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Solutions Manual for Electromechanical Dynamics
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FIELDS AND MOVING MEDIA
PROBLEM 6.1
Part a
From Fig. 6P.1 we see the geometric relations
e - Pt,
r' = r, e' =
(a)
z' = z, t' = t
There is also a set of back transformations
e = 8' + st',
r = r',
(b)
t = t'
z = z',
Part b
Using the chain rule for partial derivatives
r)
+ (2j)
atf
ae
, = ()
at
ar
(L
) +
2(-
at'
3za)
+ ()
9 (
at
(-)
(c)
at
From (b) we learn that
,=
SO,
' =0O
,
(d)
= 1
,
Hence,
at'
at
0
+
(e)
ao
p are
We note that the remaining partial derivatives of
4, = a*
2t = *
ae '
ae'
ar '
3r'
az'
(f)
a
PROBLEM 6.2
Part a
The geometric transformation laws between the two inertial systems are
x1 = x
- Vt, x'
x 2, x
= x 3, t'
= t
(a)
The inverse transformation laws are
1 = x' + Vt',
x,
x=
x2
x3 = x
t
(b)
t'
The transformation of the magnetic field when there is no electric field
present in the laboratory faame is
(c)
P'= W
Hence the time rate of change of the magnetic field seen by the moving
observer is
aB'
=3B
a
ax
B
1
+
2
3B
2
+
B
)x
a3B
at
3
(d)
FIELDS AND MOVING MEDIA
PROBLEM 6.2' (Continued)
From (b) we learn that
atx'
V,
ax1
3x2
2
ax3
t
0,
= 0,I,
at
= 1
(e)
While from the given field we learn that
aB
. kB
cos kx
kBoo
ax
aB
aB
x3
l -ax
2
=
aB
B
t
t
0
(f)
C
Combining these results
aB'
B'
at,
aB
B, = V•
aB
= VVkB
t,
'
o cos
kx1
1
(g)
which is just the convective derivative of B.
Part b
Now (b) becomes
S
x2
= x'
2
x 3 = x;, t = t'
x ' + Vt,
(h)
When these equations are used with (d) we learn that
aB' S=aB , = V aB
at,
because both
aB
aB
and -
x2
Tt
+ aBt = 0
ax2
(i)
at
The convective derivative is zero.
are naught.
at
PROBLEM 6.3
Part a
The quasistatic magnetic field transformation is
B'
(a)
= B
The geometric transformation laws are
x = x'
+ Vt'
y',
,
Z
z',
t = t'
(b)
This means that
'=
E(t,x) = B(t', x' + Vt') = iyoB
= i B
yo
cos (wt' - k(x' + Vt'))
cos[(w - kV)t' - kx']
(c)
From (c) it is possible to conclude that
w' = w - kV
(d)
Part b
If w' = 0 the wave will appear stationary in time, although it will
still have a spacial distribution; it will not appear to move.
FIELDS AND MOVING MEDIA
PROBLEM 6.3 (Continued)
w' = 0 = w - kV; V = w/k = v
(e)
The observer must move at the phase velocity v
to make the wave appear
stationary.
PROBLEM 6.4
These three laws were determined in an inertial frame of reference, and
since there is no a priori reason to prefer one inertial frame more than
another, they should have the same form in the primed inertial frame.
