirvaluestoeachotherV1+V2= IR +IR
C. The sign of IR dependsonthe directionof the current
 If the currentpassesinthe same directionof mydirectionthenputitin +ve (+IR)
 If the currentpassesinthe opposite directionof mydirectionthenputitin –ve (-IR)
6- Currentthat leavesthe batteryisthe same currententersthe battery
7- Potential difference betweentwopointsisequal tothe potential (orvoltage) atthe pointof higher
potential the potential (orvoltage) atthe pointof lowerpotential
8- Solve the 3 equationsusing calculato

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Chapter 2
G. Definitions
16- MagneticFlux:
- Numberof magneticfieldlinesof amagnetfromnorthpole to southpole
17- Magnetic fluxdensityat a point:
- It’sthe magneticflux forunitareawhichisnormal to magneticlinesaroundthispoint
Or
- The magneticforce affectingawire of 1 m lengthplacednormal to magneticflux and a current of 1 A
passesthroughit
18- Tesla
- The magneticflux densitythatgeneratesaforce of 1 Newtonona wire of 1 m lengthanda currentof 1A
passesthroughitwhenthiswire isnormal to the flux lines
19- Permeabilitycoefficientofa medium:
- The abilityof mediumtopermitthe magneticflux throughit
20- Dipole Moment:
- It’sthe magnetictorque affectingacoil placedparallel tomagneticfieldof 1teslawhenan electrical current
passesthroughit.
21- MovingCoil Galvanometer(sensitive Galvanometer):
- A device usedtodetectaveryweak current to measure itsintensityanddetermine itsdirection
22- Galvanometersensitivity:
- The deviationangle of itspointerfromzeropositionwhenacurrentof 1 A passesthroughit
23- Shunt Resistance
- A small resistance connectedinparallelwithgalvanometertoconvertitto an ammetertomeasure higher
currents
24- AmmeterSensitivity
- Ratiobetweenmaximumcurrentmeasuredbygalvanometertomaximumcurrentmeasuredafter
convertingittoammeter
25- MultiplierResistance:
- Large resistance connectedinserieswithgalvanometertoconvertitto voltmetertomeasure higher
voltages
H. What’s meant by?
7- Fluxdensityat a point is 0.4 Tesla
- It meansthat a magneticforce of 1 N isaffectingawire of 1m lengthplacednormal tothe magneticflux at
thispointand a currentof 1 A passesthroughit
8- Dipole Momentis 0.7 N.m/T
- It meansthat magnetictorque of 0.7 N.m isaffectingacoil whena currentpassesthroughit andthiscoil is
placedparallel toa magneticflux of 1 tesla

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I. Figures
J. Conditions needed to :
Attraction force bet. two wireshaving currents pass
through them
The two currentsshouldbe inthe same direction
Repulsionforce bet. two wireshaving currents pass
through them
The two currentsshouldbe inopposite directions
Fluxdensityvanishesat a point betweentwo parallel
wireshaving currents pass through them
The two currentsshouldbe inthe same direction
Neutral point existsbetweentwostraight parallel wires
in a mid-pointbetweenthem
The two currentsshouldbe equal valuesandin the same
direction
Impossibilityofexistence ofa neutral pointfor two
straight parallel wires havingcurrents pass through
them
The two currentsare equal valuesandare inopposite
directions
Vanishingthe Force affectinga wire have a current pass
through it inside a magnetic field
Wire is parallel twotothe magneticflux
Vanishingthe torque affectinga coil in which a current
passesand isplaced in a magnetic field
Whenthe plane of the coil isperpendiculartothe
magneticflux
K. Devices
Device Usage ScientificIdeaand explanation
MovingCoil Galvanometer to detectverylow DC currentsand
measuresitsvaluesanddetermines
itsdirection
Idea:
The torque affectingarotatingcoil having
a current passesthroughitinside a
magneticfield
Explanation:
Whencurrentpassesthroughthe coil,two
equal parallel forcesare generatedin
opposite directionsontwo ribs‫ضلعين‬ of
the coil whichcausesrotationtorque on
the coil
Shunt resistance in Ammeter Convertsthe galvanometerto
Ammetertomeasure higher
currents
When a small resistance is connected
in parallel with the galvanometer coil
this leads to decreasing the total
resistance of the ammeter which avoid
affecting the current needed to be
measured

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Multiplierresistance inVoltmeter Convertsthe galvanometerto
Voltmetertomeasure higher
potential differences
Connecting high resistance in series
with a galvanometer leads to
increasing the total resistance of it and
when connecting this Galvanometer in
parallel with the circuit it consumes a
very small current then it doesn’t
affect the potential difference needed
to be measured
L. Usages
Pair of spiral springsin Galvanometer 1- It’susedas connectorsto current
2- Control the pointermovementusingreverse torque to
indicate the rightvalue
3- Returnthe pointertoits zeroposition
The concave polesinGalvanometer Theyremainthe flux densityconstantinthe space inwhichcoil moves,
thisway ensuresthatflux linesare alwaysinaradiusform and they
are parallel tothe coil plane (normal to twocertainrips)
Iron core inside Galvanometer Collectingandconcentratingthe fluxlinesinside the coil
JeweledbearingsinGalvanometer Coil standson themandtheyfacilitatesitsrotation
The standards and variable resistance in
Ohmmeter
Controlsthe currentintensitytobe maximumvalue whichmovesthe
pointertothe endof reading(zeroohm) before addingthe resistance
neededtobe measured
M.Deductions
1- Magnetic forcethat affectsa wire through which electrical current passesin a magneticfield:
-
- F α B (FluxDensity) And F αI (current) And F α L (lengthaffectedinwire)
- F α BIL
-
Where  is the angle betweenthe wire and the flux
Newton= Tesla.ampere.m,Tesla= Newton /ampere.m
F = BIL sinNewton

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2- Torque affectinga rectangle coil through which current passes and is placedinside a magnetic field
- whena coil isplacedina magneticfieldandacurrentpassesthroughitand the plane of the coil isparallel
to the plane of the field
- there are 2 ripsare parallel toflux soforce iszero,and the other2 ripsare perpendiculartothe fieldsothey
are affectedby2 equal andopposite forcesnotonthe same line,thiswill cause rotation
- rotationhappenedincause of torque
- Torque = force * distance
- Force = BIL sin 
- Distance = L (the normal distance betweenthe poleandthe affectedrip) thisdistance isthe wide of the
rectangle
- Torque = B I (Length* width) sin 
- Ʈ = B I A sin
- If we have numberof turnsN inthe coil so:
And
 = B |md| sin 
Where |md| is the dipole moment|md|= I A N
3- Shunt Resistance
- Rs and Rg are connectedinparallel so
- Vg = Vs
- Then Ig Rg = Is Rs
- Rs= Ig Rg / Is but Is= I - Is
- Then
4- MultiplierResistance Rm
- Rm and Rg are connectedin seriesso
- V = Vg + Vm = Ig Rg + Ig Rm
-
-
 = B I A N sin 
[Cite your source here.]
Rs= Ig Rg / (I - Ig)
Rm= (V – Ig Rg) / Ig

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N. Factors that depends on
MagneticFlux Density
generatedfromwire at a point
B = d)
1- CurrentIntensity(directlyproportional)
2- Distance bet.the pointandwire (inversely
proportional)
3- Permeabilitycoefficientof the medium(directly
proportional)
generatedfromcoil at the
centerpoint
B = r)
3- Numberof turns (directlyproportional)
5- Radiusof coil (inverselyproportional)
proportional)
generatedfromSolenoidata
pointon itsaxis
B = L
5- Axislength (inverselyproportional)
proportional)
MagneticForce
F = BIL sin
1- Magneticflux densityB(directlyproportional)
2- Lengthof affectedpartof the wire (directly
proportional)
3- Angle betweenthe fieldandwire (directly
proportional)
4- CurrentIntensity (directlyproportional)
Torque affectinga coil
 = B I A N sin 
1- Numberof turns (directly proportional)
3- Magneticflux densityB(directlyproportional)
4- Angle betweenthe fieldandwire (directly
proportional)
5- Areaof the rectangularcoil (directlyproportional)
Dipole Moment
|md|= I A N
3- Areaof the rectangularcoil (directlyproportional)

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G. Comparisons
Two wires having a current passes through them
In the same direction In opposite directions
Resultant of flux density at a
point in between
B total = B1- B2 where B1> B2 B total = B1+ B2
Resultant of flux density at a
point outside them
B total = B1+ B2 B total = B1- B2 where B1> B2
Neutral Point B1 = B2 Falls between the two wires
I1 / (x-d) = I2 /d
Where x isthe distance bet. Wires
and d isthe distance bet.the point
and wire of lower current
Falls outside the wires
I1 / ( x + d ) = I2 /d
Where x is the distance bet. Wires and
d is the distance bet. the point and
wire of lower current
Force between the two wires Attraction Repulsion
- Rules
Ampere Right Hand Rule Right Screw Clock Rule Fleming Left Hand rule
Figure Straight wire Solenoid
Usage Specifyingthe
directionof
magneticflux
generatedbya
currentpasses
througha wire
Specifyingthe
polarityof the
field
Specifyingthe
directionof
magneticflux at
centerof a coil
or solenoidaxis
Specifyingthe
pole type
(North-South)
at the face of a
coil or solenoid
Specifyingthe directionof
magneticforce affectinga
wire whichplaced
perpendiculartoa
magneticfieldandcurrent
passesthroughit
Working
Method
Thumbfinger
referstocurrent
and other
fingersrounding
the wire will
referto
magneticflux
Thumbrefers
to magnetic
flux andother
fingers
roundingthe
solenoidwill
referto the
current
Whenscrew
rotatesinclock-
wise direction
itsrotation
referstothe
currentand its
movement
directionrefers
to magneticflux
If the direction
of currentina
specificface is
clock-wise then
thisface is
Southpole and
if anti-clock
thenit will be
Northpole
Middle fingerrefersto
current,index fingerrefers
to magneticflux thenthe
thumbreferstothe force
Notice:magnetic fluxlinesisdirectedfrom north to south outside the coil and from south to north inside the coil

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Ammeter,Voltmeter,Ohmmeter
Ammeter Voltmeter Ohmmeter
Function Measuring high currents Measuring potential
difference bet. two points
Measuring resistance
Resistance connected on
the galvanometer coil
Galvanometercoil is
connectedinparallel to
small resistance called
Shuntresistance Rs
Galvanometercoil is
connectedinseriestolarge
resistance calledMultiplier
Resistance Rm
Galvanometercoil is
connectedinseriesto
standardresistance Rc and
variable resistance Rv anda
battery
Idea Of Work Torque affectingacoil that
has a current passes
throughit,thiscoil is
rotatinginside magnetic
field
Torque affectingacoil that
has a current passes
throughit,thiscoil is
rotatinginside magnetic
field
Dependsonthe inverse
relationbet.currentand
resistance atconstant
potential difference
Law Rs= Ig Rg / (I- Ig) Rm= (V – Ig Rg) / Ig I = V / (Rg+ Rc+ Rv+Rx+r)
How it’s connected in
circuit
In series In parallel Device terminalsis
connectedto terminalsof
the external resistance
Graduation Regular as I α  Regular as V α  Not regular as I α R`+Rx)
Shunt and MultiplierResistances
Shunt Resistance MultiplierResistance
Methodof connection In parallel with galvanometercoil In serieswith galvanometercoil
Function Convertthe galvanometertoAmmeterto
measure highercurrents
Convertthe galvanometertoVoltmeterto
measure higherpotential difference
Idea of work By connectingshuntresistance inparallel
withthe galvanometerthisleadsto
reduce the total resistance of the
device(ammeter) which avoid affecting
the current needed to be measured
Connecting high resistance in series
with a galvanometer leads to increasing
the total resistance of it and when
connecting this Galvanometer in
parallel with the circuit it consumes a
very small current then it doesn’t affect
the potential difference needed to be
measured
Analog Measuringdevices Digital Measuringdevices
 Its ideadependson Torque affectingacoil that
has a current passesthroughit,thiscoil is
rotatinginside magneticfield
 Valuesare appearedona graduationonwhich
pointerismoving
 As Galvanometer,Ammeter,Voltmeter
 Dependsondigital electronics
 Valuesare appearedasdigitsdisplayedonthe
screenof the device
 As devicesof measuringDCand ACcurrents

