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6.641 Electromagnetic Fields, Forces, and Motion
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Spring 2005
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6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2005
Problem Set 11 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 11.1
A
¯ and H
¯ have the given
The given equations follow by writing out Maxwell’s equations and assuming E
directions and dependences.
B
The force equation for an incremental volume element is
F̄ = īx mne
∂vx
∂t
where F¯ is the force density due to electrical forces on the electrons
F¯ = −i¯x ene Ex
Thus,
−ene Ex = mne
∂vx
∂t
(1)
C
As the electrons move, they give rise to the current density
Jx ≈ −ene vx
(linearized)
(2)
D
Assume ej(ωt−kx) dependence and (1) and (2) require
2
e ne
Êx
Jˆx = −j
ωm
� �
ωp2
Êx
= −jωε0
ω2
�
2
where ωp = emεn0e is called the plasma frequency. (See page 600 in Woodson and Melcher, Electromechanical Dynamics,
vol. 2). Combining this with Maxwell’s equations:
�
�
ωp2
ω2
1
2
k = 2 1− 2 ;
c= √
c
ω
ε0 μ0
1
Problem Set 11
6.641, Spring 2005
E
We have a dispersion which yields evanescent waves below the plasma (cutoff) frequency. Below this fre­
quency, the electrons respond to the electric field associated with the wave in such a way as to reflect rather
than transmit an incident electromagnetic wave.
F
Waves impinging on a boundary between free space and plasma will be totally reflected if the wave frequency
ω < ωp . The plasma frequency for the ionosphere is typically fp ≈ 10 MHz. This result explains why
AM broadcasts (500 kHz < f < 1500 kHz) can commonly be monitored all over the world, whereas FM
(88 MHz < f < 108 MHz) has a range limited to “line-of-sight.”
Problem 11.2
A
The equation of motion for the string is
m
∂2ξ
∂2ξ
=
f
+ S − mg
∂t2
∂x2
(3)
where, for small deflections ξ in the “ 1r ” field from Q,
�
�
qQ
ξ
S≈
1+
2πε0 d
d
In static equilibrium, ξ = 0 and from (3)
qQ = 2πdε0 · mg
(4)
B
The perturbation equation of motion remains
�
�
qQ
∂2ξ
∂2ξ
m 2 =f 2 +
ξ
∂t
∂x
2πd2 ε0
Assume ej(ωt−kx) dependence and (5) requires (vs =
ω 2 = vs2 k2 −
(5)
�
f
m)
qQ
2πd2 ε0 m
or from (4),
ω 2 = vs2 k2 −
g
d
The boundary conditions require k =
vs2
� π �2
m<
nπ
l ,
and for stability the mot critical mode is n = 1; thus
g
l
d
�
f d π �2
g
>
l
2
Problem Set 11
6.641, Spring 2005
C
Increase f, d, or decrease l.
Problem 11.3
A
This problem is very similar to that of problem 10.7. Using the same reasoning as in that problem, we obtain
σm
∂ 2 ξ1
∂ 2 ξ1
ε0 V 2
= S 2 + 30 (2ξ1 − ξ2 )
2
∂t
∂x
d
σm
∂ 2 ξ2
∂ 2 ξ2
ε0 V 2
= S 2 + 30 (2ξ2 − ξ1 )
2
∂t
∂x
d
B
Assuming sinusoidal solutions in time and space, the dispersion relation is
−σm ω 2 + Sk2 −
2ε0 V02
ε0 V 2
= ± 30
3
d
d
We have a dispersion relation that factors into two parts. The odd mode, ξ1 = −ξ2 has the dispersion
relation
�
Sk2
3ε0 V02
ω=
−
σm
σm d3
� 12
The even mode, ξ1 = ξ2 has the dispersion relation
�
Sk2
ε0 V02
ω=
−
σm
σm d3
� 12
C
A plot of the dispersion relation appears in Figure 1.
D
π
The lowest allowed value of k is k = L
since the membranes are fixed at x = 0 and x = L. Therefore the
first mode to go unstable is the odd mode. This happens as
�
�
3ε0 V02
π2
=
Sd3
L2
or
� 2
�1
� π Sd3 � 2
�
V0 = �� 2
L ε0 3 �
Problem 11.4
We may take the results of Prob. 10.13, replacing
∂
∂t
by
3
∂
∂t
∂
+ U ∂x
and replacing ω by ω − kU .
Problem Set 11
6.641, Spring 2005
ω
2 1/2
0 0
3
m
( 3σε dV )
odd
2 1/2
0 0
3
( εSVd )
(
3ε0V02
Sd 3
1/2
k
)
ωi
2 1/2
( εσ0mVd03 )
even
ωr
Figure 1: Plot of the dispersion relation for two membranes. (Image by MIT OpenCourseWare.)
