Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P36-1
efore Starting All of your grades should now be posted (with possible exception of last problem set). If this is not the case contact me immediately. P36 -
Final Exam Topics Maxwell s Equations: 1. Gauss s Law (and Magnetic Gauss s Law ). Faraday s Law 3. Ampere s Law (with Displacement Current) & iot-savart & Magnetic moments Electric and Magnetic Fields: 1. Have associated potentials (you only know E). Exert a force 3. Move as waves (that can interfere & diffract) 4. Contain and transport energy Circuit Elements: Inductors, Capacitors, Resistors P36-3
Test Format Six Total Questions One with 10 Multiple Choice Questions Five Analytic Questions 1/3 Questions on New Material /3 Questions on Old Material P36-4
Maxwell s Equations P36-5
Maxwell s Equations C C S S Qin E da= (Gauss's Law) ε0 dφ E d s = (Faraday's Law) dt da= 0 (Magnetic Gauss's Law) ds = µ I + µ ε 0 enc 0 0 dφ dt E (Ampere-Maxwell Law) P36-6
Gauss s Law: E d A = q in S ε 0 Spherical Symmetry Planar Symmetry Cylindrical Symmetry P36-7
Maxwell s Equations C C S S Qin E da (Gauss's Law) 0 d E d s (Faraday's Law) dt da 0 (Magnetic Gauss's Law) ds I 0 enc 0 0 d dt E (Ampere-Maxwell Law) P36 -
ε Faraday s Law of Induction d Φ = E d s = N dt = N d dt Moving bar, entering field ( Acosθ ) Lenz s Law: Ramp Rotate area in field Induced EMF is in direction that opposes the change in flux that caused it P36-9
Maxwell s Equations C C S S Qin E da (Gauss's Law) 0 d E d s (Faraday's Law) dt da 0 (Magnetic Gauss's Law) ds I 0 enc 0 0 d dt E (Ampere-Maxwell Law) P36-3
ds = µ 0 I Ampere s Law:. enc Long Circular Symmetry (Infinite) Current Sheet I Torus/Coax Solenoid = Current Sheets P36-11
Displacement Current Q E = Q= ε0ea= ε0φ ε A dq dt 0 ε dφ dt E = 0 I d E C d s = µ ( I + I ) 0 encl d = µ I + µ ε 0 encl 0 0 Capacitors, EM Waves dφ dt E P36-1
Maxwell s Equations C C S S Qin E da (Gauss's Law) 0 d E d s (Faraday's Law) dt da 0 (Magnetic Gauss's Law) ds I 0 enc 0 0 d dt E (Ampere-Maxwell Law) I am nearly certain that you will have one of each They are very standard know how to do them all P36-4
EM Field Details P36-14
Electric Potential V = E d s = V V A = Ed A (if E constant e.g. Parallel Plate C) Common second step to Gauss Law E = V = e.g. dv ˆi dx Less Common Give plot of V, ask for E P36-15
Force Lorentz Force: F= q E+ v ( ) Single Charge Motion Cyclotron Motion Cross E & for no force Magnetic Force: df = Ids F = I L ( ) Parallel Currents Attract Force on Moving ar (w/ Faraday) P36-16
The iot-savart Law Current element of length ds carrying current I (or equivalently charge q with velocity v) produces a magnetic field: = µ 4 o π q v r x rˆ d = µ I ds rˆ 0 4 r π P36-17
Magnetic Dipole Moments µ IAnˆ IA Generate: Feel: 1) Torque aligns with external field ) Forces as for bar magnets τ = µ P36-18
Traveling Sine Wave i i i i i i i E Wavelength: λ Frequency : f 0 π Wave Number: k = λ Angular Frequency: ω = π f 1 π Period: T = = f ω ω Speed of Propagation: v= = λ f k Direction of Propagation: + x Eˆ E sin( kx t) = ω Good chance this will be one question! P36-19
EM Waves Travel (through vacuum) with speed of light v = c= 1 = µε 0 0 E 8 3 10 m At every point in the wave and any instant of time, E and are in phase with one another, with E 0 = = 0 E and fields perpendicular to one another, and to the direction of propagation (they are transverse): Direction of propagation = Direction of E c s P36-0
