工程电磁场第八版课后答案第09章

1、CHAPTER 9 9.1. In Fig. 9.4, let B = 0.2cos120t T, and assume that the conductor joining the two ends of the resistor is perfect. It may be assumed that the magnetic fi eld produced by I(t) is negligible. Find: a) Vab (t): Since B is constant over the loop area, the fl ux is ? = (0.15)2B = 1.41 10?2c

2、os120t Wb.Now, emf = Vba(t) = ?d?/dt = (120)(1.41 10?2)sin120t. Then Vab(t) = ?Vba(t) = ?5.33sin120t V. b) I(t) = Vba(t)/R = 5.33sin(120t)/250 = 21.3sin(120t) mA 9.2. In the example described by Fig. 9.1, replace the constant magnetic fl ux density by the time- varying quantity B = B0sin!t az. Assum

3、e that v is constant and that the displacement y of the bar is zero at t = 0. Find the emf at any time, t. The magnetic fl ux through the loop area is ?m= Z s B dS = Z vt 0 Z d 0 B0sin!t (az az)dxdy = B0vtdsin!t Then the emf is emf = I E dL = ?d?m dt = ?B0dv sin!t + !tcos!t V 9.3. Given H = 300azcos

4、(3 108 t ? y) A/m in free space, fi nd the emf developed in the general a?direction about the closed path having corners at a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic fl ux will be: ? = Z 1 0 Z 1 0 3000cos(3 108t ? y)dxdy = 3000sin(3 108t ? y)|1 0 = 3000 sin(3 108t ? 1) ? sin(3 108t) Wb

5、 Then emf = ?d? dt = ?300(3 108)(4 10?7) cos(3 108t ? 1) ? cos(3 108t) = ?1.13 105 cos(3 108t ? 1) ? cos(3 108t) V b) corners at (0,0,0), (2,0,0), (2,2,0), (0,2,0): In this case, the fl ux is ? = 2 3000sin(3 108t ? y)|2 0 = 0 The emf is therefore 0. 164 9.4. A rectangular loop of wire containing a h

6、igh-resistance voltmeter has corners initially at (a/2,b/2,0), (?a/2,b/2,0), (?a/2,?b/2,0), and (a/2,?b/2,0). The loop begins to rotate about the x axis at constant angular velocity !, with the fi rst-named corner moving in the az direction at t = 0. Assume a uniform magnetic fl ux density B = B0az.

7、 Determine the induced emf in the rotating loop and specify the direction of the current. The magnetic fl ux though the loop is found (as usual) through ?m= Z s B dS, where S = nda Because the loop is rotating, the direction of the normal, n, changing, and is in this case given by n = cos!taz? sin!t

8、ay Therefore, ?m= Z b/2 ?b/2 Z a/2 ?a/2 B0az (cos!taz? sin!tay) dxdy = abB0cos!t The integral is taken over the entire loop area (regardless of its immediate orientation). The important result is that the component of B that is normal to the loop area is varying sinusoidally, and so it is fi ne to t

9、hink of the B fi eld itself rotating about the x axis in the opposite direction while the loop is stationary. Now the emf is emf = I E dL = ?d?m dt = ab!B0sin!t V The direction of the current is the same as the direction of E in the emf expression. It is easiest to picture this by considering the B

10、fi eld rotating and the loop fi xed. By convention, dL will be counter-clockwise when looking down on the loop from the upper half-space (in the opposite direction of the normal vector to the plane). The current will be counter-clockwise whenever the emf is positive, and will be clockwise whenever t

11、he emf is negative. 9.5. The location of the sliding bar in Fig. 9.5 is given by x = 5t + 2t3, and the separation of the two rails is 20 cm. Let B = 0.8x2azT. Find the voltmeter reading at: a) t = 0.4 s: The fl ux through the loop will be ? = Z 0.2 0 Z x 0 0.8(x0)2dx0dy = 0.16 3 x3= 0.16 3 (5t + 2t3

12、)3Wb Then emf = ?d? dt = 0.16 3 (3)(5t+2t3)2(5+6t2) = ?(0.16)5(.4)+2(.4)325+6(.4)2 = ?4.32 V b) x = 0.6 m: Have 0.6 = 5t + 2t3 , from which we fi nd t = 0.1193. Thus emf = ?(0.16)5(.1193) + 2(.1193)325 + 6(.1193)2 = ?.293 V 165 9.6. Let the wire loop of Problem 9.4 be stationary in its t = 0 positio

