(12a x +14a y )= 8.5a x 9.9a y A/m

Transcription

1 Chapter 11 Odd-Numbered Show that E xs Ae jk 0z+φ is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (16), for k 0 ω µ 0 ɛ 0 and any φ and A:We take d dz Aejk 0z+φ (jk 0 ) Ae jk 0z+φ k 0E xs An H field in free space is given as H(x, t) 10 cos(10 8 t βx)a y A/m. Find a) β:since we have a uniform plane wave, β ω/c, where we identify ω 10 8 sec 1.Thus β 10 8 /( )0.33 rad/m. b) λ:we know λ π/β 18.9 m. c) E(x, t)atp (0.1, 0., 0.3) at t 1 ns:use E(x, t) η 0 H(x, t) (377)(10) cos(10 8 t βx) cos(10 8 t βx). Then the vector direction of E will be a z, since we require that S E H, where S is x-directed. At the given point, the relevant coordinate is x 0.1. Using this, along with t 10 9 sec, we finally obtain E(x, t) cos[(10 8 )(10 9 ) (0.33)(0.1)] cos( ) V/m A 150-MHz uniform plane wave in free space is described by H s (4+j10)(a x + ja y )e jβz A/m. a) Find numerical values for ω, λ, and β:first, ω π π 10 8 sec 1. Second, for a uniform plane wave in free space, λ πc/ω c/f ( )/( )m. Third, β π/λ π rad/m. b) Find H(z,t) att 1.5 ns, z 0 cm:use H(z,t) Re{H s e jωt } Re{(4 + j10)(a x + ja y )(cos(ωt βz)+jsin(ωt βz)} [8 cos(ωt βz) 0 sin(ωt βz)] a x [10 cos(ωt βz) + 4 sin(ωt βz)] a y. Now at the given position and time, ωt βz (3π 10 8 )( ) π(0.0) π/4. And cos(π/4) sin(π/4) 1/. So finally, H(z 0cm,t1.5ns) 1 (1a x +14a y ) 8.5a x 9.9a y A/m c) What is E max?have E max η 0 H max, where H max H s H s [4(4 + j10)(4 j10)+(j)( j)(4 + j10)(4 j10)] 1/ 4.1A/m Then E max 377(4.1) 9.08 kv/m. 119

2 11.7. The phasor magnetic field intensity for a 400-MHz uniform plane wave propagating in a certain lossless material is (a y j5a z )e j5x A/m. Knowing that the maximum amplitude of E is 1500 V/m, find β, η, λ, v p, ɛ R, µ R, and H(x, y, z, t):first, from the phasor expression, we identify β 5m 1 from the argument of the exponential function. Next, we evaluate H 0 H H H Then η E 0 /H / Ω. Then λ π/β π/5.5m5cm. Next, v p ω β π m/s Now we note that And µr η µ R ɛ R ɛ R v p c µr ɛ R µ R ɛ R 8.79 We solve the above two equations simultaneously to find ɛ R 4.01 and µ R.19. Finally, H(x, y, z, t) Re { (a y j5a z )e j5x e jωt} cos(π t 5x)a y + 5 sin(π t 5x)a z cos(8π 10 8 t 5x)a y + 5 sin(8π 10 8 t 5x)a z A/m A certain lossless material has µ R 4 and ɛ R 9. A 10-MHz uniform plane wave is propagating in the a y direction with E x0 400 V/m and E y0 E z0 0atP (0.6, 0.6, 0.6) at t 60 ns. a) Find β, λ, v p, and η:for a uniform plane wave, β ω µɛ ω µr ɛ R c π (4)(9) 0.4π rad/m Then λ (π)/β (π)/(0.4π) 5m. Next, v p ω β π 107 4π m/s Finally, η µ ɛ η µr ɛ R 9 51 Ω b) Find E(t) (at P ):We are given the amplitude at t 60 ns and at y 0.6 m. Let the maximum amplitude be E max, so that in general, E x E max cos(ωt βy). At the given position and time, E x 400 E max cos[(π 10 7 )( ) (4π 10 1 )(0.6)] E max cos(0.96π) 0.99E max So E max (400)/( 0.99) 403 V/m. Thus at P, E(t) 403 cos(π 10 7 t)v/m. 10

