Chapter 1: Introduction: Waves and Phasors

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2 Chapter : Introduction: Waves and Phasors Lesson # Chapter Section: Chapter Topics: EM history and how it relates to other fields Highlights: EM in Classical era: 000 BC to 900 Examples of Modern Era Technology timelines Concept of fields (gravitational, electric, magnetic) Static vs. dynamic fields The EM Spectrum Special Illustrations: Timelines from CD-ROM Timeline for Electromagnetics in the Classical Era ca. 900 BC ca. 600 BC Legend has it that while walking across a field in northern Greece, a shepherd named Magnus experiences a pull on the iron nails in his sandals by the black rock he was standing on. The region was later named Magnesia and the rock became known as magnetite [a form of iron with permanent magnetism]. Greek philosopher Thales describes how amber, after being rubbed with cat fur, can pick up feathers [static electricity]. ca. 000 Magnetic compass used as a navigational device. 75 Benjamin Franklin (American) invents the lightning rod and demonstrates that lightning is electricity. 785 Charles-Augustin de Coulomb (French) demonstrates that the electrical force between charges is proportional to the inverse of the square of the distance between them. 800 Alessandro Volta (Italian) develops the first electric battery. 80 Hans Christian Oersted (Danish) demonstrates the interconnection between electricity and magnetism through his discovery that an electric current in a wire causes a compass needle to orient itself perpendicular to the wire.

3 Lessons # and 3 Chapter Sections: - to -6 Topics: Waves Highlights: Wave properties Complex numbers Phasors Special Illustrations: CD-ROM Modules.-.9 CD-ROM Demos.-.3

4 CHAPTER 3 Chapter Section -3: Traveling Waves Problem. A -khz sound wave traveling in the x-direction in air was observed to have a differential pressure p x t 0 N/m at x 0 and t 50 µs. If the reference phase of p x t is 36, find a complete expression for p x t. The velocity of sound in air is 330 m/s. Solution: The general form is given by Eq. (.7), πt p x t Acos T πx λ φ 0 where it is given that φ From Eq. (.6), T f ms. From Eq. (.7), u λ p 330 f m Also, since π p x 0 t 50 µs 0 (N/m ) π rad Acos Acos 6 rad 0 3A it follows that A N/m. So, with t in (s) and x in (m), t 500 π 0 3 x p x t 3 36cos π (N/m ) 3 36cos 4π 0 t 3 πx 36 (N/m ) Problem. For the pressure wave described in Example -, plot (a) p x t versus x at t 0, (b) p x t versus t at x 0. Be sure to use appropriate scales for x and t so that each of your plots covers at least two cycles. Solution: Refer to Fig. P.(a) and Fig. P.(b).

5 4 CHAPTER Amplitude (N/m ) p(x,t=0) Amplitude (N/m ) p(x=0,t) Distance x (m) (a) Time t (ms) (b) Figure P.: (a) Pressure wave as a function of distance at t 0 and (b) pressure wave as a function of time at x 0. Problem.3 A harmonic wave traveling along a string is generated by an oscillator that completes 80 vibrations per minute. If it is observed that a given crest, or maximum, travels 300 cm in 0 s, what is the wavelength? Solution: f Hz u p 300 cm 0 3 m/s 0 s u λ p m 0 cm f 3 Problem.4 Two waves, y t and y t, have identical amplitudes and oscillate at the same frequency, but y t leads y t by a phase angle of 60. If y t 4cos π 0 t 3 write down the expression appropriate for y t and plot both functions over the time span from 0 to ms. Solution: y t 4cos π 0 3 t 60

6 CHAPTER 5 4 y (t) y (t) ms ms.5 ms ms z - -4 Figure P.4: Plots of y t and y t. Problem.5 The height of an ocean wave is described by the function y x t 5sin 0 5t 0 6x (m) Determine the phase velocity and the wavelength and then sketch y x t at t s over the range from x 0 to x λ. Solution: The given wave may be rewritten as a cosine function: y x t 5cos 0 5t 0 6x π By comparison of this wave with Eq. (.3), y x t Acos ωt βx φ 0 we deduce that π ω π f 0 5 rad/s β λ m/s λ 6 u p ω β 0 5 π β π 0 6 rad/m m 6

7 6 CHAPTER y (π/, t) λ Figure P.5: Plot of y x versus x. x At t s, y x 5sin 0 6x (m), with the argument of the cosine function given in radians. Plot is shown in Fig. P.5. Problem.6 A wave traveling along a string in the x-direction is given by y x t Acos ωt βx where x 0 is the end of the string, which is tied rigidly to a wall, as shown in Fig. - (P.6). When wave y x t arrives at the wall, a reflected wave y x t is generated. Hence, at any location on the string, the vertical displacement y s will be the sum of the incident and reflected waves: y s x t y x t y x t (a) Write down an expression for y x t, keeping in mind its direction of travel and the fact that the end of the string cannot move. (b) Generate plots of y x t, y x t and y s x t versus x over the range λ x 0 at ωt π 4 and at ωt π. Solution: (a) Since wave y x t was caused by wave y x t, the two waves must have the same angular frequency ω, and since y x t is traveling on the same string as y x t,

8 CHAPTER 7 Incident Wave y x x = 0 Figure P.6: Wave on a string tied to a wall at x 0 (Problem.6). the two waves must have the same phase constant β. Hence, with its direction being in the negative x-direction, y x t is given by the general form y x t Bcos ωt βx φ 0 () where B and φ 0 are yet-to-be-determined constants. The total displacement is y s x t y x t y x t Acos ωt βx Bcos ωt βx φ 0 Since the string cannot move at x 0, the point at which it is attached to the wall, y s 0 t 0 for all t. Thus, y s 0 t Acosωt Bcos ωt φ 0 0 () (i) Easy Solution: The physics of the problem suggests that a possible solution for () is B A and φ 0 0, in which case we have y x t Acos ωt βx (3) (ii) Rigorous Solution: By expanding the second term in (), we have Acosωt B cosωt cosφ 0 sinωt sin φ 0 0 or A Bcosφ 0 cos ωt Bsinφ 0 sin ωt 0 (4) This equation has to be satisfied for all values of t. At t 0, it gives A Bcosφ 0 0 (5)

9 8 CHAPTER and at ωt π, (4) gives Bsinφ 0 0 (6) Equations (5) and (6) can be satisfied simultaneously only if A B 0 (7) or A B and φ 0 0 (8) Clearly (7) is not an acceptable solution because it means that y x t 0, which is contrary to the statement of the problem. The solution given by (8) leads to (3). (b) At ωt π 4, π πx y x t Acos π 4 βx Acos 4 λ y x t Acos ωt π πx βx Acos λ Plots of y, y, and y 3 are shown in Fig. P.6(b). 4 y (ωt, x) s.5a y (ωt, x) A -λ 0 x y (ωt, x) -A -.5A At ωt π, ωt=π/4 Figure P.6: (b) Plots of y, y, and y s versus x at ωt π 4. y x t Acos π βx Asinβx Asin πx λ

10 CHAPTER 9 y x t Acos π βx Asinβx Asin πx λ Plots of y, y, and y 3 are shown in Fig. P.6(c). y (ωt, x) s A y (ωt, x) y (ωt, x) A -λ 0 x -A -A ωt=π/ Figure P.6: (c) Plots of y, y, and y s versus x at ωt π. Problem.7 Two waves on a string are given by the following functions: y x t 4cos 0t 30x (cm) y x t 4cos 0t 30x (cm) where x is in centimeters. The waves are said to interfere constructively when their superposition y s y y is a maximum and they interfere destructively when y s is a minimum. (a) What are the directions of propagation of waves y x t and y x t? (b) At t π 50 s, at what location x do the two waves interfere constructively, and what is the corresponding value of y s? (c) At t π 50 s, at what location x do the two waves interfere destructively, and what is the corresponding value of y s? Solution: (a) y x t is traveling in positive x-direction. x-direction. y x t is traveling in negative

11 0 CHAPTER (b) At t π 50 s, y s y y 4 cos 0 4π 30x cos 0 4π 3x. Using the formulas from Appendix C, sin xsin y cos x y cosx y we have Hence, y s 8sin 0 4π sin 30x 7 6sin 30x π 60 y s max 7 6 π and it occurs when sin 30x, or 30x nπ, or x n 0 nπ (c) y s min 0 and it occurs when 30x nπ, or x 30 cm. nπ 30 cm, where Problem.8 Give expressions for y x t for a sinusoidal wave traveling along a string in the negative x-direction, given that y max 40 cm, λ 30 cm, f 0 Hz, and (a) y x 0 0 at x 0, (b) y x 0 0 at x 7 5 cm. Solution: For a wave traveling in the negative x-direction, we use Eq. (.7) with ω π f 0π (rad/s), β π λ π 0 3 0π 3 (rad/s), A 40 cm, and x assigned a positive sign: y x t 40cos 0πt 0π 3 x φ 0 (cm) with x in meters. (a) y cos φ 0. Hence, φ 0 π, and y x t 40cos 0πt 0π x π 3 40sin 0πt 0π 40sin 0πt x 0π 3 (cm), if φ 0 π x 3 (cm), if φ 0 π and (b) At x 7 5 cm = m, y 0 40cos π φ 0. Hence, φ 0 0 or π, 40cos y x t 0πt 0π x 3 (cm), if φ cos 0πt 0π x 3 (cm), if φ 0 π

12 CHAPTER Problem.9 An oscillator that generates a sinusoidal wave on a string completes 0 vibrations in 50 s. The wave peak is observed to travel a distance of.8 m along the string in 50 s. What is the wavelength? Solution: 50 T s 0 5 u p m/s 5 λ u p T m Problem.0 function: The vertical displacement of a string is given by the harmonic y x t 6cos 6πt 0πx (m) where x is the horizontal distance along the string in meters. Suppose a tiny particle were to be attached to the string at x 5 cm, obtain an expression for the vertical velocity of the particle as a function of time. Solution: y x t 6cos 6πt 0πx (m) dy u 0 05 t x t dt x πsin 6πt π 96πsin 6πt (m/s) 96πsin 6πt 0πx x 0 05 Problem. Given two waves characterized by y t 3cos ωt y t 3sin ωt 36 does y t lead or lag y t, and by what phase angle? Solution: We need to express y t in terms of a cosine function: y t 3sin ωt 36 3cos π ωt 36 3cos 54 ωt 3cos ωt 54

13 CHAPTER Hence, y t lags y t by 54. Problem. The voltage of an electromagnetic wave traveling on a transmission line is given by v z t 5e αz sin 4π 0 9 t 0πz (V), where z is the distance in meters from the generator. (a) Find the frequency, wavelength, and phase velocity of the wave. (b) At z m, the amplitude of the wave was measured to be V. Find α. Solution: (a) This equation is similar to that of Eq. (.8) with ω 4π 0 9 rad/s and β 0π rad/m. From Eq. (.9a), f ω π 0 9 Hz GHz; from Eq. (.9b), λ π β 0 m. From Eq. (.30), u p ω β 0 8 m/s (b) Using just the amplitude of the wave, 5e α α m ln Np/m. Problem.3 A certain electromagnetic wave traveling in sea water was observed to have an amplitude of 98.0 (V/m) at a depth of 0 m and an amplitude of 8.87 (V/m) at a depth of 00 m. What is the attenuation constant of sea water? Solution: The amplitude has the form Ae αz. At z 0 m, Ae 0α 98 0 and at z 00 m, The ratio gives or e 0α Taking the natural log of both sides gives Hence, Ae 00α α e α e e 00α ln e 0α ln e 00α 0α ln 00α 90α ln 0 8 α (Np/m)

14 CHAPTER 3 Section -5: Complex Numbers Problem.4 result in rectangular form: (a) z 4e jπ 3, (b) z 3 e j3π 4, (c) z 3 6e jπ, (d) z 4 j 3, (e) z 5 j 4, (f) z 6 j 3, (g) z 7 j. Solution: Evaluate each of the following complex numbers and express the (Note: In the following solutions, numbers are expressed to only two decimal places, but the final answers are found using a calculator with 0 decimal places.) (a) z 4e jπ 3 4 cos π 3 j sinπ 3 0 j3 46. (b) z 3e j3π 4 3 cos 3π 4 j sin 3π 4 j j (c) z 3 6e jπ 6 cos π j sin π j6. (d) z 4 j 3 j j j, or z 4 j 3 e jπ 3 e j3π cos 3π j sin 3π j (e) z 5 j 4 e jπ 4 e jπ. (f) (g) 3 z 6 j jπ 3 e 4 e 3 j3π 4 3 3π 4 cos j sin 3π 4 j j z 7 j e jπ 4 4 e jπ j j0 45 Problem.5 Complex numbers z and z are given by z 3 j z 4 j3

15 4 CHAPTER (a) Express z and z in polar form. (b) Find z by applying Eq. (.4) and again by applying Eq. (.43). (c) Determine the product z z in polar form. (d) Determine the ratio z z in polar form. (e) Determine z 3 in polar form. Solution: (a) Using Eq. (.4), z 3 j 3 6e j33 7 z 4 j3 5e j43 (b) By Eq. (.4) and Eq. (.43), respectively, z 3 j z 3 j 3 j (c) By applying Eq. (.47b) to the results of part (a), z z 3 6e j33 7 5e j43 8e j09 4 (d) By applying Eq. (.48b) to the results of part (a), j33 z z 3 6e 7 j76 0 7e 8 5e j43 (e) By applying Eq. (.49) to the results of part (a), z 3 3 6e j e j e j0 Problem.6 (a) z, (b) z 3, (c) z, (d) z, (e) z. If z j4, determine the following quantities in polar form: Solution: (Note: In the following solutions, numbers are expressed to only two decimal places, but the final answers are found using a calculator with 0 decimal places.)

16 CHAPTER 5 (a) j6 j e 4 47 j6 6 j6 6 e 0 e j4 j4 3 j6 4 47e j350 e 0 89 j0 44e. (c) z z z j4 j (d) z j4 4. (e) z j4 4 4e jπ. z (b) z 3 Problem.7 Find complex numbers t z z and s z z, both in polar form, for each of the following pairs: (a) z j3, z j3, (b) z 3, z j3, (c) z 3 30, z 3 30, (d) z 3 30, z Solution: (a) (b) (c) (d) t z z j3 j3 3 s z z j3 j3 j6 6 08e j80 5 t z z 3 j3 4 4e j45 s z z 3 j3 4 4e j45 t z z e j30 j30 3e 6 j 5 6 j 5 5 s z z 3e j30 j30 3e 6 j 5 6 j 5 j3 3e j90 t z z j 5 6 j 5 0 s z z 6 j 5 6 j 5 5 j3 6e j30 Problem.8 Complex numbers z and z are given by z 5 60 z 45

17 6 CHAPTER (a) Determine the product z z in polar form. (b) Determine the product z z in polar form. (c) Determine the ratio z z in polar form. (d) Determine the ratio z z in polar form. (e) Determine z in polar form. Solution: (a) z z 5e j60 j45 0e j5. (b) z z 5e j60 e j45 0e j05. z (c) z 5e j45 j60 5 j05. e z z (d) z j05 5. z (e) z 5e j60 5e j30. Problem.9 If z 3 j5, find the value of ln z. Solution: z θ tan z z e 5 83e j59 jθ ln z ln 5 83e j59 ln 5 83 ln e j59 76 j59 j π 76 j 03 Problem.0 If z 3 j4, find the value of e z. Solution: e z e 3 j4 e 3 j4 e e cos4 3 j sin e and 4 rad 4 π Hence, e z 0 08 cos 9 8 j sin j5 0.

18 CHAPTER 7 Section -6: Phasors Problem. A voltage source given by v s t 5cos π 0 3 t 30 (V) is connected to a series RC load as shown in Fig. -9. If R MΩ and C 00 pf, obtain an expression for v c t, the voltage across the capacitor. Solution: In the phasor domain, the circuit is a voltage divider, and V c V s jωc R jωc V s jωrc Now V s 5e j30 V with ω π 0 3 rad/s, so V c 5e j30 V j π 0 rad/s 3 0 Ω 00 0 F 6 5e j30 V jπ 5 57e j8 5 V. 5 Converting back to an instantaneous value, v c t V c e jωt 5 57e j ωt 8 5 where t is expressed in seconds. V 5 57cos π 0 3 t 8 5 V Problem. Find the phasors of the following time functions: (a) v t 3cos ωt π 3 (V), (b) v t sin ωt π 4 (V), (c) i x t e 3x sin ωt π 6 (A), (d) i t cos ωt 3π 4 (A), (e) i t 4sin ωt π 3 3cos ωt π 6 (A). Solution: (a) V 3e jπ 3 V. (b) v t sin ωt π 4 cos π ωt π 4 cos ωt π 4 V, V e jπ 4 V. (c) i t e 3x sin ωt π 6 A e 3x cos π ωt π 6 A e 3x cos ωt π 3 A I e 3x e jπ 3 A

19 8 CHAPTER (d) (e) i t cos ωt 3π 4 I e j3π 4 e jπ e j3π 4 e jπ 4 A i t 4sin ωt π 3 3cos ωt π 6 4cos π ωt π 3 3cos ωt π 6 4cos ωt π 6 3cos ωt π 6 4cos ωt π 6 3cos ωt π 6 7cos ωt π 6 jπ I 7e A 6 Problem.3 Find the instantaneous time sinusoidal functions corresponding to the following phasors: (a) V 5e jπ 3 (V), (b) V jπ j6e 4 (V), (c) I 6 j8 (A), (d) Ĩ 3 j (A), (e) Ĩ j (A), (f) Ĩ e jπ 6 Solution: (a) (b) (c) (d) (A). V 5e V jπ 3 5e j π 3 π jπ V 5e V 3 v t 5cos ωt π 3 V jπ V j6e V 4 6e j π 4 π V 6e V jπ 4 v t 6cos ωt π 4 V I 6 j8 A 0e j53 A i t 0cos ωt 53 A. I 3 j 3 6e j46 3 i t 3 6e j46 3 e jωt 3 6 cos ωt 46 3 A

20 CHAPTER 9 (e) (f) I j e jπ i t e jπ e jωt cos ωt π sinωt A I e jπ 6 i t e jπ 6 e jωt cos ωt π 6 A Problem.4 A series RLC circuit is connected to a generator with a voltage v s t V 0 cos ωt π 3 (V). (a) Write down the voltage loop equation in terms of the current i t, R, L, C, and v s t. (b) Obtain the corresponding phasor-domain equation. (c) Solve the equation to obtain an expression for the phasor current I. R L i V s (t) C Figure P.4: RLC circuit. Solution: (a) v s t Ri L di dt C i dt (b) In phasor domain: V s RĨ jωlĩ Ĩ V (c) Ĩ s R V j ωl ωc 0 e jπ 3 R ωcv j ωl ωc 0 e jπ 3 ωrc j ω LC jωc Problem.5 A wave traveling along a string is given by y x t sin 4πt 0πx (cm)

21 0 CHAPTER where x is the distance along the string in meters and y is the vertical displacement. Determine: (a) the direction of wave travel, (b) the reference phase φ 0, (c) the frequency, (d) the wavelength, and (e) the phase velocity. Solution: (a) We start by converting the given expression into a cosine function of the form given by (.7): π (cm) y x t cos 4πt 0πx Since the coefficients of t and x both have the same sign, the wave is traveling in the negative x-direction. (b) From the cosine expression, φ 0 π. (c) ω π f 4π, f 4π π Hz (d) π λ 0π, λ π 0π 0 m. (e) u p f λ (m/s). Problem.6 A laser beam traveling through fog was observed to have an intensity of (µw/m ) at a distance of m from the laser gun and an intensity of 0. (µw/m ) at a distance of 3 m. Given that the intensity of an electromagnetic wave is proportional to the square of its electric-field amplitude, find the attenuation constant α of fog. Solution: If the electric field is of the form E x t E 0 e αx cos ωt βx then the intensity must have a form I x t E 0 e αx cos ωt βx E0e αx cos ωt βx or where we define I 0 I 0 e αx. Hence, I x t I 0 e αx cos ωt βx E0. We observe that the magnitude of the intensity varies as at x m I 0 e 4α 0 6 at x 3 m I 0 e 6α (W/m ) (W/m )

22 CHAPTER I 4α 6 0 e I 6α 0 e e 4α e 6α e α 5 α 0 8 (NP/m) Problem.7 Complex numbers z and z are given by z 3 j z j Determine (a) z z, (b) z z, (c) z, and (d) z z, all all in polar form. Solution: (a) We first convert z and z to polar form: z 3 j 3 e j tan 3 j e 3 e j e j46 3 z j 4 e j tan 5 e j63 4 (b) (c) z z 3 e j46 3 j e 65 e j8 9 z 3 e j e j e j8 9 z 3e j360 z 3 e j46 3 3e j9 6 3e j67 4 e j9 6

23 CHAPTER (d) z z 3 e j46 3 j e 3 Problem.8 If z 3e jπ 6, find the value of e z. Solution: z 3e jπ 6 3cos π 6 j3sinπ 6 6 j 5 e z e 6 j 5 e 6 e j 5 e cos 6 5 j sin j j3 43 Problem.9 The voltage source of the circuit shown in the figure is given by v s t 5cos t 45 (V) Obtain an expression for i L t, the current flowing through the inductor. R i A + i R i L v s (t) - R L R = 0 Ω, R = 30 Ω, L = 0.4 mh Solution: Based on the given voltage expression, the phasor source voltage is V s 5e j45 (V) (9) The voltage equation for the left-hand side loop is R i R i R v s (0)

24 CHAPTER 3 For the right-hand loop, and at node A, R i R L di L dt () Next, we convert Eqs. () (4) into phasor form: i i R i L () R I R I R V s (3) R I R jωli L (4) I I R I L (5) Upon combining (6) and (7) to solve for I R in terms of I, we have: Substituting (8) in (5) and then solving for I leads to: I I R jωl R jωl I (6) R I jr ωl R jωl I V s jr I R R jωl V s R R jr ωl jr ωl R V jωl s R jωl R R jωl R R V s (7) I Combining (6) and (7) to solve for I L in terms of I gives I (8) Combining (9) and (0) leads to R I L R jωl R I L R R jωl R R jωl R R V s R jωl R R jωl R R V s

25 4 CHAPTER Using () for V s and replacing R, R, L and ω with their numerical values, we have Finally, 30 I L 0 30 j j e j800 e j j8 e j45 7 5e j45 0e j e j98 i L t I L e jωt 0 75cos t 98 (A) (A)

26 5 Chapter : Transmission Lines Lesson #4 Chapter Section: -, - Topics: Lumped-element model Highlights: TEM lines General properties of transmission lines L, C, R, G

27 6 Lesson #5 Chapter Section: -3, -4 Topics: Transmission-line equations, wave propagation Highlights: Wave equation Characteristic impedance General solution Special Illustrations: Example -

28 7 Lesson #6 Chapter Section: -5 Topics: Lossless line Highlights: General wave propagation properties Reflection coefficient Standing waves Maxima and minima Special Illustrations: Example - Example -5

29 8 Lesson #7 Chapter Section: -6 Topics: Input impedance Highlights: Thévenin equivalent Solution for V and I at any location Special Illustrations: Example -6 CD-ROM Modules.-.4, Configurations A-C CD-ROM Demos.-.4, Configurations A-C

30 9 Lessons #8 and 9 Chapter Section: -7, -8 Topics: Special cases, power flow Highlights: Sorted line Open line Matched line Quarter-wave transformer Power flow Special Illustrations: Example -8 CD-ROM Modules.-.4, Configurations D and E CD-ROM Demos.-.4, Configurations D and E

31 30 Lessons #0 and Chapter Section: -9 Topics: Smith chart Highlights: Structure of Smith chart Calculating impedances, admittances, transformations Locations of maxima and minima Special Illustrations: Example -0 Example -

32 3 Lesson # Chapter Section: -0 Topics: Matching Highlights: Matching network Double-stub tuning Special Illustrations: Example - Technology Brief on Microwave Oven (CD-ROM) Microwave Ovens Percy Spencer, while working for Raytheon in the 940s on the design and construction of magnetrons for radar, observed that a chocolate bar that had unintentionally been exposed to microwaves had melted in his pocket. The process of cooking by microwave was patented in 946, and by the 970s microwave ovens had become standard household items.

33 3 Lesson #3 Chapter Section: - Topics: Transients Highlights: Step function Bounce diagram Special Illustrations: CD-ROM Modules.5-.9 CD-ROM Demos.5-.3 Demo.3

34 CHAPTER 33 Chapter Sections - to -4: Transmission-Line Model Problem. A transmission line of length l connects a load to a sinusoidal voltage source with an oscillation frequency f. Assuming the velocity of wave propagation on the line is c, for which of the following situations is it reasonable to ignore the presence of the transmission line in the solution of the circuit: (a) l 0 cm, f 0 khz, (b) l 50 km, f 60 Hz, (c) l 0 cm, f 600 MHz, (d) l mm, f 00 GHz. Solution: A transmission line is negligible when l λ 0 0. l l f (a) λ u p 0 0 m 0 0 Hz (negligible). m/s l l f (b) λ u p 50 0 m Hz (borderline) m/s l l f (c) λ u p 0 0 m Hz (nonnegligible) m/s l l f (d) λ u p 0 m Hz (nonnegligible) m/s Problem. Calculate the line parameters R, L, G, and C for a coaxial line with an inner conductor diameter of 0 5 cm and an outer conductor diameter of cm, filled with an insulating material where µ µ 0, ε r 4 5, and σ 0 3 S/m. The conductors are made of copper with µ c µ 0 and σ c S/m. The operating frequency is GHz. Solution: Given combining Eqs. (.5) and (.6) gives R π π π f µ c σ c Ω/m a 0 5 cm m b 0 cm m b a π 0 Hz 4π 0 H/m S/m m m

35 34 CHAPTER From Eq. (.7), From Eq. (.8), From Eq. (.9), µ L π ln b a 4π 0 7 H/m π ln 39 nh/m πσ G ln b a π 0 3 S/m 9 ms/m ln πε πε C ln b a r ε 0 ln b a π F/m ln 36 pf/m Problem.3 A -GHz parallel-plate transmission line consists of.-cm-wide copper strips separated by a 0.5-cm-thick layer of polystyrene. Appendix B gives µ c µ 0 4π 0 7 (H/m) and σ c (S/m) for copper, and ε r 6 for polystyrene. Use Table - to determine the line parameters of the transmission line. Assume µ µ 0 and σ 0 for polystyrene. R R s π f µ c w w σ c π 0 4π µd L 4π w (H/m) G 0 because σ 0 εw C d ε w 0ε r d 36π Solution: 38 (Ω/m) (F/m) Problem.4 Show that the transmission line model shown in Fig. -37 (P.4) yields the same telegrapher s equations given by Eqs. (.4) and (.6). Solution: The voltage at the central upper node is the same whether it is calculated from the left port or the right port: v z z t z t v z t R z i L z z t t i v z z t R z i z z t L z t i z z t

36 CHAPTER 35 + i(z, t) R' z L' z R' z L' z i(z+ z, t) + v(z, t) G' z C' z v(z+ z, t) - z Figure P.4: Transmission line model. - Recognizing that the current through the G C branch is i z t i z z t (from Kirchhoff s current law), we can conclude that i z t i z z t G z v z z t C z t v z z t From both of these equations, the proof is completed by following the steps outlined in the text, ie. rearranging terms, dividing by z, and taking the limit as z 0. Problem.5 Find α β u p, and Z 0 for the coaxial line of Problem.. Solution: From Eq. (.), γ R jωl G jωc Ω/m j π 0 s 39 0 H/m S/m 3 j π 0 s 36 0 F/m j44 5 m Thus, from Eqs. (.5a) and (.5b), α 0 09 Np/m and β 44 5 rad/m. From Eq. (.9), R jωl Z Ω/m j π 0 s 39 0 H/m G jωc 9 0 S/m j π 0 s 36 0 F/m j0 030 Ω From Eq. (.33), ω u p π 0 9 β m/s

37 36 CHAPTER Section -5: The Lossless Line Problem.6 In addition to not dissipating power, a lossless line has two important features: () it is dispertionless (µ p is independent of frequency) and () its characteristic impedance Z 0 is purely real. Sometimes, it is not possible to design a transmission line such that R ωl and G ωc, but it is possible to choose the dimensions of the line and its material properties so as to satisfy the condition R C L G (distortionless line) Such a line is called a distortionless line because despite the fact that it is not lossless, it does nonetheless possess the previously mentioned features of the loss line. Show that for a distortionless line, α R C R G β ω L C Z 0 L Solution: Using the distortionless condition in Eq. (.) gives γ α jβ R jωl G jωc L C L C L C R L R L R jω L jω jω R G C R L jω jω L C C jω L C L Hence, C ω α γ R β γ ω L C u p L β L C Similarly, using the distortionless condition in Eq. (.9) gives R jωl L R Z 0 L jω G jωc C G C jω L C Problem.7 For a distortionless line with Z 0 50 Ω, α 0 (mnp/m), u p (m/s), find the line parameters and λ at 00 MHz.

38 CHAPTER 37 Solution: The product of the expressions for α and Z 0 given in Problem.6 gives R αz (Ω/m) and taking the ratio of the expression for Z 0 to that for u p ω β L C gives Z L 0 50 u p 5 0 (H/m) With L known, we use the expression for Z 0 to find C : L C Z0 (nh/m) (F/m) 80 (pf/m) The distortionless condition given in Problem.6 is then used to find G. G R C L (S/m) 400 (µs/m) and the wavelength is obtained by applying the relation λ µ p f m Problem.8 Find α and Z 0 of a distortionless line whose R Ω/m and G 0 4 S/m. Solution: From the equations given in Problem.6, L R Z 0 C G α 4 R G Ω (Np/m) Problem.9 A transmission line operating at 5 MHz has Z 0 40 Ω, α 0 0 (Np/m), and β 0 75 rad/m. Find the line parameters R, L, G, and C. 0, the line is Solution: Given an arbitrary transmission line, f 5 MHz, Z 0 40 Ω, α 0 0 Np/m, and β 0 75 rad/m. Since Z 0 is real and α distortionless. From Problem.6, β ω L C and Z 0 L C, therefore, βz L ω π nh/m

39 38 CHAPTER Then, from Z 0 L C, C L Z 0 From α R G and R C L G, R R R G R G G and 38 nh/m pf/m L C α G 0 0 Np/m R 0 8 Ω/m αz Np/m 40 Ω 0 6 Ω/m 0 5 ms/m Problem.0 Using a slotted line, the voltage on a lossless transmission line was found to have a maximum magnitude of.5 V and a minimum magnitude of 0.6 V. Find the magnitude of the load s reflection coefficient. Solution: From the definition of the Standing Wave Ratio given by Eq. (.59), S V max V min Solving for the magnitude of the reflection coefficient in terms of S, as in Example -4, Γ S S Problem. Polyethylene with ε r 5 is used as the insulating material in a lossless coaxial line with characteristic impedance of 50 Ω. The radius of the inner conductor is. mm. (a) What is the radius of the outer conductor? (b) What is the phase velocity of the line? Solution: Given a lossless coaxial line, Z 0 50 Ω, ε r 5, a mm: (a) From Table -, Z 0 60 ε r ln b a which can be rearranged to give b ae Z 0 ε r 60 mm e mm

40 CHAPTER 39 (b) Also from Table -, u p c ε r m/s m/s Problem. A 50-Ω lossless transmission line is terminated in a load with impedance Z L 30 j50 Ω. The wavelength is 8 cm. Find: (a) the reflection coefficient at the load, (b) the standing-wave ratio on the line, (c) the position of the voltage maximum nearest the load, (d) the position of the current maximum nearest the load. Solution: (a) From Eq. (.49a), Γ Z L Z 0 Z L Z 0 30 j j e j79 8 (b) From Eq. (.59), (c) From Eq. (.56) l max θ r λ 4π S Γ Γ nλ 80 n 8 cm cm π rad 4π 0 89 cm 4 0 cm 3 cm (d) A current maximum occurs at a voltage minimum, and from Eq. (.58), l min l max λ 4 3 cm 8 cm 4 cm Problem.3 On a 50-Ω lossless transmission line, the following observations were noted: distance of first voltage minimum from the load 3 cm; distance of first voltage maximum from the load 9 cm; S 3. Find Z L. Solution: Distance between a minimum and an adjacent maximum λ 4. Hence, 9 cm 3 cm 6 cm λ 4

41 40 CHAPTER or λ 4 cm. Application of Eq. (.57) with n 0 gives which gives θ r π. Hence, Γ 0 5e jπ j0 5. Finally, Accordingly, the first voltage minimum is at min 3 cm λ 8. Z L Z 0 θ r π λ λ 8 π Γ S S Γ Γ 50 j0 5 j j0 Ω Problem.4 Using a slotted line, the following results were obtained: distance of first minimum from the load 4 cm; distance of second minimum from the load 4 cm, voltage standing-wave ratio 5. If the line is lossless and Z 0 50 Ω, find the load impedance. Solution: Following Example.5: Given a lossless line with Z 0 50 Ω, S 5, l min 0 4 cm, l min 4 cm. Then or and From this we obtain Also, l min l min 0 λ λ l min l min 0 0 cm π β π rad/cycle λ 0 cm/cycle 0π rad/m θ r βl min n n π rad 0π rad/m 0 04 π rad m 0 π rad 36 0 Γ S S 5 5 0

42 CHAPTER 4 So Z L Γ Z 0 Γ 50 0 e j e j j6 4 Ω Problem.5 A load with impedance Z L 5 j50 Ω is to be connected to a lossless transmission line with characteristic impedance Z 0, with Z 0 chosen such that the standing-wave ratio is the smallest possible. What should Z 0 be? Solution: Since S is monotonic with Γ (i.e., a plot of S vs. Γ is always increasing), the value of Z 0 which gives the minimum possible S also gives the minimum possible Γ, and, for that matter, the minimum possible Γ. A necessary condition for a minimum is that its derivative be equal to zero: 0 Z 0 Γ Z 0 R L jx L Z 0 R L jx L Z 0 Therefore, Z 0 R L X L or Z 0 Z L R Z 0 L Z 0 R L Z 0 XL XL 4R L Z 0 R L X L R L Z 0 X L Ω A mathematically precise solution will also demonstrate that this point is a minimum (by calculating the second derivative, for example). Since the endpoints of the range may be local minima or maxima without the derivative being zero there, the endpoints (namely Z 0 0 Ω and Z 0 Ω) should be checked also. Problem.6 A 50-Ω lossless line terminated in a purely resistive load has a voltage standing wave ratio of 3. Find all possible values of Z L. Solution: Γ S S For a purely resistive load, θ r 0 or π. For θ r 0, Z L Γ Z 0 Γ Ω For θ r π, Γ 0 5 and Z L Ω

43 4 CHAPTER Section -6: Input Impedance Problem.7 At an operating frequency of 300 MHz, a lossless 50-Ω air-spaced transmission line.5 m in length is terminated with an impedance Z L 40 j0 Ω. Find the input impedance. Solution: Given a lossless transmission line, Z 0 50 Ω, f 300 MHz, l 5 m, and Z L 40 j0 Ω. Since the line is air filled, u p c and therefore, from Eq. (.38), ω β u p π π rad/m Since the line is lossless, Eq. (.69) is valid: Z in Z 0 Z L jz 0 tanβl Z 0 jz L tanβl j0 j50tan π rad/m 5 m 50 j 40 j0 tan π rad/m 5 m 40 j0 j j 40 j j0 Ω Problem.8 A lossless transmission line of electrical length l 035λ is terminated in a load impedance as shown in Fig. -38 (P.8). Find Γ, S, and Z in. l = 0.35λ Z in Z 0 = 00 Ω Z L = (60 + j30) Ω Figure P.8: Loaded transmission line. Solution: From Eq. (.49a), Γ Z L Z 0 Z L Z 0 60 j j e j3 5 From Eq. (.59), S Γ Γ

44 CHAPTER 43 From Eq. (.63) Z in Z 0 Z L jz 0 tanβl Z 0 jz L tanβl π rad j30 j00tan λ 0 35λ 00 j 60 π rad j30 tan λ 0 35λ 64 8 j38 3 Ω Problem.9 Show that the input impedance of a quarter-wavelength long lossless line terminated in a short circuit appears as an open circuit. Solution: Z in Z 0 Z L jz 0 tanβl Z 0 jz L tanβl For l λ 4, βl π λ λ 4 π. With Z L 0, we have Z in Z 0 jz 0 tanπ Z 0 j (open circuit) Problem.0 Show that at the position where the magnitude of the voltage on the line is a maximum the input impedance is purely real. Solution: From Eq. (.56), l max θ r nπ β, so from Eq. (.6), using polar representation for Γ, Z in l max Γ e jθr Z 0 e jβlmax Γ e jθ r e jβl max Γ e jθ r Z 0 e j θ r nπ Γ e jθ r e j θ r Γ Z nπ 0 Γ which is real, provided Z 0 is real. Problem. A voltage generator with v g t 5cos π 0 t 9 V and internal impedance Z g 50 Ω is connected to a 50-Ω lossless air-spaced transmission line. The line length is 5 cm and it is terminated in a load with impedance Z L 00 j00 Ω. Find (a) Γ at the load. (b) Z in at the input to the transmission line. (c) the input voltage V i and input current Ĩ i.

45 44 CHAPTER Solution: (a) From Eq. (.49a), Γ Z L Z 0 Z L Z 0 00 j j e j9 7 (b) All formulae for Z in require knowledge of β ω u p. Since the line is an air line, u p c, and from the expression for v g t we conclude ω π 0 9 rad/s. Therefore Then, using Eq. (.63), β π 0 9 rad/s m/s Z in Z 0 Z L jz 0 tanβl Z 0 jz L tanβl 0π 3 rad/m 0π 00 j00 j50tan 3 rad/m 5 cm 50 0π j 00 j00 tan 3 rad/m 5 cm π 00 j00 j50tan rad 3 50 π j 00 j00 tan 3 rad 5 j 7 Ω An alternative solution to this part involves the solution to part (a) and Eq. (.6). (c) In phasor domain, V g 5 V e j0. From Eq. (.64), V i V g Z in Z g Z in 5 5 j j 7 40e j34 0 (V) and also from Eq. (.64), I i V i Z in 4e j j e j 5 (ma) Problem. A 6-m section of 50-Ω lossless line is driven by a source with v g t 5cos 8π 0 7 t 30 (V) and Z g 50 Ω. If the line, which has a relative permittivity ε r 5, is terminated in a load Z L 50 j50 Ω find (a) λ on the line, (b) the reflection coefficient at the load, (c) the input impedance,

46 CHAPTER 45 (d) the input voltage V i, (e) the time-domain input voltage v i t. Solution: ~ V g Ω Z g Generator ~ V g (a) Z g ~ I i ~ V i ~ I i v g t 5cos 8π 0 t 30 V 7 j30 V g 5e V + ~ Z in Z Z 0 = 50 Ω V L L - Transmission line l = 6 m z = -l z = 0 + ~ V i - Z in Figure P.: Circuit for Problem.. c u p ε r (m/s) 5 08 u λ p πu p π 0 8 f ω 8π m ω β u p 8π π (rad/m) βl 0 4π 6 4π (rad) + - ~ I L Load (50-j50) Ω

47 46 CHAPTER Since this exceeds π (rad), we can subtract π, which leaves a remainder βl 04π (rad). Z (b) Γ L Z 0 50 Z L Z 0 50 j50 j (c) (d) (e) Z Z in L jz Z 0 0 tanβl Z 0 jz L tan βl 50 v i t V i V g Z in j j50 0 6e j j50 j50tan 0 4π 50 j 50 j50 tan 0 4π 5 70 j7 4 Ω j30 Z g Z in 5e 5 7 j j7 4 j30 5e 5 7 j j7 4 j30 5e 0 44e e j7 44 j 56 V i e jωt e j 56 (V) e jωt cos 8π 0 7 t 56 V Problem.3 Two half-wave dipole antennas, each with impedance of 75 Ω, are connected in parallel through a pair of transmission lines, and the combination is connected to a feed transmission line, as shown in Fig..39 (P.3(a)). All lines are 50 Ω and lossless. (a) Calculate Z in, the input impedance of the antenna-terminated line, at the parallel juncture. (b) Combine Z in and Z in in parallel to obtain ZL, the effective load impedance of the feedline. (c) Calculate Z in of the feedline. Solution: (a) Z in Z 0 50 Z L jz 0 tan βl Z 0 jz L tan βl 75 j50tan π λ 0 λ 50 j75tan π λ 0 35 λ 0 j8 6 Ω

48 CHAPTER 47 0.λ 75 Ω (Antenna) 0.3λ Z in Z in Z in 0.λ 75 Ω (Antenna) Figure P.3: (a) Circuit for Problem.3. (b) (c) Z L Z in Z in Z in Z in 35 0 j j j4 3 Ω l = 0.3 λ Z in Z L ' Figure P.3: (b) Equivalent circuit. Z in j4 3 j50tan π λ 0 3λ 50 j 7 60 j4 3 tan π λ 0 3λ j56 7 Ω

49 48 CHAPTER Section -7: Special Cases Problem.4 At an operating frequency of 300 MHz, it is desired to use a section of a lossless 50-Ω transmission line terminated in a short circuit to construct an equivalent load with reactance X 40 Ω. If the phase velocity of the line is 0 75c, what is the shortest possible line length that would exhibit the desired reactance at its input? Solution: β ω u p π rad/cycle cycle/s m/s l β tan 8 38 rad/m On a lossless short-circuited transmission line, the input impedance is always purely imaginary; i.e., Zin sc jxin sc. Solving Eq. (.68) for the line length, Xin sc Z rad/m tan 40 Ω 50 Ω nπ rad 8 38 rad/m for which the smallest positive solution is 8 05 cm (with n 0). Problem.5 A lossless transmission line is terminated in a short circuit. How long (in wavelengths) should the line be in order for it to appear as an open circuit at its input terminals? Solution: From Eq. (.68), Z sc Hence, This is evident from Figure.5(d). in jz 0 tanβl. If βl π nπ, then Zin sc j Ω. l λ π π nπ λ 4 nλ Problem.6 The input impedance of a 3-cm-long lossless transmission line of unknown characteristic impedance was measured at MHz. With the line terminated in a short circuit, the measurement yielded an input impedance equivalent to an inductor with inductance of µh, and when the line was open circuited, the measurement yielded an input impedance equivalent to a capacitor with capacitance of 40 pf. Find Z 0 of the line, the phase velocity, and the relative permittivity of the insulating material. Solution: Now ω π f rad/s, so Z sc in jωl jπ j0 4 Ω

50 CHAPTER 49 and Z oc From Eq. (.74), Z 0 Eq. (.75), ω u p ωl β tan Z sc in jωc jπ j4000 Ω. Zin sczoc in j0 4 Ω j4000 Ω 40 Ω Using in Zin oc tan j0 4 j nπ m/s where n 0 for the plus sign and n for the minus sign. For n 0, u p m/s 0 65c and ε r c u p For other values of n, u p is very slow and ε r is unreasonably high. Problem.7 A 75-Ω resistive load is preceded by a λ 4 section of a 50-Ω lossless line, which itself is preceded by another λ 4 section of a 00-Ω line. What is the input impedance? Solution: The input impedance of the λ 4 section of line closest to the load is found from Eq. (.77): Z0 Z in 50 Z L Ω The input impedance of the line section closest to the load can be considered as the load impedance of the next section of the line. By reapplying Eq. (.77), the next section of λ 4 line is taken into account: Z0 Z in 00 Z L Ω Problem.8 A 00-MHz FM broadcast station uses a 300-Ω transmission line between the transmitter and a tower-mounted half-wave dipole antenna. The antenna impedance is 73 Ω. You are asked to design a quarter-wave transformer to match the antenna to the line. (a) Determine the electrical length and characteristic impedance of the quarterwave section. (b) If the quarter-wave section is a two-wire line with d 5 cm, and the spacing between the wires is made of polystyrene with ε r 6, determine the physical length of the quarter-wave section and the radius of the two wire conductors.

