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
Qin
∫∫ E ⋅ dA = ε
S
(Gauss's Law)
0
dΦB
∫C E ⋅ d s = − dt
(Faraday's Law)
∫∫ B ⋅ dA = 0
(Magnetic Gauss's Law)
dΦE
∫C B ⋅ d s = µ0 I enc + µ0ε 0 dt
(Ampere-Maxwell Law)
F = q (E + v × B)
(Lorentz force Law)
S
P31- 4
Gauss’s Law:
E
⋅
A
=
d
∫∫
S
qin
ε0
Gaussian
Pillbox
Spherical
Symmetry
Planar
Symmetry
Cylindrical
Symmetry
P31- 5
Ampere’s Law:
B
⋅
d
s
=
µ
I
0
enc
∫
.
B
Long
Circular
Symmetry
I
B
(Infinite) Current Sheet
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Torus/Coax
Solenoid
=
2 Current
Sheets
B
X
X
X
X
X
X
X
X
X
X
X
X
P31- 6
Faraday’s Law of Induction
ε=∫
dΦB
E ⋅ ds = − N
Moving bar,
dt
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 21
P31- 8
Self Inductance & Inductors
L
I
NΦ
L=
I
When traveling in
direction of current:
ε
dI
= −L
dt
Notice: This is called “Back EMF”
It is just Faraday’s Law!
P31- 9
Energy Stored in Inductor
UL = L I
1
2
2
Energy is stored in the magnetic field:
2
B
uB =
2µo
: Magnetic Energy Density
P31- 10
LR Circuit
Readings on Voltmeter
Inductor (a to b)
Resistor (c to a)
c
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:
Value(t ) = Value Final (1 − e − t /τ )
Value(t ) = Value0 e − t /τ
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)
1
ω0 =
LC
P31- 14
Damped LC Oscillations
Q =πn =
ωL
R
Resistor dissipates
energy and system
rings down over time
P31- 15
PRS Questions:
Undriven RLC Circuits
Class 25
P31- 16
AC Circuits: Summary
ResistanceReactance
(Impedance)
Element
V vs I0
Current vs.
Voltage
Resistor
V0 R = I 0 R
In Phase
R=R
I0
=
ωC
Leads (90º)
1
XC =
ωC
V0 L = I 0ω L
Lags (90º)
XL = ωL
Capacitor
Inductor
V0C
L32 - 17
Driven RLC Series Circuit
VL 0
I(t)
VS
VC 0
VS 0
I0
VR 0
Now Solve: VS = VR + VL + VC
Now we just need to read the phasor diagram!
P31- 18
Driven RLC Series Circuit
VL 0
ϕ
I (t ) = I 0 sin(ω t − ϕ )
VC 0
VS = V0 S sin ( ω t )
2
VS 0
2
I0
2
VR 0
2
VS 0 = VR 0 + (VL 0 − VC 0 ) = I 0 R + ( X L − X C ) ≡ I 0 Z
VS 0
I0 =
Z
2
Z = R + (X L − XC )
Impedance
2
φ = tan
−1
⎛ X L − XC ⎞
⎜
⎟
R
⎝
⎠
P31- 19
I
I (t ) = I 0 sin ( ω t )
0
VR (t ) = I 0 R sin ( ω t )
0
VL (t ) = I 0 X L sin ( ω t + π2 )
VL
0
VR
Plot I, V’s vs. Time
VS
VC
+π/2
VC (t ) = I 0 X C sin ( ω t − π2 )
-π/2
0
VS (t ) = VS 0 sin ( ω t + ϕ )
0
0
+φ
1
2
Time (Periods)
3
⎛ X L − XC ⎞
⎟
R
⎝
⎠
φ = tan − 1 ⎜
P31- 20
Resonance
V0
I0 = =
Z
V0
R 2 + ( X L − X C )2
C-like:
φ<0
I leads
;
L-like:
φ>0
I lags
ω0 = 1
LC
1
X L = ω L, X C =
ωC
On resonance:
I0 is max; XL=XC; Z=R;
φ=0; Power to R is max
P31- 21
Average Power: Resistor
2
< P > =< I (t ) R >
=< I sin (ω t − ϕ ) R >
2
0
2
= I R < sin (ω t − ϕ ) >
2
0
2
= I 0 R ( 12 )
2
P31- 22
PRS Questions:
Driven RLC Circuits
Class 26
P31- 23
Displacement Current
Q
E=
⇒ Q = ε 0 EA = ε 0 Φ E
ε0 A
dΦE
dQ
= ε0
≡ Id
dt
dt
B
⋅
d
s
=
µ
(
I
+
I
)
0
encl
d
∫
C
= µ 0 I encl
Capacitors,
EM Waves
dΦE
+ µ 0ε 0
dt
P31- 24
Energy Flow
Poynting vector: S =
E×B
µ0
• (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
source: V(t)=V0sin(ωt) 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.
