Electromagnetics With Solutions

UNIT 9 Electromagnetics

2011

ONE MARK

MCQ 9.1

Consider the following statements regarding the complex Poynting vector Pv for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(Pv ) denotes the real part of Pv, S denotes a spherical surface whose centre is at the point source, and nt denotes the unit surface normal on S . Which of the following statements is TRUE? (A) Re(Pv ) remains constant at any radial distance from the source (B) Re(Pv ) increases with increasing radial distance from the source (C) ## Re (Pv) : nt dS remains constant at any radial distance from s the source (D) ## Re (Pv) : nt dS decreases with increasing radial distance from s the source MCQ 9.2

A transmission line of characteristic impedance 50 Ω is terminated by a 50 Ω load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be π/4 radians. The phase velocity of the wave along the line is (B) 1.2 # 108 m/s (A) 0.8 # 108 m/s (C) 1.6 # 108 m/s

(D) 3 # 108 m/s

MCQ 9.3

The modes in a rectangular waveguide are denoted by TE mn where TM mn GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY

Visit us at: www.nodia.co.in

ISBN: 9788192276236

www.gatehelp.com Chap 9 Electromagnetics

m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE? (A) The TM 10 mode of the waveguide does not exist (B) The TE 10 mode of the waveguide does not exist (C) The TM 10 and the TE 10 modes both exist and have the same cut-off frequencies (D) The TM 10 and the TM 01 modes both exist and have the same cut-off frequencies

2011

TWO MARKS

MCQ 9.4

A current sheet Jv = 10uty A/m lies on the dielectric interface x = 0 between two dielectric media with εr 1 = 5, μr 1 = 1 in Region-1 (x < 0) and εr2 = 2, μr2 = 2 in Region-2 (x 2 0). If the magnetic field in Region-1 at x = 0− is Hv1 = 3utx + 30uty A/m the magnetic field in Region-2 at x = 0+ is

(A) Hv2 = 1.5utx + 30uty − 10utz A/m (B) Hv2 = 3utx + 30uty − 10utz A/m (C) Hv2 = 1.5utx + 40uty A/m (D) Hv2 = 3utx + 30uty + 10utz A/m MCQ 9.5

A transmission line of characteristic impedance 50 Ω is terminated in a load impedance ZL . The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of λ/4 from the load. The value of ZL is (A) 10 Ω

(B) 250 Ω

(C) (19.23 + j 46.15) Ω

(D) (19.23 − j 46.15) Ω

Page 590

GATE Previous Year Solved Paper By RK Kanodia & Ashish Murolia Published by: NODIA and COMPANY


ISBN: 9788192276236


MCQ 9.6

The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity εr and relative permeability μr = 1 are given by Ev = E p e j (ωt − 280πy) utz V/m and Hv = 3e j (ωt − 280πy) utx A/m . Assuming the speed of light in free space to be 3 # 108 m/s , the intrinsic impedance of free space to be 120π , the relative permittivity εr of the medium and the electric field amplitude E p are (A) εr = 3, E p = 120π

(B) εr = 3, E p = 360π

(C) εr = 9, E p = 360π

(D) εr = 9, E p = 120π

2010

ONE MARK

MCQ 9.7

If the scattering matrix [S ] of a two port network is 0.2+0c 0.9+90c , then the network is [S ] = > 0.9+90c 0.1+90cH (A) lossless and reciprocal

(B)

lossless but not reciprocal

(C) not lossless but reciprocal

(D)

neither lossless nor reciprocal

MCQ 9.8

A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω/m . If the line is distortion less, the attenuation constant(in Np/m) is (A) 500

(B) 5

(C) 0.014

(D) 0.002

MCQ 9.9

The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2 ) is (A) 1 (B) 1 30π 60π (C)

1 120π

(D)

1 240π Page 591



ISBN: 9788192276236


2010

TWO MARKS

MCQ 9.10

v = xyatx + x 2 aty , then If A

# Av $ dlv over the path shown in the figure is

o

C

(A) 0

(B) 2 3

(C) 1

(D) 2 3

MCQ 9.11

A plane wave having the electric field components Evi = 24 cos ^3 # 108 − βy h atx V/m and traveling in free space is incident normally on a lossless medium with μ = μ0 and ε = 9ε0 which occupies the region y $ 0 . The reflected magnetic field component is given by (A) 1 cos (3 # 108 t + y) atx A/m 10π (B)

1 cos (3 108 t + y) at A/m # x 20π

(C) − 1 cos (3 # 108 t + y) atx A/m 20π (D) − 1 cos (3 # 108 t + y) atx A/m 10π MCQ 9.12

In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 Ω line is Page 592



ISBN: 9788192276236


(A) 1.00

(B) 1.64

(C) 2.50

(D) 3.00

2009

ONE MARK

MCQ 9.13

Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y − z plane and parallel to the y − axis. The other wire is in the x − y plane and parallel to the x − axis. Which components of the resulting magnetic field are non-zero at the origin ?

(A) x, y, z components

(B) x, y components

(C) y, z components

(D) x, z components

MCQ 9.14

Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ? Page 593



ISBN: 9788192276236


(A) Only P has no cutoff-frequency (B) Only Q has no cutoff-frequency (C) Only R has no cutoff-frequency (D) All three have cutoff-frequencies

2009

TWO MARKS

MCQ 9.15

If a vector field V is related to another vector field A through V = 4# A , which of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . ) (A)

#CV $ dl = #S #C A $ d S

(C)

#C Δ # V $ dl = #S #C Δ # A $ d S (D) #C Δ # V $ dl = #S #CV $ d S

(B)

#C A $ dl = #S #CV $ d S

MCQ 9.16

A transmission line terminates in two branches, each of length λ , 4 as shown. The branches are terminated by 50Ω loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zi as seen by the source.

(A) 200Ω

(B) 100Ω

(C) 50Ω

(D) 25Ω

MCQ 9.17

A magnetic field in air is measured to be Page 594



ISBN: 9788192276236


B = B0 c

x yt − y xt m x + y2 x2 + y2 2

What current distribution leads to this field ? [Hint : The algebra is trivial in cylindrical coordinates.] t t (B) J =− B0 z c 2 2 2 m, r ! 0 (A) J = B0 z c 2 1 2 m, r ! 0 μ0 x + y μ0 x + y t (C) J = 0, r ! 0 (D) J = B0 z c 2 1 2 m, r ! 0 μ0 x + y 2008

ONE MARK

MCQ 9.18

For a Hertz dipole antenna, the half power beam width (HPBW) in the E -plane is (A) 360c (B) 180c (C) 90c

(D) 45c

MCQ 9.19

For static electric and magnetic fields in an inhomogeneous sourcefree medium, which of the following represents the correct form of Maxwell’s equations ? (B) 4$ E = 0 , 4$ B = 0 (A) 4$ E = 0 , 4# B = 0 (C) 4# E = 0 , 4# B = 0

(D) 4# E = 0 , 4$ B = 0

2008

TWO MARKS

MCQ 9.20

A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is (A) 6.25 GHz (B) 6.0 GHz (C) 5.0 GHz

(D) 3.75 GHz

MCQ 9.21

One end of a loss-less transmission line having the characteristic impedance of 75Ω and length of 1 cm is short-circuited. At 3 GHz, Page 595



ISBN: 9788192276236


the input impedance at the other end of transmission line is (A) 0

(B) Resistive

(C) Capacitive

(D) Inductive

MCQ 9.22

A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant ε = 9 ). The magnitude of the reflection coefficient is (A) 0

(B) 0.3

(C) 0.5

(D) 0.8

MCQ 9.23

In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if (A) radius as well as operating wavelength are halved (B) radius as well as operating wavelength are doubled (C) radius is halved and operating wavelength is doubled (D) radius is doubled and operating wavelength is halved MCQ 9.24

At 20 GHz, the gain of a parabolic dish antenna of 1 meter and 70% efficiency is (A) 15 dB

(B) 25 dB

(C) 35 dB

(D) 45 dB

2007

ONE MARK

MCQ 9.25

A plane wave of wavelength λ is traveling in a direction making an angle 30c with positive x − axis and 90c with positive y − axis. The E field of the plane wave can be represented as (E0 is constant) t 0 e j c ωt − (A) E = yE

