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Lecture25

# Lecture25 - Lecture 25 Chapter 23 Faraday’s Law...

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Unformatted text preview: Lecture 25 Chapter 23. Faraday’s Law Inductance Constant voltage – constant I, no curly electric field. Increase voltage: dB/dt is not zero emf For long solenoid: B = µ0 NI d Change current at rate dI/dt: emfbat R emfcoil Inductance ENC emfbat emf = Inductance ENC emfbat R L emfind emf ind = L dI dt R emfcoil EC EC µ0 N 2 2 dI πR d dt dI dt Increasing I increasing B emf ind = L emf = − dΦ mag dt emfbat R emfind L – inductance, or self-inductance Unit of inductance L: Henry = Volt.second/Ampere Increasing the current causes ENC to oppose this increase 1 Inductance: Decrease Current ENC EC emfbat R L emfind emf ind = L dI dt RL Circuits •  At t=0, the switch is closed and the current I starts to flow. a I b R I # L emf = − dΦ mag dt Conclusion: Inductance resists changes in current Orientation of emfind depends on sign of dI/dt 4/19/11 Initially, an inductor acts to oppose changes in current through it. A long time later, it acts like an ordinary connecting wire. 6 RL Circuits •  Find the current as a function of time. I= a I b R I RL Circuits ( on) ε 1 − e− Rt / L R Current /1 R L/R 2L/R ε ε 1 − e− Rt / L = 1 − e− t /τ RL R R ( ) ( ) # L I= ( ) Q f( x ) 0.5 I 0 0 •  What about potential differences? 0 1 2 x t/RC t 3 4 Voltage on L VL = L dI = ε e− Rt / L dt 1 1 # f( x V ) 0.5 L 0.0183156 4/19/11 7 4/19/11 0 0 0 1 2 x t 3 4 4 8 2 RL Circuits •  Why does RL increase for larger L? a I b R I RL Circuits After the switch has been in position for a long time, redefined to be t=0, it is moved to position b. a I b R I # L # L •  Why does RL decrease for larger R? 4/19/11 9 4/19/11 10 RL Circuits ( off) ε − Rt / L e R (on) /R 11 L/R 2L/R /1 R L/R 2L/R 1 (off) /R 1 L/R 2L/R Current I= f( x ) 0.5 I Q f( x ) 0.5 I I= ε 1 − e− Rt / L R ( ) 4 f( x ) 0.5 I I= ε − Rt / L e R 0.0183156 0 0 1 2 x t 3 4 4 1 0 0 0 1 2 x t/RC t 0.0183156 3 0 0 1 2 x t 3 4 4 Voltage on L VL = L dI = −ε e− Rt / L dt 1 00 #1 1 00 Q f( x ) V 0.5 L f( xV ) 0.5 L VL = L dI = ε e− Rt / L dt Q f( x ) 0.5 VL VL = L dI = −ε e− Rt / L dt -0 # 4/19/11 0.0183156 0 1 2 x t/RC t 3 4 0 11 0 1 2 x t 3 4 4 0 -# 4/19/11 0 0 1 2 x t/RC t 3 4 12 3 Inductor in Series i L1 i L2 Inductor in Parallel L1 i i1 i2 L2 4/19/11 13 4/19/11 14 LC Circuits •  Consider the LC and RC series circuits shown: •  Suppose that at t=0 the capacitor is charged to a value of Q. ++++ ---- LC Oscillations Kirchoff’s loop rule CR ++++ ---- I Q ++ - C L dI Q VL + VC = L + = 0 dt C C L Is there is a qualitative difference in the time development of the currents produced in these two cases. Why?? 15 4/19/11 4/19/11 16 4 x 0, r1 n .. r1 1.01 1 Q f( x ) 0 Q = Q0 cos(ω0 t ) x 0, r1 n .. r1 LC Oscillations LC Oscillations: Energy Check ω0 = 1 LC 0 1.01 VC 1 f( x ) 0 1.01 0 I 1 1.01 1 0 0 2 4 6 1.01 1 6.28 I = −ωx Q0 sin(ω 0 t ) 0 f( x ) 0 0 0 1.01 1 0 2 x 4 6 6.28 VL 1.01 1 0 1.01 1 0 f( x ) 0 2 x 4 6 6.28 0 •  The other unknowns ( Q0, ) are found from the initial conditions. eg in our original example we took as given, initial values for the charge (Qi) and current (0). For these values: Q0 = Qi, = 0. •  Question: Does this solution conserve energy? dI dt 1.01 1 f( x ) 0 0 0 0 2 tx 4 6 6.28 1.01 1 0 0 2 x 4 4/19/11 6.28 t6 17 4/19/11 18 Energy Check Energy in Capacitor 12 U E (t ) = Q0 cos 2 (ω 0t + φ ) 2C x 0, r1 n .. r1 1 UB versus UE UE f( x ) 0.5 Energy in Inductor 1 2 0 U B (t ) = Lω 0 Q02 sin 2 (ω 0tx+ φ0), r1 .. 0 0 r1 2 n 1 ω0 = LC 1 2 tx 4 6 U B (t ) = Therefore, 12 Q0 sin 2 (ω 0t + φ ) 2C UB f( x ) 0.5 U E (t ) + U B (t ) = Q02 2C 19 0 0 4/19/11 0 2 x t 4 6 20 4/19/11 5 LC Oscillations with Finite R •  If L has finite R, then energy will be dissipated in R and x 0, r1 n Driven Oscillations •  An LC circuit is a natural oscillator. n 100 the oscillations will become damped. 10 r1 r1 .. r1 x 0, n .. r1 ω resonance = 1 in absence of resistive loss LC + - + - R C r1 10 n 100 L Q 1 Q f( x ) 0 0 1 f( x ) 0 0 •  In a real LC circuit, we must account for thex 0, r1 .. r1 resistance of the inductor. This resistance will n Q damp out the oscillations. 1 f( x ) 0 1 1 0 t5 x R=0 10 1 0 5 x 10 0 R=0 t5 x 10 t 21 4/19/11 4/19/11 22 6 ...
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