W13D1:
Displacement Current,
Maxwell’s Equations,
Wave Equations
Today’s Reading Course Notes: Sections 13.1-13.4
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Announcements
Math Review Week 13 Tuesday 9pm-11 pm in 32-082
PS 9 due Week 13 Tuesday April 30 at 9 pm in boxes outside 32-082 or
26-152
Next Reading Assignment W13D2 Course Notes: Sections 13.5-13.7
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Outline
Maxwell’s Equations
Displacement Current
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Maxwell’s Equations
1
òò E × n̂ da = e òòò r dV
(Gauss's Law)
òò B × n̂ da = 0
(Magnetic Gauss's Law)
d
ò E × d s = - dt òò B × n̂ da
C
S
(Faraday's Law)
ò B × d s = m òò J × n̂ da
(Ampere's Law quasi - static)
S
0
V
S
0
C
S
Is there something missing?
4
Maxwell’s Equations
One Last Modification:
Displacement Current
“Displacement Current”
has nothing to do with
displacement and nothing to do
with current
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Ampere’s Law: Capacitor
Consider a charging capacitor:
Use Ampere’s Law
to calculate the
magnetic field just
above the top plate
ò B × ds = m I
I = òò J × n̂ da
0 enc
enc
S
1) Surface S1: Ienc= I
2) Surface S2: Ienc = 0
What’s Going On?
6
Displacement Current
We don’t have current between the capacitor
plates but we do have a changing E field. Can we
“make” a current out of that?
Q
E=
Þ Q = e 0 EA = e 0 F E
e0 A
dF E
dQ
= e0
º I dis
dt
dt
This is called the “displacement current”.
It is not a flow of charge but proportional to
changing electric flux
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Displacement Current:
I dis
dF E
d
= e 0 òò E × n̂ da = e 0
dt S
dt
If surface S2
encloses all of the
electric flux due to
the charged plate
then Idis = I
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Maxwell-Ampere’s Law
d
òC B × d s = m0 òòS J × n̂ da + m0e 0 dt
òò E × n̂ da
S
= m0 (I enc + I dis )
“flow of electric charge”
I enc =
òò J × n̂da
S
“changing electric flux”
I dis
d
= e0
dt
òò E × n̂da
S
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Concept Question: Capacitor
If instead of integrating the magnetic field around the pictured
Amperian circular loop of radius r we were to integrate around an
Amperian loop of the same radius R as the plates (b) then the
integral of the magnetic field around the closed path would be
1.
2.
3.
the same.
larger.
smaller.
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Concept Q. Answer: Capacitor
Answer 2. The line integral of B is larger for larger r
As we increase the radius of our Amperian loop we
enclose more flux and hence the magnitude of the
integral will increase.
d
B × ds = m0e 0 òò E × n̂da
ò
dt disk
circle
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Sign Conventions: Right Hand Rule
d
òC B × d s = m0 òòS J × n̂da + m0e 0 dt
òò E × n̂da
S
Integration direction clockwise for
line integral requires that unit
normal points into page for surface
integral.
Current positive into the page.
Negative out of page.
Electric flux positive into page,
negative out of page.
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Sign Conventions: Right Hand Rule
d
òC B × d s = m0 òòS J × n̂da + m0e 0 dt
òò E × n̂da
S
Integration direction counter
clockwise for line integral requires
that unit normal points out page for
surface integral.
Current positive out of page.
Negative into page.
Electric flux positive out of page,
negative into page.
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Concept Question: Capacitor
Consider a circular capacitor, with an Amperian circular loop
(radius r) in the plane midway between the plates. When the
capacitor is charging, the line integral of the magnetic field
around the circle (in direction shown) is
1.
2.
3.
4.
Zero (No current through loop)
Positive
Negative
Can’t tell (need to know direction of E)
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Concept Q. Answer: Capacitor
Answer 2. The line integral of B is positive.
There is no enclosed current through the disk. When
integrating in the direction shown, the electric flux is positive.
Because the plates are charging, the electric flux is increasing.
Therefore the line line integral is positive.
d
B × ds = m0e 0 òò E × n̂da
ò
dt disk
circle
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Concept Question: Capacitor
The figures above show a side and top view of a
capacitor with charge Q and electric and magnetic
fields E and B at time t. At this time the charge Q is:
1. Increasing in time
2. Constant in time.
3. Decreasing in time.
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Concept Q. Answer: Capacitor
Answer 1. The charge Q is increasing in time
The B field is counterclockwise, which means that the
if we choose counterclockwise circulation direction,
the electric flux must be increasing in time. So positive
charge is increasing on the bottom plate.