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O. What Happens in these cases
1- Current flowsin the same directionin two parallel wires
- The resultantof flux densitiesoutside the wires will be largerthanitbetweenthe wiressoattractionforce is
generatedbetweenthem
2- Current flowsin opposite directionsin two parallel wires
- The resultantof flux densitiesbetweenthe wireswillbe largerthanitoutside themsorepulsionforce is
generatedbetweenthem
3- Cutting a solenoidoflengthL and number of turns N at a middle pointon its axis and connectingone half
to the same battery
- The solenoidresistance isreducedtohalf andcurrentintensitywill increase todouble (andnumberof turns
remainconstantinunitlength) soflux densitywill increase todouble
4- Placing wire having a current perpendicularlyto a magnetic field
- Wire is beingaffectedbyamagneticforce whichisperpendiculartocurrentdirectionandflux lines
5- Passage of dc current of highintensity(biggerthan Ig) through the galvanometercoil
- hightorque isgeneratedinthe coil whichishigherthanthe momentof the abilityof the twobearingcoilsso
theyare destroyedandthe device is brokendown
6- Passage of highfrequency(AC) current inside galvanometer
- The pointerisoscillatingatthe zeroreadingaccordingto inertiaasthe pointercannotreact withthe change
of direction
7- Reducingthe value of shunt resistance
- Ammetersensitivityisreducedandthe graduationof readingisincreased
8- Increasingthe value of multiplierresistance
- Voltmetersensitivityisreducedanditwill measurehighervoltage
9- Nonexistence ofstandard resistance
- Galvanometercoil willbe damagedif currentishighandohmmeterpointerwill notbe accurate

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P. Give reason
1- It’s recommendedinbuildingto be far away the highervoltage towers
- To reduce the effectof the magneticfieldwhichisharmful tothe healthandenvironmentasthe magnetic
flux densityisinverselyproportionaltothe distance
2- The neutral point is positioned between twowires havingcurrents pass through themin the
same direction
- Wireswill generate twoopposite magneticfieldsatall pointsbetweenthemsoneutral pointexistsbetween
themwhentheyvanisheachother
3- The neutral point is positionedoutside twowires havingcurrents pass through themin opposite
directions
- Because of generatingtwoopposite magneticfieldsoutsidethe wiressoneutral pointexistsoutsidethem
whentheyvanisheachother
4- Parallel wiresare attracted when currents pass through themin the same direction
- As the resultantflux density betweenthemislowerthanitoutside themsomagneticforce isgenerated
fromthe higherdensityregiontolowerdensityregionsotheyare attracted
5- Fluxdensityincreaseson axis of solenoidthat has current passesthrough it whenplacing ironcore
inside it
- Because permeabilitycoefficientof ironismore thanit inair so ironwill increase andconcentrate the
magneticflux lineswhichleadstoincrease the density
6- Magneticfieldin a coil or solenoidmay vanish althoughthere isa current passesthrough it
- As the coil or solenoidisdoublerounded,sothe magneticfieldof one directionisopposingthe magnetic
fieldof the seconddirection
7- Movementofa straight wire whichhas electrical current and placednormal to a magnetic field
- Thishappensaccordingto the difference betweenthe original magneticfieldandthe fieldgeneratedbythe
wire sowire will move fromhighdensitytolow density
8- wire having current may not move although it’s placedinside magneticfield
- because it’splacedparallel tomagneticfieldso(=0) and F = B I L sinwhichleadstozero
9- whencurrent passes through a solenoidanda wire placed on its axis,the wire is affectedby
magnetic field
- asthe wiresisplacedparalleltomagneticfieldthatisgeneratedbypassage of electricalcurrentinthesolenoid
And F = B I L sin 
10- torque may not be generatedon a rectangular coil which has a current and placedinto a magnetic
field
- As the coil should be placed parallel to the magnetic field in order to be affected by magnetic force on two
rips which generate the torque, so if it’s placed perpendicular to the field torque will be 0
11- Torque is reduced on the rectangular coil through its rotation starting from parallel position
- At the parallel position the angle between the coil and the normal plane to field equal 90 so torque is
maximum, while rotating the angle is decreased until it reaches 0 at which angle is zero
sin 
12- Concave poles in Galvanometer
- To keep the magnetic flux lines always parallel to the coils so at any position magnetic field densitywill be
constant and the deviation angle will always proportional to current intensity

Mr. AhmedHekal
19
13- Coil of galvanometeris connectedto couple of bearingscoils
- Theyare usedto
A. As connectorsof currententering andexitinggalvanometer
B. Theygenerate opposite torqueusedto
. Stoppointerat the rightvalue of current
. Returnpointertozerovalue afterthe current disconnected
14- Coil of Galvanometerisplaced on jewelsbearings
- To reduce frictionsandmaintain the equilibriumof coil tofacilitate the rotation
15- Existence ofiron core cylinderinside the coil of galvanometer
- To concentrate andincrease the magneticflux inside the coil whichincrease the galvanometersensitivity
16- Graduation ofgalvanometer isregular and zero reading ispositionedinthe middle
- As the deviationangle isdirectlyproportionalwiththe currentintensityandzeroispositionedinthe middle
to specifythe directionof current
17- Galvanometercannot measure AC current
- As the magneticfieldgeneratedbyACisalsoalternatingsothe directionof torque changeseachhalf cycle
so inertiawill preventthe reactiontothischange
18- Galvanometerdoesn’tmeasure highcurrent intensities
- As itscoil cannot bearhighcurrentsbecause part of this currentis convertedtothermal energywhichleads
to meltingthe coil wire andalsothe torque generatedbyhighercurrentisverystrongwhichmaydamage
the jewelsbearings
19- Ammeteris connectedin seriesinthe circuit
- To measure the total currentof the circuit
20- In Ammeter,a verysmall shunt resistance is connectedinparallel with galvanometercoil
- To reduce the total resistance of device whichavoidcurrentdecreaseandthe majorityof currentwill pass
throughthe shuntresistance whichprotectthe coil fromdamage so we can use Ammeterinmeasuringhigh
currents
21- Voltmeterisconnectedin parallel in the circuit
- To make the potential difference acrossthe voltmeterequal tothe potential difference neededtobe
measured
22- In Voltmeter,a very highmultiplierresistance isconnectedin serieswith galvanometercoil
- To enlarge the resistance of the device soaverysmall part of current will passthroughitandthe majorityof
currentwill notaffectedsopotential difference neededtobe measuredalso will notaffected
23- Electromotive force of the battery in Ohmmetercircuit shouldbe constant
- To maintainOhm’slawandmake the current alwaysinverselyproportional toresistance incase of
constante.m.f
24- High standard resistance isconnectedin Ohmmeter circuit
- To reduce the currentpassingthroughthe circuit to protectthe galvanometercoil fromdamage andmake
itspointerdeviate tothe maximumreadingbefore connectingthe unknownresistance
25- Graduation ofAmmeter isregular and graduation of Ohmmeter isnot regular
- As inAmmeterthe deviationangle isdirectlyproportional tothe current,butinOhmmeterthe currentis
inverselyproportional withthe total resistance notonlythe unknownresistance
26- Ammetergraduation opposesthe Ohmmetergraduation
- Because the currentis inverselyproportional withresistance sowhenaddingresistance the current
decreases
-

Mr. AhmedHekal
20
Laws
2- Flux densityin2 coils
A. In the same level
- Currents inthe same direction
BT = B1+ B2
- Currents inopposite direction
BT = B1- B2 (B1>B2)
B. Coilsare normal to each other
- B = √B1
2
+ B2
2
3- Fluxdensitybetweena coil and a wire
A. If the current isin the same direction
- BT = Bwire+ Bcoil
B. If the current isin opposite directions
- BT = Bbig- B small
4- To calculate numberof turns in a coil
A. If a wire of lengthL is roundedin
form of coil
- N = L /2r
B. If the coil is lessthan 1 turn (part of a
turn)
- N =  / 360
5- In case of the wire is tangentto the
coil
- N I1 = I2/
6- If a coil of N1 turns isreformedto N2
turns and connected
- B1/B2 = N1r2/N2r1= N1
2
/N2
2
=r2
2
/ r1
2
Laws
1- MagneticFlux m instraightwires
m = 0 whenfluxis parallel to area
m = B A whenfluxis normal to area
m = B A sin  whenfluxis making angle 
with the area
If the fluxrotates with angle  from
 Parallel position m = B A sin
 Normal positionm = B A sin(90- 
- If the current isin the same direction
- Neutral Point betweenthe wires
I1 / (x-d) = I2 /d
Outside the wires betweenthe wires
BT = B1+ B2 BT = B1- B2 (B1>B2)
- If the current isin the opposite direction
BT = B1- B2 (B1>B2) BT = B1+ B2
- Neutral Point is outside the wires
I1 / (x+d) = I2 /d
-

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21
Laws
7- Solenoid
- B = µ N I /L = µ n I
- If turns are tangent
So: L = N*2 r`
- Where :n isthe no. of turns per unit area
and r` is the radius of solenoid
8- Solenoidand Coil
Bwire/Bsolenoid = Lsoleoid / 2 rcoil
9- Force
F = B I L Sin 
- Betweentwo wires
- 
F = L/2d
- Force of two wiresaffectingthird wire
F3= BT I3 L3
- When equilibrium
F = Fg
B I L = mg
10-Torque
 = B I A N sin 
isthe angle betweenthe coil and
the normal plane to the magnetic field

Laws
11- Galvanometersensitivity=
=Galvanometer sensitivity * number
of divisions
12- Ammeter
- Shunt resistance
Rs= Ig Rg / (I- Ig)
- AmmeterResistance
R`=Rg. Rs / (Rg +RS)
- Ammeter Sensitivity
Ig/Is = Rs/ (Rs+ Rg)
13- Voltmeter
- Multiplier Resistance
Rm= (V – Ig Rg) / Ig
- Voltmeter Resistance
R`= Rg+ Rm
14- Ohmmeter
To modulate the Ohmmeter
Imax. = V / (Rg+ Rc+ Rv+r)
Rc: isthe standard resistance
After addingunknown resistance
I = V / (Rg+ Rc+ Rv+Rx+r)

