hw 3 solutions

hw 3 solutions - CHAPTER 11 11.1 Show that Exs = Aej k0 z...

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Unformatted text preview: CHAPTER 11 11.1. Show that Exs = Aej k0 z+φ is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (16), for √ k0 = ω µ0 0 and any φ and A: We take d2 2 Aej k0 z+φ = (j k0 )2 Aej k0 z+φ = −k0 Exs dz2 11.2. Let E(z, t) = 200 sin 0.2z cos 108 t ax + 500 cos(0.2z + 50◦ ) sin 108 t ay V/m. Find: a) E at P (0, 2, 0.6) at t = 25 ns: Obtain EP (t = 25) = 200 sin [(0.2)(0.6)] cos(2.5)ax + 500 cos [(0.2)(0.6) + 50(2π)/360] sin(2.5)ay = −19.2ax + 164ay V/m b) |E| at P at t = 20 ns: EP (t = 20) = 200 sin [(0.2)(0.6)] cos(2.0)ax + 500 cos [(0.2)(0.6) + 50(2π)/360] sin(2.0)ay = −9.96ax + 248ay V/m Thus |EP | = (9.96)2 + (248)2 = 249 V/m. c) Es at P : Es = 200 sin 0.2zax − j 500 cos(0.2z + 50◦ )ay . Thus EsP = 200 sin [(0.2)(0.6)] ax − j 500 cos [(0.2)(0.6) + 2π(50)/360] ay = 23.9ax − j 273ay V/m 11.3. An H field in free space is given as H(x, t) = 10 cos(108 t − βx)ay A/m. Find a) β : Since we have a uniform plane wave, β = ω/c, where we identify ω = 108 sec−1 . Thus β = 108 /(3 × 108 ) = 0.33 rad/m. b) λ: We know λ = 2π/β = 18.9 m. c) E(x, t) at P (0.1, 0.2, 0.3) at t = 1 ns: Use E(x, t) = −η0 H (x, t) = −(377)(10) cos(108 t − βx) = −3.77 × 103 cos(108 t − βx). The vector direction of E will be −az , 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) = −3.77 × 103 cos[(108 )(10−9 ) − (0.33)(0.1)]az = −3.77 × 103 cos(6.7 × 10−2 )az = −3.76 × 103 az V/m 11.4. In phasor form, the electric field intensity of a uniform plane wave in free space is expressed as Es = (40 − j 30)e−j 20z ax V/m. Find: a) ω: From the given expression, we identify β = 20 rad/m. Then ω = cβ = (3 × 108 )(20) = 6.0 × 109 rad/s. b) β = 20 rad/m from part a . 182 CHAPTER 11 11.1. Show that Exs = Aej k0 z+φ is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (16), for √ k0 = ω µ0 0 and any φ and A: We take d2 2 Aej k0 z+φ = (j k0 )2 Aej k0 z+φ = −k0 Exs dz2 11.2. Let E(z, t) = 200 sin 0.2z cos 108 t ax + 500 cos(0.2z + 50◦ ) sin 108 t ay V/m. Find: a) E at P (0, 2, 0.6) at t = 25 ns: Obtain EP (t = 25) = 200 sin [(0.2)(0.6)] cos(2.5)ax + 500 cos [(0.2)(0.6) + 50(2π)/360] sin(2.5)ay = −19.2ax + 164ay V/m b) |E| at P at t = 20 ns: EP (t = 20) = 200 sin [(0.2)(0.6)] cos(2.0)ax + 500 cos [(0.2)(0.6) + 50(2π)/360] sin(2.0)ay = −9.96ax + 248ay V/m Thus |EP | = (9.96)2 + (248)2 = 249 V/m. c) Es at P : Es = 200 sin 0.2zax − j 500 cos(0.2z + 50◦ )ay . Thus EsP = 200 sin [(0.2)(0.6)] ax − j 500 cos [(0.2)(0.6) + 2π(50)/360] ay = 23.9ax − j 273ay V/m 11.3. An H field in free space is given as H(x, t) = 10 cos(108 t − βx)ay A/m. Find a) β : Since we have a uniform plane wave, β = ω/c, where we identify ω = 108 sec−1 . Thus β = 108 /(3 × 108 ) = 0.33 rad/m. b) λ: We know λ = 2π/β = 18.9 m. c) E(x, t) at P (0.1, 0.2, 0.3) at t = 1 ns: Use E(x, t) = −η0 H (x, t) = −(377)(10) cos(108 t − βx) = −3.77 × 103 cos(108 t − βx). The vector direction of E will be −az , 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) = −3.77 × 103 cos[(108 )(10−9 ) − (0.33)(0.1)]az = −3.77 × 103 cos(6.7 × 10−2 )az = −3.76 × 103 az V/m 11.4. In phasor form, the electric field intensity of a uniform plane wave in free space is expressed as Es = (40 − j 30)e−j 20z ax V/m. Find: a) ω: From the given expression, we identify β = 20 rad/m. Then ω = cβ = (3 × 108 )(20) = 6.0 × 109 rad/s. b) β = 20 rad/m from part a . 182 11.6. Let µR = R = 1 for the field E(z, t) = (25ax − 30ay ) cos(ωt − 50z) V/m. a) Find ω: ω = cβ = (3 × 108 )(50) = 15.0 × 109 s−1 . b) Determine the displacement current density, Jd (z, t): ∂D = − 0 ω(25ax − 30ay ) sin(ωt − 50z) ∂t = (−3.32ax + 3.98ay ) sin(1.5 × 1010 t − 50z) A/m2 Jd (z, t) = c) Find the total magnetic flux passing through the rectangle defined by 0 < x < 1, y = 0, 0 < z < 1, at t = 0: In free space, the magnetic field of the uniform plane wave can be easily found using the intrinsic impedance: 30 25 ay + ax cos(ωt − 50z) A/m η0 η0 H(z, t) = Then B(z, t) = µ0 H(z, t) = (1/c)(25ay + 30ax ) cos(ωt − 50z) Wb/m2 , where µ0 /η0 = √ µ0 0 = 1/c. The flux at t = 0 is now 1 = 0 1 1 B · ay dx dz = 0 0 25 25 cos(50z) dz = sin(50) = −0.44 nWb c 50(3 × 108 ) 11.7. The phasor magnetic field intensity for a 400-MHz uniform plane wave propagating in a certain lossless material is (2ay − j 5az )e−j 25x A/m. Knowing that the maximum amplitude of E is 1500 V/m, find β , η, λ, vp , R , µR , and H(x, y, z, t): First, from the phasor expression, we identify β = 25 m−1 from the √ √ √ argument of the exponential function. Next, we evaluate H0 = |H| = H · H∗ = 22 + 52 = 29. √ Then η = E0 /H0 = 1500/ 29 = 278.5 . Then λ = 2π/β = 2π/25 = .25 m = 25 cm. Next, vp = ω 2π × 400 × 106 = = 1.01 × 108 m/s β 25 Now we note that η = 278.5 = 377 And µR ⇒ R c vp = 1.01 × 108 = √ µR R We solve the above two equations simultaneously to find µR R = 0.546 ⇒ µR R R = 8.79 = 4.01 and µR = 2.19. Finally, H(x, y, z, t) = Re (2ay − j 5az )e−j 25x ej ωt = 2 cos(2π × 400 × 106 t − 25x)ay + 5 sin(2π × 400 × 106 t − 25x)az = 2 cos(8π × 108 t − 25x)ay + 5 sin(8π × 108 t − 25x)az A/m 184 ...
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