We start with the geometric laws which relate the coordinates of the
two frames
r'
r - v t,
t = t',
r = r' + v t'
We recall from Chapter 6 that as a consequence of (a)
(a)
and the definitions of the
operators
t +tr
a''
t =- t'- r
In an inertial frame of reference moving with the velocity vr we expect the equation
to take the same form as in the fixed frame. Thus,
-v'
(c)
p'iat', + p'(v'*V')v' + V'p' = 0
-' +
at'
(d)
V'*p'v' = 0
(e)
p,'(p)
p',
However, from (b) these become
'a
)+Vp'
+ p'(v'+vr
+ V.p' (v'+v) = 0
=
0
(f)
(g)
(h)
p' = p'(p')
where we have used the fact that v *Vp'=V*(v p'). Comparison of (1)-(3) with (f)-(h)
shows that a self consistent transformation rthat leaves the equations invariant in
form is
p' = P; p'
=
p; vt - v
-
r
FIELDS AND MOVING MEDIA
PROBLEM 6.5
Part a
p'(r',t')
P=(r,t) =
p (1- -)=-
C(a)
o(1-
J' = p'v' = 0
Where we have chosen v
r
=v i
oz
(b)
so that
v'
v - v
= 0
(c)
Since there are no currents, there is only an electric field in the primed
frame
r
r'2
E'
= (po/o )r
H
(d)
O, B' = • •' = 0
(e)
Part b
(f)
)
p(1-
p(r,t) =
This charge distribution generates an electric field
2
r
­
3air
(P/ r
(g)
In the stationary frame there is an electric current
S= pV
=
r
po(1-
-
(h)
oiz
)v
This current generates a magnetic field
H=
oVo(
r
2
r
i
(i)
--a)io
Part c
5-= 5' - P'v
Pvr
=
po(l---)Voi
oz
o
a
E= '-vrxB' = E' = (po/Eo - 3
SH'
+ V xD
r
If we include
r'
=
oo
r
(j)
ir
(k)
' 2
3a lie
(1)
the geometric transformation r' = r,(j), (k), and (1)
become (h), (g), and (i) of part (b) which we derived without using trans­
formation laws.
The above equations apply for r<a.
the fields in each frame for r>a.
Similar reasoning gives
FIELDS AND MOVING MEDIA
PROBLEM 6.6
Part a
In the frame rotating with the cylinder
E'(r') = -,
r
Ir
(a)
H' = 0, B' = u•H' = O
(b)
But then since r' = r, Vr(r) = rwi
E=' - vr x ' = 2'=-i
r .r
V=
f
Ed
b
=
(c)
dr = K In(b/a)
(d)
a
a
r
(e)
1
V
1
V
ln(b/a) r
In(b/a) r'
r
The surface charge density is then
- =
o
1
a'a = i r *8oE'
= In(b/a)
- = aa
a
V
SE
a'= -i *E E'
r
b
(f)
o
= -
(g)
b
In(b/a)
Part b
3 = J' + Vy p'
(h)
But in this problem we have only surface currents and charges
= '+ vr
' =v r a'
O
K(a)
WE V
)
awe V
(i)
0
In(b/a)
1a
iBe
In(b/a)
(
E V
bwE V
K(b) =-
e
In(b/a)
6
b In(b/a)
i
B
(k)
Part c
WE V
S(1)0
z
In(b/a)
(
Part d
S='
H
=
+ v
r
x D
EOx
V
v
1
r'w(1n(bla) r
r
x D'
(m)
-+(n
6)(i
x ir)
(n)
FIELDS AND MOVING MEDIA
PROBLEM 6.6 (Continued)
we V
(
in(b/a) iz This result checks with the calculation of part (c).
PROBLEM 6.7
Part a
The equation of the top surface is
f(x,y,t) = y - a sin(wt) cos(kx) + d =0
(a)
The normal to this surface is then
n
Vf = v
(b)
(b)
ak sin(wt)sin(kx)ix + iy
Applying the boundary condition n*4 = 0 at each surface and keeping only linear
terms, we learn that
h (x,d,t) = -ak sin(wt)sin(kx)
A
(c)
(d)
h (x,O,t) = 0
We look for a solution for h that satisfies
Let h = V
V22,
(e)
V*h = 0
=,
V x
(f)
= 0
Now we must make an intelligent guess for a Laplacian * using the
periodicity of the problem and the boundary condition hy
y = 0.