13、n and fi nd the induced emf that results from a magnetic fl ux density given by B(y,t) = B0cos(!t ? ?y)az, where ! and ? are constants. We begin by fi nding the net magnetic fl ux through the loop: ?m= Z s B dS = Z b/2 ?b/2 Z a/2 ?a/2 B0cos(!t ? ?y)az azdxdy = B0a ? sin(!t + ?b/2) ? sin(!t ? ?b/2) N

14、ow the emf is emf = I E dL = ?d?m dt = ?B0a! ? cos(!t + ?b/2) ? cos(!t ? ?b/2) Using the trig identity, cos(ab) = cosacosbsinasinb, we may write the above result as emf = +2B0a ! ? sin(!t)sin(?b/2) V 9.7. The rails in Fig. 9.7 each have a resistance of 2.2 /m. The bar moves to the right at a constan

15、t speed of 9 m/s in a uniform magnetic fi eld of 0.8 T. Find I(t), 0 t 1 s, if the bar is at x = 2 m at t = 0 and a) a 0.3 resistor is present across the left end with the right end open-circuited: The fl ux in the left-hand closed loop is ?l= B area = (0.8)(0.2)(2 + 9t) Then, emfl= ?d?l/dt = ?(0.16

16、)(9) = ?1.44 V.With the bar in motion, the loop resistance is increasing with time, and is given by Rl(t) = 0.3+22.2(2+9t). The current is now Il(t) = emfl Rl(t) = ?1.44 9.1 + 39.6t A Note that the sign of the current indicates that it is fl owing in the direction opposite that shown in the fi gure.

17、 b) Repeat part a, but with a resistor of 0.3 across each end: In this case, there will be a contribution to the current from the right loop, which is now closed. The fl ux in the right loop, whose area decreases with time, is ?r= (0.8)(0.2)(16 ? 2) ? 9t and emfr= ?d?r/dt = (0.16)(9) = 1.44 V. The r

18、esistance of the right loop is Rr(t) = 0.3 + 22.2(14 ? 9t), and so the contribution to the current from the right loop will be Ir(t) = ?1.44 61.9 ? 39.6t A The minus sign has been inserted because again the current must fl ow in the opposite direction as that indicated in the fi gure, with the fl ux

19、 decreasing with time. The total current is found by adding the part a result, or IT(t) = ?1.44 1 61.9 ? 39.6t + 1 9.1 + 39.6t ? A 166 9.8. A perfectly-conducting fi lament is formed into a circular ring of radius a. At one point a resistance R is inserted into the circuit, and at another a battery

20、of voltage V0is inserted. Assume that the loop current itself produces negligible magnetic fi eld. a) Apply Faradays law, Eq. (4), evaluating each side of the equation carefully and inde- pendently to show the equality: With no B fi eld present, and no time variation, the right-hand side of Faradays

21、 law is zero, and so therefore I E dL = 0 This is just a statement of Kirchos voltage law around the loop, stating that the battery voltage is equal and opposite to the resistor voltage. b) Repeat part a, assuming the battery removed, the ring closed again, and a linearly- increasing B fi eld applie

22、d in a direction normal to the loop surface: The situation now becomes the same as that shown in Fig. 9.4, except the loop radius is now a, and the resistor value is not specifi ed. Consider the loop as in the x-y plane with the positive z axis directed out of the page. The a?direction is thus count

23、er-clockwise around the loop. The B fi eld (out of the page as shown) can be written as B(t) = B0taz. With the normal to the loop specifi ed as az, the direction of dL is, by the right hand convention, a?. Since the wire is perfectly-conducting, the only voltage appears across the resistor, and is g

24、iven as VR. Faradays law becomes I E dL = VR= ?d?m dt = ? d dt Z s B0taz azda = ?a2B0 This indicates that the resistor voltage, VR= a2B0, has polarity such that the positive terminal is at point a in the fi gure, while the negative terminal is at point b. Current fl ows in the clockwise direction, a

25、nd is given in magnitude by I = a2B0/R. 9.9. A square fi lamentary loop of wire is 25 cm on a side and has a resistance of 125 per meter length. The loop lies in the z = 0 plane with its corners at (0,0,0), (0.25,0,0), (0.25,0.25,0), and (0,0.25,0) at t = 0. The loop is moving with velocity vy = 50

26、m/s in the fi eld Bz= 8cos(1.5108t?0.5x) T. Develop a function of time which expresses the ohmic power being delivered to the loop: First, since the fi eld does not vary with y, the loop motion in the y direction does not produce any time-varying fl ux, and so this motion is immaterial. We can evalu

