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# Phys Scientists & Engineers II PHY 184

MSU

GPA 3.67

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This 96 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 184 at Michigan State University taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/207637/phy-184-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15

Images Chapter 35 Review EM waves o Intensity of unpolarized light after hitting a polarizing sheet 0 Intensity of polarized light after hitting a polarizing sheet 0 Peak intensity is twice the average intensity Review EM waves Fi 3417 Manual 39 Infidel Re ected 0 Law of reflection 39 my o Law of refraction Snell s law v k avei rnnl Int l ih e 0 Index of refraction 0 Nothing has n lt1 v is always lt c 0 Frequency of wave does not change but Review EM waves Fig 3424 27 I iriiitul rust o Critical angle ec refracted ray along surface c Total internal reflection no refracted ray luciden I n2 n1 a 39IJrlpnl39arizsd Rn Ht t li d ray 0 Brewster angle reflected light is fully polarized Real amp Virtual images Fig 351 0 Real images light intersects the image point 0 Virtual images light doesn t really intersect but images appears to come from that point 0 Sunny day the mirage pool of water on the road is really reflection of low section of the sky in front of you Light rm Pool illiragc Road in Vl arm Wam39l W39Vumwr Himmler 39 Faster R and Rmu l Plane mirrors Fig 352 0 Mirror surface which reflects light in one direction instead of scattering it in many directions or absorbing it o Plane mirror flat reflecting surface Mirm 139 o Extend reflected rays from O behind mirror 0 o Intersect at point of virtual image I I j 1 JL Plane mirrors Fig 353 0 Plane mirror virtual image I is as far behind the mirror as the objec O is in front of it 0 o By convention object distances p are positive image distances for virtual images are negative llt J gt i gtl Mirror eviiew W 2 quot Miterm Midterm2 0 Wednesday October 29 at 6pm Section 1 N100 BCC Business College Section 2 158 NR Natural Resources 0 Allowed one sheet of notes both sides and calculator 0 Covers Chapters 2731 and homework sets 58 0 Send an email to your professor if you have a class conflict and need a makeup exam 0 Review in class on Tuesday October 28th Current and Resistance 0 Current o If J is uniform and paralrlel to 0 24 Current and Resistance 0 Ohm s law Power lost to heat energy in a resistor Loop Rule o Graphical representation Clich uiits 4211 battew o Substituting for 39 giives Circuits o Resistors in series 0 Resistors have identical 1 currents quot 0 Sum of VS across resistors applied l E 0 Reg is sum of all resistors rc uiits o Resistors in parallem 4 1 o Resistors have ixdenticah i E 1 0 GEMR1 etc L j 0 Reg given by Jurncti ron Rule o Arbitrarin llabel currents using different subscript for each branch 0 Using Conservation Of ill HI 1133 153 R2 In charge at each junction 2quot r A I quot3quot I aa Zr 0 At point cl 0 At point a 0 At point D At point C Circuits 0 What is 1 RJR2R3R42 0E5V Circuits OWhatis i2 0 Three unknowns so we need three equations Motion in a B Field Force on a charged particle clue to a magnetic field is F3 does not change the speed magnitude of v or kinetic energy of particle Charged particles moving with v i to a B field move in a circular path with radius r Force on a current carrying wire clue to a magnetic field is Motion in a B Field 0 Righthand rule For L positive charges when f the fingers sweep V into B through the smaller in angle g the thumb will L r be pointing in the quotjquot direction of F5 o For negative charges F5 points in opposite direction is J l Motion in a B Field o Circular motion 0 Period time for one 39 39 revolution 0 Frequency the number of revolutions per unit time o Angular frequency 3 Fields from Currents o BiotSaval c law dBinto 1 Page P disuibuu39on B Fiels from Currents r i i ii39u wiih writm o B field a distance R from a long fiumpugc Straight Wire Carrying current 39Lfm if x If ih IJL E39 K at 3quot fl 39 f it gt i item i i m N f r f j x x J x o B eld is tangent to magnetic field HP lines B Fiels from Currents o right hand rule 0 Point thumb in direction of current flow 0 Fingers will curl in the direction of the magnetic field lines due to current B Fields from Currents o B field at the center of 3 an arc is R o Express in radians o For a complete loop g5 275 then B is B Fields from Currents 0 Force on a wire carrying current 1 due to B of another parallel wire with current 2 0 Force is attractive if current in both wires are in the same 39 39 0 Force is repulsive if current in both wires are in the opposite directions 39r B In in iquot B Fields from Currents o For symmetric distributions of charge use Ampere s law to calculate B field Amperian loop 0 Integral around