3 π x− π z λ λ m

t 0 e jc ωt − λ x − (B) E = yE

π

3πz λ m

t 0 e jc ωt + (C) E = yE

3 π x+ π z λ λ m

t 0 e jc ωt − λ x + (D) E = yE

π

3πz λ m

Page 596



ISBN: 9788192276236


MCQ 9.26

If C is code curve enclosing a surface S , then magnetic field intensity H , the current density j and the electric flux density D are related by (A)

##S H $ ds = ##c c j + 22Dt m $ d t

(B)

#S H $ d l = ##S c j + 22Dt m $ dS

(C)

##S H $ dS = #C c j + 22Dt m $ d t

(D)

#C H $ d l # = ##S c j + 22Dt m $ ds c

2007

TWO MARKS

MCQ 9.27

The E field in a rectangular waveguide of inner dimension a # b is given by 2 ωμ E = 2 ` λ j H0 sin ` 2πx j sin (ωt − βz) yt a h 2 Where H0 is a constant, and a and b are the dimensions along the x − axis and the y − axis respectively. The mode of propagation in the waveguide is (A) TE20 (B) TM11 (C) TM20

(D) TE10

MCQ 9.28

A load of 50 Ω is connected in shunt in a 2-wire transmission line of Z0 = 50Ω as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is

1 −1 2 (A) > 12 1H − 2 2

− 13

(C) >

2 3 2 1 − 3 3

H

0 1 (B) = 1 0G (D) > −

1 4 3 4

− 43 1 4

H Page 597



ISBN: 9788192276236


MCQ 9.29

The parallel branches of a 2-wirw transmission line re terminated in 100Ω and 200Ω resistors as shown in the figure. The characteristic impedance of the line is Z0 = 50Ω and each section has a length of λ 4 . The voltage reflection coefficient Γ at the input is

(A) − j 7 5

(B) − 5 7

(C) j 5 7

(D) 5 7

MCQ 9.30

The H field (in A/m) of a plane wave propagating in free space is given by H = xt 5 3 cos (ωt − βz) + yt`ωt − βz + π j . η0 2 The time average power flow density in Watts is η (A) 0 (B) 100 100 η0 (C) 50η20

(D) 50 η0

MCQ 9.31

An air-filled rectangular waveguide has inner dimensions of 3 cm # 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance η0 = 377 Ω ) (A) 308 Ω

(B) 355 Ω

(C) 400 Ω

(D) 461 Ω

Page 598



ISBN: 9788192276236


MCQ 9.32

A λ dipole is kept horizontally at a height of λ0 above a perfectly 2 2 conducting infinite ground plane. The radiation pattern in the lane of the dipole (E plane) looks approximately as

MCQ 9.33

A right circularly polarized (RCP) plane wave is incident at an angle 60c to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant ξr2 is.

(A)

2

(C) 2

(B)

3

(D) 3

2006

ONE MARK

MCQ 9.34

THe electric field of an electromagnetic wave propagation in the positive direction is given by E = atx sin (ωt − βz) + aty sin (ωt − βz + π/2). The wave is (A) Linearly polarized in the z −direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized Page 599



ISBN: 9788192276236


MCQ 9.35

A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is (A) 10 Watts (B) 1 Watts (C) 0.1 Watts

(D) 0.01 Watt

2006

TWO MARKS

MCQ 9.36

When a planes wave traveling in free-space is incident normally on a medium having the fraction of power transmitted into the medium is given by (B) 1 (A) 8 9 2 (C) 1 3

(D) 5 6

MCQ 9.37

A medium of relative permittivity εr2 = 2 forms an interface with free - space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by (B) π2 m 2 (A) 2π m 2 (C) π m 2 (D) π m 2 2 MCQ 9.38

A rectangular wave guide having TE10 mode as dominant mode is having a cut off frequency 18 GHz for the mode TE30 . The inner broad - wall dimension of the rectangular wave guide is (B) 5 cm (A) 5 cm 3 (C) 5 cm 2

(D) 10 cm

Page 600



ISBN: 9788192276236


MCQ 9.39

A medium is divide into regions I and II about x = 0 plane, as shown in the figure below.

An electromagnetic wave with electric field E1 = 4atx + 3aty + 5atz is incident normally on the interface from region I . The electric file E2 in region II at the interface is (A) E2 = E1

(B) 4atx + 0.75aty − 1.25atz

(C) 3atx + 3aty + 5atz

(D) − 3atx + 3aty + 5atz

MCQ 9.40

A mast antenna consisting of a 50 meter long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna is Ohms is 2 (A) 2π 5

2 (B) π 5

2 (C) 4π 5

(D) 20π2

2005

ONE MARK

MCQ 9.41

The magnetic field intensity vector of a plane wave is given by H (x, y, z, t) = 10 sin (50000t + 0.004x + 30) aty where aty , denotes the unit vector in y direction. The wave is propagating with a phase velocity. (B) − 3 # 108 m/s (A) 5 # 10 4 m/s (C) − 1.25 # 107 m/s

(D) 3 # 108 m/s Page 601



ISBN: 9788192276236


MCQ 9.42

Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency of 1014 Hz in glass. Assume velocity of light is 3 # 108 m/s in vacuum (A) 3 μm

(B) 3 mm

(C) 2 μm

(D) 1 mm

2005

TWO MARKS

MCQ 9.43

Which one of the following does represent the electric field lines for the mode in the cross-section of a hollow rectangular metallic waveguide ?

MCQ 9.44

Characteristic impedance of a transmission line is 50 Ω. Input impedance of the open-circuited line when the transmission line a short circuited, then value of the input impedance will be. (A) 50 Ω

(B) 100 + j150Ω

(C) 7.69 + j11.54Ω

(D) 7.69 − j11.54Ω

MCQ 9.45

Two identical and parallel dipole antennas are kept apart by a distance of λ in the H - plane. They are fed with equal currents 4 but the right most antenna has a phase shift of + 90c. The radiation pattern is given as. Page 602



ISBN: 9788192276236


Statement of Linked Answer Questions 9.46 & 9.47 : Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 and a resistive load is shown in the figure.

MCQ 9.46

The value of the load resistance is (A) 50 Ω

(B) 200 Ω

(C) 12.5 Ω

(D) 0

MCQ 9.47

The reflection coefficient is given by (A)− 0.6

(B) − 1

(C) 0.6

(D) 0

MCQ 9.48

Many circles are drawn in a Smith Chart used for transmission line calculations. The circles shown in the figure represent Page 603



ISBN: 9788192276236


(A) Unit circles (B) Constant resistance circles (C) Constant reactance circles (D) Constant reflection coefficient circles.

2004

ONE MARK

MCQ 9.49

The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE10 mode is (A) equal to its group velocity (B) less than the velocity of light in free space (C) equal to the velocity of light in free space (D) greater than the velocity of light in free space MCQ 9.50

Consider a lossless antenna with a directive gain of + 6 dB. If 1 mW of power is fed to it the total power radiated by the antenna will be (A) 4 mW

(B) 1 mW

(C) 7 mW

(D) 1/4 mW

2004

TWO MARKS

MCQ 9.51

A parallel plate air-filled capacitor has plate area of 10 - 4 m 2 and plate separation of 10 - 3 m. It is connect - ed to a 0.5 V, 3.6 GHz source. The magnitude of the displacement current is ( ε = 361π 10 - 9 F/m) (A) 10 mA

(B) 100 mA

(C) 10 A

(D) 1.59 mA

Page 604



ISBN: 9788192276236


MCQ 9.52

Consider a 300 Ω, quarter - wave long (at 1 GHz) transmission line as shown in Fig. It is connected to a 10 V, 50 Ω source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is

(A) 10 V

(B) 5 V

(C) 60 V

(D) 60/7 V

MCQ 9.53

In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz (A) To increase the sensitivity of measurement (B) To transmit the signal to a far-off place (C) To study amplitude modulations (D) Because crystal detector fails at microwave frequencies MCQ 9.54

If E = (atx + jaty) e jkz - kωt and H = (k/ωμ) (aty + katx ) e jkz - jωt , the timeaveraged Poynting vector is (A) null vector

(B) (k/ωμ) atz

(C) (2k/ωμ) atz

(D) (k/2ωμ) atz

MCQ 9.55

Consider an impedance Z = R + jX marked with point P in an impedance Smith chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to Page 605



ISBN: 9788192276236


(A) adding an inductance in series with Z (B) adding a capacitance in series with Z (C) adding an inductance in shunt across Z (D) adding a capacitance in shunt across Z MCQ 9.56

A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with ε > ε0 . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be (A) 120π Ω

