Hmwk11 solutions

Hmwk11 solutions - 1 A coil consists of 100 turns of wine...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 A coil consists of 100 turns of wine wrapped around a square frame of sides 0.25 m. The coil is centered at the origin with each of its sides parallel to the x— or y-axis. Find the induced emf across the open-circuited ends of the coil if the magnetic field is given by ’ (a) B = i10e'2’ m. (b) B = i lOcosx cos 103: (T). (e) B = i lOcosx sin2y cos 103t (T). Solution: Since the coil is not moving or changing shape, 3;“ = 0 V and chf = Vc‘rmf. 0.125 0.125 d d V = - -—- - t: -— —— . " emf Nd! st ds Nth/4.125 4.1sz (Rudy), where N = 100 and the surface normal was chosen to be in the +2 direction. (a) For B = ilOe‘z' (T), chf= —ioo%(ioe-2'(o.25)2) = use-2’ (V)- (b) For B = ilOcosxcos 1031‘ (T), 0.125 0.125 cosxdxdy) = 62.3 sin103: (kV). d 3 = _ _ O t vcmf IlOOd:(10cosl 54.125 F4125 . (c) For B = ilOcosxsinZycos 103t (T), d 3 0.125 0.125 . 2 d d) 0 =—- 00— 10 0510! / cosxsm y x y = . vcmf 1 dt( c x=—0.l25 y=-0.125 2 . _ p O A Circular-loop TV antenna with 0.01 m2 area is in the presence of a uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop develops an emf with a peak value of 20 (mV). What is the peak magnitude of B of the incident wave? » Solution: TV loop antennas have one turn. At maximum orientation, . . . (I) = f B - ds = :tBA for a loop of area A and a uniformlmagnetic field With magnitude 8 = lBl. Since we know the frequency of the field is f = 300 MHz, we can express‘B as B = Bocos(o)t + do) with a) = 21: x 300 ><106 rad/s and (lo-an arbitrary reference phase. I dd) (1 chr = -N-dT = —AE[Bocos(o1t +ozo)] = ABocosin((ot+ao). chf is maximum when sin(mt + 0(0) = 1. Hence, 20><10‘3 =ABoCD=lO_2XBoX61tX108, ' , _. which yields Bo = 1.06 (nA/m). ,/ K3) . A SO-cm-long metal rod rotates about the :-axis at 180 revolutions per minute, with end 1 fixed at the origin as shown in Fig. P69. Determine the induced emf v.2 ifB = 23 ><10‘4 "1'. Solution: Since B is constant, chf = Vf‘mf. The velocity u for any point on the bar is given by u = firm, where a) = 27': rad/cycle x (180 cycles/min)/(60 s/min) = 611: rad/s. l o V12= ean= 1(uxB)-dl= (961trxi3xlO“)-r"dr a I=0.5 0 = 181tx10‘4/ rdr 1:0.5 ' 0 =91tx10‘4r2l . ' 0.5 = —91t-x 10‘4 x 0.25 ='—7o7 (w). @ The plates of a parallel-plate capacitor havefiareas 10 cm2 each and are separated by 1 cm. The capacitor is filled with a dielectric material with a = 480, and the voltage across it is given by V(t) = 20cos 21: x 1061‘ (V). Find the displacement current. Solution: Since the voltage is of the form given with V0 = 20 V and co = 21: X 106 rad/s, the displacement current is given by ' Id =‘—-::Vowsintot 4 .8 4 0"12 O 0‘4 = —X—SMI_ x 20 x 27: ><10‘5 sin(21t x 1062) 1 x 10-2 : —445 sin(21t x 105:) (nA). @ The conducting cylinder shown in Fig. $.11 rotates about its axis at 1,200 revolutions per minute in a radial field given by 8:536 (T). _.