Electricity and Magnetism-Review lecture with conceptual questions

Class 36: OutlineHour 1: Concept Review /Overview PRS Questions – Possible Exam Questions Hour 2: Sample Exam Yell if you have any questionsP36 -1Before 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 -2Final Exam Topics Maxwell’s Equations: 1. Gauss’s Law (and “Magnetic Gauss’s Law”) 2. 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)2. Exert a force 3. Move as waves (that can interfere & diffract) 4. Contain and transport energy Circuit Elements: Inductors, Capacitors, Resistors P36 -3Test FormatSix Total “Questions” One with 10 Multiple Choice Questions Five Analytic Questions 1/3 Questions on New Material2/3 Questions on Old MaterialP36 -45P36 -Maxwell’s Equations 6P36 -Maxwell’s Equations 0 0 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) (Ampere-Maxwell Law) in S B C S E enc C Qd dd dt d dd I dt ε µ µε ⋅ = Φ ⋅ = − ⋅ = Φ ⋅ = + ∫∫ ∫ ∫∫ ∫ EA E s BA B s 􀁇􀁇 􀁇 􀁇 􀁇􀁇 􀁇 􀁇 􀁷 􀁶 􀁷 􀁶 7P36 -Gauss’s Law: 0S inqd ε ⋅ =∫∫ EA 􀁇􀁇 􀁷 Spherical Symmetry Cylindrical Symmetry Planar Symmetry 2P36 Maxwwell’ Equations 0 0 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) (Ampere-Maxwell Law) in S B C S E enc C Qd d d dt d d d I dt H P PH ))³³ ³ ³³ ³ EA E s BA B s GG G G GG G G w v w v Faraday’s Law of Inductionε=􀁶∫ E 􀁇 ⋅ds 􀁇 =− NdΦB Moving bar,dt entering field =−Nd (BA cos θ) Lenz’s Law: in field Induced EMF is in direction that opposes thechange in flux that caused it dt Ramp B Rotate area P36 -93P36 Maxwwell’ Equations 0 0 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) (Ampere-Maxwell Law) in S B C S E enc C Qd d d dt d d d I dt H P PH ))³³ ³ ³³ ³ EA E s BA B s GG G G GG G G w v w v I 11P36 -= 2 Current Sheets Ampere’s Law: .∫ =⋅ encId 0µsB 􀁇􀁇 B B X X X X X X X X X X X X X X X X X X X X X X X X X X X X B Long Circular Symmetry (Infinite) Current Sheet Solenoid Torus/Coax 12P36 -Displacement Current 0 E d ddQ Idt dtε Φ = ≡ 0 0 0 E QE Q EA A ε εε = ⇒ = = Φ 0 0 0 0 ( )encl d C E encl d I I dI dt µ µ µε ⋅ = + Φ = + ∫ Bs 􀁇 􀁇 􀁶 Capacitors, EM Waves 4 P36 -Maxwell’s Equations 0 0 0 0 (Gauss's Law) (Faraday's Law) 0 (Magnetic Gauss's Law) (Ampere-Maxwell Law) in S B CS E enc C Q d d d dt d d d I dt 􀁈 􀁐 􀁐􀁈 􀂘 􀀠 􀀩 􀂘 􀀠􀀐 􀂘 􀀠 􀀩 􀂘 􀀠 􀀎 􀂳􀂳 􀂳􀂳􀂳 􀂳 E A E s B A B s􀁇 􀁇 􀁇 􀁇􀁇 􀁇 􀁇 􀁇 􀁷􀁶􀁷􀁶I am nearly certain that you will have one of each They are very standard – know how to do them all14P36 -EM Field Details… Electric PotentialB 􀁇􀁇V ∫E ⋅d s =VB −VA∆=− A =−Ed (if E constant – e.g. Parallel Plate C) Common second step to Gauss’ Law E 􀁇 =−∇ V =e.g. −dV ˆidx Less Common – Give plot of V, ask for E P36 -15ForceLorentz Force: • Single Charge Motion 􀁇 􀁇􀁇􀁇 • Cyclotron Motion F = q (E + v × B) • Cross E & B for no force Magnetic Force: dF 􀁇 = Id s 􀁇 × B 􀁇 ⇒ F 􀁇 = I (L 􀁇 × B 􀁇 )BB • Parallel Currents Attract • Force on Moving Bar (w/Faraday) P36 -16The Biot-Savart LawCurrent element of length ds carrying current I (or equivalently charge q with velocity v) produces a magnetic field:􀁇 B=oµq 􀁇 4π2r 􀁇Bd0 4 µ π I = 􀁇×d2r xˆvr srˆP36 -1718P36 -Magnetic Dipole Moments Anµ 􀁇􀁇 IIA ≡≡ ˆ Generate: Feel: 1) Torque aligns with external field 2) Forces as for bar magnets =×τ µB 􀁇􀁇 􀁇 Traveling Sine Wavei