3 11.9c. Find H(t):First, we note that if E at a given instant points in the negative x direction, while the wave propagates in the forward y direction, then H at that same position and time must point in the positive z direction. Since we have a lossless homogeneous medium, η is real, and we are allowed to write H(t) E(t)/η, where η is treated as negative and real. Thus H(t) H z (t) E x(t) η cos(π 10 7 t)1.61 cos(π 10 7 t)a/m A -GHz uniform plane wave has an amplitude of E y0 1.4 kv/m at (0, 0, 0,t 0) and is propagating in the a z direction in a medium where ɛ F/m, ɛ F/m, and µ.5 µh/m. Find: a) E y at P (0, 0, 1.8cm) at 0. ns:to begin, we have the ratio, ɛ /ɛ 1.6/ So 1/ ( ) µɛ α ω ɛ 1+ ɛ 1 (.5 10 (π )( ) 11 ) [ ] 1/ 1+(.533) 1 8.1Np/m Then 1/ ( ) µɛ β ω ɛ 1+ ɛ rad/m Thus in general, E y (z,t) 1.4e 8.1z cos(4π 10 9 t 11z) kv/m Evaluating this at t 0. nsandz 1.8 cm, find E y (1.8cm, 0.ns)0.74 kv/m b) H x at P at 0. ns:we use the phasor relation, H xs E ys /η where η µ ɛ 1 1 j(ɛ /ɛ ) j(.533) 63 + j Ω So now H xs E ys η ( )e 8.1z e j11z 71e j e 8.1z e j11z e j14 A/m Then H x (z,t) 5.16e 8.1z cos(4π 10 9 t 11z 14 ) This, when evaluated at t 0. ns and z 1.8 cm, yields H x (1.8cm, 0. ns) 3.0A/m 11

4 Let jk 0.+j1.5m 1 and η j60 Ω for a uniform plane wave propagating in the a z direction. If ω 300 Mrad/s, find µ, ɛ, and ɛ :We begin with µ 1 η ɛ j60 1 j(ɛ /ɛ ) and Then jk jω µɛ 1 j(ɛ /ɛ )0.+j1.5 ηη µ ɛ 1 1+(ɛ /ɛ ) (450 + j60)(450 j60) (1) and (jk)(jk) ω µɛ 1+(ɛ /ɛ ) (0.+j1.5)(0. j1.5).9 () Taking the ratio of () to (1), (jk)(jk) Then with ω , ηη ω (ɛ ) ( 1+(ɛ /ɛ ) ) (ɛ ) ( ) (1+(ɛ /ɛ ) ) (1+(ɛ /ɛ ) ) (3) Now, we use Eqs. (35) and (36). Squaring these and taking their ratio gives α 1+(ɛ β /ɛ ) 1+(ɛ /ɛ ) (0.) (1.5) We solve this to find ɛ /ɛ Substituting this result into (3) gives ɛ F/m. Since ɛ /ɛ 0.71, we then find ɛ F/m. Finally, using these results in either (1) or () we find µ H/m. Summary: µ H/m, ɛ F/m, and ɛ F/m A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region. Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a non-magnetic material for which a) ɛ R 1 and ɛ R 0:In a non-magnetic material, we would have: α ω µ0 ɛ 0 ɛ R ( ) ɛ 1+ 1 R ɛ R 1/ and β ω µ0 ɛ 0 ɛ R ( ) ɛ R ɛ R 1/ With the given values of ɛ R and ɛ R, it is clear that β ω µ 0 ɛ 0 ω/c, and so λ π/β πc/ω / cm. It is also clear that α 0. 1

5 (continued) b) ɛ R 1.04 and ɛ R :In this case ɛ R /ɛ R << 1, and so β. ω ɛ R /c.13 cm 1.Thusλ π/β.95 cm. Then α. ωɛ µ ɛ ωɛ R Np/m µ0 ɛ 0 ɛ R ω c ɛ R ɛ R π 1010 ( ) c) ɛ R.5 and ɛ R 7.:Using the above formulas, we obtain β π ( ) and so λ π/β 1.33 cm. Then α π ( ) ( 7..5 ( 7..5 ) +1 ) 1 1/ 1/ 4.71 cm Np/m Let η 50 + j30 Ω and jk 0.+jm 1 for a uniform plane wave propagating in the a z direction in a dielectric having some finite conductivity. If E s 400 V/m at z 0, find: a) P z,av at z 0 and z 60 cm:assume x-polarization for the electric field. Then P z,av 1 Re {E s H s} 1 { Re 400e αz e jβz a x 400 } 1 { } 1 (400) e αz Re η a z e (0.)z Re 315 e (0.)z a z W/m η e αz e jβz a y { 1 50 j30 Evaluating at z 0, obtain P z,av (z 0) 315 a z W/m, and at z 60 cm, P z,av (z 0.6) 315e (0.)(0.6) a z 48 a z W/m. } a z b) the average ohmic power dissipation in watts per cubic meter at z 60 cm:at this point a flaw becomes evident in the problem statement, since solving this part in two different ways gives results that are not the same. I will demonstrate:in the first method, we use Poynting s theorem in point form (first equation at the top of p. 366), which we modify for the case of time-average fields to read: P z,av < J E > where the right hand side is the average power dissipation per volume. Note that the additional right-hand-side terms in Poynting s theorem that describe changes in energy stored in the fields will both be zero in steady state. We apply our equation to the result of part a: < J E > P z,av d dz 315 e (0.)z (0.4)(315)e (0.)z 16e 0.4z W/m 3 13

