Download PowerPoint

Physics 212
Lecture 22
Physics 212 Lecture 21, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Billy Joel
Boz Scaggs
Mark Knopfler
Donald Fagen
Michael McDonald
1976
1994
Theme of the week?
Tower of Power, Boz Scaggs –
San Francisco sounds of the ’70s
Your Comments
“I got chills. This is incredible stuff!”
“Fun stuff! but kinda hard.”
“The derivation of the wave equation lost me ”
“This prelecture was so horrible. Maybe it was just me, but I’m
not sure that was English being spoken. Please clarify these
concepts that seemingly came out of nowhere.“
Wide Range:
We will try to
make clear,
at least the
BIG IDEAS
“I got the modified Ampere’s law, but you lost me at one-dimensional waves; I’ve
never EVER seen that equation before. I don’t understand what we just derived;
why did we differentiate one with respect to time and the other with respect to z??
This all went by so fast”
The solutions of the
“All of the derivations and second derivatives
wave equation are the
became a little jumbled. I feel like this isn’t a very
important part
hard topic, just the way it’s presented is a little
overwhelming.”
“If I were Gauss, Ampere or Faraday I’d be mad that Maxwell just came along and
claimed all the formulas for himself.”
“Woah, whoah! Lecture slide speed limit: 3.00x10^8 m/s!”
05
Physics 212 Lecture 21, Slide 3
What We Knew Before Prelecture 22
06
Physics 212 Lecture 21, Slide 5
After Prelecture 22: Modify Ampere’s Law

Q
E 
0 0 A
  EA 
Q
0
Q   0
dQ
d
 0
 ID
dt
dt
10
Physics 212 Lecture 21, Slide 6
Displacement Current: Generalization
Real Current:
Charge Q passes through area A in time t:
I
Displacement Current:
Electric flux through area A changes in time
ID  0
08
dQ
dt
dE
dt
Free space Physics 212 Lecture 21, Slide 7
Calculation
Switch S has been open a long time when at t = 0, it is closed.
Capacitor C has circular plates of radius R. At time t = t1, a
current I1 flows in the circuit and the capacitor carries
charge Q1.
S
C
V
Ra
At time t1, what is the magnetic field B1 at a radius r
(point d) in between the plates of the capacitor?
●d
I1
r
R
Q1
• Conceptual and Strategic Analysis
• Charge Q1 creates electric field between the plates of C
• Charge Q1 changing in time gives rise to a changing electric flux between
the plates
• Changing electric flux gives rise to a displacement current ID in between
the plates
• Apply (modified) Ampere’s law using ID (what does Amperian loop look like)
to determine B
10
Physics 212 Lecture 21, Slide 8
Calculation
Switch S has been open a long time when at t = 0, it is closed.
Capacitor C has circular plates of radius R. At time t = t1, a
current I1 flows in the circuit and the capacitor carries
charge Q1.
c●
●d
r
I1
S
C
V
Ra
r
R
Q1
Compare the magnitudes of the B fields at points c and d.
(A) Bc < Bd
(B) Bc = Bd
What is the difference?
Apply (modified) Ampere’s Law
point c:
I(enclosed) = I1
11
r
X
(C) Bc > Bd
r
R
point d:
ID(enclosed) < I1
Physics 212 Lecture 21, Slide 9
Calculation
Switch S has been open a long time when at t = 0, it is closed.
Capacitor C has circular plates of radius R. At time t = t1, a
current I1 flows in the circuit and the capacitor carries
charge Q1.
●d
I1
E
Q1
S
C
V
Ra
r

R E
0
What is the magnitude of the electric field between the plates?
Q1
(A) E 
 R 2 0
Q1
13
(B) E 
Q1 R 2
0
E
R