Wavelength: λ 􀁇 = ˆ −i Frequency : f EEE0 sin(kx ωt)2πi Wave Number: k = λ Good i Angular Frequency: ω= 2πf chance this 12π will be one i Period: T == question!f ω ωi Speed of Propagation:v = =λfk i Direction of Propagation: + x P36 -19EM WavesTravel (through vacuum) with speed of light 18 m == = 310 ×vc µε s00 At every point in the wave and any instant of time, E and B are in phase with one another, with E = E0 = cBB 0 E and B fields perpendicular to one another, and to the direction of propagation (they are transverse):􀁇􀁇 Direction of propagation = Direction of ×EB P36 -20 21P36 -Interference (& Diffraction) ( )1 2 Constructive Interference Destructive Interference Lm L m λ λ ∆= ⇒ ∆= + ⇒ sind mθλ=sina mθλ=Likely multiple choice problem? m=1m=2 m=0 m=3 Energy Storage Energy is stored in E & B Fields εE2uE = o : Electric Energy Density2In capacitor: UC = 1 CV 22 In EM Wave B2uB = : Magnetic Energy Density2µo In inductor: UL = 1 LI 2 2In EM Wave P36 -2223P36 -Energy Flow 0µ × = EBS 􀁇 􀁇􀁇 Poynting vector: 2 2 00 0 0 0 0 0 Intensity: 2 2 2 E B E cB I S cµ µ µ ≡< >= = = For EM Radiation • (Dis)charging C, L • Resistor (always in) •EM Radiation CircuitsThere 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 -24Circuit ElementsP36 -25 NAME Value V /ε Power /Energy Resistor R A ρ = 􀁁IR2IR Capacitor C Q V = ∆ Q C 21 2CV Inductor L N I Φ = dIL dt−21 2LI CircuitsFor “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: ,, ∝ cos( t ω0 = 1VQI ω) LC P36 -2627P36 -SAMPLE EXAM F2002 #5, S2003 #3, SFB#1, SFC#1, SFD#1Problem 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 uncharged1. Find the electric field E(t) at P vs. time t (mag. & dir.)2. 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 -28 Solution 1: Gauss’s Law 1.Find the electric field E(t): Assume a charge q on the left plate (-q on the right) ⋅d 􀁇􀁷 Gauss’s Law:􀁇 ∫∫ EAQin σAEA===ε0 ε0S σ qE =Since q(t=0) = 0, q = it=R2ε πε00􀁇E = it() t 2 to the right πεR0 P36 -29Solution 1.2: Gauss’s Law2.Find the potential V(t): Since the E field is uniform, V = E * distance 􀁇 Vt()=E()t (d −d ')= it 2 (d −d ')πεR0 Check: This should be positive since its between a positive plate (left) and zero potential (right) P36 -30 31P36 -Solution 1.3: Gauss’s Law 3.Find B(t): Ampere’s Law: 2 2 0 E itEA rR ππε ⎛ ⎞Φ= =⎜ ⎟ ⎝ ⎠ 0 2( ) out of the page 2 irt R µ π =B 􀁇 0 0 0 E enc C dd I dtµ µεΦ ⋅ = +∫ B s 􀁇 􀁇 􀁶 2 2 0 Ed ri dt R ε Φ = 2 00 2 0 2 0 rirB Rπ µε ε=+ Solution 1.4: Gauss’s Law4. Find Total Field Energy between the plates εE2 ε⎛ it ⎞2 E Field Energy Density: uE = o = o ⎜ 2 ⎟22 ⎝πε0 ⎠R B Field Energy Density: uB = B µ 2 o = 1 o ⎛⎜⎝ 2 µ0ir 2 ⎞⎟⎠ 2 22µπ R Total Energy U =∫∫∫(uE +uB )dV (Integrate over cylinder) =εo ⎛⎜ it 2 ⎞⎟ 2 iπR2d + 1 ⎛⎜ 2 µ0 Ri 2 ⎞⎟ 2 ∫ r2 id ⋅2πr dr 2 πεR 2µπ ⎝ 0 ⎠ o ⎝⎠()2 d 1 µd 22o= it 2 επR + 28 πi ⎜⎛ = 2 qC +12 Li2 ⎟⎞ 02 ⎝⎠ P36 -3233P36 -Problem 2: 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 Imax of the induced current? What is the orientation of the coil when this current is achieved? 2. What power must be supplied to maintain the rotation (ignoring friction)? 