Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P36-1
Before 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) & Biot-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 d A = (Gauss's Law) ε 0 dφ B E d s = (Faraday's Law) dt B d A = 0 (Magnetic Gauss's Law) B d s = µ I + µ ε 0 enc 0 0 dφ dt E (Ampere-Maxwell Law) P36-6
Gauss s Law: E A S d = q in ε 0 Spherical Symmetry Cylindrical Symmetry Planar Symmetry P36-7
Maxwell s Equations w ³³ C C S v ³ w ³³ S v ³ G G E d A G G E d s G G B d A G G B d s Q H in 0 d ) dt B 0 enc 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) P I P H d ) dt E (Ampere-Maxwell Law) P36
Faraday s Law of Induction ε = E ds = d Φ N B dt Moving bar, entering field = N d ( BA dt cos θ ) Lenz s Law: Ramp B Rotate area in field Induced EMF is in direction that opposes the change in flux that caused it P36-9
Maxwell s Equations w ³³ C C S v ³ w ³³ S v ³ G G E d A G G E d s G G B d A G G B d s Q H in 0 d ) dt B 0 enc 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) P I P H d ) dt E (Ampere-Maxwell Law) P36 3
B d s = µ 0 I Ampere s Law:. enc Long Circular Symmetry B B I (Infinite) Current Sheet Torus/Coax Solenoid = Current Sheets B P36-11
Displacement Current Q E = Q = ε 0 EA = ε0 Φ ε A dq dt 0 ε dφ dt E = 0 I d E B s C d = µ ( I + I ) 0 encl = I + d µ µ ε 0 encl 0 0 Capacitors, EM Waves dφ dt E P36-1
Maxwell s Equations C C S S Qin EdA (Gauss's Law) 0 d B Ed s (Faraday's Law) dt BdA0 (Magnetic Gauss's Law) Bd s 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 B V = E d s = V B V A A = Ed (if E constant e.g. Parallel Plate C) Common second step to Gauss Law dv E = V = e.g. î dx Less Common Give plot of V, ask for E P36-15
Force Lorentz Force: Single Charge Motion B) Cyclotron Motion F = q (E + v Cross E & B for no force Magnetic Force: df = Id s B B B F = I (L B ) Parallel Currents Attract Force on Moving Bar (w/ Faraday) P36-16
The Biot-Savart Law Current element of length ds carrying current I (or equivalently charge q with velocity v) produces a magnetic field: B = µ q v x r ˆ o 4π r µ db = 0 4 π I d s r rˆ P36-17
Magnetic Dipole Moments µ IA nˆ IA Generate: Feel: 1) Torque aligns with external field ) Forces as for bar magnets τ = µ B P36-18
Traveling Sine Wave i Wavelength: λ = ˆ i Frequency : f E EE sin(kx ωt) 0 π i Wave Number: k = λ i Angular Frequency: ω = π f Good chance this 1 π will be one i Period: T = = question! f ω ω i Speed of Propagation: v = = λ f k i Direction of Propagation: + x P36-19
EM Waves Travel (through vacuum) with speed of light 1 v= c = = 3 10 µε s 0 0 8 m At every point in the wave and any instant of time, E and B are in phase with one another, with E = E 0 = c B B 0 E and B fields perpendicular to one another, and to the direction of propagation (they are transverse): Direction of propagation = Direction of E B P36-0
Interference (& Diffraction) L= m λ Constructive Interference ( 1 ) L = m + 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 & B Fields ε E u = o E : Electric Energy Density u B = In capacitor: U C = 1 CV B In EM Wave : Magnetic Energy Density µ o In inductor: U L = 1 LI In EM Wave P36 -
Energy Flow Poynting vector: S = E B µ 0 (Dis)charging C, L Resistor (always in) EM Radiation For EM Radiation Intensity: I 0 0 0 0 E B E cb < 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 BUT. P36-4
