ece302hw8 - 3:32 we g (a) Find H in rectangular components...

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Unformatted text preview: 3:32 we g (a) Find H in rectangular components at P(2, 3, 4) if there is a current filament on the z axis carrying 8'rnA in the aZ direction. (b) Repeat if the filament is locatedat x = —l, y z: 2. (c) Find E if both filaments are present. Two Semi-infinite filaments on the z axis lie in the regions 5—00 < z < -a and a < z ,< 00.. Each carries a current I in the aZ direction. (a) Calculate H as a function of p and d at z = O. (b) What value of a will cause the magnitude of H at p = 1, z = O, to be one-half the value obtained for an , maflsesnt? AEurrent sheet K = Sax A/m flows in theregion —2 <.- y < 2. in the plane ‘_ z=O.CalculateHatP(0,0,3).l' Aw” “Lg-g Aim \M_wn_.wn-wmflw—_n 7-4 - v. H _ V ,_ ’ _ .1 _ _ : Assume that there is a region with cylindrical symmetry in which the . conductivity is given by a = 1.56‘150PkS/m. An electric field of 303z V/rn ‘ is present. (a) Find J. .(b) Find the total current crossing the surface ,0 < pg, 2 = 0; all 45. (0) Make use of Ampere’s circuita—l law to find H. < g: ,2 Given the field H = 20/02 21¢ A/rn: (a) Determine the current density J. (b) Integrate J over the circular surface ,0 = l, 0 < <25 < 2n, z :0, to determine the tqtal eunenthnassing through that iii-the a; direction. " (c) Find the total current once more; this time by a line integral around the circularpathp = 1,0 < 1,23 < 27:,z=0. ' i\ V .Whenlx, y, and z are posrnve and less than in certain magnetic field ' mtensrty may be expressed as H = [xzyz/(y + lflax + 3262222}, — [xyz2 / (y +V 1)]az. Find the total current in the ax direction that crosses the stupx = 2,15 31 5 4, 3 5 z 5 4, by'amethod utilizing: (a) a surface (Z?) a closed line integral. ® The magnetic field intensity is'giv'e'n'm": certain regibfi'oi’épéeé‘aé n [(x '+ 2y)/2:2]ay + (2/z)az A/m. (a) Find V x H. (17) Find J. (c) Use J to find ‘ the total current passing through the surface 2 = 4, 1 5 x s 2, 3 5 z 5 5, i in the a; direction. (d) Show that the same result is obtained using the other ' __side of Stokes’ theorem. ' A long, straight, nonmagnetic conductor of 0.2 mm radius carries a '- ' ‘ uniférmly distributed current of 2 A dc. (a) Find J within the conductor. (17) Use AmPEre’s circuital law to find H and B Within the conductor. (6) Show that V x H = J within the conductor. (0!) Find H and B outside the, ’ conductor. (e) Show that V x H = J outside the conductor. ' §“ The cylindrical shell defined by 1 cm é < l.4 'ccnsists 5f nonmagnetic conducting material and carries a total current of 50 A in the az . direction. Find the total magnetic flux crossing the plane it; = 0,0 < z < 1: (a) 0 < p < 1.2 cm; ((7) 1.0 cm <_ p < 1.4 cm;v(c) 1.4 cm < p < 20 cm. @ Assume that A = 5043232 W'b/m in a certain region of free space. (a) Find and B. (27) Find J. (cl Use J- to find the total current crossing'the surface O 5 p.51, 0 5 ('15 k 221’, z = 0. (d) Use the value of [-1.15 at p =_ 1 to calCulate‘g fH-deorp=1,z=O.' - . A point charge, Q = ——O.3 [LC’andm = 3 x 10"16 kg, is moving through "W/ the field E = 30212 V/m. Use Eq. (1) and Newton’s laws