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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 9 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 9.1
A
¯
∂B
∂t
¯
= µσE
�×B
¯=−
�×E
So
¯
¯ = −µσ ∂B
�×�×B
∂t
But
� × (� × B̄) = �(� · B̄) − �2 B̄ = −�2 B̄
So
¯ = µσ
�2 B
¯
∂B
∂t
B
Since B̄ only has a z component
∂Bz
∂t
In cylindrical coordinates
�
�
∂
1 ∂2
∂2
1 ∂
r
+ 2 2+ 2
�2 =
r ∂r
∂r
r ∂θ
∂z
�2 Bz = µσ
Here Bz = Bz (r, t), so
�
�
1 ∂
∂B̂
r
+ µσαB̂ = 0
r ∂r
∂r
C
We want the magnetic field to remain finite at r = 0, hence C2 = 0.
D
At r = a
√
B(a, t) = µ0 H0 − C1 J0 ( µ0 σαa) = µ0 H0
Hence if C1 =
� 0
√
J0 ( µ0 σαa) = 0
1
Problem Set 9
6.641, Spring 2005
E
Multiply both sides of expression for B(r, t = 0) = 0 by rJ0 (vj ar ) and integrate from 0 to a. Then,
� a
a2
r
µ0 H0 rJ0 (vj )dr = µ0 H0 J1 (vj )
a
vj
0
�
0
∞
a�
r
r
a2
Ci J0 (vi )rJ0 (vj )dr = Cj J12 (vj )
a
a
2
i=1
from which it follows that
Cj =
2µ0 H0
vj J1 (vj )
The values of vj and J1 (vj ) given in the table lead to the coefficients
C1
C2
C3
= .802;
= −.535;
= .425
2µ0 H0
2µ0 H0
2µ0 H0
F
α1 =
τ1 =
1 � v1 �2
µ0 σ a
µ0 σa2
= 0.174µ0 σa2
2
v1
τ1 = (0.174)(4π × 10−7 )
104
(25) × 10−4
4π
≈ 4.35 × 10−7 seconds
Problem 9.2
a
(2)
(1)
X1 = -L
X2
X1
X3
(3)
b
X1 = L
(4)
Figure 1: Diagram of surfaces (1), (2), (3), and (4) to evaluate the force on the lower plate using the Maxwell
Stress Tensor. (Image by MIT OpenCourseWare.)
Before finding the force, we must calculate the H̄ field at x1 = L. To find this field let us use
�
B̄ · nda
¯ =0
over the dotted surface. At x1 = +L,
H̄(x1 = L) = H0 ī1
2
(1)
Problem Set 9
6.641, Spring 2005
over surface (4) H̄ = 0, and over surface (2), H̄ is in the ī1 direction, where n̄ = ī2 . Thus over surface (2),
B̄ · n̄ = 0. Hence, the integral in (1) reduces to
�
�
−
µ0 H0 da +
µ0 H(x1 = +L)da = 0
(1)
(3)
−µ0 H0 a + µ0 Hb = 0
per unit length
Thus:
H̄(x1 = +L) = (a/b)H0 ī1
δij
µ0 Hk Hk
2
Hence, the stress tensor over surfaces (1), (2), and (3) is:
⎡ µ0 2
⎤
0
0
2 H1
⎦
− µ20 H12
0
Tij = ⎣ 0
0
0
− µ20 H12
Tij = µ0 Hi Hj −
Over surface (4)
Tij = [0]
Thus the force in the 1 direction is
�
f1 = Tij nj · da
f1 = −
�
T11 da +
(1)
�
T11 da +
(3)
�
T12 da
(2)
Thus, since the last integral makes no contribution,
�
µ0
µ0 2 � a �2
µ0 2 � a
f1 = − H02 (a) +
·b=
H0
H0 a
−1
2
2
b
2
b
(2)
Since Tij = 0 over surface (4) there is no contribution to the force from this surface and, by symmetry, there
is no contribution to the force from the surfaces perpendicular to the xj axis. Thus, the force per unit depth
in 1 direction is (2).
Problem 9.3
First, let us note the Ē fields on each of the surfaces of the figure over surfaces (1), (3), (5), and (7), E1 = 0.
Over surface
(6) E2 =
V0
a
V0
b
V0
(2) E2 =
c
From Eq. 8.3.10,
(4) E2 =
Tij = ε0 Ei Ej −
E1 = 0
E1 = 0
E1 = 0
δij
ε0 Ek Ek
2
3
Problem Set 9
6.641, Spring 2005
depth D
A
a
- V0
a-b
7
6
4
5
2
3
b
+
c
1
B
X2
X3
X1
8
Figure 2: Diagram of surfaces (1)-(8) used to find force on the lower electrode using the Maxwell Stress
Tensor. (Image by MIT OpenCourseWare.)
