hw8s - Meg/u £34 P5#3 gums PROBLEM 8.1 I The identity to...

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Unformatted text preview: Meg/u £34, P5 #3 gums PROBLEM 8.1 I The identity to be veriQied 18 v-(wx) h wv-K + Z-vw . (a) First express the identity in index notation. ' 3A a m aw 3x [WAm].- W 3x + Am 3x (b) m m m The repeated subscript indicates summation. Thus, expanding the first term on the left yields: 3A a __E 32. = .' '. I v axm + An ax” _ wv A + A Vu (c) PROBLEM 8.2 We wish to show that §£V(wZ) - wE-VK + Kfi-vw (a) First. the identity is expressed in index notation, considering the nth component of this vector equation. Note that the equation relates tuo vectors. (E-IV(wK)1)m - (wE-[vx})m + Amfide (b) Now, consider each term separately (E-r"<wz>1> -ah-a-<A)-A nal+ 12.3h () ' . m ’ k axk W m m k 3xk W k axk c 3A («afi-t\7itl)"u'1 - ”£327: . (d) Amfiovw - Amsktvak - AmBE 39— ' (e) k axk The sun or (d) and (e) give (c) so the: the identity is verified. PROBLEH 8. lo The [nth component of the force density at a point: 13 (Eq. 8.1.10) 31‘ - _11 ' ' F1 3:: (a) .1 -. Thus in the 11 direction. 9 2 3T 3T 31' I" P _ 11 12 13 _ _q o 1:1 [3x1 + 3x, + 3x3 } (32 x1 ‘ 3 x1 + 0) - O 0’) Similarly in chew}; and I3 direction: we find . F _ [.3721 + "22 + 31.23) _ 0 , (c) 2 3x1 3x2 3x3 F - (3?” + 3732 31'”) . n (d 3 3x1 33:, 3x3 ) no. ‘ ~ _ -.-.-..--.. -v, .,,j 13 I o I I I I 5 Hut. 1-: m net. an I: (1.14: on «ch of tho tart-cu 0! an 114:. I ovu’ nurl'acu (1). (3). (3) Incl (1). £1 I 0 (I) on: Iurhca V . = (6) :1 - 7°- '1 - o _ m v . (6) :1 II ‘3 I; I o ‘ ’ - ' (c) v - - . - (fizz-33:1”: . (a) from In. I. 3.10. - w 'cot‘lsl - 1-145.“ - (a) - Rue... 0V,“ sun“; (1). (J). (5) and (7) 713 u 0 ' ’ - . '. ' ' ‘ (l) and on: “when I: v 3- m ’11 ' ’ I: (:2) m "' t v 3 u) . n - ' ° (53) m . c. v 3 (2) Tu - - (‘ m Now; ‘ t1 - {Tulljda - 1?“:di O [tunzdn + I'u'f" ' (1) {771154. - 0 bouuu flu prob!“ 1; two dihutotul. ° (k) Lulu: consider .16: of fin och" integral" frunza - o - <1) humu- du “than which Mn urn-Al I: In (I). (I). (5) and (7) and by m U. have show that '13 C II mt than luttuu. use. no u: u codutbuua to the (out m “the. (8).. luau“ I v 0 (an: than th- It‘ll . .0 ” but tho ulcuuc'ioo a! a. Voice nduul is PROBLEH 8.11 (Continued) :1 - I Iii) deb - I Ti?) de‘ + I Iii) dez . (I) (6) (6) (2) 1 6 DV’ 0 o 1 1 _1_ g! . - __i—- i: . b + c. (n) Note: by aynmctry. there 1: no contribution to the torce iron the surfacee perpendicular to the 1) exit. \ PROSLFH 8.18 " ._ ._ Choose the surface sheen in the lieure. 'u‘. (3/ F—"—-‘—‘1 F" "-7 ;| I n ' g I ‘ I p ' {(9 o l I I G? ' . I u..._\ v.-L~ I; © X; 5&8: v.ew " > ‘\ f1 - {T‘jnjde - I Tllnl‘. + I lenzde + I TIJHJd‘ (e) 3.‘ 5.6 1.2 Since the pieces are perfectly conducting. 31 - O et eurfecee (5) and (6) and hence T - 0 on surfaces (5) and (6). Surfecce (1). (2). (3) and (‘) 12 are far from the body to V0 - ..n . . E a 3-— 1‘ (h) at each of thee. and thua. on surfaces (1) and (3). Tl) - 0. Therefore. 13 - - I TudaJ + I Tild'b (c) (J) (‘) t V 2 (3). (‘)_ _ _g _g T11 T11 2 ‘a ’ - w and eJ - e‘ (arena). Hence. :1 - 0 ~ (e) “10an 8.21 . '- Par: a Pro. Zn. 8.1.11. 3 I 32) .lLl 0 Y “a ’ 52“!) o ( ) 33:" x y ‘ 1 2 2 o 2u° "x'3y) where the component: of i are given in the problen. Part b The appropriate eurface_of integretion. which is fixed with resbect to the fixed (rune. is shown in the figure. Ue cqmpuce the time average force. and hence contributions from eurfecel (1) end (3) cancel. Fieida go to zero on surface (2). which is at y-.' Thus. there reeaina the stress on surface (é). The.t1ne average I value at the eurfece force density 1‘ PROBLEM 8.21 (Conttnued) 1- independent of 1. Hence. 1-, .. - <1- yy(y-o» . . (b) 1- ---—‘-<-a" +n;> (e) y 21:. Observe that -ijt fit": 1 “lei. hie >r1-n2‘ia . (d) where 3' in complex conjugate of i. and (c) hem-(re, K ) . (5k : ) -jku u _ o 0 3k: 0 o . 1k!) T, " " 1‘17; ‘fl'(uotoejkx) ("ff-1”” a I a ' - (e) u I: 2 . .25: (1 - ‘53.) (1) Finally. use the given definition of a :0 write (I) u no x0 I 1'" ——-‘3[1- )] - (a) «I + (":0 Note that Ty ta positive to thatk the tretu 1a eupported by the lune“: field. Hoyever. as ylJ-~o (the tutu in atomic“ the levitatsou torce sou to zero. Part C For the force per unit aree in the x direction: . , 1 Tx - -‘—-: <any(y-o)> (h) / '1 31:1: 3“” -jkx ‘ —-/;/ u. lle[uoxoe (firs; . ‘ . . (1) Thus 1- ' ”o‘o _ x J 1 ducal!) ( ) I - I ' ' - " until! 2 1/2 h ’ 1 2n + ( x 1 - A.- umat be expected. the force on the train in the x direction: vanishes as 0-00. h'ote that in any cue the force always tends to retard the lotion and hence could hardly be used to propel the train. The identity slaw/2) - 4- #u- «3575 is helpful in reductn; (J) to the torn -uxz ' you rx- ——-——-°°--—-——~ %(1+(——:)—1) (k) u au 2 1/2 2[1 + (z—)] ...
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