Class 31: Outline Hour 1: Concept Review / Overview PRS Questions possible exam questions Hour : Sample Exam Yell if you have any questions P31 1
Exam 3 Topics Faraday s Law Self Inductance Energy Stored in Inductor/Magnetic Field Circuits LR Circuits Undriven (R)LC Circuits Driven RLC Circuits Displacement Current Poynting Vector NO: B Materials, Transformers, Mutual Inductance, EM Waves P31
General Exam Suggestions You should be able to complete every problem If you are confused, ask If it seems too hard, you aren t thinking enough Look for hints in other problems If you are doing math, you re doing too much Read directions completely (before & after) Write down what you know before starting Draw pictures, define (label) variables Make sure that unknowns drop out of solution Don t forget units! P31 3
Maxwell s Equations C C S S Qin E d A = (Gauss's Law) ε dφ B E d s = (Faraday's Law) dt B d A = (Magnetic Gauss's Law) B d s = µ I + µ ε enc dφ dt E (Ampere-Maxwell Law) F = q ( E+ v B) (Lorentz force Law) P31 4
Gauss s Law: E A S d = q in ε Spherical Symmetry Cylindrical Symmetry Gaussian Pillbox Planar Symmetry P31 5
B d s = µ I Ampere s Law:. enc Long Circular Symmetry B B I (Infinite) Current Sheet Torus/Coax Solenoid = Current Sheets B P31 6
Faraday s Law of Induction ε = E d s = N d Φ dt B Moving bar, entering field d = N BA ( cos θ ) dt Lenz s Law: Ramp B Rotate area in field Induced EMF is in direction that opposes the change in flux that caused it P31 7
PRS Questions: Faraday s & Lenz s Law Class 1 P31 8
Self Inductance & Inductors I L L = When traveling in direction of current: di L dt ε = NΦ I Notice: This is called Back EMF It is just Faraday s Law! P31 9
Energy Stored in Inductor U = L L 1 I Energy is stored in the magnetic field: B u B = µ o : Magnetic Energy Density P31-1
c di ε IR L = dt LR Circuit Readings on Voltmeter Inductor (a to b) Resistor (c to a) t= + : Current is trying to change. Inductor works as hard as it needs to in order to stop it t= : Current is steady. Inductor does nothing. P31-11
General Comment: LR/RC All Quantities Either: Value( t ) = Value 1 Final ( t / τ e ) Value( t ) = Value e t / τ τ can be obtained from differential equation (prefactor on d/dt) e.g. τ = L/R or τ = RC P31-1
PRS Questions: Inductors & LR Circuits Classes 3, 5 P31-13
Undriven LC Circuit Oscillations: From charge on capacitor (Spring) to current in inductor (Mass) ω = 1 LC P31-14
Damped LC Oscillations Q ωl = π n = R Resistor dissipates energy and system rings down over time P31-15
PRS Questions: Undriven RLC Circuits Class 5 P31-16
AC Circuits: Summary Element V vs I Current vs. Voltage Resistance- Reactance (Impedance) Resistor V R = I R In Phase R = R Capacitor V C I = Leads (9º) ωc C = 1 ωc V Inductor L = I ω L Lags (9º) L = ωl L3-17
Driven RLC Series Circuit V L V S I(t) V S V C I V R Now Solve: V = V + V + V S R L C Now we just need to read the phasor diagram! P31-18
Driven RLC Series Circuit It () = I sin( ωt ϕ) V S = V S sin ωt ( ) V L V C V S ϕ I V R V = V + ( V V ) = I R + ( ) I Z S R L C L C I = V S Z Z = R + ( ) L C Impedance 1 L C φ = tan R P31-19
Plot I, V s vs. Time It () = I sin ω t I( ) V R V () t = I R sin ω t R ( ) V L ( π ω ) V () t = I sin t + L L V C +π/ -π/ ( π ω ) V () t = I sin t C C V S 1 3 +φ Time (Periods) V () t = V sin ω t + ϕ S φ S ( ) 1 L C = tan R P31-
Resonance V V = = = = I ;, L ω L C Z R + ( L C ) 1 ω C C-like: φ < I leads L-like: φ > I lags On resonance: I is max; L = C ; Z=R; φ=; Power to R is max ω = 1 LC P31-1
Average Power: Resistor < P >=< I () t R > =< I sin ( ωt ϕ ) R > = I R < sin ( ωt ϕ ) > = ( ) 1 I R P31-
PRS Questions: Driven RLC Circuits Class 6 P31-3
Displacement Current Q E = Q = ε EA = ε Φ ε A dq dt ε dφ dt E = I d E B s C d = µ ( I + I ) encl = I + d µ µ ε encl Capacitors, EM Waves dφ dt E P31-4
Energy Flow Poynting vector: S = E B µ (Dis)charging C, L Resistor (always in) EM Radiation P31-5
PRS Questions: Displacement/Poynting Class 8 P31-6
SAMPLE EAM: The real exam has 8 concept, 3 analytical questions P31-7
Problem 1: RLC Circuit Consider a circuit consisting of an AC voltage source: V(t)=V sin(ωt) connected in series to a capacitor C and a coil, which has resistance R and inductance L. 1. Write a differential equation for the current in this circuit.. What angular frequency ω res would produce a maximum current? 3. What is the voltage across the capacitor when the circuit is driven at this frequency? P31-8
Solution 1: RLC Circuit V sin S = V ω t ( ) 1. Differential Eqn: di Q V S IR L = dt C di d I I d R + L + = dt dt C dt = ω V cos ω t ( ) V S. Maximum current on resonance: ω = res 1 L C P31-9