51 50 CHAPTER Solution: (a) For a match condition, the input impedance of a load must match that of the transmission line attached to the generator. A line of electrical length λ 4 can be used. From Eq. (.77), the impedance of such a line should be Z 0 Z in Z L Ω (b) and, from Table -, λ 4 u p 4 f c 4 ε r f m 0 Z 0 ε ln d a d a Ω Hence, ln d a d a which leads to d a d a 7 3 and whose solution is a d cm mm. Problem.9 A 50-MHz generator with Z g 50 Ω is connected to a load Z L 50 j5 Ω. The time-average power transferred from the generator into the load is maximum when Z g ZL where ZL is the complex conjugate of Z L. To achieve this condition without changing Z g, the effective load impedance can be modified by adding an open-circuited line in series with Z L, as shown in Fig. -40 (P.9). If the line s Z 0 00 Ω, determine the shortest length of line (in wavelengths) necessary for satisfying the maximum-power-transfer condition. Solution: Since the real part of Z L is equal to Z g, our task is to find l such that the input impedance of the line is Z in j5 Ω, thereby cancelling the imaginary part of Z L (once Z L and the input impedance the line are added in series). Hence, using Eq. (.73), j00cot βl j5

52 CHAPTER 5 Z 0 = 00 Ω l 50 Ω ~ + Vg - Z L (50-j5) Ω Figure P.9: Transmission-line arrangement for Problem.9. or which leads to cotβl βl 36 or 86 Since l cannot be negative, the first solution is discarded. The second solution leads to l β π λ 0 9λ Problem.30 A 50-Ω lossless line of length l 0 375λ connects a 300-MHz generator with V g 300 V and Z g 50 Ω to a load Z L. Determine the time-domain current through the load for: (a) Z L 50 j50 Ω (b) Z L 50 Ω, (c) Z L 0 (short circuit). Solution: π (a) Z L 50 j50 Ω, βl λ 0 375λ 36 (rad) 35. Z Γ L Z 0 50 Z L Z 0 50 j50 j50 50 j50 50 j e j63 43 Application of Eq. (.63) gives: Z in Z 0 Z L jz 0 tanβl Z 0 jz L tanβl j50 j50tan j 50 j50 tan j50 Ω

53 5 CHAPTER 50 Ω Transmission line ~ + V g Zin Z Z L 0 = 50 Ω - (50-j50) Ω Generator ~ V g + - Z g ~ I i l = λ z = -l z = 0 + ~ V i - Z in Load Figure P.30: Circuit for Problem.30(a). Using Eq. (.66) gives V V 0 g Z in Z g Z in j j50 50e j35 I L V 0 Z 0 i L t I L e jωt e jβl Γe jβl (V) e j35 Γ 50e j35 50 j e e j6π 08 t 68cos 6π 0 8 t (A) 0 j63 43 j35 45e e j63 43 j e 68e (A)

54 CHAPTER 53 (b) Z L 50 Ω Γ 0 Z in Z 0 50 Ω V e j e j35 (A) (V) V I L 0 50 Z 0 e j35 j35 3e 50 j35 i L t 3e e j6π 08 t 3cos 6π 0 t 35 (A) 8 (c) Z L 0 Γ 0 Z in jz 0 Z 0 tan35 Z 0 jz 0 tan35 j50 0 (Ω) 300 V 0 j50 j35 j50 e 50 j35 e j35 50e (V) V j35 I L 0 Z 0 Γ 50e j35 6e (A) 50 i L t 6cos 6π 0 t 35 (A) 8 Section -8: Power Flow on Lossless Line Problem.3 A generator with V g 300 V and Z g 50 Ω is connected to a load Z L 75 Ω through a 50-Ω lossless line of length l 0 5λ. (a) Compute Z in, the input impedance of the line at the generator end. (b) Compute I i and V i. (c) Compute the time-average power delivered to the line, P in V i I i. (d) Compute V L, I L, and the time-average power delivered to the load, P L V L IL. How does P in compare to P L? Explain. (e) Compute the time average power delivered by the generator, P g, and the time average power dissipated in Z g. Is conservation of power satisfied? Solution:

55 54 CHAPTER 50 Ω Transmission line ~ + V g Zin Z 0 - = 50 Ω 75 Ω Generator ~ V g + - Z g ~ I i l = 0.5 λ z = -l z = 0 + ~ V i - Z in Load Figure P.3: Circuit for Problem.3. (a) (b) π βl λ Z in Z 0 I i V i 0 5λ 54 Z L jz 0 tanβl Z 0 jz L tanβl j50tan j75tan j6 35 Ω V g Z g Z in j e j0 6 (A) I i Z in 3 4e j j e j 46 (V)

56 CHAPTER 55 (c) (d) P in V i I i Z Γ L Z 0 50 Z L Z V 0 V L V 0 I L V e j e j0 6 0 V i e jβl Γe jβl 43 6e j 46 e j54 0 e j54 Z 0 P L cos 6 6 (W) j54 50e (V) (A) j54 Γ 50e j e Γ 50e j54 j e j54 80e 4e j54 6 (W) V L IL P L P in, which is as expected because the line is lossless; power input to the line ends up in the load. (e) Power delivered by generator: P g V g I i e j cos (W) Power dissipated in Z g : P Z g I i V Z g Note : P g P Z g P in W. Problem.3 (V) I i I i Z g I i Z g (W) If the two-antenna configuration shown in Fig. -4 (P.3) is connected to a generator with V g 50 V and Z g 50 Ω, how much average power is delivered to each antenna? Solution: Since line is λ in length, the input impedance is the same as Z L 75 Ω. The same is true for line 3. At junction C D, we now have two 75-Ω impedances in parallel, whose combination is Ω. Line is λ long. Hence at A C, input impedance of line is 37.5 Ω, and V I i g 50 Z g Z in (A) 5

57 56 CHAPTER 50 Ω λ/ A + 50 V Z in Line - B Generator C D λ/ Line Line 3 Z L = 75 Ω (Antenna ) λ/ Z L = 75 Ω (Antenna ) P in Figure P.3: Antenna configuration for Problem.3. I i V i This is divided equally between the two antennas (W). I i I i Zin (W) Hence, each antenna receives Problem.33 For the circuit shown in Fig. -4 (P.33), calculate the average incident power, the average reflected power, and the average power transmitted into the infinite 00-Ω line. The λ line is lossless and the infinitely long line is slightly lossy. (Hint: The input impedance of an infinitely long line is equal to its characteristic impedance so long as α 0.) Solution: Considering the semi-infinite transmission line as equivalent to a load (since all power sent down the line is lost to the rest of the circuit), Z L Z 00 Ω. Since the feed line is λ in length, Eq. (.76) gives Z in Z L 00 Ω and βl π λ λ π, so e jβl. From Eq. (.49a), Z Γ L Z 0 50 Z L Z

58 CHAPTER Ω λ/ + V Z 0 = 50 Ω Z = 00 Ω - i P av t P av r P av Figure P.33: Line terminated in an infinite line. Also, converting the generator to a phasor gives V g j0 e (V). Plugging all these results into Eq. (.66), 00 V 0 V g Z in Z g Z in e jβl Γe jβl j80 e (V) From Eqs. (.84), (.85), and (.86), Pav i V 0 e Z 0 j mw Pav r Γ Pav i 0 mw mw Pav t P av Pav i Pav 0 r 0 mw mw 8 9 mw 3 Problem.34 An antenna with a load impedance Z L 75 j5 Ω is connected to a transmitter through a 50-Ω lossless transmission line. If under matched conditions (50-Ω load), the transmitter can deliver 0 W to the load, how much power does it deliver to the antenna? Assume Z g Z 0.

59 58 CHAPTER V V 0 g Z in Z g Z in e jβl Γe jβl V g Z 0 Γe jβl Γe jβl Z 0 Z 0 Γe jβl Γe V jβl g e Γe jβl Γe jβl V jβl g e Γe jβl Γe jβl Solution: From Eqs. (.66) and (.6), Thus, in Eq. (.86), V ge jβl P av V 0 Z 0 Γ V ge jβl Z 0 Γ V g e jβl Γe jβl 8Z 0 Γ Under the matched condition, Γ 0 and P L 0 W, so V g 8Z 0 0 W. When Z L 75 j5 Ω, from Eq. (.49a), Z Γ L Z 0 75 Z L Z 0 j5 50 Ω Ω 75 j5 Ω Ω 77e j33 so P av 0 W Γ 0 W W. Section -9: Smith Chart Problem.35 to a load impedance: (a) Z L 3Z 0, (b) Z L j Z 0, (c) Z L jz 0, (d) Z L 0 (short circuit). Solution: Refer to Fig. P.35. Use the Smith chart to find the reflection coefficient corresponding (a) Point A is z L 3 j0. Γ 0 0 5e 9 (b) Point B is z L j. Γ 0 6e 7 53 (c) Point C is z L j. 0 Γ 0e (d) Point D is z L 0 80 j0. Γ 0e 0

60 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < D INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) A B C ANGLE OF REFLECTION COEFFICIENT IN DEGREES Figure P.35: Solution of Problem.35. Problem.36 Use the Smith chart to find the normalized load impedance corresponding to a reflection coefficient: (a) Γ 0 5, (b) Γ , (c) Γ, (d) Γ , (e) Γ 0, (f) Γ j. Solution: Refer to Fig. P.36.

61 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < C INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) F E B D A ANGLE OF REFLECTION COEFFICIENT IN DEGREES Figure P.36: Solution of Problem.36. (a) Point A is Γ 0 5 at z L 3 j0. (b) Point B is Γ 0 5e j60 at z L j 5. (c) Point C is Γ at z L 0 j0. (d) Point D is Γ 0 3e j30 at z L 60 j0 53. (e) Point E is Γ 0 at z L j0. (f) Point F is Γ j at z L 0 j. Problem.37 On a lossless transmission line terminated in a load Z L 00 Ω, the standing-wave ratio was measured to be.5. Use the Smith chart to find the two possible values of Z 0.

62 CHAPTER 6 Solution: Refer to Fig. P.37. S 5 is at point L and the constant SWR circle is shown. z L is real at only two places on the SWR circle, at L, where z L S 5, and L, where z L S 0 4. so Z 0 Z L z L 00 Ω 5 40 Ω and Z 0 Z L z L 00 Ω Ω ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) L RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) L ANGLE OF REFLECTION COEFFICIENT IN DEGREES Figure P.37: Solution of Problem.37. Problem.38 A lossless 50-Ω transmission line is terminated in a load with Z L 50 j5 Ω. Use the Smith chart to find the following: (a) the reflection coefficient Γ, (b) the standing-wave ratio, (c) the input impedance at 0 35λ from the load,

63 CHAPTER (d) the input admittance at 0 35λ from the load, (e) the shortest line length for which the input impedance is purely resistive, (f) the position of the first voltage maximum from the load ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Z-IN RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) Z-LOAD 0.36 θ r SWR λ λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES Figure P.38: Solution of Problem.38. Solution: Refer to Fig. P.38. The normalized impedance is at point Z-LOAD. (a) the point θ r. z L 50 j5 Ω 50 Ω j0 5 Γ 0 4e j76 0 The angle of the reflection coefficient is read of that scale at

64 CHAPTER 63 (b) At the point SWR: S 64. (c) Z in is 0 350λ from the load, which is at 0 44λ on the wavelengths to generator scale. So point Z-IN is at 0 44λ 0 350λ 0 494λ on the WTG scale. At point Z-IN: Z in z in Z j Ω 30 5 j 09 Ω (d) At the point on the SWR circle opposite Z-IN, Y in y in Z 0 64 j Ω 3 7 j 7 ms (e) Traveling from the point Z-LOAD in the direction of the generator (clockwise), the SWR circle crosses the x L 0 line first at the point SWR. To travel from Z-LOAD to SWR one must travel 0 50λ 0 44λ 0 06λ. (Readings are on the wavelengths to generator scale.) So the shortest line length would be 0 06λ. (f) The voltage max occurs at point SWR. From the previous part, this occurs at z 0 06λ. Problem.39 A lossless 50-Ω transmission line is terminated in a short circuit. Use the Smith chart to find (a) the input impedance at a distance 3λ from the load, (b) the distance from the load at which the input admittance is Y in j0 04 S. Solution: Refer to Fig. P.39. (a) For a short, z in 0 j0. This is point Z-SHORT and is at 0 000λ on the WTG scale. Since a lossless line repeats every λ, traveling 3λ toward the generator is equivalent to traveling 0 3λ toward the generator. This point is at A : Z-IN, and Z in z in Z 0 0 j Ω j54 Ω (b) The admittance of a short is at point Y -SHORT and is at 0 50λ on the WTG scale: y in Y in Z 0 j0 04 S 50 Ω j which is point B : Y -IN and is at 0 34λ on the WTG scale. Therefore, the line length is 0 34λ 0 50λ 0 074λ. Any integer half wavelengths farther is also valid.

65 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) Z-SHORT RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) λ B:Y-IN A:Z-IN λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES Y-SHORT Figure P.39: Solution of Problem.39. Problem.40 Use the Smith chart to find y L if z L 5 j0 7. Solution: Refer to Fig. P.40. The point Z represents 5 j0 7. The reciprocal of point Z is at point Y, which is at 0 55 j0 6.

66 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Y Z ANGLE OF REFLECTION COEFFICIENT IN DEGREES Figure P.40: Solution of Problem.40. Problem.4 A lossless 00-Ω transmission line 3λ 8 in length is terminated in an unknown impedance. If the input impedance is Z in j 5 Ω, (a) use the Smith chart to find Z L. (b) What length of open-circuit line could be used to replace Z L? Solution: Refer to Fig. P.4. z in Z in Z 0 j 5 Ω 00 Ω 0 0 j0 05 which is at point Z-IN and is at 0 004λ on the wavelengths to load scale. (a) Point Z-LOAD is 0 375λ toward the load from the end of the line. Thus, on the wavelength to load scale, it is at 0 004λ 0 375λ 0 379λ. Z L z L Z 0 0 j Ω j95 Ω

67 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < Z-IN INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Z-LOAD λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES Z-OPEN λ Figure P.4: Solution of Problem.4. (b) An open circuit is located at point Z-OPEN, which is at 0 50λ on the wavelength to load scale. Therefore, an open circuited line with Z in j0 05 must have a length of 0 50λ 0 004λ 0 46λ. Problem.4 Smith chart to find Γ, Z L, and Z in. A 75-Ω lossless line is 0 6λ long. If S 8 and θ r 60, use the Solution: Refer to Fig. P.4. The SWR circle must pass through S 8 at point SWR. A circle of this radius has Γ S S 0 9

68 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Z-IN λ Z-LOAD SWR Figure P.4: Solution of Problem θ r ANGLE OF REFLECTION COEFFICIENT IN DEGREES The load must have a reflection coefficient with θ r 60. The angle of the reflection coefficient is read off that scale at the point θ r. The intersection of the circle of constant Γ and the line of constant θ r is at the load, point Z-LOAD, which has a value z L 5 j0 6. Thus, Z L z L Z 0 5 j Ω 86 5 j46 6 Ω A 0 6λ line is equivalent to a 0 λ line. On the WTG scale, Z-LOAD is at 0 333λ, so Z-IN is at 0 333λ 0 00λ 0 433λ and has a value z in 0 63 j0 9

69 CHAPTER Therefore Z in z in Z j Ω 47 0 j 8 Ω. Problem.43 Using a slotted line on a 50-Ω air-spaced lossless line, the following measurements were obtained: S 6, V max occurred only at 0 cm and 4 cm from the load. Use the Smith chart to find Z L ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < λ INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Z-LOAD SWR ANGLE OF REFLECTION COEFFICIENT IN DEGREES Figure P.43: Solution of Problem.43. Solution: Refer to Fig. P.43. The point SWR denotes the fact that S 6. This point is also the location of a voltage maximum. From the knowledge of the locations of adjacent maxima we can determine that λ 4 0 cm cm 8 cm. 0 cm Therefore, the load is λ 0 357λ from the first voltage maximum, which is at 0 50λ on the WTL scale. 8 cm Traveling this far on the SWR circle we find point Z-LOAD

70 CHAPTER 69 at 0 50λ 0 357λ 0 500λ 0 07λ on the WTL scale, and here z L 0 8 j0 39 Therefore Z L z L Z j Ω 4 0 j9 5 Ω. Problem.44 At an operating frequency of 5 GHz, a 50-Ω lossless coaxial line with insulating material having a relative permittivity ε r 5 is terminated in an antenna with an impedance Z L 50 Ω. Use the Smith chart to find Z in. The line length is 30 cm. Solution: To use the Smith chart the line length must be converted into wavelengths. Since β π λ and u p ω β, π λ πu p c m/s β ω ε r f Hz 0 04 m 9 Hence, l 0 30 m 0 04 m λ 7 5λ. Since this is an integral number of half wavelengths, Z in Z L 50 Ω Section -0: Impedance Matching Problem.45 A 50-Ω lossless line 0 6λ long is terminated in a load with Z L 50 j5 Ω. At 0 3λ from the load, a resistor with resistance R 30 Ω is connected as shown in Fig. -43 (P.45(a)). Use the Smith chart to find Z in. Z in Z 0 = 50 Ω 30 Ω Z 0 = 50 Ω Z L 0.3λ 0.3λ Z L = (50 + j5) Ω Figure P.45: (a) Circuit for Problem.45.

71 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < λ Y-IN INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Y-LOAD Z-LOAD 0.36 A λ B Z-IN Figure P.45: (b) Solution of Problem ANGLE OF REFLECTION COEFFICIENT IN DEGREES Solution: Refer to Fig. P.45(b). Since the 30-Ω resistor is in parallel with the input impedance at that point, it is advantageous to convert all quantities to admittances. z L Z L Z 0 50 j5 Ω 50 Ω j0 5 and is located at point Z-LOAD. The corresponding normalized load admittance is at point Y -LOAD, which is at 0 394λ on the WTG scale. The input admittance of the load only at the shunt conductor is at 0 394λ 0 300λ 0 500λ 0 94λ and is denoted by point A. It has a value of y ina 37 j0 45

72 CHAPTER 7 The shunt conductance has a normalized conductance 50 Ω g Ω The normalized admittance of the shunt conductance in parallel with the input admittance of the load is the sum of their admittances: y inb g y ina j j0 45 and is located at point B. On the WTG scale, point B is at 0 4λ. The input admittance of the entire circuit is at 0 4λ 0 300λ 0 500λ 0 04λ and is denoted by point Y-IN. The corresponding normalized input impedance is at Z-IN and has a value of z in 9 j 4 Thus, Z in z in Z 0 9 j 4 50 Ω 95 j70 Ω Problem.46 A 50-Ω lossless line is to be matched to an antenna with Z L 75 j0 Ω using a shorted stub. Use the Smith chart to determine the stub length and the distance between the antenna and the stub. Solution: Refer to Fig. P.46(a) and Fig. P.46(b), which represent two different solutions. z L Z L Z 0 75 j0 Ω 50 Ω 5 j0 4 and is located at point Z-LOAD in both figures. Since it is advantageous to work in admittance coordinates, y L is plotted as point Y -LOAD in both figures. Y -LOAD is at 0 04λ on the WTG scale. For the first solution in Fig. P.46(a), point Y -LOAD-IN- represents the point at which g on the SWR circle of the load. Y -LOAD-IN- is at 0 45λ on the WTG scale, so the stub should be located at 0 45λ 0 04λ 0 04λ from the load (or some multiple of a half wavelength further). At Y-LOAD-IN-, b 05, so a stub with an input admittance of y stub 0 j0 5 is required. This point is Y -STUB-IN- and is at 0 43λ on the WTG scale. The short circuit admittance

73 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < λ INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Y-STUB-IN Y-LOAD Y-LOAD-IN Z-LOAD λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES Y-SHT Figure P.46: (a) First solution to Problem.46. is denoted by point Y -SHT, located at 0 50λ. Therefore, the short stub must be 0 43λ 0 50λ 0 73λ long (or some multiple of a half wavelength longer). For the second solution in Fig. P.46(b), point Y -LOAD-IN- represents the point at which g on the SWR circle of the load. Y -LOAD-IN- is at 0 355λ on the WTG scale, so the stub should be located at 0 355λ 0 04λ 0 34λ from the load (or some multiple of a half wavelength further). At Y-LOAD-IN-, b 0 5, so a stub with an input admittance of y stub 0 j0 5 is required. This point is Y -STUB-IN- and is at 0 077λ on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at 0 50λ. Therefore, the short stub must be 0 077λ 0 50λ 0 500λ 0 37λ long (or some multiple of a half wavelength

74 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < Y-STUB-IN- INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) 0.37 λ Y-LOAD Z-LOAD Y-LOAD-IN λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES Y-SHT Figure P.46: (b) Second solution to Problem.46. longer). Problem.47 Repeat Problem.46 for a load with Z L 00 j50 Ω. Solution: Refer to Fig. P.47(a) and Fig. P.47(b), which represent two different solutions. z L Z L Z 0 00 j50 Ω 50 Ω j and is located at point Z-LOAD in both figures. Since it is advantageous to work in admittance coordinates, y L is plotted as point Y -LOAD in both figures. Y -LOAD is at 0 463λ on the WTG scale.

75 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < λ INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) 0.44 RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Y-LOAD Y-STUB-IN Y-LOAD-IN Z-LOAD λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES Y-SHT Figure P.47: (a) First solution to Problem.47. For the first solution in Fig. P.47(a), point Y -LOAD-IN- represents the point at which g on the SWR circle of the load. Y -LOAD-IN- is at 0 6λ on the WTG scale, so the stub should be located at 0 6λ 0 463λ 0 500λ 0 99λ from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-, b, so a stub with an input admittance of y stub 0 j is required. This point is Y-STUB-IN- and is at 0 375λ on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at 0 50λ. Therefore, the short stub must be 0 375λ 0 50λ 0 5λ long (or some multiple of a half wavelength longer). For the second solution in Fig. P.47(b), point Y -LOAD-IN- represents the point at which g on the SWR circle of the load. Y -LOAD-IN- is at 0 338λ on the

76 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) λ RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) Y-LOAD Y-STUB-IN λ Z-LOAD Y-LOAD-IN ANGLE OF REFLECTION COEFFICIENT IN DEGREES Y-SHT Figure P.47: (b) Second solution to Problem.47. WTG scale, so the stub should be located at 0 338λ 0 463λ 0 500λ 0 375λ from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-, b, so a stub with an input admittance of y stub 0 j is required. This point is Y-STUB-IN- and is at 0 5λ on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at 0 50λ. Therefore, the short stub must be 0 5λ 0 50λ 0 500λ 0 375λ long (or some multiple of a half wavelength longer). Problem.48 Use the Smith chart to find Z in of the feed line shown in Fig. -44 (P.48(a)). All lines are lossless with Z 0 50 Ω.

77 76 CHAPTER Z = (50 + j50) Ω 0.3λ Z 0.3λ Z in 0.7λ Z Z = (50 - j50) Ω Figure P.48: (a) Circuit of Problem.48. Solution: Refer to Fig. P.48(b). and is at point Z-LOAD-. and is at point Z-LOAD-. z Z Z 0 50 j50 Ω 50 Ω j z Z Z 0 50 j50 Ω 50 Ω j Since at the junction the lines are in parallel, it is advantageous to solve the problem using admittances. y is point Y-LOAD-, which is at 0 4λ on the WTG scale. y is point Y-LOAD-, which is at 0 088λ on the WTG scale. Traveling 0 300λ from Y -LOAD- toward the generator one obtains the input admittance for the upper feed line, point Y-IN-, with a value of 97 j 0. Since traveling 0 700λ is equivalent to traveling 0 00λ on any transmission line, the input admittance for the lower line feed is found at point Y-IN-, which has a value of 97 j 0. The admittance of the two lines together is the sum of their admittances: 97 j 0 97 j j0 and is denoted Y-JUNCT λ from Y -JUNCT toward the generator is the input admittance of the entire feed line, point Y-IN, from which Z-IN is found. Z in z in Z 0 65 j Ω 8 5 j89 5 Ω

78 CHAPTER ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < λ INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) Y-LOAD- Y-IN RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) λ Y-LOAD Z-LOAD- Z-LOAD Y-IN- Y-IN Z-IN Y-JUNCT ANGLE OF REFLECTION COEFFICIENT IN DEGREES λ Figure P.48: (b) Solution of Problem.48. Problem.49 Repeat Problem.48 for the case where all three transmission lines are λ 4 in length. Solution: Since the transmission lines are in parallel, it is advantageous to express loads in terms of admittances. In the upper branch, which is a quarter wave line, Y0 Y in Z Y Z0

79 78 CHAPTER and similarly for the lower branch, Thus, the total load at the junction is Y0 Y in Z Y Z0 Y JCT Y in Y in Z Z Z 0 Therefore, since the common transmission line is also quarter-wave, Z in Z 0 Z JCT Z 0Y JCT Z Z 50 j50 Ω 50 j50 Ω 00 Ω Section -: Transients on Transmission Lines Problem.50 Generate a bounce diagram for the voltage V z t for a -m long lossless line characterized by Z 0 50 Ω and u p c 3 (where c is the velocity of light) if the line is fed by a step voltage applied at t 0 by a generator circuit with V g 60 V and R g 00 Ω. The line is terminated in a load Z L 5 Ω. Use the bounce diagram to plot V t at a point midway along the length of the line from t 0 to t 5 ns. Solution: From Eq. (.4b), Also, Γ g R g Z 0 R g Z Γ L Z L Z 0 Z L Z V V g Z 0 R g Z V 50 T l u p 3 l c ns The bounce diagram is shown in Fig. P.50(a) and the plot of V t in Fig. P.50(b).

80 CHAPTER 79 Γ = Γ g = 3 z = 0 Voltage Γ = Γ L = - 3 z = 0.5 m z = m + V = 0V 0 ns V -. V 5 ns 0 ns 0.74 V 0.5 V 5 ns V 5 ns t t Figure P.50: (a) Bounce diagram for Problem.50. V(0.5 m, t) 0 V 0 V 3.34 V. V.86 V.0 V t (ns) Figure P.50: (b) Time response of voltage.

81 80 CHAPTER Problem.5 Solution: From Eq. (.4a), Repeat Problem.50 for the current I on the line. Γ g R g Z 0 R g Z Γ L Z L Z 0 Z L Z V I g 60 R g Z A 50 The bounce diagram is shown in Fig. P.5(a) and I t in Fig. P.5(b). Γ = -Γ g = - 3 z = 0 Current Γ = -Γ L = 3 z = 0.5 m z = m + I = 0.4 A 0.33 A 5 ns 0 ns A A 5 ns 0 ns A 5 ns t t Figure P.5: (a) Bounce diagram for Problem.5.

82 CHAPTER 8 I(0.5 m, t) 0.4 A A A A A t (ns) Figure P.5: (b) Time response of current. Problem.5 In response to a step voltage, the voltage waveform shown in Fig. -45 (P.5) was observed at the sending end of a lossless transmission line with R g 50 Ω, Z 0 50 Ω, and ε r 5. Determine (a) the generator voltage, (b) the length of the line, and (c) the load impedance. V(0, t) 5 V 3 V 0 6 µs z Figure P.5: Observed voltage at sending end. Solution: (a) From the figure, V which gives V g V 0 V. 5 V. Applying Eq. (.4b), V V g Z 0 R g Z 0 V g Z 0 Z 0 Z 0 V g

83 8 CHAPTER (b) u p c 5 08 m/s. The first change in the waveform occurs at t 6 µs. But t l u p. Hence, tµ l p m ε r and the change in level from 5 V down (c) Since R g Z 0, Γ g 0. Hence V to 3 V is due to V V. But V Γ L V or Γ L V V From Z L Γ L Z 0 Γ L Ω Problem.53 In response to a step voltage, the voltage waveform shown in Fig..46 (P.53) was observed at the sending end of a shorted line with Z 0 50 Ω and ε r 4. Determine V g, R g, and the line length. V(0, t) V 0 3 V 7 µs 4 µs 0.75 V z Figure P.53: Observed voltage at sending end. Solution: Hence, l 55 m. c u p ε r m/s µs 7 0 s 6 l l u p 5 0 8

84 CHAPTER 83 From the voltage waveform, V is V. At t 7µs, the voltage at the sending end V z 0 t 7µs V Γ L V Γ g Γ L V Γ g V (because Γ L Hence, 3 V Γ g V, or Γ g 0 5. From Eq. (.8), Also, which gives V g 9 V. R g Γ g Z 0 Γ g Ω V V g Z 0 R g Z 0 or V g Problem.54 Suppose the voltage waveform shown in Fig. -45 was observed at the sending end of a 50-Ω transmission line in response to a step voltage introduced by a generator with V g 5 V and an unknown series resistance R g. The line is km in length, its velocity of propagation is 0 8 m/s, and it is terminated in a load Z L 00 Ω. (a) Determine R g. (b) Explain why the drop in level of V 0 t at t 6 µs cannot be due to reflection from the load. (c) Determine the shunt resistance R f and the location of the fault responsible for the observed waveform. Solution: V(0, t) 5 V 3 V 0 6 µs z Figure P.54: Observed voltage at sending end.

85 84 CHAPTER (a) From Fig. -45, V 5 V. Hence, V V g Z 0 R g Z R g 50 which gives R g 00 Ω and Γ g 3. (b) Roundtrip time delay of pulse return from the load is l T u p µs which is much longer than 6 µs, the instance at which V 0 t drops in level. (c) The new level of 3 V is equal to V plus V plus V, V V V 5 5Γ f 5Γ f Γ g 3 (V) which yields Γ f 0 3. But Z Γ f Lf Z 0 Z Lf Z which gives Z Lf 6 9 Ω. Since Z Lf is equal to R f and Z 0 in parallel, R f Ω. Problem.55 A generator circuit with V g 00 V and R g 5 Ω was used to excite a 75-Ω lossless line with a rectangular pulse of duration τ 0 4 µs. The line is 00 m long, its u p 0 8 m/s, and it is terminated in a load Z L 5 Ω. (a) Synthesize the voltage pulse exciting the line as the sum of two step functions, V g t and V g t. (b) For each voltage step function, generate a bounce diagram for the voltage on the line. (c) Use the bounce diagrams to plot the total voltage at the sending end of the line. Solution: (a) pulse length 0 4 µs. with V g V g V g t V g t V g t t 00U t (V) t 00U t 0 4 µs (V)

86 CHAPTER V Ω t = 0 Z 0 = 75 Ω 5 Ω 00 m Figure P.55: (a) Circuit for Problem V V(t) V g (t) t -00 V 0.4 µs V g (t) Figure P.55: (b) Solution of part (a). (b) l 00 T u p 0 8 µs We will divide the problem into two parts, one for V g t and another for V g t for V g t will mimic the solution for V g t and then we will use superposition to determine the solution for the sum. The solution, except for a reversal in sign and a delay by 0 4 µs. For V g t 00U t : Γ g R g Z 0 R g Z Z Γ L L Z 0 75 Z L Z

87 86 CHAPTER V V Z 0 R g Z V 75 V V g Z L R g Z L V 5 (i) V 0 t at sending end due to V g t Γ = Γ g = - z = 0 t = 0 : V g (t) Γ = Γ L = 4 z = 00 m + V = 50V µs 37.5V -8.75V µs 4 µs -4.69V.34V 3 µs 0.56V 5 µs 6 µs -0.8V t t Figure P.55: (c) Bounce diagram for voltage in reaction to V g t.

88 CHAPTER 87 (ii) V 0 t at sending end due to V g t Γ = Γ g = - z = 0 t = 0.4 µs : V g (t) Γ = Γ L = 4 z = 00 m + V = -50V.4 µs -37.5V 8.75V.4 µs 4.4 µs 4.69V -.34V 3.4 µs -0.56V 5.4 µs 6.4 µs 0.8V t t Figure P.55: (d) Bounce diagram for voltage in reaction to V g t.

89 88 CHAPTER (b) (i) V 0 t at sending end due to V g t : V ( 0, t ) 50V t (µs) (ii) V 0 t at sending end: Figure P.55: (e) V 0 t. V ( 0, t ) t (µs) -50V Figure P.55: (f) V 0 t.

90 CHAPTER 89 (iii) Net voltage V 0 t V 0 t V 0 t : V ( 0, t ) 50V t (µs) Figure P.55: (g) Net voltage V 0 t. Problem.56 For the circuit of Problem.55, generate a bounce diagram for the current and plot its time history at the middle of the line. Solution: Using the values for Γ g and Γ L calculated in Problem.55, we reverse their signs when using them to construct a bounce diagram for the current. V I 50 Z 0 A 75 V I 50 Z 0 75 A V I Z L 33 A

91 90 CHAPTER Γ = -Γ g = I (t) z = 0 t = 0-0.5A µs -0.5A Γ = -Γ L = - 4 z = 00 m A µs 6.5mA 3 µs 4 µs 3.5mA -7.79mA 5 µs 6 µs -3.90mA t t Figure P.56: (a) Bounce diagram for I t in reaction to V g t.

92 CHAPTER 9 Γ = -Γ g = I (t) z = 0 t = 0.4 µs 0.5A.4 µs 0.5A Γ = -Γ L = - 4 z = 00 m -A.4 µs -6.5mA 3.4 µs 4.4 µs -3.5mA 7.79mA 5.4 µs 6.4 µs 3.90mA t t Figure P.56: (b) Bounce diagram for current I t in reaction to V g t.

93 9 CHAPTER (i) I l t due to V g t : I ( 00, t ) A t (µs) (ii) I l t due to V g t : Figure P.56: (c) I l t. I ( 00, t ) t (µs) A Figure P.56: (d) I l t.

94 CHAPTER 93 (iii) Net current I l t I l t I l t : I ( 0, t ) A t (µs) Figure P.56: (e) Total I l t. Problem.57 For the parallel-plate transmission line of Problem.3, the line parameters are given by: Find α, β, u p, and Z 0 at GHz. R (Ω/m) L 67 (nh/m) G 0 C 7 (pf/m)

95 94 CHAPTER Solution: At GHz, ω π f π 0 9 rad/s. Application of (.) gives: γ R jωl G jωc jπ jπ j049 j 049 e j tan 049 e j90 j e j90 Hence, 049e j89 95 e j90 54e j e j cos j34sin j34 α 0 06 Np/m β 34 rad/m ω u p π f π 0 9 β β R jωl Z 0 G jωc j89 049e 95 e j m/s 954e j0 05 3e j j0 0 Ω Problem.58 Z 0 = 300 Ω R = 600Ω L = 0.0 mh

96 CHAPTER 95 A 300-Ω lossless air transmission line is connected to a complex load composed of a resistor in series with an inductor, as shown in the figure. At 5 MHz, determine: (a) Γ, (b) S, (c) location of voltage maximum nearest to the load, and (d) location of current maximum nearest to the load. Solution: (a) (b) (c) Z L R jωl 600 jπ j68 Ω Z Γ L Z 0 Z L Z j j j j68 63e j9 6 θ l max r λ for θ r 4π 9 6 π m S Γ Γ π λ m (d) The locations of current maxima correspond to voltage minima and vice versa. Hence, the location of current maximum nearest the load is the same as location of voltage minimum nearest the load. Thus λ l min l max m l max λ 4 5 m Problem.59

97 96 CHAPTER Z 0 = 50 Ω R L = 75 Ω C =? A 50-Ω lossless transmission line is connected to a load composed of a 75-Ω resistor in series with a capacitor of unknown capacitance. If at 0 MHz the voltage standing wave ratio on the line was measured to be 3, determine the capacitance C. Solution: Noting that: Γ S S Z L R L jx C where X C Z Γ L Z 0 Z L Z 0 Z Γ L Z 0 Z Z L Z 0 ZL 0 Z ZL 0 Z Γ L Z ZL 0 Z 0 Z L ZL Z L Z ZL 0 Z 0 Z L ZL ωc Z L ZL R L jx C R L jx C RL XC Z 0 Z L ZL Z 0 R L jx C R L jx C Z 0 R L R Γ L XC Z0 Z 0 R L RL XC Z0 Z 0 R L Upon substituting Γ L 0 5, R L 75 Ω, and Z 0 50 Ω, and then solving for X C, we have X C 66 Ω Hence C ωx C π pf Problem.60 A 50-Ω lossless line is terminated in a load impedance Z L 30 j0 Ω.

98 CHAPTER 97 Z 0 = 50 Ω Z L = (30 - j 0) Ω (a) Calculate Γ and S. (b) It has been proposed that by placing an appropriately selected resistor across the line at a distance l max from the load (as shown in the figure below), where l max is the distance from the load of a voltage maximum, then it is possible to render Z i Z 0, thereby eliminating reflections back to the sending end. Show that the proposed approach is valid and find the value of the shunt resistance. A l max Z 0 = 50 Ω R Z L = (30 - j 0) Ω Z i Solution: (a) Z Γ L Z 0 50 Z L Z 0 30 j0 S Γ Γ j0 30 j j j0 j e j (b) We start by finding l max, the distance of the voltage maximum nearest to the load. Using (.56) with n, θ l max r λ λ 4π π λ 80 4π λ 0 33λ Applying (.63) at l l max 0 33λ, for which βl π λ 0 33λ 07 radians,

99 98 CHAPTER the value of Z in before adding the shunt resistance is: Z Z in L jz Z 0 0 tanβl j0 j50tan 50 j 30 j0 tan 07 0 j0 Ω Z 0 jz L tanβl Thus, at the location A (at a distance l max from the load), the input impedance is purely real. If we add a shunt resistor R in parallel such that the combination is equal to Z 0, then the new Z in at any point to the left of that location will be equal to Z 0. Hence, we need to select R such that or R 98 Ω. R 0 50 Problem.6 For the lossless transmission line circuit shown in the figure, determine the equivalent series lumped-element circuit at 400 MHz at the input to the line. The line has a characteristic impedance of 50 Ω and the insulating layer has ε r 5. Z in Z 0 = 50 Ω 75 Ω. m Solution: At 400 MHz, λ u p f βl π λ l π c f ε r π 5 Subtracting multiples of π, the remainder is: βl 0 8π rad 0 5 m

100 CHAPTER 99 Using (.63), Z Z in L jz Z 0 0 tanβl 75 j50tan j75tan 0 8π 5 38 j0 75 Ω Z 0 jz L tanβl Z in is equivalent to a series RL circuit with R Z in L or which is a very small inductor. R 5 38 Ω ωl π f L 0 75 Ω L 0 75 π H Problem.6 R g ~ V g + - Z 0 = 00 Ω Z L = (50 + j 00) Ω The circuit shown in the figure consists of a 00-Ω lossless transmission line terminated in a load with Z L 50 j00 Ω. If the peak value of the load voltage was measured to be V L V, determine: (a) the time-average power dissipated in the load, (b) the time-average power incident on the line, and (c) the time-average power reflected by the load.

101 00 CHAPTER Solution: (a) Γ Z L Z 0 Z L Z 0 50 j j j00 50 j00 0 6e j8 9 The time average power dissipated in the load is: (b) Hence, (c) P av I L R L V L Z L R L V L Z L R L W P i av P av P i av Γ P av 0 9 Γ W P r av Γ P i av W Problem.63 C l = 3λ/8 l B = 5λ/8 B l B r A Z in Z 0 = 00 Ω Z 0 = 50 Ω Z L = (75 - j 50) Ω Use the Smith chart to determine the input impedance Z in configuration shown in the figure. of the two-line

102 CHAPTER λ ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) λ RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) Br SWR Circle A ANGLE OF REFLECTION COEFFICIENT IN DEGREES λ Smith Chart Solution: to Z 0 : Starting at point A, namely at the load, we normalize Z L with respect z L Z L Z 0 75 j j (point A on Smith chart ) From point A on the Smith chart, we move on the SWR circle a distance of 5λ 8 to point B r, which is just to the right of point B (see figure). At B r, the normalized input impedance of line is: z in 0 48 j0 36 (point B r on Smith chart) Next, we unnormalize z in : Z in Z 0 z in j j8 Ω

103 CHAPTER λ ± > WAVELENGTHS TOWARD GENERATOR > < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) Bl CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) SWR Circle Smith Chart C λ λ ANGLE OF REFLECTION COEFFICIENT IN DEGREES To move along line, we need to normalize with respect to Z 0. We shall call this z L : z L Z in Z 0 4 j j0 8 (point B on Smith chart ) After drawing the SWR circle through point B, we move 3λ 8 towards the generator, ending up at point C on Smith chart. The normalized input impedance of line is: which upon unnormalizing becomes: z in 0 66 j 5 Z in 66 j5 Ω

104 CHAPTER 03 Problem.64 B l =? A Z 0 = 75 Ω Z =? Z L = 5 Ω A 5-Ω antenna is connected to a 75-Ω lossless transmission line. Reflections back toward the generator can be eliminated by placing a shunt impedance Z at a distance l from the load. Determine the values of Z and l. Solution: 0.50 λ ± > WAVELENGTHS TOWARD GENERATOR > 0.0 < WAVELENGTHS TOWARD LOAD < INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) SWR Circle λ A B λ RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) ANGLE OF REFLECTION COEFFICIENT IN DEGREES

105 04 CHAPTER The normalized load impedance is: z L (point A on Smith chart) The Smith chart shows A and the SWR circle. The goal is to have an equivalent impedance of 75 Ω to the left of B. That equivalent impedance is the parallel combination of Z in at B (to the right of the shunt impedance Z) and the shunt element Z. Since we need for this to be purely real, it s best to choose l such that Z in is purely real, thereby choosing Z to be simply a resistor. Adding two resistors in parallel generates a sum smaller in magnitude than either one of them. So we need for Z in to be larger than Z 0, not smaller. On the Smith chart, that point is B, at a distance l λ 4 from the load. At that point: z in 3 which corresponds to y in 0 33 Hence, we need y, the normalized admittance corresponding to the shunt impedance Z, to have a value that satisfies: y in y y y in z y Z Ω In summary, l λ 4 Z 5 Ω Problem.65 In response to a step voltage, the voltage waveform shown in the figure below was observed at the midpoint of a lossless transmission line with Z 0 50 Ω and u p 0 8 m/s. Determine: (a) the length of the line, (b) Z L, (c) R g, and (d) V g.

106 CHAPTER 05 V V (l/, t) -3 V t (µs) Solution: (a) Since it takes 3 µs to reach the middle of the line, the line length must be l u p m (b) From the voltage waveform shown in the figure, the duration of the first rectangle is 6 µs, representing the time it takes the incident voltage V to travel from the midpoint of the line to the load and back. The fact that the voltage drops to zero at t 9 µs implies that the reflected wave is exactly equal to V in magnitude, but opposite in polarity. That is, V V This in turn implies that Γ L, which means that the load is a short circuit: Z L 0 (c) After V arrives at the generator end, it encounters a reflection coefficient Γ g. The voltage at 5 µs is composed of: V V V V Γ L Γ L Γ g V V V Γ g From the figure, V V 3 4. Hence, Γ g 4

107 06 CHAPTER which means that (d) R g Γ g Γ g Z Ω V V g Z 0 R g Z 0 V g R g Z 0 Z V

108 07 Chapter 3: Vector Analysis Lesson #4 Chapter Section: 3- Topics: Basic laws of vector algebra Highlights: Vector magnitude, direction, unit vector Position and distance vectors Vector addition and multiplication - Dot product - Vector product - Triple product Special Illustrations: CD-ROM Module 3.

109 08 Lessons #5 and 6 Chapter Section: 3- Topics: Coordinate systems Highlights: Commonly used coordinate systems: Cartesian, cylindrical, spherical Choice is based on which one best suits problem geometry Differential surface vectors and differential volumes Special Illustrations: Examples 3-3 to 3-5 Technology Brief on GPS (CD-ROM) Global Positioning System The Global Positioning System (GPS), initially developed in the 980s by the U.S. Department of Defense as a navigation tool for military use, has evolved into a system with numerous civilian applications including vehicle tracking, aircraft navigation, map displays in automobiles, and topographic mapping. The overall GPS is composed of 3 segments. The space segment consists of 4 satellites (A), each circling Earth every hours at an orbital altitude of about,000 miles and transmitting continuous coded time signals. The user segment consists of hand-held or vehicle-mounted receivers that determine their own locations by receiving and processing multiple satellite signals. The third segment is a network of five ground stations, distributed around the world, that monitor the satellites and provide them with updates on their precise orbital information. GPS provides a location inaccuracy of about 30 m, both horizontally and vertically, but it can be improved to within m by differential GPS (see illustration).

110 09 Lesson #7 Chapter Section: 3-3 Topics: Coordinate transformations Highlights: Basic logic for decomposing a vector in one coordinate system into the coordinate variables of another system Transformation relations (Table 3-) Special Illustrations: Example 3-8

111 0 Lesson #8 Chapter Section: 3-4 Topics: Gradient operator Highlights: Derivation of T in Cartesian coordinates Directional derivative T in cylindrical and spherical coordinates Special Illustrations: Example 3-0(b) CD-ROM Modules 3.5 or 3.6 CD-ROM Demos (any )

112 Lesson #9 Chapter Section: 3-5 Topics: Divergence operator Highlights: Concept of flux Derivation of.e Divergence theorem Special Illustrations: CD-ROM Modules (any ) CD-ROM Demos (any or )

113 Lesson #0 Chapter Section: 3-6 Topics: Curl operator Highlights: Concept of circulation Derivation of x B Stokes s theorem Special Illustrations: Example 3-

114 3 Lesson # Chapter Section: 3-7 Topics: Laplacian operator Highlights: Definition of Definition of V E Special Illustrations: Technology Brief on X-Ray Computed Tomography X-Ray Computed Tomography Tomography is derived from the Greek words tome, meaning section or slice, and graphia, meaning writing. Computed tomography, also known as CT scan or CAT scan (for computed axial tomography), refers to a technique capable of generating 3-D images of the x-ray attenuation (absorption) properties of an object. This is in contrast with the traditional x-ray technique which produces only a -D profile of the object. CT was invented in 97 by British electrical engineer Godfrey Hounsfield, and independently by Allan Cormack, a South African-born American physicist. The two inventors shared the 979 Nobel Prize for Physiology or Medicine. Among diagnostic imaging techniques, CT has the decided advantage in having the sensitivity to image body parts on a wide range of densities, from soft tissue to blood vessels and bones.

