Class 31: Outline
Hour 1: Concept Review / Overview PRS Questions – possible exam questions
Hour 2: 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- 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!
P31- 3 Maxwell’s Equations
Q EA⋅=d in (Gauss's Law) ∫∫ ε S 0 dΦ ∫ Es⋅=−d B (Faraday's Law) C dt ∫∫ BA⋅=d 0 (Magnetic Gauss's Law) S dΦ Bs⋅=dI + E (Ampere-Maxwell Law) ∫ 000enc µµε C dt
FEvB)=+×q( (Lorentz force Law) P31- 4 q Gauss’s Law: EA⋅=d in ∫∫ ε S 0
Gaussian Spherical Pillbox Symmetry Planar Symmetry Cylindrical
Symmetry P31- 5
Ampere’s Law: B ⋅ d s = µ I . ∫ 0 enc B Long Circular I Symmetry B (Infinite) Current Sheet
X X X X X X Solenoid B X X X X X
X X = X
X X
X X
X
X X
X X 2 Current X X X Sheets X X Torus/Coax P31- 6 Faraday’s Law of Induction d Φ ε =⋅=−EsdNB ∫ Moving bar, dt entering field d =−NBA()cosθ dt Ramp B Rotate area Lenz’s Law: 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 21
P31- 8 Self Inductance & Inductors NΦ L = L I I When traveling in direction of current: dI ε = −L dt Notice: This is called “Back EMF” It is just Faraday’s Law! P31- 9 Energy Stored in Inductor = 1 2 ULIL 2
Energy is stored in the magnetic field: B2 u = : Magnetic Energy Density B µ 2 o P31-10 LR Circuit Readings on Voltmeter c Inductor (a to b) Resistor (c to a) dI ε −−IR L = 0 dt
t=0+: 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:
=−()−t /τ = −t /τ Value(te ) ValueFinal 1 Value(te ) Value0
τ can be obtained from differential equation (prefactor on d/dt) e.g. τ = L/R or τ = RC P31-12 PRS Questions: Inductors & LR Circuits
Classes 23, 25
P31-13 Undriven LC Circuit Oscillations: From charge on capacitor (Spring) to current in inductor (Mass) ω = 1 0 LC
P31-14 Damped LC Oscillations
π ωL Qn== R
Resistor dissipates energy and system rings down over time
P31-15 PRS Questions: Undriven RLC Circuits
Class 25
P31-16 AC Circuits: Summary
Resistance- Current vs. Element V vs I Reactance 0 Voltage (Impedance) = = Resistor VIR00R In Phase R R
I 1 Capacitor V = 0 Leads (90º) X = 0C ωC C ωC
Inductor = ω Lags (90º) = ω VIL00L X L L
L32 - 17 Driven RLC Series Circuit
VL0 VS 0
I(t) I0 VR0 VC0 VS =++ Now Solve: VVVVSRLC
Now we just need to read the phasor diagram!
P31-18 Driven RLC Series Circuit
VL0 VS 0 =−ω ϕ ϕ It() I0 sin( t ) I0 VR0 V = ()ω C0 VVSS0 sin t =+−=2222 +−≡ VVVVIRXXIZSRLC00000() () LC 0
− V 22− ⎛⎞XX = S 0 ZRXX=+−()φ = 1 LC I0 LC tan ⎜⎟ Z ⎝⎠R Impedance P31-19 Plot I, V’s vs. Time
It()= I sin()ω t I 0 0 () = R VtR () IR0 sin t V 0 ω
Vt()=+ IX sin() t π L 0 LL0 2 V ω +π/2 -π/2 Vt()=− IX sin() t π C CC0 2
V 0 ω =+() S VtSS() V0 sin t V 0 0123 φ − XX− ωϕ + φ = 1 ⎛⎞LC Time (Periods) tan ⎜⎟ ⎝⎠R P31-20 Resonance VV 1 IXLX==00;, =ω = 0 ZC22+− LC RXX()LC ω On resonance: C-like: I0 is max; XL=XC; Z=R; φ < 0 L-like: φ=0; Power to R is max I leads φ > 0 I lags
ω = 0 1 LC P31-21 Average Power: Resistor
<>=
P31-22 PRS Questions: Driven RLC Circuits
Class 26
P31-23 Displacement Current ε = Q ==Φεε EQEA⇒ 00E 0 A dQ dΦ =≡ε E I dt0 dt d Capacitors, Bs⋅=dIIµ () + EM Waves ∫ 0 encl d C dΦ =+µµεI E 000encl dt P31-24 Energy Flow EB× Poynting vector: S = µ 0
• (Dis)charging C, L • Resistor (always in) • EM Radiation
P31-25 PRS Questions: Displacement/Poynting
Class 28
P31-26 SAMPLE EXAM:
The real exam has 8 concept, 3 analytical questions
P31-27 Problem 1: RLC Circuit
Consider a circuit consisting of an AC voltage ω source: V(t)=V0sin( t) connected in series to a capacitor C and a coil, which has resistance R and inductance L0.