•l/ay = 0 at
Try
A
cosh(ky)sin(kx)sin(wt)
+ sin(kx)sinh(ky) y]
h = A sin(wt)[cos(kx)cosh(ky)i
(g)
(h)
Equation (c) then requires the constant A to be
-ak
A
sinh(kd)0od
Part b
S •x
=
VE
(
-
E
~-)-iy(---z)=
+ sin(kx)sinh(ky)iy]
p o A cos(wt)[cos(kx)cosh(ky)ix
y
x
0t
(j)
(-)
(k)
FIELDS AND MOVING MEDIA
PROBLEM 6.7 (Continued)
-_
E = - wU
A
cos(wt)[cos(kx)sinh(ky)]i
0 K
Now we check the boundary conditions.
Because v(y=O) = 0
nx E = (n-v)B = 0 (y=0)
But E(y=O) = 0, so (m) is satisfied.
If a particle is on the top surface, its coordinates x,y,t must satisfy
(a).
It follows that
Df
D•
3f
-
=
+ v*V f = 0
Vf
-F we have that
Since n =
(n.v) =
-1
af
i1 t
awcos(wt)cos(kx)
Now we can check the boundary condition at the top surface
--
nxiE f - o
A­
o
cos(wt)cos(kx)sinh(kd)[i -ak sin(wt)sin(kx)i ]
x
y
(n.v)B = awcos(wt)cos(kx)
inh(kd)
ak
Ai
x
+
poA sin(wt)sin(kx)sinh(kd)i
Comparing (p)
]
and (q) we see that the boundary condition is satisfied at the top
surface.
PROBLEM 6.8
Part a
Since the plug is perfectly
conducting we expect that the current
I will return as a surface current on
the left side of the plug.
Also E', H'
will be zero in the plug and the transformation laws imply that E,H will then
also be zero.
Using ampere's law
-I
2 ir
I,
Nt
FIELDS AND MOVING MEDIA
PROBLEM 6.8 (Continued)
Also we know that
E
V*E = 0, Vx
-
0
0 < z < 5
(b)
We choose a simple Laplacian E field consistent with the perfectly conduct­
ing boundary conditions
E
=-
r
i
(c)
r
K can be evaluated from
dt
E"ddi =
da
(d)
C
S
If we use the deforming contour shown above which has a fixed left leg at z = z
and a moving right leg in the conductor.
The notation E" means the electric
field measured in a frame of reference which is stationary with respect to the
local element of the deforming contour.
E"(C+A) = E'(C+A)
E"(z) = E(z),
E"*d = -
J
Here
= 0
E(z,r)dr = -K In(b/a)
(e)
(f)
The contour contains a flux
JB-da = (E-z)
%oHedr = - V
I• ln(b/a)(E-z)
(g)
S
So that
-K
n(b/a)
=
-K
In(b/a)
a = +
dt
n(b/a) d-dt
(h)
= Since v -
'dt
vI
2w
r
r
<z
0
Part b
The voltage across the line at z = 0 is
b
V= -
vpo0 I
Erdr =
a
ln(b/a)
(k)
FIELDS AND MOVING MEDIA
PROBLEM 6.8 (Continued)
vi
o
I(R +
o In(b/a))
= V
2o
(1)
V
I =
vi
R + 27r
V
(m)
o
In(b/a)
2wR
(n)
+1
vp ° ln(b/a)
_V S0
R + 2
< z <
n(b/a)
(o)
H:
Vo
f
E=
1<z
i
+
vy2R
V11V
o+(inb/a)
0<z<
r
r
rr
(p)
<z
<
0
Part c
Since E = 0 to the right of the plug the voltmeter reads zero.
The terminal
voltage V is not zero because of the net change of magnetic flux in the loop
connecting these two voltage points.
Part d
Using the results of part (b)
Pin
1
i
SVI= In(b/a)
= 0
27r
Rn +
lIn(b/a)
Tr
dWm
d=
v fa
H
H 2
(r) 21Tr dr
R+ V
n(b/a)
V2
0
FIELDS AND MOVING MEDIA
PROBLEM 6.8 (Continued)
There is a net electrical force on the block, the mechanical system that keeps
the block traveling at constant velocity receives power at the rate
2
2
1
1 V1o
-- In(b/a)
27
v
[
O
In(b/a)
V2
o
2w
from the electrical system.
