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Solutions Manual for Electromechanical Dynamics
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FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.1
The identity to be verified is
V-(iýA)
= 4IV.A + R-*VI
First express the identity in index notation..
aA
= ý
[PA]
I
m
x
m
x
+ A
m
m ax
m
The repeated subscript indicates summation.
Thus, expanding the first term on
the left yields:
3A
m + A
m ax
m
m
"x
0iV.A
+ A.V
(c)
PROBLEM 8.2
We wish to show that
B-v(ipA) = iBviA + AB.V
First, the identity is expressed in index notation, conssidering the mth
Note that the equat:ion relates two vectors.
component of this vector equation.
(B.[V(pA)])
m
= (B*.[VAI)
m
+ A B.Vm
(b)
Now, consider each term separately
(A
(BE[V(1A)])m = Bk
m )
= A m Bk ax k +PBk
aA
(iB*[VA])
Am*V
=B B
= AmBk[Vi k = AmBk
The sum of (d) and (e) give (c) so that the identity is verified.
PROBLEM 8.3
Part a
aik is
the cosine of the angle between the x
(see p
age 435).
Thus for our geometry
/1
1
aik
o
o
-41­
-4
axis and the xk axis
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.3 (Continued)
Now,
we may apply the transformation law for vectors (Eq. 8.2.10)
(b)
Ai = aikAk
where the components of A in the (xl,x 2 ,x3 ) system are given as
Al = 1; A 2 = 2; A 3
- 1
(c)
Thus:
=allA
A' =alk
(d)
+ al2A 2 + a13A3
A' = 1/2 + /Y
(e)
2
A
a2 kAk
A3
a3kAk = -
+
(f)
1
1
(g)
That is:
Using matrix alegbra, we can write a more concise solution.
A1
A2
=
L3j
A1yl
a11
12
13
a21
a2 2
a23
A2
La31
32
a 33
3
- !
+ 1)
(h)
1)
Part b
The tensor aik is associated with coordinate transforms involving the
direction of force while the tensor a
is associated with coordinate trans­
forms involving the direction of the area normal vectors.
The tensor
transformation is (Eq. 8.2.17), page 437;
T'j = aik ajTk£
(i)
For example,
T11 = allk
Tkt = allallT1
+ al2allT21 + al3allT 31
+ alla12T12 + a12a12T22 + a13a12 T32
(
+ alla13T13 + a12a13T23 + a13a13T33
Thus:
T 11=
11
4
+
(k)
4
-42­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.3 (Continued)
Similarly
3 +/3
T2
12
2
4
= 0
T'
13
3 +r
21
2
5
3/
22
4
T' = 0
23
T'
= 0
T'
= 0
T'3
33
4
2
1
Written in matrix algebra, the problem is solve d below:
T'
11
T'
T]
Tl
T'
T2'a
T'
T'
33
21
31
12
a
11
13
T23
T22
32
21
a
a
12
T12
a2 1
a3 1
a
22
T22
a2 2
a32
T32
a23
a33
31 a 32
Note that the third matrix on the right i.s the transpose of aij.
Matrix
multiplication of (t) gives
7
6
+
3
3
2 )
(- 2 + r3
4
5
3r
ij
PROBLEM 8.4
The m
th
component of the force density at a point is (Eq. 8.1.10)
F
=
i dax
Thus in the 11 direction,
-43­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.4
(Continued)
aF
F
1
aT12 + T3T
+ X2
2
2
=
3
(
a
2
xI
-0 o x
+0
= 0
a
Similarly in the 12 and 13 directions we find
aT
aT
2
(axi
F=
1
2
a2
aT
aT
Fj3 =
aT
22
x1
=0
X3
aT
3
32
1
3x1 +x2
+x3
Hence, the total volume force density resulting from the given stress tensor is
zero.
PROBLEM 8.5
I__.e i~>
in region (1) E=Eo
in region (2) E = 0
ii>
I 0b)
3
Tij =E EiE
-
EoEEk
Thus in region (2)
Tij = 10]
in region (1)
-44­
+
3
1 i
)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.5 (Continued)
5
8
Tij
E2
o0
3
2
2
oo
3
E2
5
2
oo
8
2
Cc)
o0
0
8
3 E 2
oo
The total contribution to the forces found by integrating the stress tensor
over surface (c) is zero, because surface (c) lies in region (2) where the
stress tensor is zero.
By symmetry the sum of contributions to the force
resulting from integrations over the two surfaces perpendicular to the x 3 axis
is
zero.
Now let us note the fact that:
area (a) = 2
(d)
area (b) = 3
(e)
Thus:
Ti j n
fi =
f
(f)
da
= fT 1 1 da + fT1 2 da + fT1 3da
(a)
(b)
=5
8
1
f= 4
E23) + 3 E E2 (2)
o
2 oo
8
(g)
E2
(h)
o0o
f2 = fT2 1 da + fT2 2 da + fT2 3 da
(b)
2
3
(a)
oo
2
8
5
oo
E2
o o
f
2
34
f-3
fT 3 1 da + fT 3 2 da + fT 3 3 da
(j)
=0
(k)
Hence, the total force is:
7 2 +
84E E i i + 3
8
0 0
1
2
cEo i()
oo 2
-45­
)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.6
Part a
At point A, the electric field intensity is a superposition of the imposed
field and the field due to the surface charges; E = (/0f/ )1
.