to develop the appropriate difierential equations and solve them1 subject to the initial conditions at t = 0, v = 3 x 105aI m/s at the origin. At t = 3 us, find: (a) the position P(x, 3:, z) of the charge; (b) the velocity v; (c) the kinetic energy of the charge. _ ~ A point charge for which Q: 2 x 10"16 C and m = 5 x 10‘26 kg is moving in the combined fields E = 100:1: — 20an + 30032 V/m and B = —33x + 2ay —— az mT. If the charge velocity at t = O is v(0) = (23K —— 333, —- 43;)105 m/s: (a) give the unit vector shoWing the direction in which the ’ charge is accelerating at t = 0; (27) find the kinetic energy of the charge at t = O. - @ éiyen‘a material for which Xm = 3:1 and Withjn WhjCh B = OAyaz T, find: @H; (b) u; (0) gr; (d) M; (e) J; (f) 11;; (3) Jr- Let [id = 2 region 1, defined by 2:: + 3y —— 4z > 1, while 11,; = 5 in region 2 where 2x + By — 4z_ < 1. Ingregion 1, H1 = 503Jr — 302), + 2021Z A/m. Find: (a) Hm; (19) Ha; (c) Ha; (d) HNZ'; (e) 61, the angle between ELEPWNZI; (f) 92’ the angle heme“ H2 and aNZI' . ‘ g (a) 4.833,. -— 7.242.y + 9.66aZ A/m ' ' ‘ ' ‘ “ ‘ " £9m‘5 ‘ (b) 54.83ax 7 22.7633, + 10.3421z A/m / mg (1:; (a) ~29421,r + 196a,, nA/m (c) —42lax + 578a,. uA/m (g) (a) (b) 1/45 @ (a)45e_150pazkA/.m2 1 ' (b) 12.6[1 — (1 + 150p0)e-150P01 A (a) hoop/110345 A/m, —100pa¢ Wb/mz (c) 2—-p@[1 — (1 + 150p)e~15091 A/m (406010121Z A/m2 (b) 4cm A (6)4017 A (a) —259 A (b) —259 A (a) 21:: + 2y>/z3ax + We A/m (b) same as part (a) (c) 1/8 A £33") (@159 x 107aZ A/m2 (b) 7.96 x 106pa¢ A/m, 10p2¢ Wh/m2 (c) as expected (d) 1/(er)a¢ A/m, uo/(rrp)a¢ Wh/rn2 (e) as expected 1 (a) 0.392 ,u.Wb_ (b) 1.49 awn (c) 27 “W13 @ (b) _§§az A/m2 (c) —500 MA (d) —500 MA it" (a) (-90, 0, —.135) (b) 3 x 105a; * 9 x 10432 111/5 13 r (a) 77.6yaZ kA/m (b) 5.15 x 10-6 H/m (c) 1.5 x 10‘5 I 12‘" (a) .70ax + .703}, ~ .1232 (b) 7.25 f] (c) 4.1 (d) 241 yaz kA/m (e) 77.63,; kA/m2 (f) 2413.. 19:1ij (g) 318ax kA/m2 I‘V’ .a. Find H in cartesian components at P (2, 3, 4) if therei‘s’a dfiiéht‘tiéfiéht on the z axis carrying 8 mA in the a: direction: I l Applying the ‘Biot—Savart Law, we obtain _ , _ muting” °°Iczzaz><[23a+3ay+(4—z)az1—/m ———Idzl2ay—Baa] H“ ’ Loo W ‘ -00 _ -00 477022 — 812+29W2 Using integral tables, this evaluates as I 2(22 — sway 43am) °° I ‘ = —— —-———.——-—~——— = —— 2 — 3 Ha 4w [ 52(22 -— 82 + 29)”2 _ 26w< a‘y am) Then with I = 8 mA,we finally obtain Ha = ~—29'/-la,I + 196ay uA/m b. Repeat if the filament is located ata: = —1, y = 2: In this case the .Biot—Savart integral- b’ecomes - ‘ H _? . °° Idzaz >< [(2+1)ax+(3 --2)ay+(4-—z)azl = b _ _oo 477(22—82+26)3/2 I °°‘ I dz[3ay — am] hm 47r(z2 — 82 + 26)“? i . Evaluating as before, we obtain with I = 8 i __I 2(22—3)(3ay“aw)m -1. _ =_127’ ’382 A—m 40(z2 - 82 + 26)1/2' _ c. Find E if both filaments are present: This will be just the sum, of the results of parts a and b, or y . HT = Ha + Hz, = ~421am _+ 57831, [JA/m This problem can also be done (semewhat more simply) by using the known result for H from I ‘ The Biot-Savart method was used here for thesalre of illustration. \ Two semi-infinite filaments on the z axis lie in the regions —oo < error in problem statement) and a < z < on. Each carries a. current I in the az direction. , a) Calculate H as a function of p and 915 at z = 0: One way to do this is tense the field from' an infinite line and subtract from it that portion of the field that would arise onm the current segment at —a < z < a, found-from the Biot-Savart law. Thus, ~ ’ H___;I_a __/f‘ Idz_az><[pap—zaz] i _ 2,173 e _a 4flp2 +z213/2 ..:r 4» x1”; ' ‘- The integral part simplifies and is evaluated: a. _ Idzpaqg _Ipa -2 “ _ Id _.