(B) 60π Ω

(C) 600π Ω

(D) 24π Ω

MCQ 9.57

A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage of the power that is reflected back is (A) 57.73

(B) 33.33

(C) 0.11

(D) 11.11

2003

ONE MARK

MCQ 9.58

The unit of 4# H is (A) Ampere

(B) Ampere/meter

(C) Ampere/meter 2

(D) Ampere-meter

Page 606



ISBN: 9788192276236


MCQ 9.59

The depth of penetration of electromagnetic wave in a medium having conductivity σ at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be (A) 6.25 dm

(B) 12.50 cm

(C) 50.00 cm

(D) 100.00 cm

2003

TWO MARKS

MCQ 9.60

Medium 1 has the electrical permittivity ε1 = 1.5ε0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity ε2 = 2.5ε0 farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux − 3uy + 1uz ) volt/m, then E2 in medium 2 is (A) (2.0ux − 7.5uy + 2.5uz ) volt/m (B) (2.0ux − 2.0uy + 0.6uz ) volt/m (C) (2.0ux − 3.0uy + 1.0uz ) volt/m (D) (2.0ux − 2.0uy + 0.6uz ) volt/m MCQ 9.61

If the electric field intensity is given by E = (xux + yuy + zuz ) volt/m, the potential difference between X (2, 0, 0) and Y (1, 2, 3) is (A) + 1 volt

(B) − 1 volt

(C) + 5 volt

(D) + 6 volt

MCQ 9.62

A uniform plane wave traveling in air is incident on the plane boundary between air and another dielectric medium with εr = 4 . The reflection coefficient for the normal incidence, is (A) zero

(B) 0.5+180c

(B) 0.333+0c

(D) 0.333+180c Page 607



ISBN: 9788192276236


MCQ 9.63

If the

electric field intensity associated with a uniform plane

electromagnetic wave traveling in a perfect dielectric medium is given by E (z, t) = 10 cos (2π107 t − 0.1πz) V/m, then the velocity of the traveling wave is (A) 3.00 # 108 m/sec

(B) 2.00 # 108 m/sec

(C) 6.28 # 107 m/sec

(D) 2.00 # 107 m/sec

MCQ 9.64

A short - circuited stub is shunt connected to a transmission line as shown in fig. If Z0 = 50 ohm, the admittance Y seen at the junction of the stub and the transmission line is

(A) (0.01 − j0.02) mho

(B) (0.02 − j0.01) mho

(C) (0.04 − j0.02) mho

(D) (0.02 + j0) mho

MCQ 9.65

A rectangular metal wave guide filled with a dielectric material of relative permittivity εr = 4 has the inside dimensions 3.0 cm # 1.2 cm. The cut-off frequency for the dominant mode is (A) 2.5 GHz

(B) 5.0 GHz

(C) 10.0 GHz

(D) 12.5 GHz

Page 608



ISBN: 9788192276236


MCQ 9.66

Two identical antennas are placed in the θ = π/2 plane as shown in Fig. The elements have equal amplitude excitation with 180c polarity difference, operating at wavelength λ. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for φ = 0 , is

(A) 2 cos b 2πs l λ

(B) 2 sin b 2πs l λ

(C) 2 cos a πs k λ

(D) 2 sin a πs k λ

2002

ONE MARK

MCQ 9.67

The VSWR can have any value between (A) 0 and 1 (B) − 1 and + 1 (C) 0 and 3

(D) 1 and 3

MCQ 9.68

In in impedance Smith movement along a constant resistance circle gives rise to (A) a decrease in the value of reactance (B) an increase in the value of reactance (C) no change in the reactance value (D) no change in the impedance MCQ 9.69

The phase velocity for the TE10 -mode in an air-filled rectangular waveguide is (c is the velocity of plane waves in free space) (A) less than c (B) equal to c (C) greater than c

(D) none of these Page 609



ISBN: 9788192276236


2002

TWO MARKS

MCQ 9.70

t jπ/2) e jωt − jkz . This wave A plane wave is characterized by E = (0.5xt + ye is (A) linearly polarized (B) circularly polarized (C) elliptically polarized

(D) unpolarized

MCQ 9.71

Distilled water at 25c C is characterized by σ = 1.7 # 10 - 4 mho/m and ε = 78εo at a frequency of 3 GHz. Its loss tangent tan δ is ( ε = 10 36π F/m) (A) 1.3 # 10−5

(B) 1.3 # 10−3

(C) 1.3 # 10−4 /78

(D) 1.3 # 10−5 /78ε0

-9

MCQ 9.72

THe electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with ε = 80εo . The surface charge density on the conductor is ( ε = 10 36π F/m) 2 (A) 0 C/m (B) 2 C/m 2 -9

(C) 1.8 # 10 - 11 C/m 2

(D) 1.41 # 10 - 9 C/m 2

MCQ 9.73

A person with receiver is 5 Km away from the transmitter. What is the distance that this person must move further to detect a 3-dB decrease in signal strength (A) 942 m (B) 2070 m (C) 4978 m

(D) 5320 m

2001

ONE MARK

MCQ 9.74

A transmission line is distortonless if (B) RL = GC (A) RL = 1 GC (C) LG = RC

(D) RG = LC

Page 610



ISBN: 9788192276236


MCQ 9.75 2 2 If a plane electromagnetic wave satisfies the equal δ E2x = c2 δ E2x , δZ δt the wave propagates in the

(A) x − direction (B) z − direction (C) y − direction (D) xy plane at an angle of 45c between the x and z direction MCQ 9.76

The plane velocity of wave propagating in a hollow metal waveguide is (A) grater than the velocity of light in free space (B) less than the velocity of light in free space (C) equal to the velocity of light free space (D) equal to the velocity of light in free MCQ 9.77

The dominant mode in a rectangular waveguide is TE10 , because this mode has (A) the highest cut-off wavelength (B) no cut-off (C) no magnetic field component (D) no attenuation

2001

TWO MARKS

MCQ 9.78

A material has conductivity of 10 - 2 mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is (A) 45 MHz

(B) 90 MHz

(C) 450 MHz

(D) 900 MHz Page 611



ISBN: 9788192276236


MCQ 9.79

A uniform plane electromagnetic wave incident on a plane surface of a dielectric material is reflected with a VSWR of 3. What is the percentage of incident power that is reflected ? (A) 10%

(B) 25%

(C) 50%

(D) 75%

MCQ 9.80

A medium wave radio transmitter operating at a wavelength of 492 m has a tower antenna of height 124. What is the radiation resistance of the antenna? (A) 25 Ω

(B) 36.5 Ω

(C) 50 Ω

(D) 73 Ω

MCQ 9.81

In uniform linear array, four isotropic radiating elements are spaced λ apart. The progressive phase shift between required for forming 4 the main beam at 60c off the end - fire is : (A) − π

(B) − π2 radians

(C) − π4 radians

(D) − π8 radians

2000

ONE MARK

MCQ 9.82

The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are 100 Ω and 25 Ω respectively. The characteristic impedance of the line is, (A) 25 Ω

(B) 50 Ω

(C) 75 Ω

(D) 100 Ω

MCQ 9.83

A TEM wave is incident normally upon a perfect conductor. The E and H field at the boundary will be respectively, (A) minimum and minimum

(B) maximum and maximum

(C) minimum and maximum

(D) maximum and minimum

Page 612



ISBN: 9788192276236


MCQ 9.84

If the diameter of a λ dipole antenna is increased from λ to λ 2 100 50 , then its (A) bandwidth increases

(B) bandwidth decrease

(C) gain increases

(D) gain decreases

2000

TWO MARKS

MCQ 9.85

A uniform plane wave in air impings at 45c angle on a lossless dielectric material with dielectric constant dr . The transmitted wave propagates is a 30c direction with respect to the normal. The value of dr is (A) 1.5

(B)

1.5

(C) 2

(D)

2

MCQ 9.86

A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off frequency is (A) 5 GHz

(B) 10 GHz

(C) 15 GHz

(D) 12 GHz

MCQ 9.87

Two coaxial cable 1 and 2 are filled with different dielectric constants εr1 and εr2 respectively. The ratio of the wavelength in the cables (λ1 /λ2) is (A)

εr1 /εr2

(C) εr1 /εr2

(B)

εr2 /εr1

(D) εr2 /εr1

MCQ 9.88

For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz, the minimum distance required for far field measurement is closest to (A) 7.5 cm

