## Electromagnetics Subject wise

Question 1 |

Radiation resistance of a small dipole current element of length l at a frequency of 3 GHz is 3 ohms. If the length is changed by 1%, then the percentage change in the radiation resistance, rounded off to two decimal places, is _________%.

Fill in the Blank Type Question |

Question 1 Explanation:

Question 2 |

What is the electric flux through a quarter-cylinder of height H (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q?

Question 2 Explanation:

Question 3 |

In the table shown, List I and List II, respectively, contain terms appearing on the left-hand side and the right-hand side of Maxwell’s equations (in their standard form). Match the left-hand side with the corresponding right-hand side.

1 – Q, 2 – R, 3 – P, 4 – S | |

1 – Q, 2 – S, 3 – P, 4 – R | |

1 – R, 2 – Q, 3 – S, 4 – P | |

1 – P, 2 – R, 3 – Q, 4 – S |

Question 4 |

A rectangular waveguide of width w and height h has cut-off frequencies for TE

_{10}and TE_{11}modes in the ratio 1 :2 . The aspect ratio w/h, rounded off to two decimal places, is ____________.Fill in the Blank Type Question |

Question 4 Explanation:

Question 5 |

Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the magnetic field between the wires at a distance r from W1 is

Question 5 Explanation:

Question 6 |

The dispersion equation of a waveguide, which relates the wavenumber k to the frequency ω, is . where the speed of light c = 3 × 10

^{8}m/s, and ω_{o}is a constant. If the group velocity is 2 × 10^{8}m/s, then the phase velocity is1.5 × 10 ^{8} m/s | |

2 × 10 ^{8} m/s | |

4.5 × 10 ^{8} m/s | |

3 × 10 ^{8} m/s |

Question 6 Explanation:

Question 7 |

The solid angle subtended by the sun as viewed from the earth is f/ = 4x10

^{-5}steradian. A microwave antenna designed to be used for studying the microwave radiation from the sun has a very narrow beam whose equivalent solid angle is approximately equal to that subtended by the sun. What is the approximate directivity, D?10 ^{5} | |

x10 ^{5} | |

x10 ^{6} | |

10 ^{6} |

Question 7 Explanation:

Given=>σ= 4 x 10

Directivity D= 4π/σ =>

=4π/(4×10

=π×10

^{-5}, and we have to find the approximate directivity (D)Directivity D= 4π/σ =>

=4π/(4×10

^{−5})=π×10

^{−5}Question 8 |

The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is V(

*l*) = volts, where ‘I’ is the distance along the length of the cable in meters. is the complex propagation constant , and is the angular frequency. The absolute value of the attenuation in the cable in dB/meter is ____Fill in the Blank Type Question |

Question 8 Explanation:

Question 9 |

Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link

increases by a factor of 2 | |

decrease by a factor 2 | |

remains unchanged | |

decreases by a factor of |

Question 9 Explanation:

C= Blog

Channel capacity remain same.

_{2}Channel capacity remain same.

Question 10 |

An optical fiber is kept along the z direction. The refractive indices for the electric fields along x and y directions in the fiber are n

_{x}=1.5000 and n_{y}=1.5001, respectively (n_{x}≠ n_{y}due to the imperfection in the fiber cross-section). The free space wavelength of a light wave propagating in the fiber is 1.5μm. If the light wave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimeter, isFill in the Blank Type Question |

Question 10 Explanation:

Question 11 |

The expression for an electric field in free space is where x, y, z represent the spatial coordinates, t represents time, and ω, k are constants. This electric field

does not represent a plane wave | |

represents a circular polarized plane wave propagating normal to the z-axis | |

represents an elliptically polarized plane wave propagating along x-y plane. | |

represents a linearly polarized plane wave |

Question 11 Explanation:

**Given**

Question 12 |

A half wavelength dipole is kept in the x-y plane and oriented along 45

^{o }from the x-axis. Determine the direction of null in the radiation pattern for 0≤θ≤π. Here the angle θ (0≤θ≤π) is measured from the z-axis, and the angle ∅(0≤∅≤2π) is measured from the x-axis in the x-y plane.Θ=90 °, ∅=45 ° | |

Θ=45 °, ∅=90 ° | |

Θ=90 °, ∅=135 ° | |

Θ=45 °, ∅=135 ° |

Question 12 Explanation:

The Correct Answer Among All the Options is A

Null occurs along axis of the antenna which is θ=90 °, ∅=45 °

Null occurs along axis of the antenna which is θ=90 °, ∅=45 °

Question 13 |

The points and shows on the Smith chart (normalized impedance chart) in the following figure represent:

Open Circuit, Short Circuit, Matched Load | |

Open Circuit, Matched Load, Short Circuit | |

Short Circuit, Matched Load, Open Circuit | |

Short Circuit, Open Circuit, Matched Load |

Question 13 Explanation:

For

For

For

**Short circuit**z = Re(z) = Im(z) = 0 => Point "P"For

**Open circuit**z = ∞ => Point "R"For

**Matched load**z = Re(z) = 1 and Im(z) = 0 => Point "Q"**P : Short Circuit, Q : Matched Load, R : Open Circuit**Question 14 |

A lossy transmission line has resistance per unit length The line is distortionless and has characteristic impendace of The attenuation constant (in Np/m, correct to three decimal places) of the line is _____________.

Fill in the Blank Type Question |

Question 14 Explanation:

Question 15 |

A uniform plane wave traveling in free space and having the electric field

is incident on a dielectric medium (relative permittivity > 1, relative permeability = 1) as shown in the figure and there is no reflected wave.

The relative permittivity (correct to two decimal places) of the dielectric medium is _____________.

is incident on a dielectric medium (relative permittivity > 1, relative permeability = 1) as shown in the figure and there is no reflected wave.