0
(C) E 

Q1
0
Q1
Q
 12
A R
(D) E 
E
Q1
r
Q1
 0 R 2
Physics 212 Lecture 21, Slide 10
Calculation
Switch S has been open a long time when at t = 0, it is closed.
Capacitor C has circular plates of radius R. At time t = t1, a
current I1 flows in the circuit and the capacitor carries
charge Q1.
●d
I1
E
S
C
V
Ra
E
r
Q1
 R 2 0
R
Q1
What is the electric flux through a circle of radius r in between the plates?
(What does the Amperian loop look like?)
(A)  E 
Q1
0
r
r
2
(B)  E 
R
Q1
0
R
E  E  A
2
Q1r 2
(C)  E 
 0R2
Q1
2
E 

r
 0 R 2
Q1 r 2
(D)  E 
 0R2
Q1 r 2
E 
 0 R2
Physics 212 Lecture 21, Slide 11
Calculation
Switch S has been open a long time when at t = 0, it is closed.
Capacitor C has circular plates of radius R. At time t = t1, a
current I1 flows in the circuit and the capacitor carries
charge Q1.
●d
I1
E
S
Ra
Q1r 2
E 
 0R2
r
R
Q1
C
V
ID  0
dE
dt
What is the displacement current enclosed by circle of radius r ?
R2
(A) I D  I1 2
r
r
17
r
(B) I D  I1
R
R
r2
(C) I D  I1 2
R
(D) I D  I1
R
r
d  E dQ1 r 2
r2
ID  0

 I1 2
2
dt R
dt
R
r2
I D  I1 2
R
Physics 212 Lecture 21, Slide 12
Calculation
Switch S has been open a long time when at t = 0, it is closed.
Capacitor C has circular plates of radius R. At time t = t1, a
current I1 flows in the circuit and the capacitor carries
charge Q1.
●d
I1
E
Q1
S
Ra
r2
I D  I1 2
R
r
R
C
V
 
 B  d    o I  I D 
What is the magnitude of the B field at radius r ?
(A) B 
0 I1
2 R
r
19
0 I1
2 r
(C) B  0 I1 R2
2 r
 