Mr. AhmedHekal
22
Q. Experiments
1- Magnetic fielddue toa current passes througha straight wire
1) Use iron filingssprinkledona horizontal cartoon boards
2) Use a wire to penetrate the board vertically
3) Use battery to pass a current in the wire
4) You will notice that iron filingsare alignedin circlesaround a wire
2- Torque producedin a coil placed ina magnetic field
1) Rectangular coil abcd is placedparallel to a regular magnetic field
2) Ribs ad,bc are parallel to fluxlines,sothere isno force affectingthem
3) Ribs ab,cd are normal to fluxlinesso theyare affectedby two equal and opposite magneticforces
equal B I L
4) As a resultfor these two forces a torque is generated sowe noticedthat coil will rotate in a
directionspecifiedbyusingFlemingLeftHand rule for each force
- Middle fingerrefersto the current
- Index(forefront) fingerrefersto magneticfield
- Thumb will referto the force
5) Torque is specifiedbythisrelation
=B I Lab  Lbc = B I A
WhencoilhasN number of turnsso:
= B I A N sin 
Where istheanglebetweenthecoilandthenormalplan on themagnetic field

Mr. AhmedHekal
23
Chapter 3
R. Definitions
1- Electromagneticinduction
- The phenomenaofproducing inducede.m.fand inducedcurrent in a conductor as a result of
changing the magneticfluxthat’s intersectedby this conductor
2- Lenz’sRule :
- The inducedcurrent is indirectionto oppose the change that is causedby it
3- Faradays Law:
- Inducede.m.fproduced in a coil by electromagneticinductionisdirectlyproportional with the
time in rate of change magnetic fluxintersectedbythe coil and also withnumber of turns
4- Weber:
- It’s the magneticfluxthat is normallypenetratinga coil of 1 turn and whenit’s vanishedgradually
in 1 secondan inducede.m.fof 1v is generatedinthe coil
5- Mutual induction
- The electromagneticeffectbetween2adjacent coilsone ofthem has AC current which affect the
other coil so an inducedcurrent is generatedinthe 2nd
coil to oppose the change happenedto it
6- Mutual induction coefficient:
- It’s the inducede.m.fgeneratedina one coil whenchanging the current in the other coil by rate
of 1 A/S
7- Self-Induction:
- It’s the electromagneticeffecthappenedinthe same coil to oppose the changing of its current
8- self-inductioncoefficient:
- it’s the inducede.m.fgeneratedin the same coil whenits current changes by rate of 1 A.S
9- Henry:
- It’s the mutual inductioncoefficientbetween2coils whencurrent of one ofthem changes by rate
of 1 A/S an inducede.m.f of1v is generatedinthe other coil
Or
- It’s the self-inductioncoefficientofa coil when its current changesin rate of 1 A/S an induced
e.m.fof 1v isgenerated in the same coil
10- Eddy Currents:
- It’s inducedcurrents generatedina metal core as a result of its motioninside a magnetic fieldor if
it’s exposedtoa changing magneticfield
11- Generator(Dynamo)
- A device usedto convert mechanical (motion) energyto electrical energy
12- AC current:
- It’s the electrical currentthat changes its value and directionperiodicallywithtime
- It’s the current that changes its value from 0 to max. and returns back to 0 in half cycle then it
reversesitsdirectionand reachesthe max. value in opposite directionthenreturns to 0 again in
another half cycle

Mr. AhmedHekal
24
13- Effective value of AC current:
- It’s the dc current intensitythat generatesthe same amount ofthermal energythat’s generated
by AC current in the same resistance in the same time
- It’s the dc current intensitythat generatesthe same electrical power that’s generatedby the AC
current in the same resistance
14- Transformer:
- A device usedto step up or step down the alternatingvoltage
15- Transformer Efficiency:
- It’s the ratio of secondary coil power to primary coil power
- It’s the ration of the electrical energyproduced inthe secondary coil to the electrical energy
consumedin the primary coil inthe same time
16- Electrical Motor:
- A device usedto convert electrical energyto mechanical (motion) energy
S. What’s meant by
T. Figures
U. Conditions needed to :
Generatingofdirectlyinducede.m.for directlyinduced
current in the secondarycoil
1- Movingthe primarycoil away fromsecondary
coil
2- Decreasingthe currentinthe primarycoil
3- Openthe circuitof the primarycoil while itis
nearto or inside the secondarycoil
Generatingofinverselyinducede.m.for inversely
inducedcurrent in the secondarycoil
1- Movingthe primarycoil near to or inside
secondarycoil
2- Increasingthe currentinthe primarycoil
3- Close the circuitof the primarycoil while itis
nearto or inside the secondarycoil
GeneratingEddycurrents 1- Moving(rotating) apiece of metal core inside a
constantmagneticfield
2- Exposingapiece of metal core to a variable
magneticfield
Generatinga unifieddirectional currentbut variable in
its value in Dynamo (Generator)
1- Replacingthe two metal ringswithone metal
cylindercrackedintotwohalvesthiscylinder
calledcurrentrectifier
GeneratingDC current in Dynamo (Generator) 1- Usingseveral coils separatedbysmall equal
angles
2- Replacingthe twometal ringswithone metal
cylindercrackedintoseveral partswhere
No.of parts = double No.of coils

Mr. AhmedHekal
25
Improving the transformer efficiency 1- Fabricatingthe coilsfrommetal wireswhichhave
a verysmall resistance toreduce losingof
electrical energyinformof heat
2- Fabricatingthe ironcore frominsulatedslicesof
wroughtironas it has highresistivityusedto
reduce Eddycurrents
3- The wroughtiron ischaracterizedbythe ease of
movingof itsmagneticmoleculesitleads to
reduce losingenergyinformof mechanical
energy
4- The secondarycoil isrolledaroundthe primary
coil to preventleakage ‫تسرب‬ ‫منع‬ of magneticflux
linesof primarycoil awayfromthe secondarycoil
Improving efficiencyofrotatingthe electrical motor
(improvingthe torque of rotation)
1- Usingseveral coils separatedbysmall equal
angles
2- Replacingthe twometal ringswithone metal
cylindercrackedintoseveral partswhere
No.of parts= double No.of coils
V. Applications of electromagnetic Induction
FluorescentLamp Lighting Idea:
Self-inductionincoil
Explanation:
the magneticenergystoredinthe coil will
be transferredtoa vacuum tube which
has inertgas ‫خامل‬ ‫غاز‬ thisenergycauses
collisionsbetweengasatomssotheyare
ionizedandcollidewiththe innersurface
of lampwhichisplatedwithfluorescent
material sovisible lightisreleased
Inductionfurnaces Meltingmetals Idea:
Eddy Currents
Explanation:
When changing magnetic that’s
penetrated by an iron core, induced
currents are produced in this core
which leads to increase its
temperature until melting degree

Mr. AhmedHekal
26
Dynamo (Generator) Convertmechanical (motion)
energyto electrical energy
Idea:
Electromagnetic Induction
Explanation:
When rotating the coil bet. magnet
poles, it intersects variable number of
magnetic lines so an induced e.m.f and
induced AC currents is generated
inside the coil
AC current changes its value and
direction gradually with time
Electrical Transformer 1- step up or step downthe
alternatingvoltage
2- reduce losingenergy
during its transferfrom
generatorsto consumption
placesthrough far
distances
3- In some home devices
Idea:
Mutual Induction bet. 2 coils
Explanation:
When primary coil is connected to AC
source, so the changing in magnetic
field will generate induced e.m.f in the
secondary coil that will be bigger than
or smaller than e.m.f of source
according to number of turns in the
two coils
It enlarges the e.m.f when NP > Ns
It reduces the e.m.f when NP < Ns
Electrical Motor Convertelectrical energyto
mechanical (motion) energy
Idea:
Torque results in passage of electrical
current in a coil inside a magnetic field
Explanation:
When electrical current passes
through a coil, two equal and opposite
directions forces will affect the two
ribs normal to magnetic field so torque
is generated which rotates the coil in
one direction around its axis
W.Usages
Fleming’sRightHand Rule  Specifiesthe directionof inducedcurrentinastraightwire
movesnormallytomagneticfield
Unifieddirectionandvariable intensity
electrical current
Preparingsome metalsbyelectrical analysisof its compounds
Unifieddirectionandunifiedintensity
electrical current
Mobile phoneschargers

Mr. AhmedHekal
27
Deductions
5- Faraday’s Law:
Inducede.m.f producedinacoil by electromagneticinductionisdirectlyproportional withthe time rate
of change inmagneticflux intersectedbythe coil andalsowithnumberof turns
e.m.f m /t and e.m.f N
So
-
m /t:change inmagnetic fluxwith time, N: Number ofturns ofcoil
6- Induced e.m.f in a straight wire
- when a wire of length L moves with velocity V normally on a regular magnetic field of density B
An induced e.m.f is produced in the wire
e.m.finduced =  m /t= B t
Where isthe change inthe area the wire movedthrough
- if it movesa distance of x so L x
e.m.finduced = BL x /t
And x /t=V
e.m.finduced =  BL V
- And if wire makes angle withfieldso
7- E.m.f generatedby mutual induction
- Whencurrent intensitychangesin the primary coil ofrate 1 /t,andinduced(e.m.f)2 isgeneratedin
the secondarycoil which isdirectly proportional with the rate of change inmagnetic flux m /t
- So (e.m.f)2 m /t
- And m /t 1 /t
- Then (e.m.f)2  /t
- Where isthe mutual inductioncoefficient
e.m.finduced =  N m /t
e.m.finduced = BL V sin 
(e.m.f)2 = /t

Mr. AhmedHekal
28
8- E.m.f generatedby self-induction
- Inducede.m.fproduced is directlyproportional with the time rate of change in magneticflux so
e.m.f m /t
- The time rate of changing in magneticflux isdirectly proportional with time range of changing in current
m /t/t
So e.m.f /t
e.m.f = - L /t
- Where Listhe self-inductioncoefficient
9- The inducedinstantaneouse.m.f.generatedinDynamo (Generator)
- Whencoil rotates with a linearvelocityV, where the longitudinal ribs intersectthe magnetic fluxlines,so
if the angle betweenthe directionoflinear velocityand the plane of magneticfieldis theninducede.m.f
generatedin the each rib is
e.m.finduced = B L V sin 
- And e.m.fin one turn (two ribs)
 e.m.finduced = 2 B L V sin 
V = distance / time = (circumference ofcircle for 1 rotation) / time for 1 rotation
V = 2 r/t and frequencyf= 1 / t
 V = 2 rf take 2 f =
 V = r where  isthe angular velocity,r isradius of rotation and V is linearvelocity
- e.m.finduced = 2 B L rsin 
- We can take ( 2 r L = A ) where A is the Area of coil (L is the lengthand 2r is the width)
- e.m.finduced = B A  sin 
- And if coil has number of turns = N
- e.m.finduced = N A B  sin 
- =2 =t =2 ft
= 2 f =t
e.m.finduced = N A B  sin 
e.m.finduced = N A B  sin  t

Mr. AhmedHekal
29
10- Relationsfor the electrical transformer
A. Relationbetweenthe two e.m.f.of the two coils of ideal transformer and theirnumber ofturns
- Whenprimary coil isconnectedto source and circuit of the secondary coil is open,an inducede.m.fis
generatedin the primary coil by self-inductionwhichis
- VP =  NP m /t 1)
- Whenclosingthe secondarycoil circuit, an inducede.m.f.is generatedinthe secondary coil by mutual
inductionsbecause secondary coil intersectsthe fluxlinesof primary coil
- VS =  NS m /t 2)
- We suppose there is no lose in magneticflux then
Divide 2) by 1)
B. Relationbetweenthe two currents of the two coils of ideal transformer and theirnumber ofturns
- We suppose that there is no energylose in the transformer so
Energy consumedin the primary coil in a specifictime = energygeneratedin the secondarycoil in the same
time
VP P t = VS S t
VP P = VS S
- Power input(primary coil) = Poweroutput (secondary coil)
VS / VP = P /S
And VS / VP = NS / NP

