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# hw9 - m In Fig 5-46(PS 32 the ’ plane deﬁned b I um I...

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Unformatted text preview: m\ In Fig- 5-46 (PS 32) the ' . . ' ’ plane deﬁned b _ _ . I -. um I Ofpcrmeabllltypl from mediumZOfpermeabinty y fo y — 1 separates ; ts on the boundary and #2- no surface current ‘ ﬁnd B; and then evaluate your result for ' , . p1 = 5112. Hint: Start ' ' {equation for the umt vector normal to the given plane. 01“ by denwng the Solution: We need to ﬁnd ﬁz. To do so, we start by ﬁnding any two vectors in the plane x — y = 1, and to do that, we need three non-collinear points in that plane. We choose (0,—1,0),(1,0,0),and (1,0,1). VectorA1 is from (0,-1,0) to (1,0,0): 1., ectorAz is from (1,0,0) to (1,0,1): A2 = 21. if we take the cross product A2 x A1, we end up in a direction normalito‘ the ‘ lane, from medium 2 to medium 1, , ' :, was: r ‘ «V‘ 13532 Magnelic media sepllﬂed by the plane x -y = 1 (Problem 5.32). _ ﬂ _ #2 411 31’: — ”-1325 + 92.5). §5-8 and 5-9: Inductance and Magnetic Energy ., 5.35 Obtain an expression for the self-inductance per unit length for the wire transmission line of Fig. 5-27(a) in terms of a, d, and p, where a is of the wires, d is the axis-to-axis distance between the wires, and p is the " 'ty of the mdium in which they reside. on: Let us place the two wires in the x-z plane and orient the current in one then to be along the +z-direction and the current in the other one to be along the fir-direction, as shown in Fig. P535. From Eq. (5.30), the magnetic ﬁeld at point ~P('x,0,z) due to wire 1 is Figure P535: Parallel wire transmission line Tb’erefote, the total magnetic ﬁeld in the mgion between the wines is ' l Jr==y) uId B=Bl+32=yﬂ(: +d— Md_ 1) 1 s 1 Eq (5 93) states that the ﬂux linkag i» - ,_ = d9 Then Eq (5. 94) gives a total inductance ovc, recognizes that the wires are thin compared to the d > a). This has been an implied condition from the lthe ﬂux passing through the wires themselves have ‘ ' limit' in Table 2-1 for the two wire line. conductor is 10 cm? Solution: From Eq. (5.99) cable is given by \$ 2 x 41: ‘7 - L — 2L’ = 2: 10 In (g) = 277 x10‘9 (H). Eq. (5.104), Wm = L12/2 = 13912 (nJ), where Wm is in nanojoules when I is in . Alternatively, we can use Eq. (5.106) to compute Wm: 1 2 :- H d‘V. Wm Z/vvm (6.87) to ﬁnd the magnetic ﬁeld: 1 . 2 x 10-2 a _. __ 1‘an -1cop ¢ R sme 8R (e ) 2 ’2 1t 2 x 10 sin 9e42,“ instantaneous value, this is I ,OEI) = 6 % sinGcos (61: x 108: — 21tR) (uA/m). . e of the loop, 6 becomes —i and r becomes y. -, the loop is along —2, and the magnitude of the ﬂux is =5; 5cm —; - —i(30cm xdy) j - - ofwire wrapped around n square framc 0! 19° :3: origin with each of its Sldcs parallel to , ‘9‘ Led. - - ‘s the open-circuited ends of the coil if the maﬁa), frsinZy cos 103: (1'). me coil is not moving or changing shape, 3,,- = 0V 'and .. @466), E c d d 0.125 0.125B (‘dxd ) ‘ =’ -_ _ . = _N___ . z , 9‘” th/SB ds dt -0425 —o.125 y surface normal was chosen to be in the +2 direction. d 0.125 0.125 5 1 (10cos103t/ / cosxdxdy) =62.3sin1o3:. (15y), “-0.125 y=-O.125 cg: L For B = ilOcosxsin 2ycos 1031 (T), ‘ d 0.125 0.125 ' = -lOO-— (lOcos 103t/ / cosxsin2ydxdy). = u53:'2. .dt x=-0.125 y=—0.125 - ~ 7 :5. V 777777 W *7 Problem 6.5 A circular-loop TV antenna with 0 01 m2 area ' ' unif - - ' Is in th arm-amplitude BOO-MHz Signal. When oriented for maximum resmex: 10f a , 00p develops an emf with a peak value of 20 (mV) What is the - ' Peak magm the incident wave? tude °f B of Solution: TV loop antennas have one turn. At maximum orientation, Eq. (5 5) evaluates to (D = f B - ds = iBA for a loop of area A and a uniform magnetic ﬁeld with magnitude B = IBI. Since we know the frequency of the ﬁeld is f = 300m wecan expressB as B = Bocos(o)t+oco) withm: 21tx300x105radlsandaoan arbitrary reference phase. From Eq. (6.6), d . Vang: —N%:2 = —A2;[Bocos(0)t+0lo)] =ABotosm(mt+ao). V.