Explanation Given data: The electric permittivity is ε = 9 ε 0 . The magnetic permeability is μ = μ 0 . The magnetic field intensity is H ( z , t ) = x ^ 0.3 cos ( 10 8 t − k z + π 4 ) ( A / m ) (1) Calculation: The general form for magnetic field intensity is given as, H ( z , t ) = x ^ A cos ( ω t − k z + ϕ ) ( A / m ) (2) Here, H ( z , t ) is the magnetic field intensity, A is the amplitude, ω is the angular frequency, ϕ is the phase angle. Compare equation (1) and (2), the value is, ω = 10 8 rad / sec The value of k is written as, k = ω μ ε = ω ε r μ 0 ε 0 = ω ε r c Here, k is the wave number, ε r is the relative permittivity, c is the speed of light ( c = 3 × 10 8 m / s ) . Substitute 10 8 for ω , 3 × 10 8 for c and 9 for ε r in the above expression. k = 10 8 9 3 × 10 8 = 10 8 × 3 3 × 10 8 = 1 The phasor of magnetic field intensity is written as, H ˜ ( z ) = x ^ 0.3 e − j k z e j π / 4 The product is given as, ∇ × H ˜ = | x ^ y ^ z ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z 0.3 e − j k z e j π / 4 0 0 | = [ − y ^ ( 0 − ∂ ∂ z ( 0.3 e − j k z e j π / 4 ) ) + z ^ ( 0 − ∂ ∂ y ( 0.3 e − j k z e j π / 4 ) ) ] = [ y ^ ( − 0.3 j k e − j k z e j π / 4 ) ] The electric field intensity is given as, E ˜ = 1 j ω ε ∇ × H ˜ Here, E ˜ is the electric field intensity

Fundamentals of Applied Electromagnetics (7th Edition)

7th Edition

ISBN: 9780133356984

Author: ULABY

Publisher: PEARSON

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Chapter 6.11, Problem 7E

To determine

The value of E(z,t) and k.

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Chapter 6.10, Problem 6E

Chapter 6, Problem 1P

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Chapter 6 Solutions

Fundamentals of Applied Electromagnetics (7th Edition)

Ch. 6.2 - Explain Faradays law and the function of Lenzs...Ch. 6.2 - Prob. 2CQ Ch. 6.2 - Prob. 3CQ Ch. 6.2 - For the loop shown in Fig. 6-3, what is Vemftr if...Ch. 6.2 - Suppose that the loop of Example 6-1 is replaced...Ch. 6.4 - Suppose that no friction is involved in sliding...Ch. 6.4 - Is the current flowing in the rod of Fig. 6-10 a...Ch. 6.4 - For the moving loop of Fig. 6-9, find I when the...Ch. 6.4 - Suppose that we turn the loop of Fig. 6-9 so that...Ch. 6.5 - Contrast the operation of an ac motor with that of...

Ch. 6.5 - Prob. 7CQ Ch. 6.5 - Prob. 8CQ Ch. 6.7 - A poor conductor is characterized by a...Ch. 6.8 - When conduction current flows through a material,...Ch. 6.8 - Verify that the integral form of Ampres law given...Ch. 6.10 - Explain how the charge continuity equation leads...Ch. 6.10 - How long is the relaxation time constant for...Ch. 6.10 - Determine (a) the relaxation time constant and (b)...Ch. 6.11 - Prob. 7E Ch. 6 - The switch in the bottom loop of Fig. P6.1 is...Ch. 6 - The loop in Fig. P6.2 is in the xy plane and B =...Ch. 6 - A coil consists of 100 turns of wire wrapped...Ch. 6 - A stationary conducting loop with an internal...Ch. 6 - A circular-loop TV antenna with 0.02 m2 area is in...Ch. 6 - The square loop shown in Fig. P6.6 is coplanar...Ch. 6 - The rectangular conducting loop shown in Fig. P6.7...Ch. 6 - Prob. 8P Ch. 6 - Prob. 9P Ch. 6 - A 50 cm long metal rod rotates about the z axis at...Ch. 6 - The loop shown in P6.11 moves away from a wire...Ch. 6 - The electromagnetic generator shown in Fig. 6-12...Ch. 6 - The circular, conducting, disk shown in Fig. P6.13...Ch. 6 - The plates of a parallel-plate capacitor have...Ch. 6 - A coaxial capacitor of length l = 6 cm uses an...Ch. 6 - The parallel-plate capacitor shown in Fig. P6.16...Ch. 6 - In wet soil, characterized by = 102 (S/m), r = 1,...Ch. 6 - An electromagnetic wave propagating in seawater...Ch. 6 - At t = 0, charge density v0 was introduced into...Ch. 6 - If the current density in a conducting medium is...Ch. 6 - Prob. 21P Ch. 6 - If we were to characterize how good a material is...Ch. 6 - The electric field of an electromagnetic wave...Ch. 6 - The magnetic field in a dielectric material with ...Ch. 6 - Given an electric field E=xE0sinaycos(tkz), where...Ch. 6 - The electric field radiated by a short dipole...Ch. 6 - A Hertzian dipole is a short conducting wire...Ch. 6 - In free space, the magnetic field is given by...Ch. 6 - The magnetic field in a given dielectric medium is...

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