Part e
L(x) =
U0
0H(r,I)x dr
IT n (b/a)x
I
d=
1
2
e awlm W
fe w •x ;W' =2 L(x)i
fe
f=fi
1 3L
T
2 3x
1
2
o
22 2
In(bla)i
In(b/a)T
2
U
The power converted from electrical to mechanical is then
V
v
f' = f v
e dt
e
2 2w
In(b/a)
[
v
R +
2
]
o
o In(b/a)
as predicted in Part (d).
PROBLEM 6.9
The surface current circulating in the system must remain
B
(a)
K =
o
Hence the electric field in the finitely conducting plate is
B
E' (b)
o
oOs
But then
E= E' - V x
= B (
-
(c)
v)
os
v must be chosen so that E = 0 to comply with the shorted end, hence
(d)
v -
os
FIELDS AND MOVING MEDIA
PROBLEM 6.10
Part a
Ignoring the effect of the induced field we must conclude that
= 0
everywhere in the stationary frame.
(a)
But then
E' =E+ V xB=
(b)
xB
Since the platejis conducting
J' = J
aV x
(c)
The force on the plate is then
3 x B dv = DWd(oV x B)x B
F =
F
x
= - DWd av B 2
(d)
(e)
o
Part b
M -v + (DWdoB )v = 0
dt
o
(f)
DWdoB 2 t
o
v = v
M
e
(g)
Part c
The additional induced field must be small.
From (e)
(h)
J' - OB v
oo
Hence K'
= oB o dv o
(i)
The induced field then has a magnitude
B'
o
oK
--
ad <<
o
'
=Iadv
1
1
0a
0poorly
It
be
a
very
thin
or
plate
conducting one.
must
It must be a very thin plate or a poorly conducting one.
()
i)<<
(k)
FIELDS AND MOVING MEDIA
PROBLEM 6.11
Part a
The condition
i
- << H
W
o
means that the field
induced by the current can be ignored.
K(I
Then the
magnetic field in the stationary frame is
H = -H i everywhere outside the perfect
conductors
The surface currents on the sliding conductor are such that
K1 + K 2 = i/W
The force on the conductor is then
F
[(- +)
=
x B
p H di i
Part b
The circuit equation is
dX
=
Ri +d-
dt
dX
-dt
V
o
H dv
oo
Since F = M dv
dt
MR
dv
(---H d-)+ (o H d)v = V
00
(o H d)
v =
V
o
MR
(1 - e
t
)u_ (t)
oo
PROBLEM 6.12
Part a
We assume the simple magnetic field
0 <x
i
0
A(x)
W
x B dv = [(K1 + K2 )i x Bi ]WD
1 2 yoz
=
x< x1
= fE*a =
i
Part b
L(x) = X(x,)
i
< x
=
D
H+
FIELDS AND MOVING MEDIA
PROBLEM 6.12 (Continued)
Since the system is linear
f
1lo1Wx
2
1
W'(i,x)
1 L(x)i2
W
=
o
ex
1
2
i
i
2 D
Part d
The mechanical equation is
M
dt
2
+B
2
dt
D
i
The electrical circuit, equation is
0WX
d
dX
dtdt ~
(-5-- i) = Vo
Part e
From (f) we learn that
dx
dt
o
2
i
2BD
= const
while from (g)we learn that
P Wi
D
dx
dt
=
Vo
Solving these two simultaneously
[DV2
2 oWBEJ
0
dt
Part f
From (e)
2BD dx
w dt
o
D
= (ý
2/3
(2B)
1/3
0
Part g
As in part (a)
i
- i(t)i
3
3
O<X
1
x < x
Part h
The surface current K is
<X
V
1/3
FIELDS AND MOVING MEDIA
PROBLEM 6.12 (Continued)
i(t)
D
K=
(m)
12
The force on the short is
S
dv = DW ix ( 0
f3x
(n)
2
poW 2
21
01
20 (t)'l
2
Part i
E
xdVx
-7ýx
1
(o)
- 5t
D
dt 13
_o
di _
E2 = [D- x A + C]i
di
dt
=o
D
(P)
V(t)I
W
3
Part i
Choosing a contour with the right leg in the moving short, the left leg
fixed at xl
=
0'
a*d
(q)
B*da
dt
S
C
Since E' = 0 in the short and we are only considering quasistatic fields
H
'*dt
o
Wx
= V(t)
oxat
o2
=d
P Wx
(
i(t))
(Eb)
= V
dx
Wd
dt
(r)
H
oo
(s)
Part k
nx
(t)
b
Here
n
l
=
b
dt
.o x di
D
dxW
dt D
dt
D
V(t)
dx
=
dx
dt
W
3
oo
D
i)i2
)
(v)
o
0D
Part 1
Equations (n) and (e) are identical.
