This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MEEN 634 PS #6 Sabra“ front}! 6. 3 Part a
'
p"<':".t')  96.x)  941 {o pou f) . m
3' . pt;v . 0 (b)
where we have chosen ‘7:  VOI‘ so that 
‘v"  5  ‘7:  o (c) Since there ere no currenta. there in only en electric field in the primed
frame '2 I. '
E'  (pole 125' 35—31, (‘0
ﬁo,'§'u°ﬁ'o . (e)
Part
9km)   00(1 {9 . (t) This charge dietribution generatea on electric field . 2 '
z  (pony;  an}. . (a) In the etetionery 'frame there in an electric current   r
J  9v  00(1 19v}: . (h)
“this current generatea a nagnetic field
ﬁ  v (5  352 a)
po 0 2 3a 0
Part c L {.
/m,j ' .
 .. .   . EL .
J J p‘ r p00. . )VoI: , (.1)
" 'e " "I '0 I r. t. 
284!th IIB  (Doe TT)ir (k)
. 2
" .. . ' . . 5' s.’_
I! a + 'r’?” V0999 3a ’19 , (1) I: we include the geometric transformation r'  :30). (k). and (1) become (11);. (g), and (i) of part (b) which we “1:33:35“ min; tune
t’ornation lava. ,‘l'he above equetiona apply for r<e. similar reasoning givea
the ti‘elda in each frame for r>a. ' PROBLEH 6.6 Part a
In the frame rotlting with the cylinder t
ﬁ'o.§'u°ﬁ'o 3' (2') ' %I But then since r'  r, ;r(r)  rule  I ' I  " I '  E
z a v x'B a r KI:
V ﬁnd 1:Ef édr 1n(b/s)
.. 1 v _1_ +
3" ln(+/s) ? Ir' '1'<b7> r' ‘
The surface charge density is then
C V '  + 0 ' I o —1 o r coE 1n(b7s) I s
+ _ E V Part b' But in this problem we have only surface currents and charges r r
sue V , weo V 2
11(3) ——c'——*l.e ———
s 1n(b/s) 1n(b/s)16
hue °V at V 4 " O
W Wie' mica Psrt c
wcov+
1n(b7s5 1:
Part d i "+" "‘ _"
 H vr x D vr x D 60V  ' r ”(——YE7—3';u)(1e X 1r) (n)
(h) (f) (3) (h) (i) (.1) (k) (1) (m) (n) 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 3', R' , I V K
will be zero in the plug and the trans 0
formation laws imply that 2,“ will then , 1 7 also be zero.
Using ampere's law I
m. 0<Z<E
ﬁ (‘>
o £<z
Also we know that
’J ' VE  O. V x E   %% O O < z < E (b) Eh choose a simple Laplacian E field consistent with the perfectly conduct—
ing boundary conditions _ K +
memmzulr, (°)
/' K can be evaluated from
u. " .. _ i. ‘.
{1: at. d: In da (d)
cc». ,
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. Here E"<z)  E(z). E"(£+A)  E‘(£+A)  o (e)
1 b
$E"d1   I E(z,r)dr  —K ln(b/a) (f)
J a
The contour contains a flux
b
JEd'é  (El2)] uolledr   ”o :7 1n(b/a)(£z) (g)
S e
50 that
K ln(b/a)   g; I 3'3§< +’H° %; ln(b/a) %% (h)
Since v  gé , S
vu I
E . _ 2: % Ir 0 < z < E
(J) PRCBLE‘H 6.8 (Continued) Part b
The voltage across the line at z  0 is b w I
0
v .. .. L 2rd;  2n 1n(b/e) 0:)
W0
m: + 5—“— 1n(b/a))  v0 (1)
v0
I  W0 (In)
R 4 ET 1n(b/a)
1
V ' [T77 ] Vo W
 vuo 1n(b7a) 
v
o 1 "
vuo 71?; 1e 0 < = ‘ 5
R + 31? 1n(b/e)
ii: (0)
0 E < 2
v
 __.__._1 ] if 0 < z < E
[éul:+ (In b/a) ’ r
E  (P)
0 E < 2 Part c ., ‘
Since E  O to the right of the plug the voltmeter reads zero. The terminal voltage V 13 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 (1:)
W nub/a) 2
p . v1 .. ._°___..___ 1 V2
in 21! vuo o
., R 4* ~21]? 1n(b/a)
dw b u
m o 2
—  v] —H (r)21rr dr
dt ‘ 2
. _1_ [w (______1_____.)2 V2]
2 21! v11 0 O
R + ‘2'; 1n(b/a) PROBLEM 6 . 8 (Continued) There in a net electrical force on the block. the mechanical sistem that keeps the block traveling at constant velocity receives power ct the rate
vuo ln(b/a) 1 2 v2 2n vu 1n(b a o
R+ °
21! .1.