Thus at A,
a
E= i(E
x
0
) + i (E
y
0
+ --)
(a)
Eo
while at B,
E= ix
(Eo)
(b)
Cy(Eo)
+
Thus, from Eq. 8.3.10, at A,
T
ij
=
-(E
So[E
af
12
) ] CE (E
+
0
af
CE
0
a
+-)
0
+ -)
2
e[(E
[( +-)
0
- E]
(c)
0
a 2
S
E 0 [E2+(E0 + -- ) I
0
o
while at B the components are given by (c) with of
+
ý
0.
Part b
In the x direction, because the fields are independent of x and z,
fx = cb-a)[(TxyA -(T
y
(d)
D = (b-a)DEo f
or simply the area multiplied by the surface charge density and x component
of electric field intensity.
In the y direction
2
of
f
y
= (b-a)(T
- T
jA
)D = (b-a)D[E
Y B
f + 2]
(e)
0o
Note that both (d) and (e) could be found by multiplying the surface
charge density by the average electric field intensity and the area, as
shown by Eq.
8.4.8.
-46­
FIELD DESCRIPTION OF MAGNETIC AND ELECT IC FORCES
PROBLEM 8.7
I
, =L
(4)
Before finding the force, we must calculate the H field at xl = L.
To find
this field let us use
Bnda
= 0
J
over the dotted surface.
At x1 = + L,
H(xl=L) = Hoi1
(b)
over surface (4) H = 0, and over surface (2), H is in the
n = 12.
Thus over surface (2) B'n = 0.
Hence, the integral in (a) reduces to
I
-
(1)
(3)
oHo a +
-
pfoH(xl = + L)da = 0
H0da +
per unit depth
oHb = 0
Thus:
H(x, = + L) = ýa/b)Ho i
Tij
Hi
o
-j 2
I
o kHk
Hence, the stress tensor over surfaces (1),
-OH2
2 1.
0
1o
T
ij
0
0
(2) and (3) is:
2
0
0
2
H
1
0
o 2
- H
2 1
-47­
1idirection,
where
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.7 (Continued)
over surface (4)
T
Thus
= [01
the force in
fl
=
the 1 direction is
Tij n-da
fl=-f
T 11da+
f T 1 1 da+
(2)
(3)
(1)
f T 1 2 da
Thus, since the last integral makes no contribution,
11
o
f
1
-
2
Ho (a)
Vo 2 a 2
+ -2 H2 ()
*b =
0o
2 o b2
2
o
a
b
-
1}
Since Tij = 0 over surface (4) there is no contribution to the force from
this surface. and by symmetry,
surfaces perpendicular
there is
to the x 3 axis.
no contribution to the force from the
Thus,
the force per unit depth in
1 direction is (k).
PROBLEM 8.8
The appropriate surface of integration is
shown in
the figure
'I)
tI
-I
~
I
The stresses acting in the x direction on the respective surfaces are as
shown.
Because the plates are perfectly conducting, all shear stresses
required to complete the integration of Eq. 8.1.17 vanish.
contributions are from surfaces (i), (ii),
-48­
The only
(iii) and (iv), where the fields
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.8 (Continued)
are known to be
V
E =
a
V
i
(i)
y
;E =
b
V
-s•i
b
V
-i=
a
(ii) ; E
y
=
i
(iii)
y
(a)
(iv)
y
Thus,
f
= (T1 1)
ad + (T11 )
= dV2
ad
(T
­
ii
i
1 -
oob
)
iii
(b)
bd ­ (T1 1 ) bd
iv
]
(c)
a
The plate tends to be drawn to the right, where the fields are greater.
PROBLEM 8.9
A-
- -
)Z.
The volume enclosing the half of the plate is arbitrary so lo6ng as it is
defined so that it does not include additional charge.
Thus the volume shown
in the figure encloses no more than the desired distribution of charge.
over, surfaces (i) and (ii
More­
pass through the fringing fields half way
between the plates where by symmetry there is no x2 component of E.
(i) and (iii) support no shear stress T 2 1 .
Thus surfaces
There is no field at surface (iv)
and hence the only contribution is from surface (i), where the square of the
field is known to be
2
E21
V2o
2
(a)
s
and it follows that because T 2 2 on (i)
negative
f
2
4wse
o
22
1
is - 2 CeoE
2
and the normal vector is
V2
o
(b)
s
-49­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.9 (Continued)
The fringing field tends to pull the end of the plate in the + x
2
direction.
PROBLEM 8.10
-----------'1
-----------
I:
IB~
;I(7
Part a
Consider the surface shown in Figure 1.
The total force in the x
direction is:
da -
f T
f=
T
da +
8
4
5,7
1,3
f
TT
da
T
da
(a)
2,6
The first four integrals disappear because:
T
xy
= CE E
xy
= 0 on 1, 3, 5 and 7 because we are next
to the conducting plates (Ex = 0)
T
xx
= 0 an 4 and 8 because the E field = 0 there
Hence
T
f=
x
xx
da
2,6
where Tij is evaluated
E =
y
2
E2 da
y
(b)
2,6
using Eq. 8.3.10.
