Lecture 41: FRI 01 MAY Final Exam Review

1. Physics 2113 Jonathan Dowling Lecture 41: FRI 01 MAYFinal Exam Review

2. The Grading • Midterms  -  100 points each • Final Exam  -  200 points • Homework   -  75 points • In Class PP - 25 pointsTOTAL:600 points • Numerical Grade: Total Points / 6 • A: 90–100      B: 75–89      C: 60–74      D: 50–59      F: <50

3. Final Exam • 10:00AM-12:00N SAT 09 MAY • LOCKETT 6 • ≥ 50 % of Exam From HW01-14 • 100 PTS: CH 13, 21–30 • 100 PTS: CH 30–33

4. Final Exam • 2 Problems from CHS30–33 • 4 Multiple Choice Questions from CHS30–33 • 10 Multiple Choice Questions from CHS13, 21–29. • No partial credit for multiple choice questions. • Must show your work to get partial credit on problems.

5. What do you need to make on the final to get an A, B, C, etc.? A: 90–100      B: 75–89      C: 60–74      D: 50–59      F: <50 Solve this simple equation for x: Where mt1=exam1, mt2=exam2, mt3=exam3, hw=total points on your hws01–14 (out of 450), icppc=checks, icppx=X’s, icpp=number of times you were called on, max is the binary maximum function, and y is your desired cutoff number, y = 90, 75, 60, or 50. Then x is the score out of 200 you need on the final to make that cutoff grade y. This assumes no curve.

6. Example: John D’oh wants to know what he needs to make on the final in order to get an A = 90 in this class. It is very likely impossible for John to get an A as he’d need better than a perfect score on the final. How good does he need to do to avoid a C = 74? John is extremely unlikely to get an A, and is unlikely to get a C, so the most probable outcome is that John will get a B in this class.

8. Lenz’s Law

9. i Induction and Inductance • Faraday’s law: or • Inductance: L=NΦ/i • For a solenoid: L=μ0n2Al=μ0N2A/l • Inductor EMF: EL= -L di/dt • RL circuits: i(t)=(E/R)(1–e–tR/L) or i(t)=i0e–tR/L • RL Time Constant: τ = L/R Units: [s] • Magnetic energy: U=Li2/2 Units: [J] • Magnetic energy density: u=B2/2m0 Units: [J/m3]

10. Changing B-Flux Induces EMF

11. Flux Up RL Circuits Flux Down

12. RL Circuits E/2 t=? Make or Break? At t=0 all L’s make breaks so throw out all loops with a break and solve circuit. At t=∞ all L’s make solid wires so replace L’s with wire and solve circuit.

13. Largest L? R,L or 2R,L or R, 2L or 2R,2L? Checkpoints/Questions Magnitude/direction of induced current?Magnitude/direction of magnetic field inducing current? Magnitude of induced emf/current? Given |∫E∙ds| , direction of magnetic field? Given B, dB/dt, magnitude of electric field? Current inducing EL? Current through the battery? Time for current to rise 50% of max value? Largest current?

14. When the switch is closed, the inductor begins to get fluxed up, and the current is i=(E/R)(1-e–t/τ). When the switch is opened, the inductors begins to deflux. The current in this case is then i= (E/R) e–t/τ

17. LC Circuits

18. Capacitor discharges completely, yet current keeps going. Energy is all in the B-field of the inductor all fluxed up. The magnetic field on the coil starts to deflux, which will start to recharge the capacitor. Finally, we reach the same state we started with (with opposite polarity) and the cycle restarts. PHYS2113 An Electromagnetic LC Oscillator Capacitor initially charged. Initially, current is zero, energy is all stored in the E-field of the capacitor. A current gets going, energy gets split between the capacitor and the inductor.

20. Electric Oscillators: the Math Amplitude = ? Voltage as Function of Time Energy as Function of Time

21. Example • In an LC circuit, L = 40 mH; C = 4 μF • At t = 0, the current is a maximum; • When will the capacitor be fully charged for the first time? • ω= 2500 rad/s • T = period of one complete cycle • T =2π/ω= 2.5 ms • Capacitor will be charged after T=1/4 cycle i.e at • t = T/4 = 0.6 ms

22. E=10 V 1 mH 1 mF a b Example • In the circuit shown, the switch is in position “a” for a long time. It is then thrown to position “b.” • Calculate the amplitude ωq0 of the resulting oscillating current. • Switch in position “a”: q=CV = (1mF)(10 V) = 10mC • Switch in position “b”: maximum charge on C = q0 = 10mC • So, amplitude of oscillating current = 0.316 A

23. 1 . 0 0 . 8 0 . 6 U E 0 . 4 0 . 2 0 . 0 0 4 8 1 2 1 6 2 0 t i m e ( s ) Damped LCR Oscillator C L R Ideal LC circuit without resistance: oscillations go on forever; ω = (LC)–1/2 Real circuit has resistance, dissipates energy: oscillations die out, or are “damped” Math is complicated! Important points: • Frequency of oscillator shifts away from ω = (LC)-1/2 • Peak CHARGE decays with time constant = • τQLCR=2L/R • For small damping, peak ENERGY decays with time constant • τULCR= L/R

25. Q(t) t(s)

28. B ! B B Displacement “Current” Maxwell proposed it based on symmetry and math — no experiment! i i E Changing E-field Gives Rise to B-Field!