Circuit Elements NAME Value V / ε Power / Energy Resistor R = ρ IR I R A Capacitor C Q Q = V C 1 CV Inductor L = N Φ I di L dt 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( ω t ) ω = 0 1 LC P36-6
SAMPLE EAM P36-7
F00 #5, S003 #3, SFB#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 B(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) Gauss s Law: E d A Q in σ A = EA= = σ q E = = Since q(t=0) = 0, q = it ε πr 0 ε0 it E ( t ) = to the right π ε S ε 0 ε 0 R 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 π ε R 0 (d d ') 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 B(t): Ampere s Law: B d s = I enc + C µ µ ε 0 0 0 dφ dt E EA it Φ E = = πr ε 0 d Φ dt E = ri R ε 0 π r B t π rb = 0 + µ ε µ ir 0 ( ) = 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 ε E ε it o o E Field Energy Density: u E = = πr ε 0 B Field Energy Density: u B = B 1 µ = 0 ir µ µ π R Total Energy U = (u E + u B )dv o o (Integrate over cylinder) = ε o it 1 iπ R µ d + 0 i R r id π r dr πr ε0 µ o π ( ) d 1 µ od = it επr + 8 π i = qc + 1 Li 0 P36-3
Problem : Faraday s Law A simple electric generator rotates with frequency f about the y-axis in a uniform B 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 C s dφ = dt ε 1 d Φ B 1 d nbs I = = = nbs t = t R R dt R dt R B ( cos ( ω )) ω sin ( ω ) I nbs R nbs 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! nbs nbs ( ) ( ) P = I R = π f sin ω t R = R π f sin ωt R R P R nbs = π f R P36-35
ẑ d d d Problem 3: Ampere s Law L J 0 Consider the two long current sheets at left, each J 0 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 d = µ I B = µ J z By symmetry, above the top and below the bottom the B field must be 0. Elsewhere B is to right B s 0 enc 0 0 Region : d = µ I B = µ J d B s 0 enc 0 0 Region 3: B = µ ( J d 0 0 0 0 0 J ( z d )) B = µ J d z B B = µ J = µ J 0 0 3 0 0 z d ( ) P36-37
ẑ Solution 3.: Ampere s Law d d J s Why is there an electric field? Changing magnetic field Æ d J Faraday s Law! E d s = C dφ B dt Use rectangle of sides d, s to find E at bottom of top plate J is decreasing Æ B to right is decreasing Æ induced field wants to make B to right Æ E out of page se = d ( Bsd ) = sd d (µ dj ) = sd µ dj dt dt E = 1 d µ 0 a out of page 0 0 dt P36-38
ẑ Solution 3.3: Ampere s Law d d d E B J J Recall E = 1 d 4 µ 0 a out of page B = µ0 Jd to the right Calculate the Poynting vector (at bottom of top plate): 1 1 1 S = E B = ( 4 d µ 0 a)(µ 0Jd µ 0 µ 0 ) ẑ 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: 9 (( -1) ( 8 ) ) ˆ -1 B = 10 cos π m y + 3π 10 s t i Tesla (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
Solution 4.1: EM Wave =10 9 cos (( π m -1 y + 3π 10 s -1 t î Tesla B ) ( 8 (a) Travels in the - ĵ direction (-y) ) ) (b) k = π m -1 λ = π = m k ω 8-1 3 ω π 8-1 =3 10 s f = = 10 s π ω 8 m v = = λ f = 3 10 k s (c) 1-1 8-1 E = 3 10 cos ˆ ((π m ) y + (3π 10 s )t)k V/m P36-41
B Solution 4.: EM Wave (( ) ( -1 y ) t) π π = + 10 9 cos m -1 3 10 8 s i Tesla ˆ (d) 1 S = E B µ = 0 1 1 µ 0 EB 0 0 S points along direction of travel: 1 1 ( 1 )( 9 ) W = 3 10 10 7 4 π 10 m s ˆ - j P36-4
Problem 5: Interference In an experiment you shine red laser light (O=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 mλ y = L tan θ Lsin θ = L d mλ 1 600nm) d = L = ( 1m ) ()( -80-60 -40-0 0 0 40 60 80 y (0mm) Horizontal Location on Screen (mm) 7 (6 10 ) 5 = ( 1m ) ( 10 ) = 3 10 m At 60 mm a sinθ = ()λ 1 a 1 = d sinθ = ( 3 )λ d 3 a =10 5 m P36-45
Why is the sky blue? 400 nm Wavelength 700 nm 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