The relative permittivity (correct to two decimal places) of the dielectric medium is _____________.

Fill in the Blank Type Question |

Question 15 Explanation:

Question 16 |

The cut-off frequency of mode of an air filled rectangular waveguide having inner dimensions is twice that of the dominant. mode. When the waveguide is operated at a frequency which is 25% higher than the cut-off frequency of the dominant mode, the guide wavelength is found to be 4 cm. The value of b (in cm, correct to two decimal places) is ____________.

Fill in the Blank Type Question |

Question 16 Explanation:

Question 17 |

The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.1 m so that the amplitude of the wave attenuates by 20 dB, is

0.12 | |

0.23 | |

0.46 | |

2.3 |

Question 17 Explanation:

**Attenuation constant is related with skin depth**as follows

According to given condition of

**20 dB Attenuation**

we can get required depth by following calculation

Question 18 |

Two conducting spheres S1 and S2 of radii a and b (b>A. respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure.

For some charge placed on this structure, the potential and surface electric field on S1 are V

For some charge placed on this structure, the potential and surface electric field on S1 are V

_{a}and E_{a}and that on S2 are V_{b}and E_{b}respectively, which of the following is CORRECT?and | |

and | |

and | |

and |

Question 18 Explanation:

two spheres are joined with a conducting wire, the voltage on two spheres is same.

The capacitance of sphere radius

We know Q = CV

The capacitance of sphere radius

We know Q = CV

Question 19 |

A two – wire transmission line terminates in a television set. The VSWR measured on the line is 5.8. The percentage of power that is reflected from the television set is ______

Fill in the Blank Type Question |

Question 19 Explanation:

Percentage of power reflected is = ×100

% Power reflected =×100=49.82%

% Power reflected =×100=49.82%

Question 20 |

An electron (q1) is moving in free space with velocity towards a stationary electron (q2) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is _____.

[Given, mass of electron m

charge of electron , and permittivity -

]

[Given, mass of electron m

charge of electron , and permittivity -

]

Fill in the Blank Type Question |

Question 20 Explanation:

Work done due to field and external agent must be zero

qV=

Answer Range :

qV=

Answer Range :

**(4.55 to 5.55)**Question 21 |

Standard air – filled rectangular waveguides of dimensions a = 2.29 cm and b= 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominant TE

_{10}mode with the operating frequency atleast 25% above the cutoff frequency of TE_{10}mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is.Question 21 Explanation:

Question 22 |

The permittivity of water at optical frequencies is

**1.75****ε****.It is found that an isotropic light source at a distance d under-water forms an illuminated circular area of radius 5m, as shown in the figure. The critical angle is θ**_{o }_{c}. The value of d (in meter) is _____Fill in the Blank Type Question |

Question 22 Explanation:

Question 23 |

Concentric spherical shells of radii 2 m, 4 m, and 8 m carry uniform surface charge densities of respectively. The value of required to ensure that the electric flux density at radius 10 m is ______.

Fill in the Blank Type Question |

Question 23 Explanation:

Consider a Gaussian surface a sphere of radius 10m

To ensure at radius 10m, the total charge enclosed by Gaussian surface is zero.

Q

=20 x 2

Therefore,

To ensure at radius 10m, the total charge enclosed by Gaussian surface is zero.

Q

_{enc}= 0=20 x 2

^{2}- 4 x 4^{2}-P_{s}x 8^{2}= 0Therefore,

**P**_{s}= -0.25nC/m^{2}Question 24 |

Fill in the Blank Type Question |

Question 24 Explanation:

**Correct Answer is 4.714**

Question 25 |

The current density in a medium is given by

The total current and the average current density flowing through the portion of a spherical surface

are given, respectively, by

The total current and the average current density flowing through the portion of a spherical surface

are given, respectively, by

none of these |

Question 25 Explanation:

The Correct Answer Among All the Options is D

Refer the Topic Wise Question for Electrostatics Electromagnetics

Refer the Topic Wise Question for Electrostatics Electromagnetics

Question 26 |

An antenna pointing in a certain direction has a noise temperature of 50K. The ambient temperature is 290 K. The antenna is connected to a pre-amplifier that has a noise figure of 2 dB and an available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noise temperature Te for the amplifier and the noise power Paoat the output of the preamplifier, respectively, are

Question 26 Explanation:

Question 27 |

Two lossless X-band horn antennas are separated by a distance of 200. The amplitude reflection coefficients at the terminals of the transmitting and receiving antennas are 0.15 and 0.18, respectively. The maximum directivities of the transmitting and receiving antennas (over the isotropic antenna) are 18 dB and 22 dB, respectively. Assuming that the input power in the lossless transmission line connected to the antenna is 2 W, and that the antennas are perfectly aligned and polarization matched, the power (in mW) delivered to the load at the receiver is ___.

Fill in the Blank Type Question |

Question 27 Explanation:

Given Lossless horn antennas

ηT = ηR = 1

ηT = ηR = 1

Question 28 |

The electric field of a uniform plane wave travelling along the negative z direction is given by the following equation:

This wave is incident upon a receiving antenna placed at the origin and whose radiated electric field towards the incident wave is given by the following equation:

The polarization of the incident wave, the polarization of the antenna and losses due to the polarization mismatch are, respectively,

This wave is incident upon a receiving antenna placed at the origin and whose radiated electric field towards the incident wave is given by the following equation:

The polarization of the incident wave, the polarization of the antenna and losses due to the polarization mismatch are, respectively,

Linear, Circular (clockwise), —5dB | |

Circular (clockwise), Linear, —5dB | |

Circular (clockwise), Linear, —3dB | |

Elliptical, Linear, —3dB |

Question 28 Explanation:

Question 29 |

The far-zone power density radiated by a helical antenna is approximated as:

The radiated power density is symmetrical with respect to (I) and exists only in the upper hemisphere: is a constant. The power radiated by the antenna (in watts) and the maximum directivity of the antenna, respectively, are

The radiated power density is symmetrical with respect to (I) and exists only in the upper hemisphere: is a constant. The power radiated by the antenna (in watts) and the maximum directivity of the antenna, respectively, are

1.5C _{O}, 10dB | |

1.256C _{0}, 10dB | |

1.256C _{O.} 12dB | |

1.5C _{O,} 12dB |

Question 29 Explanation:

Question 30 |

The electric field component of a plane wave traveling in a lossless dielectric medium is given by V/m. The wavelength (in m) for the wave is ___________.