__a. I ' 4 _a 47T[,02+2.2l‘°"/2 —.47V ¢p2 p2+z2 -a_2wpx/p2+<12 45 I Finally, I a H=—_ 1_——————-— a Am 21rp‘: /p2+al2] / I b) What valueof a will cause the magnitude of H at p = 1, z = O, to be one~half the value. . obtained for an infinite filament? We require ' 11_____.___a :1 =>7 'a 2:l : a=1/\/§ \/'p2+a.2 pzl 2- 1+1; 2 -—-— an infinitely-long wire in cylindrical components, and transforming to cartesian components. «1.? !‘, “Hf,- .'_._.-_.-'._‘,.\. __ ,'- |‘. J. < —-a (note typographical 1" " "i ' ' A current sheet K = 8a; A/m flows in the-region —2 < y < 2 in the plane 2 = 0. Calculate I H at PG), 0,, Using the Riot-Savart law, we write K x aRdz cly 2 °° 8a,, >< (~93am —ya +33”) H = ————————— = —————————-——————' y a ‘ d d P (“R2 [2100 47T(232+y2+9)o/2 a: y Taking the cross product gives: 2 co __ __ HP=/ / 8( ya; Sawdrdy -- -2 m 41r(2:2+y2+9)°/2 We note that the 2 component is anti-symmetric in 3; about the origin (odd parity). Since the. ~ l limits are symmetric, the integral of the 2 component over 3,: is Zero. We are left ’with H _/2 /” ~24aydxdy _ 6 f2 9,— - i re d P -__2. __oo 4W($2+y2+9)3/2 Way _2 (y2+9) r—————$2+y2+9 ice y - I ' 6 2 2 4 12 1 _, y 2 4' ' _. —;r—ay [-2 y2 +9 dy -— ——?ay gtan |_2 —-I-—;1_-(2)(O.59)ay —- —-l.50ay :1 ® AsSume that there is a region with cylindrical symmetry in which the conductivity is given by r 0' = 155-1509 kS/m. An electric field of'3_0 az V/m is present. - 3.) Find 3: Use ‘ - 2 J = «7E = 4562—150" 3., kA/m b) Find the total current crossing the surface p < pg, z = 0, all qfi: 27" P0 _ '_150 PO . __ —150p _. P ....]_ . kA I=ffJ-dS—/o f0 45e‘ pdpfict (150)26 [ i p In - =12.6[1-(1+150pg)e-15DP°] A 1 c) Make use of Ampere’s circuital law to find H: Symmetry suggests that H will be ct— directed only, and so we consider a circular path of integration, centered on and perpen— dicular to the z sods. Ampere’s law becomes: 271'qu5 = 1mg, where I and IS the current found in part b, except with pa replaced by the variable, ,0. We obtain H Given the field E = ZOPZ aé A/ugciz ' Ar_ __; ,. V , , a) Determine the current density J: This is found through the curl of 1-1,. which simplifies to a single term7 since H varies only with p and has only a 45 component: ' 1 d(pH¢) _ 1 d ~3=VxH=~ az___ 203 az=60 az'Amz p dp de< P) __P___L_ i q ' b) Integrate I over the circular surface p = 1, O < qfi < 271', z = 0, to determine the total current passing through that surface in the 21; direction: The integral is: 1 “Bar 1 V I=//J-dS=/ ffiflpaz-pdpdqbaz=407rA o o . c) Find the total current once more, this time by a line integral around the circular path p=1,0<¢<27r,z=0: ‘ - 2w 21r I = ffi - dL =f 20,32 a¢IP=1-(I)d¢a¢, =/ 20dq§ = 4071'. A o o I When 1;, y, and z are positive and less than 5, a certain magnetic field intensity may be expressed as H = [m2yz/(y + + 3392231, — [ryzz/(y—l— 1)]az. Find the total current in the _ ‘ 343 direction that crosses the strip, :2: = 2, 1 _<_ y S 4, 3 5 2 S 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 _ 8H; dilly) __ _‘( $22 3;, " 6:: (y + 1).2 Jm=(V><H)m=< +6mzz) The current through the strip is then 4 4 ' 2Z2 . 