This meansthat the current intensityinany coil is inverselyproportional withits number ofturns
VS / VP = NS / NP
P /S = NS / NP

Mr. AhmedHekal
30
C. Factors that depends on
Inducede.m.fgeneratedin a
coil
1- Time inrate ofchange magnetic fluxintersectedbythe coil
2- Numberof turns
Induced e.m.f in a straight
wire
1- Flux Density
2- Lengthof the wire
3- Velocityof wire intersectingtoflux
4- Sinof angle betweenthe wire andthe field
Mutual coefficientbetween
two coils
1- Permeabilitycoefficientof medium
2- Volume of the coils(lengthandare of eachturn)
3- Numberof turnsof the coils
4- The distance betweenthem
Self-inductionofa coil 5- Geometricshape of the coil
6- Numberof turns
7- Lengthof the coil
8- Permeabilitycoefficientof medium
The inducedinstantaneous
e.m.f.generatedin Dynamo
(Generator)
1- Numberof turns
2- MagneticFlux Density
3- Areaof the coil
4- Angularvelocitythatcoil rotateswith,orfrequencyof
rotation
5- Sinof angle betweenthe coil andthe normal plane of
the magneticfield
Consumedenergyin a wire E = I V t
1- Currentintensity
2- Potential difference
3- Time of current passage
e.m.finduced = BL V sin 
e.m.finduced = N A B  sin  t

Mr. AhmedHekal
31
H. Comparisons
AC current DC current
How it’s obtained AC Generator - DC generators
- Batteriesandcells
Characteristics 1- it changesitsvalue anddirectionwith
time
2- can be transferredforlongdistances
withoutlosingenergy,bysteppingupits
voltage usingtransformers
3- can be convertedtoDC current
1- it’sconstantin value anddirection
2- cannot be transferredbecause itlose
itsenergyinformof heat
3- cannot be convertedtoACcurrent
Usage 1- Lighting
2- Heating
3- Operatingmachines
1- Lighting
2- Heating
3- Electrical plating
4- Chargingbatteries
Step-upTransformer Step-downTransformer
Usage Raisingvoltage atgeneratingstations Step-downvoltageatconsumptionplaces
Numberof turns NS > NP NP > NS
Electromotive Force Vs > VP VP > VS
Current intensity IP > IS IS > IP

Mr. AhmedHekal
32
Generator(Dynamo) Motor Transformer
Usage convertmechanical (motion)
energytoelectrical energy
A device usedtoconvert
electrical energyto
mechanical (motion)energy
1- stepup or stepdown
the alternatingvoltage
2- reduce losingenergy
duringitstransferfrom
generatorsto
consumptionplaces
throughfar distances
3- In some home devices
Structure Rectangularcoil of copper
wire rolledaroundcore of
wroughtiron,the coil and
core can be rotatedeasilyand
placedinside amagneticfield
Rectangularcoil of copper
wire rolledaroundcore of
wroughtiron,the coil and
core can be rotatedeasily
and placedinside amagnetic
field
Two coils(primary,secondary)
are rolledaround acore of
wroughtironformedof
insulatedslicestoavoideddy
currents
Operation Terminalsof coil are
connectedtotwo splitrings
rotate withthe coil and each
ringtouchesa fixedgraphite
brushwhichconnect the
inducedcurrentof coil to
external circuit
Terminalsof coil are
connectedtotwo halvesof a
cracked cylinder,eachhalf is
connectedtoa fixed
graphite brush,these
brushesare connectedtoDC
source (Cell)
Primarycoil isconnectedtothe
source of ACcurrent neededto
raise or reduce itsvoltage VP
and secondarycoil is
connectedtothe external
circuitthat needsa specific
value of voltage Vsf

Mr. AhmedHekal
33
D. Experiments
Experiment 1 Faraday’s Experiment
OR
- Experiment to generate induced current in a coil
OR
- Experiment to convert mechanical energy to electrical energy
Stepsand Notices
1- Connectterminalsof coil of copperwire toa sensitive galvanometerwhichhaszeroinmiddle grade
2- Place a magnetinside the coil
Notice
- Galvanometerpointermovesinaspecificdirection
3- Pull the magnetoutside the coil
Notice
- Galvanometerpointermovesinopposite direction
4- Fix the magnetand move the coil nearand far the magnet
Notice
- Galvanometerpointermovesinthe twodirectionsaccordingtomovementof the coil
Deduction- Results
An inducede.m.f andinducedcurrentisgeneratedinacoil as a resultof changingthe magneticflux aroundthis
coil (whenintersectingthe magneticlinesbythe coil),the directionof thisinducedcurrentdependsonthe
movementdirectionof the magnet nearorfar the coil
- Experiment 2 Mutual induction between two coils
Stepsand Notices
1- Connecta coil to a circuit has(battery,keyandrheostat) tobe the primarycoil and connectanothercoil
(secondarycoil) toa galvanometerhaszeroinits middle graduation
2- Close the circuitof the primarycoil andmove it nearto secondarycoil
Notice
3- Move the primarycoil far awaythe secondarycoil
Notice
4- Fix the primarycoil inside the secondarycoil andincrease currentintensityinthe primarycoil
Notice

Mr. AhmedHekal
34
5- Decrease currentintensityinthe primarycoil
Notice
Deduction
We can generate inducede.m.fand inducedcurrent in a secondary coil by effectofa primary coil where
- Reverse inducede.m.fand reverse inducedcurrent : by increasingthe magneticfieldof primarycoil sothe
inducedcurrentinthe secondary coil isindirectiontooppose the change causingit(inthe primarycoil) to
resistthe increase of magneticfield
- Directedinducede.m.fand directedinducedcurrent : by decreasingthe he magneticfieldof primarycoil so
the inducedcurrentinthe secondarycoil isin directiontooppose the change causingit(inthe primarycoil)
to resistthe decrease inthe magneticfield
-
Experiment 3: self- induction
Stepsand Notices
1- connecta coil of a strong magnet(of highnumberof turns) inserieswith(abatteryof 6 volts,key) andin
parallel withalampwhichworksundervoltage of 180 v
2- close the circuit
Notice
- Lamp doesn’twork
3- Openthe circuit
Notice
- An electrical sparkbetweenthe terminalsof the keyandlampis lighteningforaveryshort time
Deduction
1- Whenclosingthe circuita reverse inducede.m.fandareverse inducedcurrentisgeneratedinthe coil which
delaysthe maincurrentto reach itsmax.value anda strong magneticfieldisgeneratedinthe coil because each
turn isconsideredasmall magnet
2- Whenopeningthe circuitthe currentisattenuated ‫يضمحل‬ (decreases) soaninducede.m.f andinducedcurrent
isgeneratedinthe coil byself-inductionthisinducede.m.f ishighbecause numberof turnsisbigand the time
rate of changingthe current isalsobig e.m.f /tandinducedcurrentishighsoit generatesanelectrical
spark

Mr. AhmedHekal
35
E. Some Explanations
1- Specifying the direction of induced current using Lenz’s rule:
- Whenmovingthe northpole of a magnetnearto a coil whichhas a current passesthroughit,the face of the
coil whichisnear to the magnetwill be alsoa northpole andan inducedcurrentwill passthroughthe coil in
the anti-clockwisedirectiontoresistthe change causingitN  N
- Whenmovingthe northpole of a magnetfarfrom a coil whichhas a currentpassesthroughit,the face of
the coil whichisnear to the magnetwill be a southpole andan inducedcurrentwill passthroughthe coil in
a clock-wise directiontoresistthe change causingitN  S
2- Fleming’s Right Hand Rule
Usage:
- Specifyingthe directionof inducedcurrentpassinginastraightwire whichmovesperpendiculartoa
magneticfield
How it’sused
- make the thumb,index (forefront) andmiddle fingersare perpendicular,so
- Thumb: refersto directionof movement
- Index(forefront) : referstomagneticfield
- Middle:referstothe inducedcurrentdirection
3- Eddy currents disadvantages and how to avoid them :
- Disadvantages: large part of electrical currentislostinthermal energy
- How to avoid them? Made the ironcore from thininsulatedslicesof silicon-wroughtironwhichhashigh
resistivity
4- Transformer Operation:
- The primarycoil is connectedtoan AC source (neededtobe transformed) andthe secondarycoil is
connectedtothe external circuitwhichwill use the transformedcurrent
- Whenclosingthe circuitof the secondarycoil andAC currentpassesthroughthe primarycoil,an alternating
magneticfieldisgeneratedand the ironcore will collectandconcentrate itinthe secondarycoil turns
- An inducede.m.f andinducedcurrentisgeneratedinthe secondarycoil whichislargerorlessthanthe
source accordingto the ratiobetweennumberof turnsinboth coils
5- How does Motor works along complete cycle
- In the firsthalf cycle
 Whenthe coil is parallel tothe magneticfielditsterminals(the twohalvesof crackedcylinder) touchthe
graphite brushes,socurrentwill passthroughcoil andtwoopposite forcesactingontworibs sotheyproduce a
torque causingrotationforthe coil
 Torque decreasesgraduallywiththe rotationof the coil until itvanisheswhenitsplaneisperpendicularto
magneticflux,butthe coil continue rotatingaccordingtoinertiauntil itreturnsback toits original position
(parallel tothe field)
- In the secondhalf cycle
 Coil isparallel tothe fieldsotorque will be generatedinthe same directionsocoil will continue rotationinthe
same direction
 Torque decreasesgraduallyuntilitvanishes,butthe coil continue rotatingaccordingto inertiauntil itreturns
back to itsoriginal position(parallel tothe field) thiswill happeneachcycle

Mr. AhmedHekal
36
Current flowsin the same directionin two parallel wires
F. What Happens in these cases
1- Movinga coil has electrical current near to another coil connectedto a galvanometer
- A reverse inducede.m.f isgeneratedinthe secondcoil bythe mutual induction
2- Openan electrical circuit containsa magnetic coil connectedin parallel with a battery
- An electrical sparkhappensacrossthe keyterminals
3- Openthe primary coil circuit whenit’s inside the secondary coil
- A directinducede.m.f.isgeneratesinthe secondarycoil toresistthe shortage inthe primarycoil current
4- Increasingvalue of current in the primary coil placed inside a secondarycoil which isconnectedto a
galvanometer
- Galvanometerpointerdeflectsinone directionsbecauseof generatingreverseinducede.m.f.onthe
secondarycoil bymutual induction
5- A high frequencycurrent passesthrough a coil rolledaround a piece of metal
- Its temperature will increase accordingtoeddycurrents
6- Growth of current ina coil rolledon a wrought iron core inside it to the time of current growth
- The time of currentgrowthwill increase inthe coil asa resultof generatingabig reverse inducede.m.f
because permeabilitycoefficientof wroughtironishighandself-inductioncoefficientof itwill be alsohigh
7- Wiresof electrical resistance are doubledrolled
- Self-inductionwill vanishsothe main currentwill be onlyaffectedbyOhmicresistance,becausethe
magneticfieldproducedbyonturnwill cancel the magneticfieldproducedbythe nextturn
8- Increasingnumber of turns of dynamo to double and number of cyclesin 1 secondto double
- Inducede.m.f.willincrease4-times
9- Increasingnumber of turns of Dynamo coil to double and decreasingits angular velocity4 times
- Instantaneousinducede.m.f.will be reducedtohalf
10- Replacingthe two splitrings of Dynamo withcylindercracked to 2 halves
- AC currentwill be convertedtounifieddirectioncurrentof alternatingvalues
11- Dividingthe cracked cylinderin Dynamo to numberof piecesequal double numberof coils
- Thiswill completelyconvertthe ACcurrentto DC current whichisunifiedindirection andconstantvalue
12- Connectingprimary coil ina transformer with a cell (battery)
- The magneticflux resultingonitwill be constantso,no mutual inductionwill happenbetweenthe coilsand
the transformerwill notwork
13- Openingcircuit of secondarycoil in the transformer and connectingits primary coil withan AC source
- A reverse inducede.m.f.isgeneratedinthe primarycoil,it’sequaltoe.m.f. source so currentis approximately
zero
14- Transferring AC current for longdistances withoutraising voltage before transfer
- A lotof energyislostinthe wiresinformof thermal energy

