Chapters 21-29Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Optics and Modern PhysicsSlide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Mass Increase and KESlide 33Slide 34Photoelectric EffectSlide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44What have we learned after all?I wish all of you a bright future!Chapters 21-29•Chapter 21:45,63 •Chapter 22:25,49•Chapter 23:35,38,53,55,58,59•Chapter 24:17,18,20,42,43,44,50,52,53.59,63 •Chapter 26:27,33,34,39,54•Chapter 27:17,18,34,43,50,51,53,56•Chapter 28: 10,11,28,47,52(a) 6.8 xMagnetic Force Magnetic forces arise from the Magnetic forces arise from the interactions of interactions of movingmoving charges in charges in magnetic fieldsmagnetic fields. (The electrical force is between two charges at restat rest)F=BQvsin for a moving charge F=BIlsin for a current••••••••••FE=qE, FM=qvBStage 1: qE=qvB thus v=E/BStage 2: qvB’=mv2/r so m=qB’r/v=qBB’r/v2r1. Complete the following statement: The magnitude of the magnetic force that acts on a charged particle in a magnetic field is independent of(a) the sign of the charge.(b) the magnitude of the charge.(c) the magnitude of the magnetic field.(d) the direction of motion of the particle.(e) the velocity components of the particle.2. A charged particle is moving in a uniform, constant magnetic field. Which one of the following statements concerning the magnetic force exerted on the particle is false?(a) It does no work on the particle.(b) It increases the speed of the particle.(c) It changes the velocity of the particle.(d) It can act only on a particle in motion.(e) It does not change the kinetic energy of the particle.XXWhich one of the following statements best explains why a constant magnetic field can do no work on a moving charged particle?(a) The magnetic field is conservative.(b) The magnetic force is a velocity dependent force.(c) The magnetic field is a vector and work is a scalar quantity.(d) The magnetic force is always perpendicular to the velocity of the particle.(e) The electric field associated with the particle cancels the effect of the magnetic field on the particle.XA beam consisting of five types of ions labeled A, B, C, D, and E enters a region that contains a uniform magnetic field as shown in the figure below. The field is perpendicular to the plane of the paper, but its precise direction is not given. All ions in the beam travel with the same speed. The table below gives the masses and charges of the ions. Note: 1 mass unit = 1.67 × 10–27 kg and e = 1.6 × 10–19 C.Ion Mass ChargeA 2 units +eB 4 units +eC 6 units +eD 2 units –eE 4 units –e1. Which ion falls at position 2?(a) A (c) C (e) E(b) B (d) D2. What is the direction of the magnetic field?(a) toward the right (c) into the page (e) toward the bottom(b) toward the left (d) out of the page of the page�XXMagnetic field at the center of a Magnetic field at the center of a current loopcurrent loop B=oI/2rIr×BBrI>><<>B=oI/2rF21.31.jpgFaraday’s LawFaraday’s Law EMF= - EMF= - BB//ttEMF Induced in Moving Conductor Motional EMFMotional EMF==BB//t=Bt=BA/A/t=Bt=Blvlvt/t/t=t=BBlvlvOptics and Modern PhysicsOptics and Modern PhysicsTransverse electromagnetic (EM) Waves c=fThe Lens Equation 1/do+1/di=1/fF’ FO2F’dodifhohiSnell’s Law n1sin i =n2sin rOr sin i /sin r =v1/v2dExtra distance msin=m/d or dsin=mm Constructive inferencem=1/2,3/2,5/2, . . . Destructive inferenceA prism also disperses lightA prism also disperses lightn depends on The shorter the greater nPolarization I=Iocos2Two Principles of Relativity: •The laws of physics are the samesame for all uniformly moving observers. •The speed of light is the same for all observers. Consequences: Consequences: •Different observers measure differentdifferent times, lengths, and masses. •Only spacetimespacetime is observer-independent.Time DilationTime DilationMoving clocks runs slowly (as compared to clock at rest)t = T[1 - (v/c)2]1/2 (v has to be reasonably close to c)Length ContractionLength ContractionMoving objects are shorter in the direction of motion than when at rest l=L [1 - (v/c)2]1/2Mass Increase and KEMass Increase and KEThe mass of a body is measured to increase with speedm = mo/[1 - (v/c)2]1/2 Kinetic EnergyKE=(m-mo)c2Example: Find the speed for which the length of a meterstick is 0.5 m. Solution: Given: l=0.5 m L=1 m v=? Since l=L[1 - (v/c)2]1/2 v=c[1-(L/Lo)2]1/2=c[1-(0.5/1)2]1/2=0.866cExample: An observer watching a high-speed spaceship pass by notices that a clock on board runs slow by a factor of 1.5. If the rest mass of the clock is 0.32 kg, what is its kinetic energy?Solution: Since t = T[1 - (v/c)2]1/2 v=c[1 - (T/t)2]1/2 =c[1 - (1/1.5)2]1/2 =0.745 c KE=(m-mo)c2=moc2{1/[1 - (v/c)2]1/2-1} =1.44x1016 JPhotoelectric Effect h f = KEh f = KEmaxmax + W + Woo or KE or KEmaxmax = h f – W = h f – Woo WWoo is the work function, minimum energy required to get an electron out through the surface.Particle Nature of Light h f = KEh f = KEmaxmax + W + Woo Wave-Particle Duality =h/mvBohr ModelBohr ModelElectron’s angular momentum L=I=mvrn=nh/2n=1,2,3n is called quantum number of the orbitRadius of a circular orbit rn=(n2/Z)r1where r1=5.29x10-11 m (n=1)r1 is called Bohr radius, the smallest orbit in HTotal energy for an electron in the nth orbit: En=(Z2/n2)E1 where E1=-13.6 eV (n=1)E1 is called Ground State of the hydrogenBoth orbits and energies depend on n, the quantum number1/=R(1/22-1/n2), n=3,4,… for Balmer series where Rydberg constant R=1.097x107 m-1Example: A beam of white light containing frequencies between 4.0x1014 Hz and 7.9x1014 Hz is incident on sodium surface, which has a work function of 2.28 eV. (a) What is the range of frequencies in this beam of light for which electrons are ejected from the sodium surface? (b) Find the maximum kinetic energy of photoelectrons ejected from this surface. Solution: (a) fmin=Wo/h=2.28eV(1.6x10-19 J)/6.63x10-34 J•s=5.5x1014 Hz Frequencies that eject electrons: 5.5x1014-7.9x1014 Hz(b) Kmax=hf-Wo
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