7-4 _222 24 ’4 d I'=/Jada=—f/ +242)dclz=—- C + 2y) z s z " 3- 1 (y+1)2 y _/3 (y‘i'l) 1 4 ‘4 3 r, 1 = —-/ (—z‘ + 722') dz = —- (—23 + 3622) = —259 3 5 5 ' 3 7 ‘_“~‘r-“‘-—.._._ ,.__ _.__‘-.__._._‘......~_.,..h .,. VV.\_‘_\ ~‘n b.) a. closed line integral: We integrate counter—clockwise around the strip boundary (using the right-hand convention), Where the path normal is positive am. The current is then 4 4 . 1v 3 r '2 ‘ I=][H.dL=/l 3(2)2(3)2dyf/3 {fig dz-l—L 3(2)2§4)2dy_+/ 72:221.) dz s = 108(3) - 1-5-(43 — 33) + 192(1 — 4) — $33 — 43) = ~259 ,.~._.,____v.... . i ‘_<~_,_§‘ _ s The magnetic fi€1d ifltensitr is given in a certain region of space as _ ‘ ' _ - ‘i , ' a: + 2 ' 2 z" z ‘ a) Find V x E: For this field, the general curl expression in rectangular coordinates simplifies . to . t 8H,, 3H,, __ 2(3: + 23/) 1 .__._. ._L._. am-I- 81: az 23 aw .. zzaz A/m V H:- E X dz b) Find 1: This will be the answer of part 0,, since V X H = J. c) Use J to find the total current passing through the surface z = 4, l < a: < 2, 3 < y < 5, in the az direction: This will be 5 2 i ll . I=//le=4'azdmdy=f / Edmdy=l/8A i - , a 1 __. -—.._.._.._n_..-.y.__..,_,_. ,... . _ 1 Show that the same result is obtained-amusing the other side of Stokes’ theorem: We take l 39 H - dL over the square path at z = 4 as defined in part c. This involves two integrals of the I y component of H over the range 3 < y < 5. Integrals over as, to complete the loop, do not exist Since there is no :1: component of H. We have i , I = frag, - dL f5” 3 16 M .1, A long straight non-magnetic conduotor of 0.2 mm radius carries a. uniformly-distributed ‘ current of 2 A do. - _ , v i r a) Find J within the conductor: Assuming the current is +2 directed, E 2 J:———————————- ,=1.59><107 ,Am2 7,—(02 X 10—3)2 a“ #3:; b) Use Ampere’s circuital law to find H and B Within the conductor: Inside, at radius ,9, we have pJ 3— Then B = sol-I = (477 x 10—7)(7.96 X 105)pa¢ F 10pa¢ Wb/mg. I 27rpH¢J = 7rsz => H: age. = 7.96 x 105p aqs A/m c) Show that V x H = J Within the conductor: Using the result of part b, we find, 1 d 1 d 1.59 x107p2 7 2 =—-—- =—-—- -———————- =1.59X10 2A =3 V XI-I pdp(,ofi’¢)az pdp( 2 )a2 a /m d) ’Find H and B outside the conductor (note typo in boolc): Outside, the entire current is I enclosed by a closed path at radius p, and so ' I 1 H—fia¢—;—f;a¢ A/m New B = MOI-I = [Jo/(WP) a4, Win/mg. e) Show that'V X H = J outside the conductor: Here we use H outside the conductor and write: . 1d 1d 1 V X H = -——(PH¢)'az = -—-' (9*) at: = Q (fitS expected) ,, _ Pdp -. .de 7”” The cylindrical shell defined by 1 cm < p < 1.4 cm consists of a non-magnetic conducting material and carries a total current of 50 A in the a: direction. Find the total magnetic flux crossing the plane 9b = O, O < z < 1: a) 0 < p < 1.2c1n: We first need to find J, H, and B: The current density Will be: 50 5 I a = m 7 = . 