115 4 CHAPTER 3 Chapter 3 Section 3-: Vector Algebra Problem 3. a unit vector in the direction of A. Solution: Vector A starts at point 3 and ends at point 0. Find A ˆx ŷ ẑ 0 3 ˆx ẑ3 A A ˆx â ẑ3 A 3 ˆx0 6 3 ẑ0 95 Problem 3. Given vectors A ˆx ŷ3 ẑ, B ˆx ŷ ẑ3, and C ˆx4 ŷ ẑ, show that C is perpendicular to both A and B. Solution: A C ˆx ŷ3 ẑ ˆx4 ŷ ẑ B C ˆx ŷ ẑ3 ˆx4 ŷ ẑ Problem 3.3 In Cartesian coordinates, the three corners of a triangle are P 0 4 4, P 4 4 4, and P 3 4. Find the area of the triangle. Solution: Let B P P ŷ8 and ˆx4 C P P 3 ẑ8 represent two sides of ˆx ŷ the triangle. Since the magnitude of the cross product is the area of the parallelogram (see the definition of cross product in Section 3-.4), half of this is the area of the triangle: A B C ˆx4 ŷ8 ˆx ŷ ẑ8 ˆx 8 8 ŷ 4 8 ẑ 4 8 ˆx64 ŷ3 ẑ where the cross product is evaluated with Eq. (3.7). Problem 3.4 Given A ˆx ŷ3 ẑ and B ˆxB x ŷ ẑb z : (a) find B x and B z if A is parallel to B; (b) find a relation between B x and B z if A is perpendicular to B

116 CHAPTER 3 5 Solution: (a) If A is parallel to B, then their directions are equal or opposite: â A â B, or From the y-component, A A B B ˆxB x ŷ ẑb z 4 Bx B z ˆx ŷ3 ẑ B x B z which can only be solved for the minus sign (which means that A and B must point in opposite directions for them to be parallel). Solving for Bx B z, Bx B 0 z From the x-component, and, from the z-component, B x B 56 x B z 3 This is consistent with our result for B x B z. These results could also have been obtained by assuming θ AB was 0 or 80 and solving A B A B, or by solving A B 0. (b) If A is perpendicular to B, then their dot product is zero (see Section 3-.4). Using Eq. (3.7), 0 A B B x 6 B z or B z 6 B x There are an infinite number of vectors which could be B and be perpendicular to A, but their x- and z-components must satisfy this relation. This result could have also been obtained by assuming θ AB 90 and calculating A B A B. Problem 3.5 Given vectors A ˆx ŷ ẑ3, B ˆx ŷ4, and C ŷ ẑ4, find

117 6 CHAPTER 3 (a) A and â, (b) the component of B along C, (c) θ AC, (d) A C, (e) A B C, (f) A B C, (g) ˆx B, and (h) A ŷ ẑ. Solution: (a) From Eq. (3.4), and, from Eq. (3.5), A 3 4 â A ˆx ŷ ẑ3 4 (b) The component of B along C (see Section 3-.4) is given by (c) From Eq. (3.), Bcosθ BC B C C θ AC cos A C AC cos cos (d) From Eq. (3.7), A C ˆx 4 3 ŷ ẑ 0 ˆx ŷ4 ẑ (e) From Eq. (3.7) and Eq. (3.7), A B C A ˆx6 ŷ8 ẑ Eq. (3.30) could also have been used in the solution. Also, Eq. (3.9) could be used in conjunction with the result of part (d). (f) By repeated application of Eq. (3.7), A B C A ˆx6 ŷ8 ẑ4 ˆx3 ŷ5 ẑ4 Eq. (3.33) could also have been used.

118 CHAPTER 3 7 (g) From Eq. (3.7), ˆx B ẑ4 (h) From Eq. (3.7) and Eq. (3.7), A ŷ ẑ ˆx3 ẑ ẑ Eq. (3.9) and Eq. (3.5) could also have been used in the solution. Problem 3.6 magnitude is 9 and whose direction is perpendicular to both A and B. Given vectors A ˆx ŷ ẑ3 and B ˆx3 ẑ, find a vector C whose Solution: The cross product of two vectors produces a new vector which is perpendicular to both of the original vectors. Two vectors exist which have a magnitude of 9 and are orthogonal to both A and B: one which is 9 units long in the direction of the unit vector parallel to A B, and one in the opposite direction. C A 9 A B ˆx ŷ ẑ3 9 ˆx3 ẑ B ŷ ˆx ẑ3 ˆx3 ẑ 9 ˆx ŷ3 ẑ3 ˆx ŷ8 67 ẑ 0 Problem 3.7 Given A ˆx x y ŷ y 3z ẑ 3x y, determine a unit vector parallel to A at point P. Solution: The unit vector parallel to A ˆx x y ŷ y 3z ẑ 3x y at the point P is A ŷ5 A ˆx ẑ4 ŷ5 ˆx 5 4 ẑ4 4 ˆx0 5 ŷ0 77 ẑ0 6 Problem 3.8 By expansion in Cartesian coordinates, prove: (a) the relation for the scalar triple product given by (3.9), and (b) the relation for the vector triple product given by (3.33). Solution: (a) Proof of the scalar triple product given by Eq. (3.9): From Eq. (3.7), A B ˆx A y B z A z B y ŷ A z B x A x B z ẑ A x B y A y B x

119 8 CHAPTER 3 B C ˆx B y C z B z C y ŷ B z C x B x C z ẑ B x C y B y C x C A ˆx C y A z C z A y ŷ C z A x C x A z ẑ C x A y C y A x Employing Eq. (3.7), it is easily shown that A B C A x B y C z B z C y A y B z C x B x C z A z B x C y B y C x B C A B x C y A z C z A y B y C z A x C x A z B z C x A y C y A x C A B C x A y B z A z B y C y A z B x A x B z C z A x B y A y B x which are all the same. (b) Proof of the vector triple product given by Eq. (3.33): The evaluation of the left hand side employs the expression above for B C with Eq. (3.7): A B C A ˆx B y C z B z C y ŷ B z C x B x C z ẑ B x C y B y C x ˆx A y B x C y B y C x A z B z C x B x C z ŷ A z B y C z B z C y A x B x C y B y C x ẑ A x B z C x B x C z A y B y C z B z C y while the right hand side, evaluated with the aid of Eq. (3.7), is B A C C A B B A x C x A y C y A z C z C A x B x A y B y A z B z ˆx B x A y C y A z C z C x A y B y A z B z ŷ B y A x C x A z C z C y A x B x A z B z ẑ B z A x C x A y C y C z A x B x A y B y By rearranging the expressions for the components, the left hand side is equal to the right hand side. Problem 3.9 Find an expression for the unit vector directed toward the origin from an arbitrary point on the line described by x and z. Solution: An arbitrary point on the given line is y. The vector from this point to is: A ˆx 0 ŷ 0 y ẑ 0 ˆx ŷy ẑ A y 4 5 y â A ẑ ˆx ŷy A 5 y

120 CHAPTER 3 9 Problem 3.0 Find an expression for the unit vector directed toward the point P located on the z-axis at a height h above the x y plane from an arbitrary point Q x y 3 in the plane z 3. Solution: Point P is at 0 0 h. Vector A from Q x y 3 to P 0 0 h is: A ˆx 0 x ŷ 0 y ẑ h 3 ŷy ˆxx ẑ h 3 A x y h 3 A ŷy â ˆxx ẑ h 3 A x y h 3 Problem 3. Find a unit vector parallel to either direction of the line described by x z 4 Solution: First, we find any two points on the given line. Since the line equation is not a function of y, the given line is in a plane parallel to the x z plane. For convenience, we choose the x z plane with y 0. For x 0, z 4. Hence, point P is at For z 0, x. Hence, point Q is at 0 0. Vector A from P to Q is: A ˆx 0 ŷ 0 0 ẑ 0 4 ˆx ẑ4 A ẑ4 â ˆx A 0 Problem 3. Two lines in the x y plane are described by the expressions: Line x y 6 Line 3x 4y 8 Use vector algebra to find the smaller angle between the lines at their intersection point. Solution: Intersection point is found by solving the two equations simultaneously: x 4y 3x 4y 8

121 0 CHAPTER (0, ) (0, -3) -0-5 A B (0, -3) θ AB Figure P3.: Lines and. The sum gives x 0, which, when used in the first equation, gives y 3. Hence, intersection point is 0 3. Another point on line is x 0, y 3. Vector A from 0 3 to 0 3 is A ˆx 0 ŷ 3 3 ˆx0 ŷ0 A A point on line is x 0, y. Vector B from 0 to 0 3 is B ˆx 0 ŷ 3 ˆx0 ŷ5 B Angle between A and B is Problem 3.3 θ AB cos A B A B cos A given line is described by x y Vector A starts at the origin and ends at point P on the line such that A is orthogonal to the line. Find an expression for A.

122 CHAPTER 3 Solution: We first plot the given line. Next we find vector B which connects point P 0 to P 4 0, both of which are on the line: B ˆx 4 0 ŷ 0 ˆx4 ŷ Vector A starts at the origin and ends on the line at P. If the x-coordinate of P is x, y P (0,) (0,0) A B P (4,0) x Figure P3.3: Given line and vector A. then its y-coordinate has to be 4 x in order to be on the line. Hence P is at x 4 x. Vector A is A ˆxx x ŷ 4 But A is perpendicular to the line. Hence, Hence, A B 0 ˆxx x ŷ 4 ˆx4 ŷ 0 4x 4 x 0 or 4 x A ˆx0 8 ŷ ˆx0 8 ŷ 6 Problem 3.4 Show that, given two vectors A and B,

123 CHAPTER 3 (a) the vector C defined as the vector component of B in the direction of A is given by C â B â A B A A where â is the unit vector of A, and (b) the vector D defined as the vector component of B perpendicular to A is given by D B A B A A Solution: (a) By definition, B â is the component of B along â. The vector component of B â along A is C â B â A A B A A B A A A (b) The figure shows vectors A, B, and C, where C is the projection of B along A. It is clear from the triangle that or B C D D B C B A B A A A C B D Figure P3.4: Relationships between vectors A, B, C, and D.

124 CHAPTER 3 3 Problem 3.5 A certain plane is described by x 3y 4z 6 Find the unit vector normal to the surface in the direction away from the origin. Solution: Procedure:. Use the equation for the given plane to find three points, P, P and P 3 on the plane.. Find vector A from P to P and vector B from P to P Cross product of A and B gives a vector C orthogonal to A and B, and hence to the plane. 4. Check direction of ĉ. Steps:. Choose the following three points:. Vector A from P to P P at P at P 3 at A ˆx 8 0 ŷ 0 0 ẑ 0 4 ẑ4 ˆx8 Vector B from P to P 3 6 B ˆx 0 0 ŷ 3 0 ẑ 0 4 ŷ 6 ẑ C A B ˆx A y B z A z B y ŷ A z B x A x B z ẑ A x B y A y B x 6 ˆx ŷ ẑ ˆx 64 ŷ3 3 ẑ

125 4 CHAPTER 3 Verify that C is orthogonal to A and B 64 8 A C B C C ˆx 64 3 ŷ3 ẑ 8 3 C ĉ C ˆx ŷ3 ẑ ˆx ŷ0 56 ẑ ĉ points away from the origin as desired. Problem 3.6 Given B ˆx z 3y ŷ x 3z ẑ x y, find a unit vector parallel to B at point P 0. Solution: At P 0, B ˆx ŷ 3 ẑ ˆx ŷ5 ẑ B ˆx ˆb ẑ ŷ5 B 5 ˆx ẑ ŷ5 7 Problem 3.7 When sketching or demonstrating the spatial variation of a vector field, we often use arrows, as in Fig. 3-5 (P3.7), wherein the length of the arrow is made to be proportional to the strength of the field and the direction of the arrow is the same as that of the field s. The sketch shown in Fig. P3.7, which represents the vector field E ˆrr, consists of arrows pointing radially away from the origin and their lengths increase linearly in proportion to their distance away from the origin. Using this arrow representation, sketch each of the following vector fields: (a) E ˆxy, (b) E ŷx, (c) E 3 ˆxx ŷy, (d) E 4 ˆxx ŷy, (e) E 5 ˆφr, (f) E 6 ˆrsinφ.

126 CHAPTER 3 5 y E E x E E Figure P3.7: Arrow representation for vector field E ˆrr (Problem 3.7). Solution: (a) y E E x E E P.3a: = - x^ y E

127 6 CHAPTER 3 (b) y E E x E E (c) E P3.7b: E ŷx y E E x E E P.3c: E 3 = x^ x + y^ y

128 CHAPTER 3 7 (d) E y E x E E (e) P.3d: E 4 = x^ x + y^ y y E E x E E P.3e: E 5 = φ^ r

129 8 CHAPTER 3 (f) y E E x E E P.3f: E 6 = r^ sinφ Problem 3.8 Use arrows to sketch each of the following vector fields: (a) E ŷy, ˆxx (b) E ˆφ, (c) E 3 ŷ x, (d) E 4 ˆrcosφ. Solution:

130 CHAPTER 3 9 (a) y E E E E x (b) P.4a: E = x^ x - y^ y y E E x E E P.4b: = - φ^ E

131 30 CHAPTER 3 (c) y E x E Indicates E is infinite (d) P.4c: E 3 = y^ (/x) y E E E E E x E E E E P.4d: E 4 = r^ cosφ

132 CHAPTER 3 3 Sections 3- and 3-3: Coordinate Systems Problem 3.9 Convert the coordinates of the following points from Cartesian to cylindrical and spherical coordinates: (a) P 0, (b) P 0 0, (c) P 3 3, (d) P 4. Solution: Use the coordinate variables column in Table 3-. (a) In the cylindrical coordinate system, P tan rad In the spherical coordinate system, P 0 tan 0 tan 5 π rad 07 rad Note that in both the cylindrical and spherical coordinates, φ is in Quadrant I. (b) In the cylindrical coordinate system, P In the spherical coordinate system, 0 0 tan rad 0 0 P 0 0 tan 0 0 tan rad 0 rad 0 0 Note that in both the cylindrical and spherical coordinates, φ is arbitrary and may take any value. (c) In the cylindrical coordinate system, P 3 tan 3 π 4 rad In the spherical coordinate system, P 3 3 tan 3 tan 0 44 rad π 4 rad Note that in both the cylindrical and spherical coordinates, φ is in Quadrant I.

133 3 CHAPTER 3 (d) In the cylindrical coordinate system, P 4 In the spherical coordinate system, P 4 tan 3π 4 rad tan 3 87 rad 3π 4 rad tan Note that in both the cylindrical and spherical coordinates, φ is in Quadrant II. Problem 3.0 Convert the coordinates of the following points from cylindrical to Cartesian coordinates: (a) P π 4, (b) P 3 0, (c) P 3 4 π 3. Solution: (a) P x y z P r cos φ r sin φ z P cos π 4 sin π 4 P 4 4 (b) P x y z P 3cos 0 3sin 0 P 3 0. (c) P 3 x y z P 3 4cos π 4sin π 3 P Problem 3. Convert the coordinates of the following points from spherical to cylindrical coordinates: (a) P 5 0 0, (b) P 5 0 π, (c) P 3 3 π 0. Solution: (a) P r φ z P Rsinθ φ Rcos θ P 5sin 0 0 5cos 0 P (b) P r φ z P 5sin 0 π 5cos 0 P 0 π 5.

134 CHAPTER 3 33 (c) P 3 r φ z P 3 3sin π 0 3cos π P Problem 3. Use the appropriate expression for the differential surface area ds to determine the area of each of the following surfaces: (a) r 3; 0 φ π 3; z, (b) r 5; π φ π; z 0, (c) r 5; φ π 4; z, (d) R ; 0 θ π 3; 0 φ π, (e) 0 R 5; θ π 3; 0 φ π. Also sketch the outlines of each of the surfaces. Solution: Φ = π/3 3 y 5 5 x (a) (b) (c) (d) (e) Figure P3.: Surfaces described by Problem 3.. (a) Using Eq. (3.43a), A z π 3 φ 0 r r 3 dφ dz 3φz π 3 φ 0 z 4π

135 34 CHAPTER 3 (b) Using Eq. (3.43c), A 5 r (c) Using Eq. (3.43b), A π z (d) Using Eq. (3.50b), A π 3 θ 0 π φ 0 φ π 5 r r z 0 dφ dr r φ 5 r π φ π π 4 φ π 4 dr dz rz z 5 r R sinθ dφ dθ R (e) Using Eq. (3.50c), 5 π Rsinθ dr θ π 3 dφ A R 0 Problem 3.3 φ 0 Find the volumes described by (a) r 5; π φ π; 0 z, (b) 0 R 5; 0 θ π 3; 0 φ π. Also sketch the outline of each volume. Solution: 4φcos θ π 3 θ 0 π φ 0 π R φsin π 3 π φ 0 5 R 0 5 3π z z 5 5 y y x (a) x (b) Figure P3.3: Volumes described by Problem 3.3. (a) From Eq. (3.44), π V z 0 φ π 5 r r dr dφ dz r φz 5 r π φ π z 0 π

136 CHAPTER 3 35 (b) From Eq. (3.50e), V π φ 0 π 3 θ 0 5 R 0 cosθ R3 3 φ R sin θ dr dθ dφ 5 R 0 π 3 π θ 0 φ 0 5π 3 Problem 3.4 A section of a sphere is described by 0 R, 0 θ 90 and 30 φ 90. Find: (a) the surface area of the spherical section, (b) the enclosed volume. Also sketch the outline of the section. Solution: z y x φ=30 o Figure P3.4: Outline of section.

137 36 CHAPTER 3 Problem 3.5 π S π 4 φ π 6 π θ 0 V R 0 R 3 π 3 0 R sinθ dθ dφ R π π cosθ π π 3 3 π φ π 6 π θ 0 R sin θ dr dθ dφ π π 8 π cosθ 0 6 8π (m ) (m 3 ) A vector field is given in cylindrical coordinates by E ˆrr cos φ ˆφr sinφ ẑz Point P π 3 is located on the surface of the cylinder described by r. At point P, find: (a) the vector component of E perpendicular to the cylinder, (b) the vector component of E tangential to the cylinder. Solution: (a) E n ˆr ˆr E ˆr ˆr ˆrr cosφ ˆφr sin φ ẑz ˆrr cos φ. At P π 3, E n ˆrcos π ˆr. (b) E t E E n ˆφr sinφ ẑz. At P π 3, E t ˆφsin π ẑ3 ẑ9. Problem 3.6 coordinates by Find: At a given point in space, vectors A and B are given in spherical A ˆR4 ˆθ ˆφ B ˆR ˆφ3 (a) the scalar component, or projection, of B in the direction of A, (b) the vector component of B in the direction of A, (c) the vector component of B perpendicular to A. Solution:

138 CHAPTER 3 37 (a) Scalar component of B in direction of A: A C B â B A (b) Vector component of B in direction of A: ˆR ˆφ ˆR4 ˆθ ˆφ C âc A C ˆR4 A ˆθ ˆφ 4 ˆR 09 ˆθ 05 ˆφ0 5 (c) Vector component of B perpendicular to A: Problem 3.7 D B C ˆR ˆφ3 ˆR 09 ˆθ 05 ˆφ0 5 ˆR0 09 ˆθ 05 ˆφ 48 Given vectors A ˆr cos φ 3z ˆφ r 4sin φ ẑ r z B ˆrsinφ ẑcosφ find (a) θ AB at π 0, (b) a unit vector perpendicular to both A and B at π 3. Solution: It doesn t matter whether the vectors are evaluated before vector products are calculated, or if the vector products are directly calculated and the general results are evaluated at the specific point in question. (a) At π 0, A ˆφ8 ẑ and B ˆr. From Eq. (3.), 0 θ AB cos A B AB cos AB 90 (b) At π 3, A ˆr 7 ˆφ4 3 B and ˆr 3 ẑ. Since A B is perpendicular to both A and B, a unit vector perpendicular to both A and B is given by A B A B ˆr ˆφ ẑ ˆr0 487 ˆφ0 8 ẑ 0 843

139 38 CHAPTER 3 Problem 3.8 Find the distance between the following pairs of points: (a) P 3 and P 3 in Cartesian coordinates, (b) P 3 π 4 3 and P 4 3 π 4 4 in cylindrical coordinates, (c) P 5 4 π 0 and P 6 3 π 0 in spherical coordinates. Solution: (a) d (b) (c) d r r r r cos φ φ z z 9 π π 3 cos d R R R R cosθ cosθ sinθ sinθ cos φ φ cos πcos π sin π 0 0 sinπcos Problem 3.9 Determine the distance between the following pairs of points: (a) P and P 0 3, (b) P 3 π 3 and P 4 4 π 3, (c) P 5 3 π π and P 6 4 π π. Solution: (a) From Eq. (3.66), (b) From Eq. (3.67), d d cos π π

140 CHAPTER 3 39 (c) From Eq. (3.68), d cos π cos π sinπsin π cos π π 5 Problem 3.30 Transform the following vectors into cylindrical coordinates and then evaluate them at the indicated points: (a) A ˆx x y at P 3, (b) B ˆx y x ŷ x y at P 0, (c) C ˆxy x y ŷx x y ẑ4 at P 3, (d) D ˆRsinθ ˆθcosθ ˆφcos φ at P 4 π π 4, (e) E ˆRcosφ ˆθsinφ ˆφsin θ at P 5 3 π π. Solution: From Table 3-: (a) A ˆrcosφ ˆφsinφ r cos φ r sinφ ˆrr cos φ cos φ sinφ ˆφr sin φ cos φ sinφ P tan A P ˆr0 447 ˆφ ˆr 34 ˆφ 68 (b) B ˆrcosφ ˆφsinφ r sin φ r cosφ ˆφcosφ ˆrsinφ r cos φ r sinφ ˆrr sin φcos φ ˆφr cos φ sin φ ˆrr sin φ ˆφr cos φ P 0 tan 0 0 B P ˆr ˆφ (c) (d) r C ˆrcosφ ˆφsinφ sin φ ˆφcos r φ r ˆrsinφ cos φ r ˆrsinφcos φ sin φ cosφ ˆφ sin 3 φ cos φ 3 ẑ4 P 3 tan 45 C P 3 ˆr0 707 ẑ4 ẑ4 D ˆrsinθ ẑcosθ sinθ ˆrcos θ ẑsin θ cos θ ˆφcos φ ˆr ˆφcos φ

141 40 CHAPTER 3 P 4 sin π π 4 cos π 45 0 D P 4 ˆr ˆφ (e) E ˆrsin θ ẑcos θ cos φ ˆrcosθ ẑsinθ sin φ ˆφsin θ P 5 3 π π E P 5 ˆrsin π ẑcos π cos π ˆrcos π ẑsin π sinπ ˆφsin π ˆr ˆφ Problem 3.3 Transform the following vectors into spherical coordinates and then evaluate them at the indicated points: (a) A ˆxy ŷxz ẑ4 at P, (b) B ŷ x y z ẑ x y at P 0, (c) C ˆrcos ˆφsin φ φ ẑcos φsinφ at P 3 π 4, and (d) D ˆxy x y ŷx x y ẑ4 at P 4. Solution: From Table 3-: (a) A ˆRsin θcosφ ˆθcosθcos φ ˆφsinφ Rsinθsinφ ˆRsinθsin φ ˆθcosθsin φ ˆφcos φ Rsinθcos φ Rcosθ ˆRcosθ ˆθsinθ 4 ˆR R sin θsin φcosφ sin θsin φ cosθ 4cosθ ˆθ R sinθcos θsinφcos φ sin θsin φ cosθ 4sinθ ˆφR sinθ cos θcos φ sinθsin φ 3 P tan tan (b) A P ˆR 856 ˆθ 888 ˆφ 3 B ˆRsinθsinφ ˆθcosθsin φ ˆφcosφ R ˆRcosθ ˆRR sinθ sin φ sinθcosθ ˆθR cos θsinφ sin θ 3 ˆφR cosφ 0 tan 0 tan P ˆθsinθ R sin θ

142 CHAPTER 3 4 (c) B P ˆR0 896 ˆθ0 449 ˆφ5 C ˆRsin θ ˆθcos θ ˆφsin φ φ ˆRcosθ ˆθsinθ cosφsin φ ˆRcosφ sin θ cosθsinφ ˆθcos φ cos sinθsin θ φ ˆφsinφ P 3 tan π C P 3 ˆR0 854 ˆθ0 46 ˆφ0 707 (d) R D ˆRsinθcos φ ˆθcosθcos ˆφsin φ φ sin θsin φ R sin θsin φ R sin θcos φ ˆRsinθsin φ ˆθcos θsinφ R ˆφcosφ sin θcos φ ˆRcosθ ˆθsinθ 4 ˆR sinθcos φsin φ sinθsin φcos φ 4cosθ ˆθ cosθcos φsin φ cosθsinφcos φ 4sinθ ˆφ cos 3 φ sin φ 3 R sin θsin φ R sin θcos φ P 4 P 4 4 tan tan P D P 4 ˆR sin sin 45 sin35 6 sin 45 cos 45 4cos35 6 ˆθ cos35 6 cos45 sin 45 cos35 6 sin 45 cos 45 4sin35 6 ˆφ cos 45 3 sin 45 ˆR3 67 ˆθ 73 ˆφ Sections 3-4 to 3-7: Gradient, Divergence, and Curl Operators Problem 3.3 (a) T 3 x z, (b) V xy z 4, Find the gradient of the following scalar functions:

143 4 CHAPTER 3 (c) U zcos φ r, (d) W e R sin θ, (e) S 4x e z y 3, (f) N r cos φ, (g) M Rcosθsinφ. Solution: (a) From Eq. (3.7), (b) From Eq. (3.7), (c) From Eq. (3.8), (d) From Eq. (3.83), (e) From Eq. (3.7), T ˆx 6x x z ẑ 6z x z V ˆxy z 4 ŷxyz 4 ẑ4xy z 3 U ˆr rzcos φ r ˆφ zsinφ r r ẑ cosφ r W ˆRe R sinθ ˆθ e R R cos θ S 4x e z y 3 S ˆx S ŷ x S ẑ y S ˆx8xe z ŷ3y z ẑ4x e z (f) From Eq. (3.8), N r cos φ N ˆr N ˆφ r N ẑ r φ N z φ ˆrr cos ˆφr sin φcos φ (g) From Eq. (3.83), M Rcosθsin φ M M ˆR R ˆθ M ˆφ R θ Rsinθ M ˆRcosθsin φ φ ˆθsinθsin φ ˆφ cosφ tanθ

144 CHAPTER 3 43 Problem 3.33 The gradient of a scalar function T is given by T ẑe 3z If T 0 at z 0, find T z. Solution: T ẑe 3z By choosing P at z 0 and P at any point z, (3.76) becomes z z T z T 0 T dl Hence, 0 T z T 0 3z ẑe ˆx dx ŷ dy ẑ dz z dz e 3z 3 0 z 3z e 0 e 3z 0 e 3z e 3z Problem 3.34 Follow a procedure similar to that leading to Eq. (3.8) to derive the expression given by Eq. (3.83) for in spherical coordinates. Solution: From the chain rule and Table 3-, T ˆx T ŷ x T ẑ y T z T R T θ ˆx R x θ x T R T θ ŷ R y θ y T R T θ ẑ R z θ z T ˆx R T ŷ R T ẑ R T φ φ x x x y z y x y z z x y z T φ φ y T φ φ z T θ tan x x y z tan x y y z tan x z y z T θ T θ T φ tan y x x tan y x y T φ T φ z tan y x

145 44 CHAPTER 3 ˆx T R x x y z T θ z x y z x x y T φ y x y T y T z y T x ŷ R x y z θ x y z x y φ x y T z T ẑ R x y z θ x y z x y T φ 0 T Rsinθcosφ T Rcosθ Rsinθcosφ T ˆx R R θ R Rsinθ Rsinθsin φ φ R sin θ T Rsinθsin φ T Rcosθ Rsinθsin φ T Rsinθcos φ ŷ R R θ R Rsinθ φ R sin θ T Rcosθ T ẑ R R Rsinθ θ R T ˆx R sinθcos φ T cosθcos φ T θ R sinφ φ Rsinθ T ŷ R sinθsin φ T cosθsin φ T cosφ θ R φ Rsinθ T ẑ R cosθ T sinθ θ R ˆxsinθcos φ ŷsinθsin φ T ẑcosθ R ˆxcosθcos φ ŷcosθsin φ ẑsinθ ˆxsinφ ŷcosφ ˆR T ˆθ R R which is Eq. (3.83). T Rsinθ φ T T ˆφ θ Rsinθ φ Problem 3.35 For the scalar function V xy z, determine its directional derivative along the direction of vector A ˆx ŷz and then evaluate it at P 4. Solution: The directional derivative is given by Eq. (3.75) as dv dl V â l, where the unit vector in the direction of A is given by Eq. (3.): R â l ˆx ŷz T θ z

146 CHAPTER 3 45 and the gradient of V in Cartesian coordinates is given by Eq. (3.7): V ˆxy ŷxy ẑz Therefore, by Eq. (3.75), At P 4, dv dv y xyz dl z dl Problem 3.36 For the scalar function T e r 5 cosφ, determine its directional derivative along the radial direction ˆr and then evaluate it at P π 4 3. Solution: dt dl π 4 3 T e r cosφ 5 T ˆr T ˆφ r T ẑ r φ T ˆr z e r 5 cosφ ˆφ 0 e r 5 sinφ r dt r T ˆr e dl 5 cosφ 0 e 5 cos π Problem 3.37 For the scalar function U R sin θ, determine its directional derivative along the range direction ˆR and then evaluate it at P 5 π 4 π. Solution: du dl 5 π 4 π U R sin θ U U ˆR ˆθ R R du U ˆR dl U ˆφ θ Rsinθ sin θ R sin π U sin ˆR φ θ sin θcos θ ˆθ R R

147 46 CHAPTER 3 Problem 3.38 Vector field E is characterized by the following properties: (a) E points along ˆR, (b) the magnitude of E is a function of only the distance from the origin, (c) E vanishes at the origin, and (d) E, everywhere. Find an expression for E that satisfies these properties. Solution: According to properties (a) and (b), E must have the form E ˆRER where E R is a function of R only. Hence, and 0 R E R R R E R R R R E R dr 0 R R E R R 0 R 3 3 R E R 4R 3 E R 4R E ˆR4R R R E R R dr R 0 Problem 3.39 For the vector field E ŷyz ˆxxz ẑxy, verify the divergence theorem by computing: (a) the total outward flux flowing through the surface of a cube centered at the origin and with sides equal to units each and parallel to the Cartesian axes, and (b) the integral of E over the cube s volume. Solution: (a) For a cube, the closed surface integral has 6 sides: E ds F top F bottom F right F left F front F back

148 CHAPTER 3 47 F top F bottom F right F left F front F back x x x x x x x x y y y y y y y y z ˆxxz ŷyz ẑxy z ẑ dy dx xy dy dx x y 4 ˆxxz ŷyz ẑxy z xy dy dx x y 4 y ẑ dy dx y ˆxxz ŷyz ẑxy y ŷ dz dx z z dz dx z xz 3 3 ˆxxz ŷyz ẑxy y z z dz dx z z z z xz 3 3 x x z ŷ dz dx z ˆxxz ŷyz ẑxy x ˆx dz dy z dz dy yz z ˆxxz ŷyz ẑxy x z dz dy yz z 3 E ds y x x 0 ˆx dz dy y

149 48 CHAPTER 3 (b) E dv x x y y xy z z z ˆxxz ŷyz ẑxy dz dy dx z 3 3 z z dz dy dx z y x 8 3 Problem 3.40 For the vector field E ˆr0e ẑ3z, verify the divergence theorem r for the cylindrical region enclosed by r, z 0, and z 4. Solution: E ds π r 0 φ 0 π 4 φ 0 z 0 π r 0 φ 0 0 π 4 60πe 4 E dv z 0 8π r ˆr0e ẑ3z r ˆr0e ẑr dr dφ z 0 ẑ3z ˆrr dφ dz r ẑ3z ẑr dr dφ z 4 r ˆr0e 0e dφ dz φ 0 z 0 48π 8 77 π r 0 r 0 r 8π 0e 60πe 48π 0e r r φ 0 r r 0e r 3r dr 0e r r 8 77 r 0 π φ 0 r dr dφ 3 r dφ dr dz 3r r 0 Problem 3.4 A vector field D ˆrr 3 exists in the region between two concentric cylindrical surfaces defined by r and r, with both cylinders extending between z 0 and z 5. Verify the divergence theorem by evaluating: (a) D ds, S

150 CHAPTER 3 49 (b) D V dv. Solution: (a) D ds F inner F outer F bottom F top F inner F outer π φ 0 π φ 0 π φ 0 π φ 0 F bottom F top r r 5 z 0 5 z 0 5 z 0 5 z 0 π φ 0 π φ 0 ˆrr 3 ˆrr dz dφ r r 4 dz dφ r 0π ˆrr 3 ˆrr dz dφ r r 4 dz dφ 60π r ˆrr 3 ẑr dφ dr 0 z 0 ˆrr 3 ẑr dφ dr 0 z 5 Therefore, D ds 50π. (b) From the back cover, D r r rr 3 4r. Therefore, 5 π D dv 4r r dr dφ dz r 4 π 5 r z 0 φ 0 r φ 0 z 0 50π Problem 3.4 For the vector field D ˆR3R, evaluate both sides of the divergence theorem for the region enclosed between the spherical shells defined by R and R. Solution: The divergence theorem is given by Eq. (3.98). Evaluating the left hand side: π π D dv V φ 0 θ 0 R R R R 3R R sinθ dr dθ dφ π π cosθ θ 0 3R 4 80π R

151 50 CHAPTER 3 The right hand side evaluates to π D ds S π φ 0 θ 0 π π π π φ 0 θ 0 θ 0 ˆRR sinθ dθ dφ R ˆR3R ˆRR sinθ dθ dφ ˆR3R 3sin θ dθ π π R 48sin θ dθ 80π θ 0 Problem 3.43 (a) (b) For the vector field E ˆxxy ŷ x y, calculate E dl around the triangular contour shown in Fig. P3.43(a), and C E ds over the area of the triangle. S Solution: In addition to the independent condition that z 0, the three lines of the triangle are represented by the equations y 0, x, and y x, respectively. y y L 3 L 0 L x 0 (a) L 3 L L (b) x Figure P3.43: Contours for (a) Problem 3.43 and (b) Problem (a) E dl L L L 3 L ˆxxy ŷ x y ˆx dx ŷ dy ẑ dz x 0 xy y 0 z 0 dx 0 y 0 x y z 0 dy 0 z 0 0 y 0 dz 0

152 CHAPTER 3 5 Therefore, L ˆxxy ŷ x y ˆx dx ŷ dy ẑ dz xy z dx 0 x y x y 0 x z 0 dy y 0 y y L 3 ŷ x ˆxxy y ˆx dx ŷ dy ẑ dz 0 x x 3 3 xy y x z 0 dx 0 0 y x y 3 0 y 0 E dl 0 0 z 0 x y x y z 0 dy (b) From Eq. (3.05), E ẑ3x so that E ds Problem 3.44 x 0 x 0 x 3 y 0 ẑ3x ẑ dy dx z 0 x 3x dy dx y 0 x 0 0 x dz 0 z 0 0 y x dz 3x x 0 dx x 3 0 Repeat Problem 3.43 for the contour shown in Fig. P3.43(b). Solution: In addition to the independent condition that z 0, the three lines of the triangle are represented by the equations y 0, y x, and y x, respectively. (a) E dl L L L 3 L ŷ x ˆxxy y ˆx dx ŷ dy ẑ dz xy dx 0 y 0 z 0 x y x 0 y 0 z 0 dy 0 0 y 0 dz 0 z 0 L ŷ x ˆxxy y ˆx dx ŷ dy ẑ dz xy z 0 y x x dx x y y 0 x y z 0 dy 0 0 y x z 0 dz x x x 4y y y 3 y 0 0

153 5 CHAPTER 3 Therefore, L 3 ˆxxy ŷ x y ˆx dx ŷ dy ẑ dz 0 x 3 x 3 xy y x z 0 dx 0 0 y x y 3 0 y 0 E dl 0 x y x y z 0 dy 3 3 (b) From Eq. (3.05), E ẑ3x so that Problem 3.45 by evaluating: (a) (b) E ds x z 0 0 y x dz x 0 y 0 ẑ3x ẑ dy dx z 0 x x y 0 ẑ3x ẑ dy dx z 0 x x 3x dy 3x dy dx x 0 y 0 dx x y 0 3x x 0 x 0 3x dx x x 0 dx x 3 0 3x x 3 x 3 Verify Stokes s theorem for the vector field B ˆrr cos φ ˆφsinφ B dl over the semicircular contour shown in Fig. P3.46(a), and C B ds over the surface of the semicircle. S Solution: (a) B dl B dl B dl B L dl L L 3 B dl ˆrr cos φ ˆφsin φ ˆr dr ˆφr dφ ẑ dz r cos φ dr r sinφ dφ L B dl r r 0 r cos φ dr r 0 0 φ 0 z 0 0 φ 0 r sin φ dφ z 0

154 CHAPTER 3 53 y y L - L 3 0 L (a) x 0 L3 L L 4 L (b) x Figure P3.46: Contour paths for (a) Problem 3.45 and (b) Problem (b) L B dl L 3 B dl 0 π cosφ φ r r cos φ dr r r cos φ dr z 0 r 0 r 0 B dl 4 8 B ds φ 0 π r 0 φ π z 0 π φ 0 r sinφ dφ π φ π r sin φ dφ B ˆrr cos φ ˆφsinφ 0 sinφ ˆr ˆφ r φ z z φ r cos ẑ r r r sin φ φ φ r cos ˆr0 ˆφ0 ẑ r sinφ r sin φ ẑsinφ π ẑsinφ r ẑr dr dφ φ 0 r 0 r z 0 z 0 r 0 r sinφ r dr dφ cosφ r r π r 0 8 φ 0 Problem 3.46 Solution: Repeat Problem 3.45 for the contour shown in Fig. P3.46(b).

155 54 CHAPTER 3 (a) (b) B dl B dl B dl B dl B L dl L L 3 L 4 B dl ˆrr cos φ ˆφsin φ ˆr dr ˆφr dφ ẑ dz r cos φ dr r sinφ dφ L B dl L B dl L 3 B dl L 4 B dl B dl 3 r r r cos φ dr r 0 3 r r cos φ dr 0 φ 0 z 0 π z 0 π φ π z 0 0 z 0 0 π cosφ φ 0 r r cos φ dr 0 0 cosφ φ π 5 0 B ds r r cos φ dr φ 0 r sin φ dφ φ 0 r sinφ dφ z 0 r z 0 r sin φ dφ φ π r sin φ dφ φ π r z 0 B ˆrr cos φ ˆφsin φ 0 sinφ ˆr ˆφ r φ z z φ r cos ẑ r r r sin φ φ φ r cos ˆr0 ˆφ0 ẑ r sinφ r sin φ ẑsinφ π ẑsin φ φ 0 r r ẑr dr dφ π φ 0 r cosφ sinφ r dr dφ r r π r 5 φ 0 z 0 r 0 0 r

156 CHAPTER 3 55 Problem 3.47 evaluating it on the hemisphere of unit radius. Verify Stokes s Theorem for the vector field A ˆRcosθ ˆφsinθ by Solution: A ˆRcos θ ˆφsinθ ˆRAR Hence, A R cosθ, A θ 0, A φ sinθ. A ˆR Rsinθ θ A φ sinθ ˆR θ Rsinθ θ sin ˆθ R ˆR cos θ R ˆθ sin θ R ˆφ sinθ R For the hemispherical surface, ds ˆRR sinθ dθ dφ. π φ 0 π θ 0 π φ 0 A ds π θ 0 4πR sin θ ˆRcos θ π 0 R R ˆθ sinθ R π ˆθA θ ˆφA φ ˆθ R Rsinθ ˆφ R R R RA φ ˆφ R θ A R θ cos θ ˆφ sin θ R ˆRR sinθ dθ dφ R The contour C is the circle in the x y plane bounding the hemispherical surface. π A dl ˆRcosθ π ˆφsinθ ˆφR dφ θ π Rsinθ dφ θ π π C φ 0 R 0 R Problem 3.48 Determine if each of the following vector fields is solenoidal, conservative, or both: (a) A ˆxx ŷyxy, (b) B ˆxx ŷy ẑz, (c) C ˆr sin φ r ˆφ cosφ r, (d) D ˆR R, r (e) E ˆr 3 r ẑz, (f) F ˆxy ŷx x y, (g) G ˆx x z ŷ y x ẑ y z, (h) H ˆR Re R.

157 56 CHAPTER 3 Solution: (a) A ˆxx ŷxy x x A ˆxx ŷxy 0 ˆx y z xy ŷ ˆx0 ẑ ŷ0 y 0 xy x x 0 y z x The field A is solenoidal but not conservative. (b) B ˆxx ŷy ẑz x x B ˆxx ŷy ẑz y z ŷ z ˆx y ˆx0 ŷ0 ẑ0 y y z x The field B is conservative but not solenoidal. (c) C ˆr sinφ ˆφ r cosφ r sinφ r r r sinφ r C 3 ˆr r ẑ r x 0 ẑ x xy z z x y x z ẑ cosφ r r φ r sin φ 0 r 3 sinφ r 3 ˆr sinφ r 0 φ r ˆr0 ˆφ0 ẑ r ˆφ cos φ r cosφ z r cosφ r r cosφ ˆφ φ r The field C is neither solenoidal nor conservative. z 0 sinφ z r sinφ r cosφ r 0 x y r 0 ẑ cosφ r 3 y x y x

158 CHAPTER 3 57 (d) D D ˆR R R R ˆR R ˆR Rsinθ ˆφ R θ θ 0sin 0 R R R R θ φ 0 Rsinθ ˆθ R The field D is conservative but not solenoidal. (e) r E ˆr E r r r r E ˆr 3 r ẑz r r re E φ r φ r 3r r r θ θ 0sin sinθ φ R ˆr0 ˆθ0 ˆφ0 E z z R Rsinθ 0 R R r 3 r r r 3 3r r 6r r r r r 4r 3 r r 0 E z E φ E r φ r E z ˆφ z z ẑ r r Hence, E is conservative, but not solenoidal. (f) ˆxy F ŷx y x x y ˆx x y ŷ x y y x x y y x y xy xy x y x y 0 F x φ r re 0 φ R E r 0 r φ

159 58 CHAPTER 3 x y F ˆx 0 0 ŷ 0 0 ẑ x x y y x y x y ẑ x y x y ẑ y x x y 0 Hence, F is neither solenoidal nor conservative. (g) x y G ˆx x z ŷ y x ẑ y z G x x z y y x z y z z x y 0 G ˆx y y z z y x ŷ ẑ x y x y x z ˆxy ẑx ŷz 0 Hence, G is neither solenoidal nor conservative. (h) x y z x z H ˆR Re R H e R e R R R3 R 3R R R e R e 3 R 3 R H 0 Hence, H is conservative, but not solenoidal. x y z 0 Problem 3.49 (a) V 4xy z 3, (b) V xy yz zx, (c) V 3 x y, (d) V 5e r cosφ, (e) V 0e R sinθ. Find the Laplacian of the following scalar functions: Solution: (a) From Eq. (3.0), 4xy z 3 8xz 3 4xy z

160 CHAPTER 3 59 (b) xy yz zx 0 (c) From the inside back cover of the book, 3 x y 3r r 4 x y (d) (e) Problem e r cosφ 5e r cosφ 0e R sinθ 0e R sinθ R r r cos θ sin θ R sinθ Find a vector G whose magnitude is 4 and whose direction is perpendicular to both vectors E and F, where E ˆx ŷ ẑ and F ŷ3 ẑ6. Solution: The cross product of two vectors produces a third vector which is perpendicular to both of the original vectors. Two vectors exist that satisfy the stated conditions, one along E F and another along the opposite direction. Hence, Problem 3.5 G E 4 E F 4 ˆx ŷ ẑ ŷ3 ẑ6 F ˆx ŷ ẑ ŷ3 ẑ6 ˆx6 4 ŷ6 ẑ ˆx6 9 ŷ6 ẑ3 ˆx 3 8 ŷ 8 ẑ A given line is described by the equation: y x Vector A starts at point P 0 and ends at point P on the line such that A is orthogonal to the line. Find an expression for A. Solution: We first plot the given line.