Fill in the Blank Type Question |

Question 30 Explanation:

Question 31 |

A vector is given by . Which one of the following statements is TRUE?

is solenoidal, but not irrotational | |

is irrotational, but not solenoidal | |

is neither solenoidal nor irrotational | |

is both solenoidal and irrotational |

Question 31 Explanation:

The Correct Answer Among All the Options is A

For solenoidal

is solenoidal

For solenoidal

is solenoidal

Question 32 |

The longitudinal component of the magnetic field inside an air-filled rectangular waveguide made of a perfect electric conductor is given by the following expression

(A/m)

The cross-sectional dimensions of the waveguide are given as a = 0.08 m and b = 0.033 m. The mode of propagation inside the waveguide is

(A/m)

The cross-sectional dimensions of the waveguide are given as a = 0.08 m and b = 0.033 m. The mode of propagation inside the waveguide is

TM _{12} | |

TM _{21} | |

TE _{21} | |

TE _{12} |

Question 32 Explanation:

Axial component is H ⇒ the propagating

Mode is TE

_{mn}, m, n can be found by

The mode of propagation is TE

_{21}.

Question 33 |

The electric field intensity of a plane wave traveling in free space is given by the following expression

In this field, consider a square area 10 cm x 10 cm on a plane x + y = 1. The total time-averaged power (in mW) passing through the square area is __________.

In this field, consider a square area 10 cm x 10 cm on a plane x + y = 1. The total time-averaged power (in mW) passing through the square area is __________.

Fill in the Blank Type Question |

Question 33 Explanation:

Power density

Time average power

Question 34 |

Consider a uniform plane wave with amplitude (E

The magnitude of the transmitted electric field component (in V/m) after it has travelled a distance of 10 cm inside the dielectric region is ________.

_{0}) of 10 V/m and 1.1 GHz frequency travelling in air, and incident normally on a dielectric medium with complex relative permittivity and permeability as shown in the figure.The magnitude of the transmitted electric field component (in V/m) after it has travelled a distance of 10 cm inside the dielectric region is ________.

Fill in the Blank Type Question |

Question 34 Explanation:

Given,

Attenuation constant of the medium is given by

Where

At a distance of 10 cm |E| is given by

Attenuation constant of the medium is given by

Where

At a distance of 10 cm |E| is given by

Question 35 |

A uniform and constant magnetic field exists in the direction in vacuum. A particle of mass m with a small charge q is introduced into this region with an initial velocity
Given that are all non-zero, which one of the following describes the eventual trajectory of the particle?

Helical motion in the direction. | |

Circular motion in the xy plane. | |

Linear motion in the direction. | |

Linear motion in the direction. |

Question 36 |

Let the electric field vector of a plane electromagnetic wave propagating in a homogenous medium be expressed as , where the propagation constant is a function of the angular frequency . Assume that and are known and are real. From the information available, which one of the following CANNOT be determined?

The type of polarization of the wave. | |

The group velocity of the wave. | |

The phase velocity of the wave. | |

The power flux through the z = 0 plane. |

Question 36 Explanation:

**Option (A)**The polarization is linear

**Option(B)**

**Option(C)**

**Option(D)**It is not possible to find the intrinsic impedance of the medium. So, it is not possible to find power flux.

Question 37 |

Light from free space is incident at an angle 0, to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are n

_{1}=1.5 and n_{2}= 1.4, respectively. The maximum value of (in degrees) for which the incident light will be guided in the core of the fibre is ______Fill in the Blank Type Question |

Question 37 Explanation:

Given n

The maximum angle over which the incident light rays entering the fiber is called acceptance angle, Θ

_{1}= 1.5, n_{2}= 1.4The maximum angle over which the incident light rays entering the fiber is called acceptance angle, Θ

_{i}Question 38 |

The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that d the plate separation becomes 2d.
At this new plate separation, what is an effect stored in the capacitor, neglecting fringing the energy?

2E | |

E | |

E/2 |

Question 38 Explanation:

If capacitor is electrically isolated then charge is same

We know and

If `d' is doubled then C will be C/2 and V will be 2V

Given

Question 39 |

A microwave circuit consisting of lossless transmission lines T1 and T2 is shown in the figure.
The plot shows the magnitude of the input reflection coefficient Γ as a function of frequency f. The phase velocity of the signal in the transmission lines is

**2 × 10**The length L(in meters) of T^{8}m/s._{2}is_________Fill in the Blank Type Question |

Question 39 Explanation:

Question 40 |

A positive charge q is placed at x = 0 between two infinite metal plates placed at x = −d and at
x = +d respectively. The metal plates lie in the yz plane.
The charge is at rest at t = 0, when a voltage +V is applied to the plate at −d and voltage −V is
applied to the plate at x = +d. Assume that the quantity of the charge q is small enough that it
does not perturb the field set up by the metal plates. The time that the charge q takes to reach the
right plate is proportional to

d/V | |

√d/V | |

d/√V | |

√(d/V) |

Question 40 Explanation:

When there is no external field,

Change at rest having potential energy only

P.E = qv

By an application of an external field, change carries acquire some kinetic energy, with velocity v.

