Class 36: Outline
Hour 1:
Concept Review / Overview
PRS Questions – Possible Exam Questions
Hour 2:
Sample Exam
Yell if you have any questions
P36 - 1
Before Starting…
All of your grades should now be posted (with
possible exception of last problem set). If this is
not the case contact me immediately.
P36 - 2
Final Exam Topics
Maxwell’s Equations:
1. Gauss’s Law (and “Magnetic Gauss’s Law”)
2. Faraday’s Law
3. Ampere’s Law (with Displacement Current)
& Biot-Savart & Magnetic moments
Electric and Magnetic Fields:
1. Have associated potentials (you only know E)
2. Exert a force
3. Move as waves (that can interfere & diffract)
4. Contain and transport energy
Circuit Elements: Inductors, Capacitors, Resistors
P36 - 3
Test Format
Six Total “Questions”
One with 10 Multiple Choice Questions
Five Analytic Questions
1/3 Questions on New Material
2/3 Questions on Old Material
P36 - 4
Maxwell’s Equations
P36 - 5
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)
S
P36 - 6
Gauss’s Law:
E
⋅
A
=
d
∫∫
S
Spherical
Symmetry
qin
ε0
Planar
Symmetry
Cylindrical
Symmetry
P36 - 7
Maxwell’s Equations
G G
w
³³ E ˜ dA
Qin
G G
v³ E ˜ d s
d)B
dt
S
C
G G
w
³³ B ˜ dA
H0
0
(Gauss's Law)
(Faraday's Law)
(Magnetic Gauss's Law)
S
G G
v³ B ˜ d s
C
d)E
P 0 I enc P 0H 0
dt
(Ampere-Maxwell Law)
P36 - 2
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
P36 - 9
Maxwell’s Equations
G G
w
³³ E ˜ dA
Qin
G G
v³ E ˜ d s
d)B
dt
S
C
G G
w
³³ B ˜ dA
H0
0
(Gauss's Law)
(Faraday's Law)
(Magnetic Gauss's Law)
S
G G
v³ B ˜ d s
C
d)E
P 0 I enc P 0H 0
dt
(Ampere-Maxwell Law)
P36 - 3
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
P36 - 11
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
P36 - 12
Maxwell’s Equations
G G
w
³³ E ˜ dA
Qin
G G
v³ E ˜ d s
d)B
dt
S
C
G G
w
³³ B ˜ dA
H0
0
(Gauss's Law)
(Faraday's Law)
(Magnetic Gauss's Law)
S
G G
v³ B ˜ d s
C
d)E
P 0 I enc P 0H 0
dt
(Ampere-Maxwell Law)
I am nearly certain that you will have one of each
They are very standard – know how to do them all
P36 - 4
EM Field Details…
P36 - 14
Electric Potential
B
∆ V = − ∫ E ⋅ d s = VB − VA
A
= − Ed
(if E constant – e.g. Parallel Plate C)
Common second step to Gauss’ Law
dV ˆ
E = −∇ V = e.g. −
i
dx
Less Common – Give plot of V, ask for E
P36 - 15
Force
Lorentz Force:
(
F = q E + v×B
)
• Single Charge Motion
• Cyclotron Motion
• Cross E & B for no force
Magnetic Force:
(
dFB = Id s × B ⇒ FB = I L × B
)
• Parallel Currents Attract
• Force on Moving Bar (w/ Faraday)
P36 - 16
The Biot-Savart Law
Current element of length ds carrying current I
(or equivalently charge q with velocity v)
produces a magnetic field:
µo q v x rˆ
B=
2
4π r
µ 0 I d s × rˆ
dB =
2
4π r
P36 - 17
Magnetic Dipole Moments
µ ≡ IAnˆ ≡ IA
Generate:
Feel:
1) Torque aligns with external field τ = µ × B
2) Forces as for bar magnets
P36 - 18
Traveling Sine Wave
i Wavelength: λ
i Frequency : f
ˆ E sin( kx − ω t )
E=E
0
i Wave Number: k =
2π
λ
i Angular Frequency: ω = 2π f
1 2π
i Period: T = =
ω
f
ω
i Speed of Propagation: v =
Good
chance this
will be one
question!
