工程电磁场第八版课后答案第09章.pdf

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?2cos120t 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. Assume 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(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 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 high-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. 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!tay 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 think 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 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 the 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)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 position 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) Now 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 constant speed of 9 m/s in a uniform magnetic fi eld of 0.8 T. Find I(t), 0 a, where the region between them is a material for which = r0, = 0, and ? = 0. Find the total displacement current through 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 ? 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 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 are 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 originally 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 eld intensity in the region 0 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) = 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 displacement current crossing the surface x = 0.5d, 0 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)(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) 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 surface 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) In 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?3Acos(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 = 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, 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 Now, E = ?rV ? A t ) rV = ? A t + E ? or rV = A0! 1 ? k2 !2 ? sin(!t ? kz)ay= V y ay Integrating over y we fi nd V = A0!y 1 ? k2 !2 ? sin(!t ? kz) + C where C, the integration constant, can be taken as zero. In part b, it will be shown that k = !p, which means that V = 0. b) Specify k in terms of A0, !, and the constants of the lossless medium, and . Use the other Maxwell curl equation: r E = ?B t = ?H t so that H t = ? 1 r E = 1 Ey z ax= ?k 3A0 !2 cos(!t ? kz)ax Integrate over t (and set the integration constant to zero) and require the result to be consistant with part a: H = ? k3A0 !22 sin(!t ? kz)ax= ?kA0 sin(!t ? kz)ax |z from part a We identify k = !p 177 9.25. In a region where r= r= 1 and ? = 0, the retarded potentials are given by V = x(z ? ct) V and A = x(z/c) ? tazWb/m, where c = 1/p00. a) Show that r A = ?(V/t): First, r A = Az z = x c = xp00 Second, V t = ?cx = ? x p 00 so we observe that rA = ?00(V/t) in free space, implyin