7 A‘ 4 J 7r[(1.4 x 10“?)2 .— (1,0 x 10—2)2] al— 1 56 X 10 ad /m Next we find H¢ at radius p between 1.0 and 1.4 cm, by applying Ampere’s circuital law, and noting that the current density is Zero at radii less than 1 cm: 27:" p ' ’ 27rpH¢ = and = f f 1.66 X 10°pl'dp' dgb o 10-2 A (P2 ‘ 10—4) 2 2 =;~» H45 = 8.30 x 10‘—7—- A/m (10' m < p < 1.4 x 10— m) Then B = 11011, or ‘ p2 — 10‘4) B = 0.104 p a4; Wis/m2 New, - 1 1.2><1o—2 10-4 @a=//B-dS=// O.104[—-——]d’pdz ' 0 10-2 P 2 —2 2 _ “'4 I I w .. 10—4 m = 3.92 x 10—7Wb = 0.392 qu = 0104 l 2 1.0 b) 1.0 cm < p < 1.4 cm (note typo in book): This is part a over again, except we change the upper limit of the radial integration: —-2 1.4x10 10—4 1 @b=//B-dS=// ]dde o 10-2 P _2 2 __ —4 v W9... _ 10—4-13 = 1.49 x lO‘GWb _== 1-49 PWb 0.104 [p —— = 0.104 [ c) 1.4 cm < p < 20 cm: This is entirely outside the current distribution, so We need B there: We modify the Ampere’s circuital law result of part a to find: - 1.4. —2 2 — -4 10’5 ‘ Bout = 0.104K—X—10—l—10—l- a¢ = 21¢ Wb/m2 P P We now find 1 20x10“2 10—5 20 @czf / dpdz=10-51n(——)=2.7><10“5Wb=27qu o 1.4x10—2 P 1-4 @ nwu Assume that A = 50,028.: Wb/In in a certain region of fine space, I i ' 2.) Find H and B: Use ~ , B = \7 x A =‘—%a¢ = ~100pa¢ W‘b/mL’ I Then H = B/flo = “mop/#0 345 A/m- . I l . i‘ .13) Find J: Use II 1 a 1 3 4100,92) ' 200 - 2 J=VxH=-— H ,=-— a=———-azAm _ r pap”) - ‘03” papl #0 - z #0 / a / | c) Use J to find the total current crossing the surface 0 S p S l, 0 S <35 <I27r, z :0: The 2‘ current is ~ ' 21: 1__ —200 ' ; I=//J-ds;/ / 200a,.azpdpdq5: WA=—500MA ‘0 ’0 #0 _ .U'O . d) Use the value of H45 at p = 1 to calculate 3513- 03L for p = 1, z = 0: Have = 400“ A = —500 MA #0 . . {I ' A point charge, Q = -—0.3 MG and m = 3 X 10*15 kg, is moving through the field E = 30 a, - v Use Eq. (1) and Newton’s laws to developthe appropriate differential equations and solve them, i subject to the initial conditions at t = 0: v = 3 X 105 a.” m/s at the origin. At t = 3us,.find: a) the position P(a:, y, z) of the charge: The force on the charge is given by F = qE, and Newton’s l i l I l dgz dig : 3 second law becomes: I i F = ma = m i '= qE = (—0.3 x 10—6)(‘30 a2) describing motion of the charge in the z direction. The initial velocity in z is constant, and so no force is applied in that direction. We integrate once: - dz qE . ' at — “z - F? t 01 The initial velocity along 2, 212(0) is zero, and so 01 = 0. Integrating a second time yields the z coordinate: ' . The charge lies at the origin at t = O, and so 02 = 0. Introducing the given values, we find i - = (—0.3 x 10-5)(30) 2 X 3 X 10% = —1.5 x 10101:2 m Z At t = 3 ,u.s, z = ——(1.5 X 101°)(3 x 10‘6)2 = .—.l35 cm. NOW, considering the initial constant velocity in w, the charge in 3 us attains an a; coordinate of a: = ct = (3X 105)(3 X 10‘6) = .90 m. In summary, att = 3 ps we have P(a:, y, z) = (90,0, —.135). '«. ~ ‘ b) the velocity, v: After the first integration in part 0., We find 1;, = 37% = —(3 x 101°)(3 x 10—6) = «9 x 104 m/s including the intial m-directed velocity, we finally obtain v = 3 x 105 a3 — 9 ix 104212 m/s. I . ' = -mlv!2'= .