- Z / Sliding contact Figure P6.1 1: Rotating cylinder in a magnetic field I The cylinder, whose radius is 5 cm and height 10 cm, has sliding contacts at its top and bottom connected to a voltmeter. Determine the induced voltage. Solution: The surface of the cylinder has velocity u given by 1,200 60 L .1 V12=/0 (uxB)-dl=/ofl ($21txj6)-2dz=—3.77 (V). u = (imp: $21: x x 5 xio-2 = «321: (m/s), The loop showu in Fig. P6.8 moves away from a wire carrying a current I, = 10 (A) at a constant velocity u 2 5'5 (m/s). If R = 10 K2 and the direction of (2 is as defined in the figure, find 12 as a function of yo. the distance between the wire and the loop. Ignore the internal resistance of the loop. 7 L. 4—- CHI—O! R I]: 10A T “ 20cm [1 u R YO Figure P6.8: Moving loop Solution: Assume that the wire carrying current 11 is in the same plane as the loop. The two identical resistors are in series, so [2 = chf/ZR, where the induced voltage is due to motion of the loop and is given by vfl=vgrnf= j£(uxB)-d1. C The magnetic field B is created by the wire carrying [1. Choosing i to coincide with the direction of II , the external magnetic field of a long wire is e “01: B = —— . 21v For positive values of yo in the y-z plane, 3? = i“, so Holt“ 21W ' uxn=ylu|xa=fiu|x$%"=2 r Integrating around the four sides of the loop with dl = idz and the limits of integration chosen in accordance with the assumed direction of 12, and recognizing that only the two sides without the resistors contribute to Vc'“mf. we have 0.2 I 0 1 vgg,=/ (2“ W) -(idz)+ (2“0 W) -(idz) 0 21" r=y0 0-2 znr r=yo+0.l _41t><10‘7><10><5><0.2(l 1 ) — 27‘ Yo Yo + 0 1 1 1 =2x10'6<—— ) (V), Yo Yo + 0.1 and therefore ’1 Q A coaxial capacitor of length l = 6 cm uses an insulating dielectric material with e, = 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If the voltage applied across the capacitor is V(t) = 100 sin(1201rr) (V), what is the displacement current? I j. 23 T 1 Solution: To find the displacement current, we need to know E in the dielectric space between the cylindrical conductors. = —f' Q D = 8E 21:31 ’ - _ E - Q In 2 - — 21:81 a ' Hence, . = —f V 1 . E = _e b = 4% = 4144 3 sin(1201tt) (V/m), rm (E) ran r The displacement current flows between the conductors through an imaginary cylindrical surface of length l and radius r. The current flowing from the outer conductor to the inner conductor along ——x" crosses surface S where S = -—i"21trl. Hence, —8 [d = 5% .s = 4% sin(1201tt)) -(—f2an) = 1.15 x 10‘8 x1201: x 21tlcos(1201tt) = 1.63 cos(1201tt) (11A). Alternatively, since the coaxial capacitor is lossless, its displacement current has to be equal to the conduction current flowing through the wires connected to the voltage sources. The capacitance of a coaxial capacitor is given by (4.116) as _21tel c_ln(). hlb‘ The current is W 211:8! I=C—= m mm a [1201: x 100cos(1201tt)] = 1.63cos(1201tt) (,uA), which is the same answer we obtained before. lemme-w»! = —r-9 x 8.85 ><10‘12 x 1443 sin(120m‘) , l //-/ "I C *7 I l 1 An electromagnetic wave propagating in seawater has an electric field with :1 time variation given by E = iEocosrm‘. If the permittivity of water is I 8180 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction canent density to displacement