161 60 CHAPTER 3 y P (0, ) A B P 4 (, 0) P (x, x-) x P 3 (0, -) Next we find a vector B which connects point P 3 0 to point P 4 0, both of which are on the line. Hence, B ˆx 0 ŷ 0 ˆx ŷ Vector A starts at P 0 and ends on the line at P. If the x-coordinate of P is x, then its y-coordinate has to be y x, per the equation for the line. Thus, P is at x x, and vector A is A ˆx x 0 ŷ x ˆxx ŷ x 3 Since A is orthogonal to B, Finally, A B 0 ˆxx ŷ x 3 ˆx ŷ 0 x x 3 0 x 3 A ˆxx ŷ x 3 ˆx 3 3 ŷ 3 ˆx 3 ŷ 3

162 CHAPTER 3 6 Problem 3.5 Vector field E is given by E ˆR 5Rcos θ ˆθ R sin θcosφ ˆφ3sin φ Determine the component of E tangential to the spherical surface R at point P Solution: At P, E is given by E ˆR 5 cos 30 ˆR 8 67 ˆθ 5 ˆφ 6 ˆθ sin30 cos60 ˆφ3sin 60 The ˆR component is normal to the spherical surface while the other two are tangential. Hence, E t ˆθ 5 ˆφ 6 Problem 3.53 Transform the vector A ˆRsin θcos φ ˆθcos φ ˆφsinφ into cylindrical coordinates and then evaluate it at P π π. Solution: From Table 3-, A ˆr sinθ ẑcosθ sin θcosφ ˆr cos ẑsin θ θ cos φ ˆφsinφ ˆr sin 3 θcosφ cosθcos φ ˆφsinφ ẑ cosθsin θcos sinθcos φ φ At P π π, A ˆφ Problem 3.54 Evaluate the line integral of E ˆxx ŷy along the segment P to P of the circular path shown in the figure. y P (0, 3) P (-3, 0) x

163 6 CHAPTER 3 Solution: We need to calculate: P P E d Since the path is along the perimeter of a circle, it is best to use cylindrical coordinates, which requires expressing both E and d in cylindrical coordinates. Using Table 3-, E ˆxx ŷy ˆr cos φ ˆφsin φ r cos φ ˆr sinφ ˆφcosφ r sin φ ˆr r cos φ sin φ ˆφr sin φcos φ The designated path is along the φ-direction at a constant r 3. From Table 3-, the applicable component of d is: d ˆφ r dφ Hence, P φ 80 E d P ˆrr cos sin φ 90 φ φ ˆφ r sin φcosφ ˆφ r dφ r r sinφcos φ dφ r 3 r sin φ 80 φ 90 r 3 9 Problem 3.55 evaluating: Verify Stokes s theorem for the vector field B ˆr cosφ ˆφsinφ by (a) B d over the path comprising a quarter section of a circle, as shown in the C figure, and (b) S B ds over the surface of the quarter section. y L (-3, 0) L 3 (0, 3) L x

164 CHAPTER 3 63 Solution: (a) B d B d B d B d C L L L 3 Given the shape of the path, it is best to use cylindrical coordinates. B is already expressed in cylindrical coordinates, and we need to choose d in cylindrical coordinates: d ˆr dr ˆφr dφ ẑ dz Along path L, dφ 0 and dz 0. Hence, d ˆr dr and r 3 B d L ˆr cosφ ˆφsinφ ˆr dr r 0 φ cosφ dr r cos φ r 0 r 0 φ 90 φ 90 0 Along L, dr dz 0. Hence, d ˆφr dφ and 80 B d L ˆr cosφ ˆφsinφ ˆφr dφ φ 90 r cos φ 90 3 Along L 3, dz 0 and dφ 0. Hence, d ˆr dr and 0 B d L 3 ˆr cosφ ˆφsinφ ˆr dr r 3 φ 80 Hence, (b) S 0 r 3 B d C r ẑ r 0 cosφ dr φ 80 r B ẑ B r rb r φ φ φ r r sin φ φ cos ẑ r sinφ sinφ ẑ sin φ r B ds 3 80 r 0 φ 90 3 r r 0 cosφ 80 ẑ r sinφ ẑr dr dφ φ 90 6

165 64 CHAPTER 3 Hence, Stokes s theorem is verified. Problem 3.56 Find the Laplacian of the following scalar functions: (a) V 0r 3 sinφ (b) V R cos θsin φ Solution: (a) (b) V r V R R 4 R r r r V r V r φ r sinφ r 0r3 V z r sinφ r r 30r3 r 0r3 4sin φ 90r sin 40r sin φ φ 50r sin φ R R R V R R R R sinθ θ R sinθ θ R cosθsin φ sinθ θ R sin θ φ R cos θsinφ 4 cos cos 4 θsinφ R4 θsinφ cosθsin φ R 4 sin θ sinφ r φ 0r3 0 sinθ V θ R cosθsinφ cosθ R 4 sin θ sin φ R sin θ V φ

166 65 Chapter 4: Electrostatics Lesson # Chapter Section: 4- to 4-3 Topics: Charge and current distributions, Coulomb s law Highlights: Maxwell s Equations reduce to uncoupled electrostatics and magnetostatics when charges are either fixed in space or move at constant speed. Line, surface and volume charge distributions Coulomb s law for various charge distributions Special Illustrations: Examples 4-3 and 4-4 CD-ROM Modules CD-ROM Demos

167 66 Lesson #3 Chapter Section: 4-4 Topics: Gauss s law Highlights: Gauss s law in differential and integral form The need for symmetry to apply Gauss s law in practice Coulomb s law for various charge distributions Special Illustrations: Example 4-6 CD-ROM Module 4.6 CD-ROM Demos 4.9 and 4.0

168 67 Lesson #4 Chapter Section: 4-5 Topics: Electric potential Highlights: Concept of potential Relation to electric field Relation to charges Poisson s and Laplace s equations Special Illustrations: Example 4-7

169 68 Lesson #5 Chapter Section: 4-6 and 4-7 Topics: Electrical materials and conductors Highlights: Conductivity ranges for conductors, semiconductors, and insulators Ohm s law Resistance of a wire Joule s law Special Illustrations: Example 4-9 Technology Brief on Resistive Sensors (CD-ROM) Resistive Sensors An electrical sensor is a device capable of responding to an applied stimulus by generating an electrical signal whose voltage, current, or some other attribute is related to the intensity of the stimulus. The family of possible stimuli encompasses a wide array of physical, chemical, and biological quantities including temperature, pressure, position, distance, motion, velocity, acceleration, concentration (of a gas or liquid), blood flow, etc. The sensing process relies on measuring resistance, capacitance, inductance, induced electromotive force (emf), oscillation frequency or time delay, among others. This Technology Brief covers resistive sensors. Capacitive, inductive, and emf sensors are covered separately (in this and later chapters). Piezoresistivity According to Eq. (4.70), the resistance of a cylindrical resistor or wire conductor is given by R = l/σa), where l is the cylinder s length, A is its cross-sectional area, and σ is the conductivity of its material. Stretching the wire by an applied external force causes l to increase and A to decrease. Consequently, R increases (A). Conversely, compressing the wire causes R to decrease. The Greek word piezein means to press, from which the term piezoresistivity is derived. This should not be confused with piezoelectricity, which is an emf effect (see EMF Sensors).

170 69 Lesson #6 Chapter Section: 4-8, 4-9 Topics: Dielectrics, boundary conditions Highlights: Relative permittivity and dielectric strength Electrostatic boundary conditions for various dielectric and conductor combinations Special Illustrations: Example 4-0

171 70 Lesson #7 Chapter Section: 4-0 Topics: Capacitance Highlights: Capacitor as charge accumulator General expression for C Capacitance of parallel-plate and coaxial capacitors Joule s law Special Illustrations: Examples 4- and 4- Technology Brief on Capacitive Sensors (CD-ROM) Capacitive Sensors To sense is to respond to a stimulus (see Resistive Sensors). A capacitor can function as a sensor if the stimulus changes the capacitor s geometry usually the spacing between its conductive elements or the dielectric properties of the insulating material situated between them. Capacitive sensors are used in a multitude of applications. A few examples follow. Fluid Gauge The two metal electrodes in (A), usually rods or plates, form a capacitor whose capacitance is directly proportional to the permittivity of the material between them. If the fluid section is of height H f and the height of the empty space above it is (H H f ), then the overall capacitance is equivalent to two capacitors in parallel: where w is the electrode plate width, d is the spacing between electrodes, and ε f and ε a are the permittivities of the fluid and air, respectively.

172 7 Lesson #8 Chapter Section: 4- Topics: Energy Highlights: A charged capacitor is an energy storage device Energy density Special Illustrations: Technology Brief on Non-Contact Sensors (CD-ROM) Non-Contact Sensors Precision positioning is a critical ingredient of semiconductor device fabrication, as well as the operation and control of many mechanical systems. Non-contact capacitive sensors are used to sense the position of silicon wafers during the deposition, etching, and cutting processes, without coming in direct contact with the wafers. They are also used to sense and control robot arms in equipment manufacturing and to position hard disc drives, photocopier rollers, printing presses, and other similar systems. Basic Principle The concentric plate capacitor (A) consists of two metal plates, sharing the same plane, but electrically isolated from each other by an insulating material. When connected to a voltage source, charges of opposite polarity will form on the two plates, resulting in the creation of electric-field lines between them. The same principle applies to the adjacent-plates capacitor in (A). In both cases, the capacitance is determined by the shapes and sizes of the conductive elements and by the permittivity of the dielectric medium containing the electric field lines between them.

173 7 Lesson #9 Chapter Section: 4- Topics: Image method Highlights: Image method useful for solving problems involving charges next to conducting planes Remove conducting plane and replace with mirror images for the charges (with opposite polarity) Special Illustrations: Example 4-3 CD-ROM Demos

174 CHAPTER 4 73 Chapter 4 Sections 4-: Charge and Current Distributions Problem 4. A cube m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density is given by ρ v xy e z (mc/m 3 ). Solution: For the cube shown in Fig. P4., application of Eq. (4.5) gives Q V ρ v dv x y 3 e z x 0 y 0 z 0 x 0 y 0 z 0 xy e z dx dy dz 8 3 e 4 6 mc z m 0 m y m x Figure P4.: Cube of Problem 4.. Problem 4. Find the total charge contained in a cylindrical volume defined by r m and 0 z 3 m if ρ v 0rz (mc/m 3 ). Solution: For the cylinder shown in Fig. P4., application of Eq. (4.5) gives Q 3 π z 0 φ r3 φz r 0 0rz r dr dφ dz π 3 r 0 φ 0 z 0 480π (mc) 5 C

175 74 CHAPTER 4 z 3 m x m 0 m y Figure P4.: Cylinder of Problem 4.. Problem 4.3 Find the total charge contained in a cone defined by R m and 0 θ π 4, given that ρ v 0R cos θ (mc/m 3 ). Solution: For the cone of Fig. P4.3, application of Eq. (4.5) gives Q π φ 0 8π 3 π 4 θ 0 R 0 3 R5 φcos θ 3 0R cos θ R sinθ dr dθ dφ π 4 π R 0 θ 0 φ (mc)

176 CHAPTER 4 75 z m π/4 0 y x Figure P4.3: Cone of Problem 4.3. Problem 4.4 If the line charge density is given by ρ l 4y (mc/m), find the total charge distributed on the y-axis from y 5 to y 5. Solution: Q 5 ρ 5 l dy 5 4y 5 dy 4y mc C 5 Problem 4.5 Find the total charge on a circular disk defined by r a and z 0 if: (a) ρ s ρ s0 cos φ (C/m ), (b) ρ s ρ s0 sin φ (C/m ), (c) ρ s ρ s0 e r (C/m ), (d) ρ s ρ s0 e r sin φ (C/m ), where ρ s0 is a constant. Solution: (a) Q ρ s ds (b) Q a r 0 π φ 0 a r 0 π φ 0 ρ s0 cosφ r dr dφ ρ s0 r ρ s0 sin r φ r dr dφ ρ s0 ρ s0 a 4 a 0 φ 0 π a 0 sinφ π 0 0 cosφ dφ sinφ π 0 πa ρ s0

177 76 CHAPTER 4 (c) (d) Q a Q r 0 a π φ 0 π ρ s0 e r r dr dφ πρ s0 ρ s0 e r sin φ r dr dφ a re r dr 0 r πρ s0 re r a e 0 a πρ s0 e a r 0 φ 0 a ρ s0 re r π dr sin φ dφ r 0 φ 0 a ρ s0 e a a π πρ s0 e a Problem 4.6 If J ŷ4xz (A/m ), find the current I flowing through a square with corners at 0 0 0, 0 0, 0, and 0 0. Solution: Using Eq. (4.), the net current flowing through the square shown in Fig. P4.6 is I S J ds x 0 z 0 ŷ4xz y 0 ŷ dx dz x z x 0 z 0 6 A z m J 0 y x m Figure P4.6: Square surface.

178 CHAPTER 4 77 Problem 4.7 If J ˆR5 R (A/m ), find I through the surface R 5 m. Solution: Using Eq. (4.), we have π π I J ds ˆR 5 ˆRR S R sinθ dθ dφ φ 0 θ 0 5Rφcosθ π π R 5 θ 0 φ 0 00π 34 (A) Problem 4.8 An electron beam shaped like a circular cylinder of radius r 0 carries a charge density given by ρ v ρ 0 r (C/m 3 where ρ 0 is a positive constant and the beam s axis is coincident with the z-axis. (a) Determine the total charge contained in length L of the beam. (b) If the electrons are moving in the z-direction with uniform speed u, determine the magnitude and direction of the current crossing the z-plane. Solution: (a) Q (b) r 0 r 0 L z 0 ρ v dv r 0 L r 0 z 0 πρ 0 L r 0 0 ρ 0 J ρ v u ẑ uρ 0 r (A/m ) I J ds r 0 π r 0 φ 0 r 0 πuρ 0 Current direction is along ẑ. 0 r πr dr dz r dr r πρ 0 Lln r0 ẑ uρ 0 r ẑr dr dφ r dr r πuρ 0 ln r0 (A)

179 78 CHAPTER 4 Section 4-3: Coulomb s Law Problem 4.9 A square with sides m each has a charge of 40 µc at each of its four corners. Determine the electric field at a point 5 m above the center of the square. z P(0,0,5) Q 3 (-,-,0) R 3 R 4 R R Q (-,,0) Q 4 (,-,0) Q (,,0) y x Figure P4.9: Square with charges at the corners. Solution: The distance R between any of the charges and point P is

180 CHAPTER 4 79 R 5 7. Q R E R R 3 R 4 4πε 0 R 3 R 3 R 3 R 3 Q ŷ 4πε 0 ˆx ẑ5 ŷ 7 3 ˆx ẑ5 5Q ẑ 7 3 πε µc ẑ πε 0 ˆx 7 3 ŷ ẑ5 7 3 πε 0 4 ˆx 7 3 ŷ ẑ (V/m) ẑ5 6 (kv/m) Problem 4.0 Three point charges, each with q 3 nc, are located at the corners of a triangle in the x y plane, with one corner at the origin, another at cm 0 0, and the third at 0 cm 0. Find the force acting on the charge located at the origin. Solution: Use Eq. (4.9) to determine the electric field at the origin due to the other two point charges [Fig. P4.0]: 3 nc E ˆx0 0 3 nc ŷ0 0 4πε ˆx ŷ (kv/m) at R Employ Eq. (4.4) to find the force F qe 0 ˆx ŷ (µn) cm y Q R = -x ^ cm R = -y ^ cm R Q R Q cm x Figure P4.0: Locations of charges in Problem 4.0. Problem 4. Charge q 6 µc is located at cm cm 0 and charge q is located at cm. What should q be so that E at 0 cm 0 has no y-component? Solution: For the configuration of Fig. P4., use of Eq. (4.9) gives

181 80 CHAPTER 4 z 4 cm q R R = -x ^ + y(-) ^ = (-x ^ + ^y) cm R = (y ^ - z4) ^ cm cm cm cm 0 y q R E E x Figure P4.: Locations of charges in Problem 4.. 6µC ˆx ŷ E R ŷcm 4πε 6 4πε ˆx 0 (V/m) ẑ0 447q If E y 0, then q Problem 4. q ŷ ẑ q ŷ (µc) A line of charge with uniform density ρ l 8 (µc/m) exists in air along the z-axis between z 0 and z 5 cm. Find E at (0,0 cm,0). Solution: Use of Eq. (4.c) for the line of charge shown in Fig. P4. gives ρ E l dl ˆR 4πε 0 l R R ŷ0 ẑz πε ŷ0 ẑz z 0 0 z 3 dz ŷ0z 0 05 ẑ 4πε 0 z ŷ4 47 ẑ 3 06 ŷ ẑ76 0 (V/m) 3 0 z

182 CHAPTER 4 8 z 5 cm dz R' = ^y0. ^- zz 0 y 0 cm x Figure P4.: Line charge. Problem 4.3 Electric charge is distributed along an arc located in the x y plane and defined by r cm and 0 φ π 4. If ρ l 5 µc/m), find E at 0 0 z and then evaluate it at (a) the origin, (b) z 5 cm, and (c) z 5 cm. Solution: For the arc of charge shown in Fig. P4.3, dl r dφ 0 0 dφ and R ˆx0 0cos φ ŷ0 0sin φ ẑz. Use of Eq. (4.c) gives E 4πε 0 l ˆR ρ l dl R π 4 ρ l 4πε 0 ˆx0 0cos φ ŷ0 0sin φ ẑz φ z z 3 ˆx0 04 ŷ0 006 ẑ0 78z (a) At z 0, E ˆx 6 ŷ0 66 (MV/m) (b) At z 5 cm, E ˆx8 4 ŷ33 7 ẑ6 (kv/m) (c) At z 5 cm, E ˆx8 4 ŷ33 ẑ6 7 (kv/m) 0 0 dφ (V/m)

183 8 CHAPTER 4 z z R' = ^ -r ^ zz cm x π/4 dz ^r cm = y ^ r 0.0 m Figure P4.3: Line charge along an arc. Problem 4.4 A line of charge with uniform density ρ l extends between z L and z L along the z-axis. Apply Coulomb s law to obtain an expression for the electric field at any point P r φ 0 on the x y plane. Show that your result reduces to the expression given by Eq. (4.33) as the length L is extended to infinity. Solution: Consider an element of charge of height dz at height z. Call it element. The electric field at P due to this element is de. Similarly, an element at z produces de. These two electric fields have equal z-components, but in opposite directions, and hence they will cancel. Their components along ˆr will add. Thus, the net field due to both elements is de de de ˆr ρ l cosθ dz 4πε 0 R ˆrρ l cosθ dz πε 0 R where the cosθ factor provides the components of de and de along ˆr. Our integration variable is z, but it will be easier to integrate over the variable θ from θ 0 to θ 0 sin L r L

184 CHAPTER 4 83 z dz L/ z R de r θ 0 θ θ θ θ x-y plane -z de -L/ Figure P4.4: Line charge of length L. Hence, with R r cos θ, and z r tan θ and dz r sec θ dθ, we have L θ E de 0 θ de 0 ˆr ρ l r sec θ dθ For L r, z 0 θ 0 cos 3 θ 0 πε 0 r ρ θ l 0 ˆr cos θ dθ πε 0 r 0 ρ l ˆr πε 0 r sinθ ρ 0 l ˆr πε 0 r L r L L r L

185 84 CHAPTER 4 and ρ E l ˆr πε 0 r (infinite line of charge) Problem 4.5 Repeat Example 4-5 for the circular disk of charge of radius a, but in the present case assume the surface charge density to vary with r as where ρ s0 is a constant. ρ s ρ s0 r (C/m ) Solution: We start with the expression for de given in Example 4-5 but we replace ρ s with ρ s0 r : h de ẑ 4πε 0 r h 3 πρ s0r dr 3 E ẑ ρ a s0h r 3 dr ε 0 0 r h 3 To perform the integration, we use R r h R dr r dr E ẑ ρ s0h ε 0 a h Problem 4.6 h ẑ ρ s0h a h ε 0 h dr ẑ ρ s0h ε 0 a h R h dr R a h h h R dr h h a h Multiple charges at different locations are said to be in equilibrium if the force acting on any one of them is identical in magnitude and direction to the force acting on any of the others. Suppose we have two negative charges, one located at the origin and carrying charge 9e, and the other located on the positive x-axis at a distance d from the first one and carrying charge 36e. Determine the location, polarity and magnitude of a third charge whose placement would bring the entire system into equilibrium. Solution: If F force on Q

186 CHAPTER 4 85 Q = -9e Q 3 Q = -36e x=0 x x (d-x) then equilibrium means that d Figure P4.6: Three collinear charges. F force on Q F 3 force on Q 3 F F F 3 The two original charges are both negative, which mean they would repel each other. The third charge has to be positive and has to lie somewhere between them in order to counteract their repulsion force. The forces acting on charges Q, Q, and Q 3 are respectively ˆR Q F Q ˆR 3 Q Q 3 4πε 0 R 4πε 0 R ˆx 3 34e 4πε 0 d ˆx 9eQ 3 4πε 0 x ˆR Q F Q ˆR 3 Q 3 Q 4πε 0 R 4πε 0 R ˆx 34e 36eQ 3 3 4πε 0 d ˆx 4πε 0 d x ˆR3 Q F 3 Q 3 ˆR 3 Q Q 3 4πε 0 R 3 4πε 0 R ˆx 3 9eQ 3 36eQ 3 4πε 0 x ˆx 4πε 0 d x Hence, equilibrium requires that 34e d 9Q 3 x 34e d Solution of the above equations yields Q 3 4e 36Q 3 d x x d 3 9Q 3 x 36Q 3 d x Section 4-4: Gauss s Law Problem 4.7 Three infinite lines of charge, all parallel to the z-axis, are located at the three corners of the kite-shaped arrangement shown in Fig. 4-9 (P4.7). If the

187 86 CHAPTER 4 two right triangles are symmetrical and of equal corresponding sides, show that the electric field is zero at the origin. y -ρ l ρ l ρ l x Figure P4.7: Kite-shaped arrangment of line charges for Problem 4.7. Solution: The field due to an infinite line of charge is given by Eq. (4.33). In the present case, the total E at the origin is E E E E 3 The components of E and E along ˆx cancel and their components along ŷ add. Also, E 3 is along ŷ because the line charge on the y-axis is negative. Hence, E ŷ ρ l cos θ ρ l ŷ πε 0 R πε 0 R But cosθ R R. Hence, E ŷ ρ l R ŷ πε 0 R R ρ l πε 0 R 0 Problem 4.8 Three infinite lines of charge, ρ l 3 (nc/m), ρ l 3 (nc/m), and ρ l3 3 (nc/m), are all parallel to the z-axis. If they pass through the respective points

188 CHAPTER 4 87 y (0,b) ρ l 3 R 3 E ρ l E P (a,0) x E 3 (0,-b) ρ l Figure P4.8: Three parallel line charges. 0 b, 0 0, and 0 b in the x y plane, find the electric field at a 0 0. Evaluate your result for a cm and b cm. Solution: ρ l 3 (nc/m) ρ l 3 (nc/m) ρ l3 ρ l E E E E 3 Components of line charges and 3 along y cancel and components along x add. Hence, using Eq. (4.33), with cos θ a E ˆx ρ l πε 0 R cosθ ˆx ρ l πε 0 a a b and R a b, E ˆx3 πε 0 a a b a 0 9 (V/m)

189 88 CHAPTER 4 For a cm and b cm, E ˆx 6 (kv/m) Problem 4.9 A horizontal strip lying in the x y plane is of width d in the y-direction and infinitely long in the x-direction. If the strip is in air and has a uniform charge distribution ρ s, use Coulomb s law to obtain an explicit expression for the electric field at a point P located at a distance h above the centerline of the strip. Extend your result to the special case where d is infinite and compare it with Eq. (4.5). z de de P(0,0,h) θ 0 y R ρ s θ θ d -y y x Figure P4.9: Horizontal strip of charge. Solution: The strip of charge density ρ s (C/m ) can be treated as a set of adjacent line charges each of charge ρ l ρ s dy and width dy. At point P, the fields of line charge at distance y and line charge at distance y give contributions that cancel each other along ŷ and add along ẑ. For each such pair, de ẑ ρ s dycos θ πε 0 R

190 CHAPTER 4 89 With R h cos θ, we integrate from y 0 to d, which corresponds to θ 0 to θ 0 sin d h d. Thus, d d θ E 0 0 de ẑ ρ s πε 0 0 cosθ R dy ẑ ρ s πε 0 ẑ ρ s 0 θ 0 πε 0 cos θ h h cos θ dθ For an infinitely wide sheet, θ 0 π and E ẑ ρ s ε 0, which is identical with Eq. (4.5). Problem 4.0 Given the electric flux density D ˆx x y ŷ 3x y (C/m ) determine (a) ρ v by applying Eq. (4.6), (b) the total charge Q enclosed in a cube m on a side, located in the first octant with three of its sides coincident with the x-, y-, and z-axes and one of its corners at the origin, and (c) the total charge Q in the cube, obtained by applying Eq. (4.9). Solution: (a) By applying Eq. (4.6) ρ v D x x y 3x y 0 y (b) Integrate the charge density over the volume as in Eq. (4.7): Q V DdV 0 dx dy dz 0 (c) Apply Gauss law to calculate the total charge from Eq. (4.9) x 0 y 0 z 0 Q D ds F front F back F right F left F top F bottom ˆx x y ŷ 3x y ˆx dz dy F front y 0 y 0 z 0 z 0 x y x dz dy z x y y z 0 y 0 4

191 90 CHAPTER 4 F back F right F left F top F bottom Thus Q y 0 x 0 x 0 x 0 x 0 x 0 x 0 x 0 y 0 x 0 z 0 z 0 z 0 z 0 z 0 z 0 z 0 z 0 z 0 0 z 0 0 ˆx x y ŷ 3x y x y x 0 dz dy ˆx x y ŷ 3x y 3x y y dz dx ˆx x y ŷ 3x y 3x y y 0 dz dx ˆx x y ŷ 3x y z dy dx 0 ˆx x y ŷ 3x y z 0 dy dx 0 D ds z x 0 y zy ˆx dz dy z 0 ŷ dz dx 3 x 4x y 0 z z 0 z y 0 ŷ dz dx 3 x ẑ dy dx ẑ dy dx z 0 z 0 8 x 0 x 0 4 Problem 4. Repeat Problem 4.0 for D ˆxxy 3 z 3 (C/m ). Solution: (a) From Eq. (4.6), ρ v D x xy3 z 3 y 3 z 3 (b) Total charge Q is given by Eq. (4.7): Q V DdV z 0 y 0 x 0 y 3 z 3 dx dy dz xy 4 z 4 6 x 0 y 0 z 0 3 C

192 CHAPTER 4 9 (c) Using Gauss law we have D ds F front F S back F right F left F top F bottom Note that D ˆxD x, so only F front and F back (integration over ẑ surfaces) will contribute to the integral. F front ˆxxy 3 z 3 ˆx dy dz z 0 y 0 x xy 3 z 3 y dy dz 4 z 4 3 z 0 y 0 6 x y 0 z 0 F back ˆxxy 3 z 3 xy 3 z 3 dy dz 0 x 0 Thus Q ˆx dy z 0 y 0 dz x 0 D ds C Problem 4. Charge Q is uniformly distributed over a thin spherical shell of radius a, and charge Q is uniformly distributed over a second spherical shell of radius b, with b a. Apply Gauss s law to find E in the regions R a, a R b, and R b. Solution: Using symmetry considerations, we know D ˆRDR. From Table 3., ds ˆRR sinθ dθ dφ for an element of a spherical surface. Using Gauss s law in integral form (Eq. (4.9)), D ds Q tot S where Q tot is the total charge enclosed in S. For a spherical surface of radius R, π φ 0 π θ 0 z 0 y 0 ˆRD R ˆRR sinθ dθ dφ Q tot D R R π π cos θ 0 Q tot Q D R tot 4πR From Eq. (4.5), we know a linear, isotropic material has the constitutive relationship D εe. Thus, we find E from D.

193 9 CHAPTER 4 (a) In the region R a, Q tot 0 (b) In the region a R b, Q tot Q (c) In the region R b, E ˆRER ˆRQtot 4πR ε 0 E ˆRER ˆRQ 4πR ε Q tot Q Q E ˆRER ˆR Q Q 4πR ε (V/m) (V/m) (V/m) Problem 4.3 The electric flux density inside a dielectric sphere of radius a centered at the origin is given by D ˆRρ0 R (C/m ), where ρ 0 is a constant. Find the total charge inside the sphere. Solution: Q π π D ds ˆRρ 0 R ˆRR sinθ dθ dφ S θ 0 φ 0 πρ 0 a 3 π 0 R a sinθ dθ πρ 0 a 3 cosθ π 0 4πρ 0 a 3 (C) Problem 4.4 In a certain region of space, the charge density is given in cylindrical coordinates by the function: r ρ v 50re Apply Gauss s law to find D. Solution: (C/m 3

194 CHAPTER 4 93 z L r Figure P4.4: Gaussian surface. Method : Integral Form of Gauss s Law Since ρ v varies as a function of r only, so will D. Hence, we construct a cylinder of radius r and length L, coincident with the z-axis. Symmetry suggests that D has the functional form D ˆrD. Hence, D ds Q S ˆrD ds D πrl Q πl 50re r r r dr D ˆrD ˆr50 Method : Differential Method with D r being a function of r. r 0 00πL r e r r e r r e r r r re D ρ v D ˆrD r r 50re r r rd

195 94 CHAPTER 4 0 r r r rd 50r e r r dr r rd r 50r e dr r 0 r rd r 50 e r r e r D ˆrrD r ˆr50 r e r r r re Problem 4.5 An infinitely long cylindrical shell extending between r m and r 3 m contains a uniform charge density ρ v0. Apply Gauss s law to find D in all regions. Solution: For r m, D 0. For r 3 m, ˆrD r ds Q S For r 3 m, D r πrl ρ v0 πl r D ˆrD r ˆr ρ v0πl r ˆr ρ v0 r r 3 m πrl r D r πrl ρ v0 πl 3 8ρ v0 πl D ˆrD r ˆr 4ρ v0 r r 3 m

196 CHAPTER 4 95 z m L 3m r Figure P4.5: Cylindrical shell. Problem 4.6 If the charge density increases linearly with distance from the origin such that ρ v 0 at the origin and ρ v 40 C/m 3 at R m, find the corresponding variation of D. Solution: ρ v R a br ρ v 0 a 0

197 96 CHAPTER 4 ρ v b 40 Hence, b 0. ρ v R 0R (C/m 3 ) Applying Gauss s law to a spherical surface of radius R, D ds S V D R 4πR R ρ v dv 0 D R 5R (C/m ) 0R 4πR dr 80π R4 4 D ˆRDR ˆR5R (C/m ) Section 4-5: Electric Potential Problem 4.7 A square in the x y plane in free space has a point charge of Q at corner a a and the same at corner a a and a point charge of Q at each of the other two corners. (a) Find the electric potential at any point P along the x-axis. (b) Evaluate V at x a. Solution: R R and R 3 R 4. with V At x a, Q 4πε 0 R Q Q Q Q 4πε 0 R 4πε 0 R 3 4πε 0 R 4 πε 0 R R 3 a R a 5 R 3 Q V πε 0 a x a x a a a 5a 0 55Q πε 0 a R R 3

198 CHAPTER 4 97 y -Q a/ Q R 3 R -a/ a/ P(x,0) x R 4 R -Q -a/ Q Figure P4.7: Potential due to four point charges. Problem 4.8 The circular disk of radius a shown in Fig. 4-7 (P4.8) has uniform charge density ρ s across its surface. (a) Obtain an expression for the electric potential V at a point P 0 0 z on the z-axis. (b) Use your result to find E and then evaluate it for z h. Compare your final expression with Eq. (4.4), which was obtained on the basis of Coulomb s law. Solution: (a) Consider a ring of charge at a radial distance r. width dr is The potential at P is dq ρ s πr dr πρ s r dr dv dq 4πε 0 R The potential due to the entire disk is a ρ a V dv s 0 ε 0 0 πρ s r dr 4πε 0 r z a r dr ρ r z s r z ε 0 0 The charge contained in ρ s a z ε 0 z

199 98 CHAPTER 4 z E P(0,0,h) h ρ s dq = π ρ s r dr r a y x a dr Figure P4.8: Circular disk of charge. (b) E V ˆx V x ŷ V y ẑ V z ẑ ρ s ε 0 The expression for E reduces to Eq. (4.4) when z h. z a z Problem 4.9 A circular ring of charge of radius a lies in the x y plane and is centered at the origin. If the ring is in air and carries a uniform density ρ l, (a) show that the electrical potential at 0 0 z is given by V ρ l a ε 0 a z, and (b) find the corresponding electric field E. Solution: (a) For the ring of charge shown in Fig. P4.9, using Eq. (3.67) in Eq. (4.48c) gives ρ V R l dl 4πε 0 l R 4πε 0 π φ 0 ρ l a r ar cos φ φ z a dφ Point 0 0 z in Cartesian coordinates corresponds to r φ z 0 φ z in cylindrical coordinates. Hence, for r 0, π ρ V 0 0 z l 4πε 0 φ 0 a z a dφ ρ l a ε 0 a z

200 CHAPTER 4 99 z z R' = a + z x a 0 y a dφ' dl' = a dφ' ρ l Figure P4.9: Ring of charge. (b) From Eq. (4.5), E V ẑ ρ la ε 0 z a z ẑ ρ la ε 0 z a z 3 (V/m) Problem 4.30 Show that the electric potential difference V between two points in air at radial distances r and r from an infinite line of charge with density ρ l along the z-axis is V ρ l πε 0 ln r r. Solution: From Eq. (4.33), the electric field due to an infinite line of charge is ρ E ˆrE r l ˆr πε 0 r Hence, the potential difference is V r r E dl r r ˆrρ l dr πε 0 r ˆr ρ l r ln πε 0 r Problem 4.3 Find the electric potential V at a location a distance b from the origin in the x y plane due to a line charge with charge density ρ l and of length l. The line charge is coincident with the z-axis and extends from z l to z l.

201 00 CHAPTER 4 z l/ R' = z + b l dz z R' b V(b) y -l/ Figure P4.3: Line of charge of length. Solution: From Eq. (4.48c), we can find the voltage at a distance b away from a line of charge [Fig. P4.3]: V b 4πε l ρ l R dl ρ l 4πε l l dz z b ρ l 4πε ln l l 4b l l 4b Problem 4.3 For the electric dipole shown in Fig. 4-3, d cm and E 4 (mv/m) at R m and θ 0. Find E at R m and θ 90. Solution: For R m and θ 0, E 4 mv/m, we can solve for q using Eq. (4.56): qd E 4πε 0 R ˆRcos 3 θ ˆθsin θ Hence, qd E 4πε 0 4 mv/m at θ 0 3 q 0 8πε πε 0 d 0 0 8πε 0 (C) Again using Eq. (4.56) to find E at R m and θ 90, we have E 0 8πε 0 0 ˆR 4πε 0 0 ˆθ ˆθ (mv/m) 3 4

202 CHAPTER 4 0 Problem 4.33 For each of the following distributions of the electric potential V, sketch the corresponding distribution of E (in all cases, the vertical axis is in volts and the horizontal axis is in meters): Solution: 30 V x -30 E 0 x -0 (a) V x -4 E x -4.0 (b)

203 0 CHAPTER 4 4 V x -4 E x -.6 (c) Figure P4.33: Electric potential distributions of Problem Problem 4.34 Given the electric field E ˆR 8 R (V/m) find the electric potential of point A with respect to point B where A is at m and B at 4 m, both on the z-axis. Solution: V AB V A V B B A E dl Along z-direction, ˆR ẑ and E ẑ 8 z for z 0, and ˆR ẑ and E ẑ 8 z z 0. Hence, V AB ˆR 8 dz 4 z ẑ 0 4 ẑ 8 z ẑ dz 0 ẑ 8 z ẑ dz 4 V for

204 CHAPTER 4 03 A z = m B z = -4m Figure P4.34: Potential between B and A. Problem 4.35 An infinitely long line of charge with uniform density ρ l 9 (nc/m) lies in the x y plane parallel to the y-axis at x m. Find the potential V AB at point A 3 m 0 4 m in Cartesian coordinates with respect to point B by applying the result of Problem Solution: According to Problem 4.30, V ρ l πε 0 ln r where r and r are the distances of A and B. In this case, r m r m Hence, V AB π ln V r

205 04 CHAPTER 4 z 4m A(3, 0, 4) r m r B y x 3m Figure P4.35: Line of charge parallel to y-axis. Problem 4.36 The x y plane contains a uniform sheet of charge with ρ s 0 (nc/m and a second sheet with ρ s 0 (nc/m ) occupies the plane z 6 m. Find V AB, V BC, and V AC for A m, B 0 0 0, and C 0 m m. Solution: We start by finding the E field in the region between the plates. For any point above the x y plane, E due to the charge on x y plane is, from Eq. (4.5), E ẑ ρ s ε 0 In the region below the top plate, E would point downwards for positive ρ s on the top plate. In this case, ρ s ρ s. Hence, E E E ẑ ρ s ẑ ε 0 ρ s ε 0 ẑ ρ s ε 0 ẑ ρ s ε 0 Since E is along ẑ, only change in position along z can result in change in voltage. V AB 0 6 ẑ ρ s ε 0 ẑ dz 6 ρ s z ε 0 0 6ρ s ε V

206 CHAPTER 4 05 z ρ s = - 0. (nc/m) A 6 m C (0, -, ) ρ s = 0. (nc/m) B 0 y x Figure P4.36: Two parallel planes of charge. The voltage at C depends only on the z-coordinate of C. Hence, with point A being at the lowest potential and B at the highest potential, V BC 6 V AB V 3 V AC V AB V BC V Section 4-7: Conductors Problem 4.37 A cylindrical bar of silicon has a radius of 4 mm and a length of 8 cm. If a voltage of 5 V is applied between the ends of the bar and µ e 0 3 (m /V s), µ h 0 05 (m /V s), N e electrons/m 3, and N h N e, find (a) the conductivity of silicon, (b) the current I flowing in the bar, (c) the drift velocities u e and u h, (d) the resistance of the bar, and (e) the power dissipated in the bar.

207 06 CHAPTER 4 Solution: (a) Conductivity is given in Eq. (4.65), σ N e µ e N h µ h e (b) Similarly to Example 4.8, parts b and c, I JA σea (c) From Eqs. (4.6a) and (4.6b), 5 u e µ e E u h µ h E V 0 08 π E 0 08 E 8 5 E (m/s) E E E (m/s) 08 E E (d) To find the resistance, we use what we calculated above, V R 5V I 3 (MΩ) µa (S/m) (e) Power dissipated in the bar is P IV 5V 36 µa 6 8 µw (µa) Problem 4.38 Repeat Problem 4.37 for a bar of germanium with µ e 0 4 (m /V s), µ h 0 (m /V s), and N e N h electrons or holes/m 3. Solution: (a) Conductivity is given in Eq. (4.65), σ N e µ e N u µ u e (S/m) (b) Similarly to Example 4.8, parts b and c, 5V I JA σea 3 (c) From Eqs. (4.6a) and (4.6b), 5 u e µ e E u h µ h E π E E E (m/s) E E (m/s) E (ma)

208 CHAPTER 4 07 (d) To find the resistance, we use what we calculated above, V R 5V I 7 0 (kω) 69 5 ma (e) Power dissipated in the bar is P IV 5V 75 ma 365 (mw) Problem 4.39 A 00-m-long conductor of uniform cross section has a voltage drop of 4 V between its ends. If the density of the current flowing through it is (A/m ), identify the material of the conductor. Solution: We know that conductivity characterizes a material: J σe (A/m ) 4 (V) σ 00 (m) σ (S/m) 7 From Table B-, we find that aluminum has σ (S/m). Problem 4.40 A coaxial resistor of length l consists of two concentric cylinders. The inner cylinder has radius a and is made of a material with conductivity σ, and the outer cylinder, extending between r a and r b, is made of a material with conductivity σ. If the two ends of the resistor are capped with conducting plates, show that the resistance between the two ends is R l π σ a σ b a. Solution: Due to the conducting plates, the ends of the coaxial resistor are each uniform at the same potential. Hence, the electric field everywhere in the resistor will be parallel to the axis of the resistor, in which case the two cylinders can be considered to be two separate resistors in parallel. Then, from Eq. (4.70), or R R inner R outer σ A l σ A l σ πa l R l π σ a σ b a (Ω) σ π b a l Problem 4.4 Apply the result of Problem 4.40 to find the resistance of a 0-cmlong hollow cylinder (Fig. P4.4) made of carbon with σ (S/m). Solution: From Problem 4.40, we know that for two concentric cylinders, R l π σ a σ b a (Ω)

209 08 CHAPTER 4 3 cm cm Carbon Figure P4.4: Cross section of hollow cylinder of Problem 4.4. For air σ 0 (S/m), σ (S/m); hence, R 0 3π (mω) Problem 4.4 A 0 3 -mm-thick square sheet of aluminum has 5 cm 5 cm faces. Find: (a) the resistance between opposite edges on a square face, and (b) the resistance between the two square faces. (See Appendix B for the electrical constants of materials). Solution: (a) R l σa For aluminum, σ (S/m) [Appendix B]. l 5 cm A 5 cm 0 mm m R (mω) 7 (b) Now, l 0 3 mm and A 5 cm 5 cm m. R pω

210 CHAPTER 4 09 Section 4-9: Boundary Conditions Problem 4.43 With reference to Fig. 4-9, find E if E ŷ ˆx3 ẑ (V/m), ε ε 0, ε 8ε 0, and the boundary has a surface charge density ρ s (C/m ). What angle does E make with the z-axis? Solution: We know that E t E t for any media. Hence, E t E t ŷ. ˆx3 Also, D D ˆn ρ s (from Table 4.3). Hence, ε E ˆn ε E ˆn ρ s which gives ρ E z s ε E z ε ε (V/m) Hence, E ŷ ˆx3 ẑ0 (V/m). Finding the angle E makes with the z-axis: E ẑ E cos θ 9 4 4cosθ θ cos 6 7 Problem 4.44 An infinitely long conducting cylinder of radius a has a surface charge density ρ s. The cylinder is surrounded by a dielectric medium with ε r 4 and contains no free charges. If the tangential component of the electric field in the region r a is given by E t ˆφcos φ r, find ρ s. Solution: Let the conducting cylinder be medium and the surrounding dielectric medium be medium. In medium, E ˆrE r ˆφ r cos φ with E r, the normal component of E, unknown. The surface charge density is related to E r. To find E r, we invoke Gauss s law in medium : or which leads to r r r re r r re φ r Integrating both sides with respect to r, D 0 φ r cos φ φ 0 r cos r re r dr sin φcos φ re r r sinφcos φ sinφcos φ r r dr

211 0 CHAPTER 4 or Hence, According to Eq. (4.93), E r r sinφcosφ E ˆr φ r sinφcos ˆφ φ r cos ˆn D D ρ s where ˆn is the normal to the boundary and points away from medium. Hence, ˆn ˆr. Also, D 0 because the cylinder is a conductor. Consequently, ρ s ˆr D r a ˆr ε E r a ˆr ε r ε 0 ˆr r sin φcosφ ˆφ r cos φ 8ε 0 a sinφcos φ (C/m ) r a Problem 4.45 A -cm conducting sphere is embedded in a charge-free dielectric medium with ε r 9. If E ˆR3cos θ ˆθ3sinθ (V/m) in the surrounding region, find the charge density on the sphere s surface. Solution: According to Eq. (4.93), ˆn D D ρ s In the present case, ˆn ˆR and D 0. Hence, ρ s ˆR D r cm ˆR ε ˆR3cos θ ˆθ3sin θ 7ε 0 cosθ (C/m ) Problem 4.46 If E ˆR50 (V/m) at the surface of a 5-cm conducting sphere centered at the origin, what is the total charge Q on the sphere s surface? Solution: From Table 4-3, ˆn D D ρ s. E inside the sphere is zero, since we assume it is a perfect conductor. Hence, for a sphere with surface area S 4πa, ρ D R ρ s E R s Q ε 0 Sε 0

212 CHAPTER 4 Q E R Sε π 0 05 ε 0 3πε 0 (C) Problem 4.47 Figure 4-34(a) (P4.47) shows three planar dielectric slabs of equal thickness but with different dielectric constants. If E 0 in air makes an angle of 45 with respect to the z-axis, find the angle of E in each of the other layers. z 45 E 0 ε 0 (air) ε = 3ε 0 ε = 5ε 0 ε 3 = 7ε 0 ε 0 (air) Figure P4.47: Dielectric slabs in Problem Solution: Labeling the upper air region as region 0 and using Eq. (4.99), ε θ tan tanθ ε 0 0 tan 3tan ε θ tan tanθ ε 5 tan tan ε θ 3 tan 3 7 ε tanθ tan In the lower air region, the angle is again tan Sections 4-0 and 4-: Capacitance and Electrical Energy Problem 4.48 A 5 cm, d cm, and ε r 4 if the voltage across it is 50 V. Determine the force of attraction in a parallel-plate capacitor with

213 CHAPTER 4 Solution: From Eq. (4.3), F ẑ εa E ẑε ẑ (N) Problem 4.49 Dielectric breakdown occurs in a material whenever the magnitude of the field E exceeds the dielectric strength anywhere in that material. In the coaxial capacitor of Example 4-, (a) At what value of r is E maximum? (b) What is the breakdown voltage if a cm, b cm, and the dielectric material is mica with ε r 6? Solution: (a) From Eq. (4.4), E ˆrρ l πεr for a r b. Thus, it is evident that E is maximum at r a. (b) The dielectric breaks down when E ρ E l πεr ρ l π 6ε (MV/m) (see Table 4-), or (MV/m) which gives ρ l 00 MV/m π µc/m). From Eq. (4.5), we can find the voltage corresponding to that charge density, ρ V l πε ln b a 667 6µC/m π F/m ln 39 Thus, V 39 (MV) is the breakdown voltage for this capacitor. (MV) Problem 4.50 An electron with charge Q e C and mass m e kg is injected at a point adjacent to the negatively charged plate in the region between the plates of an air-filled parallel-plate capacitor with separation of cm and rectangular plates each 0 cm in area Fig (P4.50). If the voltage across the capacitor is 0 V, find (a) the force acting on the electron, (b) the acceleration of the electron, and (c) the time it takes the electron to reach the positively charged plate, assuming that it starts from rest. Solution: (a) The electric force acting on a charge Q e is given by Eq. (4.4) and the electric field in a capacitor is given by Eq. (4.). Combining these two relations, we have (N) F Q e E Q e V d

214 CHAPTER 4 3 cm Q e - + V 0 = 0 V Figure P4.50: Electron between charged plates of Problem The force is directed from the negatively charged plate towards the positively charged plate. (b) a F m (m/s ) (c) The electron does not get fast enough at the end of its short trip for relativity to manifest itself; classical mechanics is adequate to find the transit time. From classical mechanics, d d 0 u 0 t at, where in the present case the start position is d 0 0, the total distance traveled is d cm, the initial velocity u 0 0, and the acceleration is given by part (b). Solving for the time t, t Problem 4.5 d a s 0 7 (ns) In a dielectric medium with ε r 4, the electric field is given by E ˆx x z ŷx ẑ y z (V/m) Calculate the electrostatic energy stored in the region m x m, 0 y m, and 0 z 3 m.