2. What angular frequency ωres 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:
VS = V0 sin ( ω t )
dI Q
− =0
VS − IR − L
dt C
dI
d 2I I
d
R+L 2 + =
VS
dt
dt
C dt
= ωV0 cos ( ω t )
2. Maximum current on resonance:
ω res = 1
L0C
P31- 29
Solution 1: RLC Circuit
3. Voltage on Capacitor
VC 0 = I 0 X C What is I0, XC?
VS = V0 sin ( ω t )
VC 0
V0
= I0 X C =
R
L0
C
V0 V0
I0 = =
(resonance)
Z
R
L0C
L0
1
XC =
=
=
ωC
C
C
V = VC 0 ( − cos ( ω t ) )
= VC 0 sin ( ω t − π2 )
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?
6. Now drive the new circuit with the original ωres.
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
X
X
X
X
X
X
X
X
X
X
a
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Inside Inner Solenoid:
B
⋅
d
s
=
Bl
=
µ
nlI
+
3
nlI
(
)
0
∫
⇒ B = 4 µ 0 nI
Between Solenoids:
B = µ 0 nI
2
B
U=
iVolume
2µo
4 µ 0 nI )
(
=
π a2
2
n
3n
2µo
µ 0 nI )
(
+
π ( b2 − a 2 )
2
2µo
P31- 34
Solution 2: Self-Inductance
b
X
X
X
X
X
X
X
X
X
X
n
a
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
3n
µ 0 nI )
(
U=
π
2
2µo
{15a
U = LI
1
2
µ0 n )
(
π
⇒L=
2
µo
2
+b
2
}
2
{15a
NΦ
Could also have used: L =
I
2
+b
2
}
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?
2⎛ θ
A =πa ⎜
⎝ 2π
2
θ
a
⎞
⎟=
2
⎠
dΦB d
d θ a2
= ( BA ) = B
dt
dt
dt 2
Ba 2 dθ
=
2 dt
P31- 37
Problem 3, Part 2: Pie Wedge
3.
4.
What is the magnetic force on
the bar (magnitude and
direction – indicated on figure)
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?
F = IaB
dF = Id s × B
ε 1 d Φ B 1 Ba 2 dθ
I= =
=
R 2 dt
R R dt
B 2 a 3 dθ
F=
2 R dt
(Dir. as pictured)
4) Torque?
B 2 a 4 dθ
a
τ = r×F ⇒τ = F =
4 R dt
2
(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?
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
P31- 40
capacitor with time.
Solution 4: RLC Circuit
1. Energy Stored in Inductor
I
+Q
dI Q
−L − = 0
dt C
1 2 1 ⎛ε ⎞
U = L I = L⎜ ⎟
2
2 ⎝R⎠
2
2. Write Differential Equation
2
dQ
d Q Q
I=
⇒L 2 + =0
dt
dt
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
3.
Solution 4: RLC Circuit
Solution for Q(t): Q (t ) = Qmax sin (ω t )
ω=
1
LC
ω Qmax = I 0 =
ε
R
⇒
Qmax =
ε LC
R
4. Time to charge capacitor
T=
2π
ω
= 2π LC ⇒ TCharge
Energy in inductor = 0
T π LC
= =
4
2
1 ⎛ε ⎞
Energy in capacitor = Initial Energy: U = L ⎜ ⎟
2 ⎝R⎠
2
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<a?
P31- 44
Solution 5: Cut Circuit
1. Cut looks like capacitor.
Use Gauss to find electric field:
+Q0
A
-Q0
∫∫ E ⋅ dA = EA =
Qenc
ε0
σA
=
ε0
Q0
σ
E= =
down
2
ε0 π a ε0
P31- 45
Solution 5: Cut Circuit
Q0
E=
π a 2ε 0
+Q0
r
-Q0
2. Find B field using Ampere’s Law
dΦE
d
2
2 dE
Id = ε0
= ε 0 Eπ r = ε 0π r
dt
dt
dt
2
⎛
⎞
Q0
r dQ0
2 d
= ε 0π r
⎜ 2 ⎟ = 2
dt ⎝ π a ε 0 ⎠ a dt
2
r dQ0
∫ B ⋅ d s = B 2π r = µ0 ( I enc + I d ) = µ0 I d = µ0 a 2 dt
µ 0 r dQ0
clockwise
B=
2
2π a dt
P31- 46