1. Write a differential equation for the current in this circuit. ω 2. What angular frequency res would produce a maximum current? 3. What is the voltage across the capacitor when the circuit is driven at this frequency?
P31-28 Solution 1: RLC Circuit 1. Differential Eqn: dI Q VIRL−− −=0 S dt C dI d2 I I d R ++=LV dt dt2 C dt S = ()ω = ωω() VVS 0 sin t Vt0 cos
ω = 1 2. Maximum current on resonance: res L0C
P31-29 Solution 1: RLC Circuit 3. Voltage on Capacitor = VIXCC00 What is I0, XC? VV I ==00(resonance) 0 ZR
= ()ω 1 LC0 L VVS 0 sin t X == =0 C ωCC C
VLVV=−()cos ()ω t VIX==00 C0 CC00 R C =−()ω π VtC0 sin 2
P31-30 Problem 1, Part 2: RLC Circuit
Continue considering that LRC circuit. Insert an iron bar into the coil. Its inductance changes by a factor of 5 to L=Lcore
4. Did the inductance increase or decrease? 5. Is the new resonance frequency larger, smaller or the same as before? ω 6. Now drive the new circuit with the original res. Does the current peak before, after, or at the same time as the supply voltage?
P31-31 Solution 1, Part 2: 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.
P31-32 Problem 2: 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 2: Self-Inductance b Inside Inner Solenoid: a Bs⋅==d Blµ () nlI +3 nlI X X ∫ 0 X X X X X ⇒ BnI= 4 X µ0 X X X Between Solenoids: X X X X X BnI= µ X X 0 X X 2 X B X X U = iVolume X X µ X 2 o 22 ()4 µµnI() nI n 3n =+−00aba222 () 22oo µµππP31-34 Solution 2: Self-Inductance
b 2 a ()µ nI =+0 {}22 X X Uab 15 X X X 2 µo π X X X X X 1 2 X = X X ULI2 X X X X X 2 X X ()µ X 0n 22 X X ⇒ L =+ {}15ab X X X µo π n 3n NΦ Could also have used: L = 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? 2. 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) 2) θ = θ(t), rate of change of magnetic flux? π θθ2 Φ θ 2 2 ⎛⎞a d B dda Aa==⎜⎟ ==()BA B ⎝⎠22π dt dt dt 2 2 θ = Ba d 2 dt P31-37 Problem 3, Part 2: 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 2: Pie Wedge 3) Magnetic Force? dIdFsB=× F = IaB ε 1 dΦ 1 Ba2 dθ I == B = RRdt Rdt2 Ba23 dθ F = (Dir. as pictured) 2Rdt
4) Torque? a Ba24 dθ τ =×rF⇒τ = F = (out of page) 2 4Rdt 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? 2. At time t = 0, 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. P31-40 Solution 4: RLC Circuit 1. Energy Stored in Inductor ε 2 112 ⎛⎞ ULIL==⎜⎟ I 22⎝⎠R
+Q 2. Write Differential Equation dI Q dQ d2 Q Q −−=L 0 IL= ⇒ +=0 dt C dt dt2 C
P31-41 Problem 4, Part 2: 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 = 0 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?
P31-42 Solution 4: RLC Circuit = ()ω 3. Solution for ωQ(t): Qt() Qmax sin t 1 ε ε ω = == = LC QImax 0 ⇒ Qmax LC R R
4. Time to charge capacitor 2ππTLC TLCT==2 ⇒ == ω π Charge 42
Energy in inductor = 0 2 1 ⎛⎞ε Energy in capacitor = Initial Energy: UL= ⎜⎟ 2 ⎝⎠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=0, a thin break is made in the wire (thickness d).
1. After a time t = t0, a charge Q = Q0 accumulates at the top of the break and Q = -Q0 at the bottom. What is the electric field inside the break? 2. What is the magnetic field, B, inside the break as a function of radius r P31-44 Solution 5: Cut Circuit 1. Cut looks like capacitor. Use Gauss to find electric field: +Q0 σ A -Q Qenc A 0 ∫∫ EA⋅==dEAεε = 00 σ Q == 0 E 2 down επε00a P31-45 Solution 5: Cut Circuit Q E = 0 πε2 2. Find B field using Ampere’s Law a 0 dΦ ==εεπεπE ddE22 = IErrd 00 0 +Q dt dt dt 0 d ⎛⎞Q r 2 dQ r = επ 2 0 = 0 -Q0 0 r ⎜⎟2 2 dt⎝⎠ a 0 adt πε r 2 dQ Bs⋅=dBr2πµ = µ() I µ + I = I = 0 ∫ 000enc d d adt2 µ r dQ B = 00clockwise π 2 2 adt P31-46