Class 28: Outline
Hour 1:
Displacement Current
Maxwell’s Equations
Hour 2:
Electromagnetic waves
P28- 1
Finally:
Bringing it All Together
P28- 2
Displacement Current
P28- 3
Ampere’s Law: Capacitor
Consider a charging capacitor:
I
Use Ampere’s Law to calculate the
magnetic field just above the top plate
Ampere's law:
∫
B ⋅ d s = µ 0 I enc
1) Red Amperian Area, Ienc= I
2) Green Amperian Area, I = 0
What’s Going On?
P28- 4
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=
⇒ Q = ε 0 EA = ε 0 Φ E
ε0 A
dΦE
dQ
= ε0
≡ Id
dt
dt
This is called (for historic reasons)
the Displacement Current
P28- 5
Maxwell-Ampere’s Law
B
⋅
d
s
=
µ
(
I
+
I
)
0
encl
d
∫
C
= µ 0 I encl
dΦE
+ µ 0ε 0
dt
P28- 6
PRS Questions:
Capacitor
P28- 7
Maxwell’s Equations
P28- 8
Electromagnetism Review
• E fields are created by:
(1) electric charges
(2) time changing B fields
• B fields are created by
(1) moving electric charges
(NOT magnetic charges)
(2) time changing E fields
Gauss’s Law
Faraday’s Law
Ampere’s Law
Maxwell’s Addition
• E (B) fields exert forces on (moving) electric charges
Lorentz Force
P28- 9
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
P28- 10
Electromagnetic Radiation
P28- 11
A Question of Time…
http://ocw.mit.edu/ans7870/8/
8.02T/f04/visualizations/light/
05-CreatingRadiation/05pith_f220_320.html
P28- 12
P28- 13
Electromagnetic Radiation:
Plane Waves
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/light/07-EBlight/07-EB_Light_320.html
P28- 14
Traveling Waves
Consider f(x) =
x=0
What is g(x,t) = f(x-vt)?
t=0
t=t0
t=2t0
x=0
x=vt0 x=2vt0
f(x-vt) is traveling wave moving to the right!
P28- 15
Traveling Sine Wave
Now consider f(x) = y = y0sin(kx):
Amplitude (y0)
2π
Wavelength (λ ) =
wavenumber (k )
x
What is g(x,t) = f(x+vt)? Travels to left at velocity v
y = y0sin(k(x+vt)) = y0sin(kx+kvt)
P28- 16
Traveling Sine Wave
y = y0 sin ( kx + kvt )
At x=0, just a function of time: y = y0 sin( kvt ) ≡ y0 sin(ω t )
Amplitude (y0)
1
Period (T ) =
frequency (f )
2π
=
angular frequency (ω )
P28- 17
Traveling Sine Wave
i Wavelength: λ
i Frequency : f
i Wave Number: k =
y = y0 sin(kx − ω t )
2π
λ
i Angular Frequency: ω = 2π f
1 2π
i Period: T = =
ω
f
ω
i Speed of Propagation: v =
=λf
k
i Direction of Propagation: + x
P28- 18
Electromagnetic Waves
Hz
Remember:
λf =c
P28- 19
Electromagnetic Radiation: Plane Waves
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/light/07-EBlight/07-EB_Light_320.html
Watch 2 Ways:
1) Sine wave
traveling to
right (+x)
2) Collection of
out of phase
oscillators
(watch one
position)
Don’t confuse vectors with heights – they
are magnitudes of E (gold) and B (blue)
P28- 20
PRS Question:
Wave
P28- 21
Group Work:
Do Problem 1
P28- 22
Properties of 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
P28- 23
Direction of Propagation
ˆ E sin(k ( pˆ ⋅ r ) − ω t ); B = Bˆ B sin(k ( pˆ ⋅ r ) − ω t )
E=E
0
0
ˆ × Bˆ = pˆ
E
ˆ
E
ˆi
Bˆ
ˆj
pˆ
kˆ
( pˆ ⋅ r )
ˆj
kˆ
kˆ
ˆi
ˆi
ˆj
x
ˆj
kˆ
ˆi
ˆj
−z
ˆi
−kˆ
−ˆi
kˆ
−ˆj
z
y
−x
−y
P28- 24
PRS Question:
Direction of Propagation
P28- 25
In Class Problem:
Plane EM Waves
P28- 26
Energy & the Poynting Vector
P28- 27
Energy in EM Waves
Energy densities:
Consider cylinder:
1
1
2
2
uE = ε 0 E , uB =
B
2
2µ0
2
⎞
1⎛
B
2
dU = (u E + u B ) Adz = ⎜ ε 0 E +
⎟ Acdt
2⎝
µ0 ⎠
What is rate of energy flow per unit area?
2
⎞
B
1 dU c ⎛
2
= ⎜ ε0 E +
S=
⎟=
2⎝
µ0 ⎠
A dt
EB
EB
2
=
ε 0 µ0c + 1 =
µ0
2µ0
(
c⎛
EB ⎞
⎜ ε 0 cEB +
⎟
2⎝
cµ 0 ⎠
)
P28- 28
Poynting Vector and Intensity
Direction of energy flow = direction of wave propagation
S=
E×B
µ0
: Poynting vector
units: Joules per square meter per sec
Intensity I:
2
0
2
0
E0 B0
E
cB
=
=
I ≡<S >=
2µ0 2µ0c 2µ0
P28- 29
Energy Flow: Resistor
S=
E×B
µ0
On surface of resistor is INWARD
P28- 30
PRS Questions:
Poynting Vector
P28- 31
Energy Flow: Inductor
S=
E×B
µ0
On surface of inductor with increasing
current is INWARD
P28- 32
Energy Flow: Inductor
S=
E×B
µ0
On surface of inductor with decreasing
current is OUTWARD
P28- 33