Mr. AhmedHekal
37
Give reason
1- An inducede.m.f.is generatedina wire moves normallyon magnetic fluxlines
- Because the magneticflux affectsthe free electronsof the movingwire sothese electronswill releasefrom
one terminal of the wire(+ve terminal) toanotherterminal (-ve terminal)soa potential difference is
producedbetweenthe twoterminalsof wire
2- Inducede.m.fmay not be generatedina movingwire ina magnetic field
- Because the directionof wire movementisparallel tothe magneticflux lines,soaccordingto law
e.m.finduced = BLV sin angle betweenfieldandwire will be 0ande.m.f.will be zero
3- Inducede.m.f.produced in a coil ishigher ifthe core of the coil is made of wrought iron
- As the wroughtironhas a higherpermeabilitycoefficientandthiswillincrease concentrationof magnetic
flux linesthatare intersectedbythe coil whichleadstoincrease the inducede.m.f
4- Wiresof standard resistance are double rolled
- To avoidthe self-inductionbecause the magneticfieldproducedbyaturn is cancelledbythe magneticfield
producedbythe nextturn,so current will onlyaffectedbythe Ohmicresistance
5- A piece of wrought iron doesn’tmagnetizedifit’s caught by double rolledwire has a current passing through
it
- Because the directionof currentinthe firstwire opposesthe directionof currentinthe secondwire so
magneticfieldgeneratedbyone of themwill cancel the other,soresultantmagneticfieldequals zero
6- The directe.m.f. inducedina coil by self-inductionalwaysbiggerthan the reverse inducede.m.f
- Because the collapse rate of currentis alwaysbiggerthanthe growthrate of current
7- Current intensitydoesn’treach maximumvalue incoil in the moment closingthe circuit,and doesn’tvanish
also in the moment of openingthe circuit, it takes time
- Thishappensbecause of generatingareverse e.m.f.inthe momentof closingandopeningthe circuitthis
reverse e.m.f will opposethe maincurrentwhetherwhenitincreasesorwhenitdecreases
8- Current is growingin a straight wire faster than a coil
- Because the magneticfieldproducedaroundthe wire isnotintersectedbythe wire itself sothere isno
reverse inducedcurrentproducedinit,butincase of coil the magneticfieldproducedincoil isintersected
by the coil itself soa reverse inducede.m.f isgeneratedincoil byself-inductionandalsoareverse induced
currentso it resistthe maincurrentinthe coil
9- Vanishingthe inducedcurrent ina straight wire isfaster than it ina coil of air core whichis faster than it in a
coil rolledon an iron core
- There isno reverse inducede.m.f generatesinthe wire becausethe wire doesn’tintersectitsmagneticfield
- In the coil of air core, there isa reverse inducedcurrentisgeneratedincoil tooppose the shortage in
currentits value ishigh
- In the coil of iron core,the reverse inducedcurrentishigherbecause the magneticfield isbiggerthanitin
the coil of air core according to the permeabilitycoefficient
10- Whenopeningcircuit ofelectrical magnet there is a spark appears in the positionofcutting the current
- Because the rate of vanishingthe currentisveryhighsothe rate of changingthe magneticflux isveryhigh
whichleadstogeneratinghighinducedcurrentinthe same directionof the maincurrenttooppose its
shortage

Mr. AhmedHekal
38
11- WhenAC current passesin a coil rolledon a piece ofmetal its temperature increases
- Because of the eddycurrentsproducedinit whichleadstomeltthe metal
12- Eddy currents don’t producedin a fixedpiece ofmetal otherwise the magneticfieldaround it is variable
- As inthe variable magneticfield,the magneticflux linesintersectedbythe metal is changingsoeddy
currentsare generated
13- Temperature of an iron cylinderis raised ifit’s rolledby a coil connectedto AC source
- As the ACcurrent changesitsvalue anddirectionperiodically,sothe magneticfieldresultingbyitisalso
changing,soeddycurrentsare producedinthe cylinder
14- Inducede.m.fin the Dynamo (generator) coil ismax. value whenits plane is parallel to the magnetic field
- Accordingto thisrelation
e.m.finduced = N A B  sin 
isthe angle betweenthe coil andthe normal plane tothe magneticfieldsointhiscase  equals90 so
Value of e.m.f will be maximum= N A B 
15- E.m.f (average) in Dynamo through ¼ cycle = E.m.f (average) in Dynamo through ½ cycle
In ¼ cycle:
m =BA , t= ¼ T
e.m.f = 4 N B A F
In ½ cycle:
m =2 BA , t= ½ T
e.m.f = 4 N B A F
16- E.m.f (average) in Dynamo coil through a complete cycle = 0
- As the average value of e.m.f inone direction( ½cycle) isequal toaverage value of e.m.f inthe opposite
direction(inthe second½cycle) sothe resultantequal zero
17- the cracked cylinderin dynamo produce a unifieddirectioncurrent
- whencoil isrotatedinhalf cycle theirterminalsrotate withitandthe current will passinthe same direction
inthe external circuitwhentouchingthe graphite brushes
18- Terminalsof Dynamo coilsare connectedto number of cracked pieceswhichis double number ofcoils
- To guarantee thatbrushesare alwaystouchedtothe coil whichisparallel tothe magneticfieldtofinally
produce a current of constantvalue DC current
19- The core of transformer ismade of slicesof wrought iron whichare insulated
- Permeabilitycoefficientof wroughtironishighsoit helpsinconcentratingthe magneticfluxlines
- Resistivityof wroughtironishighandwhenit’smade of insulatedslices,thiswill increaseitsresistance to
resistthe eddycurrentsandreduce the energylossinformof heat
20- In Ammeteror Galvanometerthe iron core isNOT dividedintoinsulatedslices
- Because AmmeterandGalvanometerusedtomeasure dccurrents,sothere is noeddycurrentsare
producedinthe core , no change ishappenedinthe magneticflux
21- Transformer coilsare made of copper wires
- Because itsresistivityisverylowsothe coilsresistance isverylow whichavoid energylossinformof heat

Mr. AhmedHekal
39
22- There is no transformer with 100% efficiency
- Because there isenergylossinformsof:
A. Loss inmagneticflux linesfromprimarycoil tosecondarycoil
B. Loss inheatthroughthe wires
C. Mechanical energyaccordingtothe movementof the ironcore magneticmolecules
23- Transformers are not used instep-upor step-downa DC e.m.f.
Or Transformers don’t work if the primary coil is connectedto DC source
- Because the magneticflux producedby the DCcurrent isconstantso no inducede.m.f isproducedinthe
secondarycoil bymutual induction(there isnomutual inductionhappens)
24- Transformer doesn’tconsume power whenopeningcircuit of secondary coil,although its primary coil is
connectedto electrical source
- Whenopeningthe secondarycoil circuitareverse inducede.m.f.isproducedinthe primarycoil (byself -
induction) andit’sequal toe.m.f (source) sothere isnotpotential difference andthere isnocurrentpasses
inthe primarycoil andno powerisconsumed
25- Transformer works when closingits secondary coil circuit
- Whenclosingsecondarycoil circuit,currentpassesthroughit,the magneticflux resultingbyitwill be
intersectedbythe primarycoil anditvanishesthe reverse induced currentinit,sothe currentof source will
pass throughthe primarycoil and continue working
26- Electrical energyistransferred from generatorsstations to consumers undera highvoltage and low current
- To reduce the consumedenergyinthe wiresbecause powerisdirectlyproportionalwithsquare of current
intensity
27- Use of step-uptransformersat the generatingstations
- Because the step-uptransformersraise the voltage atthe generatingstationswhichleadstoreducingthe
currentintensityinthe transformerandthisisuseful inavoidinglossinenergyconsumedintransferring
wires
28- The step-uptransformersare current step-downand vice versa
- Because the powerisconstantand thismakesthe voltage inverselyproportionalwithcurrent
V = Pw/ I
29- Motor continue in rotation although it passesthrough the positionwhich is normal to magnetic field
Or
The coil of electrical motor doesn’tstop when the graphite brushestouchesthe insulationpart of the cracked
cylinderhalves
- Because the inertiamakesthe coil tocontinue rotatingandthe twohalvesexchange theirpositionsandalso
currentexchange itsdirectionsotorque will be inthe same direction
30- To increase Motor power,we use several coils separatedby small angles
- To increase the torque byguarantee thatthe coil is alwaysparallel tothe magneticflux sothe torque is
alwaysmaximumvalue ancoilsrotateswithhighangularvelocity,thisleadstoenhance the motorefficiency

Mr. AhmedHekal
40
Laws
E.m.f in Dynamo (Generator)
A. Instantaneous e.m.f.
e.m.f= e.m.fmax. Sin
e.m.f= N A B  sin  t
V/r = t=2T=2  f
inside(sin) =180
outside (sin) =22/7
B. Maximum e.m.f.
E.m.f max. = N A B 
C. Effective e.m.f.
E.m.f effective = e.m.fmax. Sin 45
D. Average e.m.f.
In ¼ cycle and ½ cycle
e.m.f= 4 N B A F
In ¾ cycle
e.m.f= 4/3 N B A F
In 1 cycle
e.m.f= 0
Instantaneous induced current
Instantaneous= IMAX sin 
Numberof reachingthe ACcurrent to maximumvalue
in1 second=2 f
Numberof reachingthe ACcurrent to zero in1 second
= 2 f +1
Laws
Faraday’s Law
- If area changed
e.m.finduced =  N B /t
- If flux density changed
e.m.finduced =  N   B / t
- If coil rotates
A. ¼ cycle (90)
e.m.finduced =  N  B / t
B. ½ cycle (180)
e.m.finduced =  2 N  B / t
C. 1 complete cycle
e.m.finduced = because m=0
E.m.f. induced in a coil
Self-induction:
e.m.f = - L  /t =  N m/ t
(Self-inductioncoefficient)
L = 
/L (length)
Mutual Induction
(e.m.f) 2 = /t =  N2 (m )2/t

Mr. AhmedHekal
41
Laws
Electrical Transformer
A. Ideal Transformer:
VP P t= VS S t
VS / VP = P /S
P /S = NS / NP
In case of 2 secondarycoils:
(Pw) p = (Pw) s1 +(Pw) s2
B. Non-Ideal Transformer:
ɳ = (Vs Is/VP Ip) * 100
ɳ = (Vs Np/Vp Ns) * 100
(Pw) p > (Pw) s
Pw (consumer) =Pw (station) - Pw (loss in wires)
1- Power at generating stations= I V
2- Power consumed in wires = I2
R
3- Shortage in Volt= I R
Laws
Electrical Motor