27、ate the fl ux at the original loop position to obtain: ?(t) = Z .25 0 Z .25 0 8 10?6cos(1.5 108t ? 0.5x)dxdy = ?(4 10?6) sin(1.5 108t ? 0.13) ? sin(1.5 108t) Wb Now, emf = V (t) = ?d?/dt = 6.0 102 cos(1.5 108t ? 0.13) ? cos(1.5 108t), The total loop resistance is R = 125(0.25 + 0.25 + 0.25 + 0.25) =

28、 125. Then the ohmic power is P(t) = V 2(t) R = 2.9 103 cos(1.5 108t ? 0.13) ? cos(1.5 108t)2 Watts 167 9.10 a) Show that the ratio of the amplitudes of the conduction current density and the displacement current density is ?/! for the applied fi eld E = Emcos!t. Assume = 0. First, D = E = Emcos!t.

29、Then the displacement current density is D/t = ?!Emsin!t. Second, Jc= ?E = ?Em cos!t. Using these results we fi nd |Jc|/|Jd| = ?/!. b) What is the amplitude ratio if the applied fi eld is E = Eme?t/, where is real? As before, fi nd D = E = Eme?t/, and so Jd= D/t = ?(/)Eme?t/. Also, Jc= ?Eme?t/. Fina

30、lly, |Jc|/|Jd| = ?/. 9.11. Let the internal dimension of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. The homogeneous material inside the capacitor has the parameters = 10?11F/m, = 10?5 H/m, and ? = 10?5 S/m. If the electric fi eld intensity is E = (106/)cos(105t)a V/m, fi nd: a) J: U

31、se J = ?E = 10 cos(105t)aA/m2 b) the total conduction current, Ic, through the capacitor: Have Ic= Z Z J dS = 2lJ = 20lcos(105t) = 8cos(105t) A c) the total displacement current, Id , through the capacitor: First fi nd Jd= D t = t(E) = ? (105)(10?11)(106) sin(105t)a= ? 1 sin(105t) A/m Now Id= 2lJd=

32、?2lsin(105t) = ?0.8sin(105t) A d) the ratio of the amplitude of Idto that of Ic, the quality factor of the capacitor: This will be |Id| |Ic| = 0.8 8 = 0.1 168 9.12. Find the displacement current density associated with the magnetic fi eld (assume zero con- duction current): H = A1sin(4x)cos(!t ? ?z)

33、ax+ A2cos(4x)sin(!t ? ?z)az The displacement current density is given by D t = r H = (4A2+ ?A1)sin(4x)cos(!t ? ?z)ayA/m2 9.13. Consider the region defi ned by |x|, |y|, and |z| a, where the region between them is a material for which = r0, = 0, and ? = 0. Find the total displacement current through

34、the dielectric and compare it with the source current as determined from the capacitance (Sec. 6.3) and circuit analysis methods: First, solving Laplaces equation, we fi nd the voltage between spheres (see Eq. 39, Chapter 6): V (t) = (1/r) ? (1/b) (1/a) ? (1/b)V0 sin!t Then E = ?rV = V0sin!t r2(1/a

35、? 1/b)ar )D = r0V0sin!t r2(1/a ? 1/b)ar Now Jd= D t = r0!V0cos!t r2(1/a ? 1/b) ar The displacement current is then Id= 4r2Jd= 4r0!V0cos!t (1/a ? 1/b) = C dV dt where, from Eq. 6, Chapter 6, C = 4r0 (1/a ? 1/b) 9.15. Let = 310?5H/m, = 1.210?10F/m, and ? = 0 everywhere. If H = 2cos(1010t?x)az A/m, use

36、 Maxwells equations to obtain expressions for B, D, E, and ?: First, B = H = 6 10?5cos(1010t ? ?x)azT. Next we use r H = ?H x ay= 2? sin(1010t ? ?x)ay= D t from which D = Z 2? sin(1010t ? ?x)dt + C = ? 2? 1010 cos(1010t ? ?x)ayC/m2 where the integration constant is set to zero, since no dc fi elds a

37、re presumed to exist. Next, E = D = ? 2? (1.2 10?10)(1010) cos(1010t ? ?x)ay= ?1.67? cos(1010t ? ?x)ayV/m Now r E = Ey x az= 1.67?2sin(1010t ? ?x)az= ?B t So B = ? Z 1.67?2sin(1010t ? ?x)azdt = (1.67 10?10)?2cos(1010t ? ?x)az We require this result to be consistent with the expression for B original