Hence, over surfaces (1), (3), (5) and (7)
T12 = 0
(3)
and over surfaces
ε0
= −
2
�
V0
a
�2
(4) T11 = −
ε0
2
�
V0
b
�2
(2) T11 = −
ε0
2
�
V0
c
�2
Tij nj da =
�
T11 n1 da +
(6) T11
Now
f1 =
�
�
�
T12 n2 da +
�
T13 n3 da
T13 n3 da = 0 because the problem is two dimensional
Let us consider each of the other integrals:
�
T12 n2 da = 0
because the surfaces that have normal n2 are (1),(3),(5), and (7) and by (3) we have shown that T12 = 0
over these surfaces. Also, we get no contribution to the force over surface (8), because Ē → 0 faster than
the area → ∞. Hence the calculation of the force reduces to
�
�
�
(2)
(6)
(4)
f1 =
T11 da6 −
T11 da4 −
T11 da2
(6)
f1 = −
ε0 DV02
2
(4)
�
1 1 1
− +
a
b
c
(2)
�
Note: by symmetry, there is no contribution to the force from the surfaces perpendicular to the x3 axis.
4
Problem Set 9
6.641, Spring 2005
a
͚
σ
͚
1
σ
n
φ = φ0 sin π x2
a
σ 0
ε0 , μ0
n
X2
0
X1
X3
Figure 3: Diagram of grounded electrodes and distributed electric potential source at x1 = 0. (Image by
MIT OpenCourseWare.)
Problem 9.4
A
From elementary field theory, we find that
πx2 − πxa1
φ = φ0 sin
e
a
satisfies �2 φ = 0 in the region between the plates and the required boundary conditions. The distribution
of Ē follows from
Ē = −�φ
Hence,
Ē =
πφ0 − πx1 �
πx2
πx2 �
e a sin
ī1 − cos
ī2
a
a
a
The sketch of the Ē field is obtained by recognizing that Ē is directed perpendicular to contours of constant
φ.
Figure 4: Sketch of the Ē field and equipotential lines. (Image by MIT OpenCourseWare.)
B
¯ = 0 everywhere except on the upper side where
To find the force at the bottom plate, we use surface (2). E
the normal n̄ = ī2 and the field is
Ē = −
πφ0 − πx1
e a ī2
a
5
Problem Set 9
6.641, Spring 2005
Hence,
f1 =
f2 =
�
Tij nj da = 0
�
T2j nj da =
�
T22 n2 da2
per unit x3 . This reduces to
� ∞
f2 =
T22 dx1
0
π 2 φ2
but, T22 = 12 ε0 E2 E2 = 21 ε0 a2 0 e−
�
ε0 π 2 φ20 ∞ − 2πx1
f2 =
e a dx1
2a2
0
f2 =
2πx1
a
and thus
ε0 πφ20
4a
C
On the top plate, use surface (1). Only the sign of the normal changes, and the result is
f1 = 0
ε0 πφ20
4a
or the force is equal and opposite to that on the bottom plate.
f2 = −
Problem 9.5
A
Since J � = J
K = iz K0 cos(kU t − kx)
= iz K0 cos(ωt − kx);
ω = kU
B
The track can be taken as large in the y direction when it is many skin depths thick
�
�
2
2
L = track thickness � δ =
=
ωµ0 σ
kU µ0 σ
In the track we have the diffusion equation
∂B
1
�2 B =
µ0 σ
∂t
ˆ exp j(ωt − kx),
or, with B = ReB
1
µ0 σ
�
∂ 2 B̂x
− k2 B̂x
∂y 2
6
�
= jωB̂x
Problem Set 9
6.641, Spring 2005
Let B̂x (y) = Ceαy , then
1 2
k2
α = jω +
µ0 σ
µ0 σ
�
ωµ0 σ
U µ0 σ
α = k 1 + jS; S =
=
k2
k
Since the track is modeled as infinitely thick
Bx = Ceαy ej(ωt−kx)
The gap between track and train is very thin; thus,
−iy ×
B
= K = K0 ej(ωt−kx) iz
µ0
which yields
Bx (x, y, t) = µ0 K0 eαy ej(ωt−kx)
We must also have � · B =
∂Bx
∂x
+
∂By
∂y
= 0 or
By =
jk
Bx (x, y, t)
α
To compute the current in the track we note that
�
�
∂By
∂Bx
� × B = iz
= µ0 J
−
∂x
∂y
�
�
S 2 Bx
J =− j k
(x, y, t)iz
α
µ0
C
The time average force density in the track is
1
Re(Jz Bx∗ )
2
Hence the time average lifting force per unit x − z area on the train is
� 0
� 0
1
�Ty � = −
�Fy � dy = −Re
Jz Bx∗ dy
2
−∞
−∞
�√
�
2−1
1
+
S
1
√
>0
= µ0 K02
4
1 + S2
�Fy � =
D
The time average force density in the track in the x direction is
1
�Fx � = − Re(Jz By∗ )
2
The force on the train in the x direction is then
� 0
� 0
1
�Tx � = −
�Fx � dy = Re
Jz By∗ dy
2
−∞
−∞
=−
S
µ0 K02
√
<0
√
2
4
1 + S Re 1 + jS
The problem is that this force drags the train instead of propelling it in the x direction. To make matters
worse, if the train stops, the magnetic levitation force becomes zero.
