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Physics 227: Lecture 21 Mutual Inductance, LC Circuits • Lecture 20 review: • • • • • Self-inductance of coil of wire: ξinduced = -N dΦB/dt = -L di/dt, with L = NΦB/i. Solenoid: L = μ0n2V with n the turns/unit length and V the volume. RL circuit time constant τ = L/R. Energy stored in an inductor: U = ½LI2. Energy density of magnetic field filling volume: u = ½B2/μ0. Monday, November 28, 2011 Exam 2 Results • • • Average: 60%, same as exam 1 Easiest problems: RC circuit time constant(90%), radius formula for safe wire (88%), drift speed formula of electrons in wire (84%), ... Hardest problems: energy change when current carrying coil in B field flipped 180o (22%), field between two infinite current sheets (22%), relative magnitude of electric and magnetic forces between two equal charges (37%), ... Monday, November 28, 2011 Review RL iClicker The switch S1 is closed so that a current flows through an inductor of inductance L and a resistor of resistance R as shown. The final value of the current with S1 closed and S2 open is... A. proportional to RL. B. proportional to a ratio of R and L. C. proportional to 1/LR. D. independent of L. E. 0. Monday, November 28, 2011 Mutual Inductance • • • • The changing field in a coil leads to a self-inductance of: ξinduced = -N dΦB/dt = -L di/dt, with L = NΦB/i. Changing external magnetic fields lead to emf as well, as we know: ξ2 = -N2 dΦB2/dt. If the flux ΦB2 arises from an electromagnet with current i1, then we can write N2 ΦB2 = M21 i1, where M21 is the mutual inductance. M21, like L, is a constant that depends on geometry and number of coils. This leads to ξ2 = -N2 dΦB2/dt = -M21 di1/dt. Monday, November 28, 2011 Mutual Inductance • • If the analysis is redone for the induced emf in loop 1 and the current in loop 2, it it is found that M21 = N2ΦB2/i1 = M12 = N1ΦB1/i2 ≡ M. Thus ξ2 = -M di1/dt, and ξ1 = -M di2/dt. Monday, November 28, 2011 Applications & Issues • • Applications: • • • Cordless power • Unintended cross talk in circuits Transformers ... Issues Monday, November 28, 2011 Example: Calculating M • • • Monday, November 28, 2011 Field from loop 1: B1 = μ0N1I1/L = μ0n1I1. ΦB2 = AB1 = μ0N1I1A/L. M = N2ΦB2/I1 = μ0N1N2A/L. LC Circuit Math • • • • • • Monday, November 28, 2011 Kirchoff’s Voltage Law: • • Capacitor: Q = CV → V = Q/C • I = dQ/dt Inductor: V = L dI/dt KVL → Q/C + L dI/dt = 0 Q/C + L d2Q/dt2 = 0 d2Q/dt2 + Q/LC = 0 This should look familiar. Recall the solution to d2x/dt2 + ω2x = 0 is x = A cos(ωt+φ) - simple harmonic motion with a frequency ω2 = 1/LC. Thus: Q = Q0 cos(ωt+φ), and I = -ωQ0 sin(ωt+φ) Energy in an LC Circuit • • • • • • Monday, November 28, 2011 Thus: Q = Q0 cos(ωt+φ), and I = -ωQ0 sin(ωt+φ) Ucapacitor = U = ½CV2 = ½Q2/C. Uinductor = U = ½LI2. ULC = (½Q02/C) cos2(ωt+φ) + (½L) (ωQ0)2 sin2(ωt+φ) ULC = (½Q02/C) [cos2(ωt+φ) + sin2(ωt+φ)] = (½Q02/C) The energy in the circuit is constant - there is no resistor to dissipate it. It shifts between the capacitor and the inductor. LC Circuits - Inductor + Capacitor Monday, November 28, 2011 LRC Series Circuit • • • • • • Monday, November 28, 2011 Charge up the capacitor with the emf. Flip the switch and discharge the capacitor through the resistor and inductor. What happens? Qualitatively the same as with pendulum with friction, etc.: a damped harmonic oscillator. There are oscillations in the current of decreasing magnitude. It is good to have some intuition about the answer before doing the math - we have it from LC before. RLC Ciruit Monday, November 28, 2011 RLC Circuit Math • • • • • • Monday, November 28, 2011 Kirchoff’s Voltage Law → Q/C + L dI/dt + IR = 0 d2Q/dt2 + (R/L) dQ/dt + Q/LC = 0 This is the damped harmonic oscillator with damping proportional to the ``velocity’’: the solution is Q = A e-(R/2L)t cos(ω’t+φ) - simple harmonic motion with a frequency ω’2 = (1/LC) - (R2/4L2). R small (< 4L/C): damped oscillation R2 = 4L/C: ω’ = 0, critical damping Q → A’ e-(R/2L)t R2 > 4L/C: ω’2 < 0, overdamped LC iClicker An inductor of inductance L and a capacitor of capacitance C are connected as shown. The value of the charge q on the capacitor oscillates between positive and negative. The potential difference between the capacitor plates is... A. proportional to q. B. proportional to dq/dt. C. proportional to d2q/dt2. D. both A. and C. E. all of A., B., and C. Monday, November 28, 2011 LC iClicker An inductor of inductance L and a capacitor of capacitance C are connected as shown. The inductance L and capacitance C are both doubled. What is the change in the time for an oscillation cycle in the circuit? A. 4x longer. B. 2x longer. C. Unchanged. D. 1/2 as long. E. 1/4 as long. Monday, November 28, 2011 Have a nice Holiday. See you Monday after Thanksgiving. Monday, November 28, 2011