29. 32.3: Induced Magnetic Fields: Here B is the magnetic field induced along a closed loop by the changing electric flux FEin the region encircled by that loop. Fig. 32-5 (a) A circular parallel-plate capacitor, shown in side view, is being charged by a constant current i. (b) A view from within the capacitor, looking toward the plate at the right in (a).The electric field is uniform, is directed into the page (toward the plate), and grows in magnitude as the charge on the capacitor increases. The magnetic field induced by this changing electric field is shown at four points on a circle with a radius r less than the plate radius R.

31. Example, Magnetic Field Induced by Changing Electric Field:

32. Example, Magnetic Field Induced by Changing Electric Field, cont.:

33. 32.4: Displacement Current: Comparing the last two terms on the right side of the above equation shows that the term must have the dimension of a current. This product is usually treated as being a fictitious current called the displacement current id: in which id,encis the displacement current that is encircled by the integration loop. The charge q on the plates of a parallel plate capacitor at any time is related to the magnitude E of the field between the plates at that time by in which A is the plate area. The associated magnetic field are: AND

34. Example, Treating a Changing Electric Field as a Displacement Current:

37. Electric Field: Magnetic Field: Wavenumber: Wave Speed: Angular frequency: Vacuum Permittivity: Vacuum Permeability: Fig. 33-5 Amplitude Ratio: Magnitude Ratio: Mathematical Description of Traveling EM Waves All EM waves travel a c in vacuum EM Wave Simulation (33-5)

38. The Poynting Vector: Points in Direction of Power Flow Electromagnetic waves are able to transport energy from transmitter to receiver (example: from the Sun to our skin). The power transported by the wave and its direction is quantified by the Poynting vector. John Henry Poynting (1852-1914) For a wave, since E is perpendicular to B: In a wave, the fields change with time. Therefore the Poynting vector changes too!! The direction is constant, but the magnitude changes from 0 to a maximum value. Units: Watt/m2

40. EM Wave Intensity, Energy Density A better measure of the amount of energy in an EM wave is obtained by averaging the Poynting vector over one wave cycle. The resulting quantity is called intensity. Units are also Watts/m2. The average of sin2 over one cycle is ½: Both fields have the same energy density. The total EM energy density is then

41. EM Spherical Waves The intensity of a wave is power per unit area. If one has a source that emits isotropically (equally in all directions) the power emitted by the source pierces a larger and larger sphere as the wave travels outwards: 1/r2Law! So the power per unit area decreases as the inverse of distance squared.

42. Example A radio station transmits a 10 kW signal at a frequency of 100 MHz. Assume a spherical wave. At a distance of 1km from the antenna, find the amplitude of the electric and magnetic field strengths, and the energy incident normally on a square plate of side 10cm in 5 minutes.

43. Radiation Pressure Waves not only carry energy but also momentum. The effect is very small (we don’t ordinarily feel pressure from light). If light is completely absorbed during an interval Δt, the momentum Transferred Δp is given by F and twice as much if reflected. A Newton’s law: I Now, supposing one has a wave that hits a surface of area A (perpendicularly), the amount of energy transferred to that surface in time Δtwill be therefore Radiation pressure: [Pa=N/m2]

45. EM waves: polarization Radio transmitter: If the dipole antenna is vertical, so will be the electric fields. The magnetic field will be horizontal. The radio wave generated is said to be “polarized”. In general light sources produce “unpolarized waves”emitted by atomic motions in random directions.

47. EM Waves: Polarization Completely unpolarized light will have equal components in horizontal and vertical directions. Therefore running the light through a polarizer will cut the intensity in half: I=I0/2 When polarized light hits a polarizing sheet, only the component of the field aligned with the sheet will get through. And therefore:

48. Example Initially unpolarized light of intensity I0 is sent into a system of three polarizers as shown. What fraction of the initial intensity emerges from the system? What is the polarization of the exiting light? • Through the first polarizer: unpolarized to polarized, so I1=½I0. • Into the second polarizer, the light is now vertically polarized. Then, I2 = I1cos2(60o)= 1/4 I1= 1/8 I0. • Now the light is again polarized, but at 60o. The last polarizer is horizontal, so I3 = I2cos2(30o) = 3/4 I2 =3 /32 I0 = 0.094 I0. • The exiting light is horizontally polarized, and has 9% of the original amplitude.

49. Completely unpolarized light will have equal components in horizontal and vertical directions. Therefore running the light through first polarizer will cut the intensity in half: I=I0/2 When the now polarized light hits second polarizing sheet, only the component of the field aligned with the sheet will get through.