6 At z 60 cm, this becomes < J E > 99.1 W/m 3. In the second method, we solve for the conductivity and evaluate < J E > σ<e >. We use and We take the ratio, Identifying σ ωɛ,wefind { } jk σ Re η jk jω µɛ 1 j(ɛ /ɛ ) η [ jk η jωɛ 1 j Now we find the dissipated power per volume: µ ɛ 1 1 j(ɛ /ɛ ) ( )] ɛ jωɛ + ωɛ ɛ { } 0.+j Re S/m 50 + j30 σ<e > ( 1 ) (400e 0.z ) At z 60 cm, this evaluates as 109 W/m 3. One can show that consistency between the two methods requires that { } 1 Re η σ α This relation does not hold using the numbers as given in the problem statement and the value of σ found above. Note that in Problem 11.13, where all values are worked out, the relation does hold and consistent results are obtained using both methods Perfectly-conducting cylinders with radii of 8 mm and 0 mm are coaxial. The region between the cylinders is filled with a perfect dielectric for which ɛ 10 9 /4π F/m and µ R 1. IfE in this region is (500/) cos(ωt 4z)a V/m, find: a) ω, with the help of Maxwell s equations in cylindrical coordinates:we use the two curl equations, beginning with E B/ t, where in this case, So Then E E z a φ 000 sin(ωt 4z)a φ B φ t a φ 000 B φ H φ B φ µ 0 sin(ωt 4z)dt 000 ω We next use H D/ t, where in this case cos(ωt 4z) T 000 (4π 10 7 cos(ωt 4z) A/m )ω H H φ z a + 1 (H φ ) a z 14

7 11.19a. (continued) where the second term on the right hand side becomes zero when substituting our H φ. So H H φ z a 8000 (4π 10 7 )ω sin(ωt 4z)a D t a And 8000 D (4π 10 7 )ω sin(ωt 4z)dt 8000 (4π 10 7 )ω cos(ωt 4z) C/m Finally, using the given ɛ, E D ɛ 8000 (10 16 )ω cos(ωt 4z) V/m This must be the same as the given field, so we require 8000 (10 16 )ω 500 ω rad/s b) H(, z, t):from part a, wehave H(, z, t) 000 (4π 10 7 )ω cos(ωt 4z)a φ 4.0 cos(4 108 t 4z)a φ A/m c) P(, φ, z):this will be P(, φ, z) E H 500 cos(4 108 t 4z)a 4.0 cos(4 108 t 4z)a φ cos ( t 4z)a z W/m d) the average power passing through every cross-section 8 <<0 mm, 0 <φ<π. Using the result of part c, we find P avg ( )/ a z W/m. The power through the given cross-section is now P π ddφπ 10 3 ln ( ) 0 5.7kW The cylindrical shell, 1 cm 1. cm, is composed of a conducting material for which σ 10 6 S/m. The external and internal regions are non-conducting. Let H φ 000 A/m at 1. cm. a) Find H everywhere:use Ampere s circuital law, which states: H dl π(000) π(1. 10 )(000) 48π AI encl 15

8 11.1. (continued) Then in this case J I Area a z 48 ( ) 10 4 a z a z A/m With this result we again use Ampere s circuital law to find H everywhere within the shell as a function of (in meters): H φ1 () 1 π π 0.01 Outside the shell, we would have ddφ 54.5 (104 1) A/m (.01 <<.01) H φ () 48π 4/ A/m (>.01) π Inside the shell ( <.01 m), H φ 0 since there is no enclosed current. b) Find E everywhere:we use E J σ a z 1.09 a z V/m which is valid, presumeably, outside as well as inside the shell. c) Find P everywhere:use Outside the shell, P E H 1.09 a z 54.5 (104 1) a φ 59.4 (104 1) a W/m (.01 <<.01 m) P 1.09 a z 4 a φ 6 a W/m (>.01 m) A hollow tubular conductor is constructed from a type of brass having a conductivity of S/m. The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter length at a frequency of a) dc:in this case the current density is uniform over the entire tube cross-section. We write: R(dc) L σa 1 ( )π( ) Ω/m b) 0 MHz:Now the skin effect will limit the effective cross-section. At 0 MHz, the skin depth is δ(0mhz) [πfµ 0 σ] 1/ [π( )(4π 10 7 )( )] 1/ m 16

