20075ee1_1_HW8

20075ee1_1_HW8 - 8.21. 8.21. 8.24. Points A, B, C, D, E,...

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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 = —a@ 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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This homework help was uploaded on 04/01/2008 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.

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