Unformatted text preview: Physics 212
Lecture 29
Course Review
• The Topics You Want to See
– – – – – – – – Electric Fields/Gauss’ Law/Potential (33%) Faraday’s Law (14%) RC/RL Circuits (12%) AC Circuits (10%) Geometric Optics (10%) Magnetic Fields & Forces (9%) Electromagnetic Waves/Polarization (9%) DC Circuits (3%) Physics 212 Lecture 29, Slide 1 Music
Who are the Artists?
BB BB A) B) C) D) E) Whitney Houston and Tina Turner Nina Simone and Patti LaBelle Nina LaBelle Etta James and Bonnie Raitt Etta Raitt The Dixie Cups Marcia Ball, Irma Thomas, Tracy Nelson THANKS FOR YOUR REQUESTS I WILL BE CHECKING THEM OUT… Why? Why? Marcia Ball was requested ! Marcia I’ve been thinking about New Orleans… New Orleans Jazzfest Poster from last year New Jazzfest Poster The Sweet Soul Queen of New Orleans: Irma Thomas
Physics 212 Lecture 29, Slide 2 Physics 212
Lecture 29
Course Review
• The Topics You Want to See
– – – – – – – – Electric Fields/Gauss’ Law/Potential (33%) Faraday’s Law (14%) RC/RL Circuits (12%) AC Circuits (10%) Geometric Optics (10%) Magnetic Fields & Forces (9%) Electromagnetic Waves/Polarization (9%) DC Circuits (3%) Physics 212 Lecture 29, Slide 3 1
I I1 I – I1 L dI1 + IR − V = 0 dt 2 1 L dI1 − ( I − I1 ) R = 0 dt IR = V − L dI1 dt 30% 2 L dI1 dI − V + L 1 + I1R = 0 dt dt 2L dI1 + I1R = V dt Strategy: Back to First Principles
• The time constant is determined from The a differential equation for the current through the inductor. through • Equation for current through inductor Equation obtained from Kirchhoff’s Rules obtained τ= " L" 2 L = " R" R Physics 212 Lecture 29, Slide 4 BB Horizontal components cancel E from top arc points down E from bottom arc points up Etotal points down Top arc produces smaller horizontal components Calculation:
Etop = ∫ dq 4πε 0 r
2
θ cos θ Etop = 1
2 Q 4πε 0 r 2rθtop 2 ∫ rdθ cos θ
0 θ top sinθ θ Etop = Q sin θtop 4πε 0 r 2 θtop θ
Physics 212 Lecture 29, Slide 5 BB Potential Energy is a measure of work done by E field Spherical symmetry & Gauss’ law law E = 0 inside shell inside E = 0 inside shell inside No work done to move q No No change in potential energy !
Physics 212 Lecture 29, Slide 6 BB ALWAYS START ALWAYS FROM DEFINITION OF POTENTIAL OF
∆V = − ∫ E ⋅ dr
(A) (B)
1 2Q 3Q − 4πε 0 a b 1 3Q 2Q − 4πε 0 b a a (C) 0 (D) (E) 1 2Q 2Q − 4πε 0 a b 1 2Q 2Q − 4πε 0 b a b Spherical symmetry & Gauss’ law determines E Gauss 1 2Q a < r < b: E = 4πε 0 r 2
a dr 2Q ∆V = − 4πε 0 ∫2 br ∆V = 2Q 1 1 − 4πε 0 a b Physics 212 Lecture 29, Slide 7 BB (A) (B) 1 2Q 4πε 0 a − 1 2Q 4πε 0 a (C) 0 (D) (E) 1 3Q 4πε 0 b − 1 3Q 4πε 0 b Spherical symmetry & Spherical Gauss’ law determines E Gauss r < a:
E=0 ∆V = − ∫ E ⋅ dr = 0
0 a Physics 212 Lecture 29, Slide 8 Charge must be Charge induced to insure E = 0 within conducting shell conducting BB Spherical symmetry & Gauss’ law determines E Gauss Q E ⋅ dA = enclosed ∫ ε0
(A) (B)
− Q1 4πa 2 + Q1 4πa
2 (C) Q2 − Q1 4π (b − a )
2 2 (D) (E) − Q1 4πb 2 + Q1 4πb 2 E ⋅ 4πr 2 = 0
Qenclosed = Q1 + (−Q1 ) σ= − Q1 4πb 2 Physics 212 Lecture 29, Slide 9 BB Physics 212 Lecture 29, Slide 10 10 BB ALWAYS START ALWAYS FROM DEFINITION OF POTENTIAL OF
∆V = − ∫ E ⋅ dr
0
b Break integral into two pieces Break
∆V = − ∫ E ⋅ dr − ∫ E ⋅ dr
0
conductor: = 0 insulator: ∫ 0 insulator: a b a
same for same conductor & insulator insulator Physics 212 Lecture 29, Slide 11 11 BB 76% Current induced because flux is changing Flux is changing beause B is changing Flux beause At t = 5 seconds, B is positive, but decreasing Lenz’ law: emf induced to oppose change that brought it into being law: emf Induced current must produce positive B field Positive B field produced by counterclockwise current
Physics 212 Lecture 29, Slide 12 12 BB 62% Flux definition: Φ = ∫ B ⋅ dA = BA = Bwh Flux Faraday’s law: ε = −
dB = −2T / s dt dΦ dB = − wh dt dt I= ε
R = 9/5 = .012 A 150
Physics 212 Lecture 29, Slide 13 13 BB 65% Current is determined by time rate of change of the flux dΦ/dt is determined by dB/dt dB/dt (6s) = dB/dt (5 s) = 2 T/s (5 The induced currents at t = 5s and t = 6s are equal
Physics 212 Lecture 29, Slide 14 14 Phasor diagram at t = 0 What is VC at t = π/(2ω) at
(A)
+ VC max sin α − VC max sin α (C) (D) + VC max cos α − VC max cos α α BB (B) Phasor diagram at t = 0
VR VC α
VL Voltage is equal to Voltage projection of phasor phasor along vertical axis
Physics 212 Lecture 29, Slide 15 15 ...
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This note was uploaded on 09/22/2011 for the course PHYSICS 212 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.
 Fall '10
 Selig
 Electric Fields, Force, Polarization

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