Equations (s) and (g) are
identical if V(t) = V . Since we used (e) and (g) to solve the first part
o
we would get the same answer using (n) and (s) in the second part.
FIELDS AND MOVING MEDIA
PROBLEM 6.12 (Continued)
Part m
di
Sinced= 0,
dt
V(t) 4.
E 2(x) =
Vo
tiy
= -
i-
(x)
PROBLEM 6.13
Part a
2
dt
K
Te()
+ T 2( )
(a)
Part b
ii
1Ir
D2r ;1
1
oHio
001
H
1
D2aR
J1
e
(b)
Similarly
oHoi
S
F
0 0
2
2
D2R
*
i
(c)
(c)
Part c
= [f(r xf)dv]z =
T
Te
oHo(R2-Rl)il
oHo(R2-R1)i 2
(d)
(e)
Part d
1 = E 1 (R2 -R 1);
V2 = E2 (R2 -RI)
(
Part e
1
E
=
1
V
=
J
GE
-2aDR
a
=
-ii
2a R
H (E1+iB) =
(EI+RUoHo -)
d
(h)
0Ho dt
1
(g)
- pH R(R -R1 ) --
(i)
1R2 - R
2
a 2aRD
2-
HR(R -R
2
1) dt
Part f
KoHo(R
2
dt
2 -R1 )i0
K(t)=(R-2 -R1 )
v2 (t) =
u-l(t)
(k)
t2
(1)
0
2R
(poHo(R2R1))2
7200 2 1m)
u-(t)
t u-(t) (t)
(m)
FIELDS AND MOVING MEDIA
PROBLEM 6.13 (Continued)
1 (R2 -R1)
+
v 1 (t) = [a 2RD
2aRD
Part g
2R
(t)
K t]ioUH (R22-R 1) K
( 0oo
0 -1
2
K
-
2
dt
dt2
oHo (R 2 -R 1 )i
1
S oHo(R2 -R1 )O2aRD
d22 + KI ddt
dt dt2
K 1 = [(
Hd
(R2-R 1) o
SR2R1)
K2 l(t )
2
HoR)2 2aD(R 2 -R )a]/K
p H 2aDR Y
2
K
Find the particular solution
-JK2
2
P (, t) = R
o
e
K
K2 o
2o
-=
jt]
(t)
sin(wt+tan
w
K+
(t) = A
B
-Kt
-K1e
+
p (w,t)
We must choose A and B so that
(o0)
K2
A =KIW
(0) = 0
= 0
K2m
v
o
1
B =
(K2 + 2 )
(K~1 +
v
FIELDS AND MOVING MEDIA
PROBLEM 6.13(Continued)
Wct)~
Part h
The secondary terminals are constrained so that v 2 =-i2 R 2.