Question 41 |

If a right-handed circularly polarized wave is incident normally on a plane perfect conductor, then the reflected wave will be

Right-handed circularly polarized | |

Left-handed circularly polarized | |

Elliptically polarized with a tilt angle of 45 ° | |

Horizontally polarized |

Question 41 Explanation:

If incident wave is right handed polarized then the reflected wave is left handed polarized.

Question 42 |

Faraday’s law of electromagnetic induction is mathematically described by which one of the following equations?

Question 42 Explanation:

Differential from of Faraday’s law in given by

Question 43 |

Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. At 10 GHz operating frequency, the value of the propagation constant (per meter) of the corresponding propagating mode is _________.

Fill in the Blank Type Question |

Question 43 Explanation:

Question 44 |

Consider an air-filled rectangular waveguide with dimensions

*a*= 2.286 cm and*b*= 1.016 cm. The increasing order of the cut-off frequencies for different modes isQuestion 44 Explanation:

Question 45 |

A radar operating at 5 GHz uses a common antenna for transmission and reception. The antenna has a gain of 150 and is aligned for maximum directional radiation and reception to a target 1 km away having radar cross-section of . If it transmits 100 kW, then the received power (in )is _____.

Fill in the Blank Type Question |

Question 45 Explanation:

Question 46 |

The directivity of an antenna array can be increased by adding more antenna elements, as a larger number of elements

improves the radiation efficiency | |

increases the effective area of the antenna | |

results in a better impedance matching | |

allows more power to be transmitted by the antenna |

Question 46 Explanation:

As increasing no. of antennas, increases the effective area.

Effective area ( A

Therefore, the directivity of an antenna array can be increased by increasing the effective area of the antenna.

Effective area ( A

_{e}) and directivity (D) are related by,Therefore, the directivity of an antenna array can be increased by increasing the effective area of the antenna.

Question 47 |

A coaxial cable is made of two brass conductors. The spacing between the conductors is filled with Teflon (e

_{r}= 2.1, tan = 0). Which one of the following circuits can represent the lumped element model of a small piece of this cable having length z?Question 47 Explanation:

It is given that

So,

G -> Conductivity of the dielectric material

Hence, conductivity is

So

Since Loss tangent is zero between conductors hence loss will be zero in between medium.

So,

G -> Conductivity of the dielectric material

Hence, conductivity is

So

*G*= 0**Exp-2**Since Loss tangent is zero between conductors hence loss will be zero in between medium.

Question 48 |

In the circuit using an ideal opamp, the 3-dB cut-off frequency (in Hz) is __________.

Fill in the Blank Type Question |

Question 48 Explanation:

Question 49 |

Consider the 3 m long lossless air-filled transmission line shown in the figure. It has a characteristic impedance of 120pW, is terminated by a short circuit, and is excited with a frequency of 37.5 MHz. What is the nature of the input impedance (Z

_{in})?Open | |

Short | |

Inductive | |

Capacitive |

Question 49 Explanation:

Hence, input impedance is capacitive in nature.

Question 50 |

Right handed circular | |

Right handed elliptical | |

Left handed circular | |

Left handed elliptical |

Question 50 Explanation:

Question 51 |

In a source free region in vacuum, if the electrostatic potential = 2x

^{2}+ y^{2}+ cz^{2}, the value of constant c must be________________-2 | |

-3 | |

-4 | |

-5 |

Question 51 Explanation:

Given electrostatic potential

So, the electric field is obtained as

In source free region,

Substituting equation (1), we get

or – 4 –2 – 2C = 0

or C = – 3

So, the electric field is obtained as

In source free region,

Substituting equation (1), we get

or – 4 –2 – 2C = 0

or C = – 3

Question 52 |

Two half-wave dipole antennas placed as shown in the figure are excited with sinusoidally varying currents of frequency 3 MHz and phase shift of π/2 between them (the element at the origin leads in phase). If the maximum radiated E-field at the point P in the x-y plane occurs at an azimuthal angle of 60

^{o}the distance d (in meters) between the antennas is ____________.25 | |

30 | |

50 | |

100 |

Question 52 Explanation:

For maximum electric field, we have

Where

θ = Azimuthal angle = 60 º

α = Phase shift

Substituting these values in equation (1), we get

Where

θ = Azimuthal angle = 60 º

α = Phase shift

Substituting these values in equation (1), we get

Question 53 |

An air-filled rectangular waveguide of internal dimensions a cm × b cm(a > b) has a cutoff frequency of 6 GHz for the dominant TE

_{10}mode. For the same waveguide, if the cutoff frequency of the TM_{11}mode is 15 GHz, the cutoff frequency of the TE_{01}mode in GHz is ___________.27 | |

12.5 | |

15 | |

13.74 |

Question 53 Explanation:

Question 54 |

The force on a point charge +q kept at a distance d from the surface of an infinite grounded metal plate in a medium of permittivity is

0 | |

Question 54 Explanation:

Question 55 |

In spherical coordinates, let denote until vectors along the directions.

represent the electric and magnetic field components of the EM wave of large distances r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell located at

represent the electric and magnetic field components of the EM wave of large distances r from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell located at

45.5 | |

55.5 | |

65.5 | |

75.5 |

Question 55 Explanation:

Question 56 |

For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of the spacing between the plates by a factor of two results in

Halving of v and no change in Z | |

No changes in v and halving of Z | |

No change in both v and Z | |

Halving of both v and Z |

Question 56 Explanation:

d distance between the two plates

so, z

_{0}– changes, if the spacing between the plates changes.

independent of spacing between the plates

Question 57 |

The input impedance of a λ/8 section of a lossless transmission line of characteristic impedance 50Ω is found to be real when the other end is terminated by a load Z

_{L}(=R + jX)Ω. If X is 30Ω, the value of R (in Ω ) is _______30 | |

40 | |

50 | |

60 |

Question 57 Explanation:

Question 58 |

To maximize power transfer, a lossless transmission line is to be matched to a resistive load impedance via a transformer as shown.