=λf
k
i Direction of Propagation: + x
P36 - 19
EM Waves
Travel (through vacuum) with
speed of light
v=c=
1
m
= 3 ×10
s
µ 0ε 0
8
At every point in the wave and any instant of time,
E and B are in phase with one another, with
E E0
=
=c
B B0
E and B fields perpendicular to one another, and to
the direction of propagation (they are transverse):
Direction of propagation = Direction of E × B
P36 - 20
Interference (& Diffraction)
∆L = mλ
⇒ Constructive Interference
∆L = ( m + 12 ) λ ⇒ Destructive Interference
Likely multiple choice problem?
a sin θ = mλ
m=0
d sin θ = mλ
m=3
m=2
m=1
P36 - 21
Energy Storage
Energy is stored in E & B Fields
uE =
εo E
2
2
: Electric Energy Density
In capacitor:
In EM Wave
UC = C V
1
2
2
2
B
uB =
2µo
: Magnetic Energy Density
In inductor: U L
In EM Wave
= LI
1
2
2
P36 - 22
Energy Flow
Poynting vector: S =
E×B
µ0
• (Dis)charging C, L
• Resistor (always in)
• EM Radiation
For EM Radiation
2
0
2
0
E0 B0
E
cB
Intensity: I ≡ < S > =
=
=
2µ0 2µ0c 2µ0
P36 - 23
Circuits
There will be no quantitative circuit
questions on the final and no questions
regarding driven RLC Circuits
Only in the multiple choice will there be
circuit type questions
BUT….
P36 - 24
Circuit Elements
NAME
Resistor
Value
R=
ρ
A
Capacitor
Q
C=
∆V
Inductor
NΦ
L=
I
V/ε
Power /
Energy
2
IR
Q
C
dI
−L
dt
I R
1
2
CV
1
2
2
LI
2
P36 - 25
Circuits
For “what happens just after switch is thrown”:
Capacitor: Uncharged is short, charged is open
Inductor: Current doesn’t change instantly!
Initially looks like open, steady state is short
RC & RL Circuits have “charging” and “discharging”
curves that go exponentially with a time constant:
LC & RLC Circuits oscillate:
V , Q, I ∝ cos(ω t )
ω0 =
1
LC
P36 - 26
SAMPLE EXAM
P36 - 27
F2002 #5, S2003 #3, SFB#1, SFC#1, SFD#1
Problem 1: Gauss’s Law
A circular capacitor of
spacing d and radius R is
in a circuit carrying the
steady current i shown.