~dt
R3 i ; R
RK4 2
3
dt
Then, it follows from (a),
di
2
dt
from which it
=
R
0
+ 1 (R2-RI)
RD
2 RD
K
4
=
pH (R -Rl)
2 1
Thus, (j) becomes
(w)
(d) and (e) that
K2 Ri
4 o
cos wt
KR3
RK2
4
+
i
KR 3 2
follows that
ji21
S K2R
KR4 R
I
RK
0
2
2
2
KR
R3
3
PROBLEM 6.14
Part a
The electric field in the moving laminations is
E'
J'
a
J
.
a
i *
i
OA z
(a)
The electric field in the stationary frame is
E = E'-VxB
1 Ni
B
y
2D
V = (A -
i
­
(+ rwB )i
y z
CA
(b)
(c)
S
o12Dz•N
-)i
(d)
FIELDS AND MOVING MEDIA
PROBLEM 6.14 (Continued)
The device is in series with an
Now we have the V-i characteristic of the device.
inductance and a load resistor Rt=RL+Rint.
j N2 aD
p2DrN
2D
]i +
d
[R + _Y
o
uA
S
t
S
dt
(e)
0
Part b
Let
2
2D
1 =I
Pd
If
e
= i/
, L
(f)
S
~ 1/L) t
-
2D
R
ON 2 a D
rNw S
S
R1 = Rt +AaA
(g)
t
[e
2DjorNw
= R + 2D
< 0
o
the power delivered is unbounded as t
4
(h)
o.
Part c
As the current becomes large, the electrical nonlinearity of the magnetic
circuit will limit the exponential growth and determine a level of stable
steady state operation (see Fig. 6.4.12).
PROBLEM 6.15
After the switch is closed, the armature circuit equation is#
diL
(RL + Ra)i
L
+ La -
i
(a)
(a)
= GOi
Since Ghif is a constant and iL(0) = 0 we can solve for the load current and
shaft torque
(RL+R a)
Gif
iL(t) = (R+Ra)
L
(l-e
Te(t) = iL(t) Gif
2
,
=
(R+Rai) (l-e
(RL+R )
L
(RL+R a )
a
)u_(t)
(b)
)u_(t)
(c)
t
1
FIELDS AND MOVING MEDIA
PROBLEM 6.15 (Continued)
From the data given
T = La/RL+Ra= 2.5 x 10 -
3
sec
(d)
Gef
iL
= R+R
max
RL+R a
628 amps
T
(Gif)
max RL+Ra
£
(e)
1695 newton-meters
(f)
-j
428­
-1~
1~7
cl~o·,
-
t
-
~)
/67s~
PROBLEM 6.16
Part a
With S1 closed the equation of the field circuit is
di
Rfif + Lf dt
Vf
Since if(0) = 0
(a)
R
f
if(t)
=RP
f
(1-e
(b)
)u_1(t)
Since the armature circuit is open
Rf
a
Gif
Vf C
S
R
(1-e
f
t
)u-1 (t)
(c)
FIELDS AND MOVING MEDIA
PROBLEM 6.16 (Continued)
From the given data
T = Lf/Rf
= 0.4 sec
Vf G6
V
=
a
max
= 254 volts
R
f
t
o. 4-
.
Part b
Since there is no coupling of the armature circuit to the field circuit
if is still given by (b).
Because S2 is closed, the armature circuit equation is
dVL
(d)
f (RL+Ra)VL + La - - = RLG
Since the field current rises with a time constant
(e)
T = 0.4 sec
while the time constant of the armature circuit is
T = La/RL+Ra = 0.0025 sec (f)
we will only need the particular solution for VL(t)
RG
VL(t)
RL G
= RL+Ra i = (
VL max
=
RL
L a
RL
a)G
Vf
(1-e
R
f t
Lf
(g)
)ul(t)
(h)
(j-)Vf = 242 volts
f
4?_
1!l~
0.4 sec
fY
FIELDS AND MOVING MEDIA
PROBLEM 6.17
The equation of motion of the shaft is
T
o
W
+
Jr
r dt
=
T
o
o
+ T (t)
e
(a)
If Te(t) is thought of as a driving term, the response time of the mechanical
circuit is
J
T =
ro
T
= 0.0785 sec
(b)
o
In Probs. 6.15 to 6.16 we have already calculated the armature circuit time
constant to be
L
a -=2.5
x 10 -3 sec
R +R
s
Ra+R L
T
•
(c)
We conclude that therise time of the armature circuit may be neglected, this is
equivalent to ignoring the armature inductance.