A. Current Intensity
- Before operation
I = e.m.f.(source)/ R
- During Running
I = e.m.f.(motor)/ R
B. Electromotive force
E.m.f motor = e.m.fsource– e.m.freverse

Mr. AhmedHekal
42
Chapter 4
X. Definitions
1- AC current
- It’sthe current that changesitsmagnitude graduallyfrom0to maximumafterquartercycle and
changesitsdirectionafterhalf cycle
2- AC frequency:
- It’sthe numberof complete cyclesof ACcurrentin1 second
3- PeriodicTime of AC:
- It’sthe time takenbyACcurrent to make a complete cycle
4- Hot Wire Ammeter:
- Device usedtomeasure ACor DC currentand it depends onthe expansionof awire made of alloyof
platinumandiridiumbythe thermal effectof electrical current
5- Inductive Reactance:
- It’sthe resistance of acoil causedby itsself-inductionXl
6- Electrical Capacitor:
- Two parallel insulatedmetal plates,usedtostore electrical energyinformof electrical field
7- Capacitance:
- The ratio betweenthe charge placedonone plate andthe potential difference betweenthe twoplates
8- Capacitive Reactance:
- It’sthe resistance of acapacitor causedby itscapacitance Xc
9- Farad:
- It’sthe capacitance of a capacitor that ,if it’schargedby a charge of 1 coulombthe potential difference
betweenitsplatesis1volt
10- Impedance:
- It’sthe equivalentforthe ohmicresistance,inductive reactance andthe capacitive reactance foranAC
circuit
11- OscillatorCircuit:
- It’san electrical circuitinwhichthere isanexchange forenergystoredininductive coil informof
magneticfieldandthe energystoredinacapacitor informof electrical field
12- Resonant Circuit:
- It’san oscillatorcircuitcontainsa resistance,inductivecoil,capacitorandACsource and itonlyallows
passage of AC currenthas frequencyequal orveryclose toitsfrequency

Mr. AhmedHekal
43
Y. What’s meant by
1- AC frequencyis50 Hz
- it meansthatnumberof complete cyclesthatmade byAC currentin one secondis50 cycles
2- PeriodicTime of AC current is 0.02 second
- Thismeansthat time takenbyAC currentto complete 1 cycle is0.02 second
3- Inductive reactance of coil is 100 ohm
- It meansthat the resistance forthe coil resultinginitsself-inductionis100 ohm
4- Capacitance for a capacitor is 5 micro-farad
- It meansthat the quantityof chargesplacedonone plate is5 *10-6
coulombswhenthe potential
difference betweenisplatesis1volt
5- Capacitive reactance is 100 ohm
- It meansthat the resistance of the capacitoraccordingto its capacityis100 ohm
6- Impedance is 50 ohm
- it meansthatthe equivalentresistance for(ohmicresistance.Inductive reactance,capacitive reactance)
is50 ohm
7- Phase angle for a circuit has inductive coil and resistance is 45o
- It meansthat the total voltage leadscurrentbyangle 45o
Tan  = VL/VR = XL/R = 1 , XL =R, VL=VR
8- Phase angle for a circuit has capacitor and resistance is 45o
- It meansthat the total voltage lagscurrentby angle 45o
- Tan  = - VC/VR = -XC/R = - 1 , XC =R, VC=VR
9- Resonant circuit frequencyis104
Hz
- It meansthat the oscillatorcircuitfrequencyequalssource frequency= 104
Hz and it onlyallowsthe
currentof thisfrequencytopassthroughit andinductive reactance equalsthe capacitivereactance only
at frequency 104
Hz
Z. Devices
1- Hot Wire Ammeter Measuresthe effectivevalue of AC
currentand it alsomeasuresDC
current
Idea:
Thermal effectof electrical current
Explanation:
Whencurrent(DC or AC) passesthrough
an ohmicresistance itgeneratesa
quantityof heatwhichdependsonthe
effectivevalue of thiscurrent
2- Antennas Radiochannels Idea:
Resonance circuit
Explanation:
When we change the channel on radio
device, the frequency of the resonance
circuit changes to a specific value
which equals the frequency of the
desired channel current (because
electromagnetic wave of the channel is
converted to AC current)

Mr. AhmedHekal
44
AA. Usages
1- Platinum-IridiumWire  It’sheatedupand expandswhenelectrical currentpasses
throughit so we can measure the effective value of current
2- Silk thread inHot-Wire Ammeter  It is pulledbythe platinum-iridiumwiresothe rollerwill
rotatesand pointerwill move andstopsonthe effective value
of the current
3- The board on which the platinum
wire is tensed
 Get ridof zeroerror
4- Rollerin Hot-Wire Ammeter  It rotateswhenit’spulledbythe silkthread,sopointerwill
deviate until itreachesthe effective value
5- Coil in Hot-Wire Ammeter  It pullsthe silkwire torotate the rollerandmove the pointer
to the effective value of the current
6- The resistance connectedto the
platinum-Iridiumwire
 It dividesthe total currenttoallow a suitable currenttopass
throughthe wire
7- The variable-capacitycapacitor in
RLC circuit whichworks as resonant
circuit
 whenchangingthe capacitorcapacitance its capacitive
reactance will change until itequalsthe inductive reactance of
coil whichmeansthatthe impedance will equalthe ohmic
resistance only(minimumimpedance) andcurrentwill be
maximumvalue (it’susedinreceiverdevices)
8- Resonance Circuit  Usedin receiverstoreceive aspecificwave
BB. Figures
CC. Deductions
1- Frequencyof current in resonantcircuit
- In resonantcircuit,currentwill be maximumwheninductivereactance equal capacitivereactance
- XL= XC
- 2fL= 1 / 2fC
- f 2
=1/4
LC
f = 1/2√LC

Mr. AhmedHekal
45
DD. Factors dependent on :
4- Angle of deviationinHot-Wire Ammeter - Square of current intensity 
5- Inductive reactance of a coil XL= 2 f L
4- Self-inductioncoefficient
5- Frequencyof current
6- Capacitive Reactance of a capacitor XC= 1/ 2 f C
1- Capacityof capacitor
2- Frequencyof current
7- Total impedance Z = √ R2
+(XL-XC)2
1- OhmicResistance
2- Inductive reactance
3- Capacitive reactance
8- Frequencyof resonantcircuit f = 1/2√LC
1- Square root of capacity
2- Square root of self-induction coefficient
I. Comparisons
Hot-Wire Ammeter MovingCoil Ammeter
Idea Of Work Expansionresultingonthermal effectof
electrical current
Torque resultingonthe magneticeffectof
electrical current
Usage Measuringintensityof DCcurrentand effective
value of AC current
MeasuringDC currentonly
Scale (graduation) Non-regular Regular
Effectof room
temperature
It’saffectedbyroomtemperature It doesn’taffected
Pointermove It movesslowlywhenpassingthe currentorcut-
off
It moves fasterwhenpassingthe currentor
cut-off the current

Mr. AhmedHekal
46
Circuit ofR and XL (RL)
circuit
Circuit ofR and XC
(RC) circuit
Circuit ofXL, XC,R
(RCL) Circuit
Resonant Circuit
Circuit
Total Voltage V = √ VR
2
+ VL
2
V = √ VR
2
+ VC
2
V = √ VR
2
+ (VL- VC)2
VL=VC
V = VR
Impedance Z = √ R2
+XL
2
Z = √ R2
+XC
2
Z = √ R2
+(Xl-XC)2
XL= XC
Z = R
Phase Angle Tan =VL/VR =XL/XR
 is Positive
Tan =- VC/VR
= -XC/XR
 is Negative
Tan =(VL-VC) /VR
=(XL-XC)/R
When XL> XC  is Positive
When XL< XC  is Negative
Tan =0


Mr. AhmedHekal
47
J. Important Explanations
1- Disadvantages of Hot-Wire Ammeter
A. Its pointermovesslowlyuntilstopateffectivevalue andalsoit returnsbackto zeroslowly
B. Platinum-Iridiumwireisaffectedbythe roomtemperature whichcausessome errorsinreadings
calledzeroerror
2- AC circuit contains non-inductive resistance
A. V = Vmax. sinω t
B. I = V / R So I = Imax. sinω t
C. From A. and B. we foundthat inthiscircuit voltage andcurrent are in phase itmeanstheyreach 0
togetherandmaximumvalue together
3- AC circuit contains an inductive (non-resistive) coil
A. Whenclosingthe circuit,voltage betweenterminalsof coil reaches Vmax. ,currentgrowsgraduallyand
voltage decreasesgradually accordingtothe reverse inducede.m.f. until currentreachesmaximumat
the momentinwhichvoltage iszero
B. Inducedcurrentisgeneratedinthe coil andit resistthe change causingitthisis the cause of lag1-ging
currentin reachingmaximumwithvoltage
C. Vinduced = - L /t
D. Currentlags ‫عن‬ ‫يتأخر‬ voltage by90o
or ¼ cycle
E. XL = 2πfL

Mr. AhmedHekal
48
4- AC circuit contains a capacitor
A. In 1st
quarter,current reachesmaximum,where –ve chargesare transferredonplate A and itspotential is
decreased,these chargesaffectplate Bandrepulsedwith –ve chargesonplate B so plate B has only+ve charges,at
thismomentcapacitoris chargedand currentstops(equal 0)because voltage oncapacitor= voltage of source =
maximumvalue
B. In 2nd
quarter e.m.f.source decreasessopotential difference acrossthe capacitorishigherthansource so it discharges
insource,current will reachmaximumandvoltage of capacitorreaches0
C. In 3rd
quarter,capacitorwill charge againbut inthe opposite direction(plate Bis –ve and Plate A is+ve) until its
voltage reachese.m.f.source socurrentstops=0 and voltage ismaximum
D. In 4th
quarter,capacitordischargesanditsvoltage will be 0and current ismaximum
I = C V/t
E. Currentleads‫يسبق‬ voltage by90o or ¼ cycle
F. XC= 1 / 2πfC
5- RL Circuit
A. In the coil V leads by 90o
or ¼ cycle
B. In Ohmicresistance V and  are inphase
C. Currentisthe same because theyare connectedinseries
D. Voltage of coil leadsvoltage of resistanceby90o
E. V = √ VR
2
+ VL
2
F. Z = √ R2
+XL
2
6- RC Circuit
A. In capacitorV lags  by 90o
or ¼ cycle
C. Currentisthe same because theyare connectedinseries
D. Voltage of resistance leadsvoltage of capacitorby90o
E. V = √ VR
2
+ VC
2
F. Z = √ R2
+XC
2
7- RCL circuit
A. In the coil V leads by 90o
or ¼ cycle
C. In capacitorV lags  by 90o
or ¼ cycle
D. Currentisthe same because theyare connectedinseries
E. V = √ VR
2
+ (VL- VC)2
F. Z = √ R2
+(Xl-XC)2