The characteristic impedance (in) of the transformer is _________.

The characteristic impedance (in) of the transformer is _________.

60.85 | |

67.50 | |

70.70 | |

75.20 |

Question 58 Explanation:

Given the input impedance,

Z

Load impedance

Z

For a λ/4 transformer, the input impedance is given as

Z

Here impedance is matched by using QWT

Z

_{in}= 50ΩLoad impedance

Z

_{L}= 100ΩFor a λ/4 transformer, the input impedance is given as

Z

_{in}= (Z_{o})^{2}/Z_{L}Here impedance is matched by using QWT

Question 59 |

Which one of the following field patterns represents a TEM wave travelling in the positive x direction?

Question 59 Explanation:

For TEM wave

Electric field (E), Magnetic field (H) and

Direction of propagation (P) are orthogonal to each other.

Here P = + a

By verification

i.e. the TEM wave travelling in positive x -direction.

Electric field (E), Magnetic field (H) and

Direction of propagation (P) are orthogonal to each other.

Here P = + a

_{x}By verification

i.e. the TEM wave travelling in positive x -direction.

Question 60 |

Left elliptical | |

Left circular | |

Right elliptical | |

Right circular |

Question 60 Explanation:

Question 61 |

In the transmission line shown, the impedance Z

_{in}(in ohms) between node A and the ground is _________.22.22 | |

33.33 | |

44.44 | |

55.55 |

Question 61 Explanation:

Question 62 |

For a rectangular waveguide of internal dimensions a × b(a > b) , the cut-off frequency for the mode is the arithmetic mean of the cut-off frequencies for mode and mode. If the value of b (in cm) is _____.

1 cm | |

2 cm | |

4 cm | |

8 cm |

Question 62 Explanation:

Question 63 |

Consider an air filled rectangular waveguide with a cross-section of 5 cm × 3 cm. For this waveguide, the cut-off frequency (in MHz) of mode is _________.

7420 MHz | |

7640 MHz | |

7810 MHz | |

8410 MHz |

Question 63 Explanation:

Question 64 |

In the following figure, the transmitter Tx sends a wideband modulated RF signal via a coaxial cable to the receiver Rx. The output impedance of Tx, the characteristic impedance of the cable and the input impedance of Rx are all real.

Which one of the following statements is TRUE about the distortion of the received signal due to impedance mismatch?

Which one of the following statements is TRUE about the distortion of the received signal due to impedance mismatch?

The signal gets distorted if irrespective of the value of | |

The signal gets distorted if irrespective of the value of | |

Signal distortion implies impedance mismatch at both ends: | |

Impedance mismatches do NOT result in signal distortion but reduce power transfer efficiency |

Question 64 Explanation:

- If either of Z
_{R}and Z_{T}is matched with Z_{o}, it will cause complete absorption of signal travelling on the line. So, for distorted signal, We have both conditions Z_{T}≠ Z_{o}and Z_{R}≠ Z_{o} - Signal distortion implies impedance mismatch at both ends. i.e.,

Question 65 |

Given the vector
where denote unit vectors along x, y directions, respectively. The magnitude of curl of A is ___

0 | |

1 | |

2 | |

sinx |

Question 65 Explanation:

Question 66 |

A region shown below contains a perfect conducting half-space and air. The surface current on the surface of the perfect conductor is amperes per meter. The tangential field in the air just above the perfect conductor is

Question 66 Explanation:

Question 67 |

Assume that a plane wave in air with an electric field V/m is incident on a non-magnetic dielectric slab of relative permittivity 3 which covers the region. Z > 0 The angle of transmission in the dielectric slab is _________________ degrees.

30 ^{0} | |

45 ^{0} | |

60 ^{0} | |

90 ^{0} |

Question 67 Explanation:

Question 68 |

For an antenna radiating in free space, the electric field at a distance of 1 km is found to be 12mV/m. Given that intrinsic impedance of the free space is , the magnitude of average power density due to this antenna at a distance of 2 km from the antenna is________________.

50.7 | |

48.7 | |

45.7 | |

47.7 |

Question 68 Explanation:

Question 69 |

Match column A with column B

Question 69 Explanation:

A Yagi–Uda antenna, commonly known as a Yagi antenna, is a directional antenna consisting of multiple parallel elements in a line
Also called a "beam antenna", or "parasitic array", the Yagi is very widely used as a high-gain antenna on the HF, VHF and UHF bands. It has moderate to high gain which depends on the number of elements used, typically limited to about 20 dBi, linear polarization, unidirectional (end-fire) beam pattern with high front-to-back ratio of up to 20 dB. and is lightweight, inexpensive and simple to construct.

Figure: Yagi-uda antenna

**Refer**: https://en.wikipedia.org/Uda_antenna- Point electromagnetic source, can radiate fields in all directions equally, so isotropic.
- Dish antenna → highly directional
- Yagi – uda antenna → End fire

Figure: Yagi-uda antenna

Question 70 |

The electric field (assumed to be one-dimensional) between two points A and B is shown. Let be the electrostatic potentials at A and B, respectively. The value of in Volts is _________.