At time t=0 it is uncharged
1. Find the electric field E(t) at P vs. time t (mag. & dir.)
2. Find the potential at P, V(t), given that the potential at
the right hand plate is fixed at 0
3. Find the magnetic field B(t) at P
4. Find the total field energy between the plates U(t) P36 - 28
Solution 1: Gauss’s Law
1.Find the electric field E(t):
Assume a charge q on the
left plate (-q on the right)
Gauss’s Law:
σA
∫∫S E ⋅ dA = EA = ε 0 = ε 0
σ
q
E= =
2
ε0 π R ε0
Qin
Since q(t=0) = 0, q = it
it
E(t ) =
to the right
2
π R ε0
P36 - 29
Solution 1.2: Gauss’s Law
2.Find the potential V(t):
Since the E field is uniform,
V = E * distance
it
V (t ) = E(t ) ( d − d ') =
( d − d ')
2
π R ε0
Check: This should be positive since its between a
positive plate (left) and zero potential (right)
P36 - 30
Solution 1.3: Gauss’s Law
3.Find B(t):
Ampere’s Law:
dΦE
∫C B ⋅ d s = µ0 I enc + µ0ε 0 dt
r 2i
2π rB = 0 + µ 0ε 0 2
R ε0
⎛ it ⎞ 2
Φ E = EA = ⎜
⎟π r
2
⎝ π R ε0 ⎠
2
µ 0ir
dΦE
ri
B(t ) =
out of the page
= 2
2
2π R
dt
R ε0
P36 - 31
Solution 1.4: Gauss’s Law
4. Find Total Field Energy between the plates
it ⎞
= ⎜
⎟
2
2 ⎝ π R ε0 ⎠
2
1 ⎛ µ 0ir ⎞
B
=
B Field Energy Density: u B =
⎜
2 ⎟
2 µ o 2 µ o ⎝ 2π R ⎠
2
E Field Energy Density: u E =
εo E
2
2
εo ⎛
2
Total Energy U = ∫∫∫ ( uE + uB ) dV
2
ε o ⎛ it ⎞
1
2
= ⎜
⎟ iπ R d +
2
2 ⎝ π R ε0 ⎠
2µo
it )
(
=
2
1 µo d 2
d
+
i
2
2 ε 0π R
2 8π
(Integrate over cylinder)
⎛ µ 0i ⎞
⎜ 2π R 2 ⎟
⎝
⎠
2
2
r
∫ id ⋅ 2π r dr
⎛ q2 1 2 ⎞
+ Li ⎟
⎜=
⎝ 2C 2
⎠
P36 - 32
Problem 2: Faraday’s Law
A simple electric generator rotates with frequency f
about the y-axis in a uniform B field. The rotor
consists of n windings of area S. It powers a lightbulb
of resistance R (all other wires have no resistance).
1. What is the maximum value Imax of the induced
current? What is the orientation of the coil when this
current is achieved?
2. What power must be supplied to maintain the
rotation (ignoring friction)?
P36 - 33
Solution 2: Faraday’s Law
dΦB
Faraday's Law: ∫ E ⋅ d s = −
dt
C
nBS
ε
1 dΦB
1 d
=−
I = =−
nBS cos (ω t ) ) =
ω sin (ω t )
(
R
R dt
R dt
R
I max
nBS
nBS
=
ω=
2π f
R
R
Max when flux is
changing the most – at
90º to current picture
P36 - 34
Solution 2.2: Faraday’s Law
2. Power delivered?
Power delivered must equal power dissipated!
2
⎛ nBS
⎞
⎛ nBS
2
P= I R=⎜
2π f sin (ω t ) ⎟ R = R ⎜
2π
⎝ R
⎠
⎝ R
R ⎛ nBS
P = ⎜
2π
2⎝ R
⎞
f⎟
⎠
2
⎞
f ⎟ sin 2 (ω t )
⎠
2
P36 - 35
ẑ
Problem 3: Ampere’s Law
d
d
d
L
J0 Consider the two long
current sheets at left, each
carrying a current density J0
XJ
0
(out the top, in the bottom)
a) Use Ampere’s law to find the magnetic field for all z.
Make sure that you show your choice of Amperian
loop for each region.
At t=0 the current starts decreasing: J(t)=J0 – at
b) Calculate the electric field (magnitude and direction)
at the bottom of the top sheet.
c) Calculate the Poynting vector at the same location
P36 - 36
ẑ
d
d
d
Solution 3.1: Ampere’s Law
3
2
1
J0
X
z=0
Region 1:
∫ B ⋅ ds = µ I
0 enc
J0
By symmetry, above the
top and below the bottom
the B field must be 0.
Elsewhere B is to right
⇒ B = µ0 J 0 z ⇒ B = µ0 J 0 z
Region 2:
∫ B ⋅ ds = µ I
0 enc
⇒ B = µ0 J 0 d ⇒ B = µ0 J 0 d
Region 3:
B = µ 0 ( J 0 d − J 0 ( z − 2d ) ) ⇒ B = µ 0 J 0 ( 3d − z )
P36 - 37
Solution 3.2: Ampere’s Law
ẑ
d
d
d
J
X
J
dΦB
∫C E ⋅ d s = − dt
s
Why is there an electric field?
Changing magnetic field Æ
Faraday’s Law!