current density at each of the following frequencies: (a) 1 kHz, (b) 1 MHz. (c) 1 GHz, (d) 100 GHz. Solution: the displacement current density is given by a a = -—D = -——E 3“ a: 88: and, the conduction current is J = 0E. Converting to phasors and 'I taking the ratio of the magnitudes, j _ GE _ 0' 3d (0580 . (a)Atf= 1 kHz, to: 21cx 103 rad/seand 4 =-————-—-—=88 13, 21rx 103x81x8.854x10-12 8* 0 l Jd The displacement current is negligible. (b)Atf= 1 MHz. (0: 21tx 106 rad/s, and an I T; The displacement current is practically negligible. (c)Atf= 1 GHz,(o= 21rx109rad/s,and 4 = _—_= 88.- 21!:x10ls x 81 x 8.854 x 10-12 8 4 2 21: x 109 x 81 x 8.854 x 10—12 : 0'888' i 1. Neither the displacement current nor the conduction current are negligible. (d) Atf: lOOGHz, a): 21: x10”rad/s, and 3 4 3 ~— _—_ ———-————_— = . 8 10" . 1d 2nx1011x81x8854xio-12 88 X The conduction current is practically negligible. @ At t = 0, charge density pvo was introduced into the interior of a material with a relative permittivity e, = 480. If at t = 1 us the charge density has dissipated down to 10‘3pvo, what is the conductivity of the material? Solution: We start by find 1.: mm = pvoe‘m', Of _6 10—3on = one—lo m, which gives 0—6 rum-3 = —1 , Tr OI' ' 6 10‘ _ 1,: —ln10_3 = 1.45x 10 7 (8). But 1, = 2/6 = 480/0“. Hence 450 4 x 8.854 ><10“2 62—: t 145Xm_7 =2.44x1o-“ (S/m). , . If the current density in a conducting medium is given by “LEE-t) = - + izx>COS (DI, determine the corresponding charge distribution pv(x, y,z;t). Solution: . V-J = 33%. (6) t The divergence OM is- A 3 . a A a . . 2 A V J_l(x§;+y-a—y+za—) (L—y3y +z2x)cosu)t a a = —3— 'cosmt = —6 coscut ayo ) y Using this result in Eq. (6) and then integrating both sides with respect to t gives pV= —/(V-J)dt= —/ —6ycosa)tdt= 66ysinmt-l-CO, where Co isaconstant of integration. . ' '________,_‘__,_, ._ .. t . g V ___ _ fie“‘éie¢:fic field of an electromagnetic wave prooagating-in air is : given by E(z,t) = R4cos(6 x 108: — 2z) + 5'3sin(6 x 103: — 22) (Win). Find the associated magnetic field H(z,t). Solution: Converting to phasor form, the electric field is given by ' Em = sate-121 — meg-1'21 (V/m), H(z) = _ VxE . I 2‘ 1 i y 2 _—.-— a/ax a/ay a/az ‘1‘” 4w?! -j3e-fiz o 1 -J'(Dt1 (fiée'nz — §j8e'flz) — 6x108 x41cx10—7 Converting back to instantaneous values, this is H(t,z) = —i8.0sin(6 x 108: — 22) + 910.6cos(6 x103t— Zz) (mA/m). ' j (:26 — way-1'22 .—_ Jame-fix + muse-12"- (mA/m). @ i L2 0 = O is given by The magnetic field in a dielectric material with e = 450, u : no. and 4 H(y,t) = :25 cos(21t x 107t+ky) (A/m). Find k and the associated electric field E. Solution: In phasor form, the magnetic field is given by fl = :‘(Sefl‘y (A/m). which, together with the original phasor expression for f1, implies that 7 4 wfi_2nx10fl n (rad/m). have”: c f 3x103 ~30 Inserting this value in the expression for E above, ~ . 