38、ly found. So (1.67 10?10)?2= 6 10?5)? = 600 rad/m 170 9.16. Derive the continuity equation from Maxwells equations: First, take the divergence of both sides of Amperes circuital law: r r H |z 0 = r J + tr D = r J + v t = 0 where we have used r D = v, another Maxwell equation. 9.17. The electric fi e

39、ld intensity in the region 0 x 5, 0 y /12, 0 z 0: Take r H = ?Hy z ax= ?5? sin(109t ? ?z)ax= 200 E t So E = Z ?5? 200 sin(109t ? ?z)axdt = ? (4 109)0 cos(109t ? ?z)ax Then r E = Ex z ay= ?2 (4 109)0 sin(109t ? ?z)ay= ?0 H t So that H = Z ?2 (4 109)00 sin(109t ? ?z)axdt = ?2 (4 1018)00 cos(109t ? ?z)

40、 = 5cos(109t ? ?z)ay where the last equality is required to maintain consistency. Therefore ?2 (4 1018)00 = 5)? = 14.9 m?1 b) the displacement current density at z = 0: Since ? = 0, we have r H = Jd= ?5? sin(109t ? ?z) = ?74.5sin(109t ? 14.9z)ax = ?74.5sin(109t)axA/m at z = 0 c) the total displaceme

41、nt current crossing the surface x = 0.5d, 0 y b, and 0 z 0.1 m in the ax direction. We evaluate the fl ux integral of Jdover the given cross section: Id= ?74.5b Z 0.1 0 sin(109t ? 14.9z)ax axdz = 0.20 cos(109t ? 1.49) ? cos(109t) A 9.19. In the fi rst section of this chapter, Faradays law was used t

42、o show that the fi eld E = ?1 2kB0e kta? results from the changing magnetic fi eld B = B0ektaz. a) Show that these fi elds do not satisfy Maxwells other curl equation: Note that B as stated is constant with position, and so will have zero curl. The electric fi eld, however, varies with time, and so

43、rH = D t would have a zero left-hand side and a non-zero right-hand side. The equation is thus not valid with these fi elds. b) If we let B0= 1 T and k = 106s?1 , we are establishing a fairly large magnetic fl ux density in 1 s. Use the r H equation to show that the rate at which Bzshould (but does

44、not) change with is only about 5 10?6T/m in free space at t = 0: Assuming that B varies with , we write r H = ?Hz a?= ? 1 0 dB0 d ekt= 0 E t = ?1 20k 2B0ekt Thus dB0 d = 1 200k 2B0 = 1012(1) 2(3 108)2 = 5.6 10?6 which is near the stated value if is on the order of 1m. 172 9.20. Given Maxwells equati

45、ons in point form, assume that all fi elds vary as estand write the equations without explicitly involving time: Write all fi elds in the general form A(r,t) = A0(r)est, where r is a position vector in any coordinate system. Maxwells equations become: r E0(r)est= ? t ?B 0(r)est ? = ?sB0(r)est r H0(r

46、)est= J0(r)est+ t ?D 0(r)est ? = J0(r)est+ sD0(r)est r D0(r)est= 0(r)est r B0(r)est= 0 In all cases, the estterms divide out, leaving: r E0(r) = ?sB0(r) r H0(r) = J0(r) + sD0(r) r D0(r) = 0(r) r B0(r) = 0 9.21. a) Show that under static fi eld conditions, Eq. (55) reduces to Amperes circuital law. F

47、irst use the defi nition of the vector Laplacian: r2A = ?r r A + r(r A) = ?J which is Eq. (55) with the time derivative set to zero. We also note that r A = 0 in steady state (from Eq. (54). Now, since B = r A, (55) becomes ?r B = ?J)r H = J b) Show that Eq. (51) becomes Faradays law when taking the

48、 curl: Doing this gives r E = ?r rV ? tr A The curl of the gradient is identially zero, and r A = B. We are left with r E = ?B/t 173 9.22. In a sourceless medium, in which J = 0 and v= 0, assume a rectangular coordinate system in which E and H are functions only of z and t. The medium has permittivi

49、ty and permeability . a) If E = Exaxand H = Hyay, begin with Maxwells equations and determine the second order partial dierential equation that Exmust satisfy. First use r E = ?B t ) Ex z ay= ?Hy t ay in which case, the curl has dictated the direction that H must lie in. Similarly, use the other Max