9 11.3. (continued) This is much less than the outer radius of the tube. Therefore we can approximate the resistance using the formula: R(0MHz) L σa 1 πbδ 1 ( )(π(.01))( ) Ω/m c) GHz:Using the same formula as in part b, we find the skin depth at GHz to be δ m. The resistance (using the other formula) is R(GHz) Ω/m A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0.3 mm and a velocity of m/s. Assuming the conductor is non-magnetic, determine the frequency and the conductivity:first, we use Next, for a good conductor, δ λ π 1 πfµσ σ 4π λ fµ f v λ Hz 1 GHz 4π ( )(10 9 )(4π 10 7 ) S/m The planar surface at z 0 is a brass-teflon interface. Use data available in Appendix C to evaluate the following ratios for a uniform plane wave having ω rad/s: a) α Tef /α brass :From the appendix we find ɛ /ɛ.0003 for Teflon, making the material a good dielectric. Also, for Teflon, ɛ R.1. For brass, we find σ S/m, making brass a good conductor at the stated frequency. For a good dielectric (Teflon) we use the approximations: α. σ µ ɛ ( )( ɛ 1 ɛ ( ɛ β. ω µɛ [ ) ω µɛ 1 ɛ ( ɛ ɛ ) ω ɛ R c )]. ω ω µɛ ɛ R c For brass (good conductor) we have α. β. ( ) 1 πfµσ brass π (4 10 π 10 )(4π 10 7 )( ) m 1 Now α Tef 1/(ɛ /ɛ )(ω/c) α brass ɛ R (1/)(.0003)( / ).1 πfµσbrass b) λ Tef (π/β Tef ) λ brass (π/β brass ) β brass β Tef c πfµσbrass ω ɛ R Tef (3 108 )( ) ( )

10 11.7. (continued) c) v Tef (ω/β Tef ) v brass (ω/β brass ) β brass as before β Tef Consider a left-circularly polarized wave in free space that propagates in the forward z direction. The electric field is given by the appropriate form of Eq. (80). a) Determine the magnetic field phasor, H s : We begin, using (80), with E s E 0 (a x + ja y )e jβz. We find the two components of H s separately, using the two components of E s. Specifically, the x component of E s is associated with a y component of H s, and the y component of E s is associated with a negative x component of H s. The result is H s E 0 η 0 (a y ja x ) e jβz b) Determine an expression for the average power density in the wave in W/m by direct application of Eq. (57):We have P z,avg 1 Re(E s H s) 1 ( Re E 0 (a x + ja y )e jβz E ) 0 (a y ja x )e +jβz η 0 E 0 η 0 a z W/m (assuming E 0 is real) A linearly-polarized uniform plane wave, propagating in the forward z direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along y (ɛ Ry ) differs from that seen by waves polarized along x (ɛ Rx ). Suppose ɛ Rx.15, ɛ Ry.10, and the wave electric field at input is polarized at 45 to the positive x and y axes. Assume free space wavelength λ. a) Determine the shortest length of the material such that the wave as it emerges from the output end is circularly polarized:with the input field at 45, the x and y components are of equal magnitude, and circular polarization will result if the phase difference between the components is π/. Our requirement over length L is thus β x L β y L π/, or L With the given values, we find, π (β x β y ) πc ω( ɛ Rx ɛ Ry ) L (58.3)πc ω 58.3 λ λ b) Will the output wave be right- or left-circularly-polarized? With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the output. The field can thus be written as E E 0 (a y ja x ), which is left circular polarization. 18

11 Given a wave for which E s 15e jβz a x +18e jβz e jφ a y V/m, propagating in a medium characterized by complex intrinsic impedance, η. a) Find H s :With the wave propagating in the forward z direction, we find: H s 1 [ 18e jφ ] a x +15a y e jβz A/m η b) Determine the average power density in W/m :We find P z,avg 1 Re {E s H s} 1 Re { (15) } η + (18) η { } 1 75 Re η W/m 19