A
The equations of motion are
�
σm
∂
∂
+U
∂t
∂x
�2
ξ1 = S
∂ 2 ξ1
ε0 V02
+
(2ξ1 − ξ2 )
∂x2
d3
(6)
and
�
σm
∂
∂
+U
∂t
∂x
�
ξ2 = S
∂ 2 ξ2
ε0 V 2
+ 30 (2ξ2 − ξ1 )
2
∂x
d
(7)
B
The dispersion relation is biquadratic, and factors into
−σm (ω − kU )2 + Sk2 −
2ε0 V02
ε0 V02
=
±
d3
d3
(8)
The (±) signs correspond to the cases ξ1 = −ξ2 and ξ1 = ξ2 respectively, as will be seen in part (d).
C
�
The dispersion relations are plotted in figures (2) and (3) for U >
D
Let ξ1 = ξ2 . Then (6) and (7) become
�
σm
∂
∂
+U
∂t
∂x
�2
ξ1 = S
∂ 2 ξ1
ε0 V 2
+ 30 ξ1
2
∂x
d
4
S
σm .
Problem Set 11
6.641, Spring 2005
ω
U-vs
U+vs
vs =
ω=
[U/(U2- σSm ]
3ε0V02
Sd 3
1/2
]
-1
3ε0V02
k=
σmd3
[
[(U2- σSm )
S
σm
1/2
] [U
k
1/2
2
- v s2
]
Figure 2: Plot of the dispersion relation for odd motions (ξ1 = −ξ2 ). (Image by MIT OpenCourseWare.)
ω
U+vs
U-vs
vs =
ω=
[(U2- σSm )
[U/(U2- σSm ]
k=
3ε V 2
1/2
3ε0V02
Sd 3
S
σm
1/2
]
-1
1/2
k
[ σm0d30 ] [U 2 - v s2 ]
Figure 3: Plot of the dispersion relation for even motions (ξ1 = ξ2 ). (Image by MIT OpenCourseWare.)
and
�
σm
∂
∂
+U
∂t
∂x
�2
ξ2 = S
ε0 V02
∂ 2 ξ2
+
ξ2
∂x2
d3
These equations are identical for ξ1 = −ξ2 ; the dispersion equation is (8) with the + sign.
5
Problem Set 11
6.641, Spring 2005
E
ˆ jωt = −ξ2 (0, t)
ξ1 (0, t) = Reξe
∂ξ2
∂ξ1
=
= 0 at x = 0
∂x
∂x
The odd mode is excited. Hence, we use the + sign in (8)
−σm (ω − kU )2 + Sk2 −
3ε0 V02
=0
d3
k2 (S − σm U 2 ) + 2σm ωkU − σm ω 2 −
3ε0 V02
=0
d3
Solving for k, we obtain
k± = α ± β
ωU
U 2 −vs2
where α =
�
β=
ω 2 vs2 −
3ε0 V02 (U 2 −vs2 )
σm d3
� 1
2
U 2 − vs2
with vs2 = σSm .
Therefore
��
�
�
ξ1 = Re Ae−j(α+β)x + Be−j(α−β)x ejωt
Applying the boundary conditions, we obtain
(β − α)
A = ξˆ
2β
B=
(α + β)ξˆ
2β
Therefore, if ξˆ is real
α
ξ1 (x, t) = −ξ2 (x, t) = ξˆ cos βx cos(ωt − αx) − ξˆ sin βx sin(ωt − αx)
β
F
We can see that β can be imaginary, for which we will have spatially growing curves. This can happen when
ω 2 vs2 −
3ε0 V02 2
(U − vs2 ) < 0
σm d3
or
V02 >
σm d3 ω 2 vs2
3ε0 (U 2 − vs2 )
(9)
G
With V0 = 0 and v > vs : (see Figure 4)
Amplifying waves are obtained as (9) is satisfied (see Figure 5)
6
Problem Set 11
6.641, Spring 2005
ξ1
x
π /β
ξ2
x
ω/ α
Figure 4: ξ1 and ξ2 with V0 = 0 and v > vs . (Image by MIT OpenCourseWare.)
ξ1
x
ξ2
x
ω/ α
Figure 5: Amplifying waves. (Image by MIT OpenCourseWare.)
Problem 11.5
A
The equation of motion is
�
σm
with T =
2
ε0 V0
2 (d−ξ )2
�2
∂2ξ
− σm g + T
∂x2
�
�
≈ ε20 V02 d12 + d2ξ3 .
∂
∂
+U
∂t
∂x
(10)
ξ=S
7
Problem Set 11
6.641, Spring 2005
For equilibrium, ξ = 0 and from (10)
ε0 V02
= σm g
2d2
or
�
2σm gd2
V0 =
ε0
� 12
B
With solutions of the form ej(ωt−kx) the dispersion relation is
(ω − kU )2 =
S 2 ε0 V02
k −
σm
σm d3
Solving for k, we obtain
�
ωU ±
k=
�
For U >
S
2
σm ω
�
− U2 −
(U 2 −
S
σm ,
S
σm
��
ε0 V02
σm d3
�
S
σm )
and not to have spatially growing waves
�
��
�
S 2
ε0 V02
S
2
ω − U −
> 0
σm
σm
σm d3
or
ω2 >
��
�
�
ε0 V02
S
U2 −
σm Sd3
8