(B) 15 cm

(C) 15 m

(D) 150 m Page 613



ISBN: 9788192276236


1999

ONE MARK

MCQ 9.89

An electric field on a place is described by its potential V = 20 (r−1 + r−2) where r is the distance from the source. The field is due to (A) a monopole (B) a dipole (C) both a monopole and a dipole

(D) a quadruple

MCQ 9.90

Assuming perfect conductors of a transmission line, pure TEM propagation is NOT possible in (A) coaxial cable (B) air-filled cylindrical waveguide (C) parallel twin-wire line in air (D) semi-infinite parallel plate wave guide MCQ 9.91

Indicate which one of the following will NOT exist in a rectangular resonant cavity. (A) TE110 (B) TE 011 (C) TM110

(D) TM111

MCQ 9.92

Identify which one of the following will NOT satisfy the wave equation. (B) sin [ω (10z + 5t)] (A) 50e j (ωt − 3z) (C) cos (y2 + 5t)

(D) sin (x) cos (t)

1999

TWO MARKS

MCQ 9.93

In a twin-wire transmission line in air, the adjacent voltage maxima are at 12.5 cm and 27.5 cm . The operating frequency is (B) 1 GHz (A) 300 MHz (C) 2 GHz

(D) 6.28 GHz

Page 614



ISBN: 9788192276236


MCQ 9.94

A transmitting antenna radiates 251 W isotropically. A receiving antenna, located 100 m away from the transmitting antenna, has an effective aperture of 500 cm2 . The total received by the antenna is (A) 10 μW

(B) 1 μW

(C) 20 μW

(D) 100 μW

MCQ 9.95

In air, a lossless transmission line of length 50 cm with L = 10 μH/m , C = 40 pF/m is operated at 25 MHz . Its electrical path length is (A) 0.5 meters

(B) λ meters

(C) π/2 radians

(D) 180 deg rees

MCQ 9.96

A plane wave propagating through a medium [εr = 8, vr = 2, and σ = 0] t − (z/3) sin (108 t − βz) V/m . The has its electric field given by Ev = 0.5Xe wave impedance, in ohms is (A) 377

(B) 198.5+180c

(C) 182.9+14c

(D) 133.3

1998

ONE MARK

MCQ 9.97

The intrinsic impedance of copper at high frequencies is (A) purely resistive (B) purely inductive (C) complex with a capacitive component (D) complex with an inductive component MCQ 9.98

The Maxwell equation V # H = J + 2D is based on 2t (A) Ampere’s law (B) Gauss’ law (C) Faraday’s law

(D) Coulomb’s law Page 615



ISBN: 9788192276236


MCQ 9.99

All transmission line sections shown in the figure is have a characteristic impedance R 0 + j 0 . The input impedance Zin equals

(A) 2 R 0 3

(B) R 0

(C) 3 R 0 2

(D) 2R 0

1998

TWO MARKS

MCQ 9.100

The time averages Poynting vector, in W/m2 , for a wave with Ev = 24e j (ωt + βz) avy V/m in free space is (A) − 2.4 avz π

(B) 2.4 avz π

(C) 4.8 avz π

(D) − 4.8 avz π

MCQ 9.101

The wavelength of a wave with propagation constant (0.1π + j0.2π) m−1 is (A)

2 m 0.05

(C) 20 m

(B) 10 m (D) 30 m

Page 616



ISBN: 9788192276236


MCQ 9.102

The depth of penetration of wave in a lossy dielectric increases with increasing (A) conductivity (B) permeability (C) wavelength

(D) permittivity

MCQ 9.103

The polarization of wave with electric field vector Ev = E 0 e j^ωt + βz h ^avx + avy h is (A) linear (B) elliptical (C) left hand circular

(D) right hand circular

MCQ 9.104

The vector H in the far field of an antenna satisfies (A) d $ Hv = 0 and d # Hv = 0 (B) d $ Hv ! 0 and d # Hv ! 0 (C) d $ Hv = 0 and d # Hv ! 0 (D) d $ Hv ! 0 and d # Hv = 0 MCQ 9.105

The radiation resistance of a circular loop of one turn is 0.01 Ω. The radiation resistance of five turns of such a loop will be (B) 0.01 Ω (A) 0.002 Ω (C) 0.05 Ω

(D) 0.25 Ω

MCQ 9.106

An antenna in free space receives 2 μW of power when the incident electric field is 20 mV/m rms. The effective aperture of the antenna is (B) 0.05 m2 (A) 0.005 m2 (C) 1.885 m2

(D) 3.77 m2

MCQ 9.107

The maximum usable frequency of an ionospheric layer at 60c incidence and with 8 MHz critical frequency is (B) 16 MHz (A) 16 MHz 3 (C) 8 MHz

(D) 6.93 MHz Page 617



ISBN: 9788192276236


MCQ 9.108

A loop is rotating about they y -axis in a magnetic field Bv = B 0 cos (ωt + φ) avx T. The voltage in the loop is (A) zero (B) due to rotation only (C) due to transformer action only (D) due to both rotation and transformer action MCQ 9.109

The far field of an antenna varies with distance r as (A) 1 (B) 12 r r (C) 13 r

(D) 1 r

1997

ONE MARK

MCQ 9.110

A transmission line of 50 Ω characteristic impedance is terminated with a 100 Ω resistance. The minimum impedance measured on the line is equal to (A) 0 Ω

(B) 25 Ω

(C) 50 Ω

(D) 100 Ω

MCQ 9.111

A rectangular air filled waveguide has cross section of 4 cm #10 cm . The minimum frequency which can propagate in the waveguide is (A) 0.75 GHz

(B) 2.0 GHz

(C) 2.5 GHz

(D) 3.0 GHz

MCQ 9.112

A parabolic dish antenna has a conical beam 2c wide, the directivity of the antenna is approximately (A) 20 dB

(B) 30 dB

(C) 40 dB

(D) 50 dB

Page 618



ISBN: 9788192276236


1997

TWO MARKS

MCQ 9.113

A very lossy, λ/4 long, 50 Ω transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately (A) 0

(B) 50 Ω

(C) 3

(D) None of the above

MCQ 9.114

The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity of an electromagnetic wave in the conductor at 1, 000 MHz is about (A) 6 # 106 m/ sec

(B) 6 # 107 m/ sec

(C) 3 # 108 m/ sec

(D) 6 # 108 m/ sec

1996

ONE MARK

MCQ 9.115

A lossless transmission line having 50 Ω characteristic impedance and length λ/4 is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is (A) 0

(B) 0.02 A

(C) 3

(D) none of these

MCQ 9.116

The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Z 0 respectively. The velocity of a travelling wave on the transmission line is 1 Z0 C

(A) Z 0 C

(B)

(C) Z 0 C

(D) C Z0 Page 619



ISBN: 9788192276236


MCQ 9.117

A transverse electromagnetic wave with circular polarization is received by a dipole antenna. Due to polarization mismatch, the power transfer efficiency from the wave to the antenna is reduced to about (A) 50% (B) 35.3% (C) 25%

(D) 0%

MCQ 9.118

A metal sphere with 1 m radius and a surface charge density of 10 Coulombs/m2 is enclosed in a cube of 10 m side. The total outward electric displacement normal to the surface of the cube is (B) 10π Coulombs (A) 40π Coulombs (C) 5π Coulombs

(D) None of these

1996

TWO MARKS

MCQ 9.119

A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incident power that is reflected from the air-glass interface is (A) 0% (B) 4% (C) 20%

(D) 100%

MCQ 9.120

The critical frequency of an ionospheric layer is 10 MHz. What is the maximum launching angle from the horizon for which 20 MHz wave will be reflected by the layer ? (B) 30c (A) 0c (C) 45c

(D) 90c

MCQ 9.121

A 1 km long microwave link uses two antennas each having 30 dB gain. If the power transmitted by one antenna is 1 W at 3 GHz, the power received by the other antenna is approximately (B) 76.8 μW (A) 98.6 μW (C) 63.4 μW

(D) 55.2 μW

Page 620



ISBN: 9788192276236


MCQ 9.122

Some unknown material has a conductivity of 106 mho/m and a permeability of 4π # 10−7 H/m . The skin depth for the material at 1 GHz is (A) 15.9 μm

(B) 20.9 μm

(C) 25.9 μm

(D) 30.9 μm

***********

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ISBN: 9788192276236


SOLUTION SOL 9.1

Power radiated from any source is constant. Hence (C) is correct option. SOL 9.2

We have d = 2 mm and f = 10 GHz Phase difference = 2π d = π ; 4 λ = λ = 8d = 8 # 2 mm = 16 mm v = fλ = 10 # 109 # 16 # 10−3 = 1.6 # 108 m/ sec Hence (C) is correct option.

