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Lecture 25. Overview Final Exam: Secs. 501, 503, 504, 526, 527: Dec. 10, M., 10:30am - 12:30pm Secs. 516-520: Dec, 12, W., 10:30am - 12:30pm This week off. hours: Today: 1:30-2:30pm; Th: 12:30-1:30pm; F: 2:00pm-4:00 pm Evaluation is helpful! • You should all have received an email with a link to evaluate your PHYS 208 class. • I encourage you all to fill this out • It gives a feedback to me and TAs • This year it’s especially important in view of of coming revolutionary changes in teaching: 1) big classes (1 lecturer for all),2) flipped lectures (pre-lectures for learning the concepts and no derivations by lecturer on board), 3)multiple choice exams. The Course Grade (CG): CG=(3E+2xF+R+L/2+HW+CP)/7. 3 Exams (3E) 300 Final Exam (F) 100 Homework (HW) Laboratory (L) 50 100 Recitation (R) 100 Class Points (CP) not limited A: >90, B: 80-90, C: 70-80, D: 55-70, F<55 Check your Midterm Exams Grades on elearning! Average E1=61+5 (curve) Average E2=56+5 (curve) Average E3=56+5 (curve) Average A3E=(E1+E2+E3+15)/3=63 Average lab grade: (AL)=92 Average rec. grade (AR)=82 For students with A3E=63, AL=92, AR=82, HW=50 Before Final Exam: ACG=(3xA3E+R+HW+L/2)/5=73. For students with A3E=63, L=100, R=100, HW=50 Before Final Exam: ACG=(3xA3E+R+HW+L/2)/5=78. If FE>min {E1,E2,E3} then min{E1,E2,E3} FE Then the course grade is calculated as: The Advanced Course Grade (ACG): ACG=(2E+3xF+R+L/2+HW+CP)/7. If before curve of FE FE>90 then CG is A; FE>80 and FE>max{CG,ACG} then CG=B FE>70 and FE>max{CG,ACG} then CG=C FE>55 and FE>max{CG,ACG} then CG=D Webct homework (HW) is due by Dec.10. To get full credit (50p) complete HW 1-12 Last HW (N13) is not required! You get 25p if complete >6HW but <12HW If you complete <6HW you get no credit Final Exam: (Ch.21-24, 27-29, 32,33,35,36) Two Parts: 2 problems (50p) old material : Chs. 21-24, 27-30 2 problems (50p) new material : Chs. 32, 33, 35,36 Students may be excused from the first part if 1) Each midterm exam is higher or =85 (with a curve) 2) An average of 3 exams is higher or = 90 (with a curve) In this case the Course Grade: CG= (3E+F+R+L/2+HW+CP)/6. Preparation to the final exam • • • • • • • • This lecture Formula sheet Old Final Exams Old 1st and 3d exams Lectures notes (especially, Examples) Homework Textbook Sleep well! Ch.21-23 (Lect.1-6) 1.Electric field, electric force kQ E 2 r0 FQq qEQ r 2. Potential (voltage), potential energy b Vab E dl U (r ) q0V V=kq/r a 3. Superposition principle 4. Gauss’s Law E dA Qencl surface 0 5. Energy conservation law, work Ka+Ua= Kb+Ub Wab =Ua-Ub kdq kdq dV (r ) , 2 2 r a x k kQ V dV dq 2 2 2 2 a x a x ring ring Example. A uniformly charged thin ring has a radius10 cm and a total charge 12 nC. An electron is placed on the ring axis a distance 25 cm from the center of the ring. The electron is then released from rest. Find the speed of the electron when it reaches the center of the ring. U ( x) U (0) K (0) K (0) U ( x) U (0) q[V ( x) V (0)] kQ kQ V ( x) ; V ( x 0) a a2 x2 2keQ 1 1 v [ ] 2 2 m a a x m v2 1 keQ[ 2 a a x 2 2 ] 1 1 ] 2 2 0.1m (0.1m) (0.25m) 7 1 . 