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Lecture 18-1 Eddy Current A current induced in a solid conducting object, due to motion of the object in an external magnetic field. • The presence of eddy current in the object results in dissipation of electric energy that is derived from mechanical motion of the object. • The dissipation of electric energy in turn causes the loss of mechanical energy of the object, i.e., the presence of the field damps motion of the object. Lecture 18-2 Self-Inductance, L • As current i through coil increases, magnetic flux through coil increases. This in turn induces counter EMF in the coil itself (Lenz) • When current i is decreasing, EMF is induced again in the coil itself in such a way as to slow the decrease maintain the field. Self-induction B L i NB L (if flux linked) i L T m2 / A Wb / A H (henry) Faraday’s Law: dB dI L dt dt Lecture 18-3 DEMO SELF INDUCTANCE 6D-10 Lecture 18-4 Inductance and Faraday’s Law B L I + - L T m2 / A Wb / A H I dQ dt (henry) d 2Q L 2 dt INERTIA d 2x compare with f m 2 dt This could have been a battery, instead. With the simplifying advantage that the applied ε is then ~constant. Lecture 18-5 Qualitative discussion of currents in L Note that if the initial current in an inductor, L, is ZERO, then it cannot instantly jump to a finite value when a potential difference is applied to it. Current I will “accelerate” from zero in a continuous way, just as the velocity of a mass at rest will accelerate in a continuous way under the influence of a force. At large times, the transients have died out, and since I is no longer changing, no voltage drop exists across the inductor L Lecture 18-6 Reading Quiz 1 Which of the following statements is incorrect? A| The inductance of a coil with N turns is proportional to N2 turns. B| Like a capacitance (C) an inductor (L) depends on geometric factors and the nature of the material inside C and L. C| The potential across an inductor depends only on the magnitude of the current through the inductor. D| The magnetic energy stored in an inductor is proportional to the square of the current through the inductor. Lecture 18-7 Solenoid: Archetypical Inductor Current i flows through a long solenoid of radius r with N turns in length l r l B 0 N i l For each turn N A r B BA 0 i r 2 l For the solenoid NB N2 2 N L 0 r 0 l r 2 i l l or 2 2 L 0n 2 Al Al is the volume of the solenoid Inductance, like capacitance, only depends on geometry (if made of conductor and air). Add a magnetically polarizable material, increase L, just as adding an electrically polarizable material (dielectric) increases C. Lecture 18-8 Potential Difference Across Inductor VL o I r +V ΔV I internal resistance must be symbolically separated from L !! • “Analogous” to a battery • An ideal inductor has r =0 V=0 • All dissipative effects are to be included in the external + internal resistances (i.e., including those of the iron core if any) dI IR L 0 dt dI IR L 0 dt Lecture 18-9 1. 2. Energy Stored By Inductor Switch on at t=0 As the current tries to begin flowing, self-inductance induces back EMF, thus opposing the increase of I. Loop Rule: IR L + dI 0 dt - 3. Multiply through by I dI I I R LI dt 2 Rate at which battery is supplying energy Rate at which energy is dissipated by the resistor Rate at which energy is stored in inductor L dU m dI LI dt dt Um 1 2 LI 2 Lecture 18-10 6C07 ENERGY STORED IN AN INDUCTOR Lecture 18-11 Where is the Energy Stored? • Energy must be stored in the magnetic field! Energy stored by a capacitor is stored in its electric field • Consider a long solenoid where 2 B 0nI , L 0n Al 2 1 2 1 1 B U m LI 0n 2 Al I 2 Al 2 2 2 0 • So energy density of the magnetic field is 2 Um 1 B um Al 2 0 1 uE 0 E 2 2 area A (Energy density of the electric field) length l Lecture 18-12 1. RL Circuits – Starting Current Switch to ε at t=0 As the current tries to begin flowing, self-inductance induces back EMF, thus opposing the increase of I. + I0 0 2. Loop Rule: - dI IR L 0 dt 3. Solve this differential equation I R 1 et /( L / R ) , VL L τ=L/R is the inductive time constant dI e t /( L / R ) dt T m2 / A T m2 / A s L / R V/A Lecture 18-13 Warm-up quiz 2 The circuit is turned on at t=0. Which of the following statements is correct? R1 V R2 L A| At t = 0, the potential drop across the inductor is V; When t = ∞, the current through R1 is V/R1 yes B| At t = 0, the potential drop across the inductor is 0; When t = ∞, the current through R1 is V. C| At t = 0, the potential drop across the inductor is V; t = ∞, the current through R1 is V/(R1+R2) D| At t = 0, the potential drop across the inductor is V; When t = ∞, the current through R1 is V/R2 X No fully correct answer is When offered See next slide Lecture 18-14 Warm-up quiz 2 The circuit is turned on at t=0. Which of the following statements is correct? R1 V R2 L Analysis: At t=0, no current has yet started flowing in L. BUT: Resistor R2 “short-circuits” L, in the sense that resistors are NONINDUCTIVE and current can “instantly” start flowing in them. So there IS an instant current path available at t=0, and we have a simple resistive voltage divider chain. VL = V R2/(R1+R2) At t= I is no longer changing, so L acts like a wire and completely shortcircuits R2. All the voltage drop is across R1, and the current is V/R1, as in the SECOND part of answer (a) Lecture 18-15 GROWTH AND DECAY OF CURRENT OF AN RL CIRCUIT 6C-05 Lecture 18-16 Remove Battery after Steady I already exists in RL Circuits 1. Initially steady current Io is flowing: I 0 R1 R 2. - Switch from e to f at t=0, causing back EMF to oppose the change. dI IR L 0 dt 3. Loop Rule: 4. Solve this differential equation I R + e t /( L / R ) dI VL L e t /( L / R ) dt like discharging a capacitor I cannot instantly become zero! Self-induction Lecture 18-17 Starting Current through Inductor vs Charging Capacitor q C (1 e t / RC ) IR 1 e t /( L / R ) R R I R e t / RC dI VL L e t /( L / R ) dt Note: when “charging”, in L voltage leads current, in C, current leads voltage Lecture 18-18 Qualitative discussion of currents in L Note that if the initial current in an inductor, L, is ZERO, then it cannot instantly jump to a finite value when a potential difference is applied to it. Current I will “accelerate” from zero in a continuous way, just as the velocity of a mass at rest will accelerate in a continuous way under the influence of a force. At large times, the transients have died out, and since I is no longer changing, no voltage drop exists across the inductor L Lecture 18-19 Behavior of Inductors • Increasing Current – Initially, the inductor behaves like a battery connected in reverse. – After a long time, the inductor behaves like a conducting wire. • Decreasing Current – Initially, the inductor behaves like a “reinforcement” battery. – After a long time, the inductor behaves like a conducting wire. Lecture 18-20 Physics 241 March 22, 2011 9:30 –Quiz 3 The switch in this circuit is initially open for a long time, and then closed at t = 0. What is the magnitude of the voltage across the inductor just after the switch is closed? a) zero b) V c) R / L d) V / R e) 2V Lecture 18-21 Physics 241 March 22 2011 10:30 Quiz 3 The switch in this circuit is closed at t = 0. What is the magnitude of the voltage across the resistor a long time after the switch is closed? a) zero b) V c) R / L d) V / R e) 2V Lecture 18-22 Physics 241 March 22, 2011 11:30–Quiz 3 The switch in this circuit has been open for a long time. Then the switch is closed at t = 0. What is the magnitude of the current through the resistor immediately after the switch is closed? a) zero b) V / L c) R / L d) V / R e) 2V / R