or

SOL 9.3

TM11 is the lowest order mode of all the TMmn modes. Hence (A) is correct option. SOL 9.4

From boundary condition Bn1 = Bn2 μ1 Hx1 = μ2 Hx2 or Hx2 = Hx1 = 1.5 2 or Hx2 = 1.5utx Further if H z = 1.5utx + Auty + Buz Then from Boundary condition 10uty Jv =− 30utz =− Autz + Buty + 10uty Comparing we get A = 30 and B =− 10 So H z = 1.5utx + 30uty − 10utz A/m Hence (A) is correct option. (3utx + 30uty) utx = (1.5utx + Auty + Butz ) xt +

Page 622



ISBN: 9788192276236


SOL 9.5

Since voltage maxima is observed at a distance of λ/4 from the load and we know that the separation between one maxima and minima equals to λ/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive. Now Γ = s−1 s+1 also RL = R 0 (since voltage maxima is formed at the load) s RL = 50 = 10 Ω 5 Hence (A) is correct option. SOL 9.6

From the expressions of Ev & Hv , we can write, β = 280 π 2 π or = 280 π & λ = 1 140 λ v E Wave impedance, Zw = E = p = 120 π 3 εr Hv again, f = 14 GHz 8 3 Now λ = C = 3 # 10 9 = 140 εr εr f εr 14 # 10 3 or = 1 140 140 εr εr = 9 Ep Now = 120π = E p = 120π 3 9 Hence (D) is correct option. or

SOL 9.7

For a lossless network S11 2 + S21 2 = 1 For the given scattering matrix S11 = 0.2 0c , S12 = 0.9 90c S21 = 0.9 90c , S22 = 0.1 90c Here, (not lossless) (0.2) 2 + (0.9) 2 ! 1 Reciprocity : S12 = S21 = 0.9 90c (Reciprocal) Hence (C) is correct option. Page 623



ISBN: 9788192276236


SOL 9.8

For distortion less transmission line characteristics impedance Z0 = R G Attenuation constant α = RG So, α = R = 0.1 = 0.002 50 Z0 Hence (D) is correct option. SOL 9.9

Intrinsic impedance of EM wave μ μ0 = η = = 120π = 60π ε 2 4ε0 Time average power density 2 1 Pav = 1 EH = 1 E = = 1 2 # 60π 120π 2 2 η Hence (C) is correct option. SOL 9.10

v = xyatx + x 2 aty A v = dxatx + dyaty dl v = # (xyatx + x 2 aty) : (dxatx + dyaty) = # (xydx + x 2 dy) # Av : dl

C

C

C

=

2/ 3

#1/

3

xdx +

1/ 3

#2/

3

3xdx +

#1

3

4 dy + 3

#3

1

1 dy 3

= 1 : 4 − 1 D + 3 :1 − 4 D + 4 [3 − 1] + 1 [1 − 3] = 1 2 3 3 2 3 3 3 3 Hence (C) is correct option. SOL 9.11

In the given problem

Page 624



ISBN: 9788192276236


Reflection coefficient η − η1 τ = 2 = 400π − 120π =− 1 η2 + η 1 2 40π + 120π τ is negative So magnetic field component does not change its direction Direction of incident magnetic field atE # atH = atK atZ # atH = aty atH = atx ( + x direction) So, reflection magnetic field component Hr = τ # 24 cos (3 # 108 + βy) atx , y $ 0 η =

1 # 24 cos (3 108 + βy) at , y $ 0 # x 2 # 120π

8 β = ω = 3 # 108 = 1 vC 3 # 10 Hr = 1 cos (3 # 108 + y) atx , y $ 0 10π

So,

Hence (A) is correct option. SOL 9.12

For length of λ/4 transmission line Z + jZo tan βl Z in = Zo ; L Zo + jZL tan βl E ZL = 30 Ω , Zo = 30 Ω, β = 2π , l = λ 4 λ tan βl = tan b 2π : λ l = 3 4 λ R ZL V S tan βl + jZo W 2 W = Z 0 = 60 Ω Z in = Zo S S Zo + jZL W ZL S tan βl W T X For length of λ/8 transmission line Z + jZo tan βl Z in = Zo ; L Zo + jZL tan βl E So,

Zo = 30 Ω, ZL = 0 (short) tan βl = tan b 2π : λ l = 1 8 λ Z in = jZo tan βl = 30j Circuit is shown below. Page 625



ISBN: 9788192276236


Reflection coefficient 60 + 3j − 60 = τ = ZL − Zo = 60 + 3j + 60 ZL + Zo VSWR =

1 17

1+ τ = 1 + 17 = 1.64 1− τ 1 − 17

Hence (B) is correct option. SOL 9.13

Due to 1 A current wire in x − y plane, magnetic field be at origin will be in x direction. Due to 1 A current wire in y − z plane, magnetic field be at origin will be in z direction. Thus x and z component is non-zero at origin. Hence (D) is correct option. SOL 9.14

Rectangular and cylindrical waveguide doesn’t support TEM modes and have cut off frequency. Coaxial cable support TEM wave and doesn’t have cut off frequency. Hence (A) is correct option. SOL 9.15

We have V = 4# A By Stokes theorem

# A $ dl

## (4 # A) $ ds

# A $ dl

##

= From (1) and (2) we get

...(1) ...(2)

= V $ ds Hence (B) is correct option. Page 626



ISBN: 9788192276236


SOL 9.16

The transmission line are as shown below. Length of all line is λ 4

2 2 Zi1 = Z01 = 100 = 200Ω ZL1 50 2 2 Zi2 = Z02 = 100 = 200Ω ZL2 50

ZL3 = Zi1 Zi2 = 200Ω 200Ω = 100Ω 2 2 Zi = Z0 = 50 = 25Ω ZL3 100

Hence (D) is correct option. SOL 9.17

We have

Bv = B0 c

x a − y a y xm x + y2 x2 + y2 2

To convert in cylindrical substituting x = r cos φ and y = r sin φ ax = cos φar − sin φaφ and ay = sin φar + cos φaφ In (1) we have Bv = Bv0 aφ v Bv a Now Hv = B = 0 φ μ0 μ0 Jv = 4# Hv = 0

...(1)

constant since H is constant

Hence (C) is correct option. SOL 9.18

The beam-width of Hertizian dipole is 180c and its half power beamwidth is 90c. Hence (C) is correct option. Page 627



ISBN: 9788192276236


SOL 9.19

Maxwell equations 4− B = 0 4$ E = ρ/E 4# E =− B 4# Ht = D + J For static electric magnetic fields 4$ B = 0 4$ E = ρ/E 4# E = 0 S 4# H = J Hence (D) is correct option. SOL 9.20

Cut-off Frequency is fc = c 2

m 2 n 2 ` a j +`b j

For TE11 mode, 3 # 1010 fc = 2

1 2 1 2 ` 4 j + ` 3 j = 6.25 GHz


Zin = Zo

ZL + iZo tan (βl) Zo + iZL tan (βl)

For ZL = 0 , Zin = iZo tan (βl) The wavelength is 8 λ = c = 3 # 109 = 0.1 m or 10 cm f 3 # 10 βl = 2π l = 2π # 1 = π λ 10 5 Thus

Zin = iZo tan π 5

Thus Zin is inductive because Zo tan π is positive 5 Hence (D) is correct option. Page 628



ISBN: 9788192276236


SOL 9.22

We have

μ ε

η =

Reflection coefficient Γ=

η2 − η1 η2 + η1

Substituting values for η1 and η2 we have μ − με τ = εμε = 1 − εr = 1 − μ 1 + εr 1+ εε + ε =− 0.5 Hence (C) is correct option. o

o

o r

o

o

o

o r

o

9 9

since εr = 9

SOL 9.23

In single mode optical fibre, the frequency of limiting mode increases as radius decreases Hence r \ 1 f So. if radius is doubled, the frequency of propagating mode gets halved, while in option (D) it is increased by two times. Hence (C) is correct option. SOL 9.24

Gain

8 λ = c = 3 # 10 9 = 3 f 200 20 # 10 2 2 Gp = ηπ2 ` D j = 0.7 # π2 c 13 m = 30705.4 λ 100

= 44.87 dB Hence (D) is correct option. SOL 9.25

γ = β cos 30cx ! β sin 30cy = 2π 3 x ! 2π 1 y λ 2 λ 2 = π 3 x! πy λ λ E = ay E0 e j (ωt - γ) = ay E0 e j;ωt - c Hence (A) is correct option.

