**Unformatted text preview: **8.21. 8.21. 8.24. Points A, B, C, D, E, and F are each 2 mm from the origin on the coordinate axes indicated
in Fig. 8.23. The value of H at each point is given. Calculate an approximate value for V x H
at the origin: Vile use the approximation: 39H-dL . 1H 2'
cur AG where no limit as Aa —’ 0 is taken (hence the approximation), and where Aa : 4 mmg. Each
curl component is found by integrating H over a square path that is normal to the component
in question. (continued) Each of the four segments of the contour passes through one of the given points.
Along each segment, the ﬁeld is assumed constant, and so the integral is evaluated by summing
the products of the ﬁeld and segment length (4 mm) over the four segments. The 1‘. component
of the curl is thus: (HZ.C‘ _ HyJS‘ _ Hz,D ‘1‘ Hy,F)(4 X 10—3)
(4 x 10-3)2
= (15.69 W 13.88 7 14.3.5 713.10)(250) = 530 A/m’2 (VX lei The other components are:
(H23 "Hag — Hz..4 — Ems x 10-3) (4 x 10’3)2 = [15.82 —— 11.11 — 14.21— 10.88)(250) = 460 A/mg (V xH)y; and 3
(HM * ch e HUBHLDJH x 10- ) (4 x 10-3)2
2 (—13.78 —10.49 + 12.19 + 11.49)(250) 2 Finally we assemble the results and write: (V x szi —148 A/m2 V X H='530ax+460ay —148az Evaluate both sides of Stokes‘ theorem for the ﬁeld G = 10 sinﬂatp and the surface 1“ = 3,
0 6 90°. 0 ah 90°. Let the surface have the a, direction: Stokes’ theorem reads: ﬁ‘G-sz/[JVXthda Considering the given surface, the contour, C, that forms its perimeter consists ofthree joined
arcs of radius 3 that sweep out 90° in the my, .132, and zy planes. Their centers are at the
origin. Of these three, only the arc in the my plane (which lies along 81¢) is in the direction of
G: the other two [in the fag and aa directions respectively) are perpendicular to it, and so
will not contribute to the path integral. The left-hand side therefore consists of only the 33y
plane portion of the closed path, and evaluates as 7172
fG-szfo 10sinu9W/2a¢-a¢3sin6W/2dqﬁ:15v To evaluate the right—hand side, we ﬁrst ﬁnd V'sz 1 d 20 '6
—[[sin6)10sin6] 3,»: COS ar 6' d6 1‘ 7’ sin The surface over which we integrate this is the one—eighth spherical shell of radius 3 in the
ﬁrst octant. bounded by the three arcs described earlier. The right—hand side becomes rr/Q Tr/Q ,8
[/(v x G) -11da=/ / at” ar-ar(3)2sin6d9dqﬁ=15_w
S D [l '3 It would appear that the theorem works. 8.25. When 9:, y, and z are positive and less than 5, a certain magnetic ﬁeld intensity may be
expressed as H = [.rtgyz/[y—l— 1]]a;r +3.1:ngay — [myzQ/(y + 1)]az. Find the total current in the
am direction that crosses the strip, .1: = 2, 1 y 4, 3 z 4, by a method utilizing: a) a surface integral: we need to ﬁnd the current density by taking the curl of the given H.
Actually, since the strip lies parallel to the yz plane, we need only ﬁnd the a: component
of the current density, as only this component. will contribute to the requested current. This is 3H 8H 2
_ I. _ z y _ 372 I I g
Jx—(VXHJI (ay 82) ((y+1)2 .Br 2) am The current through the strip is then 4 4 4 222 4 _2ZQ
I=fJ-a da27/f(+24z) dydz=i/( +24zy) dz
3 x 3 1 (EH—1)2 3 [y‘tll l
4 3 1 4
=_/ (thmz) dz=— (7234—3622) =—259
3 O O 3 8.29. A long straight non—magnetic conductor of 0.2 mm radius carries a uniformly—distributed
current of 2 A dc.
a) Find J within the conductor: Assuming the current is +2 directed,
2
: —a
«(0.2 x 10—3)2 z J =1.59 x 107213 iii/m2 b) Use Ampere’s circuital law to ﬁnd H and B within the conductor: Inside, at radius ,0, we have J
2mm. : prJ :5 H : ﬁfty : 7.96 x 106pa¢ A/m Then B = NH = (47: x 10-7)(7.96 >< 106)pa¢a = 10mg.) “Tb/m2. c) Show that V x H : J within the conductor: Using the result of part b, we ﬁnd, 1d 1d 1."9 1072
VxH=——[pH¢,)az— ( 0 X2 ’0 a2 _ 1.59 x 10%., A m2 = J
pdp pdp ) —/ (1) Find H and B outside the conductor (note typo in book): Outside, the entire current is
enclosed by a closed path at radius ,0. and so I 1
ab = —[email protected] A/m H =
Zap mo Now B = ng = rig/(mo) 21¢ “Tb/mg. e) Show that V X H = J outside the conductor: Here we use H outside the conductor and
write: 103 1d 1
V x H = —— H. :12 = —— ,0—) a2 = Q as expected
pdprp a MA W ( J 8.31. The cylindrical shell deﬁned by 1 cm < ,0 < 1.4 cm consists of a non—magnetic conducting
material and carries a total current of 50 A in the :12 direction. Find the total magnetic ﬂux
crossing the plane 0') = 0, 0 < z < 1: a) 0 < p < 1.2 cm: we ﬁrst need to ﬁnd J, H, and B: The current density will be: J_ 50 _ 221.66 105.211 9
Tr[(1.4 x 10—2)? — (1.0 x 10—2)2] a X '1 /n1 Next we ﬁnd H¢ at radius ,0 between 1.0 and 1.41 cm, by applying Ampere’s circuital law,
and noting that the current density is zero at radii less than 1 cm: 271' ,0
21rpH¢ : few: :1; jig—2 1.66 X 105,01dede (102 — 10—4) =4 H¢ = 8.30 x104 p A/m (10-2 m < p < 1.4 x 10-9111) Then B = ng. or
(142 —10‘4)
p B = 0.104 41¢ Wb/m2 Now , 1 1.:2x10—2 10_4
(Daz/fB-dsz/f 0.104[p——]d,0dz
0 10—2 .0 (1.2 x 10*?)2 — 10*4 = 0.104
1 2 1.2
— 10—4 In = 3.92 x 10—7W'b = 0.392 11V“) 1)) 1.0 cm < ,0 < 1.4 cm [note typo in book): This is part a over again, except we change the
upper limit of the radial integration: 1 1.4x10—2 1074
@bszB-dssz 0.104[p——]dpdz
0 10—2 P . _2 2 _ _4 .
W — 10—4 In = 1.49 x 10-‘9 Wb = 1.4011Wb = 0.104
i 1.0 c) 1.4 cm < ,0 < 20 cm: This is entirely outside the current distribution, so we need B there:
we modify the Ampere‘s circuital law result of part 0: to ﬁnd: [(1.4 x 10-5?)2 — 10-4] 10-5
a¢ —
p P BM _ 0.104 41¢ Wb/m2 we now ﬁnd 1 :20x10-2 _r
10 J 4 20
(Dc:// dpdz—10_”ln(—)
o 1.4><10—2 P 1-4 : 2.7 X10_5V\7b : prﬁrb ...

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- Fall '08
- Joshi
- Electrical Engineering