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# General Physics II PHY 114

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This 113 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 114 at Wake Forest University taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/230739/phy-114-wake-forest-university in Physics 2 at Wake Forest University.

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Date Created: 10/28/15

Nature of Light and Laws of Geometric Optics I The nature of light Reflection Refraction Index of refraction Huygens Principle I Dispersion and Prisms Total Internal Reflection The Nature of Light I Light as Particles Tactile Theory Ancient Greeks Emission Theory alHaitham Newton Light as Waves a k Huygens Maxwell I Quantum Theory E hf Photons Fizeau s Method for Speed of Light Measurement Mirror wheel 602751 6VS 505x10 5s a 275 2d 27500m 8 127500 c272m297gtlt10 I l lS Ray Approximation in Geometric Optics Reflection Re ected HY Double Reflection Refraction Nonnal Incidvnl my RL HCL ICd 111 o 6 A 8111 V 1 2 2 constant sm 91 v1 Nurmu RUIi uHcd 1391 Concept Question Suppose the sprinters wish to get from point Q on the beach to point P on the parking lot as quickly as possible Which path takes the least time P parking lot 1 39 a N 2 b b Lquot d quot 3 c 4 d 5 e 6 All paths take the same amount of time Index of Refraction n Speed of light in vacuum 3 Speed of light in medium v Going from one medium to another the frequency of light does not change but its wavelength does Snell s Law of Refraction 111 Sill 91 r12 sin 62 Light Propagation Through a Slab sm 62 1 sm 61 quot2 9 K h n sm 63 2 251n62 n1 n st3 2 2 151n6l sm6 l 39 Huygens Principle Old m New NE IVG 7 quot M52 639 from fmn I From Huygens Principle Applied to Reflection and Refraction Dispersion and Prisms Crown glam Acrylit39 FUSCd IUZU39IZ Mn SI N HI m 7111 in I1ll39h Measuring n Using a Prism lt1 elzeza3 2 Angle of Deviation Total Internal Reflection Norm 21 Non 11ml Critical Angle ac 141 Sin 96 142 Sin 900 n Sln 96 2 Only for n gt n2 n1 View From a Fish Eye For 0lt49 The fish sees above the water For 4949c The fish sees the shoreline For 6Mc The fish sees the pond bottom Applications of TIR Retroreflector Periscope Fiber Optic Cable For Next Class I Reading Assignment Chapter 36 Image Formation I WebAssign Assignment 13 Sources of the Magnetic Field I Moving charges currents I Ampere s Law I Gauss Law in magnetism Magnetic materials BiotSavart Law yo 47rgtlt107TmA permeability of free space Magnetic Field Surrounding a Thin Straight Conductor d gtltf gtlt I9 dxcos6 dB 4013 60 1 47 139 dB dxcos6 47 r2 01 Z COS 39 12 cos2 6 47m x a tan 6 M6 cos2 6 If the wire is very long 91 z 900 then B 2 92 90 27m dx a sec2 6d6 H B z jcos6d6 sin61 sin6q 47m 61 47m 39 Field Due to a Circular Arc of Wire de idssin90 47 R2 ds Rd 39 m B J dB 47 T d Full Circle 425 27z Magnetic Force Between Two Parallel Conductors F1 11132 111F1012 27m Concept Question On a computer chip two conducting strips carry charge from P to Q and from R to S If the current direction is reversed in both wires the net magnetic force of strip 1 on strip 2 1 remains the same 2 reverses 3 changes in magnitude but not in direction 4 changes to some other direction 5 other Aline integral ofBds around a closed path equals 0 where l is the total continuous current passing through any surface bounded by the closed path Long Current Carrying Wire Forrgt R l d B ds B27Uquot uOI 32 27W ForrltR 8amp7 0 zuol2 1 The Toroid By symmetry B is constant over loop 1 and tangent to it N Consider loop 2 whose plane is OUtS39de the toro39d39 perpendicular to the screen The Solenoid The Magnetic Field of a Solenoid Consider loop 2 Along path 2 and 4 B is perpendicular to ds Along path 3 BO Bds lBdsB J dszBl 110111 pthl 93 Magnetic Flux For a uniform field making an angle 6with the surface normal DB BACOS9 BZQBMXZAB Gauss Law in Magnetism Unlike electrical fields magnetic field lines end on themselves forming loops no magnetic monopoles Electric Field Lines Magnetic