π 3 x! π y λ λ mE

Page 629



ISBN: 9788192276236


SOL 9.26

4# H = J + 2D 2t

Maxwell Equations

## 4# H $ ds = ## `J + 22Dt j .ds

Integral form

## `J + 22Dt j .ds

Stokes Theorem

s

# H $ dl

s

=

s

Hence (D) is correct option. SOL 9.27 2 ωμ π H sin ` 2πx j sin (ωt − βz) yt 2 `2j 0 a h This is TE mode and we know that mπy Ey \ sin ` mπx j cos ` a b j

E =

Thus m = 2 and n = 0 and mode is TE20 Hence (A) is correct option. SOL 9.28

The 2-port scattering parameter matrix is S11 S12 S == S21 S22 G (Z Z ) − Zo (50 50) − 50 S11 = L 0 = =− 1 (ZL Z0) + Zo (50 50) + 50 3 2 (ZL Zo) 2 (50 50) S12 = S21 = = =2 (ZL Zo) + Zo (50 50) + 50 3 (Z Z ) − Zo (50 50) − 50 S22 = L o = =− 1 (ZL Zo) + Zo (50 50) + 50 3 Hence (C) is correct option. SOL 9.29

The input impedance is 2 Zin = Zo ; ZL

if l = λ 4

2 2 Zin1 = Zo1 = 50 = 25 ZL1 100 2 2 Zin2 = Zo2 = 50 = 12.5 ZL2 200

Now

ZL = Zin1 Zin2

Page 630



ISBN: 9788192276236


25 12.5 = 25 3 (50) 2 = 300 25/3 Γ = ZS − Zo = 300 − 50 = 5 ZS + Zo 300 + 50 7

Zs =


We have For free space

2 2 2 = Hx2 + Hy2 = c 5 3 m + c 5 m = c 10 m ηo ηo ηo ηo H 2 E2 2 η = = o c 10 m = 50 watts P = 2 ηo ηo 2ηo 2

H

2


The cut-off frequency is fc = c 2

m 2 n 2 ` a j +`b j

Since the mode is TE20, m = 2 and n = 0 8 fc = c m = 3 # 10 # 2 = 10 GHz 2 2 2 # 0.03 ηo 377 = η' = = 400Ω 10 2 fc 2 1−c m 1 − c 10 10 m f 3 # 10 Hence (C) is correct option. SOL 9.32

Using the method of images, the configuration is as shown below

Here d = λ, α = π, thus βd = 2π Array factor is

= cos ;

βd cos ψ + α E 2 Page 631



ISBN: 9788192276236


= cos ;

2π cos ψ + π E = sin (π cos ψ) 2


The Brewster angle is tan θn = tan 60c =

εr2 εr1 εr2 1

or εr2 = 3 Hence (D) is correct option. SOL 9.34

We have E = atxx sin (ωt − βz) + aty sin (ωt − βz + π/2) Here Ex = Ey and φx = 0, φy = π2 Phase difference is π2 , thus wave is left hand circularly polarized. Hence (C) is correct option. SOL 9.35

We have or Now gain or

10 log G = 10 dB G = 10 G = Prad Pin 10 = Prad 1W

or Prad = 10 Watts Hence (A) is correct option. SOL 9.36

η − η1 = Γ= 2 η2 + η1

μo εo εr μo εo εr

− +

μo εo μo εo

= 1+ 1+

εr = 1 − εr 1+

4 =− 1 3 4

The transmitted power is

or

Pt = (1 − Γ2) Pi = 1 − 1 = 8 9 9 Pt = 8 Pi 9

Hence (A) is correct option. Page 632



ISBN: 9788192276236


SOL 9.37

or

sin θ = 1 = 1 2 εr π θ = 45c = 4

The configuration is shown below. Here A is point source.

Now AO From geometry BO Thus area Hence (D) is correct

=1m =1m = πr2 = π # OB = π m 2 option.

SOL 9.38

The cut-off frequency is fc = c 2

m 2 m 2 ` a j +` b j

Since the mode is TE30 , m = 3 and n = 0 fc = c m 2 a or or

8 18 # 109 = 3 # 10 3 2 a a = 1 m = 5 cm 40 2


We have E1 = 4ux + 3uy + 5uz Since for dielectric material at the boundary, tangential component of electric field are equal E21 = E1t = 3aty + 5atz at the boundary, normal component of displacement vector are equal i.e. Dn2 = Dn1 or ε2 E2n = ε1 E1n or 4εo E2n = 3εo 4atz Page 633



ISBN: 9788192276236


or E2n = 3atx Thus E2 = E2t + E2a = 3atx + 3aty + 5atz Hence (C) is correct option. SOL 9.40

Since antenna is installed at conducting ground, 2 2 2 50 Rrad = 80π2 ` dl j = 80π2 c = 4π Ω m 3 5 λ 0.5 # 10 Hence (C) is correct option. SOL 9.41

ω = 50, 000 and β =− 0.004 4 Phase Velocity is vP = ω = 5 # 10 - 3 = 1.25 # 107 m/s β − 4 # 10 Hence (C) is correct option. SOL 9.42

Refractive index of glass μ = 1.5 Frequency f = 1014 Hz c = 3 # 108 m/sec 8 λ = c = 3 # 10 = 3 # 10 - 6 14 f 10 wavelength in glass is -6 λg = α = 3 # 10 = 2 # 10 - 6 m 1.5 μ Hence (C) is correct option. SOL 9.43


ZZC

Zo2 = ZOC .ZSC 2 = Zo = 50 # 50 = 50 100 + j150 2 + 3j ZOC 50 (2 − 3j) = 7.69 − 11.54j = 13

Hence (D) is correct option. Page 634



ISBN: 9788192276236


SOL 9.45

The array factor is A = cos b

βd sin θ + α l 2

Here β = 2π , d = λ and α = 90c 4 λ Thus

A = cos c

2π λ λ 4

sin θ + 2

π 2

π π m = cos ` sin θ + j 4 2

The option (A) satisfy this equation. SOL 9.46

From the diagram, VSWR is s = Vmax = 4 = 4 Vmin 1 When minima is at load ZO = s.ZL or ZL = Zo = 50 = 12.5Ω s 4 Hence (C) is correct option. SOL 9.47

The reflection coefficient is Γ = ZL − ZO = 12.5 − 50 =− 0.6 ZL + ZO 125. + 50 Hence (A) is correct option. SOL 9.48

The given figure represent constant reactance circle. Hence (C) is correct option. SOL 9.49

We know that vp > c > vg . Hence (D) is correct option. SOL 9.50

We have

GD (θ, φ) =

4πU (θ, φ) Prad

For lossless antenna Page 635



ISBN: 9788192276236


Prad = Pin Here we have Prad = Pin = 1 mW and 10 log GD (θ, φ) = 6 dB or GD (θ, φ) = 3.98 Thus the total power radiated by antenna is 4πU (θ, φ) = Prad GD (θ, φ) = 1 m # 3.98 = 3.98 mW Hence (A) is correct option. SOL 9.51

The capacitance is - 12 -4 C = εo A = 8.85 # 10 - 3 # 10 = 8.85 # 10 - 13 d 10 The charge on capacitor is Q = CV = 8.85 # 10 - 13 = 4.427 # 10 - 13 Displacement current in one cycle Q I = = fQ = 4.427 # 10 - 13 # 3.6 # 109 = 1.59 mA T Hence (D) is correct option. SOL 9.52

or

VL = ZO Zin Vin VL = ZO Vin = 10 # 300 = 60 V Zin 50



Ravg = 1 Re [E # H*] 2 E # H* = (atx + jaty) e jkz − jωt # k (− jatx + aty) e−jkz + jωt ωμ

Thus

Ravg

= atz ; k − (− j) (j) k E = 0 ωμ ωμ = 1 Re [E # H*] = 0 2

Hence (A) is correct option. Page 636



ISBN: 9788192276236


SOL 9.55

Suppose at point P impedance is Z = r + j (− 1) If we move in constant resistance circle from point P in clockwise direction by an angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c from point P in clockwise direction. It’s impedance is Z1 = r − 0.5j or Z1 = Z + 0.5j Thus movement on constant r - circle by an +45c in CW direction is the addition of inductance in series with Z . Hence (A) is correct option. SOL 9.56

