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Physics 1502: Lecture 21 Today’s Agenda • Announcements: – Chap.27 & 28 • Homework 06: due next Monday • Induction / AC current Faraday's Law dS B N S v B S N v B B Induction Self-Inductance, RL Circuits I a I R XXX XXXX XX b L e L/R e1 1 VL f( x ) 0.5 0.0183156 0 0 1 2 3 4 t Recap from the last Chapter: Faraday's Law of Induction B N S v B S N v Can time varying current in a conductor induce EMF in in that same conductor ? • Time dependent flux is generated by change in magnetic field strength due motion of the magnet • Note: changing magnetic field can also be produced by time varying current in a nearby loop dI/dt B Self-Inductance • The inductance of an inductor ( a set of coils in some geometry ..eg solenoid, toroid) then, like a capacitor, can be calculated from its geometry alone if the device is constructed from conductors and air. • If extra material (eg iron core) is added, then we need to add some knowledge of materials as we did for capacitors (dielectrics) and resistors (resistivity) SI UNITS for L : Henry • Archetypal inductor is a long solenoid, just as a pair of parallel plates is the archetypal capacitor. l r N turns r << l A ++++ d ----- Calculation l • Long Solenoid: N turns total, radius r, Length l r N turns For a single turn, The total flux through solenoid is given by: Inductance of solenoid can then be calculated as: This (as for R and C) depends only on geometry (material) RL Circuits • At t=0, the switch is closed and the current I starts to flow. a I I R b • Loop rule: e Note that this eqn is identical in form to that for the RC circuit with the following substitutions: RCRL: RC: \ L Lecture 21, ACT 1 • At t=0 the switch is thrown from position b to position a in the circuit shown: 1A – What is the value of the current I a long time after the switch is thrown? I a I R b L e R (a) I = 0 (b) I = e / 2R (c) I = 2e / R 1B • What is the value of the current I0 immediately after the switch is thrown? (a) I0 = 0 (b) I0 = e / 2R (c) I0 = 2e / R RL Circuits • To find the current I as a fct of time t, we need to choose an exponential solution which satisfies the boundary condition: a I I R b e L RL = R • We therefore write: • The voltage drop across the inductor is given by: L RL Circuit (e on) Current Max = e/R L/R e/ R 2L/R I 63% Max at t=L/R 0 Voltage on L Max = e/R t e VL 37% Max at t=L/R 0 t RL Circuits • After the switch has been in position a for a long time, redefined to be t=0, it is moved to position b. • Loop rule: • The appropriate initial condition is: • The solution then must have the form: a I I R b e L RL Circuit (e off) Current Max = e/R e/R L/R 2L/R I 37% Max at t=L/R 0 Voltage on L Max = -e t 0 VL 37% Max at t=L/R -e t e on e/R L/R e off 2L/R e/R I L/R 2L/R I 0 t t 0 e VL VL 0 0 -e t t Review: RC Circuits (Time-varying currents) I a I • Discharge capacitor: C initially charged with Q=Ce Connect switch to b at t=0. Calculate current and charge as function of time. b e • Loop theorem • Convert to differential equation for q: R + + C - - Review: RC Circuits (Time-varying currents) I a I • Discharge capacitor: b e • Trial solution: R + + C - - q = C ee -t/RC • Check that it is a solution: Note that this “guess” incorporates the boundary conditions: ! Review: RC Circuits (Time-varying currents) I a • Discharge capacitor: q = C ee -t/RC b • Current is found from differentiation: I R + + C e - - Conclusion: • Capacitor discharges exponentially with time constant = RC • Current decays from initial max value (= -e/R) with same time constant Discharging Capacitor Charge on C Max = Ce Ce RC 2RC q 37% Max at t=RC 0 t 0 Current I Max = -e/R 37% Max at t=RC -e/R t Charging Ce RC 2RC Ce q 0 Discharging RC 2RC q 0 t t 0 e/R I I 0 - e/R t t Energy of an Inductor • How much energy is stored in an inductor when a current is flowing through it? • Start with loop rule: I a I R b e L • Multiply this equation by I: • From this equation, we can identify PL, the rate at which energy is being stored in the inductor: • We can integrate this equation to find an expression for U, the energy stored in the inductor when the current = I: Where is the Energy Stored? • Claim: (without proof) energy is stored in the Magnetic field itself (just as in the Capacitor / Electric field case). • To calculate this energy density, consider the uniform field generated by a long solenoid: l • The inductance L is: r N turns • Energy U: • We can turn this into an energy density by dividing by the volume containing the field: Mutual Inductance • Suppose you have two coils with multiple turns close to each other, as shown in this cross-section Coil 1 Coil 2 B • We can define mutual inductance M12 of coil 2 with respect to coil 1 as: It can be shown that : N1 N2 Inductors in Series • What is the combined (equivalent) inductance of two inductors in series, as shown ? a a L1 Note: the induced EMF of two inductors now adds: L2 b Since: And: Leq b Inductors in parallel • What is the combined (equivalent) inductance of two inductors in parallel, as shown ? Note: the induced EMF between points a and be is the same ! a L1 a L2 b Also, it must be: We can define: And finally: Leq b LC Circuits • Consider the LC and RC series circuits shown: C R C L • Suppose that the circuits are formed at t=0 with the capacitor C charged to a value Q. Claim is that there is a qualitative difference in the time development of the currents produced in these two cases. Why?? • Consider from point of view of energy! • In the RC circuit, any current developed will cause energy to be dissipated in the resistor. • In the LC circuit, there is NO mechanism for energy dissipation; energy can be stored both in the capacitor and the inductor! RC/LC Circuits i Q+++ --- i Q+++ --- C R 0 i 0 1 t L LC: current oscillates RC: current decays exponentially -i C 0 t LC Oscillations (qualitative) + + - - C L C C L L - - + + C L LC Oscillations (quantitative) • What do we need to do to turn our qualitative knowledge into quantitative knowledge? • What is the frequency w of the oscillations? + + - - C L LC Oscillations (quantitative) i • Begin with the loop rule: Q + + - - C L • Guess solution: (just harmonic oscillator!) remember: where: • w0 determined from equation • f, Q0 determined from initial conditions • Procedure: differentiate above form for Q and substitute into loop equation to find w0. Review: LC Oscillations i • Guess solution: (just harmonic oscillator!) Q where: + + - - C L • w0 determined from equation • f, Q0 determined from initial conditions which we could have determined from the mass on a spring result: 1 Lecture 21, ACT 2 • At t=0 the capacitor has charge Q0; the resulting oscillations have frequency w0. The maximum current in the circuit during these oscillations has value I0 . – What is the relation between w and w2 , the 1A frequency of oscillations when0 the initial charge = 2Q0 ? (a) w2 = 1/2 w0 (b) w2 = w0 t=0 + + Q Q0 - - C (c) w2 = 2 w0 L Lecture 21, ACT 2 t=0 • At t=0 the capacitor has charge Q0; the resulting oscillations have frequency w0. The maximum current in the circuit during these oscillations has value I0 . + + Q Q0 - - C 1B • What is the relation between I0 and I2 , the maximum current in the circuit when the initial charge = 2Q0 ? (a) I2 = I0 (b) I2 = 2 I0 (c) I2 = 4 I0 L