10 | |

-10 | |

15 | |

-15 |

Question 70 Explanation:

Question 71 |

2.14 | |

2.75 | |

3.14 | |

3.75 |

Question 71 Explanation:

Question 72 |

Consider a vector field . The closed loop line integral can be expressed as

over the closed surface bounded by the loop | |

over the closed volume bounded by the loop | |

over open volume bounded by the loop | |

over the open surface bounded by the loop |

Question 72 Explanation:

Stoke’s theorem states that the circulation a vector field around a closed path

i.e.,

Here, line integral is taken across a closed path which is denoted by a small circle on the integral notation where as, the surface integral of is taken over open surface bounded by the loop.

Hence correct option is D.

*l*is equal to the surface integral of the curl of over the open surface*S*bounded by*l*.i.e.,

Here, line integral is taken across a closed path which is denoted by a small circle on the integral notation where as, the surface integral of is taken over open surface bounded by the loop.

Hence correct option is D.

Question 73 |

The return loss of a device is found to be 20 dB. The voltage standing wave ratio (VSWR) and magnitude of reflection coefficient are respectively

1.22 and 0.1 | |

0.81 and 0.1 | |

– 1.22 and 0.1 | |

2.44 and 0.2 |

Question 73 Explanation:

Question 74 |

A plane wave propagation in air with V/m is incident on a perfectly conducting slab positioned at x ≤ 0. The field of the reflected wave is

Question 74 Explanation:

Since perfect conductor will reflect wave totally. Let reflected wave is

tangential component of incident wave is

Since at the boundary net tangential field will be zero.

For this tangential component of reflected wave and tangential component of incident wave must cancel out each other, for this

• Reflected wave will have normal component such that it will cancel out the normal component of incident wave so it will be .

• Also the direction of propagation will be in -x direction.

Question 75 |

The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction, is given by . The frequency and polarization of the wave, respectively, are

1.2 GHz and left circular | |

4 Hz and left circular | |

1.2 GHz and right circular | |

4 Hz and right circular |

Question 75 Explanation:

Question 76 |

A coacial cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is field with a dielectric of relative permittivity 10.89. Given , , the characteristic impedance of the cable is

330 Ω | |

100 Ω | |

143.2 Ω | |

15.89 Ω |

Question 76 Explanation:

Characteristics impedance of the co-axial cable is given by

Question 77 |

The radiation pattern of an antenna in spherical co-ordinates is given by

The directivity of the antenna is

The directivity of the antenna is

10 dB | |

12.6 dB | |

11.5 Db | |

18 dB |

Question 77 Explanation:

we know that directivity D is

……(1)

F(θ) is nothing but radiation intensity u(θ, ) and is radiated power.

……(2)

above equation is written from the formula

Where dΩ is solid angle and

So from (2)

…(3)

So,

In dB directivity = 10 log10 D = 10 dB

Question 78 |

A transmission line with a characteristic impedance of 100 Ω is used to match a 50 Ω section to a 200 Ω section. If the matching is to be done both at 429 MHz and 1 GHz. the length of the transmission line can be approximately

82.5 cm | |

1.05 m | |

1.58 m | |

1.75 m |

Question 78 Explanation:

Question 79 |

The direction of vector A is radially outward from the origin, with |A| = kr

^{n}where r^{2}= x^{2}+ y^{2}+ z^{2}and k is a constant. The value of n for which is–2 | |

2 | |

1 | |

0 |

Question 79 Explanation:

|A|=kr

(since it is radially outward)

in spherical coordinate is

So, will be zero if will be zero and will be zero if r

n + 2 = 0

⇒ n = -2

^{n}(since it is radially outward)

in spherical coordinate is

So, will be zero if will be zero and will be zero if r

^{n+2}will be constant and this is possible ifn + 2 = 0

⇒ n = -2

Question 80 |

An infinitely long uniform solid wire of radius a carries a uniform dc current of density .

The magnetic field at a distance r from the centre of the wire is proportional to

The magnetic field at a distance r from the centre of the wire is proportional to

r for r < a and l/r ^{2} for r > a | |

0 for r a | |

r for r < a and 1/r for r > a | |

0 for r2 for r > a |

Question 80 Explanation:

We know that magnetic flux density at a distance r from the wire is

So,

So,

for r > a

So,

So,

for r > a

Question 81 |

An infinitely long uniform solid wire of radius a carries a uniform dc current of density.

A hole of radius b(b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

The magnetic field inside the hole is

A hole of radius b(b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

The magnetic field inside the hole is

Uniform and depends only on d | |

Uniform and depends only on b | |

Uniform and depends on both b and d | |

Non uniform |

Question 81 Explanation:

Magnetic field inside hole depends on radius of hole i.e. b and also on the location of hole from center of the conductor i.e d. As hole has uniform cross section , magnetic field is uniform.

Question 82 |

The vector R

the coordinates of B will be

_{AB}extends from A(1, 2, 3) to B. If the length of R_{AB}is 10 units and its direction is given bythe coordinates of B will be

Question 82 Explanation:

Question 83 |

What is the value for the total charge enclosed in an incremental volume of 10

^{-9}m^{3}located at the origin if ?8 nC | |

4 nC | |

2 nC | |

1 nC |

Question 83 Explanation:

Question 84 |

The unit vector extending from origin toward the point G(2, -2. -1) is

Question 84 Explanation:

0(0,0,0) & G (2,–2,–1)

Question 85 |

Ground waves progress along the surface of the earth and must be polarized

horizontally | |

circularly | |

elliptically | |

vertically |

Question 85 Explanation:

Ground waves are vertically polarized waves because
Vertical polarization has considerably less attenuation than horizontally polarization.