We have or Thus Now or

1− Γ VSWR = Emax = 5 = Emin 1+ Γ Γ =2 3 Γ =− 2 3 η − η1 Γ= 2 η2 + η1 η − 120π −2 = 2 3 η2 + 120π

or η2 = 24π Hence (D) is correct option. SOL 9.57

The VSWR or Thus or

1− Γ 1+ Γ Γ =1 3 2=

Pref = Γ2= 1 9 Pinc Pref = Pinc 9

i.e. 11.11% of incident power is reflected. Hence (D) is correct option. Page 637



ISBN: 9788192276236


SOL 9.58

By Maxwells equations 4# H = 2D + J 2t Thus 4# H has unit of current density J that is A/m2 Hence (C) is correct option. SOL 9.59

We know that Thus

or

δ \ 1 f f1 δ2 = f2 δ1 δ2 = 1 4 25 1 # 25 = 12.5 cm δ2 = 4


E1 = 2ux − 3uy + 1uz E1t = − 3uy + uy and E1n = 2ux Since for dielectric material at the boundary, tangential component of electric field are equal (x = 0 plane) E1t =− 3uy + uy = E2t E1n = 2ux At the boundary the for normal component of electric field are D1n = D2n or ε1 E1n = ε2 E2n or 1.5εo 2ux = 2.5εo E2n or E2n = 3 ux = 1.2ux 2.5 We have

Thus E2 = E2t + E2n =− 3uy + uz + 1.2ux Hence (C) is correct option. SOL 9.61

We have

E = xux + yuy + zuz dl = utx dx + uty dy + utz dz VXY =−

#XE.dl Y

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ISBN: 9788192276236


=

#1

2

xdxutx +

#2 ydyutz + #3 zdzuzt 0

0

0

2 2 2 0 y2 =−= x + +z G 2 1 2 2 2 3 =− 1 [22 − 12 + 02 − 22 + 02 − 32] = 5 2


μ ε

η = Reflection coefficient τ =

η2 − η1 η2 + η1

Substituting values for η1 and η2 we have μ − με τ = εμε = 1 − εr = 1 − μ 1 + εr 1+ εε + ε = − 1 = 0.333+180c 3 o

0

o r

o

o

o

o r

o

4 4

since εr = 4


E (z, t) = 10 cos (2π # 107 t − 0.1πz) ω = 2π # 107 t β = 0.1π 7 Phase Velocity u = ω = 2π # 10 = 2 # 108 m/s β 0.1π We have where


The fig of transmission line is as shown below . [Z + jZo tan βl] We know that Zin = Zo L [Zo + jZL tan βl] For line 1, l = λ and β = 2π , ZL1 = 100Ω 2 λ [Z + jZo tan π] Thus Zin1 = Zo L = ZL = 100Ω [Zo + jZL tan π] For line 2, l = λ and β = 2π , ZL2 = 0 (short circuit) 8 λ Page 639



ISBN: 9788192276236


Thus

[0 + jZo tan π4 ] = jZo = j50Ω [Zo + 0] Y = 1 + 1 = 1 + 1 = 0.01 − j0.02 Zin1 Zin2 100 j50

Zin2 = Zo

Hence (A) is correct option. SOL 9.65 8 u = c = 3 # 10 = 1.5 # 108 2 ε0 In rectangular waveguide the dominant mode is TE10 and fC = v ` m j2 + ` n j2 2 a b 8 1 2 + 0 2 = 1.5 # 108 = 2.5 GHz = 1.5 # 10 ` 0.03 j ` b j 2 0.06


Normalized array factor = 2 cos

Now

ψ θ d φ δ ψ 2 cos 2

ψ 2

= βd sin θ cos φ + δ = 90c, = 2 s, = 45c, = 180c βd sin θ cos φ + δ = 2 cos ; E 2

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ISBN: 9788192276236


= 2 cos 8 2π 2 s cos 45c + 180 B λ. 2 2 = 2 cos 8 πs + 90cB = 2 sin ` πs j λ λ Hence (D) is correct option. SOL 9.67

VSWR

s = 1+Γ 1−Γ

where Γ varies from 0 to 1

Thus s varies from 1 to 3. Hence (D) is correct option. SOL 9.68

Reactance increases if we move along clockwise direction in the constant resistance circle. Hence (B) is correct option. SOL 9.69

Phase velocity VC

VP =

f 2 1−c c m f

When wave propagate in waveguide fc < f $ VP > VC Hence (C) is correct option. SOL 9.70

We have

π 2

t j ) e j (ωt − kz) E = (0.5xt + ye Ex = 0.5e j (ωt − kz) π

Ey = e j 2 e j (ωt − kz) Ex π = 0.5e− 2 Ey Since

Ex ! 1, it is elliptically polarized. Ey

Hence (C) is correct option. Page 641



ISBN: 9788192276236


SOL 9.71

1.7 # 10 - 4 tan α = σ = ωε 2π # 3 # 109 # 78εo -4 9 = 1.7 # 10 9# 9 # 10 = 1.3 # 10 - 5 3 # 10 # 39 Hence (A) is correct option. Loss tangent

SOL 9.72

The flux density is σ = εE = ε0 εr E = 80 # 8.854 # 10 - 12 # 2 or σ = 1.41 # 10 - 9 C/m 2 Hence (D) is correct option. SOL 9.73

P \ 12 r P1 = r22 Thus P2 r12 3 dB decrease $ Strength is halved P1 = 2 Thus P2 Substituting values we have 2 2 = r22 5 or r2 = 5 2 kM = 7071 m Distance to move = 7071 − 5000 = 2071 m Hence (B) is correct option. SOL 9.74

A transmission line is distortion less if LG = RC Hence (C) is correct option. SOL 9.75

d2 Ex = c2 d2 Ex dz2 dt2 This equation shows that x component of electric fields Ex is traveling in z direction because there is change in z direction. Hence (B) is correct option. We have

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ISBN: 9788192276236


SOL 9.76

In wave guide vp > c > vg and in vacuum vp = c = vg where vp $ Phase velocity c $ Velocity of light vg $ Group velocity Hence (A) is correct option. SOL 9.77

In a wave guide dominant gives lowest cut-off frequency and hence the highest cut-off wavelength. Hence (A) is correct option. SOL 9.78

or or or

Ic = Id σE = jω d E σ = 2πfεo εr ω = 2πf and ε = εr ε0 -2 9 σ f = = 2σ = 9 # 10 # 2 # 10 2π # εo εr 4πεo εr 4

or f = 45 # 106 = 45 MHz Hence (A) is correct option. SOL 9.79

or or Now

VSWR = 1 + Γ 1−Γ 3 = 1+Γ 1−Γ Γ = 0.5 Pr = Γ2 = 0.25 Pi

Thus 25% of incident power is reflected. Hence (B) is correct option. SOL 9.80

We have

λ = 492 m

and height of antenna = 124 m . λ 4 It is a quarter wave monopole antenna and radiation resistance is 25 Ω. Hence (A) is correct option. Page 643



ISBN: 9788192276236


SOL 9.81

The array factor is where

Thus or

ψ = βd cos θ + δ d =λ 4

Distance between elements

Because of end fire ψ =0 θ = 60c 0 = 2π # λ cos 60c + δ = π # 1 + δ 2 2 λ 4 δ =− π 4


Zo = ZOC .ZSC = 100 # 25 = 10 # 5 = 50Ω Hence (B) is correct option. SOL 9.83

As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary. Hence (C) is correct option. SOL 9.84

BW \

1 (Diameter)

As diameter increases Bandwidth decreases. Hence (B) is correct option. SOL 9.85

The fig is as shown below :

As per snell law Page 644



ISBN: 9788192276236


or

sin θt = 1 sin θi εr sin 30c = 1 sin 45c εr 1 2 1 2

= 1 εr or εr = 2 Hence (C) is correct option. SOL 9.86

Cutoff frequency

fc =

vp 2

m 2 n 2 ` a j +`b j

For rectangular waveguide dominant mode is TE01 8 v Thus For air vp = 3 # 108 fc = p = 3 # 10- 2 = 15 # 109 2a 2 # 10 = 15 GHz Hence (C) is correct option. SOL 9.87