7
Problem Set 9
6.641, Spring 2005
Problem 9.6
A
From Eq. 8.1.11,
⎡
1
2
2µ0 (Bx −
Bx By
µ0
⎢
Tij = ⎣
By2 )
Bx By
µ0
1
2
(−B
x+
2µ0
0
⎤
0
⎥
0
⎦
1
2
2
(−B
−
B
)
x
y
2µ0
By2 )
0
¯ are given in the problem.
where the components of B
B
The appropriate surface of integration, which is fixed with respect to the fixed frame, is shown in Figure (5).
We compute the time average force, and hence contributions from surfaces (1) and (3) cancel. Fields go to
zero on surface (2), which is at y → ∞. Thus, there remains the stress on surface (4). The time average
value of the surface force density T̄ is independent of x. Hence,
2
train
y
1
3
z
x
4
s
track
Figure 5: Diagram of the Maxwell Stress Tensor surface to find the levitation force on a train. (Image by
MIT OpenCourseWare.)
Ty = − < Tyy (y = 0) >
Ty = −
1
< −Bx2 + By2 >
2µ0
(4)
Observe that
�
�
ˆ −jkUt ReBe
ˆ −jkUt ≡ 1 ReÂB̂ ∗
ReAe
2
where B̂ ∗ is complex conjugate of B̂, and (4) becomes
�
�
1
(−jkµ0 K0 ) jkx (jkµ0 K0 ) −jkx
jkx
−jkx
Ty = −
Re −(µ0 K0 e )(µ0 K0 e
)+
e
e
4µ0
α
α∗
=
µ0 K02
4
�
�
k2
1−
αα∗
(5)
8
Problem Set 9
6.641, Spring 2005
Finally, use the given definition of α to write (5) as
⎡
⎤
Ty =
µ0 K02
4
⎢
⎥
1
⎢1 − �
⎥
⎣
�
�2 ⎦
1 + µ0k
σU
Note that Ty is positive so that the train is supported by the magnetic field. However, as U → 0 (the train
is stopped) the levitation force goes to zero.
C
For the force per unit area in the x direction
1
< Bx By (y = 0) >
2µ0
�
�
�
�
1
jkµ0
jkx
−jkx
K0 e
=−
Re µ0 K0 e
2µ0
α∗
Tx = −
Thus
Tx = − �
2 1+
µ0 K02
Rej
�
�2 � 12
µ0 σU
k
�
1−j
�
µ0 σU
k
�
(6)
As must be expected, the force on the train in the x direction vanishes as U → 0. Note that in any case the
force always tends to retard the
� motion and hence could hardly be used to propel the train.
The identity sin(θ/2) = ± (1 − cos θ)/2 is helpful in reducing (6) to the form
�
⎞
� ⎛�
�
�2
�
2
µ0 σU
−µ0 K0
�
1 ⎝
�
− 1⎠
1+
Tx = �
�
�2 � 2
1 2
k
2 1 + µ0kσU
Problem 9.7
A
From Ampere’s Law,
Bz =
µ0 N iF
D
B
λ = N W T Bz ≡ LiF ⇒ L =
µ0 N 2 W T
D
9
Problem Set 9
6.641, Spring 2005
C
Apply Faraday’s Law to the armature circuit and assume perfectly conducting wires.
�
�
d
→
→
−
→ −
−
→ −
B · dS = 0
E · dl =
−
dt S � �� �
C
zero
�
�
(−)
Ey dy +
(+)
(+)
(−)
� �� �
−
→
−�φ · dl = 0 ⇒ Ey W = vA
� �� �
f luid
terminals
Ohm’s Law ⇒ J = σ(E + v × B) ⇒ Ey =
Jy
σ
+ vBz
iA
+ vBz
σDT
�
�
�
�
µ0 N W
W
iA +
viF
vA =
σDT
D
� �� �
� �� �
Ey =
R
D
G
Force density =
J × B = Jy Bz x̂ =
µ0 N iF iA
x̂
TD
Power =
Jy Bz U · T
�� � =
� DW
volume
E
µ0 N W
iF iA U = GiF iA U
D
vA = RiA + GU iF
diF
vF = L
dt
vF = vA
iF = −iA
Putting everything together,
L
diF
= −RiF + GU iF
dt
Self excitation implies
GU > R ⇒ U >
1
µ0 σN T
10