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

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

  1. 1. MR.AHMED HEKAL PHYSICS 3RD SECONDARY CHAPTERS (1-7)
  2. 2. 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
  3. 3. Mr. AhmedHekal 2 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
  4. 4. Mr. AhmedHekal 3 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 2- Temperature of thisconductor Conductivityof a conductor 1- Material type of thisconductor 2- Temperature 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
  5. 5. Mr. AhmedHekal 4 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
  6. 6. Mr. AhmedHekal 5 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
  7. 7. Mr. AhmedHekal 6
  8. 8. Mr. AhmedHekal 7 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
  9. 9. Mr. AhmedHekal 8
  10. 10. Mr. AhmedHekal 9 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
  11. 11. Mr. AhmedHekal 10 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
  12. 12. Mr. AhmedHekal 11 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
  13. 13. Mr. AhmedHekal 12 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
  14. 14. Mr. AhmedHekal 13 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
  15. 15. Mr. AhmedHekal 14 N. Factors that depends on Physical quantity Factors that this quantity dependson MagneticFlux Density generatedfromwire at a point B = d) 1- CurrentIntensity(directlyproportional) 2- Distance bet.the pointandwire (inversely proportional) 3- Permeabilitycoefficientof the medium(directly proportional) MagneticFlux Density generatedfromcoil at the centerpoint B = r) 3- Numberof turns (directlyproportional) 4- CurrentIntensity(directlyproportional) 5- Radiusof coil (inverselyproportional) 6- Permeabilitycoefficientof the medium(directly proportional) MagneticFlux Density generatedfromSolenoidata pointon itsaxis B = L 3- Numberof turns (directlyproportional) 4- CurrentIntensity(directlyproportional) 5- Axislength (inverselyproportional) 6- Permeabilitycoefficientof the medium(directly 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) 2- CurrentIntensity (directlyproportional) 3- Magneticflux densityB(directlyproportional) 4- Angle betweenthe fieldandwire (directly proportional) 5- Areaof the rectangularcoil (directlyproportional) Dipole Moment |md|= I A N 1- Numberof turns (directlyproportional) 2- CurrentIntensity (directlyproportional) 3- Areaof the rectangularcoil (directlyproportional)
  16. 16. Mr. AhmedHekal 15 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
  17. 17. Mr. AhmedHekal 16 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
  18. 18. Mr. AhmedHekal 17 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
  19. 19. Mr. AhmedHekal 18 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
  20. 20. 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 -
  21. 21. 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 -
  22. 22. Mr. AhmedHekal 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)
  23. 23. 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
  24. 24. 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
  25. 25. 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
  26. 26. 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 Device Usage ScientificIdeaand explanation 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
  27. 27. 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
  28. 28. 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
  29. 29. 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
  30. 30. 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
  31. 31. Mr. AhmedHekal 30 C. Factors that depends on Physical quantity Factors that this quantity dependson 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 =  N m /t e.m.finduced = BL V sin  e.m.finduced = N A B  sin  t
  32. 32. 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
  33. 33. 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
  34. 34. 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 - Galvanometerpointermovesinaspecificdirection 3- Move the primarycoil far awaythe secondarycoil Notice - Galvanometerpointermovesinopposite direction 4- Fix the primarycoil inside the secondarycoil andincrease currentintensityinthe primarycoil Notice - Galvanometerpointermovesinaspecificdirection
  35. 35. Mr. AhmedHekal 34 5- Decrease currentintensityinthe primarycoil Notice - Galvanometerpointermovesinopposite direction 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
  36. 36. 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
  37. 37. 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
  38. 38. 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
  39. 39. 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 e.m.finduced =  N m /t 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
  40. 40. 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
  41. 41. 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 e.m.finduced =  N m /t - 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
  42. 42. 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
  43. 43. 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
  44. 44. 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 Device Usage ScientificIdeaand explanation 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)
  45. 45. 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
  46. 46. 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
  47. 47. 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 
  48. 48. 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
  49. 49. 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 B. In Ohmicresistance V and  are inphase 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 B. In Ohmicresistance V and  are inphase 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
  50. 50. 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
  51. 51. 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
  52. 52. 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 +…….)
  53. 53. 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
  54. 54. 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
  55. 55. 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
  56. 56. 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
  57. 57. 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
  58. 58. 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
  59. 59. 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
  60. 60. 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
  61. 61. 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
  62. 62. 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

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