armature is
The circuit equation for the
then
(d)
(R a + RL)iL = Gwif
-- (Gif)2
Then
T
Te
ii
=
(e)
=
f-Gif
L = Ra + RL
(e)
Plugging into (a)
J
Here
de
d+
r dt
T
=
T
-
V
); i
R +R
condition that w(0) =
T
T0
w(t)
(f)
o
(Gi )2
K (-R
W
Using the initial
Kw
+
K
f
= f
(g)
R
wo
/J)
t > 0
(w - -)e
o
K
(h)
From which we can calculate the net torque on the shaft as
T= Jrdt
r dt
= (T o -KW o )e
u
(t)
(i)
1
and the armature current iL(t)
Gi
iL(t) = (R
a
l)(t)
L*
t >0
(j)
FIELDS' AND MOVING MEDIA
PROBLEM 6.17 (Continued)
From the given data
T
w
final
=
K
= 119.0 rad/sec = 1133 RPM
Tma x = (To-Kw)
(k)
1890 newton-m
(1)
Gi
i
w0
(m)
700 amps
Gi
it
L
max
)
= (R
R +R,
a
L
fna
final
K = 134.5 newton-meters,
(n)
793 amps
T = Jr/K
=
0.09 sec
•1
1i/33
/Ood
,÷
i8 O
713
700
(o)
FIELDS AND MOVING MEDIA
PROBLEM 6.18
Part a
Let the coulomb torque be C, then the equation of motion is
d0
dt
Since w(0)
= wt
- -
w(t) - 0 (1-
t)
0 < t < O/C)
Part b
Now the equation of motion is
dw
J -+
Bw = 0
dt
w(t) =
-.
0 e•)
\
wCe
Part
c
Let C = Bwo,
the equation of motion is now
dw
J d + BLc= -Bw
o
dt
B
--
{(t)-w0o + 2w oe-JE
t
< t <
Ji
B
2
to
FIELDS AND MOVING MEDIA
PROBLEM 6.18 (Continued)
( Ct )
'r = z/0
PROBLEM 6.19
Part a
The armature circuit equation is
diL
= Gwif - Va
+ LaRi
a U-1(t)
aLL
a dt
f
Differentiating
2
dw
diL
dL
La - 2 + R -d = Gi
Vu (t)
f dt
ao
a dt
a dt
The mechanical equation of motion is
dw
J r -4-= - Gi i
dt
L f
Thus, (b) becomes 2
2
(Gif)
di
diL
L
a+R
-+-L---i
J
L
a dt
2
dt
-Vu(t)
ao
Initial conditions are
diL
iLfrom(d) = ,
and it follows from (d) that
iL(t) = (-
a
+
(0
V
a
L
a
-at
e-esinBt)ul(t)
a
where
R
a
7.5/sec
a
(Gif)
8
r
a
Ra
('-L)
a
= 19.9 rad/sec
FIELDS AND MOVING MEDIA
PROBLEM 6.19 (Continued)
a
LAB - 1160 amps
V
w(t) =
-Gi
a
Se tsin
t + (e-at cos 8t-l)]
f
V
a
Gi
= 153.3 rad/sec
I (k)
IN
Part b
Now we replace R a by R+RL in part (a).
Because of the additional
damping
Ir
iL(t)
where
2L-Y (e
a
R
=
~r\~
)u 1 (t)
+R
a L
75/sec
2L
a
R + RL2
y =
a
2L
)
(Gi )
Jr La
= 10.6/sec.
FIELDS AND MOVING MEDIA
PROBLEM 6.19 (Continued)
W(t) =
Gif
a
a
1
[2Ly
e
-(a-Y)t
+
1
-(a+y)t
+ +e
2
]
o)
c-Y
r
I
ii
AL.