Field Lines Magnetic Moments of Atoms Current carrying loop Electron in orbit 27239 v T a L mevr a r I e 60 6V T27r2727quot Electron spin L j 927 X10 34JT 8 Magnetic Moments of Atoms I The net orbital magnetic moment in most substances is zero or very small since the moments of electrons in orbit cancel each other out I Similarly the electron spin does not contribute a large magnetic moment in substances with an even number of electrons because electrons pair up with ones of opposite spin Ferromagnetism In certain crystals neighboring groups of atoms called domains have their magnetic moments aligned A ferromagnetic crystal can be given a permanent magnetization by applying an external field Examples include Fe Co Ni Unmagnetized crystal An external field As the external field has domains of aligned increases the sizes gets larger the magnetic moments of the domains unaligned domains aligned with it become smaller Paramagnetism I Paramagnetic substances have a small but positive magnetism that comes from atoms with permanent magnetic moments I An external field can align these moments for a net magnetization but the effect is weak and not permanent I Certain organic compounds such as myoglobin are paramagnetic Diamagnetism I A very small effect caused by the induction of an opposing field in the atoms by an external field Superconductors exhibit perfect diamagnetism Meissner effect Summary l BiotSavart Law gives the magnetic field due to a current carrying wire I Ampere s Law can simplify these calculations for cases of high symmetry The magnetic flux through a closed surface is zero I Solenoids and toroids can confine magnetic fields I Orbital and spin magnetic moments of electrons give rise to magnetism in matter For Next Class I Reading Assignment Chapter 31 Faraday s Law I WebAssign Assignment 8 Electrostatics Attractiverepulsive force between charges Every charge generates an electric field Electric Flux Only charges within the Gaussian surface contribute to the electric flux through the surface Electric Potential Electric Potential Energy Potential Difference In a Uniform Field 90 AV 2 Ed For a Point Charge I dc V k For a Charge Distribution Conductors in Electric Fields I There is no electric field inside a conductor I The field on the surface is perpendicular to the surface and proportional to the surface charge density I All charges reside on the surface I Electric potential is constant throughout the conductor Capacitors Stored Energy in a Capacitor Parallel Plate Capacitor O Q 1 1 2 QAV CAV 2C 2 2 Forse es capacitors Capacitor with a dielectric For parallel capacitors DC Currents and Resistive Circuits m 2 Kirchoff s Rules Magnetic Force AFB 3911 For a straight wire in IL x B a uniform fied For a closed loop of current Cyclotron Motion 2 2 m v a qB Sources of Magnetic Field uOI d gtltf quot B EJ r2 Ra syo Forrgt R B 10 27m ForrltR B IuOIO 7 I 27ZRk Magnetic Flux and Induction Magnetic Flux 436154 0 Faraday s Law The polarity of the induced emf is such that it tends to produce a current that creates a magnetic flux to oppose the change in magnetic flux through the area enclosed by the current loop Maxwell s Equations Displacement Current F qE qv x B Lorentz Force Law EM Waves 1 C Speed of Light 050 Angular Frequency 0 Wavelength E E coskx at 1113K 27 BB coskx at 167 111 3X EE max EB 1113K EM Energy Momentum and Pressure Momentum Pressure U Complete Absorption p c 2U 25 Complete Reflection p P Reflection and Refraction Normal Rc cclcd my Incide my I 39 I x e I i 4 9 w IF I v I Nm nml I Im uh39nl my 1 Rcl39hL39Lml my 91 0 4 w q B va39l39zu lwl my Images by Mirrors From Back or real 8 16 virtual sick and I pushch 1 and 1 negative Incident lighL lt V 0 lw llr Murm Re rctrd light H39H X l CUHCUVC mirror Fm H Front Buck positive 4 1169111in 1 negative I pusitivc anI gt gt Incident light Rcfi39zlcttrd light Interference I The intensity observed at any point is proportional to the time average of the sum of the fields incident on that point I Ifthe fields are coherent and