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Group Problem: Capacitor
A circular capacitor of
spacing d and radius R is
in a circuit carrying the
steady current i shown. At
time t = 0 , the plates are
uncharged
1. Find the electric field E(t) at P vs. time t (mag. & dir.)
2. Find the magnetic field B(t) at P
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Maxwell’s Equations
1
òò E ××n̂ da = e òòò r dV
(Gauss's Law)
òò B × ×n̂ da = 0
(Magnetic Gauss's Law)
d
òC E × d s = - dt òòS B × n̂ da
(Faraday's Law)
d
ò B × d s = m0 òò J × n̂ da + m0e 0 dt òò E × n̂ da
C
S
S
(Maxwell - Ampere's Law)
S
0
V
S
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Electromagnetism Review
E fields are associated with:
(1) electric charges
(2)
time changing B fields
(Gauss’s Law )
(Faraday’s Law)
B fields are associated with
(3a) moving electric charges (Ampere-Maxwell Law)
(3b) time changing E fields (Maxwell’s Addition (AmpereMaxwell Law)
Conservation of magnetic flux
(4) No magnetic charge (Gauss’s Law for Magnetism)
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Electromagnetism Review
Conservation of charge:
òò
closed
surface
d
J × dA = r dV
òòò
dt volume
enclosed
E and B fields exert forces on (moving) electric
charges:
Fq = q(E + v ´ B)
Energy stored in electric and magnetic fields
e0 2
U E = òòò uE dV = òòò
E dV
2
all space
all space
1 2
U B = òòò uB dV = òòò
B dV
2 m0
all space
all space
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Maxwell’s Equations
in Vacua
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Maxwell’s Equations
1. òò E × d A =
S
Qin0
e0
2. òò B × d A = 0
(Gauss's Law)
(Magnetic Gauss's Law)
S
dF B
3. ò E × d s = dt
C
0
dF E
4. ò B × d s = m0 I enc + m0e 0
dt
C
(Faraday's Law)
(Ampere - Maxwell Law)
What about free space (no charge or current)?
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How Do Maxwell’s Equations
Lead to EM Waves?
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Wave Equation
Start with Ampere-Maxwell Eq and closed oriented loop
d
ò B × d s = m0e 0 dt ò E × n̂ da
C
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Wave Equation
Start with Ampere-Maxwell Eq:
Apply it to red rectangle:
d
ò B × d s = m0e 0 dt ò E × n̂ da
C
ò B × d s = B (x,t)l - B (x + Dx,t)l
z
C
z
d
E × n̂ da
ò
dt
æ
¶E y (x + Dx / 2,t) ö
= m0e 0 ç l Dx
÷
¶t
è
ø
m0 e 0
¶E y (x + Dx / 2,t)
Bz (x + Dx,t) - Bz (x,t)
= m0 e 0
Dx
¶t
So in the limit that dx is very small:
¶E y
¶Bz
= m0 e 0
¶x
¶t
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Group Problem: Wave Equation
d
Use Faraday’s Law
ò E × d s = - dt ò B × n̂ da
C
and apply it to red rectangle to find the partial differential
equation in order to find a relationship between
¶E y / ¶x and ¶Bz / ¶t
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Group Problem: Wave Equation Sol.
d
ò E × d s = - dt ò B × n̂ da
C
Use Faraday’s Law:
and apply it to red rectangle:
ò E × d s = E (x + Dx,t)l - E (x,t)l
y
y
C
¶Bz
d
- ò B × n̂ da = -ldx
dt
¶t
E y (x + dx,t) - E y (x,t)
dx
¶Bz
=¶t
So in the limit that dx is very small:
¶E y
¶Bz
=¶x
¶t
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1D Wave Equation for Electric Field
¶E y
¶Bz
=¶x
¶t
¶E y
¶Bz
= m0e 0
¶x
¶t
(1)
(2)
Take x-derivative of Eq.(1) and use the Eq. (2)
¶2 E y
¶x 2
2
æ
ö
¶E
¶
Ey
æ
ö
æ
ö
¶Bz
¶
¶
¶ ¶Bz
y
= ç
=- ç
= m0e 0 2
÷ = ç÷
÷
¶x è ¶x ø ¶x è ¶t ø
¶t è ¶x ø
¶t
¶2 E y
¶x
2
= m0e 0
¶2 E y
¶t
2
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1D Wave Equation for E
¶ Ey
2
¶x
2
¶ Ey
2
= m0e 0
¶t
2
This is an equation for a wave. Let
ü
= f '' x - vt ï
2
ï
¶x
ý Þ v=
2
¶ Ey
ï
2
= v f '' x - vt ï
2
¶t
þ
¶ Ey
2
(
)
(
)
E y = f (x - vt)
1
m0 e 0
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Definition of Constants and Wave Speed
Recall exact definitions of
c º 299792458 m × s
-1
m0 º 4p ´ 10-9 N × s2 × C-2
The permittivity of free space e 0 is exactly defined by
e0 º
1
c m0
2
º 8.854187817 ´10 C × m × N
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Þv=c=
1
m0 e 0
2
-2
-1
in vacua
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Group Problem: 1D Wave Eq. for B
¶E y
¶Bz
=¶t
¶x
¶E y
¶Bz
= - m0 e 0
¶x
¶t
Take appropriate derivatives of the above equations
and show that
¶ Bz
1 ¶ Bz
= 2 2
2
¶x
c ¶t
2
2
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Wave Equations: Summary
Both electric & magnetic fields travel like waves:
¶ Ey
2
¶x
2
1 ¶ Ey
= 2
2
c ¶t
¶ Bz
1 ¶ Bz
= 2 2
2
¶x
c ¶t
2
with speed
c=
2
2
1
m0e 0
But there are strict relations between them:
¶E y
¶Bz
=¶t
¶x
¶E y
¶Bz
= - m0 e 0
¶x
¶t
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Electromagnetic Waves
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Electromagnetic Radiation: Plane
Waves
http://youtu.be/3IvZF_LXzcc
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