54 10 m/ s 31 2 2 9 10 kg C 2 9 109 Nm 2 1.6 1019 C 12109 C[ 1 Calculation of E using Q E dA the Gauss’s law encl surface Three types of charge distribution symmetry Spherical A=4 π r2, V=4 π r3/3 Cylindrical Aside=2 π rL, V= π r2L L →∞ Plane A=L1L2, V=AL3 L1, L2 →∞ 0 Example. Very long conducting cylinder, L>>R, r Q L L 1. r<R: E(r)=0→V(r)=V(R) 2. r>R: For infinitely long charged objects never choose V(r→∞)=0 !!! Choose V(R)=0. V(r) R dr R r V (r ) 2k 2k ln 2k ln r r R r r Ch.27-28 (Lect.12-15) Magnetic force on a moving charge F qv B Motion of the charged particles in B; crossed B and E Force on a segment of a current Amper’s law B dl I F I Bl Example1. At t=0 proton with v=(1.5x105 i+2x105 j)m/s enters at the origin of the coordinate system the region with B=0.5iT. Describe and plot the pass. Find the coordinates of the proton at t=T/2 where T is the period of the circle. F qv B y F m a qvy B m vy 2 R 2R 2m R , T qB vy qB m vy F y (t T / 2) 0, z (t T / 2) 2 R, x(t T / 2) v x T / 2 z y Vy z B R F Motion of q in B┴E. Velocity selector E qvB qE v B Independent on the mass and q Particles with this velocity will be undeflected. Particles with larger velocity will be deflected by B. Particles with smaller velocity will be deflected by E. Amper’s law: Conducting cylinder 1.r R I B 2r 0 Jr , J 2 R 0 Ir B 2R 2 0 I r R:B 2R 2.r R B 2r 0 I 2 0 I B 2r Lect.16,17,20 (Ch.29,32) d B Law of EM Induction (Faraday’s law) dt a Motional emf ( v B ) dl b Major characteristics of e.m. waves 0 n Absorbing plane Malus’s law u EB Prad I c EB S P S dA, U Pdt Nonabsorbing plane I out Iin cos 2 E vB c v n A Prad 2I c T P F Prad A c I Random I out I in cos 2 in polarization 2 n Km K Lenz’s law Magnetic field produced by induced current opposes change of magnetic flux Slide-wire generator vBl Origin of this emf is in separation of charges in a rod caused by its motion in B due to magnetic force F qv B The secondary magnetic force Fm ' I Bl Fext Fm ' I Bl 2 2 ( vBl ) Pel I 2 R R R dW dx (vBl) 2 Pmech Fext IBlv Blv dt dt R R Example. Find motional emf in the rod. (v B) dl L 0 I V + d L F qv B V I + - + - 0 I B 2r 0 I v 2 d L d dr 0 I d l v ln r 2 d Example. Find induced current in the loop with resistance R. I=0 Induced nonelectrostatic electric field when B(t) Find E(r). d B E dl dt 1.r R dB E 2r r dt B ni, K m 0 2 E n di 2 dt 2) r R r dB E 2r R dt n di R 2 E 2 dt r 2 R General form of Amper’s law (displacement current) d E B dl (i dt ) μ= Kmμ0, ε=Kε0, In free space K=1, Km =1 Let’s find B between the plates. 1.r R r2 B 2r 0 jd r 0ic 2 R id ic jd 2 A R 0ic r B 2 R 2 2.r R B 2r 0id 2 B r R 0ic B 2r B X X X E X X X E and B in e.m. wave E y E0 cos(t kx) Bz B0 cos(t kx) or E j E0 cos(t kx) B k B0 cos(t kx) E y E0 cos(t kx) Bz B0 cos(t kx) or E j E0 cos(t kx) B k B0 cos(t kx) This is y-polarized wave. The direction of E oscillations determines polarization of the wave.. Example. A carbon-dioxide laser emits a sinusoidal e.m. wave that travels in vacuum in the negative x direction. The wavelength is 10.6μm and the wave is z-polarized. Maximum magnitude of E is 1.5MW/m. Write vector equations for E and B as functions of time and position. Plot the wave in a figure. E k E0 cos(t kx) B j B0 cos(t kx) E0 1.5 106 V / m 3 B0 5 10 T 8 c 3 10 m / s 2 2 3.17rad 5 k 5 . 93 10 rad / m 6 10.6 10 m ck (3 108 m / s ) 5.93105 rad / m 1.781014 rad / s NB: Since B=E/c→B (in T) <<E (in V/m) Standing Waves If two conductors are placed parallel to each other on the distance L the nodes of E should be on the ends (just as on the string with fixed ends) max v v L max 2 L f min 2 max 2 L 2L vn n , n 1,2... f n f minn n 2L Example. In a microwave oven a wavelength 12.2cm (strongly absorbed by a water) is used. What is the minimum size of the oven? What are the other options? Why in the other options rotation is