Interference (& Diffraction) L= mλ Constructive Interference ( m 1 ) L= + λ Likely multiple choice problem? Destructive Interference a sin θ = m λ m=0 d m=3 sin θ = m λ m= m=1 P36-1
Energy Storage Energy is stored in E & Fields u E u ε o E = : Electric Energy Density = µ o In capacitor: In EM Wave : Magnetic Energy Density In inductor: In EM Wave UC U L = = 1 1 CV LI P36 -
Energy Flow Poynting vector: S = E µ 0 (Dis)charging C, L Resistor (always in) EM Radiation For EM Radiation Intensity: I 0 0 0 0 E E c < S>= = = c µ µ µ 0 0 0 P36-3
Circuits There will be no quantitative circuit questions on the final and no questions regarding driven RLC Circuits Only in the multiple choice will there be circuit type questions UT. P36-4
Circuit Elements NAME Value V / ε Power / Energy Resistor R ρ = IR I R A Capacitor C Q Q = V C Inductor L = NΦ I di L dt 1 CV 1 LI P36-5
Circuits For what happens just after switch is thrown : Capacitor: Uncharged is short, charged is open Inductor: Current doesn t change instantly! Initially looks like open, steady state is short RC & RL Circuits have charging and discharging curves that go exponentially with a time constant: LC & RLC Circuits oscillate: VQI,, cos( ωt) ω = 0 1 LC P36-6
SAMPLE EAM P36-7
F00 #5, S003 #3, SF#1, SFC#1, SFD#1 Problem 1: Gauss s Law A circular capacitor of spacing d and radius R is in a circuit carrying the steady current i shown. At time t=0 it is uncharged 1. Find the electric field E(t) at P vs. time t (mag. & dir.). Find the potential at P, V(t), given that the potential at the right hand plate is fixed at 0 3. Find the magnetic field (t) at P 4. Find the total field energy between the plates U(t) P36-8
Solution 1: Gauss s Law 1.Find the electric field E(t): Assume a charge q on the left plate (-q on the right) E Gauss s Law: E S Q in A = = = d EA σ q = = Since q(t=0) = 0, q = it ε0 πr ε0 it E ( t) = to the right πr ε 0 ε σ A ε 0 0 P36-9
Solution 1.: Gauss s Law.Find the potential V(t): Since the E field is uniform, V = E * distance Vt () = E () t d d' = it d d' πr ε ( ) ( ) 0 Check: This should be positive since its between a positive plate (left) and zero potential (right) P36-30
Solution 1.3: Gauss s Law 3.Find (t): Ampere s Law: d s = Ienc + C µ µ ε 0 0 0 dφ dt E it Φ E = EA = πr ε0 d dt Φ E = ri R ε0 π r π r = 0+ µ ε µ ir 0 ( t) = out of the page π R ri 0 0 R ε 0 P36-31
Solution 1.4: Gauss s Law 4. Find Total Field Energy between the plates E Field Energy Density: εoe ε o it ue = = πr ε0 Field Energy Density: 1 µ 0ir u = = µ µ πr Total Energy U = ( u ) E + u dv 1 0i iπ 0 o R ε o it µ = R d + r d πr dr πr ε µ π i o o (Integrate over cylinder) it = + ( ) d 1 µ d o επ 0 R 8 π i q 1 = + C Li P36-3
Problem : Faraday s Law A simple electric generator rotates with frequency f about the y-axis in a uniform field. The rotor consists of n windings of area S. It powers a lightbulb of resistance R (all other wires have no resistance). 1. What is the maximum value I max of the induced current? What is the orientation of the coil when this current is achieved?. What power must be supplied to maintain the rotation (ignoring friction)? P36-33
Solution : Faraday s Law Faraday's Law: E d s = C dφ dt ε 1 dφ 1 d ns I = = = ns t = t R R dt R dt R ( cos( ω )) ωsin ( ω ) I ns R ns R max = ω = π f Max when flux is changing the most at 90º to current picture P36-34
Solution.: Faraday s Law. Power delivered? Power delivered must equal power dissipated! ns ns ( ) ( ) P = I R= π f sin ωt R= R π f sin ωt R R P R ns f R π = P36-35