Ampere’s Law:  B  d   o I  I D 
(B) B 
R
(D) B 

r2 
B  2 r  0  0  I1 2 
R 

I r
B 0 1 2
Physics 212
2 R
0 I1 r
2 R 2
Lecture 21, Slide 13
Checkpoint 1b
At time t=0 the switch in the circuit shown below is closed. Points A and B lie inside the
capacitor; A is at the center and B is toward the outer edge.
A
Compare the magnitudes of the magnetic fields at points A and B just after the switch is
closed
A. BA < BB
B. BA = BB
C. BA > BB
“B has more flux enclosed, has a greater electric flux, and has greater B-field.”
“the electric field is the same, so the magnetic field should also be the same.”
“A is right in the center of the electric field, while B is further away, so A will be larger”
21
Physics 212 Lecture 21, Slide 14
Checkpoint 1b
At time t=0 the switch in the circuit shown below is closed. Points A and B lie inside the
capacitor; A is at the center and B is toward the outer edge.
A
Compare the magnitudes of the magnetic fields at points A and B just after the switch is
closed
A. BA < BB
B. BA = BB
C. BA > BB
From the
calculation we
just did:
0 I1 r
B
2 R 2
21
Physics 212 Lecture 21, Slide 15
Checkpoint 1a
At time t=0 the switch in the circuit shown below is closed. Points A and B lie inside the
capacitor; A is at the center and B is toward the outer edge.
A
After the switch is closed, there will be a magnetic field at point A which is proportional to
the current in the circuit:
A. True
B. False
“There is displacement current between the plates, this creates a magnetic field as
well”
“larger magnetic field will be produced by larger current.”
“at A, the B field is zero”
Physics 212 Lecture 21, Slide 16
Checkpoint 1a
At time t=0 the switch in the circuit shown below is closed. Points A and B lie inside the
capacitor; A is at the center and B is toward the outer edge.
A
After the switch is closed, there will be a magnetic field at point A which is proportional to
the current in the circuit:
A. True
B. False
B
0 I1 r
2 R 2
B is proportional to I
but
At A, B = 0 !!
Physics 212 Lecture 21, Slide 17
Follow-Up
Switch S has been open a long time when at t = 0, it is
closed. Capacitor C has circular plates of radius R. At
time t = t1, a current I1 flows in the circuit and the
capacitor carries charge Q1.
What is the time dependence of the magnetic field B
at a radius r between the plates of the capacitor?
(A)
(B)
S
C
V
Ra
B1 
0 I1 r
2 R 2
(C)
B at fixed r is proportional to the current I
Close switch: VC =0  I = V/Ra (maximum)
I exponentially decays with time constant t = RaC
25
Physics 212 Lecture 21, Slide 18
Follow-Up 2
Suppose you were able to charge a capacitor with
constant current (does not change in time).
Does a B field exist in between the plates of the
capacitor?
(A) YES
(B)
NO
Constant current  Q increases linearly with
time
Therefore E increases linearly with time ( E = Q/(A0)
dE/dt is not zero  Displacement current is not zero
 B is not zero !
Physics 212 Lecture 21, Slide 19
We learned about waves in Physics 211
30
Physics 212 Lecture 21, Slide 20
“How can light move at
the same velocity in
any inertial frame of
reference? That's
really trippy. ”
see PHYS 225
33
Physics 212 Lecture 21, Slide 21
35
Physics 212 Lecture 21, Slide 22
37
Physics 212 Lecture 21, Slide 23
Checkpoint 2a
An electromagnetic plane-wave is traveling in the +z direction. The illustration below shows
this wave at some instant in time. Points A, B and C have the same z coordinate.
Ex = E0sin(kz-wt)
Compare the magnitudes of the electric fields at points A and B
A. EA < EB
B. EA = EB
C. EA > EB
“A is always greater because it has 2 magnetic fields acting on it instead of just 1 like the
other points. “
“points A and B are located in the electric field wave, so both would be influenced by the
electric field. However, point C is located in the magnetic field and not the electric field
and as such, has 0 electric field.“
“E = E0 sin (kz - wt): E depends only on z coordinate for constant t. z coordinate is same
for A, B, C. “
40
Physics 212 Lecture 21, Slide 24
Checkpoint 2a
An electromagnetic plane-wave is traveling in the +z direction. The illustration below shows
this wave at some instant in time. Points A, B and C have the same z coordinate.
Ex = E0sin(kz-wt)
Compare the magnitudes of the electric fields at points A and B
A. EA < EB
B. EA = EB
C. EA > EB
E = E0 sin (kz - wt):
E depends only on z coordinate for constant t.
z coordinate is same for A, B, C.
40
Physics 212 Lecture 21, Slide 25
Checkpoint 2b
An electromagnetic plane-wave is traveling in the +z direction. The illustration below shows
this wave at some instant in time. Points A, B and C have the same z coordinate.
Ex = E0sin(kz-wt)
Compare the magnitudes of the electric fields at points A and C
A. EA < EC
B. EA = EC
C. EA > EC
“A is always greater because it has 2 magnetic fields acting on it instead of just 1 like the
other points. “
“points A and B are located in the electric field wave, so both would be influenced by the
electric field. However, point C is located in the magnetic field and not the electric field
and as such, has 0 electric field.“
“E = E0 sin (kz - wt): E depends only on z coordinate for constant t. z coordinate is same
for A, B, C. “
45
Physics 212 Lecture 21, Slide 26
Checkpoint 2b
An electromagnetic plane-wave is traveling in the +z direction. The illustration below shows
this wave at some instant in time. Points A, B and C have the same z coordinate.
Ex = E0sin(kz-wt)
Compare the magnitudes of the electric fields at points A and C
A. EA < EC
B. EA = EC
C. EA > EC
E = E0 sin (kz - wt):
E depends only on z coordinate for constant t.
z coordinate is same for A, B, C.
45
Physics 212 Lecture 21, Slide 27
Follow-up
Ex = E0sin(kz-wt)
Consider a point (x,y,z) at time t when Ex
is negative and has its maximum value.
At (x,y,z) at time t, what is By?
A)
B)
C)
D)
45
By is positive and has its maximum value
By is negative and has its maximum value
By is zero
We do not have enough information
Physics 212 Lecture 21, Slide 28