215 4 CHAPTER 4 Solution: Electrostatic potential energy is given by Eq. (4.4), W e ε E ε dv V 4ε 0 4ε 0 3 z 0 y 0 x 5 x5 yz 3 z x 3 y 4 3 z3 xy (J) 5 x 4 y z dx dy dz 3 y z x 4 x z x Problem 4.5 Figure 4-34a (P4.5(a)) depicts a capacitor consisting of two parallel, conducting plates separated by a distance d. The space between the plates y 0 z 0 d ε A A ε + - V (a) + C C V - (b) Figure P4.5: (a) Capacitor with parallel dielectric section, and (b) equivalent circuit. contains two adjacent dielectrics, one with permittivity ε and surface area A and another with ε and A. The objective of this problem is to show that the capacitance C of the configuration shown in Fig. 4-34a (P4.5(a)) is equivalent to two capacitances in parallel, as illustrated in Fig. 4-34b (P4.5(b)), with C C C (4.3)

216 CHAPTER 4 5 where C ε A d (4.33) C ε A d (4.34) To this end, you are asked to proceed as follows: (a) Find the electric fields E and E in the two dielectric layers. (b) Calculate the energy stored in each section and use the result to calculate C and C. (c) Use the total energy stored in the capacitor to obtain an expression for C. Show that Eq. (4.3) is indeed a valid result. Solution: ε ε E E d + V - (c) Figure P4.5: (c) Electric field inside of capacitor. (a) Within each dielectric section, E will point from the plate with positive voltage to the plate with negative voltage, as shown in Fig. P4-5(c). From V Ed, V E E d (b) W e But, from Eq. (4.), ε E V ε V d A d ε V A d W e A Hence C ε d. Similarly, C A ε d. (c) Total energy is W e W e W e C V V d ε A ε A CV

217 6 CHAPTER 4 Hence, C ε A d ε A d C C Problem 4.53 Use the result of Problem 4.5 to determine the capacitance for each of the following configurations: (a) conducting plates are on top and bottom faces of rectangular structure in Fig. 4-35(a) (P4.53(a)), (b) conducting plates are on front and back faces of structure in Fig. 4-35(a) (P4.53(a)), (c) conducting plates are on top and bottom faces of the cylindrical structure in Fig. 4-35(b) (P4.53(b)). Solution: (a) The two capacitors share the same voltage; hence they are in parallel. (b) (c) A C ε d ε 0 0 5ε 0 0 A C ε d 4ε ε 0 0 C C C 5ε 0 30ε ε F A C 4 ε 0 d ε ε 0 0 A C ε d 4ε ε 0 0 C C C F A C ε d 8ε 4πε F A C ε d 4ε π r r 0 0 πε F A C 3 3 ε 3 d ε π r r πε F C C C C F πr

218 CHAPTER 4 7 cm 3 cm 5 cm cm ε r = ε r = 4 (a) r = mm r = 4mm r 3 = 8mm ε 3 ε ε cm ε = 8ε 0 ; ε = 4ε 0 ; ε 3 = ε 0 (b) Figure P4.53: Dielectric sections for Problems 4.53 and 4.55.

219 8 CHAPTER 4 Problem 4.54 The capacitor shown in Fig (P4.54) consists of two parallel dielectric layers. We wish to use energy considerations to show that the equivalent capacitance of the overall capacitor, C, is equal to the series combination of the capacitances of the individual layers, C and C, namely C C C C C (4.36) where C ε A d C ε A d ε ε A d d + - V (a) C C + V - (b) Figure P4.54: (a) Capacitor with parallel dielectric layers, and (b) equivalent circuit (Problem 4.54). (a) Let V and V be the electric potentials across the upper and lower dielectrics, respectively. What are the corresponding electric fields E and E? By applying the appropriate boundary condition at the interface between the two dielectrics, obtain explicit expressions for E and E in terms of ε, ε, V, and the indicated dimensions of the capacitor. (b) Calculate the energy stored in each of the dielectric layers and then use the sum to obtain an expression for C.

220 CHAPTER 4 9 d d ε ε E E + V - + V V Figure P4.54: (c) Electric fields inside of capacitor. (c) Show that C is given by Eq. (4.36). Solution: (a) If V is the voltage across the top layer and V across the bottom layer, then V V V and E V d E V d According to boundary conditions, the normal component of D is continuous across the boundary (in the absence of surface charge). This means that at the interface between the two dielectric layers, or Hence, D n D n ε E ε E V E d E d E d which can be solved for E : V E ε d d ε Similarly, V E ε d d ε ε E ε d

221 0 CHAPTER 4 (b) W e W e ε E V ε ε E V ε W e W e W e V d V ε ε d V ε d d ε ε ε Ad ε ε Ad ε d ε d Ad V Ad V ε ε Ad ε d ε d ε ε Ad ε d ε d But W e CV, hence, C ε ε Ad ε ε Ad ε d ε d ε ε A ε d ε d ε d ε d ε ε A ε d ε d (c) Multiplying numerator and denominator of the expression for C by A d d, we have ε A ε A d C d CC ε A ε A C d C d where C ε A d C ε A d Problem 4.55 Use the expressions given in Problem 4.54 to determine the capacitance for the configurations in Fig. 4.35(a) (P4.55) when the conducting plates are placed on the right and left faces of the structure. Solution: A C ε d ε ε F A 4 C ε d 4ε F C C C C C F 8

222 CHAPTER 4 cm 3 cm 5 cm cm ε r = ε r = 4 Figure P4.55: Dielectric section for Problem Section 4-: Image Method Problem 4.56 With reference to Fig (P4.56), charge Q is located at a distance d above a grounded half-plane located in the x y plane and at a distance d from another grounded half-plane in the x z plane. Use the image method to (a) establish the magnitudes, polarities, and locations of the images of charge Q with respect to each of the two ground planes (as if each is infinite in extent), and (b) then find the electric potential and electric field at an arbitrary point P 0 y z. z P(0, y, z) d Q(0, d, d) d y Figure P4.56: Charge Q next to two perpendicular, grounded, conducting half planes. Solution: (a) The original charge has magnitude and polarity Q at location 0 d d. Since the negative y-axis is shielded from the region of interest, there might as well be a conducting half-plane extending in the y direction as well as the y direction. This ground plane gives rise to an image charge of magnitude and polarity Q at location

223 CHAPTER 4 z -Q d Q P(y, z) -d Q -d d -Q y Figure P4.56: (a) Image charges. 0 d d. In addition, since charges exist on the conducting half plane in the z direction, an image of this conducting half plane also appears in the z direction. This ground plane in the x-z plane gives rise to the image charges of Q at 0 d d and Q at 0 d d. (b) Using Eq. (4.47) with N 4, V x y z Q 4πε Q 4πε Q 4πε ˆxx ŷ y d ẑ z d ˆxx ŷ y d ẑ z d ˆxx ŷ y d ẑ z d ˆxx ŷ y d ẑ z d x y d z d x y d z d x y d z d x y d z d x y yd z zd d x y yd z zd d x y yd z zd d x y yd z zd (V) d

224 CHAPTER 4 3 From Eq. (4.5), E V Q 4πε Q 4πε x y d z d x y d z d x y d z d x y d z d ˆxx ŷ y d ẑ z d ˆxx x 3 y d z d ŷ y d ẑ z d x y 3 d z d ˆxx ŷ y d ẑ z d ˆxx x y d z 3 d ŷ y d ẑ z d x y d z 3 d (V/m) Problem 4.57 Conducting wires above a conducting plane carry currents I and I in the directions shown in Fig (P4.57). Keeping in mind that the direction I I (a) (b) Figure P4.57: Currents above a conducting plane (Problem 4.57). of a current is defined in terms of the movement of positive charges, what are the directions of the image currents corresponding to I and I? Solution: (a) In the image current, movement of negative charges downward movement of positive charges upward. Hence, image of I is same as I.

225 4 CHAPTER 4 I + t=t + t=0 (image) I - t=0 - t=t Figure P4.57: (a) Solution for part (a). (b) In the image current, movement of negative charges to right movement of positive charges to + q I + t=t - I - t=t (image) Figure P4.57: (b) Solution for part (b). Problem 4.58 Use the image method to find the capacitance per unit length of an infinitely long conducting cylinder of radius a situated at a distance d from a parallel conducting plane, as shown in Fig (P4.58). Solution: Let us distribute charge ρ l (C/m) on the conducting cylinder. Its image cylinder at z d will have charge density ρ l. For the line at z d, the electric field at any point z (at a distance of d z from the center of the cylinder) is, from Eq. (4.33), ρ E l ẑ πε 0 d z

226 CHAPTER 4 5 a d V = 0 Figure P4.58: Conducting cylinder above a conducting plane (Problem 4.58). a ρ l d z d a ρ l Figure P4.58: (a) Cylinder and its image. where ẑ is the direction away from the cylinder. Similarly for the image cylinder at distance d z and carrying charge ρ l, ρ E ẑ l πε 0 d ρ z l ẑ πε 0 d z The potential difference between the cylinders is obtained by integrating the total electric field from z d a to z d a : V E E ẑ dz a d ẑ d a ρ l πε 0 d z d z ẑ dz

227 6 CHAPTER 4 ρ l d a πε 0 d dz z ρ l πε 0 ρ l πε 0 ρ l πε 0 ln For a length L, Q ρ l L and C Q V and the capacitance per unit length is C d a d z ln d z ln d a z d d a ln a ln d a ln d a ln a a d a ρ l L ρ l πε 0 ln d a a C L πε 0 ln d a (C/m) Problem 4.59 A circular beam of charge of radius a consists of electrons moving with a constant speed u along the z direction. The beam s axis is coincident with the z-axis and the electron charge density is given by ρ v cr (c/m 3 ) where c is a constant and r is the radial distance from the axis of the beam. (a) Determine the charge density per unit length. (b) Determine the current crossing the z-plane. Solution: (a) ρ l ρ v ds a r 0 π φ 0 cr r dr dφ πc r4 4 a 0 πca 4 (C/m)

228 CHAPTER 4 7 (b) J ρ v u ẑcr u (A/m ) I J ds a πcu r 0 π φ 0 ẑcur ẑr dr dφ 0 a r 3 dr πcua 4 ρ l u (A) Problem 4.60 A line of charge of uniform density ρ l occupies a semicircle of radius b as shown in the figure. Use the material presented in Example 4-4 to determine the electric field at the origin. z b ρ l y x Solution: Since we have only half of a circle, we need to integrate the expression for de given in Example 4-4 over φ from 0 to π. Before we do that, however, we need to set h 0 (the problem asks for E at the origin). Hence, de ρ l b 4πε 0 E ˆrρ l 4πε 0 b dφ π φ 0 ˆrb ẑh b h 3 dφ h 0 de ˆrρ l 4ε 0 b Problem 4.6 A spherical shell with outer radius b surrounds a charge-free cavity of radius a b. If the shell contains a charge density given by ρ v ρ v0 R a R b where ρ v0 is a positive constant, determine D in all regions.

229 8 CHAPTER 4 r 3 r b a ρ v r Solution: Symmetry dictates that D is radially oriented. Thus, D ˆRDR At any R, Gauss s law gives (a) For R S D ds Q S ˆRD R ˆR ds Q 4πR D R Q Q D R 4πR a, no charge is contained in the cavity. Hence, Q 0, and D R 0 R a (b) For a R b, Hence, Q R R a D R ρ v dv R R a ρ v0 R 4πR dr 4πρ v0 R a ρ v0 R a R a R b

230 CHAPTER 4 9 (c) For R b, Q D R b ρ v dv 4πρ v0 b a ρ v0 b a R R b R a Problem 4.6 Two infinite lines of charge, both parallel to the z-axis, lie in the x z plane, one with density ρ l and located at x a and the other with density ρ l and located at x a. Obtain an expression for the electric potential V x y at a point P x y relative to the potential at the origin. y P(x,y) r'' r' -ρ l ρ l (-a, 0) (a, 0) x Solution: According to the result of Problem 4.30, the electric potential difference between a point at a distance r and another at a distance r from a line charge of density ρ l is ρ V l r ln πε 0 r Applying this result to the line charge at x a, which is at a distance a from the origin: V ρ l πε 0 ln a ρ l πε 0 ln r a and r r r a x a y Similarly, for the negative line charge at x a, ρ V l a πε 0 ln r a and r r r ρ l a ln πε 0 x a y

231 30 CHAPTER 4 The potential due to both lines is V V V ρ l πε 0 ln a x a y ln a x a y At the origin, V 0, as it should be since the origin is the reference point. The potential is also zero along all points on the y-axis (x 0). Problem 4.63 A cylinder-shaped carbon resistor is 8 cm in length and its circular cross section has a diameter d mm. (a) Determine the resistance R. (b) To reduce its resistance by 40%, the carbon resistor is coated with a layer of copper of thickness t. Use the result of Problem 4.40 to determine t. Solution: (a) From (4.70), and using the value of σ for carbon from Appendix B, R l σa (b) The 40%-reduced resistance is: Using the result of Problem 4.40: l σπ d π Ω R 0 6R Ω R l π σ a σ b a 04 Ω With σ S/m (carbon), σ S/m (copper), a mm m, and b unknown, we have b m and 4 t b a m µm Thus, the addition of a copper coating less than 0. µm in thickness reduces the resistance by 40%.

232 CHAPTER 4 3 Problem 4.64 A coaxial capacitor consists of two concentric, conducting, cylindrical surfaces, one of radius a and another of radius b, as shown in the figure. The insulating layer separating the two conducting surfaces is divided equally into two semi-cylindrical sections, one filled with dielectric ε and the other filled with dielectric ε. ε b a ε E + + l - + V (a) Develop an expression for C in terms of the length l and the given quantities. (b) Evaluate the value of C for a mm, b 6 mm, ε r l 4 cm., ε r 4, and Solution: (a) For the indicated voltage polarity, the E field inside the capacitor exists in only the dielectric materials and points radially inward. Let E be the field in dielectric ε and E be the field in dielectric ε. At the interface between the two dielectric sections, E is parallel to E and both are tangential to the interface. Since boundary conditions require that the tangential components of E and E be the same, it follows that: E E ˆrE

233 3 CHAPTER 4 At r a (surface of inner conductor), in medium, the boundary condition on D, as stated by (4.0), leads to or Similarly, in medium D ε E ˆnρ s ˆrε E ˆrρ s ρ s ε E ρ s ε E Thus, the E fields will be the same in the two dielectrics, but the charge densities will be different along the two sides of the inner conducting cylinder. Since the same voltage applies for the two sections of the capacitor, we can treat them as two capacitors in parallel. For the capacitor half that includes dielectric ε, we can apply the results of Eqs. (4.4) (4.6), but we have to keep in mind that Q is now the charge on only one half of the inner cylinder. Hence, Similarly, and (b) C πε l ln b a πε C l ln b a πl ε C C C ε ln b a C π ln pf.

234 33 Chapter 5: Magnetostatics Lesson #30 Chapter Section: 5- Topics: Magnetic forces and torques Highlights: Lorentz force on a charged particle Magnetic force on a current in a magnetic field Torque on a loop Special Illustrations: Examples 5-

235 34 Lesson #3 Chapter Section: 5- Topics: Biot-Savart law Highlights: Magnetic field induction by electric currents Magnetic field due to linear conductor Magnetic dipole Special Illustrations: Example 5- Example 5-3 CD-ROM Modules 5.3 and 5.4

236 35 Lesson #3 Chapter Section: 5-3, 5-4 Topics: Magnetic force, Ampère s law Highlights: Attraction and repulsion forces between currents Gauss s law for magnetics Ampère s law Special Illustrations: Example 5-6 CD-ROM Modules 5. and 5.

237 36 Lesson #33 Chapter Section: 5-5, 5-6 Topics: Vector magnetic potential, magnetic materials Highlights: Relation of A to B Vector Poisson s Eq. Magnetic permeability Ferromagnetism, hysteresis Special Illustrations: Technology Brief on Electromagnetic and magnetic switches (CD-ROM) Electromagnets and Magnetic Relays William Sturgeon developed the first practical electromagnet in the 80s. Today the principle of the electromagnet is used in motors, relay switches in read/write heads for hard disks and tape drives, loudspeakers, magnetic levitation and many other applications. Basic Principle Electromagnets can be constructed in various shapes, including the linear solenoid described in Section When an electric current generated by a power source, such as a battery, flows through the wire coiled around the central core, it induces a magnetic field with field lines resembling those generated by a bar magnet (A). The strength of the magnetic field is proportional to the current, the number of turns, and the magnetic permeability of the core material. By using a ferromagnetic core, the field strength can be increased by several orders of magnitude, depending on the purity of the iron material. When subjected to a magnetic field, ferromagnetic materials, such as iron or nickel, get magnetized and act like magnets themselves.

238 37 Lesson #34 Chapter Section: 5-7 Topics: Boundary conditions Highlights: Analogy with electric-field boundary conditions Special Illustrations: Technology Brief on Magnetic Recording (CD-ROM) Magnetic Recording Valdemar Poulsen, a Danish engineer, invented magnetic recording by demonstrating in 900 that speech can be recorded on a thin steel wire using a simple electromagnet. Magnetic tapes were developed as an alternative medium to wires in the 940s and became very popular for recording and playing music well into the 960s. Videotapes were introduced in the late 950s for recording motion pictures for later replay on television. Because video signals occupy a much wider bandwidth, tape speeds for video recording (past the magnetic head) have to be at rates on the order of 5 m/s, compared with only 0.3 m/s for audio. Other types of magnetic recording media were developed since then, including the flexible plastic disks called floppies, the hard disks made of glass or aluminum, the magnetic drum, and the magnetic bubble memory. All take advantage of the same fundamental principle of being able to store electrical information through selective magnetization of a magnetic material, as well as the ability to retrieve it (playback) when so desired.

239 38 Lesson #35 Chapter Section: 5-8 Topics: Inductance Highlights: Solenoid Self inductance Special Illustrations: Example 5-8 Technology Brief on Inductive Sensors (CD-ROM) Inductive Sensors Magnetic coupling between different coils forms the basis of several different types of inductive sensors. Applications include the measurement of position and displacement (with sub-millimeter resolution) in device fabrications processes, proximity detection of conductive objects, and other related applications. Linear Variable Differential Transformer (LVDT) A LVDT comprises a primary coil connected to an ac source, typically a sine wave at a frequency in the 0 KHz range, and a pair of secondary coils, all sharing a common ferromagnetic core (A). The magnetic core serves to couple the magnetic flux generated by the primary coil into the two secondaries, thereby inducing an output voltage across each of them. The secondary coils are connected in opposition, so that when the core is positioned at the magnetic center of the LVDT, the individual output signals of the secondaries cancel each other out, producing a null output voltage. The core is connected to the outside world via a nonmagnetic rod. When the rod moves the core away from the magnetic center, the magnetic fluxes induced in the secondary coils are no longer equal, resulting in a non-zero output voltage. The LVDT is called a linear transformer because the output voltage is a linear function of displacement over a wide operating range.

240 39 Lesson #36 Chapter Section: 5-9 Topics: Magnetic energy Highlights: Magnetic energy density Magnetic energy in a coax Special Illustrations: Example 5-9

241 40 CHAPTER 5 Chapter 5 Sections 5-: Forces and Torques Problem 5. An electron with a speed of m/s is projected along the positive x-direction into a medium containing a uniform magnetic flux density B ˆx4 ẑ3 T. Given that e C and the mass of an electron is m e kg, determine the initial acceleration vector of the electron (at the moment it is projected into the medium). Solution: The acceleration vector of a free particle is the net force vector divided by the particle mass. Neglecting gravity, and using Eq. (5.3), we have F a m e m e m e u B ŷ4 0 8 (m/s ) ˆx8 0 6 ˆx4 ẑ3 Problem 5. When a particle with charge q and mass m is introduced into a medium with a uniform field B such that the initial velocity of the particle u is perpendicular to B, as shown in Fig. 5-3 (P5.), the magnetic force exerted on the particle causes it to move in a circle of radius a. By equating F m to the centripetal force on the particle, determine a in terms of q, m, u, and B. Solution: The centripetal force acting on the particle is given by F c mu a. B u q + F m a F m u q + F m q + P + q u Figure P5.: Particle of charge q projected with velocity u into a medium with a uniform field B perpendicular to u (Problem 5.).

242 CHAPTER 5 4 Equating F c to F m given by Eq. (5.4), we have mu a qubsin θ. Since the magnetic field is perpendicular to the particle velocity, sin θ. Hence, a mu qb. Problem 5.3 The circuit shown in Fig. 5-3 (P5.3) uses two identical springs to support a 0-cm-long horizontal wire with a mass of 0 g. In the absence of a magnetic field, the weight of the wire causes the springs to stretch a distance of 0. cm each. When a uniform magnetic field is turned on in the region containing the horizontal wire, the springs are observed to stretch an additional 0.5 cm. What is the intensity of the magnetic flux density B? The force equation for a spring is F kd, where k is the spring constant and d is the distance it has been stretched. 4Ω V + - Springs B 0 cm Figure P5.3: Configuration of Problem 5.3. Solution: Springs are characterized by a spring constant k where F kd is the force exerted on the spring and d is the amount the spring is stretched from its rest configuration. In this instance, each spring sees half the weight of the wire: F mg kd k mg d (N/m) Therefore, when the springs are further stretched by an additional 0.5 cm, this amounts to an additional force of F 49 N/m m 45 mn per spring, or a total additional force of F 0 49 N. This force is equal to the force exerted on the wire by the interaction of the magnetic field and the current as described by Eq. (5.): F m I B, where and B are at right angles. Moreover B is in the

243 4 CHAPTER 5 downward direction, and I V R V 4 Ω 3 A. Therefore, F F m I B B m 0 49 I 3 0 (T) 63 Problem 5.4 The rectangular loop shown in Fig (P5.4) consists of 0 closely wrapped turns and is hinged along the z-axis. The plane of the loop makes an angle of 30 with the y-axis, and the current in the windings is 0.5 A. What is the magnitude of the torque exerted on the loop in the presence of a uniform field B ŷ 4 T? When viewed from above, is the expected direction of rotation clockwise or counterclockwise? z I = 0.5 A 0.4 m 0 turns 30 y 0. m x Figure P5.4: Hinged rectangular loop of Problem 5.4. Solution: The magnetic torque on a loop is given by T m B (Eq. (5.0)), where m ˆnNIA (Eq. (5.9)). For this problem, it is given that I 0 5 A, N 0 turns, and A 0 m 0 4 m 0 08 m. From the figure, ˆn ˆxcos 30 ŷsin30. Therefore, m ˆn0 8 (A m and T ˆn0 8 (A m ŷ 4 T ẑ 66 (N m). As the torque is negative, the direction of rotation is clockwise, looking from above. Problem 5.5 In a cylindrical coordinate system, a -m-long straight wire carrying a current of 5 A in the positive z-direction is located at r 4 cm, φ π, and m z m. (a) If B ˆr0 cos φ (T), what is the magnetic force acting on the wire?

244 CHAPTER 5 43 z m 5A 4 cm y - m Figure P5.5: Problem 5.5. (b) How much work is required to rotate the wire once about the z-axis in the negative φ-direction (while maintaining r 4 cm)? (c) At what angle φ is the force a maximum? Solution: (a) At φ π, ˆφ ˆx. Hence, (b) W π φ 0 F dl 0 π r F I B 5ẑ ˆr0 cos φ ˆφcos φ F ˆxcos π 0 ˆφ cos φ 0 π cosφ dφ ˆφ r dφ r 4 cm π r 4 cm 8 0 sinφ 0 0 The force is in the ˆφ-direction, which means that rotating it in the ˆφ-direction would require work. However, the force varies as cos φ, which means it is positive

245 44 CHAPTER 5 when π φ π and negative over the second half of the circle. Thus, work is provided by the force between φ π and φ π (when rotated in the ˆφ-direction), and work is supplied for the second half of the rotation, resulting in a net work of zero. (c) The force F is maximum when cosφ, or φ 0. Problem 5.6 placed in the y z plane as shown in Fig (P5.6). A 0-turn rectangular coil with side l 0 cm and w 0 cm is z 0-turn coil w l I φ n^ y x Figure P5.6: Rectangular loop of Problem 5.6. (a) If the coil, which carries a current I 0 A, is in the presence of a magnetic flux density B 0 ˆx ŷ (T) determine the torque acting on the coil. (b) At what angle φ is the torque zero? (c) At what angle φ is the torque maximum? Determine its value. Solution: (a) The magnetic field is in direction ˆx ŷ, which makes an angle φ 0 tan The magnetic moment of the loop is m ˆnNIA ˆn ˆn6 (A m )

246 CHAPTER 5 45 z y x φ 0 = o B Figure P5.6: (a) Direction of B. where ˆn is the surface normal in accordance with the right-hand rule. When the loop is in the negative-y of the y z plane, ˆn is equal to ˆx, but when the plane of the loop is moved to an angle φ, ˆn becomes ˆn ˆxcosφ ŷsinφ or (b) The torque is zero when T m B ˆn6 0 ˆx ŷ ˆxcos φ ŷsinφ 6 0 ˆx ŷ ẑ0 cos φ sinφ (N m) tanφ cos φ sinφ 0 φ or 6 57 Thus, when ˆn is parallel to B, T 0. (c) The torque is a maximum when ˆn is perpendicular to B, which occurs at φ or Mathematically, we can obtain the same result by taking the derivative of T and equating it to zero to find the values of φ at which T is a maximum. Thus, T φ φ 0 cos φ sinφ 0

247 46 CHAPTER 5 or which gives tan φ sin φ cosφ 0, or φ 6 57 or at which T ẑ0 7 (N m). Section 5-: Biot Savart Law Problem 5.7 An 8 cm cm rectangular loop of wire is situated in the x y plane with the center of the loop at the origin and its long sides parallel to the x-axis. The loop has a current of 50 A flowing with clockwise direction (when viewed from above). Determine the magnetic field at the center of the loop. Solution: The total magnetic field is the vector sum of the individual fields of each of the four wire segments: B B B B 3 B 4. An expression for the magnetic field from a wire segment is given by Eq. (5.9). z -6 cm 3-4 cm 4 4 cm I y x 6 cm Figure P5.7: Problem 5.7. For all segments shown in Fig. P5.7, the combination of the direction of the current and the right-hand rule gives the direction of the magnetic field as z direction at the origin. With r 6 cm and l 8 cm, µil B ẑ πr 4r l 7 4π ẑ π ẑ9 4 0 (T)

248 CHAPTER 5 47 For segment, r 4 cm and l cm, Similarly, µil B ẑ πr 4r l 4π ẑ π ẑ B 3 ẑ (T) B 4 ẑ The total field is then B B B B 3 B 4 ẑ0 60 (mt). Problem 5.8 Use the approach outlined in Example 5- to develop an expression for the magnetic field H at an arbitrary point P due to the linear conductor defined by the geometry shown in Fig (P5.8). If the conductor extends between z 3 m and z 7 m and carries a current I 5 A, find H at P φ 0. z (T) (T) P (z ) θ I P (z ) θ r P(r, φ, z) Figure P5.8: Current-carrying linear conductor of Problem 5.8. Solution: The solution follows Example 5- up through Eq. (5.7), but the expressions for the cosines of the angles should be generalized to read as cos θ z z r z z cosθ z z r z z

249 48 CHAPTER 5 instead of the expressions in Eq. (5.8), which are specialized to a wire centered at the origin. Plugging these expressions back into Eq. (5.7), the magnetic field is given as H ˆφ I 4πr For the specific geometry of Fig. P5.8, H ˆφ 4π z z r z z z z r z z ˆφ (A/m) ˆφ77 4 (ma/m) Problem 5.9 The loop shown in Fig (P5.9) consists of radial lines and segments of circles whose centers are at point P. Determine the magnetic field H at P in terms of a, b, θ, and I. I b θ a P Figure P5.9: Configuration of Problem 5.9. Solution: From the solution to Example 5-3, if we denote the z-axis as passing out of the page through point P, the magnetic field pointing out of the page at P due to the current flowing in the outer arc is H outer ẑiθ 4πb and the field pointing out of the page at P due to the current flowing in the inner arc is H inner ẑiθ 4πa. The other wire segments do not contribute to the magnetic field at P. Therefore, the total field flowing directly out of the page at P is H H outer H inner ẑ Iθ 4π a b ẑ b a Iθ 4πab Problem 5.0 An infinitely long, thin conducting sheet defined over the space 0 x w and y is carrying a current with a uniform surface current

250 CHAPTER 5 49 z P(0, 0, z) R = x + z R 0 x x w Figure P5.0: Conducting sheet of width w in x y plane. density J s ŷ5 (A/m). Obtain an expression for the magnetic field at point P 0 0 z in Cartesian coordinates. Solution: The sheet can be considered to be a large number of infinitely long but narrow wires each dx wide lying next to each other, with each carrying a current I x J s dx. The wire at a distance x from the origin is at a distance vector R from point P, with R ˆxx ẑz Equation (5.30) provides an expression for the magnetic field due to an infinitely long wire carrying a current I as H B µ 0 ˆφI πr We now need to adapt this expression to the present situation by replacing I with I x J s dx, replacing r with R x z, as shown in Fig. P5.0, and by assigning the proper direction for the magnetic field. From the Biot Savart law, the direction of H is governed by l R, where l is the direction of current flow. In the present case, l is in the ŷ direction. Hence, the direction of the field is l R l ŷ ˆxx ẑz R ŷ ˆxx ˆxz ẑz ẑx x z

251 50 CHAPTER 5 Therefore, the field dh due to the current I x is dh and the total field is H 0 0 z w J s π x 0 J s π J s π 5 π 5 π ˆxz ẑx x z ˆxz ẑx J s dx π x z w ˆxz dx ẑx x 0 x z w dx ˆxz x 0 x z ẑ tan x ˆxz z z w ˆxπtan w ˆxπtan I x ˆxz ẑx J s dx πr π x z w x 0 w x dx x z ẑ x 0 ln x z w ẑ z ln w z ln 0 z x 0 z ẑ ln w z z (A/m) for z An alternative approach is to employ Eq. (5.4a) directly. 0 0 for z Problem 5. An infinitely long wire carrying a 5-A current in the positive x-direction is placed along the x-axis in the vicinity of a 0-turn circular loop located in the x y plane as shown in Fig (P5.(a)). If the magnetic field at the center of the loop is zero, what is the direction and magnitude of the current flowing in the loop? m d = m I x Figure P5.: (a) Circular loop next to a linear current (Problem 5.). Solution: From Eq. (5.30), the magnetic flux density at the center of the loop due to

252 CHAPTER 5 5 I Figure P5.: (b) Direction of I. the wire is B ẑ µ 0 πd I where ẑ is out of the page. Since the net field is zero at the center of the loop, I must be clockwise, as seen from above, in order to oppose I. The field due to I is, from Eq. (5.35), B µ 0 H ẑ µ 0NI a Equating the magnitudes of the two fields, we obtain the result NI I a πd or ai I 5 πnd π 0 0 A Problem 5. Two infinitely long, parallel wires carry 6-A currents in opposite directions. Determine the magnetic flux density at point P in Fig (P5.). I = 6A I = 6A 0.5m P m Figure P5.: Arrangement for Problem 5.. Solution: B ˆφ µ 0I ˆφ π 0 5 µ 0I π 5 ˆφ µ 0 π 6 ˆφ 8µ 0 π (T)

253 5 CHAPTER 5 Problem 5.3 A long, East-West oriented power cable carrying an unknown current I is at a height of 8 m above the Earth s surface. If the magnetic flux density recorded by a magnetic-field meter placed at the surface is 5 µt when the current is flowing through the cable and 0 µt when the current is zero, what is the magnitude of I? Solution: The power cable is producing a magnetic flux density that opposes Earth s, own magnetic field. An East West cable would produce a field whose direction at the surface is along North South. The flux density due to the cable is B 0 5 µt 5µT As a magnet, the Earth s field lines are directed from the South Pole to the North Pole inside the Earth and the opposite on the surface. Thus the lines at the surface are from North to South, which means that the field created by the cable is from South to North. Hence, by the right-hand rule, the current direction is toward the East. Its magnitude is obtained from 6 µ 5 µt I 4π 0 7 I πd π 8 which gives I 00 A. Problem 5.4 Two parallel, circular loops carrying a current of 40 A each are arranged as shown in Fig (P5.4). The first loop is situated in the x y plane with its center at the origin and the second loop s center is at z m. If the two loops have the same radius a 3 m, determine the magnetic field at: (a) z 0, (b) z m, (c) z m. Solution: The magnetic field due to a circular loop is given by (5.34) for a loop in the x y plane carrying a current I in the ˆφ-direction. Considering that the bottom loop in Fig. P5.4 is in the x y plane, but the current direction is along ˆφ, Ia H ẑ a z 3 where z is the observation point along the z-axis. For the second loop, which is at a height of m, we can use the same expression but z should be replaced with z. Hence, Ia H ẑ a z 3

254 CHAPTER 5 53 z z = m a I 0 a y I x Figure P5.4: Parallel circular loops of Problem 5.4. The total field is H H H ẑ Ia a z 3 (a) At z 0, and with a 3 m and I 40 A, H ẑ (b) At z m (midway between the loops): H ẑ a z ẑ0 5 A/m (c) At z m, H should be the same as at z 0. Thus, H ẑ0 5 A/m A/m 9 3 ẑ 38 A/m Section 5-3: Forces between Currents Problem 5.5 The long, straight conductor shown in Fig (P5.5) lies in the plane of the rectangular loop at a distance d 0 m. The loop has dimensions

255 54 CHAPTER 5 I I b = 0.5m d = 0.m a = 0.m Figure P5.5: Current loop next to a conducting wire (Problem 5.5). a 0 m and b 0 5 m, and the currents are I 0 A and I 30 A. Determine the net magnetic force acting on the loop. Solution: The net magnetic force on the loop is due to the magnetic field surrounding the wire carrying current I. The magnetic forces on the loop as a whole due to the current in the loop itself are canceled out by symmetry. Consider the wire carrying I to coincide with the z-axis, and the loop to lie in the x side of the x-z plane. Assuming the wire and the loop are surrounded by free space or other nonmagnetic material, Eq. (5.30) gives B ˆφ µ 0I πr In the plane of the loop, this magnetic field is B ŷ µ 0I πx Then, from Eq. (5.), the force on the side of the loop nearest the wire is F m I B I ẑb ŷ µ 0I πx x d The force on the side of the loop farthest from the wire is F m I B I ẑb ŷ µ 0I πx x a d ˆx µ 0I I b πd ˆx µ 0I I b π a d

256 CHAPTER 5 55 The other two sides do not contribute any net forces to the loop because they are equal in magnitude and opposite in direction. Therefore, the total force on the loop is F F m F m ˆx µ 0I I b πd ˆx µ 0I I ab ˆx µ 0I I b π a d πd a d ˆx 4π π ˆx0 4 (mn) The force is pulling the loop toward the wire. Problem 5.6 In the arrangement shown in Fig. 5-4 (P5.6), each of the two long, parallel conductors carries a current I, is supported by 8-cm-long strings, and has a mass per unit length of. g/cm. Due to the repulsive force acting on the conductors, the angle θ between the supporting strings is 0. Determine the magnitude of I and the relative directions of the currents in the two conductors. z z x 8 cm θ = 0 F' θ F' v θ d x F' h y (a) (b) (c) Figure P5.6: Parallel conductors supported by strings (Problem 5.6). Solution: While the vertical component of the tension in the strings is counteracting the force of gravity on the wires, the horizontal component of the tension in the strings is counteracting the magnetic force, which is pushing the wires apart. According to Section 5-3, the magnetic force is repulsive when the currents are in opposite directions. Figure P5.6(b) shows forces on wire of part (a). The quantity F is the tension force per unit length of wire due to the mass per unit length m g/cm 0

257 56 CHAPTER 5 kg/m. The vertical component of F balances out the gravitational force, where g 9 8 (m/s ). But Hence, F v m g (9) F v F cos θ (0) F m g cos θ () The horizontal component of F must be equal to the repulsion magnitude force given by Eq. (5.4): where d is the spacing between the wires and in Fig. P5.6(c). From Fig. 5.6(b), F h µ 0 I πd µ 0 I π sin θ () is the length of the string, as shown F h F sin θ m g cos θ sin θ m gtan θ (3) Equating Eqs. () and (3) and then solving for I, we have I sin θ 4π m g µ 0 cos θ sin5 4π π 0 7 cos (A) Problem 5.7 An infinitely long, thin conducting sheet of width w along the x-direction lies in the x y plane and carries a current I in y-direction. Determine the (a) the magnetic field at a point P midway between the edges of the sheet and at a height h above it (Fig. 5-4 (P5.7)), and then (b) determine the force per unit length exerted on an infinitely long wire passing through point P and parallel to the sheet if the current through the wire is equal in magnitude but opposite in direction to that carried by the sheet. Solution: (a) The sheet can be considered to consist of a large number of infinitely long but narrow wires each dx wide lying next to each other, with each carrying a current

258 CHAPTER 5 57 h I P I Figure P5.7: A linear current source above a current sheet (Problem 5.7). w I x I dx w. If we choose the coordinate system shown in Fig. P5.7, the wire at a distance x from the origin is at a distance vector R from point P, with R ˆxx ẑh Equation (5.30) provides an expression for the magnetic field due to an infinitely long wire carrying a current I as H B µ 0 ˆφ I πr We now need to adapt this expression to the present situation by replacing I with I x I dx w, replacing r with R x h, and by assigning the proper direction for the magnetic field. From the Biot Savart law, the direction of H is governed by l R, where l is the direction of current flow. In the present case, l is out of the page, which is the ŷ direction. Hence, the direction of the field is l R l ŷ ˆxx ẑh R ŷ ˆxx ˆxh ẑh ẑx x h Therefore, the field dh due to current I x is and the total field is H 0 0 h I πw ˆxh dh ẑx I x x h ˆxh ẑx I dx πr πw x h w ˆxh x w I dx ẑx πw x h w ˆxh dx ẑx x h I πw x w ˆxh w x w dx x h ẑ w x w x dx x h

259 58 CHAPTER 5 I πw ˆxh tan x h h ˆx I tan w πw h (A/m) w x w ẑ ln x h w x w At P in Fig. P5.7, the field is pointing to the left. The z-component could have been assumed zero with a symmetry argument. An alternative solution is to employ Eq. (5.4a) directly. (b) From Eq. (5.9), a differential force is of the form df m I dl B or, assuming dl â d, the force per unit length is given by F m F m Iâ B Iŷ ˆx µ 0I πw tan w h ẑ µ 0I πw tan w h (N) The force is repulsive; the wire is experiencing a force pushing it up. Problem 5.8 Three long, parallel wires are arranged as shown in Fig (P5.8(a)). Determine the force per unit length acting on the wire carrying I 3. I = 0A m m I 3 = 0A m I = 0A Figure P5.8: (a) Three parallel wires of Problem 5.8. Solution: Since I and I are equal in magnitude and opposite in direction, and equidistant from I 3, our intuitive answer might be that the net force on I 3 is zero. As

260 CHAPTER 5 59 z I x into the page (y) m R B R x I 3 x m B R = R R I out of the page (-y) Figure P5.8: (b) B fields due to I and I at location of I 3. I x F' 3 F 3 ' θ x I 3 x I Figure P5.8: (c) Forces acting on I 3. we will see, that s not the correct answer. The field due to I (which is along ŷ) at location of I 3 is B ˆb µ 0 I πr where ˆb is the unit vector in the direction of B shown in the figure, which is

261 60 CHAPTER 5 perpendicular to ˆR. The force per unit length exerted on I 3 is µ F 3 0 I I 3 µ πr ŷ ˆb 0 I I 3 ˆR πr Similarly, the force per unit length excited on I 3 by the field due to I (which is along ŷ) is F 3 ˆR µ 0 I I 3 πr The two forces have opposite components along ˆx and equal components along ẑ. Hence, with R R 8 m and θ sin 8 sin 45, F3 F3 F3 ẑ ẑ µ 0 I I 3 πr µ 0 I I 3 πr sinθ 4π π 8 ẑ 0 5 N/m Problem 5.9 A square loop placed as shown in Fig (P5.9) has -m sides and carries a current I 5 A. If a straight, long conductor carrying a current I 0 A is introduced and placed just above the midpoints of two of the loop s sides, determine the net force acting on the loop. z I 4 a y a 3 I x Figure P5.9: Long wire carrying current I, just above a square loop carrying I (Problem 5.9). Solution: Since I is just barely above the loop, we can treat it as if it s in the same plane as the loop. For side, I and I are in the same direction, hence the force on

262 CHAPTER 5 6 side is attractive. That is, F ŷ µ 0I I a π a 4π ŷ π ŷ 0 5 N I and I are in opposite directions for side 3. Hence, the force on side 3 is repulsive, which means it is also along ŷ. That is, F 3 F. The net forces on sides and 4 are zero. Total net force on the loop is F F ŷ4 0 5 N Section 5-4: Gauss s Law for Magnetism and Ampère s Law Problem 5.0 Current I flows along the positive z-direction in the inner conductor of a long coaxial cable and returns through the outer conductor. The inner conductor has radius a, and the inner and outer radii of the outer conductor are b and c, respectively. (a) Determine the magnetic field in each of the following regions: 0 r a, a r b, b r c, and r c. r 0 cm, given that I 0 A, a cm, b 4 cm, and c 5 cm. (b) Plot the magnitude of H as a function of r over the range from r 0 to Solution: (a) Following the solution to Example 5-5, the magnetic field in the region r a, H ˆφ ri πa and in the region a r b, H ˆφ I πr The total area of the outer conductor is A π c b and the fraction of the area of the outer conductor enclosed by a circular contour centered at r 0 in the region b r c is π r b π c b r b c b The total current enclosed by a contour of radius r is therefore I enclosed I r b c r c b I c b

263 6 CHAPTER 5 and the resulting magnetic field is H ˆφ I enclosed πr ˆφ I πr c r c b For r c, the total enclosed current is zero: the total current flowing on the inner conductor is equal to the total current flowing on the outer conductor, but they are flowing in opposite directions. Therefore, H 0. (b) See Fig. P Magnetic field magnitude H (A/cm) Radial distance r (cm) Figure P5.0: Problem 5.0(b). Problem 5. A long cylindrical conductor whose axis is coincident with the z-axis has a radius a and carries a current characterized by a current density J ẑj 0 r, where J 0 is a constant and r is the radial distance from the cylinder s axis. Obtain an expression for the magnetic field H for (a) 0 r a and (b) r a. Solution: This problem is very similar to Example 5-5. (a) For 0 r a, the total current flowing within the contour C is π r I J ds r φ 0 r 0 ẑj 0 r ẑr dr dφ π r 0 J 0 dr πr J 0

264 CHAPTER 5 63 Therefore, since I πr H, H J 0 within the wire and H ˆφJ 0. (b) For r a, the total current flowing within the contour is the total current flowing within the wire: π a I J ds a J 0 dr πaj 0 φ 0 r 0 ẑj 0 r ẑr dr dφ π Therefore, since I πrh, H J 0 a r within the wire and H ˆφJ 0 a r. Problem 5. Repeat Problem 5. for a current density J ẑj 0 e r. z r 0 S r J a Figure P5.: Cylindrical current. Solution:

265 64 CHAPTER 5 (a) For r Hence, (b) For r a, a, Ampère s law is H dl I J ds c S r r ˆφH ˆφπr J ds ẑj 0 e r ẑπr dr 0 0 r πrh πj 0 re r dr 0 r πj 0 e r r r 0 πj 0 e r H ˆφH ˆφ J 0 r e r r for r a r πrh πj 0 e r a a 0 πj 0 e a H ˆφH ˆφ J 0 a r e a r a Problem 5.3 In a certain conducting region, the magnetic field is given in cylindrical coordinates by Find the current density J. Solution: H ˆφ 4 r 3r e 3r J H ẑ 4 r 3r 3r e r r r ẑ e r r r r e ẑ4e 3r A/m Section 5-5: Magnetic Potential Problem 5.4 With reference to Fig. 5-0, (a) derive an expression for the vector magnetic potential A at a point P located at a distance r from the wire in the x y plane, and then (b) derive B from A. Show that your result is identical with the expression given by Eq. (5.9), which was derived by applying the Biot Savart law.