(c)
s
and hence:
-50­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.10 (Continued)
f
Sd2
(
=
x
-2
E
da= -
--
s
v
2
(d)
2,6
Part b
The coenergy of the system is
W' =
where
C(x)v2
(e)
2
C(x) = 2(a-x)d(f)
s
Thus, (see Sec. 3.1.2b)
= =W'
ax
fx
221 3C(x)
v2 = ----des v 2
ax
(g)
which is the same value determined in part (a).
Part c
The equation of motion of the plate is:
2
M
2x
dt2
+ K(x-a) = f
= --
x
s
V
(h)
o
When the system reaches equilibrium with the switch closed,
K(X -a) =-dE
0
s
(i)
o
thus
X = a
o
sK
V2
()
o
After the switch is opened,
M
dt2
+ K(x-a) =
dt
-
dc v 2 (t)
(k)
s
The electrical circuit is like an R-C circuit with time varying elements
+÷
v
R(x)
v + R(x)i(t) = 0
(M)
d
v + R(x) d
(m)
v + R(x)C(x)
[C(x)v] = 0
dv
dC(x)
- + R(x) dxv
dt
dx
dx
dt
-51­
=
0
(n)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.10 (Continued)
where:
R(x)
s
2ad(a-x)
and
C(x) 2d(a-x)E
s
(o)
v = 0
(p)
Hence
v +
d-- -x a dt
La (a-x)
dt]
Part d
Dropping the inertial term from (h) leaves:
K(x-a) = - _c v2(t)
S
from (k)
(q)
K(x-a)
(r)
But we may write the identity
1
(a-x)
dx
dt
1
d
=
K(x-a) dt
and then, from (q)
1
dx
(a-x) dt
s
dev 2 t)
1
d
dt
d
2
dt v
t
2
d
2
s
2 dv
v dt
(s)
Substituting back into (p) we have
v+ E - dv + - 2E dv =
a dt
a dt
(t)
Solving we find
v = V e-(/3E) t
(u)
o
and substituting back into (q),
2a
2
x = a -• dE V e
sK o
3 E
(v)
Along relaxation time is consistent with neglecting the inertial terms, as
then x(t) varies slowly.
Part e
Proceed as in (c),
and record the time constant T of a-x(t) by measuring
the mechanical displacement.
a
Then,
= 22
(w)
3
-52­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.10 (Continued)
This problem should raise questions as to the appropriate form of Tij
used in
(b).
Note that the surface of integration encloses liquid as well
as the plate.
We want only the force on the plate, so our calculation is
correct only if there is no net force on the enclosed liquid.
force density in the liquid is given by Eq. 8.5.45.
The electrical
There is no free charge
or gradient of permittivity in the bulk of the liquid and hence the first
two of the three contributions to this force density vanish in the liquid.
there remains the electrostriction force density.
However,
Note that it
is
ignored in our calculation because the electrostriction term was not included
in the stress tensor (we used Eq. 8.3.10 rather than 8.5.46).
Our reason for
ignoring the electrostriction is this: it gives rise to a force density that
takes the form of the gradient of a pressure. Hence, it simply alters the
distribution of liquid pressure around the plate.
Because each element of the
liquid is in static equilibrium and can give way to motions of the plate without
changing its volume, the "hydrostatic pressure" of the liquid is altered by
the electric field so as to exactly cancel the effect of the electrostriction force
Hence, to correctly include the effect of electrostriction in integrat­
density.
ing the stresses over the surface, we must also include the hydrostatic pressure
of the liquid.
If this is done, the effect of the electrostriction will cancel
out, leaving the force on the plate we have derived by two alternative methods
here.
PROBLEM 8.11
B
I
L_
(C-2)I
_
_
-
--
-
­
-53­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.11 (Continued)
on each of the surfaces of the figure
First, let us note the E fields
(3),
over surfaces (1),
and (7),
(5)
(a)
E1 = 0
over surface
= 0
(b)
E 1 =0
(c)
E1 = 0
(d)
(6) E2 = -a
V
(4) E2
-
V
(2) E 2 = -
c
From Eq. 8.3.10,
Tij=
ij
Hence, over surfaces (1),
(e)
F F
0okk
oEE -E
oij
2
(3),
(5) and (7)
T12 = 0 (f)
and over surfaces
E
(6)
T 1 1 =
V
2
(
S v
2
(g)
(4) T11 = -
(h)
C
V 2
) (2)
T1 1
f=
fTij nfda =
(i
Now;
IT 1 3 n 3 da =
0
T 1 lnlda +
T 1 2 n 2 da + fT1 3 n 3 da (j)
(k)
because the problem is two dimensional.
Let us consider each of the other integrals:
(a)
fT1 2 n 2 da = 0
because the surfaces which have normal n 2 are (1),
(f) we have shown that T12 = 0 over these surfaces.
contribution to the force over surface (8),
Hence the calculation of the force reduces tc
(5)
and (7)
and by
Also, we get no
because E
area 4 m.
-54­
(3),
+
0 faster than the
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.11 (Continued)
i= fl
T6) da
T (4)
(6) da
T)
-
(4)
(m)
da2
(2)
2
f
EDDV
o
1
1
1(n)
2
a
b
c
0+ 0
1
(n)
Note: by symmetry, there is no contribution to the force from the surfaces
perpendicular to the x 3 axis.
PROBLEM 8.12
Part a
,, P
7
/
0-~
-
/
/
S, --
/
/
/
/
,-
,
,,
/
-­
T1
-­
7
From elementary field theory, we find that
wx2
=
o sin -'-
a
o
- wxl/a
e
(a)
satisfies V25 = 0 in the region between the plates and the required boundary
The distribution of E follows from
conditions.
