Solved Problems in Midterm Exam for Fundamental of Electromagnetic Fields and Waves | EE 30348, Exams of Electrical and Electronics Engineering

Download Solved Problems in Midterm Exam for Fundamental of Electromagnetic Fields and Waves | EE 30348 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! Fundamentals of Electromagnetic Fields and Waves: I Fall 2006, EE 30348, Electrical Engineering, University of Notre Dame 1st Mid Term Exam (10/12/2006) Note: Please show your steps clearly and sketch figures wherever necessary. Points will be awarded for correct steps shown in the solutions. Fundamental Constants: !0 ! 136! " 10 !9F/m, µ0 = 4" " 10!7 H/m, c = 1""0µ0 ! 3" 10 8m/s, #0 = ! µ0 "0 ! 377!. Maxwell’s Equations: (Law): [Integral form], [Di"erential form] (Gauss’s Law for Electric Field): [ " s !0E · dS = # v $vdv], [# · (!0E) = $v]. (Gauss’s Law for Magnetic Field): [ " s B · dS = 0], [# · B = 0]. (Faraday’s Law): [ " c E · dl = $ d dt # s B · dS], [#"E = $ #B #t ]. (Ampere’s Law): [ " c B µ0 · dl = # s J · dS + d dt # s !0E · dS], [#" B µ0 = J + #"0E#t ]. Charge continuity equation: # · J + #$v#t = 0. All symbols have their usual meanings. Good luck!! Problem 1 Answer the following short questions: (a) (2 Points) Show that Ampere’s law directly leads to the charge continuity equation. What is the historic significance of this relationship between Ampere’s law and charge continuity? Solution: Ampere’s law in the di"erential form is #" Bµ0 = J + #"0E #t . Taking the divergence of both sides, and making use of the identity # ·#" (...) = 0 and Gauss’s law for electric field # · (!0E) = $v, we get the continuity equation. Historically, the displacement current was introduced by Maxwell by noticing that if the #"0E#t term is left out, charge continuity is violated. (b) (2 Points) The electric field due to an unknown volume charge distribution is given by E = 2xax + 3bxay + 2e!zaz V/m, where b is a constant. Find the volume charge density at the point (0, 0, 0). Solution: Using Gauss’s law for electric field, $v(x, y, z) = # · (!0E) = !0(2$ 2e!z) C/m3. Therefore the charge density at (0, 0, 0) is $v(0, 0, 0) = 0" !0 = 0 C/m3. Page-1 (c) (2 Points) Find the value of the constant b such that the electric field in part (b) is static. Solution: For a static electric field, #"E = 3baz = 0. Therefore b = 0. (d) (3 Points) The electric field intensity of a 100 GHz uniform electromagnetic plane wave propagating in source-free space is given in V/m by the phasor form Ê = (2ax + 2ay)e!j%0z, (1) where %0 is the phase factor (wavevector). i) Find %0, ii) express the electric field in real-time form E(r, t), and identify the iii) direction of propagation and the iv) polarization of the wave. Solution: i) Since f = 1011Hz, & = 2" · 1011rad/s, and %0 = &/c = 2000"/3 m!1. ii) E(r, t) = Re[Êej&t] = (2ax + 2ay) cos(2" · 1011t$ 2000"z/3) V/m. iii) The wave propagates along the +ve z direction. iv) It is linearly polarized at 45# to the x-axis, since the x$ and y components are in phase, and Ex = Ey at all times. (e) (3 Points) Find the magnetic field corresponding the the electric field in part (d). Express it first in phasor notation (B̂), and then in real-time form B(r,t). Are the electric and magnetic fields of the EM wave perpendicular? Solution: Using Faraday’s law in the phasor notation, B̂ = $%Ê!j& = 1 c [$2ax + 2ay]e !j%0z. Clearly, the real-time form is then given by B(r, t) = Re[B̂ej&t] = 1c [$2ax + 2ay] cos(&t$ %0z) = E(r, t)/c H/m. Since E · B = 0, the electric and magnetic fields in the EM wave are perpendicular to each other, as they should be. Furthermore, E"B gives us the direction of wave propagation, the +ve z direction. (f) (2 Points) Find the value of the surface integral " s !0E · dS on the closed surface shown in Figure 1(a). The charges inside and outside the sphere are shown in Coulombs. Is the electric field constant over the surface? Solution: From Gauss’s law, the answer is $1 Coulomb. No, the electric field is changing on the surface of the sphere; it is not constant. Page-2