266 CHAPTER 5 65 Solution: (a) From the text immediately following Eq. (5.65), that equation may take the form A µ 4π (b) From Eq. (5.53), B A I µ dl 0 R 4π z µ 0 4π ẑi ln z z I z r ẑ dz r 4π ln 4r 4r ẑ µ 0I 4π ln ẑ µ 0I ẑ µ 0I 4π ln 4r 4r ˆφ µ 0I 4π r ln 4r 4r ˆφ µ 0I 4π 4r 4r ˆφ µ 0I 4r r 4r 4r 4π 4r 4r r 4r 4r z r r r 4r 4r ˆφ µ 0I 4π 4r 4r 4r 4r 4r 4r ˆφ µ 0I 4r µ 0 I ˆφ (T) 4π 4r 4r 4r which is the same as Eq. (5.9). πr Problem 5.5 In a given region of space, the vector magnetic potential is given by A ˆx5cos πy ẑ sinπx (Wb/m).

267 66 CHAPTER 5 (a) Determine B. (b) Use Eq. (5.66) to calculate the magnetic flux passing through a square loop with 0.5-m-long edges if the loop is in the x y plane, its center is at the origin, and its edges are parallel to the x- and y-axes. (c) Calculate Φ again using Eq. (5.67). Solution: (a) From Eq. (5.53), B A ẑ5πsin πy ŷπcosπx. (b) From Eq. (5.66), Φ B ds m y 0 5 m cos πy 5πx 0 5 m x 0 5 m π 0 5 cos π 8 cos x 0 5 ẑ5πsinπy ŷπcosπx ẑ dx dy π 0 5 y (c) From Eq. (5.67), Φ A d, where C is the square loop in the x-y plane with C sides of length 0.5 m centered at the origin. Thus, the integral can be written as Φ C A d S front S back S left S right where S front, S back, S left, and S right are the sides of the loop. 0 5 S front S back S left x ˆx5cos πy ẑ sinπx y 0 5 5cos πy y 0 5 dx x xcos πy y x x 0 5 x cos ˆx dx ˆx5cos πy ẑ sinπx y 0 5 ˆx dx 5cos πy y 0 5 dx 5xcos πy y y 0 5 π cos π 8 x cos π 4 8 ˆx5cos πy ẑ sinπx x 0 5 ŷ dy

268 CHAPTER 5 67 Thus, Φ S right Problem y y y x 0 5 dy 0 ˆx5cos πy ẑ sinπx x 0 5 ŷ dy 0 x 0 5 dy 0 5 A d S front S c back S left S right cos π 4 8 A uniform current density given by J ẑj 0 (A/m ) gives rise to a vector magnetic potential A ẑ µ 0J 0 x y 4 (Wb/m) 5 4 cos π (a) Apply the vector Poisson s equation to confirm the above statement. (b) Use the expression for A to find H. (c) Use the expression for J in conjunction with Ampère s law to find H. Compare your result with that obtained in part (b). Solution: (a) A ˆx A x ŷ A y ẑ A z ẑ x y J 0 µ 0 4 x y J 0 ẑµ 0 4 ẑµ 0 J 0 Hence, A µ 0 J is verified. (b) H µ 0 A ˆx µ 0 µ 0 µ 0 A z y ˆx A z ŷ y A z x ˆx y ˆx J 0y ŷ J 0x A y z ŷ A x z µ 0 J 0 4 x y ŷ x (A/m) z A z x ẑ A y x A x y µ 0 J 0 4 x y

269 68 CHAPTER 5 z r J 0 Figure P5.6: Current cylinder of Problem 5.6. (c) H dl I J ds C S ˆφH φ ˆφπr J 0 πr H r ˆφH φ ˆφJ 0 We need to convert the expression from cylindrical to Cartesian coordinates. From Table 3-, ˆφ ˆxsinφ y x ŷcosφ ˆx ŷ x y x y r x y

270 CHAPTER 5 69 Hence y H ˆx ŷ x y x x y which is identical with the result of part (b). J 0 x y ˆx yj 0 ŷ xj 0 Problem 5.7 A thin current element extending between z L and z L carries a current I along ẑ through a circular cross section of radius a. (a) Find A at a point P located very far from the origin (assume R is so much larger than L that point P may be considered to be at approximately the same distance from every point along the current element). (b) Determine the corresponding H. z L/ P θ R I -L/ Cross-section πa Figure P5.7: Current element of length L observed at distance R L. Solution: (a) Since R L, we can assume that P is approximately equidistant from all segments of the current element. Hence, with R treated as constant, (5.65) gives µ A 0 J µ dv 0 I ẑ 4π V R 4πR V πa πa dz ẑ µ 0I dz ẑ µ 0IL 4πR 4πR L L

271 70 CHAPTER 5 (b) H µ 0 A µ 0 ˆx µ 0 y IL 4π ˆx A z y ŷ A z x µ 0 IL 4π ˆxy ŷx x y z 3 x y z ŷ x µ 0 IL 4π x y z Section 5-6: Magnetic Properties of Materials Problem 5.8 In the model of the hydrogen atom proposed by Bohr in 93, the electron moves around the nucleus at a speed of 0 6 m/s in a circular orbit of radius 5 0 m. What is the magnitude of the magnetic moment generated by the electron s motion? Solution: From Eq. (5.69), the magnitude of the orbital magnetic moment of an electron is m 0 eur (A m ) Problem 5.9 Iron contains atoms/m 3. At saturation, the alignment of the electrons spin magnetic moments in iron can contribute.5 T to the total magnetic flux density B. If the spin magnetic moment of a single electron is (A m ), how many electrons per atom contribute to the saturated field? Solution: From the first paragraph of Section 5-6., the magnetic flux density of a magnetized material is B m µ 0 M, where M is the vector sum of the microscopic magnetic dipoles within the material: M N e m s, where m s is the magnitude of the spin magnetic moment of an electron in the direction of the mean magnetization, and N e is net number of electrons per unit volume contributing to the bulk magnetization. If the number of electrons per atom contributing to the bulk magnetization is n e, then N e n e N atoms where N atoms atoms/m 3 is the number density of atoms for iron. Therefore, N n e e N atoms M m s N atoms B µ 0 m s N atoms 5 4π (electrons/atom)

272 CHAPTER 5 7 Section 5-7: Magnetic Boundary Conditions Problem 5.30 The x y plane separates two magnetic media with magnetic permeabilities µ and µ, as shown in Fig (P5.30). If there is no surface current at the interface and the magnetic field in medium is H ˆxH x ŷh y ẑh z find: (a) H (b) θ and θ, and (c) evaluate H, θ, and θ for H x (A/m), H y 0, H z 4 (A/m), µ µ 0, and µ 4µ 0. z H θ µ µ x-y plane Figure P5.30: Adjacent magnetic media (Problem 5.30). Solution: (a) From (5.80), µ H n µ H n and in the absence of surface currents at the interface, (5.85) states H t H t In this case, H z H n, and H x and H y are tangential fields. Hence, µ H z µ H z H x H x H y H y

273 7 CHAPTER 5 and (b) (c) H ˆxH x ŷh y ẑ µ µ H z H t H x H y tanθ H t H z tanθ H t H z H x H y H z H x H y µ µ H z tanθ µ µ H ˆx ẑ 4 4 ˆx ẑ θ tan θ tan (A/m) Problem 5.3 Given that a current sheet with surface current density J s ˆx8 (A/m). exists at y 0, the interface between two magnetic media, and H ẑ (A/m) in medium y 0, determine H in medium y 0 Solution: J s ˆx8 A/m H ẑ A/m H is tangential to the boundary, and therefore H is also. With ˆn ŷ, from Eq. (5.84), we have ˆn H H J s ŷ ẑ H ˆx8 ˆx ŷ H ˆx8 or ŷ H ˆx3

274 CHAPTER 5 73 y n H J s x H Figure P5.3: Adjacent magnetic media with J s on boundary. which implies that H does not have an x-component. Also, since µ H y µ H y and H does not have a y-component, it follows that H does not have a y-component either. Consequently, we conclude that H ẑ3 Problem 5.3 In Fig (P5.3), the plane defined by x y separates medium of permeability µ from medium of permeability µ. If no surface current exists on the boundary and B ˆx ŷ3 (T) find B and then evaluate your result for µ 5µ. Hint: Start out by deriving the equation for the unit vector normal to the given plane. Solution: We need to find ˆn. To do so, we start by finding any two vectors in the plane x y, and to do that, we need three non-collinear points in that plane. We choose 0 0, 0 0, and 0. Vector A is from 0 0 to 0 0 : A ˆx ŷ Vector A is from 0 0 to 0 : A ẑ Hence, if we take the cross product A A, we end up in a direction normal to the given plane, from medium to medium, ˆn A A A A ẑ ˆx ŷ A A ŷ ˆx ŷ ˆx

275 74 CHAPTER 5 y n Medium n (, 0) x µ (0, -) Medium µ Figure P5.3: Magnetic media separated by the plane x y (Problem 5.3). In medium, normal component is ŷ B n ˆn B ŷ B n ˆn B n Tangential component is Boundary conditions: Hence, Finally, For µ 5µ, ˆx ˆx ŷ3 3 ˆx B t B B n ˆx ŷ3 B n B n or H t H t or ŷ µ B t µ B t µ ŷ ˆx B B n B t B ŷ ŷ ˆx ŷ ˆx B n B t B µ t µ µ ˆx 5 ŷ 5 (T) ˆx ˆx 5 ŷ 5 µ µ ˆx 5 ŷ 5

276 CHAPTER 5 75 Problem 5.33 The plane boundary defined by z 0 separates air from a block of iron. If B ˆx4 ŷ6 ẑ8 in air (z 0), find B in iron (z 0), given that µ 5000µ 0 for iron. Solution: From Eq. (5.), H B µ µ ˆx4 ŷ6 ẑ8 The z component is the normal component to the boundary at z 0. Therefore, from Eq. (5.79), B z B z 8 while, from Eq. (5.85), or H x H x µ 4 H y H y B x µ H x µ µ 4 B y µ H y where µ µ µ r Therefore, B ˆx0000 ŷ30000 ẑ8 µ 6 µ µ 6 Problem 5.34 Show that if no surface current densities exist at the parallel interfaces shown in Fig (P5.34), the relationship between θ 4 and θ is independent of µ. B 3 θ 4 µ 3 θ B θ 3 µ B θ µ Figure P5.34: Three magnetic media with parallel interfaces (Problem 5.34).

277 76 CHAPTER 5 Solution: and But B n B n and B t µ B t µ. Hence, We note that θ θ 3 and tanθ B t B n tanθ B t B n tanθ B t B n µ µ µ µ tanθ tanθ 4 µ 3 µ tanθ 3 µ 3 µ tan θ µ 3 µ µ µ tanθ µ 3 µ tanθ which is independent of µ. Sections 5-8 and 5-9: Inductance and Magnetic Energy Problem 5.35 Obtain an expression for the self-inductance per unit length for the parallel wire transmission line of Fig. 5-7(a) in terms of a, d, and µ, where a is the radius of the wires, d is the axis-to-axis distance between the wires, and µ is the permeability of the medium in which they reside. Solution: Let us place the two wires in the x z plane and orient the current in one of them to be along the z-direction and the current in the other one to be along the z-direction, as shown in Fig. P5.35. From Eq. (5.30), the magnetic field at point P x 0 z due to wire is B ˆφ µi πr ŷ µi πx where the permeability has been generalized from free space to any ˆφ substance with permeability µ, and it has been recognized that in the x-z plane, ŷ and r x as long as x 0. Given that the current in wire is opposite that in wire, the magnetic field created by wire at point P x 0 z is in the same direction as that created by wire, and it is given by µi B ŷ π d x

278 CHAPTER 5 77 z I x P d-x x a d I Figure P5.35: Parallel wire transmission line. Therefore, the total magnetic field in the region between the wires is B B B ŷ µi µid π x d x ŷ πx d x From Eq. (5.9), the flux crossing the surface area between the wires over a length l of the wire structure is z Φ B ds 0 l d a µid ŷ S z z 0 x a πx d x ŷ dx dz a µild π d ln x d x d x a µil a a ln d ln π a d a µil d π ln a µil d a π ln a a

279 78 CHAPTER 5 Since the number of turns in this structure is, Eq. (5.93) states that the flux linkage is the same as magnetic flux: Λ Φ. Then Eq. (5.94) gives a total inductance over the length l as L Λ I Φ I µl π ln d a a (H) Therefore, the inductance per unit length is L L µ d l π ln a a µ π ln d a (H/m) where the last approximation recognizes that the wires are thin compared to the separation distance (i.e., that d a). This has been an implied condition from the beginning of this analysis, where the flux passing through the wires themselves have been ignored. This is the thin-wire limit in Table - for the two wire line. Problem 5.36 A solenoid with a length of 0 cm and a radius of 5 cm consists of 400 turns and carries a current of A. If z 0 represents the midpoint of the solenoid, generate a plot for H z as a function of z along the axis of the solenoid for the range 0 cm z 0 cm in -cm steps. Solution: Let the length of the solenoid be l 0 cm. From Eq. (5.88a) and Eq. (5.88b), z atan θ and a t a sec θ, which implies that z z a sinθ. Generalizing this to an arbitrary observation point z on the axis of the solenoid, z z z z a sinθ. Using this in Eq. (5.89), B H 0 0 z µ ẑni sin θ sinθ l ẑ ni z l z l z a l z a l ẑ ni z l l z a z l (A/m) z a A plot of the magnitude of this function of z with a 5 cm, n 400 turns 0 cm turns/m, and I A appears in Fig. P5.36.

280 CHAPTER 5 79 Magnitude of Magnetic Field H (ka/m) Position on axis of solenoid z (cm) Figure P5.36: Problem Problem 5.37 In terms of the d-c current I, how much magnetic energy is stored in the insulating medium of a 3-m-long, air-filled section of a coaxial transmission line, given that the radius of the inner conductor is 5 cm and the inner radius of the outer conductor is 0 cm? Solution: From Eq. (5.99), the inductance per unit length of an air-filled coaxial cable is given by (H/m) µ L 0 π ln b a Over a length of m, the inductance is L L 3 4π 0 7 π ln From Eq. (5.04), W m LI 08I (nj), where W m is in nanojoules when I is in amperes. Alternatively, we can use Eq. (5.06) to compute W m : W m µ 0 H V dv (H)

281 80 CHAPTER 5 From Eq. (5.97), H B µ 0 I πr, and W m 3m z 0 π φ 0 b r a I µ 0 πr r dr dφ dz 08I (nj) Problem 5.38 The rectangular loop shown in Fig (P5.38) is coplanar with the long, straight wire carrying the current I 0 A. Determine the magnetic flux through the loop. z 0A 30cm 5cm 0cm y x Figure P5.38: Loop and wire arrangement for Problem Solution: The field due to the long wire is, from Eq. (5.30), B ˆφ µ 0I ˆx πr µ 0I ˆx πr µ 0I πy where in the plane of the loop, ˆφ becomes ˆx and r becomes y. The flux through the loop is along ˆx, and the magnitude of the flux is µ Φ B ds 0 I S π 0 cm 5 cm ˆx y ˆx 30 cm dy dy y µ 0 I 0 π µ π ln (Wb)

282 CHAPTER 5 8 Problem 5.39 A circular loop of radius a carrying current I is located in the x y plane as shown in the figure. In addition, an infinitely long wire carrying current I in a direction parallel with the z-axis is located at y y 0. z n^ parallel to z^ P(0, 0, h) I a y o y x I (a) Determine H at P 0 0 h. (b) Evaluate H for a 3 cm, y 0 0 cm, h 4 cm, I 0 A, and I 0 A. Solution: (a) The magnetic field at P 0 0 h is composed of H due to the loop and H due to the wire: H H H From (5.34), with z h, H ẑ I a a h 3 (A/m) From (5.30), the field due to the wire at a distance r y 0 is H ˆφ I πy 0 where ˆφ is defined with respect to the coordinate system of the wire. Point P is located at an angel φ 90 with respect to the wire coordinates. From Table 3-, ˆφ ˆxsin φ ŷcos φ ˆx at φ 90 Hence, I H a ẑ a h ˆx 3 I πy 0 (A/m)

283 8 CHAPTER 5 (b) H ẑ36 ˆx3 83 (A/m) Problem 5.40 A cylindrical conductor whose axis is coincident with the z-axis has an internal magnetic field given by H ˆφ 4r 4r e r (A/m) for r a where a is the conductor s radius. If a 5 cm, what is the total current flowing in the conductor? Solution: We can follow either of two possible approaches. The first involves the use of Ampère s law and the second one involves finding J from H and then I from J. We will demonstrate both. Approach : Ampère s law Applying Ampère s law at r a, π For a 5 cm, I 0 (A). 0 ˆφ r 4r e 4r ˆφr dφ C I 4π 4a e 4a H d r a I (A) r a I

284 CHAPTER 5 83 Approach : H J I J H ẑ φ r r rh ẑ r r 4r 4r e ẑ 4r r 8e 8 4r 4r e ẑ3e 4r a 4r I J ds ẑ3e ẑπr dr S r 0 64π re a 4r dr r 0 64π 4a 4a e 6 4π 4a 4a e (A) Problem 5.4 Determine the mutual inductance between the circular loop and the linear current shown in the figure. y y a x d I Solution: To calculate the magnetic flux through the loop due to the current in the conductor, we consider a thin strip of thickness dy at location y, as shown. The magnetic field is the same at all points across the strip because they are all equidistant

285 84 CHAPTER 5 (at r d y) from the linear conductor. The magnetic flux through the strip is Let z d y dφ B y ds ẑ L I µ 0 π dz dy. Hence, L µ 0 π µ 0 π µ 0 π where R a 0 b 0 z c 0 z and d µ 0 I π d ẑ a y y dy µ 0 I a y π d dy y dφ S a a y a a y dy d y a z d z dz z a d d a d a d z dz a z R z dz a 0 a d b 0 d c 0 4a 0 c 0 b 0 4a 0 From Gradshteyn and Ryzhik, Table of Integrals, Series, and Products (Academic Press, 980, p. 84), we have For For R d a R z dz R a 0 dz z R b 0 z dz dz R z d a d z a dz d a z d a dz z R, several solutions exist depending on the sign of a 0 and.

286 CHAPTER 5 85 For this problem, 0, also let a 0 0 (i.e., d a). Using the table of integrals, dz d a a 0 z R a a 0 a 0 sin 0 b 0 z z b0 4a 0 c 0 d a π d a sin a d dz az d z d a a z d a dz For R, different solutions exist depending on the sign of c 0 and. In this problem, 0 and c 0 0. From the table of integrals, b 0 dz b 0 z R c 0 sin c 0z d a b 0 z a d d a d sin πd Thus d z a z d a L µ 0 π πd π d a µ 0 d d a

287 86 Chapter 6: Maxwell s Equations for Time-Varying Fields Lesson #37 Chapter Section: 6-, 6- Topics: Faraday s law, stationary loop in changing magnetic field Highlights: Faraday s law EMF Special Illustrations: Example 6- Example 6- CD-ROM Demo 6. CD-ROM Modules 6. and 6.

288 87 Lesson #38 Chapter Section: 6-3, 6-4 Topics: Ideal transformer, moving conductor Highlights: Transformer voltage and current relations EMF for moving conductor Special Illustrations: CD-ROM Modules 6.3 and 6.4 CD-ROM Demo 6.

289 88 Lesson #39 Chapter Section: 6-5, 6-6 Topics: EM Generator, moving conductor in changing field Highlights: Motor and generator reciprocity EMF for combination of motional and transformer Special Illustrations: Technology Brief on EMF Sensors (CD-ROM) EMF Sensors An electromotive force (emf) sensor is a device that can generate an induced voltage in response to an external stimulus. Three types of emf sensors are profiled in this Technical Brief: the piezoelectric transducer, the Faraday magnetic flux sensor, and the thermocouple. Piezoelectric Transducers Piezoelectricity refers to the property of certain crystals, such as quartz, to become electrically polarized when the crystal is subjected to mechanical pressure, thereby exhibiting a voltage across it. The crystal consists of polar domains represented by equivalent dipoles (A). Under the absence of an external force, the polar domains are randomly oriented throughout the material (A), but when compressive or tensile (stretching) stress is applied to the crystal, the polar domains align themselves along one of the principal axes of the crystal, leading to a net polarization (electric charge) at the crystal surfaces (A and A3). Compression and stretching generate voltages of opposite polarity. The piezoelectric effect (piezein means to press or squeeze in Greek) was discovered by the Curie brothers, Pierre and Paul-Jacques, in 880, and a year later Lippmann predicted the converse property, namely that if subjected to an electric field, the crystal would change in shape. Thus, the piezoelectric effect is a reversible (bidirectional) electro-mechanical process.

290 89 Lesson #40 Chapter Section: 6-7, 6-8 Topics: Displacement current, boundary conditions Highlights: Concept of displacement current Boundary conditions for the dynamic case Special Illustrations: Example 6-7

291 90 Lesson #4 Chapter Section: 6-9, 6-0 Topics: Charge-current continuity, charge dissipation Highlights: Continuity equation Relaxation time constant Special Illustrations:

292 9 Lesson #4 Chapter Section: 6- Topics: EM potentials Highlights: Retarded potential Relation of potentials to fields in the dynamic case Special Illustrations: Example 6-8

293 9 CHAPTER 6 Chapter 6 Sections 6- to 6-6: Faraday s Law and its Applications Problem 6. The switch in the bottom loop of Fig. 6-7 (P6.) is closed at t 0 and then opened at a later time t. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? R I + - R Figure P6.: Loops of Problem 6.. Solution: The magnetic coupling will be strongest at the point where the wires of the two loops come closest. When the switch is closed the current in the bottom loop will start to flow clockwise, which is from left to right in the top portion of the bottom loop. To oppose this change, a current will momentarily flow in the bottom of the top loop from right to left. Thus the current in the top loop is momentarily clockwise when the switch is closed. Similarly, when the switch is opened, the current in the top loop is momentarily counterclockwise. Problem 6. The loop in Fig. 6-8 (P6.) is in the x y plane and B ẑb 0 sinωt with B 0 positive. What is the direction of I ( ˆφ or ˆφ) at (a) t 0, (b) ωt π 4, and (c) ωt π? Solution: I V emf R. Since the single-turn loop is not moving or changing shape with time, Vemf m 0 V and V emf tr. Therefore, from Eq. (6.8), I Vemf R tr ds R B S t If we take the surface normal to be ẑ, then the right hand rule gives positive flowing current to be in the ˆφ direction. I A R t B AB 0 sinωt 0 ω R cosωt (A)

294 CHAPTER 6 93 z R y I x Figure P6.: Loop of Problem 6.. where A is the area of the loop. (a) A, ω and R are positive quantities. At t 0, cosωt so I 0 and the current is flowing in the ˆφ direction (so as to produce an induced magnetic field that opposes B). (b) At ωt π 4, cosωt so I 0 and the current is still flowing in the ˆφ direction. (c) At ωt π, cosωt 0 so I 0. There is no current flowing in either direction. Problem 6.3 A coil consists of 00 turns of wire wrapped around a square frame of sides 0.5 m. The coil is centered at the origin with each of its sides parallel to the x- or y-axis. Find the induced emf across the open-circuited ends of the coil if the magnetic field is given by (a) B ẑ0e 3t (T), (b) B ẑ0cos x cos0 3 t (T), (c) B ẑ0cos x siny cos0 3 t (T). Solution: Since the coil is not moving or changing shape, Vemf m 0 V and V emf Vemf tr. From Eq. (6.6), V emf N d B ds N dt S d B ẑ dx dy dt where N 00 and the surface normal was chosen to be in the ẑ direction. (a) For B ẑ0e 3t (T), V emf 00 d 3t 3t 0e e (V) dt

295 94 CHAPTER 6 (b) For B ẑ0cos xcos 0 3 t (T), V emf 00 d dt 0cos 0 3 t 0 5 x 0 5 (c) For B ẑ0cos xsin ycos 0 3 t (T), V emf 00 d dt 0cos 0 3 t 0 5 y x 0 5 cosx dx dy 4 6sin 0 3 t 0 5 y 0 5 cosxsin y dx dy 0 (kv) Problem 6.4 A stationary conducting loop with internal resistance of 0.5 Ω is placed in a time-varying magnetic field. When the loop is closed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a -Ω resistor is connected across its open ends? Solution: V emf is independent of the resistance which is in the loop. Therefore, when the loop is intact and the internal resistance is only 0 5 Ω, V emf 5 A 0 5 Ω 5 V When the small gap is created, the total resistance in the loop is infinite and the current flow is zero. With a -Ω resistor in the gap, I V emf Ω 0 5 Ω 5 V 5 Ω (A) Problem 6.5 A circular-loop TV antenna with 0.0 m area is in the presence of a uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop develops an emf with a peak value of 30 (mv). What is the peak magnitude of B of the incident wave? Solution: TV loop antennas have one turn. At maximum orientation, Eq. (6.5) evaluates to Φ B ds BA for a loop of area A and a uniform magnetic field with magnitude B B. Since we know the frequency of the field is f 300 MHz, we can express B as B B 0 cos ωt α 0 with ω π rad/s and α 0 an arbitrary reference phase. From Eq. (6.6), V emf N dφ dt A d dt B 0 cos ωt α 0 AB 0 ωsin ωt α 0 V emf is maximum when sin ωt α 0. Hence, AB 0 ω 0 0 B 0 6π 0 8

296 CHAPTER 6 95 which yields B (na/m). Problem 6.6 The square loop shown in Fig. 6-9 (P6.6) is coplanar with a long, straight wire carrying a current I t 5cos π 0 4 t (A) (a) Determine the emf induced across a small gap created in the loop. (b) Determine the direction and magnitude of the current that would flow through a 4-Ω resistor connected across the gap. The loop has an internal resistance of Ω. z I(t) 5cm 0cm 0cm y x Figure P6.6: Loop coplanar with long wire (Problem 6.6). Solution: (a) The magnetic field due to the wire is B ˆφ µ 0I πr ˆx µ 0I πy where in the plane of the loop, ˆφ ˆx and r y. The flux passing through the loop

297 96 CHAPTER 6 is (b) Φ S B ds V emf 5 cm 5 cm µ 0 I 0 π ˆx µ 0I πy ln 5 5 ˆx0 (cm) dy 4π 0 7 5cos π 0 4 t 0 π 0 7 cos π 0 4 t (Wb) dφ dt π 0 4 sin π 0 4 t sin π 0 4 t (V) V I ind emf sin π 0 t 4 38sin π 0 t (ma) 4 5 At t 0, B is a maximum, it points in ˆx-direction, and since it varies as cos π 0 4 t, it is decreasing. Hence, the induced current has to be CCW when looking down on the loop, as shown in the figure. Problem 6.7 The rectangular conducting loop shown in Fig. 6-0 (P6.7) rotates at 6,000 revolutions per minute in a uniform magnetic flux density given by B ŷ50 (mt) Determine the current induced in the loop if its internal resistance is 0 5 Ω. Solution: Φ S B ds ŷ ŷ cos φ t cosφ t (rad/s) φ t ωt π t 00πt 60 Φ cos 00πt (Wb) dφ 5 V emf πsin 00πt sin 00πt (V) dt V I ind emf πt (ma) 5 7sin

298 CHAPTER 6 97 z cm 3cm φ(t) ω B B y x Figure P6.7: Rotating loop in a magnetic field (Problem 6.7). The direction of the current is CW (if looking at it along ˆx-direction) when the loop is in the first quadrant (0 φ π ). The current reverses direction in the second quadrant, and reverses again every quadrant. Problem 6.8 A rectangular conducting loop 5 cm 0 cm with a small air gap in one of its sides is spinning at 700 revolutions per minute. If the field B is normal to the loop axis and its magnitude is T, what is the peak voltage induced across the air gap? Solution: ω π rad/cycle 700 cycles/min 60 s/min 40π rad/s A 5 cm 0 cm 00 cm/m m From Eqs. (6.36) or (6.38), V emf AωB 0 sin ωt; it can be seen that the peak voltage is V peak emf AωB π µv Problem 6.9 A 50-cm-long metal rod rotates about the z-axis at 90 revolutions per minute, with end fixed at the origin as shown in Fig. 6- (P6.9). Determine the induced emf V if B ẑ 0 4 T. Solution: Since B is constant, V emf Vemf m. The velocity u for any point on the bar is given by u ˆφrω, where ω π rad/cycle 90 cycles/min 60 s/min 3π rad/s

299 98 CHAPTER 6 z B y x ω Figure P6.9: Rotating rod of Problem 6.9. From Eq. (6.4), V V m emf u B dl 0 r π 0 ˆφ3πr ẑ 0 4 ˆr dr 0 3π 0 4 r r 0 5 r dr 3π µv Problem 6.0 The loop shown in Fig. 6- (P6.0) moves away from a wire carrying a current I 0 (A) at a constant velocity u ŷ7 5 (m/s). If R 0 Ω and the direction of I is as defined in the figure, find I as a function of y 0, the distance between the wire and the loop. Ignore the internal resistance of the loop. Solution: Assume that the wire carrying current I is in the same plane as the loop. The two identical resistors are in series, so I V emf R, where the induced voltage is due to motion of the loop and is given by Eq. (6.6): V emf V u B m dl emf The magnetic field B is created by the wire carrying I. Choosing ẑ to coincide with the direction of I, Eq. (5.30) gives the external magnetic field of a long wire to be B ˆφ µ 0I πr C

300 CHAPTER 6 99 z 0 cm R I = 0 A 0 cm I u u R y 0 Figure P6.0: Moving loop of Problem 6.0. For positive values of y 0 in the y-z plane, ŷ ˆr, so u B ŷ u B ˆr u ˆφ µ 0I πr ẑµ 0I u πr Integrating around the four sides of the loop with dl ẑ dz and the limits of integration chosen in accordance with the assumed direction of I, and recognizing that only the two sides without the resistors contribute to Vemf m, we have Vemf m 0 ẑ µ 0I u 0 ẑ 0 πr r y 0 dz ẑ µ 0I u ẑ 0 πr r y 0 dz 0 4π π y 0 y y 0 y 0 0 (V) and therefore I V m emf R 50 y 0 y 0 0 (na) Problem 6. The conducting cylinder shown in Fig. 6-3 (P6.) rotates about its axis at,00 revolutions per minute in a radial field given by B ˆr6 (T)

301 300 CHAPTER 6 z 5cm ω 0cm + - V Sliding contact Figure P6.: Rotating cylinder in a magnetic field (Problem 6.). The cylinder, whose radius is 5 cm and height 0 cm, has sliding contacts at its top and bottom connected to a voltmeter. Determine the induced voltage. Solution: The surface of the cylinder has velocity u given by u ˆφωr ˆφπ ˆφπ (m/s) 60 L 0 u B dl ˆφπ ˆr6 ẑ dz 3 77 V 0 0 (V) Problem 6. The electromagnetic generator shown in Fig. 6- is connected to an electric bulb with a resistance of 50 Ω. If the loop area is 0. m and it rotates at 3,600 revolutions per minute in a uniform magnetic flux density B T, determine the amplitude of the current generated in the light bulb. Solution: From Eq. (6.38), the sinusoidal voltage generated by the a-c generator is V emf AωB 0 sin ωt C 0 V 0 sin ωt C 0. Hence, V 0 AωB 0 0 π (V) 60 V I (A) R 50 Problem 6.3 The circular disk shown in Fig. 6-4 (P6.3) lies in the x y plane and rotates with uniform angular velocity ω about the z-axis. The disk is of radius a and is present in a uniform magnetic flux density B ẑb 0. Obtain an expression for the emf induced at the rim relative to the center of the disk.

302 CHAPTER 6 30 z - V + ω a y x Figure P6.3: Rotating circular disk in a magnetic field (Problem 6.3). φ r u y x Figure P6.3: (a) Velocity vector u. Solution: At a radial distance r, the velocity is u ˆφωr where φ is the angle in the x y plane shown in the figure. The induced voltage is a a V u B dl ˆφωr ẑb 0 ˆr dr 0 0 ˆφ ẑ is along ˆr. Hence, V ωb 0 0 a r dr ωb 0 a Section 6-7: Displacement Current Problem 6.4 The plates of a parallel-plate capacitor have areas 0 cm each and are separated by cm. The capacitor is filled with a dielectric material with

303 30 CHAPTER 6 ε 4ε 0, and the voltage across it is given by V t 30cos π 0 6 t (V). Find the displacement current. Solution: Since the voltage is of the form given by Eq. (6.46) with V 0 30 V and ω π 0 6 rad/s, the displacement current is given by Eq. (6.49): εa I d d V 0ωsin ωt π 0 6 sin π 0 t sin π 0 t (ma) 6 Problem 6.5 A coaxial capacitor of length l 6 cm uses an insulating dielectric material with ε r 9. The radii of the cylindrical conductors are 0.5 cm and cm. If the voltage applied across the capacitor is V t 50sin 0πt (V) what is the displacement current? l + V(t) a b - r I d Figure P6.5: Solution: To find the displacement current, we need to know E in the dielectric space between the cylindrical conductors. From Eqs. (4.4) and (4.5), Q E ˆr πεrl Hence, E ˆr V b r ln V Q πεl ln b a 0πt ˆr a 50sin ˆr r ln 7 r sin 0πt (V/m)

304 CHAPTER D εe ε r ε 0 E ˆr r ˆr r sin 0πt sin 0πt (C/m ) The displacement current flows between the conductors through an imaginary cylindrical surface of length l and radius r. The current flowing from the outer conductor to the inner conductor along ˆr crosses surface S where S ˆrπrl Hence, I d D t S ˆr sin 0πt t r π πl cos 0πt 0 8cos 0πt (µa) ˆrπrl Alternatively, since the coaxial capacitor is lossless, its displacement current has to be equal to the conduction current flowing through the wires connected to the voltage sources. The capacitance of a coaxial capacitor is given by (4.6) as The current is I C dv πεl dt b ln a C πεl ln b a which is the same answer we obtained before. 0π 50cos 0πt 0 8cos 0πt (µa) Problem 6.6 The parallel-plate capacitor shown in Fig. 6-5 (P6.6) is filled with a lossy dielectric material of relative permittivity ε r and conductivity σ. The separation between the plates is d and each plate is of area A. The capacitor is connected to a time-varying voltage source V t. (a) Obtain an expression for I c, the conduction current flowing between the plates inside the capacitor, in terms of the given quantities.

305 304 CHAPTER 6 I A + V(t) - ε, σ d Figure P6.6: Parallel-plate capacitor containing a lossy dielectric material (Problem 6.6). (b) Obtain an expression for I d, the displacement current flowing inside the capacitor. (c) Based on your expression for parts (a) and (b), give an equivalent-circuit representation for the capacitor. (d) Evaluate the values of the circuit elements for A 4 cm, d 0 5 cm, ε r 4, σ 5 (S/m), and V t 0cos 3π 0 t 3 (V). Solution: (a) R d σa I V c V σa R d (b) V E d I D d A εa E εa V t t d t (c) The conduction current is directly proportional to V, as characteristic of a resistor, whereas the displacement current varies as V t, which is characteristic of a capacitor. Hence, R d σa and C εa d (d) R Ω

306 CHAPTER V(t) + - ε, σ R I c I I d C V(t) Actual Circuit Equivalent Circuit Figure P6.6: (a) Equivalent circuit. C F Problem 6.7 An electromagnetic wave propagating in seawater has an electric field with a time variation given by E ẑe 0 cosωt. If the permittivity of water is 8ε 0 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction current density to displacement current density at each of the following frequencies: (a) khz, (b) MHz, (c) GHz, (d) 00 GHz. Solution: From Eq. (6.44), the displacement current density is given by J d t D ε t E and, from Eq. (4.67), the conduction current is J σe. Converting to phasors and taking the ratio of the magnitudes, J J d σe jωε r ε 0 E (a) At f khz, ω π 0 3 rad/s, and σ ωε r ε 0 J 4 J d π The displacement current is negligible. (b) At f MHz, ω π 0 6 rad/s, and J 4 J d π

307 306 CHAPTER 6 The displacement current is practically negligible. (c) At f GHz, ω π 0 9 rad/s, and J 4 J d π Neither the displacement current nor the conduction current are negligible. (d) At f 00 GHz, ω π 0 rad/s, and J 4 J d π The conduction current is practically negligible. Sections 6-9 and 6-0: Continuity Equation and Charge Dissipation Problem 6.8 At t 0, charge density ρ v0 was introduced into the interior of a material with a relative permittivity ε r 9. If at t µs the charge density has dissipated down to 0 3 ρ v0, what is the conductivity of the material? Solution: We start by using Eq. (6.6) to find τ r : ρ v t ρ v0 e t τ r or 0 3 ρ v0 ρ v0 e 0 6 τ r which gives or τ r ln τ r 0 6 ln But τ r ε σ 9ε 0 σ. Hence σ 9ε 0 τ r (s) (S/m)

308 CHAPTER Problem 6.9 If the current density in a conducting medium is given by J x y z;t ˆxz ŷ4y ẑx cos ωt determine the corresponding charge distribution ρ v x y z;t. Solution: Eq. (6.58) is given by The divergence of J is J ρ v t (4) J ˆx x ŷ y ẑ z ˆxz ŷ4y ẑx cos ωt 4 cosωt y y 8ycos ωt Using this result in Eq. (4) and then integrating both sides with respect to t gives 8y ρ v J dt 8ycosωt dt ω sinωt C 0 where C 0 is a constant of integration. Problem 6.0 In a certain medium, the direction of current density J points in the radial direction in cylindrical coordinates and its magnitude is independent of both φ and z. Determine J, given that the charge density in the medium is ρ v ρ 0 r cos ωt C/m 3 Solution: Based on the given information, J ˆrJ r r With J φ J z 0, in cylindrical coordinates the divergence is given by J r r r rj From Eq. (6.54), J ρ v t t ρ 0 r cos ωt ρ 0 rωsin ωt

309 308 CHAPTER 6 Hence 0 r r r r rj ρ 0 rωsin ωt r rj r ρ 0 r ωsin ωt r rj r dr ρ 0 ωsin ωt r r rj r 0 ρ 0 ωsinωt 3 3 ρ J r 0 ωr sinωt 3 0 r r dr r 0 and J ˆrJ r ˆr ρ 0ωr 3 sinωt (A/m ) Problem 6. If we were to characterize how good a material is as an insulator by its resistance to dissipating charge, which of the following two materials is the better insulator? Dry Soil: ε r 5, σ 0 4 (S/m) Fresh Water: ε r 80, σ 0 3 (S/m) Solution: Relaxation time constant τ r ε σ. For dry soil, τ r s. 80 For fresh water, τ r s. 3 Since it takes longer for charge to dissipate in fresh water, it is a better insulator than dry soil. Sections 6-: Electromagnetic Potentials Problem 6. given by The electric field of an electromagnetic wave propagating in air is E z t ˆx4cos t z ŷ3sin t z (V/m)

310 CHAPTER Find the associated magnetic field H z t. Solution: Converting to phasor form, the electric field is given by jz jz E z ˆx4e jŷ3e (V/m) which can be used with Eq. (6.87) to find the magnetic field: H z jωµ E ˆx ŷ ẑ x y z jωµ 4e jz j3e jz 0 jωµ ˆx6e jz ŷ j8e jz j π 0 7 ˆx6 ŷ j8 e jz jˆx8 0e jz ŷ06e jz Converting back to instantaneous values, this is H t z ˆx8 0sin t z ŷ0 6cos t z (ma/m) (ma/m) Problem 6.3 σ 0 is given by The magnetic field in a dielectric material with ε 4ε 0, µ µ 0, and Find k and the associated electric field E. H y t ˆx5cos π 0 7 t ky (A/m) Solution: In phasor form, the magnetic field is given by H ˆx5e jky (A/m). From Eq. (6.86), and, from Eq. (6.87), E H jk jωε jωε H E jk ω ε k ω εµ r π c ẑ5e jky jωµ jω εµ which, together with the original phasor expression for H, implies that 4π (rad/m) 30 ˆx5e jky

311 30 CHAPTER 6 Inserting this value in the expression for E above, 4π 30 E ẑ π e j4πy 30 ẑ94e j4πy 30 (V/m) Problem 6.4 Given an electric field E ˆxE 0 sinaycos ωt kz where E 0, a, ω, and k are constants, find H. Solution: E ˆxE 0 sinaycos ωt kz E ˆxE 0 sinay e jkz H jωµ E ŷ jωµ z E 0 sin ay e jkz ẑ y E 0 sinay e jkz E 0 sinay ωµ ŷk ẑ jacos ay e jkz H He jωt E 0 E 0 E 0 ωµ ŷk sinay ẑacosay e jπ e jkz e jωt ωµ ωt kz ŷk sinaycos ẑacos aycos ωt kz ωµ ωt kz ŷk sin aycos ẑacos aysin ωt kz π Problem 6.5 The electric field radiated by a short dipole antenna is given in spherical coordinates by E R θ;t 0 ˆθ sinθ cos 6π 0 8 R t πr (V/m) Find H R θ;t. Solution: Converting to phasor form, the electric field is given by E R θ ˆθE θ 0 jπr ˆθ sinθe (V/m) R

312 CHAPTER 6 3 which can be used with Eq. (6.87) to find the magnetic field: H R θ jωµ E ˆR jωµ E θ ˆφ Rsinθ φ θ R R RE 0 jωµ ˆφ sinθ e jπr R R π 0 jπr ˆφ 6π 0 4π 0 8 sin θe 7 R (µa/m) ˆφ 53 R sinθ e jπr Converting back to instantaneous value, this is H R θ;t ˆφ 53 R sinθcos 6π 0 8 t πr (µa/m) Problem 6.6 A Hertzian dipole is a short conducting wire carrying an approximately constant current over its length l. If such a dipole is placed along the z-axis with its midpoint at the origin and if the current flowing through it is i t I 0 cos ωt, find (a) the retarded vector potential A R θ φ at an observation point Q R θ φ in a spherical coordinate system, and (b) the magnetic field phasor H R θ φ. Assume l to be sufficiently small so that the observation point is approximately equidistant to all points on the dipole; that is, assume that R R. Solution: (a) In phasor form, the current is given by I I 0. Explicitly writing the volume integral in Eq. (6.84) as a double integral over the wire cross section and a single integral over its length, µ l J R A i jkr e ds dz 4π R l where s is the wire cross section. The wire is infinitesimally thin, so that R s is not a function of x or y and the integration over the cross section of the wire applies only to the current density. Recognizing that J ẑi 0 s, and employing the relation R R, A ẑ µi 0 4π l l e jkr R dz ẑ µi 0 4π l l e jkr R dz ẑ µi 0l 4πR e jkr

313 3 CHAPTER 6 In spherical coordinates, ẑ ˆRcosθ (b) From Eq. (6.85), H µ A I 0 l ˆθsinθ, and therefore A ˆRcosθ ˆθsinθ µi 0 l 4πR e jkr 4π ˆRcosθ I 0 l 4π ˆφ R ˆφ I 0l sinθe jkr 4πR ˆθsinθ e jkr R R sinθe jkr jk R θ e jkr cosθ R Problem 6.7 The magnetic field in a given dielectric medium is given by H ŷ6cos zsin 0 7 t 0 x (A/m) where x and z are in meters. Determine: (a) E, (b) the displacement current density J d, and (c) the charge density ρ v. Solution: (a) H ŷ6cos zsin 0 t 0 x 7 ŷ6cos zcos 0 t 0 x π H 7 j0 ŷ6cos z e e x jπ j0 ŷ j6cos z e x E jωε H ˆx ŷ ẑ x y z jωε ˆx jωε ˆx 0 j6cos z e j0 x 0 j0 6 j0 j6cos z z e x ẑ e ωε sinz j0 x ẑ e ωε cos z j0 x From the given expression for H, ω 0 7 (rad/s) x j6cos z e j0 x

314 CHAPTER 6 33 β 0 (rad/m) Hence, and ε r u p ω β c u p Using the values for ω and ε, we have E E or (b) 08 (m/s) (V/m) 5 ˆx30sin z ẑ j 5cos z 0 e 3 j0 x ˆx30sin zcos 0 t 0 x 7 ẑ 5cos zsin 0 t 0 x 7 D εe ε r ε 0 E ˆx0 6sin z ẑ j0 03cos z 0 e 6 j0 x D J d t J d jωd J d J d e jωt ˆx jsin z ẑ0 6cos z e j0 x (C/m ) (kv/m) ˆxsin zsin 0 7 t 0 x ẑ0 6cos zcos 0 7 t 0 x (A/m ) (c) We can find ρ v from or from Applying Maxwell s equation, yields ρ v ε r ε 0 D ρ v J ρ v t ρ v D ε E ε r ε 0 E x x E z z 30sin zcos x 0 t 0 7 x z 5cos zsin 0 t 0 x 7 ε r ε 0 3sin zsin 0 t 0 x 7 3sin zsin 0 t 0 x 7 0

315 34 CHAPTER 6 Problem 6.8 The transformer shown in the figure consists of a long wire coincident with the z-axis carrying a current I I 0 cosωt, coupling magnetic energy to a toroidal coil situated in the x y plane and centered at the origin. The toroidal core uses iron material with relative permeability µ r, around which 00 turns of a tightly wound coil serves to induce a voltage V emf, as shown in the figure. z V emf - + N I b a c y x Iron core with µ r (a) Develop an expression for V emf. (b) Calculate V emf for f 60 Hz, µ r 4000, a 5 cm, b 6 cm, c cm, and I 0 50 A. Solution: (a) We start by calculating the magnetic flux through the coil, noting that r, the distance from the wire varies from a to b b Φ B ds ˆx µi dr S πr ˆxc µci π ln b V emf N dφ dt a µcn π ln b di a dt b ln sin ωt a µcnωi 0 π a (V)

316 CHAPTER 6 35 (b) 7 V emf π π 60 50ln 6 5 π 5 5sin 377t (V) sin 377t Problem 6.9 In wet soil, characterized by σ 0 (S/m), µ r, and ε r 36, at what frequency is the conduction current density equal in magnitude to the displacement current density? Solution: For sinusoidal wave variation, the phasor electric field is E E 0 e jωt J c σe σe 0 e jωt or Problem 6.30 (a) Determine k. (b) Determine E. (c) Determine J d. D J d ε E jωεe 0 e jωt t t J c σ J d σ ωε πε f σ f 0 πε π MHz In free space, the magnetic field is given by H ˆφ 36 r cos t kz (ma/m) Solution: (a) From the given expression, ω (rad/s), and since the medium is free space, ω k c (rad/m)

317 36 CHAPTER 6 (b) Convert H to phasor: (c) H ˆφ 36 e jkz r E jωε 0 H jωε 0 jωε 0 jωε 0 (ma/m) ˆr H φ ẑ z r ˆr 36 z ˆr j36k r ˆr 36k e jkz ωε 0 ˆr r E Ee jωt ˆr 3 6 r r rh φ e jkz r jkz e r ẑ r cos t 0z (V/m) r 36e jkz e jkz 0 3 ˆr 3 6 r e j0z (V/m) E J d ε 0 t 3 ˆr 6 r ε 0 cos t t 0z ˆr 3 6ε sin r t 0z (A/m ) ˆr sin 6 0 t 0z 9 (A/m ) 0 7 r

318 37 Chapter 7: Plane-Wave Propagation Lesson #43 Chapter Section: 7- Topics: Time-harmonic fields Highlights: Phasors Complex permittivity Wave equations Special Illustrations:

319 38 Lesson #44 Chapter Section: 7- Topics: Waves in lossless media Highlights: Uniform plane waves Intrinsic impedance Wave properties Special Illustrations: Example 7- CD-ROM Modules 7.3 and 7.4

320 39 Lesson #45 and 46 Chapter Section: 7-3 Topics: Wave polarization Highlights: Definition of polarization Linear, circular, elliptical Special Illustrations: CD-ROM Demos Liquid Crystal Display Liquid Crystal Display (LCD) LCDs are used in digital clocks, cellular phones, desktop and laptop computers, and some televisions and other electronic systems. They offer a decided advantage over other display technologies, such as cathode ray tubes, in that they are much lighter and thinner and consume a lot less power to operate. LCD technology relies on special electrical and optical properties of a class of materials known as liquid crystals, first discovered in the 880s by botanist Friedrich Reinitzer. Physical Principle Liquid crystals are neither a pure solid nor a pure liquid, but rather a hybrid of both. One particular variety of interest is the twisted nematic liquid crystal whose molecules have a natural tendency to assume a twisted spiral structure when the material is sandwiched between finely grooved glass substrates with orthogonal orientations (A). Note that the molecules in contact with the grooved surfaces align themselves in parallel along the grooves. The molecular spiral causes the crystal to behave like a wave polarizer; unpolarized light incident upon the entrance substrate follows the orientation of the spiral, emerging through the exit substrate with its polarization (direction of electric field) parallel to the groove s direction.