4"/30 I '4 /3o _ A '4 /3o 2 E = ‘12:: x 107 x 4 x 8.854x10’125ej “y “ “294161 U (Wm ____ The electric field radiated by a dipole antenna is given-in“! spherical coordinates by . . ' ' .2" 10-2 E(R,9;t)=9 x sine cos(61: x 108: — 21tR) (V/m). Find H(R,6; t). Solution: Converting to‘phasor form, the electric field is given by .2x10-2 17:(R,e) = éEe = a sin ee-W (V/m), which can be used to find the magnetic field: I ~ 1 ' .. H(R,9) = _jwuVXE= [RRsine ad, +¢E-a7é(RE9)] :¢6EX108X4EX10'7 . 3 i =¢§Esin8e"2"k (uA/m). Converting back to instantaneous value, this is H(R,e;t) = $ 51—:- sinecos(6n x108t— 21:12) (uA/m). at the internal dimension of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. l The homogeneous material inside the capacitor has the parameters 6 =.10‘11 F / m, a = 10-5 3/111, and cr = 10—5 S/rn. If the electric field intensity is E =t(106/p) cos(lO5t)ap V/rn, find: a) J: Use _ ‘ J = 0E = cos(105t)ap A/In2 b) the total conduction current, Ia, through the capacitor: Have Ic = f/J - d5 = 27rle = 207rlcos(105t) = 871' cos(105t) A c) the total displacement current, Id, through the capacitor: First find 105 1 -11 6 I ' _' —-——-———( X 0 )(10 ) sin(105t)ap = —£ sin(10°t) A/m l I I an i P P l l l i -' 3 Now I4 = 27mm = +27rlsin(105t) = '——0.81rsin(105t) A d) the ratio of the amplitude of L1 to that of IC, the quality factor 'of the capacitor: This will _ be ‘ [IA 0.8 ._ = _. = 0.1 Show that the displacement current flowing between the two conducting cylinders in a loesles's 1 coaxial capacitor is exactly the same as the conduction Current flowing in the external circuit if the applied voltage between conductors is VD coscth volts. ' From Chapter 7 ,_ we know that for a given applied voltage between the cylinders, the. electric field is _ Vcoswt - 'elficoswt I E=EgflmapV/m => D: agC/mz- Then the displacement current density is " I. I _a__13 _ —we%sinwté 8t — .pln(b/a) P Over a length If, the displacement current will be .. - 3D ' I 6D QtZwel/B sin-wt '- . V I = —-.- = —— _‘_——-—- = _ = I Id / f at d5 - 2"” at 11n(b/a) Colt c where we recall that the capacitance is given by = 21rel/ ln(b/d).'\ I Tow/a) '- 7 C Consider the region defined by ix}, lyl, and < 1. Let 6, = 5', p, = 4, and cr = 0. If .34 = 20 coe(l.5 >< 103t"— bm)ay pA/rn’"; a) find D and E: Since Jd = BD/at. we write - 20x10‘6." '3 D=/Jddt+0=msm(l.5x10 —ba:)ay = 1.33 x 10-13 ein(l.5 x 10% - may C/m2 where the integration constant is set to zero (assuming no dc fields are. present). Then. D 1.33 x 10-13 ' —=—-——-— ' st—b e (5x8.85x10'12)sm(15xm ' flay = 3.0 x 10-3sin(1.5 x 10% — bx)a.y V/m _ E: b)’ Iuse the point form of Faraday’s law and an integration with respect to time te find B' and H: In this case, .' _6E.‘J _ _3v "8- _ 8B V x E — Ea, — —b(3.0 10 )cos(1.5i>< 10 t baa-a; —- at Solve for B by integrating over time: ——— . .I .8 - = 2. I_n 1.5 Ost— '. I: 1.5x 108 s1n(15V>< 10t b:L')az ( O)b><10 sm( ><1 bm)ad T _ i _ B 2._ —11 ' ’ H = Z = 10875 - bm)az = (4.0 x 10‘6)bsin(1.5 x 1032: - bm)az A/m c) use V x H = J4 + J to find Jd: Since a = 0, there is no conduction current, so in this case i v x H = -%5:iay = 4.0 x 10-sz cos(1.5 x108t — b )ay A/m2 = Jul d) What is the numerical value of b? We. set the given expression for J 4 equal to. the result of part c to obtain: 20 x10‘6 = 4.0 x 10-61;2 2 b: V5.0 m-l _ b(3.0 x 