2 from the electrical system.
'Part e u°H(r,I)x dt no
L(x)  I I  a; 1n (bl)x .11 + —3 1n(b/a) 2x
as predicted in Part (d).
PROBLEM 6.12
Part a
we assume the simple magnetic field
1+
  < <
 D 13 0 x1 x
H .(a)
O x < :1
_ __ HOWX
1(x)  [3°da  D i (b) Part b 1 n (c) PROBLEM 6.12 (Continued)
Since the ayatem is linear u Hz
, _ l. 2  l. o 2
Wn(i,x) 2 L001 2 n 1 (:1)
Part c I
.6  3w  l now 12 (e)
3: 2 D
Part d
The mechanical equation is
I 2 u H
dx dx 1 o 2
*4 7 + 3 a? " 2 T 1 (0
dt
The electrical circuit. equation in
u Vx
d1 d 0
EE 3? ( D 1) v0 (8)
Part e
From (f) we learn that
. dx "o" 2
d_l'. " 2—35 1 . const , . (h)
while from (3) we learn that
u Hi
0 dx
D FE ' v0 (1)
Solving these two simultaneously V
3
{NZ
51‘  ° < )
dt 21: us 3
0
Part f
Prom (e) 2/3
. £2 91 . _n_ 1/3 1/3
1 u H dt (u R) (23) V0 “0
o 0
Part 5
As in part (a)
+
_ ‘ 35E?13 o < x1 < x
a I i, (1)
0' x < x1 Part h The surface current K is ’~_>"r‘. "‘ ' PROBLEM 6.12 (Continued)  1 t " ' '
""9 12 (a)
The force on the short is _ ﬁ _ ,
_ u H + U .
I?  I? x i dv  DU xx ( ° 1 ° 2) . ’ ~ .(n)
In: A ’
o 2 *
mi (”‘1
Part 1 a .» . ‘
V E 'v  1” ¢ » n, V '
" __2] / ﬂaiﬂ.
"a  [& 11”)? ' (p) ‘
a“ a "dc 3 _ n
u ,
' [52 x §% ' V;t 113 '2 (F!) .
Part 1 1 vgt LW<“ Choosing a contour with the right leg in the movin 3 short. the left leg,
' fixed at x1  0 ' "0 a. u _ d— . "' ‘ ‘ ,
3“ d‘ a: f B 6 (q)
C 5
Since 2'  0 in the short and we are only considering huasistatic fields
‘ I 3"0 dx
'0  I h ——
f3 d V(t) w x no at +u dc “030 (r)
u "x /
 —— ( ° 1m) (3,
dt D (r: , 7; , (V115 ‘.
Part E — ,
E X (Eb)  V Eh I ‘ (Vt) (1:)
Here
(n)
ssx
« ¥
(V)
'(w)
a $ 
Part 1 , V” ; Equations (n) and (e) are identical. Equations (3) and (g) are
identical if V(t) f V0. Since we used (e) and (g) to solve the first part
we would get the same ensue: using (n) and (s) in the second part. Part n ‘ . ”‘07,
di
Since a;  0, v .
..  v t + 0 +
220:) . v 1? ' ' T 11 . (x) PROBLDi 6 . 20
Part a .
The armature circuit equation is 32'. . . (b)
J dt Glfi. Which may be integrated to yield . G c ‘ . w(c)  3 ! 1‘0.) , ’09
Combining (c) with (a) (GIf)2 t va  Raia + Jr I_“ ia(t) (d)
We recognize that J
C I r 2 . (e)
(GTE) Part b J
c  r  —$9:§l———  0.22 farads (GI£)2 (1.5)zc1) PROBLEM 6.22 'From (6.4.50) 53. seg. the homopolar nachine, viewed from the disk terminals
in the steady state, has the volt ampere relation v‘ I Raia + Gwif _ lnSb/az e 2ndd For definition of v‘ and 1.
shown to the right and with the interconnection with the coil
shown in Fig. 6P.22 Then from (6.4.52) ma mu Ni
0 2 2 o
Gwif 2 (b a ) ‘4d a (b2_'2) PROBLEM 6.22 (Continued) Substitution of thie into the voltage equation yields for steady state (because the coil resistance is zero). we)“:
4d for selfexcitetion uith i‘ 3‘ 0 2 2
OR.i.+ (b e) um N
o 2 2
‘73 (b " > ' “R. _,
Because all terms on the left are positive except for m, we specify o: < 0 ' (it rotates in the direction opposite to that shown). With this provinion the
number of turns mus‘t be 4dR ‘ 4d1n(b/e) N C _—a———— I
Imluoﬂazaz) 2nodeIu°(b2e2)
u _ 21ngb/a)  napalm] (bzaz) ...
View
Full Document
 Fall '08
 WonjongKim

Click to edit the document details