E
= -
V4 (b)
Hence,
E
-
w
--
a
0o
e
wxla
1
The sketch of the E field is
In
wx
'r2
a
a
-
i
wx2
os2
a - cos -a
2
l
obtained by recognizing that E is directed
perpendicular to contours of constant 4.
-55­
(c)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.12 (Continued)
Part b
To find the force as the bottom plate, we use surface (2).
where except on the upper side where the normal n = 12
and the field is
o$
- Txl/a
= - -a e
i
2
Hence,
fl = ITi
f2 =
n
T2 j n
da = 0
da =
T 2 2 n 2 da
2
per unit x 3 , this reduces to
F
f2
T 2 2 dIx
1
1
2.2
but, T
1
22
2
EE
o 2 2
1
= -E
0 e
a
-o
a
and thus
2
f2
2a2
2
1dx
4a
-56­
E = 0 every­
(d)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.12 (Continued)
Part c
On the top plate, use surface (1).
Only the sign of the normal changes,
and the result is
fl =0
2
2
4a
or the force is equal and opposite to that on the bottom plate.
PROBLEM 8.13
Part a
Tij
Hence:
EiEj
EKEk
-
2V
2
22
3a
2V
T21
ooE2E1
3a
2
x1x
2
Part b
Consider the surface of integration shown in the figure.
D
O
f2 =T
2 jnjda
2
= f22T2 1 n 1da + f22T22 n22 da + f~l3/3da
(2)(3)
(1)(4)
-57­
by symmetry
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
Let us look at each of these integrals separately
ST22n2da
T22da
-
over surface (1), E
0
*
T2 2 da 4 = -
0 and hence, the integral is merely:
T22
Sxl=a
-
22da
(4)
(1)
(1)(4)
E°
0
Jx =-a
2
2V
(-
3a
2
)
(x2
-
xl) wdx1
x 2 = 2a
2a
E V2
44 o0 o
a
T7
Thus,
44
Snda
S22n2
(1)(4)
SV2w
oo
=27
a
Let us now evaluate:
I
T 2 1 nlda
(2)(3)
Consider the surface shown.
in this region field = 0
hence, no contribution to the
integral over this area.
i
Z•r =•
-58­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
Thus;
x2=a5d
=
IT21da
x
2a
3a
x2a
(3) 2
2V
awx2 dx
2
xl=a
0V2 w
2
oo
9
a
Over surface (2), we have essentially the same thing, except n = - i
and xl = - a.
Hence:
2
I
E V2w
oo
"a
a9
ST21 da2
(2)
Therefore, the total force in the f2 direction is
56
27
2
E V2w
00
a
Part c
0
f1
+
=
Tllnlda
(1)(4)
(2)(3)
by symmetry
over (1) we get
4
(4)
(1)(4)
0
da
f-T 1 2 da
I T 1 2 n 2 da
S
T12n2da + f1T
0 as before
x2 wdx 1 = 0
02 fa
2V
3a2
-a x
X
X2WdX
x 2=2a
Now, over surfaces, 2 and 3
Tl1 1n
da
= -
T 1 1 da2
Tllda 3
(2)
(2)(3)
because,
T1 1 12
T1113
hence fl = 0.
-59­
0
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
Part d
a
= n
E0 (o)
The functional relation,
at the lower surface of the movable conductor.
f(xl 2), for the lower surface if the movable conductor is given as
4a2 + x
f(x1 x2 ) =
- x2 = 0 (p)
the outward unit normal to this surface is
n ~Vf(xlx2)
xl
4a
- i
(q)
[ 2
l'2
at x 2 =
I
+ x12
S1/2
2 x2
of
Eo[n l E
+ n 2 E2 ] = 32
2
4
3a
2
4a 2
x2
J4a
2
12
/
2xl
+2x
The surface force density (see Eq. 8.4.8) is equal to:
-E
+ E•
2
f
where,
Eb = field just below the charge sheet
Ea = field just above the charge sheet
Since
a = 0,
T = 1 a
(t)
2
thus
T
2
e o 2V o0 2 x
2- 2
4a
)
x2
3a
+ x2
x1il-
2 i2
4a2 +
4a
1/2
2
+ x
2
(u)
+ 2x1 J
To find the total force, the surface force density must be integrated over the
surface.
Hence, we find
V
f2 =
22c
0
3a
2
a
a
j2x
-a
1/2
2 + 4a2
1/2
fx2 + 4a2} dx1
(v)
If the student wishes, he may carry out this integral, but the complexity of
the integration shows the value of the stress tensor in calculating such a
-60­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.13 (Continued)
force.
that by using the stress tensor, we have essentially
We realize
carried out this difficult integral by an integration by parts.
PROBLEM 8.14
Part a
V
i a
1
(a)
xY
2
(b)
= - V
hence,
E
(
2 x2) +
2
a
and,
from Eq.
8.3.10
Ti
= CEE
- 6j
V
a
xl)
(c)
(d)
1ýE
Tstress
ijhe
tensor becomes2
Thus, the stress tensor becomes:
V
(-)
a
2
V
-
(x2-x )
V
Tij
(X 1 X2 )
a
2
(-)
a
o(xlx2)
0
V 2 E°
-2- (x-x2)
a
(- )
V 2
-(-)
0
(e)
02
)
-•xx
2
a
Part b
Consider the surface shown, bounded by the line segment x2 = 2a, x2 = a,
and xl = a/2 and x1 = a.