34P36 -Solution 2: Faraday’s Law Faraday's Law: B C dd dt Φ ⋅ =−∫ E s 􀁇 􀁇 􀁶 ()( ) ()1 1 cos sin Bd d nBS I nBS t tR R dt R dt R ε ω ω ω Φ = =− =− = max 2nBS nBS I fR Rω π= = Max when flux is changing the most – at 90º to current picture 35P36 -Solution 2.2: Faraday’s Law () () 2 2 2 22 sin 2 sin nBS nBS P IR f t R R f tR Rπ ω π ω⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 2. Power delivered? 2 2 2 RnBS P fR π⎛ ⎞ = ⎜ ⎟⎝ ⎠ Power delivered must equal power dissipated! Problem 3: Ampere’s Law X d d d ˆz J0 Consider the two long current sheets at left, each J0 carrying a current density J0 L (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)=J0 – 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 -36Solution 3.1: Ampere’s Law X d d d J0 J0 ˆz By symmetry, above the top and below the bottom the B field must be 0. Elsewhere B is to rightz=0 1 2 3 􀁁 Region 1:􀁇􀁇 ⋅ d =µI ⇒ B􀁁 =µJz 􀁁􀁶∫ Bs 0 enc 00 ⇒ Region 2:􀁇􀁇 ⋅ d =µI ⇒ B􀁁 =µJd 􀁁􀁶∫ Bs 0 enc 00 ⇒ Region 3: B􀁁 =µ( Jd − J ( z − 2d ))􀁁 ⇒00 0 0 0B J z µ= 0 0B J d µ= 37P36 -( )0 0 3B J d zµ= − zˆSolution 3.2: Ampere’s Law X s Why is there an electric field?d J d Changing magnetic field • Æd J Faraday’s Law! 􀁶∫ E 􀁇 ⋅ d s 􀁇 =− dΦB Use rectangle of sides d, s to dt find E at bottom of top plateC J is decreasing ÆB to right is decreasing Æinduced field wants to make B to right ÆE out of page 2sE = d ( Bsd )= sd d (µdJ )= sd 2µ dJ dt dt 00 dt􀁇 ⇒ E = 12 d 2µ0a out of page P36 -38Solution 3.3: Ampere’s Law X d d d ˆz BE RecallJ 􀁇E = 14 d 2µ0a out of page J B 􀁇 =µ0Jd to the right Calculate the Poynting vector (at bottom of top plate):􀁇 1 􀁇􀁇 1 12S = E× B =( d µa)(µJd ) zˆ40 0µ0 µ0 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(1/2 LI2)P36 -39Problem 4: EM WaveThe magnetic field of a plane EM wave is:􀁇 −9 (( -1 )( 8 -1 )) ˆB = 10 cos πm 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? (µo= 4 π x l0-7 in SI units; retain fractions and the factor π in your answer.) P36 -40Solution 4.1: EM Wave−9 (( -1 )( 8 -1 ))B 􀁇 =10 cos πm y + 3π×10 s t ˆi Tesla(a) Travels in the -ˆj direction (-y) (b) k =πm-1 ⇒ λ= 2π=2m k 8-1 ω 3 8-1 ωπ=3 ×10 s ⇒ f = =×10 s 2π 2 ω 8m v == f =310 λ×k s 􀁇(c) −1 -1 8-1 ˆE =− × 3 10 cos ((πm )y +(3π×10 s )t )k V/mP36 -4142P36 -Solution 4.2: EM Wave ( )( ) 0 00 0 1 9 7 2 1(d) 11 2 1 1 W310 10 24 10 m s EB µ µ π − − − = × = ⎛ ⎞ = ×⎜ ⎟×⎝ ⎠ S E B 􀁇 􀁇 􀁇 ( ) ( )( )9 -1 8 -1 ˆ10 cos m 3 10 s Tesla y tπ π−= + ×B i 􀁇 S points along direction of travel: ˆ-j Problem 5: InterferenceIn 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 20 mm a) Are you looking at a single slit or at two slits? b) What are the relevant lengths (width, separation if 2 slits)? What is the orientation of the slits? P36 -5Solution 5.1: InterferenceFirst translate the picture to a plot: (a) Must be two slits a Intensity -80 -60 -40 -20 0 20 40 60 80Horizontal Location on Screen (mm) d P36 -6 Solution 5.2: InterferenceIntensity mλ = tan θ≈Lsin yL θ=Ld mλ 1 600nm)( )()( dL = 1= m -80-60 -40 -20 0 20 40 60 80 y (20mm)Horizontal Location on Screen (mm) ×(610 −7 )−5=( )1m (× )=3×10 m 210 −2 At 60 mm… sinθ=()λ a1a 1 ⇒ = sinθ=()λ d 3d 3 a =10−5 m P36 -45 Why is the sky blue?400 nm Wavelength 700 nmSmall particles preferentially scatter small wavelengthsYou 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 atmosphereP36 -46