hw6s - MEEN 634 PS#6 Sabra“ front 6 3 Part a...

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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 fi-o,'§'-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 nag-netic 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. - 2-8-4!th IIB - (Doe T-T)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 ' V-E - 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 JE-d'é - (El-2)] 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 + 3-1?- 1n(b/e) ii: (0) 0 E < 2 v - __.__._1 ] if 0 < z < E [é-ul:-+ (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—3-5 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) ss-x « ¥ (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 self-excitetion uith i‘ 3‘ 0 2 2 O-R.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 Imluoflaz-az) 2nodeIu°(b2-e2) u _ 21ngb/a) - napalm] (bz-az) ...
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