10-3) . ‘ ‘\ \ Ti J , a3. :1 l .4 +1. Let ,u. = 3>< 10‘s H/In, e = 1.2 X 10‘10 F/In, and 0' = O every-where. IfH = 2 cos(lOl°t—,Ba')a: A/rn, use Maxwell’s equations to obtain expressions for B, D, E, and ,3: First, B = pH = 6 X 10'5 cos(10mt — Hafiaz T. Next we use K 3H 53:) =——— = s' {1 lot— a =— V X H aw 2,8 m\ .0 ,8 )ay at from which 2 n D = /2fi sin(101°t —— 52:) dt + C' = —10—€0 cos(101°t — [390% C/m‘ where the integation constant is set to zero, since no dc fields are presumed to exist. Next, D _ I 2-6 10 _ 10 _ -r E = —€— _ —(1.2 X 10_10)(10m) cos(10 t —;39:)a., —— —1.67,8 cos(10 t [Sway V/m Now _ 3Ey _ 2 . 10 _ BB V x E -— Ea, — 1.67;? sm(10_ t —fi:r)az — -—at SQ _ , B = — f1.675?sin(101°t — fi$)azdt = (1.57- x 104%2 cos(1ol°t — flm)az We require this result to be consistent with the expression for B originally found. So (1.67 x 10-10m2 = 6 x 10-5 => s = i600 rad/m The electric field intensity in the region 0 < < 5, O < y <I1r/12, O < z < 0.06 m in free space is given by E = C sin(12y) sin(qz) cos(2‘ x 10107?) ag V/m. Beginning with the V x E relationship, use Maxwell’s equations to find a numerical value for a, if it is known that a is greater than zero: In this case we find 8E3, 8E, VXE— 62 ay— ay a.z = C [a sin(12y) cos(a;)a;y — 12 cos(12y) sin(az)az] cos(2 x 101°t) = -%§« H=—--1—/V XE n+01 #0 C' ,uo(2x101° [a sin(l2y) cos(az)ay — 12 cos(12y) sin(az)a=] 5111(2) x 101°t) A /m where the integration constant, 01 = 0, since there are'no initial conditions. Using this result, we now find C(l44 +a2) _ aHz aHy __ . . . 10 _ V x H — [ 6y m — #00 x 1010) sm(12y) sm(az) sm(2 x 10 t) — ET Now I I sin(1:2'y) sin(az) cos(2 x 10101;) a¢ 2 E=':—3-=/%V><Hdt+02= 0044+“) 0 p.060(2 X 1010)2 where Cg = O. This field must be the same as the original field as stated, and so we require IC'(144 + a2) a 1' i - w H " 10-2 " 1/2 ' “50(2 x 1010)2 = 1 Y USing #060-= (3 X 108)-2:- ' a = _. 144] = 66 m—1 (3 x 105)2 . _ I . _ w @ In a region where p,- = 5., = 1 arm 0' = O, the retarded potentials are given by V = 2(2 — ct) V' and A = J:[(z/c) —‘b]az W'b/m, where c = l/V’poeo. a) S” ow that V -.A = —-;le(8V/3t): First, A, v A = = 3 = mm Second,_ _ I 3V :1: at ' «#1060 so We obeerve that V -A = ’—/.1.geo(3V/ at) in free space, implying that the givén statement would hold true in general media. ' : b) Find‘B, H, E, and D: Use _ V .' ’__ v '_ 314a _ 2. B“VXA‘ 3:1: ‘(t'c‘)ayT Then~ : B -1 z ' -', I: 'Now, I I v l - I . ' .' BA' V - ' ‘ ' E=—VV—79-{=—(z—ct)am—zaz+zaz=(ct—z)a¢V/m Then I D = 60E = 60(Ct — z)a, C/m2 c) Show that these results satisfy MaxWell’s equations if J and [5,, are zero; i.V-D=V-eo(ct—z)a¢=0 ' V-B=V-(t-z/c)'a,y=0 ‘ 5H - 1 e ' Vxfi=--§£a¢=-—am= -°-a¢ Z #06 #0 which we require to equal BD/at: 9.12 _ 6 ca _ gnaw ,,,, _. 3t — '° — #0 iv. - ' »—-~-—~—-._---___...; ;._,__ __ V x E = 853% -—ay ...
View Full Document

This note was uploaded on 12/06/2010 for the course ECE 302 taught by Professor F during the Spring '10 term at Cal Poly.

Page1 / 11

Hmwk11 solutions - 1 A coil consists of 100 turns of wine...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online