Mr. AhmedHekal
49
8- OscillatorCircuit
A. Structure
I. Inductive coil of averysmall resistance
II. Capacitor
III. Batteryand all are connectedinparallel
B. Operation
I. Whenclose key“A”
- Currentpassesinthe capacitor
- One plate (connectedtopositivepole) ischargedwith
positive charge,andthe otherplate ischargedwith
negative charge
- Currentstopswhenpotential differenceacrosscapacitor
isequal to VBattery1
- Energyis storedincapacitorin formof electrical field
- Openkey“A” ,nowcapacitoris charged
II. Whenclosingkey“B”
- Capacitordischargesthroughcoil andcurrent flowsfromthe positiveplate tonegative plate,potential
difference betweenplateswill decrease until itvanishesand the electrical field disappears
- The coil stores energyin form of magneticfieldresultingon the current passes through it
- In the beginning,the potentialdifference throughcapacitorishighsocurrentpassingthroughthe coil is
high,afteran interval of time P.D.incapacitordecreasesandcurrentalsodecreasesinthe coil
- Thisshortage incurrent leadstogenerate a directinducede.m.f inthe coil byitsself-induction,this
inducede.m.f.attractspositivechargesfromthe positiveplate tonegativeplate,sopositive platewill be
chargedwithnegative chargesandnegative plate will be chargedwithpositivecharges
- Capacitoris nowchargedin opposite directionand thismeansthat magneticenergyis convertedagain
to electrical energy
- Capacitorwill startdischarginginopposite directioninthe coil andelectrical fieldisconvertedtoa
magneticfieldandsoon, whichcausesmany oscillationsinthe circuitaccordingtothisexchange
9- Relationbetweenfrequencyand(XL XC RZ)
- Impedance Zdecreases until itreachesminimumZ=R whenXL=XC and itincreaseswithfrequency
- Currentincreaseswithfrequencyuntil itreachesmaximumwhenXL=XC thenitdecreaseswithfrequency
increase,thisisbecause currentisinverselyproportionalwithimpedance
- Circuitisresonant(inresonance state) whenXL=XC
f = 1/2√LC

Mr. AhmedHekal
50
G. What Happens in these cases
1- Flowof AC current in an Ohm resistance to its temperature
- Its temperature increasesbecauseof energylossinformof thermal energy
2- AC or DC current pass through the Hot-Wire Ammeter
- Thermal energyisgenerated inthe Platinum-Iridiumwire soitexpandsandallowstothe silkthreadto
make the rollerto rotate and pointertodeviate soitgivesthe value of effectivecurrent
3- Cutting-offcurrentpasses in Hot-Wire Ammeter
- The wire iscooledand attracts the silkthreadwhichrotate the rollertoreturnsthe pointerto zero
4- The silkthread is cut
- The expansionof platinum-iridiumwirewillnotaffectthe rollerorthe pointerso,Ammeterwill notgive
a reading
5- Passage of ACcurrent inan inductive coil to the phase angle betweencurrentand voltage
- Voltage leadscurrentby90O
or ¼ cycle
6- if frequencyofAC current is highlyincreased
- the inductive reactance will increasesalsobyrelation XL=2 fL until itpreventsthe flow of current
7- Capacitor is connectedto DC source
- Currentflowsinthe circuitina small interval of time thenitdecreasesuntil itvanishedwhenP.D.of
capacitor = VBattery
8- Passage of ACcurrent ina circuit contains capacitor to the phase angle betweencurrent and volt
- Currentleadsvoltage by90O
or ¼ cycle
9- High Increase of frequencyof electrical current passesthrough a capacitor
- Capacitive reactance will decrease accordingtorelation XC=1/ 2 fC and circuitis consideredshort-
circuit
10- Connectinga resistance with an inductive coil and AC source to the phase angle betweencurrentand
total voltage
- Voltage leadscurrentbyangle where tan  = VL /VR = XL /R
11- Connectinga resistance with an inductive coil and AC source to the phase angle betweencurrentand
total voltage
- Currentleadsvoltage byangle where tan  = - VC/VR = - XC / R
12- Connectinga charged capacitor to inductive coil
- Capacitordischargesinthe coil and an instantaneouscurrentflows,soareverse inducede.m.f.is
generatedinthe coil inopposite directionof the maincurrent,thisoperationisreversedseveral times
causingoscillationssoit’scalledOscillatorcircuit
13- The inductive reactance equalsthe capacitive reactance in RCL circuit
- The circuit isresonant,andtotal impedance isminimum(Z=R) andcurrent ismaximum,voltage and
currentare in phase =0 and frequency f = 1/2√LC
14- ReplacingAC source with DC source of the same effective value inRL circuit to the current intensity
- Currentwill increase becausetotal impedance isdecreasedwhenACsource isreplacedbyDCsource
- In case of AC: currenthas frequencyf andcoil has inductive reactance XL=2fLsoZ ishigh Z = √ R2
+XL
2
- In case of DC : frequency=0 and XL=0 so Z is low Z = R only,current will increase

Mr. AhmedHekal
51
K. Give reason
31- Hot-Wire Ammeterisused to measure AC and DC currents
- As itgenerallydependsonthe thermal effectof anyelectrical current,itmeasuresthe effective valueof
currentaccording to the expansionof Iridium-Platinumwire byheatcausedbycurrentpassage inthe
wire.
32- An alloy of Platinum-Iridiumusedinthe Hot-Wire Ammeter
- As itexpandseasilybyheatwhencurrentflowsthroughit
33- Platinum-Iridiumwire isconnectedin parallel with a resistance R
- To work as a shuntresistance (itdividesthe current) toallow asuitable partof currentto pass through
the wire
34- Hot-Wire Ammeterisconnectedin seriesin the electrical circuit
- To measure the currentneededtobe measurednotpartof it
35- Hot-Wire Ammeterdivisionsare not regular
- Because the quantityof heatgeneratedinthe wire isdirectlyproportional withthe square of effective
currentnot the current only
36- There is an error in Hot-Wire Ammetercalledzeroerror
- Because the platinum-iridiumwire isaffectedbythe roomtemperature
37- The platinum-iridiumwire istensedon a board made of a material has the same expansioncoefficientof
the wire and insulatedfrom it
- To overcome the errorcausedby affectingthe wire bythe roomtemperature
38- Current and Volt are in phase in ohmic resistance circuit
- Because V = Vmax sint and  = V / R ,  = (Vmax sint) / R
So  = max sint
Theyhave the same phase angle so theyvanishestogetherandreachesmaximumtogether
39- At very high frequenciesACcurrentmay not pass through the inductive coil
- Because inductive reactance isveryhigh XL= 2 f L and circuitisconsideredopencircuit
40- Passage of ACin an inductive coil (non-resistive) don’tlossenergy
- Because the onlyexistentresistance isthe inductivereactance thatresultsingeneratingreversed
inducede.m.f inthe coil soitmaintainsthe energyinformof magneticfield
41- Whenincreasingnumberof turns for a coil the inductive reactance increasesifAC current of a constant
frequencypasses
- Because inductive reactance XL isdirectlyproportional withself-inductioncoefficientLwhenthe
frequencyisconstant XL=2 f L
Andself- inductionisdirectlyproportional withsquare of numberof turnsL = µ N2
A / L (length)
42- Inductive reactance increaseswhenput wrought iron core inside the coil and passingthe same AC current
- Because inductive reactance XL isdirectlyproportional withself-inductioncoefficient Lwhenthe
frequencyisconstant XL= 2 f L
- Andself- inductionisdirectlyproportional withpermeabilitycoefficient L= µ N2
A / L (length)
- Andpermeabilitycoefficient of wroughtironishigherthanAir
43- Whenconnectinggroup of inductive coilsin parallel,the resultant inductive reactance islower than the
smallestone
- Because the reciprocal of the resultantinductivereactance isequal tothe sumof the reciprocal of all of
them(1/XL = 1/XL1+1/XL2 +1/XL3 +…….)

Mr. AhmedHekal
52
44- Whenconnectinga capacitor with DC source, current flowsin a short time then it vanishes
- Currentpassesfrombatteryto capacitor and+ve chargesare placedonone plate and negative charges
placedonthe otherplate soa reverse potential difference isgeneratedonthe capacitorandincreases
withtime andcurrent decreasesuntil Vcapacitor = VBattery , at thismomentcurrentstopsbecause there isno
difference inpotential betweenbatteryandcapacitor
45- The capacitive reactance doesn’t cause loss in energy
- Because capacitorstore electrical energyinformof electrical field
46- Whenan AC current of high frequencypassesthrough a capacitor, the circuit isconsideredclosedcircuit
- Because XC= 1/ 2 fC , thismeansthatthe capacitive reactance isinverselyproportionalwithfrequency
so at highfrequencythe capacitive reactance isverylow andcurrentpassesina closedcircuit(no
resistance)
47- Whenconnectinga group ofcapacitors in parallel the capacitive reactance for the group is lessthan the
lowestcapacitive reactance for each capacitor
Because the total capacitance of a group connectedinparallel equalsthe summationof them
(CT = C1 + C2+ C3+ ….) so itwill be higherthananyof them, andthe capacitive reactance isinversely
proportional withthe capacitance XC= 1/ 2 f C
48- The inductive reactance ofa coil passesthrough it a dc current equals0
- Because DC currentis unifiedinmagnitude anddirectionsoitsfrequencyequals0andinductive
reactance XL= 2 f L so itwill be XL= 0
49- It’s impossible toproduce an inductive coil of resistance zero
- Because anycoil made of conductingwireswhichshouldhave alittle value of resistance accordingtoits
resistivity
50- If an inductive coil has ohmic resistance isconnectedto AC source,the total voltage leads current by angle
where0<  < 90
- Because currentand voltare in phase inthe ohmc resistance andvoltleadscurrentbyangle of 900
in
the coil so the total voltage leadscurrentbyangle tan=VL / VR
51- If a capacitor is connectedto an ohmicresistance and AC source of high frequencyinseries,the current
will leadthe total voltage by angle where 0<  < 90
- Because voltage andcurrentare inphase in the resistance andcurrentleadsvoltage byangle of 900
in
the capacitor so the currentwill leadthe total voltage byangle  where tan =- VC / VR
52- In Oscillator circuit,the process ofcharging/discharging stops after an interval oftime
- Because a part of electrical energyisconvertedgraduallytothermal energyconsumedinwires
accordingto its resistance soaftertime ,AC currentdecreasesandP.D.across capacitoralsodecreases
until itvanishes
53- To continue incharging/discharging process we shouldfeedthe capacitor with additional charges after
intervalsof time
- To overcome energyloss asheataccordingto wires resistance
54- Ohmic resistance has a constant value regardlessthe value of frequencybut inductive reactance and
capacitive reactance changeswith frequency
- Because ohmicresistance doesn’tdependonthe frequencybutinductive reactance andcapacitive
reactance dependonfrequencyaccordingtothese relations
- XC= 1/ 2 fC
- XL= 2 f L

Mr. AhmedHekal
53
-
55- Capacitor allows the AC current to pass through its circuit
- WhenAC currentpasses,the capacitoris chargedinthe 1st
quarteruntil voltage of itisequal to VSource
thene.m.f of source decreasesinthe 2nd
quarterso VCapacitor > VSource so capacitorwill discharge inthe
source but source continue indecreasingitse.m.f until itreacheszeroatthe momentinwhichVSource
alsoreachedzeroand thisprocessisrepeatedinthe 3rd
and4th
quarterbut inthe opposite direction
56- In resonance state current intensityis maximum
OR
In resonance state the current and total voltage are in phase
- Because the inductive reactance isequal tocapacitive reactance sothe total impedance Z= R so the
currentis maximumvalue becauseresistance isminimumvalue andthe voltage andcurrentare insame
phase accordingto these relations
- V = Vmax sint and  = V / R ,  = (Vmax sint) / R
So  = max sint
57- The average ofelectrical power consumedin a complete cycle of AC current in an inductive coil is 0
- Because coil storesenergyinthe 1st
quarterin formof magneticfield anddischargesitinthe 2nd
quarterand repeatsthisinthe secondhalf(3rd
and 4th
quarters) so after1 cycle the total power=0
58- The average of electrical power consumedin a complete cycle of AC current in a resistance isnot 0
- Because currentneedsworktotransferchargesin bothdirectionsandthisworkdoesn’tdependonthe
directionof the current
59- We don’t sum the voltages in RCL circuit to get the total voltage
- Because eachvolt has a specificdirectionsowe dealswiththemasvectors
-
V = √ VR
2
+ (VL- VC)2
60- WhenreplacingDC voltage source by an AC voltage source of the same effective e.m.f.inRL circuit, the
impedance increases
- Because an inductive reactance isgeneratedinthe coil accordingtoitsself-inductionwhichwasn’texist
incase of DC because frequencywaszero,
- It was Z = R and became Z = √ R2
+XL
2