required? Lmin 6.1cm 2 L 12.2cm onenodein the m iddle L 18.3cm... Example A radio station on the surface of the earth radiates a sinusoidal wave with an avearge total power 50kW. Assuming that transmitter radiates equally in all directions, find the amplitudes of E and B detected by a satellite at a distance 100km. P 5 104 7 W I 7.9610 2 10 2 2R 6.2810 m m2 2 E0 B0 E0 I 20 20c E0 2 0 cI 2.5 10 2 V / m E0 B0 8.2 1011 T c Malus’s law Eout In general case when linear plz wave goes through the filter only its projection on the axis of the filter goes through. Ein Eout Ein cos I out Iin cos2 Unpolarized em wave (random polarization) I out I in cos 2 I in 2 NB: After the filter em wave is always linear polarized along the axis of the filter. Lect. 23 (Ch.35) Conditions for constractive and distractive interference (i) for phase difference (ii) for path difference Double-slit experiment Interference in the thin films Phase difference and Path difference E1 E0 cos(t kr1 ) r1 E2 E0 cos(t kr2 0 ) S1 k (r2 r1 ) 0 kr 0 r2 S2 1. 2m I t 4 I 0 ( Et 0 2 E0 ) constractive int erfernce 1 2. 2 (m ) I t 0 ( Et 0 0) destructive int erference 2 3. 2 (2m 1) I t 2 I 0 ( Et 0 2 E0 ) no int erference 2 Taking into account that k In particular case when 0 0 1.r m constractive int erfernce 1 2. r (m ) destructive int erference 2 1 3.r (m ) no int erference 2 2 Max and min positions (bright and dark stripes) If R d r r2 r1 d sin , kr max : 2m d sin m 1 1 min : 2 (m ) d sin (m ) 2 2 If is sm allsin tan Rm max : ym d R (m 1 / 2) min : ym d y R Intensity distribution I t 4 I 0 cos2 ( / 2) 2 I 0 (1 cos ), Constructive int ereference: I t 4 I 0 Destructice int erefernce: I t 0 No int erference: I t 2 I 0 Example A radio station operating at a frequency of 1500kHz has two identical vertical dipole antennas spaced 400m apart, oscillating in phase. At distances much greater then 400m, in what directions is the intensity greatest in the resulting radiation pattern? If intensity produced by each antenna 400km away along y axis is 2mW/m2 what is the total intensity produced by two antennas at this point? y c 3 108 m / s 1. 200m 6 f 1.5 10 1 / s m m(200m) m max : d sin m sin d 400m 2 m 0 0, m 1 30 , m 2 90 | m | 2 sin 1 im possible. 2.I 4 I 0 8m W / m 2 Interference in the thin films Normal coincidence: k 2t , k 2 film 1 1 1. max : 2m 2t film (m ) 0 (m ) 2 n film 2 m0 1 2. min : 2 (m ) 2t m film 2 n film film n<nfilm Example The walls of a soap bubble have about the same refractive index as a plain water, n=1.33. In the point where the wall is 120nm thick what colors of incoming white light are the most strongly reflected? 0 1 max : 2t (m ) n film 2 m 0, 0 4tn film 4 120nm 1.33 640nm (orange) m 1, 0 640nm 213nm(UVradiation, nonvisible) 3 Interference at the thin wedge of air NB: Interference between the light reflected from the upper and lower surfaces of the glass plate is neglected due to the larger thickness of the plate. 1 max : 2t 0 (m ) 2 min : 2t m0 NB: the fringe at the line of contact is dark, because of phase shift produced by reflection from the second plate. Example A monochromatic light with wavelength in the air 500nm is at normal incidence on the top glass plate (see the figure). What is the spacing of interference fringes? x l tl x , min : 2t m0 t h h m0l 0l (500109 m)0.1m xm , xm1 xm 1.25m m 3 2h 2h 2(0.0210 m) The position of the first minimum in diffraction pattern 1 (a / 2) sin / 2 a Diffraction grating max: 2m d sin m E0 t Eot NE0 I ot N I 0! 2