Phase Velocity or Thus we get

β = 2π = ω με λ λ = 2π ω με λ \ 1 ε λ1 = ε2 λ2 ε1

Hence (B) is correct option. SOL 9.88 2 λ ` 2 jd = l 8 λ = c = 3 # 10 9 = 3 m f 40 4 # 10 2 3 ` 40 # 2 j d = (2.4)

or

d =

80 # (2.4) 2 . 150 m 3




ISBN: 9788192276236


SOL 9.89

We know that for a monopole its electric field varies inversely with r 2 while its potential varies inversely with r . Similarly for a dipole its electric field varies inversely as r 3 and potential varies inversely as r 2 . In the given expression both the terms a _ r1 + r1 i are present, so this potential is due to both monopole & dipole. Hence (C) is correct option. −1

−2

SOL 9.90

In TE mode Ez = 0 , at all points within the wave guide. It implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide. Hence (D) is correct option. SOL 9.91


A scalar wave equation must satisfy following relation 2 2 E − μ 22 2 E = 0 2t 2 2z 2 Where μ = ω (Velocity) β

...(1)

Basically ω is the multiply factor of t and β is multiply factor of z or x or y . In option (A) E = 50e j (ωt − 3z) μ =ω=ω 3 β We can see that equations in option (C) does not satisfy equation (1) Hence (C) is correct option. SOL 9.93

We know that distance between two adjacent voltage maxima is equal to λ/2 , where λ is wavelength. λ = 27.5 − 12.5 2 Page 646



ISBN: 9788192276236


Frequency

λ = 2 # 15 = 30 cm 10 υ = C = 3 # 10 = 1 GHz 30 λ


Power received by antenna −4 PR = PT 2 # (apeture) = 251 # 500 # 102 = 100 μW 4π r 4 # π # (100) Hence (D) is correct option. SOL 9.95

Electrical path length = βl Where β = 2π , l = 50 cm λ We know that λ =υ =1# f f

1 LC

aυ=

1 LC

1 1 6 # −6 25 # 10 10 # 10 # 40 # 10−12 7 = 5 # 10 6 = 2 m 25 # 10 Electric path length = 2π # 50 # 10−2 5 =

Hence (C) is correct option.

= π radian 2

SOL 9.96

In a lossless dielectric (σ = 0) median, impedance is given by μ 0c η = ε =

μ0 μr ε0 εr

= 120π #

μr εr

= 120π #

2 = 188.4 Ω 8




ISBN: 9788192276236


SOL 9.97

Impedance is written as jωμ σ + jωε

η =

Copper is good conductor i.e. σ >> ωε jωμ ωμ So = 45c η = σ σ Impedance will be complex with an inductive component. Hence (D) is correct option. SOL 9.98

This equation is based on ampere’s law as we can see # H $ dl = I enclosed (ampere's law) l

=

#s Jds

Applying curl theorem # (4 # H) $ ds =

#s Jds

#l H $ dl

or

s

4# H = J then it is modified to 4# H = J + 2D 2t

Based on continuity equation




Propagation constant r = α + iβ = 0.1π + j0.2π here β = 2π = 0.2π λ λ = 2 = 10 m 0.2 Hence (B) is correct option. Page 648



ISBN: 9788192276236


SOL 9.102

The depth of penetration or skin depth is defined as – 1 δ= πfμσ δ\ 1 \ λ f so depth increases with increasing in wavelength. Hence (C) is correct option. SOL 9.103

Given j (ωt + βz) v a x + ε0 e j (ωt + βz) avy ...(1) E (z, t) = Eo e Generalizing ...(2) E (z) = avx E1 (z) + avy E2 (z) Comparing (1) and (2) we can see that E1 (z) and E2 (z) are in space quadrature but in time phase, their sum E will be linearly polarized along a line that makes an angle φ with x -axis as shown below. Fig. Hence (A) is correct option.

SOL 9.104

v v Hv = 1 4 A μ # v is auxiliary potential function. where A So 4: H = 4: (4 # A) = 0 4# H = 4# (4 # A) = Y 0 Hence (C) is correct option. SOL 9.105

Radiation resistance of a circular loop is given as Rr = 8 ηπ3 :NΔ2 S D 3 λ Rx \ N 2 N " no. of turns So, Rr 2 = N 2 # Rr 1 = (5) 2 # 0.01 = 0.25 Ω Hence (D) is correct option. Page 649



ISBN: 9788192276236


SOL 9.106

We have Aperture Area

Power Re ceived Polynting vector of incident wave A =W P =

2 P = E η0 = 120π is intrinsic impedance of space η0 −6 −6 A = 2 # 10 = 2 # 10 −3 2 # 120 # 3.14 2 E (20 # 10 ) c η0 m

So

−6 3.14 = 1.884 m2 = 2 # 10 # 12 −# 6 400 # 10 Hence (C) is correct option.

SOL 9.107

Maximum usable frequency fo fm = sin Ae fm = 8MHz = sin 60c

8 = 16 MHz 3 3 c 2 m


When a moving circuit is put in a time varying magnetic field educed emf have two components. One for time variation of B and other turn motion of circuit in B . Hence (D) is correct option. SOL 9.109

Far field \ 1 r Hence (A) is correct option. SOL 9.110

Z in

min

= Z0 S

where S = standing wave ratio Page 650



ISBN: 9788192276236


S =

1 + ΓL 1 − ΓL

ΓL = reflection coefficient ΓL = ZL − Z 0 = 100 − 50 = 50 = 1 ZL + Z 0 100 + 50 150 3 1+1 3 =2 S = 1 1− 3 Z in min = 50 = 25 Ω 2 Hence (B) is correct option. SOL 9.111

The cutoff frequency μl fc = 2

is given by m 2 n 2 a a k +a2k

Here a < b , so minimum cut off frequency will be for mode TE 01 m = 0, n = 1 8 a μl = c 3 10 1 # 2 fc = * 2#2 (10 # 10−12) c = 3 # 108 8 3 10 # = 0.75 GHz = 2 # 2 # 10 # 10−2 Hence (A) is correct option. SOL 9.112


For any transmission line we can write input impedance Z + jZ 0 tanh lγ Zin = Z 0 ; L Z 0 + jZL tanh lγ E Here given ZL = 3 (open circuited at load end) R jZ tanh lγ V S1 + 0 W Z0 ZL W= so Zin = Z 0 lim S Z " 3S Z 0 j tanh lγ W S ZL + j tanh lγ W T X Hence (A) is correct option. L

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ISBN: 9788192276236


SOL 9.114

We know that skin depth is given by 1 s = = 1 # 10−2 m πf1 μσ 1 or = 10−2 6 π # 10 # 10 # μ σ −3 or μσ = 10 π

f 1 = 10 MHz

Now phase velocity at another frequency f 2 = 1000 MHz is 4π f 2 V = μσ Put

−3

μσ = 10 in above equation π V =

4 # π # 1000 # 106 # π - 6 106 m/ sec # 10−3


Input impedance of a lossless transmission line is given by Z + jZ 0 tan βl Zin = Z 0 ; L Z 0 + jZL tan βl E where

so here

and so

Z 0 = Charateristic impedance of line ZL = Load impedance β = 2π l = length λ βl = 2π λ = π 2 λ 4 ZL = 0 (Short circuited) Z 0 = 50 Ω 0 + j50 tan π/2 Zin = 50 = =3 50 + j0 tan π/2G

Thus infinite impedance, and current will be zero. Hence (A) is correct option. SOL 9.116

For lossless transmission line, we have Velocity V =ω= 1 β LC

...(1)

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ISBN: 9788192276236


Characteristics impedance for a lossless transmission line Z0 = L C

...(2)

From eqn. (1) and (2) V =

1 = 1 C (Z 0 C ) Z 0 C




Reflected power Er = ΓEi Γ = Reflection coefficient η − η1 Γ = 2 = 1.5 − 1 = 1 η 2 + η1 1.5 + 1 5

Ei " Incident power

Er = 1 # Ei 5

So

Er = 20% Ei Hence (C) is correct option. SOL 9.120

We have maximum usable frequency formulae as f0 fm = sin Ae 20 # 106 = 10 # 10 sin Ae

6

sin Ae = 1 2 Ae = 30c Hence (B) is correct option. Page 653



ISBN: 9788192276236


SOL 9.121


δ=

Skin depth

Putting the given value δ=

1 πfμσ

1 = 15.9 μm 3.14 # 1 # 109 # 4π # 10−7 # 106

Hence (A) is correct option.

***********

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ISBN: 9788192276236

Electromagnetics With Solutions

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