—(‘3 >< 104631113 x100)“ = 1.5 x 10‘“ J . .r_§____> .3 r - A point charge for which Q = 2 X 10-16 G and m = 5 x 10‘26 kg is moving in the combined fields E = lOOaw - 200a? + 300az V/m and B = -—3a.s + 22y — az mT. H the charge velocity at t = 0 is 3 . V(O) = (2a.: - 33y — 421:) x 105 111/5: . ‘ a r a.) give the unit vector showing the direction in Which the charge is accelerating at t = 0: Use 1 F t = O) = q[E + (v(0) x 13)], where d '- ‘ 1 1 v ’ v(0) X B =1 (2a.; — Say —.4az)105' >< + 23? — 3.2)10"3 = 11.0051m + 14003.5. 4 500.33 .- 1 So the force in newtons becomee - g ‘ F(o)' = (2 x 10-15)[(10b41106)am+§1400—200)ay+(3oo—500)az] = 42 10—14[6aq +5211;— az] The unit vector that-gives the acceleration-direction 1e found tom the force to be ( '63-a:+6‘3»g—az 8.1: '= = freem + .7031,g — .12az $723, . h) kdnetic energy of the charge at t = O:‘ / v. ammo)? =, %(5 x 10—26. kgj(5.'39 x 105'm/s)2 = 7.25 210-15 J = 7.255 ' t I; h . . ... _ ,...._,r. engage”, 32:143.} - - - 1’) t=<1+3-1-JHO=W- iv 1“ C) Hr_=(l+3.1)'=4_._l_, I d) M; me = (3.1)(77tyay) =. 241yaz kA/m ’ e) 3 = v x H =,(de)'/(dy) ‘am '=,77.63¢ ILA/1:12; f) Jb; v x M = (mg/(em am = 241%. kA/mz; 'j , g g) 3T e V x B/Po .= 318% kA/mg- - g? " ——..;‘_._. '- ILet #51 = 2 in region 1, define'cigby Qw-ESZI—dz nhfle 141.3 ¥ 5 inregion 2 wherev25n+3y—4z < 1. .— In region 1, H1 = 5an — soay‘+;20az'A/m. Find: . ‘ - . e) HNl (normal component of H1 at the boundary): -We first need a Limit Vector'normal to the surface, found through ‘ ’ ' ' ‘ V(2:r+.3y—-4z) 'Zar+3ay_4a . . " a = -——_* = _________.z = ._.I N [V'(2a: + 3y — 42M vie? ‘ “9’73” + 563? ‘ flat Since this vector is fonnd through the gradient, it will point in the direction” of increasing values of 2a: + 3y — 4z, and so will be directed into regionl. Thus We write em = aygl. The normal component of H1 will now be: ' ' ' HM ’= (H1 ' 3N21)a2\r21 = [(50.212 — 30a? + 20az) . (37% + 5-day — .74a,)] (.3733 + .seay — flag)- = ~4.83am — 7.24211, '+ 9.66512 A/m 41. b) ET {ta—.2321; 2::sone11t ole at the boundary): ‘ ._ v. - - § Em = E1 - HNI‘. . . = (50am — '3an + 20%) —_- («4.83:1cc —- 7.24217}, + 9.66az) = 54.83.21m — 22.76ay‘+ 10.34212, A/m :_—- ‘Vtengential Vcomponeht of H2 at the houndarjr): Since tangential components of H are 0011-. ::-;ous aeross a. boundary between two media of difi'erent permeabihties, we have HT; ; Em = 54.83am —- 22.76% + 10.342: A/m d) Hm (nofmal comfionent of Hg. at the boundery): Since normal components of B are contin— uous across a boundary between media. of different permeabflities, We write WE N1 = #313372 01' ' Mrl #122 Hm = HNl = 125—(—-4.83a-m — 7.24% + 9.662%) = ~1.93am — 2.90% + 3.8635 A/m e) 491, the angle between E1 and afim: This will be __ E1 50am —— 30%, + 202.3 I ' ~ ' COS 61 ' aNgl = ' + — .7432 — ' Therefore 61 = cos‘1(—-.2l) = 102°. ‘f) 62,. the angle between 33 and aNgll First, ' ' ‘ ' ' _’ Hg = Hm + Hm = (54.33.21,: — 22.76% +.1o.34az) + (—1.93% — 2,9021%, + 3.86az) ’ = 52.90% — 25.56% + 14.20a5'A/m‘ ‘ New H3 , 52.90343, -— 25.66% + 14,205., ' '- ' ' 9 = - = —-—————————————————- - . .56 — .74 = —U.09 005 2 ‘H2‘ aNzl [ 6049 (37am+ az) _ Therefore 62 = cos—1(——.DQ) = 9__5_°_. - .’ . 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