Solution 1: RLC Circuit 3. Voltage on Capacitor V sin S = V ω t V = I = C ( ) V L C R C V = I I C V C V What is I, C? = = (resonance) C Z R 1 LC L = = = ω C C C C ( cos ( ω ) ) V = V t ( π ω t ) = V sin C P31-3
Problem 1, Part : RLC Circuit Continue considering that LRC circuit. Insert an iron bar into the coil. Its inductance changes by a factor of 5 to L=L core 4. Did the inductance increase or decrease? 5. Is the new resonance frequency larger, smaller or the same as before? 6. Now drive the new circuit with the original ω res. Does the current peak before, after, or at the same time as the supply voltage? P31-31
Solution 1, Part : RLC Circuit 4. Putting in an iron core INCREASES the inductance 5. The new resonance frequency is smaller 6. If we drive at the original resonance frequency then we are now driving ABOVE the resonance frequency. That means we are inductor like, which means that the current lags the voltage. P31-3
Problem : Self-Inductance The above inductor consists of two solenoids (radius b, n turns/meter, and radius a, 3n turns/meter) attached together such that the current pictured goes counter-clockwise in both of them according to the observer. What is the self inductance of the above inductor? P31-33
Solution : Self-Inductance b a U = Inside Inner Solenoid: B d s = Bl = µ nli + 3 nli B = 4 µ ni Between Solenoids: B i Volume µ o B = µ ni ( ) n 3n ( 4 µ ni ) ( µ ni ) = π a + π b a µ µ ( ) o o P31-34
Solution : Self-Inductance b a ( ni ) µ U = π a + b µ { 15 } o U = 1 L ( n) µ L π a b µ I { 15 } = + o n 3n Could also have used: L = NΦ I P31-35
Problem 3: Pie Wedge Consider the following pie shaped circuit. The arm is free to pivot about the center, P, and has mass m and resistance R. 1. If the angle θ decreases in time (the bar is falling), what is the direction of current?. If θ = θ(t), what is the rate of change of magnetic flux through the pie-shaped circuit? P31-36
Solution 3: Pie Wedge 1) Direction of I? Lenz s Law says: try to oppose decreasing flux I Counter-Clockwise (B out) ) θ = θ(t), rate of change of magnetic flux? A θ θ a π = = π a d Φ B d d θ a = ( BA ) = B dt dt dt = Ba d θ dt P31-37
Problem 3, Part : Pie Wedge 3. What is the magnetic force on the bar (magnitude and direction indicated on figure) 4. What torque does this create about P? (HINT: Assume force acts at bar center) P31-38
Solution 3, Part : Pie Wedge 3) Magnetic Force? d F = Ids B I ε = = R R F = dφ dt 1 B 3 Ba d θ R dt F 1 = R = IaB Ba d θ dt (Dir. as pictured) 4) Torque? τ = r F τ = a F = 4 Ba d θ 4 R dt (out of page) P31-39
Problem 4: RLC Circuit The switch has been in position a for a long time. The capacitor is uncharged. 1. What energy is currently stored in the magnetic field of the inductor?. At time t =, the switch S is thrown to position b. By applying Faraday's Law to the bottom loop of the above circuit, obtain a differential equation for the behavior of charge Q on the capacitor with time. P31-4
Solution 4: RLC Circuit 1. Energy Stored in Inductor I 1 1 ε U = L I = L R +Q. Write Differential Equation di Q L = dt C I dq L d Q Q = + = dt dt C P31-41
Problem 4, Part : RLC Circuit 3. Write down an explicit solution for Q(t) that satisfies your differential equation above and the initial conditions of this problem. 4. How long after t = does it take for the electrical energy stored in the capacitor to reach its first maximum, in terms of the quantities given? At that time, what is the energy stored in the inductor? In the capacitor? P31-4
Solution 4: RLC Circuit Qt () = Q sin ωt 3. Solution for Q(t): ( ) ω = 1 LC ω Q = I = max 4. Time to charge capacitor ε R max π T π LC T = = π LC TCharge = = ω 4 Energy in inductor = Energy in capacitor = Initial Energy: ε LC Q max = R U 1 ε = L R P31-43
Problem 5: Cut Circuit Consider the circuit at left: A battery (EMF ε) and a resistor wired with very thick wire of radius a. At time t=, a thin break is made in the wire (thickness d). 1. After a time t = t, a charge Q = Q accumulates at the top of the break and Q = -Q at the bottom. What is the electric field inside the break?. What is the magnetic field, B, inside the break as a function of radius r<a? P31-44
Solution 5: Cut Circuit 1. Cut looks like capacitor. Use Gauss to find electric field: A +Q -Q E A Q enc d = EA = = ε σ A ε E σ Q = = ε π a ε down P31-45
E = Q πa ε +Q r -Q Solution 5: Cut Circuit. Find B field using Ampere s Law d Φ E d de I d = ε = ε E π r = ε π r dt dt dt d Q r dq επ = r dt π a = ε a dt B d s = B π r = µ ( I ) enc + I µ d = I µ d = µ r dq B = clockwise π a dt r dq a dt P31-46