monochromatic they can interfere with each other creating bright and dark areas fringes I Destructive or constructive interference depends on the relative phase of the waves with each other 6 27 1 Causes for Phase Difference Path length difference Reflection n1 gt n2 no phase Double Slit n1 lt n2 7 phase 5 2nt Thin Film r 1 r 2 Relative phase change on reflections O 7 ml Const Dest m12M Dest Const The Geometries 180quot phase No phase Change chuqu 4 f 2004 Thomson BrooksCale Diffraction I Rays going through obstacles or openings can diffract around the edges I The amount of diffraction depends on the ratio of the wavelength of the wave to the size of the openingobstacle I The diffracted beam interferes with itself to create a pattern of bright and dark fringes on a far away screen Narrow Slit Diffraction si 11 9 2 Bigquot I sin 6 Mquot I si1190 71 Hill 6 71 I 51119 21x quot39rl a For Circular apertures ViiMug u wru ivuings JL39t ii Diffraction Gratings A device for spectrally analyzing light sources Polarization of Light Waves Polarization of a wave is the direction of its electric field vector Ordinary light is unpolarized A polarizer passes light only with polarization along its transmission axis lf light is incident on a surface at Brewster s angle the reflection is also polarized 61 62 2 90 A Electric Flux Electric flux is proportional to the total number of electric field lines through a surface Concept Question A cylindrical piece of insulating material is placed in an external electric field as shown The net electric flux passing through the surface of the cylinder is 1posmve 2 negative 3 zero General Flux Definition AcDE EiAA c059 EiAAi 13E EAR szlljface Flux Through a Cube Gauss s Law Start with a single Calculate the Generanze to charge and a flux through the any surface spherical surface sphere around it Some Important Notes About Gauss Law I A closed region that contains no charge has zero net flux through it I A zero flux does not imply a zero field I Even though the net flux is determined only by the charges inside the surface the electric field at any given point on the surface is a result of all charges inside and outside Application to Charge Distributions Choosing a Gaussian Surface I The field over the surface is constant through symmetry I The field E and the surface vector dA are parallel simplifying the dot product to an algebraic product I or they are perpendicular making the dot product zero I The field is zero over the surface Spherically Symmetric Charge Distribution i f Uzi leaiym Splwrr UHIISRiEIH ljilit39i l39t Line of Charge Cylindrical Symmetry bozzom CD CD EACOS90 O 10p bozzom CD CIDside EAside cos 0 2 E 2727 Insulating Plane of Charge Ga l 55in 11 L39ylim lm39 Conductors in Equilibrium I Put a conductor in an electric field I The free electrons inside the conductor will accelerate in the opposite direction of the field lines I As the negative and positive charges separate an internal field opposing the external field will be established The acceleration of charges will continue until the internal field cancels out the external field and the conductor will reach electrostatic equilibrium The electric field is zero everywhere inside a conductor at electrostatic equilibrium Conductors and Gauss s Law I Take a Gaussian surface inside a conductor that is arbitrarily close to the surface I Since the electric field inside the conductor and hence on the Gaussian surface is zero there is no net charge inside the surface I Since the surface is arbitrary and can be made infinitesimally close to the outer surface of the conductor any net charge on a conductor will reside on the surface Conductors and Gauss s Law The flux through the top surface is EA since E is perpendicular to A electrostatic equilibrium Therefore the flux is zero through the side wall outside the conductor The field and hence the flux through the surfaces inside the conductor are also zero Conductors and Gauss Law The electric field is zero everywhere inside a conductor at electrostatic