. ir"
(A 1.t/
vcL
r
PROBLEM 6.20
Part a
The armature circuit equation is
v a = R aa
i + GIfw
f
The equation of motion is
dw
i
J i = GI fa
dt
Which may be integrated to yield
w(t)
J
G=
wt J
i(t)
'a
FIELDS AND MOVING MEDIA
PROBLEM 6.20 (Continued)
Combining (c) with (a)
a
2
S=Rai +
asa
ia(t)
Jr
(d)
We recognize that
J
C = -(e)
(GI ) 2
Part b
J
C=
(0.5)
(1.5)2(1)
(GIf)2
0.22 farads
PROBLEM 6.21
According to (6.4.30) the torque of electromagnetic origin is
Te = Gi i
fa
For operation on a-c, maximum torque is produced when if and is are in phase,
a situation assured for all loading conditions by a series connection of field
and armature.
between if
Parallel operation, on the other hand, will yield a phase relation
and is that varies with loading.
phase connecting means are employed.
This gives reduced performance unless
This is so troublesome and expensive that
the series connection is used almost exclusively.
PROBLEM 6.22
From (6.4.50) et. seq. the homopolar machine, viewed from the disk terminals
in the steady state, has the volt ampere relation
v
Ri
Ra
+ Gwif
T
iln(b/a)
2Oad
For definition of v
and i
shown to the right and with the
interconnection with the coil
snhown in rig. or.L2
1 Nia
B
­
o
2d
Then from (6.4.52)
Gwi
o
f
22 2
(b -a
BoNi
) =
o a
2
2
a (b -a)
4d
FIELDS AND MOVING MEDIA
PROBLEM 6.22 (Continued)
Substitution of this into the voltage equation yields for steady state (because
zero).
• Ni
0 = Ri
+
asa
4d
the coil resistance is
(b -a2
)
for self-excitation with i-a 0 0
W1oN
2 2
(b -a ) = -R
Because all terms on the left are positive except for w, we specify w < 0
(it rotates in the direction opposite to that shown).
With this prov4sion'ithe
number of turns must be
4dR
4dln(b/a)
-2rodwjI p(b
M1lo (b2-a2)
N =
2
-a 2 )
21n(b/a)
oralowj(b
2
-a
2
)
PROBLEM 6.23
Part a
Denoting the left disk and magnet as 1 and the right one as 2, the flux
densities defined as positive upward are
BoN
B2
-
(i+i
2)
Adding up voltage drops around the loop carrying current i
2
R =
a
where
dB
2dB
dB
ilRa
2 dBil+
1
QB
2
_B2a we have:.
M
,
­
In( )
2nah
Part b
Substitution of the expression for B1 and B2 into this voltage expression
and simplification yield
di
L d
+ il(R+Ra) - Gil + Gi
162
2
= 0
FIELDS AND MOVING MEDIA
PROBLEM 6.23 (Continued)
where
2
2
2
N2na
-o
N
)
(b -a
2Z
The equation for the circuit carrying current 12 can he written similarly as
di
L
ti
+2(R
(+Ra)-Gi2-GOil
= 0
These are linear differential equations with constant coefficients, hence, assume
i
Ilest
I2est
2
Then
[Ls + RL+Ra-GG]I + GOI
1
2
= 0
[Ls + RL+R -G]lI2 - GoI 1 = 0
EliminatCon of I1 yields
[Ls + RL+ R a- GSJ]2
+GS]
I
+ GO
0
=0
If 12 0 0 as it must be if we are to supply current to the load resistances,
then
[LI + RL+Ra-G]C2 + (Ga)2 = 0
For steady-state sinusoidal operation a must be purely imaginary.
RL + R
or
G =
- G
2 2
--U N(b -a)
21G
This requires
= 0
I n(W4e
2rh
RL +
=
This is the condition required.
Part c
-When the condition of (b)
-
e+J
+JL
+
2 2 22
PN(b
-a )j•
N2
29oN2
is
2
satisfied
GSG
b2
(-2
b2
a
-1)Q
-)
2*2N
163
,
.4
FIELDS AND MOVING MEDIA
PROBLEM 6.23 (Continued)
Thus the system will operate in the sinusoidal steady-state with amplitudes
determined by initial conditions.
With the condition of part (b) satisfied the
voltage equations show that
1 1fi
2
and the currents form a balanced two-phase set.
164