Question 86 |

For a lossless line terminated in a short circuit, the stationary voltage minima and maxima are separated by

Question 86 Explanation:

In standing waves voltage maxima are separated by λ/2

In standing waves voltage minima are separated by λ/2

In standing waves voltage minima are separated by λ/2

**Therefore voltage maxima and voltage minima are separated by λ/4**Question 87 |

The characteristic impedance of an 80 cm long lossless transmission line having L = 0.25 μH/m and C = 100 pF/m will be

25 Ω | |

40 Ω | |

50 Ω | |

80 Ω |

Question 87 Explanation:

Given
L = 0.25 μH/m and C = 100 pF/m

As it is lossless transmission line so R=G=0

Characteristic impedance = =

∴ = 50Ω

As it is lossless transmission line so R=G=0

Characteristic impedance = =

∴ = 50Ω

Question 88 |

It is required to match a 200 Ω load to a 300 Ω transmission line to reduce the SWR along the line to 1. If it is connected directly to the load, the characteristic impedance of the quarterwave transformer used for this purpose will be

275 Ω | |

260 Ω | |

245 Ω | |

230 Ω |

Question 88 Explanation:

= 200Ω, = 300Ω

∴ (√6 * 100 Ω) = 245Ω

∴ (√6 * 100 Ω) = 245Ω

Question 89 |

For a standard rectangular waveguide having an aspect ratio of 2: 1 the cutoff wavelength for TM

_{1,1}mode will be nearly0.9a | |

0.7a | |

0.5a | |

0.3a |

Question 89 Explanation:

Aspect ratio

∴ λ

∴ λ

_{C11}= 0.9 aQuestion 90 |

The irises in the rectangular metallic waveguide may be

1) inductive

2) resistive

3) capacitive

Select the correct answer using the code given below.

1) inductive

2) resistive

3) capacitive

Select the correct answer using the code given below.

1, 2 and 3 | |

1 and 2 only | |

1 and 3 only | |

2 and 3 only |

Question 90 Explanation:

The irises in the metallic wave guide can be classify as

1. Inductive

2. Capacities

3. Tune circuit (LC)

In option we are seeing inductive or capacitive so Option(C) is correct

1. Inductive

2. Capacities

3. Tune circuit (LC)

In option we are seeing inductive or capacitive so Option(C) is correct

Question 91 |

A 10 GHz signal is propagated in a waveguide whose wall separation is 6 cm. The greatest number of half-waves of electric intensity will be possible to establish between the two walls. The guide wavelength for this mode of propagation will be

6.48 cm | |

4.54 cm | |

2.48 cm | |

1.54 cm |

Question 91 Explanation:

Question 92 |

In Te

_{m, n}mode m and n are integers denoting the number ofthe wavelengths of intensity between each pair of walls | |

the wavelengths of intensity between each pair of walls | |

the wavelengths of intensity between each pair of walls | |

the wavelengths of intensity between each pair of walls |

Question 92 Explanation:

In

m, n stands for feed connection machanism and positioning of feed.

m & n are determined by number of feed connection

and also they define number of Halfcycle between guide walls

**TE**modes_{m,n}m, n stands for feed connection machanism and positioning of feed.

m & n are determined by number of feed connection

and also they define number of Halfcycle between guide walls

Question 93 |

Consider the following statements with reference to dipole arrays:

1) In broadside array, all the dipoles are fed in the same phase from the same source.

2) In end-fire array, the magnitude of the current in each element is same and there is no phase difference between these currents.

Which of the above statements is/are correct?

1) In broadside array, all the dipoles are fed in the same phase from the same source.

2) In end-fire array, the magnitude of the current in each element is same and there is no phase difference between these currents.

Which of the above statements is/are correct?

1 only | |

2 only | |

Both 1 and 2 | |

Neither 1 nor 2 |

Question 93 Explanation:

We know that E = 2Eo cos (Ψ/2)

For broad side array Q

For E to be maximum Ψ = 0

∴ Ψ = α + β d cos Q = 0

∴ α + β d cos (90) = 0

∴ α = 0

For end fire array Q

∴ Ψ = α + β d cos Q = 0

∴ α + β d cos ( 0 or 180

∴ α = ± β d

∴ Broadside array α= 0

∴Endfire array α = ± βd

∴so only statement 1 is correct

For broad side array Q

_{max}must be 90^{o}or 270^{o}For E to be maximum Ψ = 0

∴ Ψ = α + β d cos Q = 0

∴ α + β d cos (90) = 0

∴ α = 0

For end fire array Q

_{max}must be 0^{o}or 180^{o}∴ Ψ = α + β d cos Q = 0

∴ α + β d cos ( 0 or 180

^{o}) = 0∴ α = ± β d

∴ Broadside array α= 0

∴Endfire array α = ± βd

∴so only statement 1 is correct

Question 94 |

An antenna is fed with 200 W power. The efficiency of the antenna is 75%. If the radiation pattern of the antenna is pattern of the antenna is for (azimuth angle) and (elevation angle). =0 elsewhere

Find the radiation intensity in the direction of maximum radiation

Find the radiation intensity in the direction of maximum radiation

225 W/sterdian | |

150 W/steradian | |

200 W/steradian | |

250 W/steradian |

Question 94 Explanation:

The Correct Answer Among All the Options is A

= =>

Maximum radiation intensity is given by,

=> , where

Also it is given , for (azimuth angle) and (elevation angle).

For a given radiation pattern ,

=

=

=

Now, => = 225 watt/str

Refer the Topic Wise Question for Antenna Electromagnetics

= =>

Maximum radiation intensity is given by,

=> , where

Also it is given , for (azimuth angle) and (elevation angle).