equilibrium I Any net charge on a conductor will reside on the su ace The electric field just outside a conductor is perpendicular to the surface and is proportional to the charge density I The charge density is highest near parts of the conductor with the smallest radius of curvature A Sphere Inside a Spherical Shell rlta E20 altrltb bltrltc E20 FgtC Summary I Flux through any closed surface is proportional to the net charge enclosed by the surface I Use symmetry to simplify calculations I All excess charge on a conductor will reside at the outer surface The field inside the conductor is zero For Next Class I Reading Assignment Chapter 25 Electric Potential l WebAssign Assignment 2 DC Circuits I Electromotive Force Resistor Circuits u Connections in parallel and series Kirchoff s Rules Complex circuits made easy RC Circuits Charging and discharging Electromotive Force EMF EMF 6 is the work per unit charge done by a source such as a battery or generator Ideally AV But every real life source has internal resistance r If 0 then A 9 5 R Ir 939 Rr 512R12r I Resistors in Series 1 l BaLLery R7R1R2R3m W In a series connection the current through one resistor is the same as the other the potential drop on each resistor adds up to the applied potential Resistors in Parallel A V Avg M H In a parallel connection the voltage across the resistors are the same current gets divided at the junctions Concept Question Charge flows through a light bulb Suppose a wire is connected across the bulb as shown When the wire is connected 1 all the charge continues to flow through the bulb 2 half the charge flows through the wire the other half continues through the bulb 3 all the charge flows through the wire 4 none of the above Concept Question Two light bulbs A and B are connected in series to a constant voltage source When a wire is connected across B as shown bulb A 1 burns more brightly 2 burns as brightly 3 burns more dimly 4 goes out Problem 2811 A 6 V battery supplies current to the circuit shown in the figure When the doublethrow switch 8 is open as shown in the figure the current in the battery is 100 mA When the switch is closed in position 1 the current in the battery is 110 mA When the switch is closed in position 2 the current in the battery is 210 mA Find the resistances R1 R2 and R3 H111 1quot More Complicated Circuits Not really possible to reduce to a single loop with an equivalent resistance Kirchoff s Rules I The sum of currents entering a junction is equal to the sum of currents leaving it I The sum of potential differences across all elements around a closed loop is zero 21in 210m junction junction closed loop in I 1out 1 A V A V A VO I out2 Sign Conventions I The potential change across a resistor is IR if the loop is traversed along the chosen direction of current potential drops across a resistor I The potential change across a resistor is IR if the loop is traversed opposite the chosen direction of current I If an emf source is traversed in the direction of the emf the change in potential is 6 I If an emf source is traversed opposite the direction of the emf the change in potential is Using Kirchoff s Rules Draw a circuit diagram and label all known and unknown quantities Assign currents to each branch Don t worry about direction but be consistent Apply the junction rule to any junction in the circuit that provides a relationship between the various currents Apply the loop rule to as many loops in the circuit as necessary to solve for the unknowns Follow the sign rules Solve the equations simultaneously 5amp n Qampgampo Example 287 10V 691 2213 0 abcda C 10V 69 C 10V12V 2V 49Z 14V 6Q11 10V 0 befcb RC Circuits Circuit includes a resistor a capacitor possibly a battery and a switch When the battery is connected the current charges the capacitor through the resistor Without the battery the accumulated charge on the capacitor is discharged through the resistor Either way a timevarying temporary current is created Charging a Capacitor Discharging a Capacitor ChargeDischarge Example See