Mr. AhmedHekal
54
AC andDC currentComparison
Hot Wire Ammeter
Usedto measure the effectivevalue of AC
Construction:
1- Thinwire of (platinum-Iridium)alloy,thiswire isstretchedbetweenterminalsof the device,sowhenitheated
up itwill ex
pand
2- One endof silkthread attachedtothe middle of the wire,the otherendrolledoverarollerwhichisfixedona
springto the wall
3- A pointerisattachedto rollerandmovesoverascale
4- The alloyedwire isconnectedwithashuntresistance inparallel
Direct Current DC
1- It’sa constantvalue currentthat
flowsfrom+ve pole to –ve pole of a
battery(conventional direction)
2- Its sourcesare (cells,batteriesand
DC generators)
3- Its e.m.f cannotbe raisedor
reduced
4- Measuredbymovingcoil
GalvanometerorAmmeter
5- Usedin electrolysis,electroplating
and Batteriescharging
6- Transferred,butwitha biglossof
energy
Alternative CurrentAC
1- It’sa currentthat changesitsvalue
fromzero to maximumeachquarter
cycle and changesits directioneach
half cycle
2- Its source is ACgenerators
3- Its e.m.f increasesanddecreases
usingtransformers
4- Measuredbythe Hot wire Ammeter
whichmeasuresitseffectivevalue
5- Used inmost electrical devicesand
lightening
6- Transferredwithminimallossof
energy

Mr. AhmedHekal
55
Operation of Hot-Wire Ammeter:
1- It’sconnectedinseries
2- WhenAC passesthroughthe wire,temperatureincreasesandwire willexpand
3- The silkthreadpullsthe roller,sopointerwillmove onthe scale
4- Pointerdeflectionisdirectlyproportionalwiththe currentflow
5- Whencurrentstops the wire iscooledandrollerwill pull the pointertozero
Disadvantages
- Pointerisslowinmoving.
- It’saffectedbythe room temperature thatmaycause errorsin readings
To overcome thatwe stretchthe wire ona plate that has the same expandingcoefficientof the wire material and
insulate itfromwire
How to calibrate thisdevice?
- Calibratingmeanstoensure thatit’sworkinggoodandit can measure the correcteffective value
- By comparingitsreadingwiththe readingof the movingcoil Ammeterbyconnectingbothinseriesinadc circuit
withrheostat
Notices
-
The scale is not regularbecause the thermal amountgenerated
inthe wire isdirectlyproportional withI2
,notI
- It can measure bothACand DC currents because the thermal
effectof currentdoesn’trelyonthe currentdirection

Mr. AhmedHekal
56
Laws
Capacitor
A. Capacitive reactance
XC= 1 / 2πfC = 1 / C
B. Capacity
C = Q / V
C. Compare between2capacitorscapacitive
reactance
XC1/ XC2= f2 C2 / f1 C1 =  C2 /  C1
D. AC current intensity in a capacitor
 = VC / XC
E. ConnectingCapacitors
a- In parallel
1/XC= 1/XC1+1/XC2 +1/XC3 +……
CT =C1+ C2 + C3
In case of capacitors are similar
CT = C1 *n
XC = XC1 / n
b- In series
XC= XC1+ XC2 + XC3 +……
1/CT = 1/C1 + 1/C2 + 1/C3
In case of capacitorsare similar
CT = C1 /n
XC = XC1 * n
Laws
Inductive Coil
A. Inductive reactance
XL= 2 f L = L
B. Inductionor(self-inductioncoefficient)
L = 
/L(length)
C. Compare between2coilsinductive
reactance
XL1/ XL2= f1 L1 / f2 L2 =  L1 /  L2
D. AC current intensity in a coil
 = VL / XL
E. Connectinginductive Coils
a- In parallel
1/XL = 1/XL1+1/XL2 +1/XL3 +……
1/LT = 1/L1+ 1/L2 + 1/L3
In case of coilsare similar
L = L1 / n
XL = XL1 / n
b- In series
XL= XL1+ XL2 + XL3 +……
LT = L1+ L2 + L3
In case of coilsare similar
LT = L1 * n
XL = XL1* n

Mr. AhmedHekal
57
Laws
RL Circuit
V = √ VR
2
+ VL
2
Z = √ R2
+XL
2
Tan =VL/VR =XL/XR
 is Positive
RC Circuit
V = √ VR
2
+ VC
2
Z = √ R2
+XC
2
Tan =- VC/VR
= -XC/XR
 is Negative
Laws
RCL Circuit

V = √ VR
2
+ (VL- VC)2
Z = √ R2
+(Xl-XC)2
Tan =(VL-VC) /VR
Tan  =(XL-XC)/R
When XL> XC  is Positive
When XL< XC  is Negative
Resonance Circuit
A. Resonance circuit frequency
f = 1/2√LC
B. To compare betweentwo frequencies
f1 /f2 = √ (L2 C2 /L1 C1)
C. VL = VC
D. XL = XC
E. Z = R
F. I = V / R
G.  0

Mr. AhmedHekal
58
Chapter 5
Definitions
1- Black Body
Is the bodythat can absorb all kindsof radiationsof differentwave lengths(ideal absorber) andre-emitsthemin
an ideal form(Ideal emitter
2- Planck’s Curve
It’sthe curve that describesthe relationbetweenradiationintensityandwave lengthof spectrumemittedfrom
bodies
3- Wein’sLaw:
The wavelengthcorrespondingtomaximumradiation intensityisinverselyproportionalwiththe temperature
on Kelvinscale
4- Remote sensing
It’sthe technologyof discoveringthe natural resourcesbyimagingthe earthsurface usingthe different
spectrumregions(suchasthe infraredradiation)
5- Surface Potential Barrier
It’sthe attractionforce that usedto attract electronsinsidethe metal andpreventitfromexitingthe surface
6- Thermionicemission(effect)
It’sthe phenomenaof emittingelectronsfrommetal surface whenheating
7- Photo-electricemission(effect)
It’sthe phenomenaof emittingelectronsfromthe metal surface whenalightbeamof specificfrequencyfallson
the metal surface
8- Metal Work function
It’sthe minimumenergyrequiredtofree the electronfromthe metal surface
9- Critical (threshold) frequency
It’sthe minimumfrequencyrequiredtofree the electronfromthe metal surface
10- Compton Effect:
Whena highenergyphoton(Of X-rayor gamma-ray) collide withfree electron,photonfrequency will
decrease anditsdirectionwill change andelectronvelocitywill increaseanditsdirectionwill increase
11- Planck’s constant:
It’sthe ratiobetweenthe photonenergyanditsfrequency
12- Photon:
Is a quantumof energyunchargedandhas a mass duringitsmovement
13- De ’Broglie Equation:
The wave lengthforthe wave correspondingamovingparticle isequal tothe ratiobetweenthe Planck’s
constantand linearmomentumof the particle
What’smeant by?
1- Work functionfor iron equals6.89*1014
J
The minimumenergyrequiredtofree anelectronfromthe ironmetal surface is6.89*1014
J
2- Critical wave length for a metal is0.0002 m
It meansthat the highestwavelengthrequiredforthe incidentlighttofree anelectronfromthe surface of this
metal is0.0002 m

Mr. AhmedHekal
59
What are conditionsrequiredto
1- Emitting an electronfrom metal surface
Answer:
A. Electronshouldgaina thermal orlightenergybiggerthanor equal the metal workfunction
Or
B. The frequencyof the incidentlightisbiggerthanorequal the critical frequencyof the metal
2- Visionof a very small body usingthe electronicmicroscope:
Answer:
- The wave lengthcorrespondingthe electronbeamissmallerthanthe detailsof thisbody
Devices
Device Usage ScientificIdea
Remote sensingdevices 1- Militaryapplications
Like nightvisiondevices
2- Medicine:speciallyin
Oncologyand
embryology
3- Discoveringnatural
resources
Thermionicemission
Cathode Ray Tube CRT T.V.and ComputersMonitors Thermionicemission
Explain:emittingelectronsfrom
the metal surface whenheating
Photo-ElectricCell Convertingthe lightenergyto
electrical energyasin
calculatorsand automatically
(open/close) doors
Photoelectriceffect
Explain:emittingelectronsfrom
the metal surface whena light
beamwiththe critical frequency
fallsonthissurface
ElectronicMicroscope Enlargingthe verysmall objects Particle-Wave Duality
Explain:itincrease the speedof
electronsbyincreasingthe
potential difference between
cathode and Anode so
,electronswillgainlarge Kinetic
Energyand large linear
momentumwhichdecrease the
wave-lengthof the wave
correspondingthe electron
beamuntil reachesthe details
of the small objectwhich
achieve the visioncondition

Mr. AhmedHekal
60
Usages:
Infra-RedRadiation 1- Discoveringthe natural resources
2- NightVisionDevices
3- Remote Sensing
MicrometerWaves Radars
Gridin the CRT Control the electronbeamintensity
The electricand magnetic fieldsinCRT Control the directionof the electronbeamstofall
on the fluorescentscreendotbydot
RelationsDiagrams
Relationbetween Diagram UsedLaw and Slope
Square of electronvelocityv2
and Potential difference
betweenCathode andAnode V
eV = ½ mv2
Slope = Δv2
/ΔV =2e/m
PhotonsEnergyE and its
Frequency 
E= h
Slope=h
KineticEnergyemittedfrom
metal surface K.E.and
frequencyof incidentlight
hK.E.+Ew
K.E.= hEw
Slope=ΔK.E/Δ
EnergyE andMass m E= mC2
Slope=ΔE/Δm= C2

Mr. AhmedHekal
61
Force (F)Affectingbyanincident
lightbeamona surface and
Powerof the beamPw
F= 2 Pw/C
Slope=ΔF/ΔPw=2/c
Wave Lengthof wave
correspondingamovingparticle
andinvertof the linear
momentum(1/PL)
= h/ PL = h/mv
Slope=h
Deductions:
1- Force resultedinfallinga light beam of photonson a surface
Momentumof incidentbeam= mc
Momentumof reflectedbeam=-mc
Change of momentum= mc-(-mc) =2mc = 2h/c where m= hc2
Rate of fallingphotonsonthe surface=φL
Force is the time rate of change inmomentum(Newton’sLaw)
Then:
F= 2hφL /c
φL=1/Δt
F=2 h/Δt.C
PowerPw= h/ Δt
Then
F = 2 Pw/C
2- Relationbetweenthe wave length of Photon and linear momentumof it
= c/ 
= hc / h=h / (h/c)
But PL= (h/c)
= h / PL = h / mc

Physics 3rd Secondary Chapters 1-7 revision مراجعة فيزياء 3 ثانوى من 1-7 مستر أحمد هيكل

Physics 3rd Secondary Chapters 1-7 revision مراجعة فيزياء 3 ثانوى من 1-7 مستر أحمد هيكل

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