XK
1
-61­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.14 (Continued)
As before, because the geometry and fields are two-dimensional, the force in
the 13 direction is zero.
Also, since along surface (1)
the E field = 0, and hence Tij = 0 along this surface.
,
= constant, then
Thus the calculation
of the force on AB reduces to:
fl = -
T11da -
T12 da
(2)
f2 = -
(3)
(2)
2
2
2
a
a
(g)
2
x2
oD
and hence
V
-=-Eo
T 2 2 da
(3)
V
fl
f
T2 1 da -
fl
(f)
a
+
- (2 ) ]dx
a/2
(h)
2
Da3 [*•]
17
()
a
a
(i)
Similarly:
V 2a
f2
a
1
a
x(x-a
2 x2d2
2 a/2
Do
2
2
l
)dxl
(J)
and hence
V
f
2
-
Thus,
2
oD a 3 (- )
o
a
(k)
48
2
v
f = - E
o a
o
D [i
17
1 12
+
i
2
31
4
48
(-)
PROBLEM 8.15
Part a
The E field in the laboratory frame is zero since the two perfectly
conducting plates are shorted.
This can be seen by integrating E around a
fixed contour through the block and short and recognizing that the enclosed
flux is constant.
E'E+vx
Hence,
E ,
E
0
(a)
and thus
E'
v x B = - V1
oHi2
(b)
Therefore we may now calculate J in the moving block.
-62­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.15 (Continued)
aE'
J'
ap
-
VHo
2
Thus:
F
Jx
ap 2 VH2i
0
o 1
(3 x B)dV =-
-
i
0
oVH (abD)
0
11
volume
Part b
The closed surface of integration is shown in the figure below.
All,
""~~
"I'I­
-­
'
IX
I
i
(
A
XI
Since the field is uniform everywhere,
the only non-zero components of the stress
tensor are the diagonal elements
T
=T
11
22
1
2
H2
oo
T
Thus
33
f1 = i T1da3 - f Tllda2
(3)
(2)
=
H2 bD = 0
2•o
2 H2bD
o
Similarly
f2 =
T22da
-
(4)
(1)
f3
=
T 33da5
(5)
Hence:
T22da4 =0
-
T 33da6 =0
(6)
1
2
H2
oo
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.15 (Continued)
Part c
The magnetic field strength and the current density are inconsistant.
The
quasi-static magnetic field cannot be uniform and irrotational in a region where
a finite current density exists.
the aid of Ampere's
The Maxwell stress tensor was developed with
Law (quasi-static) which relates current density and magnetic
field rotation.
S=VxH
F
Vo J x H =
(k)
o(VxH)x H
(1)
For this case, we have assumed that
V x H= 0
(m)
In the limit of small magnetic Reynold's number, (Rm << 1),
the motion does not
appreciably affect the field, and the answer found in part a is a good
approximation.
There are some problems more easily handled with the stress tensor.
This problem illustrates that in other cases it is easiest to use the force
density J x B directly.
Note that we could compute the field induced by J and
then use the Maxwell stress tensor and the self-consistent fields to find the
same force as given by (e).
PROBLEM 8.16
To find the force on the block, we will use the stress tensor
surface shown in the figure.
Note
over the
that the surface is just outside the block.
X,
-64­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.16 (Continued)
In
the region to the left of the block
I
=D o 13
,
and to the right H=0
Thus:
T12 n2 da +
+
n
T11nda
f
(a)
T31 n3 da
but, since
H1 = H 2 = 0; T 1 2 = T13 = 0
(b)
f
(c)
hence,
= -
Tl1da5 + f T1dal
(5)
(1)
on surface (5),
J 12
T
o
11
2
o
(d)
d
2
D
on surface (1)
T11 f 0
(e)
therefore
2
f1
+
2
2
oo2 .Dd = +
D
(f)
(f)
2D
Similarly, f2 reduces to
f2 =
T22da2 -
T22da6
2
(g)
6
But, since T22 is a function of xl alone (1 is a function of x l alone) the
two surface integrals are identical, and hence f2 = 0.
shows that f
-
Similar reasoning
= 0 and thus the total force is
~ood
f
2D
d
i
PROBLEM 8.17
Part a
2-B
(a)
V =po-
o at
Assume a solution of the form:
H = Re [H (x)e
(b)
iWt
e-z
- joZ
2
ax
o HZ
(c)
-65­
V
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.17 (Continued)
Try
H (x) = H
Kx
eKx
(d)
where
2
K 2 =
(e)
•0
°
and hence
K = +
(f)
(1+j)
Let us define the skin depth as:
6
(g)
2
[
And thus
(+j)
iz
+He
2
e
-H=
(h)
Because the skin depth 6 is assumed to be small, and the excitation is on the
left,
Hz(large x)
0
+
-
Hence,
H(xlt) = H
which implies
12 = 0
x(l+j)
e
(i)
ejt
z But, our boundary condition at x = 0 is
i
= - Re - e
H(x=O,t) =ReHe
D
H(Ot)z
(j)
z
and thus
(x•)
D e
+J)ejt
(k)
i
3H
((6)iyx
J=Vx H=-
-
D
6
- -K(l+j)
ejt
e
i
y
(M)
Part b
=
f
x
E dV=
Jfx V RdV
(n)
e2jWt dV]
dV] + ReL 2-
f Re
(m)
Now, solving each of these integrals:
S2x
2
dV =
oaD (1
T
4
(1+j)
Da I (1+j)i
x
D
x
e
dx
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.17 (Continued)
2x
2xT(1+j)
2
e
fixi
1
4
o•a
D
2
2jwt
(p)
x
Hence, taking the real part, the force as in equation (n) is:
S1
4
o a 12(1 + cos 2wt)i
D
(q)
x
Part c
Using the Maxwell stress tensor, we choose the surface shown in the
figure,
O
f
T jnjda =
=
Txxn da +
(1)(3)
T
(r)
n da
(2)(4)
Along surfaces (2) and (4), Hx = 0 along the interface between the perfect
conductors and the finite conductivity block.