ẑ Problem 3: Ampere s Law d d d L J 0 J 0 Consider the two long current sheets at left, each carrying a current density J 0 (out the top, in the bottom) a) Use Ampere s law to find the magnetic field for all z. Make sure that you show your choice of Amperian loop for each region. At t=0 the current starts decreasing: J(t)=J 0 at b) Calculate the electric field (magnitude and direction) at the bottom of the top sheet. c) Calculate the Poynting vector at the same location P36-36
ẑ Solution 3.1: Ampere s Law d d d 3 1 z=0 Region 1: J 0 J 0 y symmetry, above the top and below the bottom the field must be 0. Elsewhere is to right d s = µ 0Ienc = µ 0J0z = µ 0J0z Region : d s = µ 0Ienc = µ 0J0d = µ 0J0d Region 3: ( ( )) ( 3 ) = µ J d J z d = µ J d z 0 0 0 0 0 P36-37
ẑ Solution 3.: Ampere s Law d d d C J J E d s = dφ dt Why is there an electric field? Changing magnetic field Faraday s Law! Use rectangle of sides d, s to find E at bottom of top plate J is decreasing to right is decreasing induced field wants to make to right E out of page 1 0 s d d ( ) ( ) dj se = sd = sd µ 0dJ = sd µ 0 dt dt dt E = d µ a out of page P36-38
ẑ Solution 3.3: Ampere s Law d d d E J J Recall E = = µ 0Jd to the right 1 4 d µ 0a out of page Calculate the Poynting vector (at bottom of top plate): 1 1 ( 1 S= E = )( ) ˆ 4 d µ 0a µ 0Jd z µ µ 0 0 That is, energy is leaving the system (discharging) If this were a solenoid I would have you integrate over the outer edge and show that this = d/dt(1/ LI ) P36-39
Problem 4: EM Wave The magnetic field of a plane EM wave is: (( ) y ( ) t) π π = 10 cos m + 3 10 s i Tesla 9-1 8-1 ˆ (a) In what direction does the wave travel? (b) What is the wavelength, frequency & speed of the wave? (c) Write the complete vector expression for E (d) What is the time-average energy flux carried in the wave? What is the direction of energy flow? (µ o = 4 π x l0-7 in SI units; retain fractions and the factor π in your answer.) P36-40
(a) Solution 4.1: EM Wave (( ) y ( ) t) π π = 10 cos m + 3 10 s i Tesla 9-1 8-1 ˆ Travels in the - ˆj direction (-y) π (b) k = πm λ = = m k ω 3 ω=3π 10 s f = = 10 s π 8 m v = ω = λ f = 3 10 k s (c) E -1 8-1 8-1 (( ) y ( ) t) π π = 3 10 cos m + 3 10 s k V/m 1-1 8-1 ˆ P36-41
Solution 4.: EM Wave (( ) y ( ) t) π π = 10 cos m + 3 10 s i Tesla 9-1 8-1 ˆ (d) 1 S = E µ = 0 1 1 µ 0 E 0 0 S points along direction of travel: 1 1 ( 1)( 9) W = 3 10 10 π 7 4 10 m s ˆ - j P36-4
Problem 5: Interference In an experiment you shine red laser light ( =600 nm) at a slide and see the following pattern on a screen placed 1 m away: You measure the distance between successive fringes to be 0 mm a) Are you looking at a single slit or at two slits? b) What are the relevant lengths (width, separation if slits)? What is the orientation of the slits? P36-5
Solution 5.1: Interference First translate the picture to a plot: (a) Must be two slits Intensity -80-60 -40-0 0 0 40 60 80 Horizontal Location on Screen (mm) a d P36-6
Intensity Solution 5.: Interference -80-60 -40-0 0 0 40 60 80 Horizontal Location on Screen (mm) mλ y = Ltanθ Lsinθ = L d mλ d = L = y ( ) ( 7 6 10 m ) ( 10 ) 1 600nm ( 1m ) ()( ) ( 0mm) = 1 = 3 10 5 m At 60 mm a sinθ = 1 d () sinθ = 3 ( ) λ a 1 = λ d 3 5 a = 10 m P36-45
Why is the sky blue? 400 nm 700 nm Wavelength Small particles preferentially scatter small wavelengths You also might have seen a red moon last fall during the lunar eclipse. When totally eclipsed by the Earth the only light illuminating the moon is diffracted by Earth s atmosphere P36-46