“ is maximum when sin(03t + a0) = 1. Hence, 20 x 10"3 =ABoaJ = 10-2 xBo x 6nx10‘, .. a. . dfthecnnentdmxwouldﬁowm sh gap. mloophnsaninmal mistanceof CHAPIER6 223 is (p B d 15cm( -IJOI ‘ —/S -s—/5m —x2ny)-[—xm(cmndy _ [.101 X 10-1 15 _ 21: 5 41: x 10‘7 x 2.5 cos(21t x 10"?) x 10‘1 21: = 0.55 x 10-7 cos(21t x 104:) (Wb). X1.1 vemf = -59 = 0.55 x 21: x 104 sin(21t x10“t)><10‘7 dt = 3.45 x 10-3 sin(21c x 104:) (V). (b) Vemf _ 3.45 x 10-3 4+1 " At t = 0, B is a maximum, it points in —i-direction, and since it V b cos(21t x 1042‘), it is decreasing. Hence, the induced current has to be CCW w en looking down on the loop, as shown in the ﬁgure. ____________________——— sin(21t x 104:) = 0.69 sin(21t x 104:) (mA). [ind = arid: as Igl/ew a ,r Problem 6.9 A 50-cm-long metal rod rotates about the z-axis at 180 revolutions per minute, with end 1 ﬁxed at the origin as shown in Fig. 6-21 (P69). Determine the induced emf V12 ifB = i3 ><10‘4 T. Solution: Since B is constant, Vemf = is given by u = ore), where 21: rad/ cle x 180 c cles/min Cy ( y ) = 6n rad/s. (0=.___.__—————————' (60 s/min) Vemmf, The velocity u for any point on the bar Flgurc P69: Rotating rod of Problem 6.9. From Eq. (6.24), V m 1 ° . 12=Vm=A (“XB)-dl=/ 5(¢61trx23x10‘4)-f'dr 7:0 = 181cx10""'/0 rdr r=0.5 0 = 91cx 10-4r2\ 0.5 ~' ~' = —91cx10‘4 x025: Problem 6.15 A coaxial capacitor of length l- _ 6 cm use: an insulating d1 material withs,=9. Theradiiofthe cylindiical cond ctors the voltage applied across the capacitor is u m 0 5 cm m V(t) = 100sin(120m) (V), what is the displacement current? Figure P6.15: Solution: To ﬁnd the displacement current, we need to know E m the dielectric space between the cylindtical conductors From Eqs. (4 114) and (4.115), __ . Q E' l.21iat-zrl’ V=a,.% (2'?) 20m 44. 3 E=_f__‘_,__=_f1925.i‘l(_l.__)_ —rl———sin(120xt)w‘ nae) rm RF .VguOmmoﬁ—mﬁuovﬁwagmagﬁ” 3,8538 Ba a bsemﬁam 932 uo €93 0383:. was a, _ _ a GEE 30 .mm 5 58% §§§ 033-35; 2:. 2.» 5.: .283 conﬁne 03 8328 25m 0.5 mm 5033, \$5 s . oo . H moo X "I H 25 Sans” m3 RES: 9: .5332” >30 N minus—30:8 .Awsuu men an A2 39 3 55% mm 8:893 158 a mo 85mg? 05. .8058 nus"? 06 8 @8858 «9:3 05 5:85 wagon 3058 nouosuuoo 05 S 3.50 on 8 an: «55o “58083”? 813332 ﬂ 5:8? 3x58 o5 35m iguana“? .93 353833 u 383303 x 52 x 1: x n: u 995\$. Q2835 k V may 1 . a 5 \$2 x24 .32: u m ’Fi‘gure P6.16: Parallel- late ' ' ' - . . 6. 16). p Capamtor comammg a [0583’ dielectric main-.31 (Pmbl. (b) Obtain an expression for 14, the displacement current ﬂowing inside the capacitor. (c) Based on your expression for parts (a) and (b), give an equivalent-dram representation for the capacitor. (d) Evaluate the values of the circuit elements forA = 2 cm2, d = 0.5 cm, a, = 4, o = 2.5 (S/m), and V(t) = lOcos(31t x 10%) (V). Solution: 0:) CHAPTER6 + 8.0“ R 6W!) Equivalent Circuit 1 Figure P6.l6: (a) Equivalent circuit. 4 x 8.85 x 10-12 x 2 x 10-4 0 5x104 = 1.42x 10-12 1=. ...
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