At surface (3),
(s)
= 0
oHHy
=
T
Thus,
the field is zero since all current filaments complete a
closed loop circuit with the source through the block.
Hence
(t)
Txx
= 0 on surface (3)
Therefore the calculation of the force reduces to
f
=-
f
T
(u)
da
o H2
T
xx
2
(v)
z
And thus,
aDoH
f
x
=
2
(w)
o H2
z
-67­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.17 (Continued)
where the field H z is evaluated on surface 1, i.e. x = 0 and is simply given
by the boundary condition (j).
al
a=4D 2
Thus it follows
(1 + cos 2wt}ix
(x)
Note that the distribution of J and H, as found in
which checks with (q).
part (a), are not required to find the total force in this problem.
Even more,
(x) is not limited to 6 << x block d1•lension, while the detailed integration is.
Note:
We have made use of the rule for products, namely of:
a(t) = Re[Ae
+ A*e
2
Ae
jwt
j
]
=
]
b(t) = Re[Be
2
=
2
2
then
a(t)b(t) =
AB* + A*B
4
4
2
ABe jwt + A*B*e
+4
AB*
- 2
jwt
4
AB
= Re[ --2 * + Re[-
2
e
2jwt
t
time varying part
avg. value
PROBLEM 8.18
Choose the surface shown in the figure.
r -----­
/
I-.
--
f fTijnjda
r-
0j
-
-
-
-­
-­
T2n2da +
= f Tlnlda+
3,4
5,6
(a)
T3n3da
1,2
Since the plates are perfectly conducting, E 1 = 0 at surfaces (5) and (6)
and .hence T12 = 0 on surfaces (5) and (6).
are far from the body so
-68­
Surfaces (1),
(2),
(3) and (4)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.18 (Continued)
V
E=
i
d
(b)
z
at each of them, and thus, on surfaces (1) and (3), T 1 3
=
fl
-
(c)
(4)
T )
11
Therefore,
11da3 + f Tllda
(3)
T(3)
0=.
o (V
(d)
2
11
and a 3 = a 4 (areas).
Hence,
(e)
fl = 0
PROBLEM 8.19
Part a
Since the system is electrically linear,
S= BI + Br
(a)
where BZ and Br are respectively the fields from the left and right wires.
The force on a unit length of the right wire is
=r
x
B
da = Jr x
, da +
x BE
da
(b)
but,
(c)
Jr x Br da = 0
ane hence,
f
=
(d)
x Bi da
J
Since, we don't need the fields near the wire,
P
I
x2 1-
(xl+a) 2
9 2 (xl+a) + x2
1l
0
r
2
-x2
e
1 + (x 1 -a)12
22
(x -a) +x 2
j
S
(f)
Hence,
fr
=
rx B da - I
p o1
r
2
2w1
(2a)i
2
(2a)
3
x Bz (xx
1 a, x2 =0) (g)
pl 2
4wa
(h)
1
-69­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.19 (Continued)
Part b
I
/­
I
Along the symmetry plane of the surface shown in the figure
-o
=2w
(-2a)
2 2
(i)
2
(a +x 2 )
The terms of Tij go as B2 , but B2a - and the surface area goes as 27R on surface
(2), hence the contributions of the
stress tensor will vanish on surface (2) as
R-o; we need only compute the integral on surface (1).
Because H1
0 in the
plane xl = 0
f=
f-T 1 1 da =
o Ia2
- -j-
2
dx 2
dx 2
22 2
(a +x2)
Solving this integral, we find
Io2
f
1
P 4ra
f2
f3
(k)
also
0
(Y)
31 = 0
(m)
=
since
T
21
T
and hence the total force is
that of (k) and it
in part (a).
-70­
agrees with that determined
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.20
ioL7
\
\
~
0
4-21
|
IF- O
•
|
!
I
.j~
-
r
X.
/
0
/
I~/
Part a
Use the contour indicated in the figure.
At infinity the fields will go
to zero, and hence there will be no contribution to the force from the semi­
circular part of the area, i.e. surface (2).