Active Figure 2816 httpwwwwebassiqnnetserpseAFAF 28165wf a 10 V R 1 M9 and C 15 F Find 10 Q 239 I How long to discharge half the maximum charge Maximum energy stored I How long to release half of the stored energy Ammeters and Voltmeters An ideal ammeter should have no resistance In practice its resistance should be very low compared to the circuit it is attached to An ideal voltmeter should have infinite resistance In practice its resistance should be very high compared to the element it is attached to Summary I EMF device as a charge pump I Power and energy in circuits I Loop and junction rules I Resistors in series and parallel I RC Circuits For Next Class I Midterm 1 Review on Wednesday I Midterm 1 on Thursday I Reading Assignment for Friday I Chapter 29 Magnetic Fields I WebAssign Assignment 6 due Wednesday 11 PM Alternating Current AC Circuits I AC Voltages and Phasors I Resistors Inductors and Capacitors in AC Circuits I RLC Circuits I Power and Resonance I Transformers AC Sources AC sources have voltages and currents that vary sinusoidally I The current in a circuit driven by an AC source also oscillates at the same frequency The standard AC frequency in the US is 60 Hz 337 rads Avt AVmaX sinat Phasors Consider a sinusoidal voltage source Ava A VmaX Sinat This source can be represented graphically as a vector called a phasor quot Ava Mm Resistors in AC Circuits Av AVR 0 AV sin or AVR 1113K V V sm or I sm or 1113K w Al sin 0 AVR ImaXR Sln or Ill R AT R Since both have the same arguments in the sine term iR and AVR are in line they are in phase Root Mean Square RMS Current and Voltage 1R 1max sm at RiRZ leax RiRZ leax 1113K Z 070711nax 2 MW 0707AV J5 Inductors in AC Circuits A 1an 39 72 IL 51n 01 0L E 2 1 Av AVL O 1113K AV sin or L dz 7 AL V ampsmmldt L XL 2 0L Inductive Reactance ALHIWX AT AV mux sin 01 AVL I ITIIIX X L sin of For a sinusoidal voltage the current in the inductor always lags the voltage across it by 90 Capacitors in AC Circuits ITIIIX Av AVC AV sin of q CA V sin of ITIIIX 1 Capacitive C 0C Reactance aquot C q oCAV cos of dz AT ZAV 2 sin 0 7 72 m iC oCAV s1n mt 1113K AVC I X C sin of ITIIIX 1113K For a sinusoidal voltage the current in the capacitor always leads the voltage across it by 90 39nmx RLC Series Circuit 394 17 gtilt All k A AV 2 AV sin cot 1113K i max AVR I Rsincot AVR sincot 1113K AVL I maxX L sinat AVL cos cot AVC I maXX C sinat AVC COS cot AV 2 AVR AVL AVC Using Phasors in a RLC Circuit AV max mm 1 VR2I VXL I ch max 1113K max AL I R2 XL XC2 7 max max ZER2XL XC2 Impedance AV 2 1113K 1113K Example 334 F A L39I gtllt mquot1 4lt A 1 R 425 Q L 125 H C 35 HF 03 377 3391 Avmax 150 V AVR MR 0292A425 2124V AVLI max XL 2 0292A471g2138V XC 0292A758 2 221V AVC 1 max AVR 124Vsin 377 X0 AVL 138Vcos377l 1T1 21X 2 AV 150V AVC 221Vcos377l Z 5139 39 XL I QXC ta1114719 7389 34 4250 Power in an AC Circuit 9 iAv max sinat AV sin wt 1113K 9 I maXA Vmax sin wt cos cos wt sin sin at 9 2 AV sin2 wtcos I 111213 1113K 1 2 9 1 AV 7 av 2 max max 96 I rm S AV sin wt cos wt sin 11133 1113K g MAIrm cos No power loss occurs in an ideal capacitor or AvR AVmaX Inductor g I AImax maxR nme For a purely resistive load 0 7V I IHS I IHS 2 AImax 2 9 1 AV 1 17 15 I IHS Resonance in Series RLC Circuits A circuit is in resonance when its current is maximum AV AV 1 H S I WIS Z 1 S R2 XL XC2 1 w 0 Resonance Frequency Power in Series RLC Circuits I77IS AV YR I FLU uH ri 2quot n Iquot Mm rm m an 1 1H7 mm 39 3 0 quot 39 l m II B I 1 ml x Transformers 5011 imu Z2 Primary conch11y inpuu mltpul Power Transmission Power is transmitted in high voltage over long distances and then gradually stepped down to 240V by the time it gets to a house g 6 2 g 29x103W287A 97g 20 MW AV 230x V For 22kV transmission cost 10 lkWh 9 12R 87A22915kW Total Cost 4100 ToralCosr 1 5kW24h0 1 36 For Next Class I Midterm 2 Review on Monday I Midterm 2 on Tuesday I Reading Assignment for Wednesday I Chapter 34 Electromagnetic Waves I WebAssign Assignment 11 due Monday 5 PM

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