321 30 Lesson #47 Chapter Section: 7-4 Topics: Waves in lossy media Highlights: Attenuation and skin depth Low loss medium Good conductor Special Illustrations: CD-ROM Demos

322 3 Lesson #48 Chapter Section: 7-5 Topics: Current flow in conductors Highlights: Skin depth dependence on frequency Surface impedance Special Illustrations:

323 3 Lesson #49 Chapter Section: 7-6 Topics: EM power density Highlights: Power density in a lossless medium Power density in a lossy medium Time-average power Special Illustrations: CD-ROM Module 7.5

324 CHAPTER 7 33 Chapter 7 Section 7-: Propagation in Lossless Media Problem 7. The magnetic field of a wave propagating through a certain nonmagnetic material is given by H ẑ30cos 0 8 t 0 5y (ma/m) Find (a) the direction of wave propagation, (b) the phase velocity, (c) the wavelength in the material, (d) the relative permittivity of the material, and (e) the electric field phasor. Solution: (a) Positive y-direction. (b) ω 0 8 rad/s, k 0 5 rad/m. ω u p 0 8 k 0 m/s 5 08 (c) λ π k π m. c (d) ε r u p (e) From Eq. (7.39b), E ηˆk H µ η 0π 0π ε ε r 5 j0 5y 3 ˆk ŷ and H ẑ30e (Ω) (A/m) Hence, and E 5 33ŷ ẑ30e j0 5y 0 3 ˆx7 54e j0 5y (V/m) E y t Ee jωt ˆx7 54cos 0 8 t 0 5y (V/m) Problem 7. Write general expressions for the electric and magnetic fields of a -GHz sinusoidal plane wave traveling in the y-direction in a lossless nonmagnetic medium with relative permittivity ε r 9. The electric field is polarized along the x-direction, its peak value is 6 V/m and its intensity is 4 V/m at t 0 and y cm.

325 34 CHAPTER 7 Solution: For f GHz, µ r, and ε r 9, ω π f π 0 9 rad/s π k π π f ε r λ λ 0 c ε r π E y t ˆx6cos π 0 t 9 0πy φ 0 (V/m) At t 0 and y cm, E 4 V/m: 4 6cos 0π 0 Hence, which gives and φ 0 0 4π cos φ 0 rad π rad/m φ 0 6cos 0 4π φ rad E y t ˆx 6cos π 0 9 t 0πy 0 9 (V/m) Problem 7.3 The electric field phasor of a uniform plane wave is given by E ŷ0e j0 z (V/m). If the phase velocity of the wave is m/s and the relative permeability of the medium is µ r 4, find (a) the wavelength, (b) the frequency f of the wave, (c) the relative permittivity of the medium, and (d) the magnetic field H z t. Solution: (a) From E ŷ0e j0 z (V/m), we deduce that k 0 rad/m. Hence, π λ π k 0 0π 3 4 m (b) (c) From u p u f p λ 3 4 c c µ r ε r ε r µ r u p Hz 4 77 MHz

326 CHAPTER 7 35 (d) µ ε µ r 0π ε r 4 0π η 45 (Ω) H η ẑ E η ẑ ŷ0e ˆx j0 z j0 z 3e (ma/m) H z t ˆx 3cos ωt 0 z (ma/m) with ω π f 9 54π 0 6 rad/s. Problem 7.4 The electric field of a plane wave propagating in a nonmagnetic material is given by E ŷ3sin π 0 7 t 0 πx ẑ4cos π 0 7 t 0 πx (V/m) Determine (a) the wavelength, (b) ε r, and (c) H. Solution: (a) Since k 0 π, But (b) Hence, π λ π k 0 0 m π ω u p π 0 7 k 0 π ε r c u p u p c ε r 5 07 m/s (c) H ˆk E ˆx ŷ3sin π 0 7 η η t 0 πx ẑ4cos π 0 t 0 πx 7 ẑ 3 π η sin 0 t 0 πx 7 ŷ 4 π η cos 0 t 0 πx (A/m) 7

327 36 CHAPTER 7 with η η 0 ε r 0π 6 0π 6 83 (Ω) Problem 7.5 A wave radiated by a source in air is incident upon a soil surface, whereupon a part of the wave is transmitted into the soil medium. If the wavelength of the wave is 60 cm in air and 0 cm in the soil medium, what is the soil s relative permittivity? Assume the soil to be a very low loss medium. Solution: From λ λ 0 ε r, ε r λ 0 λ Problem 7.6 The electric field of a plane wave propagating in a lossless, nonmagnetic, dielectric material with ε r 56 is given by E ŷ0cos 6π 0 9 t kz (V/m) Determine: (a) f, u p, λ, k, and η, and (b) the magnetic field H. Solution: (a) ω π f 6π 0 rad/s 9 f 3 0 Hz 9 3 GHz c u p ε r m/s u λ p cm f π k λ π rad/m 35 6 Ω 6 η η 0 ε r

328 CHAPTER 7 37 (b) H ˆx 0 6π η cos 0 t kz 9 0 ˆx 35 6π 6 cos 0 t 0 4z 9 ˆx cos 6π 0 t 0 4z (A/m) 9 Section 7-3: Wave Polarization Problem 7.7 An RHC-polarized wave with a modulus of (V/m) is traveling in free space in the negative z-direction. Write down the expression for the wave s electric field vector, given that the wavelength is 6 cm. y z ωt=0 x ωt=π/ Figure P7.7: Locus of E versus time. Solution: For an RHC wave traveling in ẑ, let us try the following: E ˆxacos ωt kz ŷasin ωt kz Modulus E a a a (V/m). Hence, a

329 38 CHAPTER 7 Next, we need to check the sign of the ŷ-component relative to that of the ˆx-component. We do this by examining the locus of E versus t at z 0: Since the wave is traveling along ẑ, when the thumb of the right hand is along ẑ (into the page), the other four fingers point in the direction shown (clockwise as seen from above). Hence, we should reverse the sign of the ŷ-component: E ˆx cos ωt kz ŷ sin ωt kz (V/m) with and Problem 7.8 π k λ π ω kc π λ 3 08 π 0 0 (rad/m) (rad/s) For a wave characterized by the electric field E z t ˆxa x cos ωt kz ŷa y cos ωt kz δ identify the polarization state, determine the polarization angles γ χ, and sketch the locus of E 0 t for each of the following cases: (a) a x 3 V/m, a y 4 V/m, and δ 0, (b) a x 3 V/m, a y 4 V/m, and δ 80, (c) a x 3 V/m, a y 3 V/m, and δ 45, (d) a x 3 V/m, a y 4 V/m, and δ 35. Solution: ψ 0 tan a y a x [Eq. (7.60)] tanγ tanψ 0 cos δ sinχ sinψ 0 sin δ [Eq. (7.59a)] [Eq. (7.59b)] Case a x a y δ ψ 0 γ χ Polarization State (a) Linear (b) Linear (c) Left elliptical (d) Right elliptical (a) E z t ˆx3cos ωt kz ŷ4cos ωt kz. (b) E z t ˆx3cos ωt kz ŷ4cos ωt kz. (c) E z t ˆx3cos ωt kz ŷ3cos ωt kz 45. (d) E z t ˆx3cos ωt kz ŷ4cos ωt kz 35.

330 CHAPTER 7 39 y y x x (a) (b) y y (c) x (d) x Figure P7.8: Plots of the locus of E 0 t. Problem 7.9 The electric field of a uniform plane wave propagating in free space is given by E ˆx jŷ 0e jπz 6 (V/m). Specify the modulus and direction of the electric field intensity at the z 0 plane at t 0, 5 and 0 ns.

331 330 CHAPTER 7 Solution: From At z 0, E z t Ee jωt ˆx jπz jŷ 0e 6 e jωt ˆx ŷe 0e jπ jπz 6 e jωt ˆx0cos ωt πz 6 ŷ0cos ωt πz 6 π ˆx0cos ωt πz 6 ŷ0sin ωt πz 6 (V/m) E Ex Ey 0 (V/m) E ψ tan y E x ωt πz 6 c f kc π λ π π ω π f 5π 0 rad/s Hz 0 at t 0 ψ ωt 5π 0 7 t 0 5π 45 at t 5 ns 0 5π 90 Therefore, the wave is LHC polarized. at t 0 ns Problem 7.0 expressed as the sum of an RHC polarized wave with magnitude a R and an LHC polarized wave with magnitude a L. Prove this statement by finding expressions for a R and a L in terms of a x. Solution: A linearly polarized plane wave of the form E ˆxa x e jkz can be RHC wave: LHC wave: E ˆxa x e jkz E R a R ˆx jπ jkz ŷe e a R ˆx jŷ e jkz E L a L ˆx ŷe e jπ jkz a L ˆx jŷ e jkz E E R E L ˆxa x a R ˆx jŷ a L ˆx jŷ By equating real and imaginary parts, a x a R a L, 0 a R a L, or a L a x, a R a x.

332 CHAPTER 7 33 Problem 7. The electric field of an elliptically polarized plane wave is given by (V/m) E z t ˆx0sin ωt kz 60 ŷ30cos ωt kz Determine (a) the polarization angles γ χ and (b) the direction of rotation. Solution: (a) Phasor form: E z t ˆx0sin ωt kz 60 ŷ30cos ωt kz ˆx0cos ωt kz 30 ŷ30cos ωt kz (V/m) E ˆx0e j30 ŷ30 e jkz Since δ is defined as the phase of E y relative to that of E x, δ ψ 0 tan tanγ tan ψ 0 cosδ 0 65 or γ 73 5 sin χ sin ψ 0 sinδ 0 40 or χ 8 73 (b) Since χ 0, the wave is right-hand elliptically polarized. Problem 7. Compare the polarization states of each of the following pairs of plane waves: (a) wave : E ˆxcos ωt kz ŷsin ωt kz, wave : E ˆxcos ωt kz ŷsin ωt kz, (b) wave : E ˆxcos ωt kz ŷsin ωt kz, wave : E ˆxcos ωt kz ŷsin ωt kz. Solution: (a) E ˆxcos ωt kz ŷsin ωt kz ˆxcos ωt kz ŷcos ωt kz π jkz E ˆxe ŷe e jkz jπ

333 33 CHAPTER 7 Hence, wave is RHC. Similarly, ψ 0 tan ay ax tan 45 δ π E ˆxe jkz ŷe jkz e jπ Wave has the same magnitude and phases as wave except that its direction is along ẑ instead of ẑ. Hence, the locus of rotation of E will match the left hand instead of the right hand. Thus, wave is LHC. (b) E ˆxcos ωt kz ŷsin ωt kz jkz E ˆxe ŷe jkz e jπ Wave is LHC. E ˆxe jkz ŷe jkz e jπ Reversal of direction of propagation (relative to wave ) makes wave RHC. Problem 7.3 Plot the locus of E 0 t for a plane wave with E z t ˆxsin ωt kz ŷcos ωt kz Determine the polarization state from your plot. Solution: E ˆx sin ωt kz ŷcos ωt kz Wave direction is ẑ. At z 0, E ˆx sinωt ŷcos ωt Tip of E rotates in accordance with right hand (with thumb pointing along ẑ). Hence, wave state is RHE.

334 CHAPTER y ωt=π/ z x ωt=0 Figure P7.3: Locus of E versus time. Sections 7-4: Propagation in a Lossy Medium Problem 7.4 For each of the following combination of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, u p, and η c : (a) glass with µ r, ε r 5, and σ 0 S/m at 0 GHz, (b) animal tissue with µ r, ε r, and σ 0 3 S/m at 00 MHz, (c) wood with µ r, ε r 3, and σ 0 4 S/m at khz. Solution: Using equations given in Table 7-: Case (a) Case (b) Case (c) σ ωε Type low-loss dielectric quasi-conductor good conductor α Np/m 9.75 Np/m Np/m β rad/m.6 rad/m rad/m λ.34 cm 5.69 cm 0 km u p m/s m/s m/s η c 68 5 Ω j3 7Ω 6 8 j Ω

335 334 CHAPTER 7 Problem 7.5 At each of the following frequencies, determine if dry soil may be considered a good conductor, a quasi-conductor, or a low-loss dielectric, and then calculate α, β, λ, µ p, and η c : (a) 60 Hz, (b) khz, (c) MHz, (d) GHz. Dry soil is characterized by ε r 5, µ r, and σ 0 4 (S/m). Solution: ε r 5, µ r, σ 0 4 S/m. f 60 Hz khz MHz GHz ε ε σ ωε σ π f ε r ε 0 Type of medium Good conductor Good conductor Quasi-conductor Low-loss dielectric α (Np/m) β (rad/m) λ (m) u p (m/s) η c (Ω) 54 j 6 8 j 04 8 j Problem 7.6 In a medium characterized by ε r 9, µ r, and σ 0 S/m, determine the phase angle by which the magnetic field leads the electric field at 00 MHz. Solution: The phase angle by which the magnetic field leads the electric field is θ η where θ η is the phase angle of η c. Hence, quasi-conductor. µ η c ε j ε ε σ ωε 0π ε r 0 36π π σ j ωε 0 ε r 5 67 j 7 49 j

336 CHAPTER Therefore θ η 3 7. Since H η c ˆk E, H leads E by θ η, or by 3 7. In other words, H lags E by 3 7. Problem 7.7 Generate a plot for the skin depth δ s versus frequency for seawater for the range from khz to 0 GHz (use log-log scales). The constitutive parameters of seawater are µ r, ε r 80 and σ 4 S/m. Solution: δ s α ω ω π f µε ε µε µ 0 ε 0 ε r r ε σ ε ωε ε ε c c σ ωε 0 ε r See Fig. P7.7 for plot of δ s versus frequency π π f f 09 0 Skin depth vs. frequency for seawater 0 0 Skin depth (m) Frequency (MHz) Figure P7.7: Skin depth versus frequency for seawater.

337 336 CHAPTER 7 Problem 7.8 Ignoring reflection at the air-soil boundary, if the amplitude of a 3-GHz incident wave is 0 V/m at the surface of a wet soil medium, at what depth will it be down to mv/m? Wet soil is characterized by µ r, ε r 9, and σ S/m. Solution: σ ωε Hence, medium is a low-loss dielectric. σ α αz E z E 0 e 0e αz π π µ σ ε 0π e 03z ln0 0 03z z 87 8 m ε r π 0 03 (Np/m) 9 Problem 7.9 The skin depth of a certain nonmagnetic conducting material is 3 µm at 5 GHz. Determine the phase velocity in the material. Solution: For a good conductor, α β, and for any material δ s α. Hence, ω u p π f β β π f δ s π (m/s) 4 Problem 7.0 Based on wave attenuation and reflection measurements conducted at MHz, it was determined that the intrinsic impedance of a certain medium is 8 45 Ω and the skin depth is m. Determine (a) the conductivity of the material, (b) the wavelength in the medium, and (c) the phase velocity. Solution: (a) Since the phase angle of η c is 45, the material is a good conductor. Hence, or η c α j 8 j45 8 σ e cos 45 j8 sin 45 α 8 cos σ

338 CHAPTER Since α δ s 0 5 Np/m, σ α S/m (b) Since α β for a good conductor, and α 0 5, it follows that β 0 5. Therefore, π λ π β 0 4π 57 m 5 (c) u p f λ m/s. Problem 7. The electric field of a plane wave propagating in a nonmagnetic medium is given by E ẑ5e 30x cos π 0 9 t 40x (V/m) Obtain the corresponding expression for H. Solution: From the given expression for E, From (7.65a) and (7.65b), α β ω µε ω π 0 9 (rad/s) α 30 (Np/m) β 40 (rad/m) αβ ω µε ω ω µ 0 ε 0 ε r c r ε ω c r ε Using the above values for ω, α, and β, we obtain the following: µ η c ε η 0 εr j ε ε j ε r εr εr 6 εr 5 47 j e (Ω) j

339 338 CHAPTER 7 E ẑ5e e j40x 30x H η c ˆk E 57 ˆx ẑ5e 30x j40x e 9e ŷ0 6e e 30x 40x j36 85 e j36 85 He jωt ŷ0 6e 30x cos π 0 t 40x (A/m) 9 H Section 7-5: Current Flow in Conductors Problem 7. In a nonmagnetic, lossy, dielectric medium, a 300-MHz plane wave is characterized by the magnetic field phasor H ˆx j4ẑ e y e j9y (A/m) Obtain time-domain expressions for the electric and magnetic field vectors. Solution: E η c ˆk H To find η c, we need ε and ε. From the given expression for H, α (Np/m) β 9 (rad/m) Also, we are given than f 300 MHz Hz. From (7.65a), α β ω µε 4 8 π π 0 ε r 0 36π whose solution gives εr 95 Similarly, from (7.65b), αβ ω µε 9 π π 0 7 ε r π which gives εr 0 9

340 CHAPTER µ η c j ε ε ε Hence, η 0 εr 0 9 j j0 56 9e j 6 E 56 9e ŷ ˆx j4ẑ e e j 6 y j9y ˆx j4 ẑ 56 9e e y j9y e j 6 ˆx4e jπ ẑ 56 9e e y j9y e E j 6 Ee jωt ˆx 03 0 e 3 y cos 9y ωt 0 6 ẑ56 9e y cos 9y ωt 6 (V/m) H He jωt ˆx j4ẑ e e y j9y e jωt ˆxe y cos ωt 9y ẑ4e y sin ωt 9y (A/m) Problem 7.3 A rectangular copper block is 30 cm in height (along z). In response to a wave incident upon the block from above, a current is induced in the block in the positive x-direction. Determine the ratio of the a-c resistance of the block to its d-c resistance at khz. The relevant properties of copper are given in Appendix B. w J l 30 cm Figure P7.3: Copper block of Problem 7.3.

341 340 CHAPTER 7 Solution: l d-c resistance R dc σa l a-c resistance R ac σwδ s l 0 3σw R ac R dc 0 3 δ s 0 3 π f µσ 0 3 π 0 4π Problem 7.4 The inner and outer conductors of a coaxial cable have radii of 0.5 cm and cm, respectively. The conductors are made of copper with ε r, µ r and σ S/m, and the outer conductor is 0.5 mm thick. At 0 MHz: (a) Are the conductors thick enough to be considered infinitely thick so far as the flow of current through them is concerned? (b) Determine the surface resistance R s. (c) Determine the a-c resistance per unit length of the cable. Solution: (a) From Eqs. (7.7) and (7.77b), δ s π f µσ π 0 7 4π mm Hence, Hence, conductor is plenty thick. (b) From Eq. (7.9a), (c) From Eq. (7.96), R R s π d δ s 0 5 mm 0 0 mm 5 R s σδ s Ω a b π (Ω/m)

342 CHAPTER 7 34 Section 7-6: EM Power Density Problem 7.5 The magnetic field of a plane wave traveling in air is given by H ˆx50sin π 0 t ky 7 (ma/m). Determine the average power density carried by the wave. Solution: H ˆx50sin π 0 t ky (ma/m) E 7 η 0 ŷ H ẑη 0 50sin π 0 t ky (mv/m) 7 S av ẑ ˆx η ŷ 0π ŷ0 48 (W/m ) Problem 7.6 A wave traveling in a nonmagnetic medium with ε r 9 is characterized by an electric field given by E ŷ3cos π 0 7 t kx ẑcos π 0 7 t kx (V/m) Determine the direction of wave travel and the average power density carried by the wave. Solution: η η 0 0π ε r 40π (Ω) 9 The wave is traveling in the negative x-direction. Problem 7.7 S av ˆx 3 η 3 ˆx 40π ˆx0 05 (W/m ) The electric-field phasor of a uniform plane wave traveling downward in water is given by E ˆx5e 0 z e j0 z (V/m) where ẑ is the downward direction and z 0 is the water surface. If σ 4 S/m, (a) obtain an expression for the average power density, (b) determine the attenuation rate, and (c) determine the depth at which the power density has been reduced by 40 db.

343 34 CHAPTER 7 Solution: (a) Since α β 0, the medium is a good conductor. From Eq. (7.09), S av ẑ E 0 η c η c j α σ j 0 4 j e j45 (Ω) e αz 5 cos θ η ẑ 0 e 0 4z 0 4z cos 45 ẑ5e 0707 (b) A 8 68αz z 74z (db). (c) 40 db is equivalent to 0 4. Hence, 0 4 e αz e 0 4z ln z or z 3 03 m. (W/m ) Problem 7.8 The amplitudes of an elliptically polarized plane wave traveling in a lossless, nonmagnetic medium with ε r 4 are H y0 3 (ma/m) and H z0 4 (ma/m). Determine the average power flowing through an aperture in the y-z plane if its area is 0 m. Solution: η η 0 0π ε r 60π 88 5 Ω 4 S av ˆx η y0 H Hx0 88 ˆx (mw/m ) 3 P S av A (mw) Problem 7.9 A wave traveling in a lossless, nonmagnetic medium has an electric field amplitude of 4.56 V/m and an average power density of.4 W/m. Determine the phase velocity of the wave. Solution: or S av E 0 η η E 0 S av η Ω

344 CHAPTER But Hence, η η 0 ε r 377 ε r ε r u p c ε r m/s 9 Problem 7.30 At microwave frequencies, the power density considered safe for human exposure is (mw/cm ). A radar radiates a wave with an electric field amplitude E that decays with distance as E R R (V/m), where R is the distance in meters. What is the radius of the unsafe region? Solution: E S av R η 0 (mw/cm ) 0 3 W/cm 0 W/m R 0 4 0π R R m 0 Problem 7.3 Consider the imaginary rectangular box shown in Fig. 7-9 (P7.3). (a) Determine the net power flux P t entering the box due to a plane wave in air given by E ˆxE 0 cos ωt ky (V/m) (b) Determine the net time-average power entering the box. Solution: (a) S t E H ŷ E 0 E ˆxE 0 cos ωt ky H ẑ E 0 cos η 0 ωt ky cos η 0 ωt ky P t S t A y 0 S t A y b E 0 η 0 ac cos ωt cos ωt kb

345 344 CHAPTER 7 z a b c y x Figure P7.3: Imaginary rectangular box of Problems 7.3 and 7.3. (b) where T π ω. E0 P av ac η 0 ω π P av T 0 π ω 0 T P t dt cos ωt cos ωt kb dt Net average energy entering the box is zero, which is as expected since the box is in a lossless medium (air). 0 Problem 7.3 Repeat Problem 7.3 for a wave traveling in a lossy medium in which E ˆx00e 0y cos π 0 9 t 40y (V/m) H ẑ0 64e 0y cos π 0 9 t 40y (A/m) The box has dimensions A cm, b cm, and c 0 5 cm. Solution: (a) S t E H ˆx00e 0y cos π 0 t 40y 9 ẑ0 64 e 0y cos π 0 t 40y ŷ64e 9 40y cos π 0 t 40y 9 cos π 0 t 40y

346 CHAPTER Using the identity cos θcosφ cos θ φ cos θ φ, 64 S t e 40y cos 4π 0 9 t 80y cos36 85 P t S t A y S 0 t A y b 3ac cos 4π 0 t cos b e cos 4π 0 t 80y cos36 85 (b) P av T 0 T P t dt ω π 0 π ω P t dt The average of cos ωt θ over a period T is equal to zero, regardless of the value of θ. Hence, P av 3ac e 40b cos With a cm, b cm, and c 0 5 cm, P av (W) This is the average power absorbed by the lossy material in the box. Problem 7.33 Given a wave with E ˆxE 0 cos ωt kz calculate: (a) the time-average electric energy density w e av T 0 T w e dt T (b) the time-average magnetic energy density w m av T and (c) show that w e av w m av. Solution: (a) w e av T 0 0 T T w m dt T 0 0 T T εe dt εe 0 cos ωt kz dt µh dt

347 346 CHAPTER 7 With T π ω, (b) (c) w e av ωεe 0 4π εe0 4π εe0 4 w m av T 0 0 π π ω cos ωt kz dt cos ωt kz d ωt H ŷ E 0 η cos ωt kz T µe0 4η w m av µe 0 T 0 T 0 µh dt µe0 4η µ 4 µ E 0 η cos ωt kz dt ε εe 0 4 w e av Problem 7.34 A 60-MHz plane wave traveling in the x-direction in dry soil with relative permittivity ε r 4 has an electric field polarized along the z-direction. Assuming dry soil to be approximately lossless, and given that the magnetic field has a peak value of 0 (ma/m) and that its value was measured to be 7 (ma/m) at t 0 and x 075 m, develop complete expressions for the wave s electric and magnetic fields. Solution: For f 60 MHz Hz, ε r 4, µ r, ω k c ε r π π (rad/m) Given that E points along ẑ and wave travel is along ˆx, we can write E x t ẑe 0 cos π t 0 8πx φ 0 (V/m) where E 0 and φ 0 are unknown constants at this time. The intrinsic impedance of the medium is η η 0 ε r 0π 60π (Ω)

348 CHAPTER With E along ẑ and ˆk along ˆx, (7.39) gives H η ˆk E or Hence, H x t ŷ E 0 η cos π 0 8 t 0 8πx φ 0 (A/m) E 0 η 0 (ma/m) E π π (V/m) Also, H 0 75 m cos 0 8π 0 75 φ which leads to φ Hence, E x t ẑ0 6πcos π 0 8 t 0 8πx 53 6 (V/m) H x t ŷ0cos π 0 8 t 0 8πx 53 6 (ma/m) Problem 7.35 At GHz, the conductivity of meat is on the order of (S/m). When a material is placed inside a microwave oven and the field is activated, the presence of the electromagnetic fields in the conducting material causes energy dissipation in the material in the form of heat. (a) Develop an expression for the time-average power per mm 3 dissipated in a material of conductivity σ if the peak electric field in the material is E 0. (b) Evaluate the result for meat with E (V/m). Solution: (a) Let us consider a small volume of the material in the shape of a box of length d and cross sectional area A. Let us assume the microwave oven creates a wave traveling along the z direction with E along y, as shown.

349 348 CHAPTER 7 z A y E d k^ E x Along y, the E field will create a voltage difference across the length of the box V, where V Ed Conduction current through the cross sectional area A is I JA σea Hence, the instantaneous power is P IV σe Ad σe V where V Ad is the small volume under consideration. obtained by setting V 0 3 3, P P 9 0 σe 0 9 (W/mm 3 ) The power per mm 3 is As a time harmonic signal, E E 0 cosωt. The time average dissipated power is Pav T σe0 cos 9 ωt dt 0 T 0 σe (W/mm ) 3

350 CHAPTER (b) P av (W/mm 3 ) Problem 7.36 A team of scientists is designing a radar as a probe for measuring the depth of the ice layer over the antarctic land mass. In order to measure a detectable echo due to the reflection by the ice-rock boundary, the thickness of the ice sheet should not exceed three skin depths. If εr 3 and εr 0 for ice and if the maximum anticipated ice thickness in the area under exploration is. km, what frequency range is useable with the radar? Solution: Hence, 3δ s km 00 m δ s 400 m α δ s µ ε π f εr ε 0 εr ε 0 µ 0 π f εr Since ε ε, we can use (7.75a) for α: ωε α For α f 0, (Np/m) π f c ε r Np/m f f 4 6 MHz Since α increases with increasing frequency, the useable frequency range is f 4 6 MHz

351 350 Chapter 8: Reflection, Transmission, and Waveguides Lessons #50 and 5 Chapter Section: 8- Topics: Normal incidence Highlights: Analogy to transmission line Reflection and transmission coefficient Special Illustrations: Example 8- CD-ROM Modules CD-ROM Demos 8.

352 35 Lesson #5 Chapter Section: 8- Topics: Snell s laws Highlights: Reflection and refraction Index of refraction Special Illustrations: Example 8-4 Technology Brief on Lasers (CD-ROM) Lasers Lasers are used in CD and DVD players, bar-code readers, eye surgery and multitudes of other systems and applications. A laser acronym for light amplification by stimulated emission of radiation is a source of monochromatic (single wavelength), coherent (uniform wavefront), narrow-beam light, in contrast with other sources of light (such as the sun or a light bulb) which usually encompass waves of many different wavelengths with random phase (incoherent). A laser source generating microwaves is called a maser. The first maser was built in 953 by Charles Townes and the first laser was constructed in 960 by Theodore Maiman.

353 35 Lesson #53 Chapter Section: 8-3 Topics: Fiber optics Highlights: Structure of an optical fiber Dispersion Special Illustrations: Example 8-5 Technology Brief on Bar-Code Reader (CD-ROM) Bar Code Readers A bar code consists of a sequence of parallel bars of certain widths, usually printed in black against a white background, configured to represent a particular binary code of information about a product and its manufacturer. Laser scanners can read the code and transfer the information to a computer, a cash register, or a display screen. For both stationary scanners built into checkout counters at grocery stores and handheld units that can be pointed at the bar-coded object like a gun, the basic operation of a bar-code reader is the same.

354 353 Lessons #54 and 55 Chapter Section: 8-4 Topics: Oblique incidence Highlights: Parallel and perpendicular polarizations Brewster angle Total internal reflection Special Illustrations: Example 8-6 and 8-7 CD-ROM Demos

355 354 Lesson #56 Chapter Section: 8-5 Topics: Reflectivity and transmissivity Highlights: Power relations Special Illustrations: Example 8-7

356 355 Lessons #57 59 Chapter Section: 8-6 to 8-0 Topics: Waveguides Highlights: TE and TM modes Cutoff frequency Phase and group velocities Special Illustrations: Examples 8-8, 8-9, and 8-0 Lesson #60 Chapter Section: 8- Topics: Cavity Resonators Highlights: Resonant frequency Q factor Applications

357 356 CHAPTER 8 Chapter 8 Section 8-: Reflection and Transmission at Normal Incidence Problem 8. A plane wave in air with an electric field amplitude of 0 V/m is incident normally upon the surface of a lossless, nonmagnetic medium with ε r = 5. Determine: (a) the reflection and transmission coefficients, (b) the standing-wave ratio in the air medium, and (c) the average power densities of the incident, reflected, and transmitted waves. Solution: (a) η = η 0 = 0π (Ω), η = η 0 = 0π = 4π (Ω). εr 5 From Eqs. (8.8a) and (8.9), (b) Γ = η η 4π 0π = η + η 4π + 0π = = 0.67, τ = + Γ = 0.67 = (c) According to Eqs. (8.9) and (8.0), S = + Γ Γ = = 5. S i av = Ei 0 η 0 = 400 0π = 0.5 W/m, S r av = Γ S i av = (0.67) 0.5 = 0.4 W/m, S t av = τ Ei 0 η = τ η η S i av = (0.33) 0π 4π 0.5 = 0.8 W/m. Problem 8. A plane wave traveling in medium with ε r =.5 is normally incident upon medium with ε r = 4. Both media are made of nonmagnetic, nonconducting materials. If the electric field of the incident wave is given by E i = ŷ8cos(6π 0 9 t 30πx) (V/m), (a) obtain time-domain expressions for the electric and magnetic fields in each of the two media, and

358 CHAPTER (b) determine the average power densities of the incident, reflected and transmitted waves. Solution: (a) E i = ŷ8cos(6π 0 9 t 30πx) (V/m), η = η 0 = η 0 = η 0 εr.5.5 = 377 = 5.33 Ω,.5 η = η 0 = η 0 = 377 = 88.5 Ω, εr 4 Γ = η η / /.5 = η + η / + /.5 = 0.43, τ = + Γ = 0.43 = 0.857, E r = ΓE i =.4ŷcos(6π 0 9 t + 30πx) (V/m). Note that the coefficient of x is positive, denoting the fact that E r belongs to a wave traveling in x-direction. E = E i + E r = ŷ[8cos(6π 0 9 t 30πx).4cos(6π 0 9 t + 30πx)] (A/m), H i = ẑ 8 η cos(6π 0 9 t 30πx) = ẑ3.83cos(6π 0 9 t 30πx) (ma/m), H r = ẑ.4 η cos(6π 0 9 t + 30πx) = ẑ4.54cos(6π 0 9 t + 30πx) (ma/m), H = H i + H r = ẑ[3.83cos(6π 0 9 t 30πx) cos(6π 0 9 t + 30πx)] (ma/m). Since k = ω µε and k = ω µε, ε k = k = ε 4 30π = 40π (rad/m),.5 E = E t = ŷ8τcos(6π 0 9 t 40πx) = ŷ6.86cos(6π 0 9 t 40πx) (V/m), H = H t = ẑ 8τ cos(6π 0 9 t 40πx) = ẑ36.38cos(6π 0 9 t 40πx) η (ma/m). (b) S i av 8 64 = ˆx = η 5.33 = ˆx7.3 (mw/m ), S r av = Γ S i av = ˆx(0.43) 0.7 = ˆx.6 (mw/m ),

359 358 CHAPTER 8 S t av = Et 0 η = ˆxτ (8) η = ˆx (0.86) = ˆx4.7 (mw/m ). Within calculation error, S i av + S r av = S t av. Problem 8.3 A plane wave traveling in a medium with ε r = 9 is normally incident upon a second medium with ε r = 4. Both media are made of nonmagnetic, nonconducting materials. If the magnetic field of the incident plane wave is given by H i = ẑcos(π 0 9 t ky) (A/m), (a) obtain time domain expressions for the electric and magnetic fields in each of the two media, and (b) determine the average power densities of the incident, reflected and transmitted waves. Solution: (a) In medium, u p = c εr = = 0 8 (m/s), k = ω u p = π 09 = 0π (rad/m), 08 H i = ẑcos(π 0 9 t 0πy) η = η 0 = 377 = 5.67 Ω, εr 3 η = η 0 = 377 = 88.5 Ω, εr E i = ˆxη cos(π 0 9 t 0πy) (A/m), = ˆx5.34cos(π 0 9 t 0πy) (V/m), Γ = η η = η + η = 0., τ = + Γ =., E r = ˆx cos(π 0 9 t + 0πy) = ˆx50.7cos(π 0 9 t + 0πy) (V/m),

360 CHAPTER H r = ẑ 50.7 cos(π 0 9 t + 0πy) η = ẑ0.4cos(π 0 9 t + 0πy) (A/m), E = E i + E r = ˆx[5.34cos(π 0 9 t 0πy) cos(π 0 9 t + 0πy)] (V/m), H = H i + H r = ẑ[cos(π 0 9 t 0πy) 0.4cos(π 0 9 t + 0πy)] In medium, ε 4 40π k = k = 0π = (rad/m), ε 9 3 ( E = E t = ˆx5.34τcos π 0 9 t 40πy ) 3 ( = ˆx30.6cos π 0 9 t 40πy ) (V/m), 3 H = H t = ẑ 30.6 ( cos π 0 9 t 40πy ) η 3 ( = ẑ.6cos π 0 9 t 40πy ) (A/m). 3 (b) S i av = ŷ E 0 η = ŷ (5.34) 5.67 = ŷ5.34 (W/m ), S r av = ŷ Γ (5.34) = ŷ0.05 (W/m ), S t av = ŷ( ) = ŷ4.9 (W/m ). (A/m). Problem 8.4 A 00-MHz left-hand circularly polarized plane wave with an electric field modulus of 5 V/m is normally incident in air upon a dielectric medium with ε r = 4 and occupying the region defined by z 0. (a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0. (b) Calculate the reflection and transmission coefficients. (c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z 0.

361 360 CHAPTER 8 (d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium. Solution: (a) LHC wave: Hence, (b) k = ω c π 08 = = 4π 3 rad/m, k = ω u p = ω c εr = 4π 3 4 = 8π 3 rad/m. Ẽ i = a 0 (ˆx + ŷe jπ/ )e jkz = a 0 (ˆx + jŷ)e jkz, E i (z,t) = ˆxa 0 cos(ωt kz) ŷa 0 sin(ωt kz), E i = [a 0 cos (ωt kz) + a 0 sin (ωt kz)] / = a 0 = 5 (V/m). Ẽ i = 5(ˆx + jŷ)e j4πz/3 (V/m). η = η 0 = 0π (Ω), η = η 0 = η 0 = 60π (Ω). εr Equations (8.8a) and (8.9) give (c) Γ = η η 60π 0π = η + η 60π + 0π = = 3, τ = + Γ = 3. Ẽ r = 5Γ(ˆx + jŷ)e jkz = 5 j4πz/3 (ˆx + jŷ)e (V/m), 3 Ẽ t = 5τ(ˆx + jŷ)e jkz = 0 j8πz/3 (ˆx + jŷ)e (V/m), 3 Ẽ = Ẽ i + Ẽ r = 5(ˆx + jŷ) [e j4πz/3 3 ] e j4πz/3 (V/m). (d) % of reflected power = 00 Γ = 00 9 =.%, ( 3 % of transmitted power = 00 τ η η = 00 ) 0π 60π = 88.89%.

362 CHAPTER 8 36 Problem 8.5 Repeat Problem 8.4 after replacing the dielectric medium with a poor conductor characterized by ε r =.5, µ r =, and σ = 0 4 S/m. Solution: (a) Medium : Medium : η = η 0 = 0π (Ω), k = ω c = π = 4π 3 σ π = ωε π = Hence, medium is a low-loss dielectric. From Table 7-, (rad/m). α = σ = σ µ ε 0π = σ εr 0π = 0 4 0π.5.5 =.6 0 (NP/m), β = ω µ ε = ω ε r = π (rad/m), ( c µ η = + jσ ) = 0π ( + j 0 3 ) 0π = 80π (Ω). ε ωε εr.5 LHC wave: Ẽ i = a 0 (ˆx + jŷ)e jk z, Ẽ i = a 0 = 5 (V/m), Ẽ i = 5(ˆx + jŷ)e j4πz/3 (V/m). (b) According to Eqs. (8.8a) and (8.9), (c) Γ = η η 80π 0π = = 0., τ = + Γ = 0. = 0.8. η + η 80π + 0π Ẽ r = 5Γ(ˆx + jŷ)e jkz = (ˆx + jŷ)e j4πz/3 (V/m), Ẽ t = 5τ(ˆx + jŷ)e αz e jβzz = 4(ˆx + jŷ)e.6 0 z e jπz Ẽ = Ẽ i + Ẽ r = 5(ˆx + jŷ)[e j4πz/3 0.e j4πz/3 ] (V/m). (V/m),

363 36 CHAPTER 8 (d) % of reflected power = 00 Γ = 00(0.) = 4%, % of transmitted power = 00 τ η η = 00(0.8) 0π 80π = 96%. Problem 8.6 A 50-MHz plane wave with electric field amplitude of 50 V/m is normally incident in air onto a semi-infinite, perfect dielectric medium with ε r = 36. Determine (a) Γ, (b) the average power densities of the incident and reflected waves, and (c) the distance in the air medium from the boundary to the nearest minimum of the electric field intensity, E. Solution: (a) η = η 0 = 0π (Ω), µ η = = 0π = 0π = 0π ε εr 6 (Ω), Γ = η η 0π 0π = η + η 0π + 0π = 0.7. Hence, Γ = 0.7 and θ η = 80. (b) (c) In medium (air), From Eqs. (8.6) and (8.7), S i av = Ei 0 η = (50) 0π = 3.3 (W/m ), S r av = Γ S i av = (0.7) 3.3 =.67 (W/m ). λ = c f = = 6 m. l max = θ rλ 4π = π 6 =.5 m, 4π l min = l max λ 4 =.5.5 = 0 m (at the boundary). Problem 8.7 What is the maximum amplitude of the total electric field in the air medium of Problem 8.6, and at what nearest distance from the boundary does it occur?

364 CHAPTER Solution: From Problem 8.6, Γ = 0.7 and λ = 6 m. Ẽ max = ( + Γ )E0 i = ( + 0.7) 50 = 85.5 V/m, l max = θ rλ 4π = π 6 =.5 m. 4π Problem 8.8 Repeat Problem 8.6 after replacing the dielectric medium with a conductor with ε r =, µ r =, and σ = S/m. Solution: (a) Medium : Medium : η = η 0 = 0π = 377 (Ω), λ = c f = = 6 m, σ = π ωε π =. 0 9 Hence, Medium is a quasi-conductor. From Eq. (7.70), µ η = ε (b) ( j ε ε ) / = 0π( j σ ) / ωε = 0π( j) / = 0π( ) / e j.5 = ( j.3) (Ω). Γ = η η ( j.3) 377 = η + η ( j.3) = j0. = (c) In medium (air), S i av = Ei 0 η = 50 0π = 3.3 (W/m ), S r av = Γ S i av = (0.) (3.3) = 0.6 (W/m ). λ = c f = = 6 m. For θ r = 4.5 = rad, Eqs. (8.6) and (8.7) give l max = θ rλ 4π + (0)λ = (6) = 3 m,

365 364 CHAPTER 8 l min = l max λ 4 = 3 6 = 3.5 =.5 m. 4 Problem 8.9 The three regions shown in Fig. 8-3 (P8.9) contain perfect dielectrics. For a wave in medium incident normally upon the boundary at z = d, what combination of ε r and d produce no reflection? Express your answers in terms of ε r, ε r3 and the oscillation frequency of the wave, f. d Medium Medium Medium 3 z ε r ε r ε r3 z = -d z = 0 Figure P8.9: Three dielectric regions. Solution: By analogy with the transmission-line case, there will be no reflection at z = d if medium acts as a quarter-wave transformer, which requires that and The second condition may be rewritten as d = λ 4 η = η η 3. [ ] η 0 η0 η / 0 =, or ε r = ε r ε r3, εr εr εr3 λ = λ 0 εr = c f ε r = c f (ε r ε r3 ) /4, and d = c 4 f (ε r ε r3 ) /4.