Along the line x2 = 0, E 2 = 0 by symmetry and
E1 =
2
X
(
)sin8
o
r
2
= a
2
sinG =
2
+ X
x1
x
r
2
o
Hence
X
E1 = oe
f2 =
x1
X1
2
a2+x2
T2jnda =
T21nda + t22n2da + T23n3da
(1)
(1)
(1)
first and last integrals = 0, n 1 and n3 = 0 on surface 1
2
T
22
2
2
1
2
2
2)o2
c ft (a +x )2
-71­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.20 (Continued)
Thus
SXl2
dxl
x 2 22
f2
2c
r
f =- 2
2
o (a +x )
f2
2 P4a
a
Part b
From electrostatics,
f =
AE
From the figure, we see that
E(x2=a)
= 2
(2a)
Hence,
2
4we 0 a
2
which is the same as we obtained using the stress tensor - (see equation (h)).
PROBLEM 8.21
Part a
From Eq. 8.1.11,
BB
Xy
0
lo
Tij
x
2L0
0
0
0
y
1 (-Bx2 -By)
2
2v0
x y
-
where the components of B are given in the problem.
Part b
The appropriate surface of integration, which is fixed with respect to the
fixed frame, is shown in the figure.
to
We compute the time average force,
d.h
an iUbL
ence contr
rL U
i
I
ut. ons
(1) and (3) cancel.
LUL
U
f
rom SUL aces
Fields go to
zero on surface (2), which is at
y-.
Thus, there remains the stress
on surface (4).
I
The time average
+· c c~ea
value of the surface force density T
-72­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.21 (Continued)
is independent of x.
T
T
Hence,
= - <T
y
(b)
(y=O)>
yy
y
<-B
21p
2
x
+ B2>
y
(c)
Observe that
^
-JkUt
j
<Re A -e kUt
Re B e kUt>
Re A B*
22
where B* is complex conjugate of B, and (c) becomes
T R-kx
(-jklj 0 K°) jkx (jki K )
Ty
Re-( Koe kX ) (p K e )+
e
e
(d)
jk x
0
2
=
k2
(1
4
(e)
aa*
)
(f)
Finally, use the given definition of a to write (f) as
T =y
4
-
(g)
U 2
S1+
Note that T
Y
(-10-)
is positive so that the train is supported by the magnetic field.
However, as U-O (the train is stopped) the levitation force goes to zero.
Part c
For the force per unit area in the x direction;
T
-
x
1
<B B (y=o)>
21 °
x y
1
=
2V
Thus
°K e jkx
Re[VKek
Re
(h)
-jk
( jk0 K ejkx
a*
•K2
Tx
(i)
M
o0
V aU
0
Re j
1-
j
SaU 2 1/2
2[1 + (-ý)
I
As must be expected, the force on the train in the x directions vanishes as
U-O.
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(e/2) = +
/(l - cosO)/2
is helpful in reducing (j) to
the form
K2
-
T=°
(
0
2-
11 0oU 2 1/2
--
-73­
p crU 2
1+
(-e-
1)
(k)
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.22
This problem makes the same point as Probs. 8.16 and 8.17, with the
additional effect of material motion included.
Regardless of the motion,
with the current constrained as given, the magnetic field intensity is zero
to the right of the block and uniform into the paper (z direction) to the left
of the block, where
I
H= i
zd
(a)
0
The only contribution to an integration of the stress tensor over a surface
enclosing the block is on the left surface.
f
x
= ds T
= - ds1
2
2
xx
Thus
H
0oz
(b)
I
ds 1o
0)
(c)
The magnetic force is to the right and independent of the magnetic Reynolds
number.
PROBLEM 8.23
In plane geometry, a knowledge of the charge on the upper plate is equivalent
to knowing the electric field intensity on the surface of the plate.
Thus, the
surface charge density on the upper plate is
I
a
f
=
t
A
o
E (x=a)
=
I
I
coswt dt
sin wt
0
(a)
a)
and
X
Qf
E
I
0 sin wt
AE W
o
(b)
o
Now, we enclose the upper plate with a surface just outside the electrode
surface.
The only contribution to the integration of Eq. 8.1.17 using the
stress tensor 8.3.10 is
A
= - AT
2
=
o E2(x=a)
x
xx
2
x
which we can evaluate from (b) as
f
(x=a)
AE
fx
I
o2 (
(c)
22
ow)sin20t
(d)
o
The force of attraction between the conducting slab and upper electrode is not
dependent on 01 or a o .
-74­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.24
The force on the lower electrode in the x direction is zero, as can be seen
by integrating the Maxwell stress tensor over the surface shown.
t
The fields are zero on surfaces (2), (3) and (4).
Hence, the total force per
unit depth into the paper is
f=
f
T
(a)
dx
where contributions from surfaces in the plane of the paper cancel because the
Moreover, by symmetry the electric field intensity
problem is two-dimensional.
on the surface (1), even in the fringing regions, is in the y direction only
and T
xy
=
C E E
in (a) is zero.
oxy
Thus, the total x directed force is zero.
PROBLEM 8.25
The force density in the dielectric slab is Eq. 8.5.45.
Not only is the
first term zero, but because the block moves as a rigid body (we are interested
only in the net force giving rise to a rigid body displacement) the last term,
which originates in changes in volume of the material, does not give a
Hence, the force density is
contribution.
E.Ev-
=
(a)
2
and the stress tensor is
6
T
= EEIE
Note that, from (a),
- -
c
(b)
kE
the force density in the xl direction is confined to the
right edge of the block,.where it acts as a surface force.
Thus, we obtain the
total force by simply integrating over a surface that encloses the right edge;
-75­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.25 (Continued)
fl = aD
E
ao(E2
+1 (Eb)
(c)
where a and b are to the right and left of the right edge of the slab.
