ch29_Magnetic_Induct..
Transcription
ch29_Magnetic_Induct..
Chapter 29 Magnetic Induction Dr. Ray Kwok SJSU Static Fields (stationary charges, steady current) r r Qt ∫ E ⋅ da = ε Gauss’s Law o r r E ⋅ d l = 0 Conservative ∫ r r B ⋅ da = 0 No magnetic charge ∫ r r B ⋅ d l = µ I Ampere’s Law o t ∫ linear, homogeneous, isotropic Qt = Q f + Qb = Qf εr ≤ Qf Orsted’s Discovery R 4/21/1820 Danmark V I compass S Hans Christian Ørsted (1777–1851) Danmark cgs unit for B r r ∫ B ⋅ d l = µoIt André-Marie Ampère (1775–1836) France SI unit for current Ampere’s Law, 9/18/1820, after he learned about Orsted’s discovery on 9/11/1820 !! Faraday’s Experiment With stationary magnet, no current induced (1831) Michael Faraday (1791–1867) England SI unit for capacitance Joseph Henry (1797–1878) USA SI unit for inductance Magnetic induction Faraday’s Law r r dΦ Vemf ≡ ∫ E ⋅ d l = − dt r r Φ = ∫ B ⋅ da oppose the “change” of magnetic flux x Φ = ∫ Bldx = Blx 0 B l R I x u dΦ dx = Bl = Blu = −Vemf dt dt Vemf Blu I= = R R direction given by Lenz’s Law Lenz’s Law (1834) “Back emf” to oppose the “change” of magnetic flux B R I u x Heinrich Friedrich Emil Lenz (1804-1865) Italy Example – moving magnet Example - jumping ring FB FB B I Al ring I R V S e.g. magnet in a copper tube S N Bind Eddy current - disc brake Lenz’s Law metal plate r r r FB = IL × B To reduce eddy current transformer Example – metal detector Example – guitar pickup Question Find the direction of the current in the resistor R shown in Figure at each the following steps: (a) at the instant the switch is closed, (b) after the switch has been closed for several minutes, (c) when the variable resistance r increases, (d) when the circuit containing R moving to the right, away from the other circuit, and (e) at the instant the switch is opened. r S R V Right, 0, Left, Left, Left Moving Conductor FB + r rr FB=qu×B B saturate when u FB − induced emf B R r r FE + FB = 0 separate the charges r r r qE + qu × B = 0 induced E r r r E = −u × B similar to Hall Effect V = uB l V = uBl induced current + I = |V/R| = B u/R l I u - direction agrees w/ Lenz’s Law emf is due to the FB, no E field build up. Example – rotating loop θ = ωt uniform B d r r Vemf = − ∫ B ⋅ da dt d = − ( BA cos ωt ) = ωBA sin ωt dt AC generator dΦ Vemf = − N dt = NωBA sin ωt DC generator Electromotor Project 1: Make one in two weeks. To turn faster, should we 1. 2. 3. 4. use thicker wire? use more turns? make bigger loop? use stronger magnet? Answer… (not counting friction) dΦ = NωBA sin ωt dt NωBA | I |= R r r r FB = I l × B r r r r r τ = r × FB = m × B torque r magnetic moment m = NIAn̂ r r I = moment of inertia τ = Iα Vemf = − N l resistivity A' m = ρV = ρA' l mass density R =ρ 1. use thicker wire? same. e.g. half R, double mass, same α 2. use more turns? same. double N, double R, same I, double mm, but double inertia, same α. 3. use bigger loop? (same N) same. bigger A, more Φ somewhat higher R, but still more I ~A/R~ r mm~A2/R~r3, inertia~r3, same α 4. use stronger magnet? YES more I, m, τ, α Another important application – telephony / loudspeaker (?? before him) Alexander Graham Bell (1847 –1922) UK (practical improvement) Thomas Alva Edison (1847 –1931) USA Ideal Transformer (AC) dΦ dt dΦ V2 = − N 2 dt V1 N1 = Voltage transformer V2 N 2 Φ is confined in the core (µ = ∞) V1 = − N1 P1 = P2 RL V1I1 = V2 I 2 V1 I 2 N1 = = V2 I1 N 2 2 2 V1 V2 N1 N1 = R L R in = = I1 I2 N 2 N 2 Impedance transformer Self-Inductance back emf causes I lags V NΦ L≡ I LI = NΦ dΦ dI V = −N = −L dt dt Ideal Solenoid ideal: • large N tightly wound • no end effect • uniform internal B • zero external B in the vicinity Ampere’s Law: µ o NI = µ o nI B= l r r Φ = ∫ B ⋅ da =µ o nIA NΦ Nµ o nIA L= = = µ o n 2lA I I LC Resonator (Lenz’s Law, later resonator…) Transformer - Primary / Secondary coils same cross-section? V1 V2 ∂Φ = =− N1 N 2 ∂t N 2 N1 M12 = = I1 I 2 Φ12 same equations different cross-section? Electrodynamics r r Qt ∫ E ⋅ da = ε o dΦ r r ∫ E ⋅ d l = − dt r r ∫ B ⋅ da = 0 r r ∫ B ⋅ d l = µ o I t Gauss’s Law Faraday’s Law No magnetic charge Ampere’s Law Exercise Rectangular loop of 20 turns is placed at 1 m away from a long current-carrying wire as shown. I2 N = 20 I1(t) = 2 cos(60t) (A) Total resistance of the wire in the loop is 4 Ω. 5m Find I2. I1(t) 3m 1m