Use rectangle of sides d, s to
find E at bottom of top plate
J is decreasing Æ B to right is decreasing Æ induced
field wants to make B to right Æ E out of page
d
d
dJ
2
2 sE = ( Bsd ) = sd ( µ 0 dJ ) = sd µ 0
dt
dt
dt
⇒ E = 12 d 2 µ 0 a out of page
P36 - 38
ẑ
d
d
Solution 3.3: Ampere’s Law
E
J Recall
B
d
X
E = 14 d 2 µ 0 a out of page
J
B = µ 0 Jd to the right
Calculate the Poynting vector (at bottom of top plate):
S=
1
µ0
E×B =
1
µ0
(
1
4
d µ 0 a ) ( µ 0 Jd ) zˆ
2
That is, energy is leaving the system (discharging)
If this were a solenoid I would have you integrate
over the outer edge and show that this = d/dt(1/2 LI2)
P36 - 39
Problem 4: EM Wave
The magnetic field of a plane EM wave is:
((
B = 10 cos π m
−9
-1
) y + ( 3π ×10 s ) t ) ˆi Tesla
8 -1
(a) In what direction does the wave travel?
(b) What is the wavelength, frequency & speed of the wave?
(c) Write the complete vector expression for E
(d) What is the time-average energy flux carried in the wave?
What is the direction of energy flow? (µo = 4 π x l0-7 in SI
units; retain fractions and the factor π in your answer.)
P36 - 40
Solution 4.1: EM Wave
((
B = 10 cos π m
−9
(a)
(b)
(c)
-1
) y + ( 3π ×10 s ) t ) ˆi Tesla
8 -1
Travels in the - ˆj direction (-y)
2π
-1
= 2m
k =πm ⇒ λ =
k
ω 3
8 -1
8 -1
ω =3π ×10 s ⇒ f =
= × 10 s
2π 2
ω
8 m
v = = λ f = 3 ×10
k
s
((
E = −3 ×10 cos π m
−1
-1
) y + ( 3π ×10 s ) t ) kˆ V/m
8 -1
P36 - 41
Solution 4.2: EM Wave
((
B = 10 cos π m
−9
(d)
S =
1
µ0
-1
) y + ( 3π ×10 s ) t ) ˆi Tesla
8 -1
S points along
direction of travel: - ˆj
E×B
1 1
E0 B0
=
2 µ0
1⎛
1
W
⎞
−1
−9
3 ×10 10
= ⎜
2
−7 ⎟
2 ⎝ 4π ×10 ⎠
ms
(
)(
)
P36 - 42
Problem 5: Interference
In an experiment you shine red laser light (O=600 nm)
at a slide and see the following pattern on a screen
placed 1 m away:
You measure the distance between successive
fringes to be 20 mm
a) Are you looking at a single slit or at two slits?
b) What are the relevant lengths (width, separation if
2 slits)? What is the orientation of the slits?
P36 - 5
Solution 5.1: Interference
First translate the picture to a plot:
Intensity
(a) Must be two slits
-80
-60
-40
-20
0
20
40
60
Horizontal Location on Screen (mm)
80
a
d
P36 - 6
Solution 5.2: Interference
Intensity
mλ
y = L tan θ ≈ L sin θ = L
d
-80
-60
-40
-20
0
20
40
60
Horizontal Location on Screen (mm)
80
1)( 600nm )
(
mλ
= (1m )
d=L
y
( 20mm )
= (1m )
−7
×
6
10
(
)
( 2 ×10 )
−2
−5
= 3 ×10 m
At 60 mm…
a sin θ = (1) λ
a 1
⇒ =
d sin θ = ( 3) λ d 3
a = 10−5 m
P36 - 45
Why is the sky blue?
400 nm
Wavelength
700 nm
Small particles preferentially scatter small wavelengths
You also might have seen a red moon last fall –
during the lunar eclipse.
When totally eclipsed by the Earth the only light
illuminating the moon is diffracted by Earth’s
atmosphere
P36 - 46