Ea = E
2
2
= - V/a.
0
fl
2
Hence (c) becomes
V
2
(ECo)
(o)
Also
(d)
(d)
The force acts to the right, as could be computed by the energy method.
PROBLEM 8.26
Part a
The force density for polarizable materials is:
-
F =-
1
1
-­
E*E VC + - V(E*E p -)
2
2
ap
(a)
The second term on the right side represents electrostriction.
this is
a case where the material volume must change,
electrostriction is important.
changes in permittivity and
and hence the effect of
Sincd free space and the elastic bulk are homogeneous,
occur only at the boundary where
ac/ap
the permittivity is discontinuous.
are constrained by the plates.
Note that
The upper and lower elastic bulk surfaces
Thus only the xl component of force is pertinent.
Since the left-hand edge is fixed, any stress arising from the discontinuity in
permittivity at that boundary is counterbalanced by the rigidity of the wall.
Therefore,
all of the force arises at the right-hand boundary which is
move.
The closed surface of integration is shown in the figure.
-76­
free to
FIELD DESCRIPTION ON MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.26 (Continued)
Tij
1
2
EEi J
- p
9e
EkEk
(b)
1
2- a the field at the dielectric interface is essentially
Since a/c << 1 and b
uniform.
ij
V
0o
-
E-= - i 2
(c)
aW
The relevant components of the stress tensor are:
T
ae 2
1
2
p
E2
2 +
2
2
ap 2
S
2
-
11
T12 =
E1EE
2 =0
f
T11nda
n
+
=
(d)
(e)
T 22da
(f)
(4)
(1)(3)
Hence
f
fT11da3 - fTldal
(3)
(1)
E
V 2
S 2
T)
(aD)
E 1 V 2
(1 ) (aD) +
-
P
V 2
(aD)
(g)
Thus;
(Ef
1
1
2a
Eo)V2 D
00
o
2
d
r(
V2 D
0(h)
)
a
Part b
In order to use lumped parameter energy methods, the charge on the upper
plate will be found.
The permittivity of the dielectric bulk is a junction of
the displacement of the rightýhand edge.
That is, if mass conservation is to
hold,
po abD = (po + Ap)aD(b+F)
(i)
P = Po + Ap, Ap = 0 if 5 = 0
(j)
where
Thus, if Ap << po and E << b, to first order
Ap = -p
-Po b
(see Eqs. 8.5.9 and 8.5.10)
(k)
Furthermore, to first order, using a Taylor series,
-77­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.26 (Continued)
p
a
+p
1
Po 2E
b ap
1
Also, the electric field will be assumed as uniform everywhere between the
plates.
Hence; in the block
V
;-T 22 -2a [2 +l Ap] }
(m)
to the right of the block
V
D=- 1 2a
-E
2 a
a 0 C2
(n)
(o
By employing Gauss's law, we find the charge on the upper plate as:
q = (P
dw
V
p E}(b+)D + Eo--)(c-b-Q)D
p0
V
-
= fqdv +
(o)
f dx
(p)
integrating we find
w'
e = i(a
2
Thus,
11
Thus,
e
c v=V
e
(b+ý)D + 22
1 a o(c-b-)D
--
0
2a
o
(q)
V2
(E-E)V2D
eo
f
E
b ap
1
2
o
a
_E
o
(PD
pr)
o
Second order terms have been dropped in the co-energy expression (alternatively,
first order terms can be dropped in the force expression).
Part c
If the result of part (a) is written for p = po + Ap, where po >> Ap,
then the answers to part (a) and (b) are identical to first order.
This
should be expected since the lumped parameter approach assumed a value for
permittivity which was correct only to first order.
PROBLEM 8.27
The surface force density is
T
= [Ta
- Tbn
mn
mn
m
(a)
n
For this problem, we require m = 1 and n = 12.
T1 =
(T
2
-
Thus
(b)
Tb2 )
From Eq. 8.5.46,
-78­
FIELD DESCRIPTION OF MAGNETIC AND ELECTRIC FORCES
PROBLEM 8.27 (Continued)
EEbEb
T 1 =EoEEa
(c)
Note that E2 = E2 (see Eq. 6.2.31).
coEa =
CE1 (see Eq. 6.2.33).
1
2 olI - EEl]
T 1 = Ea[EoE
Moreover, because there is no free charge
Thus, (c) becomes
(d)
f0
That the shear surface force density is zero in the x 3 direction follows the
same reasoning.
PROBLEM 8.28
The force density, Eq. 8.5.45, written in component form, is
F
i
=
i
E
ax
2 Ekk
ac
+a
1
a
ac
EkEk
(a)
)
(a)
The first term can be rewritten as two terms, one of which is in the
desired form
i
a
i
Because V x
-x
(i
= O, aDE/ax j =
and combined with the third.
a
Finally, we introduce 6
Ej/Dx i
1
De
1
2 k
ax
j
k ax
ax
2
E)
k k -5p
(b)
(b)
so that the second term can be rewritten
(Note the j is a dummy summation variable.)
a
a i
i
(see Eq. 8.1.7) to write (c) in the required form
aT
(d)
F=
i
ax
where
)
(E-p
Tij = CE This is identical
to E. 8.5.46.
This is
identical to Eq.
8.5.46.
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(e)