closed loop called Amperian loop ig Dirac inn if integration B Fields from Currents Use the righthand rule to determine the signs for the currents encircled by the Amperian loop Curl right hand around Amperian loop with fingers pointing in direction of integration Current going through loop in the same direction as thumb is positive Current going in the opposite direction is negative Direction of integration B Fields from Currents quot 7 m 1 n is turnslength o For toroid Currents from B Fields 0 Magnetic flux o Faraday s law N loops 0 Lenz s law induced emf gives rise to a current whose B field opposes the change in flux that produced it Faraday s law 0 We can change the magnetic flux through a loop or coil by 0 Changing magnitude of B field within coil 0 Changing area of coil or portion of area within 5 field 0 Changing angle between B field and area of coil eyg rotating coil Chater quotNorthng yet How about you Newtonquot Review 0 RL and RC circuits CharQE current and potential grow and decay exponentially 0 LC circuit 0 Charge current and potential change sinusoidally q gt r v39 r 2H 0 Total electromagnetic energy is u r P g a r j y 3 41 i i Review 0 Ideal LC circuit 0 Total energy conserved 0 Solved differential equation to find Review 0 RLC circuit 0 Energy is no longer conserved becomes thermal energy in resistor o Oscillations are damped 0 Solved differential equation to find If R is very small Resistive Load 0 Apply loop rule 0 Using 0 We have o Amplitude across resistor is same as across emf o Rewrite VR as Forced Oscillations Resistive load Resistive load 0 Use definition of resistance to find I 0 Voltage and current are functions of snadt with 0 so are in phase 0 No damping of VR and I39R since the generator supplies energy Resistive load 0 Compare current to general form o Minus sign for phase is tradition o For purely resistive load the phase constant O 0 Voltage amplitude is related to current amplitude Capacitive load o Use definition of capacitance 0 Use definition of current and differentiate o Replace cosine term with a phaseshifted sine term Capacitive Ioa o V0tage and current reliations o XC is called the capacitive reactance and has units of ohms Capacitive load 0 Compare V6 and ic of capacitor o Voltage and current are out of phase by 90 0 Current leads voltage by 1174 v w r vf Inductive Iota skipped in lecture o Derivatiom of current o Selfimduced emf across an inductor is quot o Relate 1 Induc veload o VOItage and current rellations o XL is called the inductive reactance has units of ohms Induc veload 0 Compare vi and L of inductor o I 3 o and VL are 90 out E gmogmmi of phase V g V g 39 39139 0 Current lags voltage by T4 IL if E l t l I 1 Summary of Forced Oscillations Element Reactance Phase of Phase Amplitude Resistance Current angleo Relation Resistor R In phase 0 VRIRR Capacitor XC 1codC Leads vC 90 VCICXC ICE Inductor XLoodL Lags vL 90 VLILXL ELI o ELI positively is the ICE man 0 Voltage or emf E before current I in an inductor L a Phase constant 1 is positive for an inductor 0 Current I before voltage or emf E in capacitor C EM Oscillations o If the driving frequency ml in a circuit is increased does the amplitude voltage and amplitude current increase decrease or remain the same 0 For purely resistive circuit a R Lit u 0 From loop rule VR 5 50 amplitude voltage I4 stays the same o R also stays the same R only depends on R EM Oscillations o If the driving frequency 0 in a circuit is increased does the amplitude voltage and amplitude current increase decrease or remain the same o For purely capacitive circuit it C 1 W o From loop rule V 6 0 So amplitude voltage lC stays the same 0 C depends on XC which depends on a by o So C increases EM Oscillations o If the driving frequency ad in a circuit is increased does the amplitude voltage and amplitude current increase decrease or remain the same i 33 a o For purely inductive circuit c From loop rule V 8m 0 So amplitude voltage V stays the same 0 139 depends on X which depends on 0 by 0 So 1 decreases 7 W l LL 77 110 1amp1 RLC Circuits 0 LC and RLC circuits with no external emf c Free oscillations with natural angular frequency a t Hill 5 l f a i Qty 0 Add external oscillating emf eg ac generator to RLC circuit a Oscillations said to be driven or forced a Oscillations occur at driving angular frequency cud a When cud i called resonance the current amplitude I39i is maximum RLC circuits 0 RLC circuit resistor capacitor and inductor in series 0 Apply alternating emf Ci o Elements are in series so same current is driven through each 0 From the loop rule at any time t the sum of the voltages across the elements must r L i equal the applied emf 5 T T RLC circuits o Taking into account the phase find current amplitude to be 0 Write amplitude voltage as o Where Z is the impedance 0 