50、well curl equation to fi nd r H = D t )?Hy z ax= Ex t ax Now, dierentiate the fi rst equation with respect to z, and the second equation with respect to t: 2Ex z2 = ? 2Hy tz and 2Hy zt = ? 2Ex t2 Combining these two, we fi nd 2Ex z2 = 2Ex t2 b) Show that Ex= E0cos(!t?z) is a solution of that equatio

51、n for a particular value of ?: Substituting, we fi nd 2Ex z2 = ?2E0cos(!t ? ?z) and 2Ex t2 = ?!2E0cos(!t ? ?z) These two will be equal provided the constant multipliers of cos(!t ? ?z) are equal. c) Find ? as a function of given parameters. Equating the two constants in part b, we fi nd ? = !p. 174

52、9.23. In region 1, z 0, 2= 1/2, 2= 21, and ?2= ?1/4. It is known that E1= (30ax+ 20ay+ 10az)cos(109t) V/m at P1(0,0,0?). a) Find EN1, Et1, DN1, and Dt1: These will be EN1= 10cos(109t)azV/mEt1= (30ax+ 20ay)cos(109t) V/m DN1= 1EN1= (2 10?11)(10)cos(109t)azC/m2= 200cos(109t)azpC/m2 Dt1= 1Et1= (2 10?11)

53、(30ax+ 20ay)cos(109t) = (600ax+ 400ay)cos(109t) pC/m2 b) Find JN1and Jt1at P1: JN1= ?1EN1= (4 10?3)(10cos(109t)az= 40cos(109t)azmA/m2 Jt1= ?1Et1= (4 10?3)(30ax+ 20ay)cos(109t) = (120ax+ 80ay)cos(109t) mA/m2 c) Find Et2, Dt2, and Jt2at P1: By continuity of tangential E, Et2= Et1= (30ax+ 20ay)cos(109t

54、) V/m Then Dt2= 2Et2= (10?11)(30ax+ 20ay)cos(109t) = (300ax+ 200ay)cos(109t) pC/m2 Jt2= ?2Et2= (10?3)(30ax+ 20ay)cos(109t) = (30ax+ 20ay)cos(109t) mA/m2 d) (Harder) Use the continuity equation to help show that JN1?JN2= DN2/t?DN1/t and then determine EN2, DN2, and JN2: We assume the existence of a s

55、urface charge layer at the boundary having density sC/m2. If we draw a cylindrical “pillbox” whose top and bottom surfaces (each of area ?a) are on either side of the interface, we may use the continuity condition to write (JN2? JN1)?a = ?s t ?a where s= DN2? DN1. Therefore, JN1? JN2= t(DN2 ? DN1) I

56、n terms of the normal electric fi eld components, this becomes ?1EN1? ?2EN2= t(2EN2 ? 1EN1) Now let EN2= Acos(109t) + B sin(109t), while from before, EN1= 10cos(109t). 175 9.23d(continued) These, along with the permittivities and conductivities, are substituted to obtain (4 10?3)(10)cos(109t) ? 10?3

57、Acos(109t) + B sin(109t) = t 10?11Acos(109t) + B sin(109t) ? (2 10?11)(10)cos(109t) = ?(10?2Asin(109t) + 10?2B cos(109t) + (2 10?1)sin(109t) We now equate coe?cients of the sin and cos terms to obtain two equations: 4 10?2? 10?3A = 10?2B ?10?3B = ?10?2A + 2 10?1 These are solved together to fi nd A

58、= 20.2 and B = 2.0. Thus EN2= 20.2cos(109t) + 2.0sin(109t) az= 20.3cos(109t + 5.6?)azV/m Then DN2= 2EN2= 203cos(109t + 5.6?)azpC/m2 and JN2= ?2EN2= 20.3cos(109t + 5.6?)azmA/m2 176 9.24. A vector potential is given as A = A0cos(!t ? kz)ay. a) Assuming as many components as possible are zero, fi nd H,

59、 E, and V ; With A y-directed only, and varying spatially only with z, we fi nd H = 1 r A = ? 1 Ay z ax= ?kA0 sin(!t ? kz)axA/m Now, in a lossless medium we will have zero conductivity, so that the point form of Amperes circuital law involves only the displacement current term: r H = D t = E t Using the magnetic fi eld as found above, we fi nd r H = Hx z ay= k2A0 cos(!t ? kz)ay= E t ) E = k2A0 ! sin(!t ? kz)ayV/m No