366 CHAPTER Problem 8.0 For the configuration shown in Fig. 8-3 (P8.9), use transmissionline equations (or the Smith chart) to calculate the input impedance at z = d for ε r =, ε r = 9, ε r3 = 4, d =. m, and f = 50 MHz. Also determine the fraction of the incident average power density reflected by the structure. Assume all media are lossless and nonmagnetic. Solution: In medium, Hence, λ = λ 0 εr = c f ε r = = m. β = π λ = π rad/m, β d =.π rad. At z = d, the input impedance of a transmission line with load impedance Z L is given by Eq. (.63) as ( ) ZL + jz 0 tanβ d Z in ( d) = Z 0. Z 0 + jz L tanβ d In the present case, Z 0 = η = η 0 / ε r = η 0 /3 and Z L = η 3 = η 0 / ε r3 = η 0 /, where η 0 = 0π (Ω). Hence, ( ) ( η3 + jη tan β d Z in ( d) = η = η 0 + j ( ) 3) tan.π η + jη 3 tan β d j ( ) = η 0 (0.35 j0.4). tan.π At z = d, Γ = Z in Z Z in + Z = η 0(0.35 j0.4) η 0 η 0 (0.35 j0.4) + η 0 = 0.49e j6.4. Fraction of incident power reflected by the structure is Γ = 0.49 = 0.4. Problem 8. Repeat Problem 8.0 after interchanging ε r and ε r3. Solution: In medium, λ = λ 0 εr = c f ε r = = m.

367 366 CHAPTER 8 Hence, β = π λ = π rad/m, β d =.π rad. At z = d, the input impedance of a transmission line with impedance Z L is given as Eq. (.63), ( ) ZL + jz 0 tanβd Z in ( d) = Z 0. Z 0 + jz L tanβd In the present case, Z 0 = η = η 0 / ε r = η 0 /3, η 0 = 0π (Ω). Hence, ( ) η3 + jη tan.π Z in ( d) = η η + jη 3 tan.π = η ( ) 0 + ( j/3)tan.π 3 (/3) + j tan.π At z = d, = η 0 ( + ( j/3)tan.π + j3tan.π Γ = Z in Z Z in + Z = Z L = η 3 = η 0 / ε r = η 0, where ) = (0.66 j0.337)η 0 = 0.43η = Fraction of incident power reflected by structure is Γ = 0.4. Problem 8. Orange light of wavelength 0.6 µm in air enters a block of glass with ε r =.44. What color would it appear to a sensor embedded in the glass? The wavelength ranges of colors are violet (0.39 to 0.45 µm), blue (0.45 to 0.49 µm), green (0.49 to 0.58 µm), yellow (0.58 to 0.60 µm), orange (0.60 to 0.6 µm), and red (0.6 to 0.78 µm). Solution: In the glass, The light would appear green. λ = λ 0 εr = = µm. Problem 8.3 A plane wave of unknown frequency is normally incident in air upon the surface of a perfect conductor. Using an electric-field meter, it was determined that the total electric field in the air medium is always zero when measured at a

368 CHAPTER distance of m from the conductor surface. Moreover, no such nulls were observed at distances closer to the conductor. What is the frequency of the incident wave? Solution: The electric field of the standing wave is zero at the conductor surface, and the standing wave pattern repeats itself every λ/. Hence, in which case λ = m, or λ = 4 m, f = c λ = = = 75 MHz. Problem 8.4 Consider a thin film of soap in air under illumination by yellow light with λ = 0.6 µm in vacuum. If the film is treated as a planar dielectric slab with ε r =.7, surrounded on both sides by air, what film thickness would produce strong reflection of the yellow light at normal incidence? Solution: The transmission line analogue of the soap-bubble wave problem is shown in Fig. P8.4(b) where the load Z L is equal to η 0, the impedance of the air medium on the other side of the bubble. That is, The normalized load impedance is η 0 = 377 Ω, η = = 87.5 Ω. z L = η 0 η =.3. For the reflection by the soap bubble to be the largest, Z in needs to be the most different from η 0. This happens when z L is transformed through a length λ/4. Hence, L = λ 4 = λ 0 4 ε r = 0.6 µm 4 = 0.5 µm,.7 where λ is the wavelength of the soap bubble material. Strong reflections will also occur if the thickness is greater than L by integer multiples of nλ/ = (0.3 n) µm. Hence, in general L = ( n) µm, n = 0,,,.... According to Section -7.5, transforming a load Z L = 377 Ω through a λ/4 section of Z 0 = 87.5 Ω ends up presenting an input impedance of Z in = Z 0 = (87.5) = 9.5 Ω. Z L 377

369 368 CHAPTER 8 Air Soap Air ε r = ε r =.7 ε r3 = Yellow Light λ = 0.6 µm L (a) Yellow light incident on soap bubble. η 0 = 377Ω η Z L = η 0 = 377 Ω (b) Transmission-line equivalent circuit Figure P8.4: Diagrams for Problem 8.4. This Z in is at the input side of the soap bubble. The reflection coefficient at that interface is Γ = Z in η = Z in + η = 0.7. Any other thickness would produce a reflection coeffficient with a smaller magnitude. Problem 8.5 A 5-MHz plane wave with electric field amplitude of 0 (V/m) is normally incident in air onto the plane surface of a semi-infinite conducting material with ε r = 4, µ r =, and σ = 00 (S/m). Determine the average power dissipated (lost) per unit cross-sectional area in a -mm penetration of the conducting medium. Solution: For convenience, let us choose E i to be along ˆx and the incident direction to be +ẑ. With k = ω π 5 06 = c = π (rad/m), 30

370 CHAPTER we have From Table 7-, E i = ˆx0cos (π 0 7 t π ) 30 z η = η 0 = 377 Ω. (V/m), ε ε = σ ωε r ε 0 = 00 36π π = 9 04, which makes the material a good conductor, for which α = π f µσ = π π = β = η c = ( + j) α σ (rad/m), = ( + j) = 0.44( + j) Ω. (Np/m), According to the expression for S av given in the answer to Exercise 8.3, ( ) S av = ẑ τ Ei 0 e α z Re. The power lost is equal to the difference between S av at z = 0 and S av at z = mm. Thus, where z = mm. τ = + Γ η c P = power lost per unit cross-sectional area = S av (0) S av (z = mm) ( ) = τ Ei 0 Re [ e α z ] η c = + η η 0.44( + j) 377 = + η + η 0.44( + j) ( + j) = e j45. ( ) ( ) Re η = Re c 0.44( + j) ( ) ( ) + j = Re = Re = 0.44( j) =.4, P = ( ) 0.4[ e ] = (W/m ).

371 370 CHAPTER 8 Problem 8.6 A 0.5-MHz antenna carried by an airplane flying over the ocean surface generates a wave that approaches the water surface in the form of a normally incident plane wave with an electric-field amplitude of 3,000 (V/m). Sea water is characterized by ε r = 7, µ r =, and σ = 4 (S/m). The plane is trying to communicate a message to a submarine submerged at a depth d below the water surface. If the submarine s receiver requires a minimum signal amplitude of 0.0 (µv/m), what is the maximum depth d to which successful communication is still possible? Solution: For sea water at 0.5 MHz, ε ε = σ ωε = 4 36π π = Hence, sea water is a good conductor, in which case we use the following expressions from Table 7-: α = π f µσ = π π =.8 (Np/m), β =.8 η c = ( + j) α σ (rad/m), Γ = η η η + η =.8 = ( + j) = 0.7( + j) Ω, 4 0.7( + j) ( + j) = ( j ), τ = + Γ = e j44.89, E t = τe i 0e α d. We need to find the depth z at which E t = 0.0 µv/m = 0 8 V/m. 0 8 = e.8d, e.8d = ,.8d = ln( ) =.8, or d = 7.54 (m).

372 CHAPTER 8 37 Sections 8- and 8-3: Snell s Laws and Fiber Optics Problem 8.7 A light ray is incident on a prism at an angle θ as shown in Fig (P8.7). The ray is refracted at the first surface and again at the second surface. In terms of the apex angle φ of the prism and its index of refraction n, determine the smallest value of θ for which the ray will emerge from the other side. Find this minimum θ for n = 4 and φ = 60. B θ A φ n θ θ 3 C Figure P8.7: Prism of Problem 8.7. Solution: For the beam to emerge at the second boundary, it is necessary that θ 3 < θ c, where sinθ c = /n. From the geometry of triangle ABC, 80 = φ + (90 θ ) + (90 θ 3 ), or θ = φ θ 3. At the first boundary, sinθ = nsin θ. Hence, ( ( )) sinθ min = nsin(φ θ 3 ) = nsin φ sin, n or ( ( ))] θ min = sin [nsin φ sin. n For n = 4 and φ = 60, [ ( )] θ min = sin 4sin(60 sin =

373 37 CHAPTER 8 Problem 8.8 For some types of glass, the index of refraction varies with wavelength. A prism made of a material with n = λ 0, (λ 0 in µm), where λ 0 is the wavelength in vacuum, was used to disperse white light as shown in Fig (P8.8). The white light is incident at an angle of 50, the wavelength λ 0 of red light is 0.7 µm and that of violet light is 0.4 µm. Determine the angular dispersion in degrees. B 50 A θ 60 θ 3 C θ 4 Angular dispersion Red Green Violet Figure P8.8: Prism of Problem 8.8. Solution: For violet, or n v = =.66, sinθ = sinθ sin 50 = n v.66, From the geometry of triangle ABC, θ = = 60 + (90 θ ) + (90 θ 3 ), or and or θ 3 = 60 θ = = 3.5, sinθ 4 = n v sinθ 3 =.66sin 3.5 = 0.89, θ 4 = 63.8.

374 CHAPTER For red, n r = =.6, [ 30 ] sin50 θ = sin = 8.,.6 θ 3 = = 3.78, θ 4 = sin [.6sin 3.78 ] = Hence, angular dispersion = = 4.6. Problem 8.9 The two prisms in Fig (P8.9) are made of glass with n =.5. What fraction of the power density carried by the ray incident upon the top prism emerges from bottom prism? Neglect multiple internal reflections. S i S t Figure P8.9: Periscope problem. Solution: Using η = η 0 /n, at interfaces and 4, Γ a = n n n + n = = 0.. At interfaces 3 and 6, Γ b = Γ a = 0..

375 374 CHAPTER 8 At interfaces and 5, θ c = sin ( n ) ( ) = sin = Hence, total internal reflection takes place at those interfaces. At interfaces, 3, 4 and 6, the ratio of power density transmitted to that incident is ( Γ ). Hence, S t S i = ( Γ ) 4 = ( (0.) ) 4 = Problem 8.0 A light ray incident at 45 passes through two dielectric materials with the indices of refraction and thicknesses given in Fig (P8.0). If the ray strikes the surface of the first dielectric at a height of cm, at what height will it strike the screen? cm n = n =.5 n 3 =.3 θ 45 h 45 h n 4 = θ 4 θ 3 h 3 3cm 4cm 5cm h 4 screen Figure P8.0: Light incident on a screen through a multi-layered dielectric (Problem 8.0). Solution: Hence, Hence, sinθ = n n sinθ =.5 sin45 = θ = 8.3, h = 3 cm tanθ = 3 cm 0.53 =.6 cm, sinθ 3 = n sinθ =.5 n 3.3 sin8.3 = θ 3 = 3.96,

376 CHAPTER Hence, h 3 = 4 cm tan3.96 =.6 cm, sinθ 4 = n 3 n 4 sinθ 3 = θ 4 = 45, h 4 = 5 cm tan45 = 5 cm. Total height = h + h + h 3 + h 4 = ( ) =. cm. Problem 8. Figure P8. depicts a beaker containing a block of glass on the bottom and water over it. The glass block contains a small air bubble at an unknown depth below the water surface. When viewed from above at an angle of 60, the air bubble appears at a depth of 6.8 cm. What is the true depth of the air bubble? 60 x d a θ 0 cm θ d t x θ 3 x d Figure P8.: Apparent position of the air bubble in Problem 8.. Solution: Let d a = 6.8 cm = apparent depth, d t = true depth. [ ] [ ] θ = sin n sinθ i = sin sin 60 = 40.6, n.33

377 376 CHAPTER 8 Hence, and [ ] [ ] θ 3 = sin n sinθ i = sin n 3.6 sin60 = 3.77, x = (0 cm) tan40.6 = 8.58 cm, Hence, d t = (0 + 5) = 5 cm. x = d a cot30 = 6.8cot 30 =.8 cm. x = x x = = 3. cm, d = x cot3.77 = (3. cm) cot3.77 = 5 cm. Problem 8. A glass semicylinder with n =.5 is positioned such that its flat face is horizontal, as shown in Fig (P8.). Its horizontal surface supports a drop of oil, as shown. When light is directed radially toward the oil, total internal reflection occurs if θ exceeds 53. What is the index of refraction of the oil? n oil oil drop θ n glass =.5 Figure P8.: Oil drop on the flat surface of a glass semicylinder (Problem 8.). Solution: sinθ c = n n = n oil.5, n oil =.5sin 53 =.. Problem 8.3 A penny lies at the bottom of a water fountain at a depth of 30 cm. Determine the diameter of a piece of paper which, if placed to float on the surface of

378 CHAPTER x d x water surface θ c 30 cm θ c Figure P8.3: Light cone bounded by total internal reflection. the water directly above the penny, would totally obscure the penny from view. Treat the penny as a point and assume that n =.33 for water. Solution: [ ] θ c = sin = 48.75,.33 d = x = [(30 cm)tanθ c ] = (60 cm) tan48.75 = 68.4 cm. Problem 8.4 Suppose the optical fiber of Example 8-5 is submerged in water (with n =.33) instead of air. Determine θ a and f p in that case. Solution: With n 0 =.33, n f =.5 and n c =.49, Eq. (8.40) gives or sinθ a = (n f n n c )/ = [ (.5) (.49) ] / = 0.3, 0.33 θ a = 3.. The data rate f p given by Eq. (8.45) is not a function of n 0, and therefore it remains unchanged at 4.9 (Mb/s).

379 378 CHAPTER 8 Problem 8.5 Equation (8.45) was derived for the case where the light incident upon the sending end of the optical fiber extends over the entire acceptance cone shown in Fig. 8-(b). Suppose the incident light is constrained to a narrower range extending between normal incidence and θ, where θ < θ a. (a) Obtain an expression for the maximum data rate f p in terms of θ. (b) Evaluate f p for the fiber of Example 8-5 when θ = 5. Solution: (a) For θ i = θ, sinθ = n f sin θ, l max = l l = = cosθ sin θ t max = l max u p = l maxn f c t min = l u p = l n f c, τ = t = t max t min = l n f c ln f =, c n f (sinθ ) n f l ln f ( ) =, sinθ n f n (sinθ ) f, n f (sinθ ) f p = τ = c n f ln f n f (sinθ ) (bits/s). (b) For: n f =.5, θ = 5, l = km, c = m/s, f p = (Mb/s).

380 CHAPTER Sections 8-4 and 8-5: Reflection and Transmission at Oblique Incidence Problem 8.6 A plane wave in air with Ẽ i = ŷ0e j(3x+4z) (V/m), is incident upon the planar surface of a dielectric material, with ε r = 4, occupying the half space z 0. Determine: (a) the polarization of the incident wave, (b) the angle of incidence, (c) the time-domain expressions for the reflected electric and magnetic fields, (d) the time-domain expressions for the transmitted electric and magnetic fields, and (e) the average power density carried by the wave in the dielectric medium. Solution: (a) Ẽ i = ŷ0e j(3x+4z) V/m. Since E i is along ŷ, which is perpendicular to the plane of incidence, the wave is perpendicularly polarized. (b) From Eq. (8.48a), the argument of the exponential is Hence, from which we determine that jk (xsin θ i + zcosθ i ) = j(3x + 4z). k sinθ i = 3, k cosθ i = 4, tanθ i = 3 4 or θ i = 36.87, and Also, (c) k = = 5 (rad/m). ω = u p k = ck = = (rad/s). η = η 0 = 377 Ω, η = η 0 = η 0 = 88.5 Ω, εr [ ] [ θ t = sin sinθi sin36.87 = sin ] = 7.46, εr 4

381 380 CHAPTER 8 Γ = η cos θ i η cosθ t η cos θ i + η cosθ t = 0.4, τ = + Γ = In accordance with Eq. (8.49a), and using the relation E r 0 = Γ E i 0, Ẽ r = ŷ8.e j(3x 4z), H r = (ˆx cosθ i + ẑsinθ i ) 8. η 0 e j(3x 4z), where we used the fact that θ i = θ r and the z-direction has been reversed. E r = Re[Ẽ r e jωt ] = ŷ8.cos( t 3x + 4z) (V/m), H r = (ˆx7.4 + ẑ3.06)cos( t 3x + 4z) (ma/m). (d) In medium, k = k ε ε = 5 4 = 0 (rad/m), and [ ] [ ] θ t = sin ε sinθ i = sin ε sin = 7.46 and the exponent of E t and H t is jk (xsin θ t + zcosθ t ) = j0(xsin zcos7.46 ) = j(3x z). Hence, Ẽ t = ŷ0 0.59e j(3x+9.54z), H t = ( ˆxcosθ t + ẑsin θ t ) e j(3x+9.54z). η E t = Re[Ẽ t e jωt ] = ŷ.8cos( t 3x 9.54z) (V/m), H t = ( ˆxcos ẑsin7.46 ) cos(.5 09 t 3x 9.54z) = ( ˆx ẑ8.78) cos( t 3x 9.54z) (ma/m). (e) S t av = Et 0 η = (.8) 88.5 = 0.36 (W/m ).

382 CHAPTER 8 38 Problem 8.7 Repeat Problem 8.6 for a wave in air with H i = ŷ 0 e j(8x+6z) (A/m), incident upon the planar boundary of a dielectric medium (z 0) with ε r = 9. Solution: (a) H i = ŷ 0 e j(8x+6z). Since H i is along ŷ, which is perpendicular to the plane of incidence, the wave is TM polarized, or equivalently, its electric field vector is parallel polarized (parallel to the plane of incidence). (b) From Eq. (8.65b), the argument of the exponential is jk (xsin θ i + zcosθ i ) = j(8x + 6z). Hence, from which we determine k sinθ i = 8, k cosθ i = 6, ( ) 8 θ i = tan = 53.3, 6 k = = 0 (rad/m). Also, (c) ω = u p k = ck = = (rad/s). η = η 0 = 377 Ω, η = η 0 = η 0 = 5.67 Ω, εr 3 [ ] [ θ t = sin sinθi sin53.3 = sin ] = 5.47, εr 9 Γ = η cosθ t η cosθ i η cosθ t + η cosθ i = 0.30, τ = ( + Γ ) cosθ i cosθ t = In accordance with Eqs. (8.65a) to (8.65d), E i 0 = 0 η and Ẽ i = (ˆxcos θ i ẑsinθ i ) 0 η e j(8x+6z) = (ˆx4.5 ẑ6.03)e j(8x+6z).

383 38 CHAPTER 8 Ẽ r is similar to Ẽ i except for reversal of z-components and multiplication of amplitude by Γ. Hence, with Γ = 0.30, E r = Re[Ẽ r e jωt ] = (ˆx.36 + ẑ.8)cos(3 0 9 t 8x + 6z) V/m, H r = ŷ 0 Γ cos(3 0 9 t 8x + 6z) (d) In medium, = ŷ0.6 0 cos(3 0 9 t 8x + 6z) A/m. ε k = k = 0 9 = 30 rad/m, ε θ t = sin [ ε ε sinθ i and the exponent of E t and H t is ] = sin [ 3 sin53.3 ] = 5.47, jk (xsin θ t + zcosθ t ) = j30(xsin zcos5.47 ) = j(8x + 8.9z). Hence, (e) Ẽ t = (ˆxcosθ t ẑsin θ t )E i 0 τ e j(8x+8.9z) = (ˆx0.96 ẑ0.7) e j(8x+8.9z) = (ˆx3.8 ẑ0.90)e j(8x+8.9z), H t = ŷ Ei 0 τ e j(8x+8.9z) η = ŷ.64 0 e j(8x+8.9z), E t = Re{Ẽ t e jωt } = (ˆx3.8 ẑ0.90)cos(3 0 9 t 8x 8.9z) V/m, H t = ŷ.64 0 cos(3 0 9 t 8x 8.9z) A/m. Sav t = Et 0 = Ht 0 η = (.64 0 ) 5.67 = 44 mw/m. η Problem 8.8 Natural light is randomly polarized, which means that, on average, half the light energy is polarized along any given direction (in the plane orthogonal

384 CHAPTER to the direction of propagation) and the other half of the energy is polarized along the direction orthogonal to the first polarization direction. Hence, when treating natural light incident upon a planar boundary, we can consider half of its energy to be in the form of parallel-polarized waves and the other half as perpendicularly polarized waves. Determine the fraction of the incident power reflected by the planar surface of a piece of glass with n =.5 when illuminated by natural light at 70. Solution: Assume the incident power is W. Hence: Incident power with parallel polarization = 0.5 W, Incident power with perpendicular polarization = 0.5 W. ε /ε = (n /n ) = n =.5 =.5. Equations (8.60) and (8.68) give Γ = cos 70.5 sin 70 cos sin 70 = 0.55, Γ =.5cos sin 70 = 0...5cos sin 70 Reflected power with parallel polarization = 0.5(Γ ) = 0.5(0.) = mw, Reflected power with perpendicular polarization = 0.5(Γ ) = 0.5(0.55) = 5.3 mw. Total reflected power = = 73.3 mw, or 7.33%.. Problem 8.9 A parallel polarized plane wave is incident from air onto a dielectric medium with ε r = 9 at the Brewster angle. What is the refraction angle? θ ε r = θ ε r = 9 Figure P8.9: Geometry of Problem 8.9. Solution: For nonmagnetic materials, Eq. (8.7) gives θ = θ B = tan ε ε = tan 3 = 7.57.

385 384 CHAPTER 8 But or θ = sin θ = sin θ = sinθ = sin7.57 = 0.3, εr 3 3 Problem 8.30 A perpendicularly polarized wave in air is obliquely incident upon a planar glass-air interface at an incidence angle of 30. The wave frequency is 600 THz ( THz = 0 Hz), which corresponds to green light, and the index of refraction of the glass is.6. If the electric field amplitude of the incident wave is 50 V/m, determine (a) the reflection and transmission coefficients, and (b) the instantaneous expressions for E and H in the glass medium. Solution: (a) For nonmagnetic materials, (ε /ε ) = (n /n ). Using this relation in Eq. (8.60) gives Γ = cos θ i (n /n ) sin θ i cos30 (.6) sin 30 = = 0.7, cos θ i + (n /n ) sin θ i cos30 + (.6) sin 30 τ = + Γ = 0.7 = (b) In the glass medium, or θ t = 8.. sinθ t = sin θ i n = sin 30.6 = 0.3, η = µ = η 0 = 0π = 75π = 35.6 (Ω), ε n.6 k = ω = π f u p c/n = π f n = π c = 6.4π 0 6 rad/m, E0 t = τ E0 i = = 36.5 V/m. From Eqs. (8.49c) and (8.49d), Ẽ t = ŷet 0e jk (xsinθ t +zcosθ t ), H t = ( ˆxcos θ t + ẑsinθ t ) Et 0 η e jk (xsinθ t +zcosθ t ),

386 CHAPTER and the corresponding instantaneous expressions are: E t (x,z,t) = ŷ36.5cos(ωt k xsin θ t k zcosθ t ) (V/m), H t (x,z,t) = ( ˆxcosθ t ẑcosθ t )0.6cos(ωt k xsin θ t k zcos θ t ) (A/m), with ω = π 0 5 rad/s and k = 6.4π 0 6 rad/m. Problem 8.3 Show that the reflection coefficient Γ can be written in the form Γ = sin(θ t θ i ) sin(θ t + θ i ). Solution: From Eq. (8.58a), Γ = η cos θ i η cosθ t η cos θ i + η cosθ t = (η /η )cos θ i cosθ t (η /η )cos θ i + cosθ t. Using Snell s law for refraction given by Eq. (8.3), we have η η = sinθ t sinθ i, we have Γ = sinθ t cos θ i cosθ t sinθ i sinθ t cos θ i + cosθ t sinθ i = sin(θ t θ i ) sin(θ t + θ i ). Problem 8.3 Show that for nonmagnetic media, the reflection coefficient Γ can be written in the form Γ = tan(θ t θ i ) tan(θ t + θ i ). Solution: From Eq. (8.66a), Γ is given by Γ = η cos θ t η cosθ i η cos θ t + η cosθ i = (η /η )cos θ t cosθ i (η /η )cos θ t + cosθ i. For nonmagnetic media, µ = µ = µ 0 and η η = ε ε = n n.

387 386 CHAPTER 8 Snell s law of refraction is Hence, Γ = sinθ t sinθ i = n n. sinθ t cos θ t cosθ i sinθ i sinθ t cosθ t sinθ i cos θ i =. sinθ t sinθ cos θ t + cosθ t cosθ t + sinθ i cos θ i i sinθ i To show that the expression for Γ is the same as Γ = tan(θ t θ i ) tan(θ t + θ i ), we shall proceed with the latter and show that it is equal to the former. Using the identities (from Appendix C): tan(θ t θ i ) tan(θ t + θ i ) = sin(θ t θ i )cos(θ t + θ i ) cos(θ t θ i )sin(θ t + θ i ). sin xcos y = sin(x + y) + sin(x y), and if we let x = θ t θ i and y = θ t +θ i in the numerator, while letting x = θ t +θ i and y = θ t θ i in the denominator, then tan(θ t θ i ) tan(θ t + θ i ) = sin(θ t) + sin( θ i ) sin(θ t ) + sin(θ i ). But sinθ = sin θcos θ, and sin( θ) = sinθ, hence, which is the intended result. tan(θ t θ i ) tan(θ t + θ i ) = sinθ t cosθ t sinθ i cos θ i sinθ t cosθ t + sinθ i cos θ i, Problem 8.33 A parallel polarized beam of light with an electric field amplitude of 0 (V/m) is incident in air on polystyrene with µ r = and ε r =.6. If the incidence angle at the air polystyrene planar boundary is 50, determine (a) the reflectivity and transmissivity, and (b) the power carried by the incident, reflected, and transmitted beams if the spot on the boundary illuminated by the incident beam is m in area.

388 CHAPTER Solution: (a) From Eq. (8.68), Γ = (ε /ε )cosθ i + (ε /ε ) sin θ i (ε /ε )cos θ i + (ε /ε ) sin θ i =.6cos sin 50 = 0.08,.6cos sin 50 R = Γ = (0.08) = , (b) T = R = P i = Ei 0 η Acosθ i = (0) 0π cos50 = 85 mw, P r = R P i = ( ) = 0.55 mw, P t = T P i = = mw. Sections 8-6 to 8- Problem 8.34 Derive Eq. (8.89b). Solution: We start with Eqs. (8.88a and e), ẽ z y + jβẽ y = jωµ h x, jβ h x h z x = jωεẽ y. To eliminate h x, we multiply the top equation by β and the bottom equation by ωµ, and then we add them together. The result is: β ẽ z y + jβ ẽ y ωµ h z x = jω µεẽ y.

389 388 CHAPTER 8 Multiplying all terms by e jβz to convert ẽ y to Ẽ y (and similarly for the other field components), and then solving for Ẽ y leads to ( ) Ẽ y = = j k c j(β ω µε) ( β Ẽ z y + ωµ H z x β Ẽ z y + ωµ H z x ), where we used the relation k c = ω µε β. Problem 8.35 A hollow rectangular waveguide is to be used to transmit signals at a carrier frequency of 6 GHz. Choose its dimensions so that the cutoff frequency of the dominant TE mode is lower than the carrier by 5% and that of the next mode is at least 5% higher than the carrier. Solution: For m = and n = 0 (TE 0 mode) and u p0 = c (hollow guide), Eq. (8.06) reduces to f 0 = c a. Denote the carrier frequency as f 0 = 6 GHz. Setting f 0 = 0.75 f 0 = GHz = 4.5 GHz, we have a = c = 3 08 = 3.33 cm. f If b is chosen such that a > b > a, the second mode will be TE 0, followed by TE 0 at f 0 = 9 GHz. For TE 0, f 0 = c b. Setting f 0 =.5 f 0 = 7.5 GHz, we get b = c f 0 = = cm.

390 CHAPTER Problem 8.36 A TE wave propagating in a dielectric-filled waveguide of unknown permittivity has dimensions a = 5 cm and b = 3 cm. If the x-component of its electric field is given by E x = 36cos(40πx)sin(00πy) determine: (a) the mode number, (b) ε r of the material in the guide, (c) the cutoff frequency, and (d) the expression for H y. sin(.4π 0 0 t 5.9πz), (V/m) Solution: (a) Comparison of the given expression with Eq. (8.0a) reveals that Mode is TE 3. (b) From sin(ωt βz), we deduce that mπ = 40π, hence m = a nπ = 00π, hence n = 3. b ω =.4π 0 0 rad/s, β = 5.9π rad/m. Using Eq. (8.05) to solve for ε r, we have [ ε r = c ( mπ ) ( nπ ) ] ω β + + a b (c) =.5. u p0 = c = 3 08 = 0 8 m/s. εr.5 ( f 3 = u ) ( ) p a b = 0.77 GHz.

391 390 CHAPTER 8 (d) Z TE = E / x = η ( f 3 / f ) H y = 377 ( ) 0.77 / = Ω. εr Hence, H y = E x Z TE = 0.063cos(40πx)sin(00πy)sin(.4π 0 0 t 5.9πz) (A/m). Problem 8.37 A waveguide filled with a material whose ε r =.5 has dimensions a = cm and b =.4 cm. If the guide is to transmit 0.5-GHz signals, what possible modes can be used for the transmission? Solution: Application of Eq. (8.06) with u p0 = c/ ε r = /.5 = 0 8 m/s, gives: f 0 = 5 GHz (TE only) f 0 = 7.4 GHz (TE only) f = 8.7 GHz (TE or TM) f 0 = 0 GHz (TE only) f =.8 GHz (TE or TM) f = 5. GHz (TE or TM). Hence, any one of the first four modes can be used to transmit 0.5-GHz signals. Problem 8.38 For a rectangular waveguide operating in the TE 0 mode, obtain expressions for the surface charge density ρ s and surface current density J s on each of the four walls of the guide. Solution:

392 CHAPTER 8 39 For TE 0, the expressions for Ẽ and H are given by Eq. (8.0) with m = and n = 0, Ẽ x = 0, Ẽ y = j ωµπh 0 k c a ( πx ) sin e jβz, a Ẽ z = 0, H x = j βπh ( 0 πx ) kca sin e jβz, a H y = 0, ( πx ) H z = H 0 cos e jβz. a The applicable boundary conditions are given in Table 6-. At the boundary between a dielectric (medium ) and a conductor (medium ), ρ s = ˆn D = ε ˆn Ẽ, J s = ˆn H, where Ẽ and H are the fields inside the guide, ε is the permittivity of the material filling the guide, and ˆn is the normal to the guide wall, pointing away from the wall (inwardly). In view of the coordinate system defined for the guide, ˆn = ˆx for side wall at x = 0, ˆn = ˆx for wall at x = a, etc. b y n^ 4 n^ n^ x a 3 n^ 0 (a) At side wall at x = 0, ˆn = ˆx. Hence, ρ s = ε ˆx ŷe y x=0 = 0 J s = ˆx (ˆx H x + ẑ H z ) x=0 = ŷ H z x=0 = ŷh 0 e jβz.

393 39 CHAPTER 8 (b) At side wall at x = a, ˆn = ˆx. Hence, ρ s = 0 J s = ŷh 0 e jβz. (c) At bottom surface at y = 0, ˆn = ŷ. Hence, ρ s = ε ŷ ŷe y y=0 = j ωεµπh 0 kca sin J s = ŷ (ˆx H x + ẑ H z ) ( πx = H 0 [ˆxcos a ( πx ) e jβz a (d) At top surface at y b, ˆn = ŷ. Hence, ρ s = j ωεµπh 0 k ca J s = H 0 [ ˆxcos ( πx ) sin a ( πx a ) ẑ j βπ ( πx kca sin a e jβz ) + ẑ j βπ ( πx kc a sin a ) ] e jβz. ) ] e jβz. Problem 8.39 A waveguide, with dimensions a = cm and b = 0.7 cm, is to be used at 0 GHz. Determine the wave impedance for the dominant mode when (a) the guide is empty, and (b) the guide is filled with polyethylene (whose ε r =.5). Solution: For the TE 0 mode, When empty, f 0 = u p 0 a = c a. ε r f 0 = 3 08 = 5 GHz. 0 When filled with polyethylene, f 0 = 0 GHz. According to Eq. (8.), Z TE = η ( f0 / f ) = η 0 εr ( f0 / f ).

394 CHAPTER When empty, When filled, Z TE = Z TE = 377 = 570 Ω. (5/0) 377 = 90 Ω..5 (0/0) Problem 8.40 A narrow rectangular pulse superimposed on a carrier with a frequency of 9.5 GHz was used to excite all possible modes in a hollow guide with a = 3 cm and b =.0 cm. If the guide is 00 m in length, how long will it take each of the excited modes to arrive at the receiving end? Solution: With a = 3 cm, b = cm, and u p0 = c = m/s, application of Eq. (8.06) leads to: f 0 = 5 GHz f 0 = 7.5 GHz f = 9.0 GHz f 0 = 0 GHz Hence, the pulse with a 9.5-GHz carrier can excite the top three modes. Their group velocities can be calculated with the help of Eq. (8.4), u g = c ( f mn / f ), which gives: 0.85c = m/s, for TE 0 u g = 0.6c = m/s, for TE 0 0.3c = m/s, for TE and TM Travel time associated with these modes is: T = d = µs, for TE 0 = 0.54 µs, for TE u g u 0 g.05 µs, for TE and TM.

395 394 CHAPTER 8 Problem 8.4 If the zigzag angle θ is 4 for the TE 0 mode, what would it be for the TE 0 mode? Solution: For TE 0, the derivation that started with Eq. (8.6) led to ( ) π θ 0 = tan, TE 0 mode. βa Had the derivation been for n = (instead of n = ), the x-dependence would have involved a phase factor (πx/a) (instead of (πx/a)). The sequence of steps would have led to ( ) π θ 0 = tan, TE 0 mode. βa Given that θ 0 = 4, it follows that Hence, π βa = tan4 = 0.90 θ 0 = tan ( 0.9) = Problem 8.4 Measurement of the TE 0 frequency response of an air-filled cubic cavity revealed that its Q is 480. If its volume is 64 mm 3, what material are its sides made of? Solution: According to Eq. (8.), the TE 0 resonant frequency of a cubic cavity is given by Its Q is given by f 0 = 3 08 a = = 53.0 GHz Q = a 3δ s = 480, which gives δ s = m. Applying δ s = π f0 µ 0 σ c, and solving for σ c leads to σ c S/m.

396 CHAPTER According to Appendix B, the material is silver. Problem 8.43 A hollow cavity made of aluminum has dimensions a = 4 cm and d = 3 cm. Calculate Q of the TE 0 mode for (a) b = cm, and (b) b = 3 cm. Solution: For the TE 0 mode, f 0 is independent of b, ( f 0 = c ) ( ) + a d ( = 3 ) 08 ( = 6.5 GHz. For aluminum with σ c = S/m (Appendix B), δ s = π f0 µ 0 σ c = m. (a) For a = 4 cm, b = cm and d = 3 cm, Q = δ s abd(a + d ) [a 3 (d + b) + d 3 (a + b)] = (b) For a = 4 cm, b = 3 cm, and d = 3 cm, Q = ) Problem 8.44 A 50-MHz right-hand circularly polarized plane wave with an electric field modulus of 30 V/m is normally incident in air upon a dielectric medium with ε r = 9 and occupying the region defined by z 0. (a) Write an expression for the electric field phasor of the incident wave, given that the field is a positive maximum at z = 0 and t = 0. (b) Calculate the reflection and transmission coefficients.

397 396 CHAPTER 8 (c) Write expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region z 0. (d) Determine the percentages of the incident average power reflected by the boundary and transmitted into the second medium. Solution: (a) k = ω π = c = π 3 rad/m, k = ω εr = π 9 = π rad/m. c 3 From (7.57), RHC wave traveling in +z direction: Ẽ i = a 0 (ˆx + ŷe jπ/ )e jkz = a 0 (ˆx jŷ)e jk z E i (z,t) = Re [Ẽi ] e jωt [ ] = Re a 0 (ˆxe j(ωt kz) + ŷe j(ωt kz π/) ) = ˆxa 0 cos(ωt k z) + ŷa 0 cos(ωt k z π/) = ˆxa 0 cos(ωt k z) + ŷa 0 sin(ωt k z) E i = [ a 0 cos (ωt k z) + a 0 sin (ωt k z) ] / = a0 = 30 V/m. Hence, (b) Ẽ i = 30(x 0 jy 0 )e jπz/3 (V/m). η = η 0 = 0π (Ω), η = η 0 εr = 0π 9 = 40π (Ω). Γ = η η η + η = 40π 0π 40π + 0π = 0.5 τ = + Γ = 0.5 = 0.5.

398 CHAPTER (c) (d) Ẽ r = Γa 0 (ˆx jŷ)e jk z = (ˆx jŷ)e jk z = 5(ˆx jŷ)e jπz/3 (V/m). Ẽ t = τa 0 (ˆx jŷ)e jk z = 5(ˆx jŷ)e jπz (V/m). Ẽ = Ẽ i + Ẽ r = 30(ˆx jŷ)e jπz/3 5(ˆx jŷ)e jπz/3 = 5(ˆx jŷ)[e jπz/3 e jπz/3 ] (V/m). % of reflected power = 00 Γ = 00 (0.5) = 5% % of transmitted power = 00 τ η η = 00 (0.5) 0π 40π = 75%. Problem 8.45 Consider a flat 5-mm-thick slab of glass with ε r =.56. (a) If a beam of green light (λ 0 = 0.5 µm) is normally incident upon one of the sides of the slab, what percentage of the incident power is reflected back by the glass? (b) To eliminate reflections, it is desired to add a thin layer of antireflection coating material on each side of the glass. If you are at liberty to specify the thickness of the antireflection material as well as its relative permittivity, what would these specifications be? Solution:

399 398 CHAPTER 8 Green Light Air Glass ε r =.56 Air 5 mm A Z = η 0 Z = η g Z L = η 0 Z i 5 mm (a) Representing the wave propagation process by an equivalent transmission line model, the input impedance at the left-hand side of the air-glass interface is (from.63): ( ) ZL + jz 0 tanβl Z i = Z 0 Z 0 + jz L tanβl For the glass, Z 0 = η g = η 0 εr = η 0.56 = η 0.6 Z L = η 0 βl = π λ l = π π εr l = λ = π. Subtracting the maximum possible multiples of π, namely 30768π, leaves a remainder of βl =.3π rad. Hence, Z i = η ( ) 0 η0 + j(η 0 /.6)tan.3π.6 (η 0 /.6) + jη 0 tan.3π ( ).6 + j tan.3π 0π = + j.6tan.3π.6 ( ).6 + j0.88 0π = + j.4.6 = = (4. j08.4) Ω.

400 CHAPTER With Z i now representing the input impedance of the glass, the reflection coefficient at point A is: Γ = Z i η 0 Z i + η 0 4. j08.4 0π = 4. j π = = % of reflected power = Γ 00 = 9.4%. (b) To avoid reflections, we can add a quarter-wave transformer on each side of the glass. Antireflection coating Air Glass ε r =.56 Antireflection coating Air d 5 mm d This requires that d be: d = λ + nλ, n = 0,,,... 4 where λ is the wavelength in that material; i.e., λ = λ 0 / ε rc, where ε rc is the relative permittivity of the coating material. It is also required that η c of the coating material be: η c = η 0 η g. Thus or Hence, η 0 ε rc = η 0 η 0 εr, ε rc = ε r =.56 =.6. λ = λ µm = = 0.4 µm, εrc.6 d = λ 4 + nλ = ( n) (µm), n = 0,,,...

401 400 CHAPTER 8 Problem 8.46 A parallel-polarized plane wave is incident from air at an angle θ i = 30 onto a pair of dielectric layers as shown in the figure. (a) Determine the angles of transmission θ, θ 3, and θ 4. (b) Determine the lateral distance d. θ i Air 5 cm 5 cm θ θ 3 µ r = ε r = 6.5 µ r = ε r =.5 d θ 4 Air Solution: (a) Application of Snell s law of refraction given by (8.3) leads to: εr sin θ = sin θ = sin30 ε r 6.5 = 0. Similarly, And, θ =.54. εr 6.5 sinθ 3 = sinθ = sin.54 ε r3.5 = 0.33 θ 3 = εr3.5 sin θ 4 = sinθ 3 = sin 9.48 ε r4 = 0.5 θ 4 = 30. As expected, the exit ray back into air will be at the same angle as θ i.

402 CHAPTER 8 40 (b) d = (5 cm)tanθ + (5 cm)tan θ 3 = 5tan tan9.48 =.79 cm. Problem 8.47 A plane wave in air with Ẽ i = (ˆx ŷ4 ẑ6)e j(x+3z) (V/m) is incident upon the planar surface of a dielectric material, with ε r =.5, occupying the half-space z 0. Determine (a) The incidence angle θ i. (b) The frequency of the wave. (c) The field Ẽ r of the reflected wave. (d) The field Ẽ t of the wave transmitted into the dielectric medium. (e) The average power density carried by the wave into the dielectric medium. Solution: x θ i θ z (a) From the exponential of the given expression, it is clear that the wave direction of travel is in the x z plane. By comparison with the expressions in (8.48a) for

403 40 CHAPTER 8 perpendicular polarization or (8.65a) for parallel polarization, both of which have the same phase factor, we conclude that: k sinθ i =, k cosθ i = 3. Hence, k = + 3 = 3.6 (rad/m) θ i = tan (/3) = Also, (b) k = k εr = = 5.4 (rad/m) [ ] θ = sin sinθ i =.7..5 k = π f c f = k c = = 7 MHz. π π (c) In order to determine the electric field of the reflected wave, we first have to determine the polarization of the wave. The vector argument in the given expression for Ẽi indicates that the incident wave is a mixture of parallel and perpendicular polarization components. Perpendicular polarization has a ŷ-component only (see 8.46a), whereas parallel polarization has only ˆx and ẑ components (see 8.65a). Hence, we shall decompose the incident wave accordingly: Ẽ i = Ẽ i + Ẽi with Ẽ i = ŷ4e j(x+3z) (V/m) Ẽ i = (ˆx ẑ6)e j(x+3z) (V/m) From the above expressions, we deduce: E i 0 = 4 V/m E i 0 = + 6 = 6.3 V/m.

404 CHAPTER Next, we calculate Γ and τ for each of the two polarizations: Γ = cos θ i (ε /ε ) sin θ i cos θ i + (ε /ε ) sin θ i Using θ i = 33.7 and ε /ε =.5/ =.5 leads to: Similarly, Γ = 0.5 τ = + Γ = Γ = (ε /ε )cos θ i + (ε /ε ) sin θ i = 0.5, (ε /ε )cos θ i + (ε /ε ) sin θ i τ = ( + Γ ) cos θ i cos 33.7 = ( 0.5) cos θ t cos.7 = The electric fields of the reflected and transmitted waves for the two polarizations are given by (8.49a), (8.49c), (8.65c), and (8.65e): Based on our earlier calculations: Ẽ r = ŷe r 0 e jk (xsinθr zcosθr) Ẽ t = ŷet 0 e jk (xsinθt+zcosθt) Ẽ r = (ˆxcosθ r + ẑsinθ r )E r 0 e jk (xsinθr zcosθr) Ẽ t = (ˆxcosθ t ẑsinθ t )E t 0 e jk (xsinθt+zcosθt) θ r = θ i = 33.7 θ t =.7 k = 3.6 rad/m, Using the above values, we have: k = 5.4 rad/m, E 0 r = Γ E 0 i = ( 0.5) ( 4) = V/m. E 0 t = τ E 0 i = 0.75 ( 4) = 3 V/m. E 0 r = Γ E 0 i = ( 0.5) 6.3 = 0.95 V/m. E 0 t = τ E 0 i = = 4.8 V/m. Ẽ r = Ẽ r + Ẽ r = ( ˆx ŷ ẑ0.53)e j(x 3z) (V/m).

405 404 CHAPTER 8 (d) (e) Ẽ t = Ẽ t + Ẽt = (ˆx4.46 ŷ3 ẑ.78)e j(x+5z) (V/m). S t = Et 0 η E0 t = (4.46) (.78) = 3.06 η = η 0 = 377 εr.5 = 5.3 Ω S t = = 63.8 (mw/m ).

406 405 Chapter 9: Radiation and Antennas Lesson #6 Chapter Section: 9- Topics: Retarded potential, short dipole Highlights: Radiation by short dipole Far-field distance Special Illustrations: Exercise 9.

407 406 Lesson #6 Chapter Section: 9- Topics: Radiation characteristics Highlights: Antenna pattern Antenna directivity Antenna gain Special Illustrations: Example 9- Example 9-3

408 407 Lesson #63 Chapter Section: 9-3 and 9-4 Topics: Half-wave dipole Highlights: Radiation pattern Directivity Radiation resistance Special Illustrations: CD-ROM Module 9. CD-ROM Demo 9.

409 408 Lesson #64 Chapter Section: 9-5, 9-6 Topics: Effective area, Friis formula Highlights: Receiving aperture of an antenna Relation of aperture to directivity Friis formula Special Illustrations: Example 9-5 Demo 9.