Like resistance and has units of Ohms o If XL gt XC the circuit is more inductive than capacitive 0 49 is positive 0 Emf is before current ELI o If X ltXc the circuit is more capacitive than induc ve 0 is negative 0 Current is before emf ICE EM Oscillations Pmiliw up i r N rugv Negative qt RLC Circuits it i a i m p o If X Xc the circuit is in A resonance emf and 39 current are in phase V 0 Current amplitude I is max when impedance Z is min EM Oscillations 0 When XL Xc the driving frequency is 7 i 7 ii o For RLC circuit resonance and the max current I occurs when cud a August 26th Electric Charge Chapter 22 Electric Charge Review from yesterday There are 2 types of charge and Like charges repel unlike charges attract Most objects are electrically neutral there are equal numbers of neg and pos charges so the net charge O An object becomes charged net charge 0 by adding or removing electrons Conductors are materials where some of the electrons can move freely Insulators are materials where none of the charges can move freely Electric Charge Charge is quantized comes in discrete values Proven by Millikan oildrop experiment section 238 Electric charge q is an integer multiple of the fundamental or elementary charge constant e qne where n 0111213 and e160x10 19C Particle Electric Charge Mass Electron e 16E1019 C Me911E1031 kg Proton e 16E1019 C Mp1672E1027 kg Neutron O M1674E1027 kg Electric Charge Net charge of an object is the difference between the number of protons and electrons in it times e Charge is conserved Net charge of any isolated system cannot change Same as energy linear and angular momentum Electric Charge An object can be given a charge by conduction or induction ln conduction the charge is transferred between objects by direct contact For example Rubbing a glass rod an insulator with silk Connecting 2 conductors through a conducting pathway such as a wire or by grounding the object Demos Van de Graaf rubber belt transfers charge to metal sphere using conduction Electric Charge An electrically neutral object can have an induced charge when some of its positive and negative charges separate due to a nearby charge Neutral object will display characteristics of a charged object even though there is no net charge Can we get an induced charge with an insulator MW Neutral co er F PP 51 C harng plastic Demos Obedient ruler meter stick glued to bottle so isolated system foil on one end silk and glass rod Charge rod bring near foil end will attract by induction What will happen at wood end Will also attract due to induction ruler still neutral Ruler is insulator so charges don t move freely but molecules align so get attraction Why can we stick a balloon to the wall by rubbing it on our hair or shirt Transfer charge to balloon by conduction Induction causes molecules to align in wall and attract baHoon Electric Force The magnitude of the electrostatic force F between 2 charged particles with charges q1 and q2 respectively and separated by a distance r is defined as F k lq 1 q 2 2 739 This is Coulomb s law where k is a constant The forces on 2 point charges are equal and opposite pointing to away from the other particle for unlike like charges Electric Force Coulomb s law should remind you of Newton s equation for the gravitational force F Gmlm2 r2 k is called the electrostatic constant k 1 899x109Nm2 C2 472290 80 is called the permittivity constant Electric Force Electrostatic force and gravitational force are both inverse square laws involving a property of the interacting particles Electrostatic force differs from gravitational Can be either attractive or repulsive Holds for all experimental tests and over all ranges Both obey the superposition principle The net force acting on any charge is the vector sum of the forces due to all other charges in a given distribution C d 0 d o d o 11 12 13 1 quot7 0 4 11 4 1 2 39 1 77 7 4 11 11 r 3 4 November 10th Electromagnetic Waves Chapter 34 gavel d head Iaging IiFalesslg in The rada39rainacl dirt Hmquot Review EM Waves 0 Electromagnetic waves 0 Beam of light is a traveling wave of E and 5 fields 0 No limit to wavelength or frequency I ll zII39ulIngrli IIIII I l IIIH III3 III IIIquot III I IIIquot In H I III J III E II Iquotquot IIIquot III III III397 III39H III III39II39III IIIquot3III39I 39IIIquotIIIIquotquot IIIquot l I I I I I l l J I Long limes Ratllnwaves Infrared Ulmviolui mes Gamma rays 1 l l l l I I l I I I l l l I I l l l l I III III IIIII IIII III III III7 III III III39 IIIquot ill IIII39I IIIH III III39 III IIIIH III IIIquot III IIIL I39 IIIquot III l I39IsIIIIVIIrI39 l li EF FM null a I TV t39llilllllt lh Mill llllliif 5 l IlIII39iIlIIII lll39I39liTillllllt ill Milrllillit and A M II rImIIIIIiI39nI I Z I39iI39I39II IIII Ixmrl Elt l39tllllllllll39ul IlstIs Milli LIIIIl IIIIIlIiII39 l39utllll I quotII