Exam 3 Topics. Displacement Current Poynting Vector. Faraday s Law Self Inductance. Circuits. Energy Stored in Inductor/Magnetic Field


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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
2 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! P313
3 Maxwell s Equations C C S S Qin E da (Gauss's Law) ε dφ B E d s (Faraday's Law) dt B da (Magnetic Gauss's Law) B ds µ I + µ ε enc dφ dt E (AmpereMaxwell Law) F q( E+ v B) (Lorentz force Law) P314
4 Gauss s Law: E d A q in S ε Spherical Symmetry Cylindrical Symmetry Gaussian Pillbox Planar Symmetry P315
5 B ds µ I Ampere s Law:. enc Long Circular Symmetry B B (Infinite) Current Sheet I Torus/Coax Solenoid Current Sheets B P316
6 ε Faraday s Law of Induction d Φ E d s N dt N d dt B Moving bar, entering field ( BAcosθ ) Lenz s Law: Ramp B Rotate area in field Induced EMF is in direction that opposes the change in flux that caused it P317
7 Self Inductance & Inductors I L L When traveling in direction of current: di ε L dt NΦ I Notice: This is called Back EMF It is just Faraday s Law! P319
8 Energy Stored in Inductor U L 1 LI Energy is stored in the magnetic field: B u B µ o : Magnetic Energy Density P311
9 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. P3111
10 General Comment: LR/RC All Quantities Either: Final ( t / ) Value( t) Value 1 e τ Value( ) Value t / t e τ τ can be obtained from differential equation (prefactor on d/dt) e.g. τ L/R or τ RC P311
11 Undriven LC Circuit Oscillations: From charge on capacitor (Spring) to current in inductor (Mass) ω 1 LC P3114
12 Damped LC Oscillations Q ωl π n R Resistor dissipates energy and system rings down over time P3115
13 AC Circuits: Summary Element V vs I Current vs. Voltage Resistance Reactance (Impedance) Resistor V I R R In Phase R R Capacitor V C I ωc Leads (9º) C 1 ωc Inductor V I ω L L Lags (9º) L ωl 17 L3 
14 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! P3118
15 Driven RLC Series Circuit V L V S It () Isin( ωt ϕ) VS V S sin ωt ( ) V C 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 P3119
16 Plot I, V s vs. Time It () I sin ωt I( ) V R V () t I R sin ωt 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 ωt+ ϕ S φ S ( ) 1 L C tan R P31
17 Resonance V V I ;, L ω L C Z R + ( L C) 1 ωc Clike: φ < I leads Llike: φ > I lags On resonance: I is max; L C ; ZR; φ; Power to R is max ω 1 LC P311
18 Average Power: Resistor < P>< I () t R> < I sin ( ωt ϕ) R> I R< sin ( ωt ϕ) > ( ) 1 I R P31
19 Displacement Current Q E Q εea εφ ε A dq dt ε dφ dt E I d E B C d s µ ( I + I ) encl d µ I + µ ε encl Capacitors, EM Waves dφ dt E P314
20 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? P318
21 Solution 1: RLC Circuit 1. Differential Eqn: di Q VS IR L dt C V sin S V ω t ( ) di d I I d + + dt dt C dt R L V ωv cos ωt ( ) S. Maximum current on resonance: ω res 1 L C P319
22 Solution 1: RLC Circuit 3. Voltage on Capacitor V sin S V ω t ( ) V I I C V C V What is I, C? (resonance) C Z R 1 LC L ωc C C V I C C V R L C C ( cos ( ω )) V V t ( ω ) π V sin t C P313
23 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 LL 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? P3131
24 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. P313
25 Problem : SelfInductance 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 counterclockwise in both of them according to the observer. What is the self inductance of the above inductor? P3133
26 Solution : SelfInductance b a U Inside Inner Solenoid: B d s Bl µ nli + 3nlI B Between Solenoids: B B ivolume µ o µ ni 4µ ni ( ) n 3n ( 4µ ni ) ( µ ni ) πa + π b a µ µ ( ) o o P3134
27 Solution : SelfInductance b a ( ni ) µ U π a + b µ o U ( n) 1 LI { 15 } µ L π a b µ o { 15 } + n 3n Could also have used: L NΦ I P3135
28 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 pieshaped circuit? P3136
29 Solution 3: Pie Wedge 1) Direction of I? Lenz s Law says: try to oppose decreasing flux I CounterClockwise (B out) ) θ θ(t), rate of change of magnetic flux? A θ θa π π a dφ B d d θ a ( BA) B dt dt dt Ba dθ dt P3137
30 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) P3138
31 Solution 3, Part : Pie Wedge 3) Magnetic Force? df Ids B I ε dφ R R dt F 1 B 3 Ba d θ R dt F 1 R IaB Ba dθ dt (Dir. as pictured) 4) Torque? τ r F τ a F Ba 4R 4 dθ dt (out of page) P3139
32 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. P314
33 Solution 4: RLC Circuit 1. Energy Stored in Inductor I 1 1 ε U LI L R +Q. Write Differential Equation di Q L dt C I dq L d Q Q + dt dt C P3141
34 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? P314
35 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 1 ε U L R P3143
36 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? P3144
37 Solution 5: Cut Circuit 1. Cut looks like capacitor. Use Gauss to find electric field: A +Q Q E Q enc d A EA ε σ A ε E σ Q ε π ε a down P3145
38 r E Q πa Solution 5: Cut Circuit ε +Q Q. Find B field using Ampere s Law dφ E d de Id ε ε Eπr ε π r dt dt dt d Q επr r dq dt π a ε a dt B d s Bπ r µ ( I ) enc + I µ d I µ d B r a µ r dq clockwise π a dt dq dt P3146
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