Iquot 4 39IIIIl IIIIIlIIII39 I39lltllll I I I I I I I IIIquot IIIquot IIIquot III7 III III IIIIll III FrI qIII IIn39 I l39I39 Review EM Waves 0 EM Waves are 0 Transverse waves E and 5 fields are J to direction of wave s travel 0 Direction of wave s travel is given by cross product o E field is J B field 0 E and 5 fields vary 0 Sinusoidally 0 With same frequency and in phase Wave traveling out of page F I CremeM magnitudes Jquot l u l 239 H H c P I Uri M r 1 nmgnimdes l I I 1 lg r i P B P M if it ll Greatesl magnitudes Review EM Waves Fi 345 0 Write E and 5 fields as sinusoidal functions of position Xalong the path of the wave and time t Fill 39 39 rl V 1 39iquot I Electric 39 Mugneut ITDHIPHIH MI E UIIIPHIHTIH o Angularfrequency a and angular wave number k o E and 8 components cannot exist independently o Maxwell s equations for induction Review EM Waves 0 A light wave requires no medium for its travel 0 Travels through a vacuum at speed of light C 0 Speed of light is the same regardless of the frame of reference from which it is measured Energy transport in EM Waves 0 EM waves can transport energy and deliver it to an object it falls on eg a sunburn o Rate of energy transported per unit area at any instant is given by Poyntin vector 5 r o and defined as 0 SI unit is Wm2 0 Direction of 5 gives wave s direction of travel Traveling EM Waves Fig 344 l 1 irvzlw l gr 39 s 5 Hi If H H 3 lt1 4p a P II M But E field is L B field so I the magnitude of S is quot Checkpoint 2 0 Have an E field shown in picture A wave is transporting energy in the negative 2 direction What is the direction of the B field of the wave A um quot i39 Z 0 Use righthand rule to find B field Energy transport in EM Waves o Rewrite 5 in terms of E since most instrument39s measure E component rather than B o Instantaneous energy ow rate is V Energy transport in EM Waves 0 Usually want timeaveraged value of 5 called intensity I ll 0 Use the rms value c Rewrite average 5 Energy transport Fi 348 0 Find intensity I of oint source which emits light isotropically 39 eCual in all directions 39t N k 1 x quot quot o Find 139 at distance r from source 0 Imagine sphere of radius r and area o I decreases with square of distance EM Waves Problem 3411 o Isotropixc point light soumce has power of 250 W You are 18 meters away Caltcutlate the rms values of the E and 3 fields 0 To find Ems EM Waves Problem 3411 o Isotropilc point light source as power of 250 W You are 18 meters away Calculate the rms values of the E and B fields 0 To find Ems EM Waves Problem 341 0 Look at sizes of Ems and 15quots o This is why most instruments measure E 0 Does not mean that E component is stronger than B component in EM wave 0 Can t compare different units 0 Average energies are equal for E and B EM Waves Energy density 0 The energy density of electric eld If is equal to energy density of magnetic field U5 z 1 in EM Waves Radiation pressure 0 Light shining on an object exerts radiation pressure on it o Object s change in momentum is related to its change in energy 0 If object absorbs all radiation from EM wave total absorption 0 Think of object struck by elastic ball tennis ball 0 If object reflects all radiation back in original direction total re ection 0 Think of object struck by inelastic ball lump of putty EM Waves Radiation pressure 0 Change in energy is amount of power P in time t 0 Power is related to intensity by c Find force is EM Waves Radiation pressure 0 For total absorption force on object is EM Waves Radiation pressure 0 Express in terms of radiation pressure p which is forcearea 0 SI unit is Nlm2 called pascal Pa 0 Total absorption 0 Total reflection EM Waves Polarization Fig 3410 If Mi39llhllimi 0 Source emits EM waves with E field always in same plane wave is polarized 0 Indicate a wave is polarized by drawing double arrow 0 Plane containing the E field is called plane of oscillation 1 EM Waves Polarization Fig 3411 L 0 Source emits EM waves with random planes of oscillation E field changes direction is unpolarized 0 Example light bulb or Sun 0 Resolve E field into components 0 Draw unpolarized light as superposition of 2 polarized waves with E fields L to each other 4 November 11th Electromagnetic Waves Chapter 341 Review EM Waves o Poynting vector 5 rate of energy transorted er unit area o Instantaneous energy flow rate Review Radiation pressure 0 For total absorption force momentum and radiation pressure on the object are o For total reflection back along original path force momentum and radiation pressure are EM Waves Polarization Fig 3410 If Mi39llhllimi 0 Source emits EM waves with E field always in same plane wave is polarized 0 Indicate a wave is polarized by drawing double arrow 0 Plane containing the E field is called plane of oscillation 1

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