For a given radiation pattern ,

=

=

=

Now, => = 225 watt/str

Refer the Topic Wise Question for Antenna Electromagnetics

Question 95 |

Two communication antennas A and B, are operating at 300 MHz and other at 3 GHz respectively and having same gain, are illuminated with identical flux density of -100 dBW/m

^{2}. What is the relation between the received powers (P_{A}: P_{B})?1:10 | |

10:1 | |

1:100 | |

100:1 |

Question 95 Explanation:

The Correct Answer Among All the Options is D

For receiving antenna

=

=

=100:1

Refer the Topic Wise Question for Antenna Electromagnetics

For receiving antenna

=

=

=100:1

Refer the Topic Wise Question for Antenna Electromagnetics

Question 96 |

Design a single section, quarter wave impedance transformer at 5 GHz from 3.75 cm x 2 cm guide to 3.75 cm x 1 cm guide. Assume air filled wave guide with transformer section having same width as that of the input and output sections.

Height = 1.414 cm, Length = 3 cm | |

Height = 1.5 cm, Length = 2.5 cm | |

Height = 1.414 cm, Length = 2.5 cm | |

Height = 1.5 cm, Length = 3 cm |

Question 96 Explanation:

The Correct Answer Among All the Options is C

height =1.414cm

Length = 2.5cm

Refer the Topic Wise Question for Waveguides Electromagnetics

height =1.414cm

Length = 2.5cm

Refer the Topic Wise Question for Waveguides Electromagnetics

Question 97 |

A cell phone transmits at a power level of 800mW with an antenna gain of 3.0 dB. The cell tower has an antenna gain of 10.0 dB and is at a distance of 5 km away. Transmission frequency is 600 MHz. Noise level at Receiver Input is -95 dBm and required Signal to Noise ratio to close the link is 5dB. Find the link margin in dB. (assume )

150 dB | |

60 dB | |

30 dB | |

35 dB |

Question 97 Explanation:

The Correct Answer Among All the Options is C

= transmitted power

=transmitter gain =3dB=10=>

=effective aperture =

R= distance from transmitter

= =10dB = 10 =>

On putting values, we get, =

Given noise(N) in dB=-95dBm =10

ratio in dB :

10 =10log - 10logN

=10 log - (-95)

=35dB

Required ratio in dB = 5dB

Link margin in dB=35dB-5dB

= 30dB

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

= transmitted power

=transmitter gain =3dB=10=>

=effective aperture =

R= distance from transmitter

= =10dB = 10 =>

On putting values, we get, =

Given noise(N) in dB=-95dBm =10

ratio in dB :

10 =10log - 10logN

=10 log - (-95)

=35dB

Required ratio in dB = 5dB

Link margin in dB=35dB-5dB

= 30dB

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Question 98 |

What is the value of magnetic flux in Weber, if it is 2000 in Maxwell?

2 x 10 ^{-5} | |

2 x 10 ^{-3} | |

2 x 10 ^{5} | |

2 x 10 ^{3} |

Question 98 Explanation:

The Correct Answer Among All the Options is A

1 Maxwell =

2000 Maxwell = 2000

= 2

Refer the Topic Wise Question for Electrostatics Electromagnetics

1 Maxwell =

2000 Maxwell = 2000

= 2

Refer the Topic Wise Question for Electrostatics Electromagnetics

Question 99 |

A monostatic pulsed radar operating at 30GHz has a transmitter with 2KW O/P power and an antenna with 30dB gain. Minimum detectable signal in the receiver is -100dBm. Determine the maximum range of the radar, if it is requidred to detect a target having radar cross section of 10sq.m (consider log

_{10}4 = 1.1). Assume EM wave propagate under ideal conditions.10 km | |

21.5 km | |

56 km | |

100 km |

Question 99 Explanation:

The Correct Answer Among All the Options is A

Given data are

F=30GHz =>

30 =>

Radar range equation :

R =(

On putting all values. We get

R = m=10km

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Given data are

F=30GHz =>

30 =>

Radar range equation :

R =(

On putting all values. We get

R = m=10km

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Question 100 |

Which of the following statement is not true about delay line cancellers?

It eliminates DC components of fixed targets and passes AC components of moving targets | |

It is used in moving target indicator radar | |

Time delay in one channel of the delay line canceller is one half of the pulse repetition period | |

It rejects any moving target whose Doppler frequency happens to be the same as the PRF or a multiple thereof |

Question 100 Explanation:

The Correct Answer Among All the Options is C

Delay lines are used in moving target indicator radar. moving target indicator removes the clutter due to stationary targets and ground objects. The basic principle of moving target indicator is to compare a set of received echoes with those received during the previous sweep and cancelling out those whose phase has remain unchanged. Stationary target will give same phase in every cycle unlike moving targey. A single pulse of the received signal at the PRF(pulse repetition frequency) i.e the reciprocal of the time between successive pulses was stored in the delay line. .

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Delay lines are used in moving target indicator radar. moving target indicator removes the clutter due to stationary targets and ground objects. The basic principle of moving target indicator is to compare a set of received echoes with those received during the previous sweep and cancelling out those whose phase has remain unchanged. Stationary target will give same phase in every cycle unlike moving targey. A single pulse of the received signal at the PRF(pulse repetition frequency) i.e the reciprocal of the time between successive pulses was stored in the delay line. .

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Question 101 |

An air filled rectangular waveguide with dimensions a = 75 mm, b = 37.5 mm has same guide wavelength at frequencies f

_{1}and f_{2}when operated at TE_{10}and TE_{20}modes respectively. If the frequency f_{1}is GHz, what is frequency f_{2}in GHz?10 | |

5 | |

Question 101 Explanation:

The Correct Answer Among All the Options is B

Given , a=75mm

b=37.5mm

As we know, guide wavelength in rectangular waveguide,

Cut-off frequency is given by ,

And Cut-off frequency in case of air-filled waveguide ,

For mode,

For mode,

Now,

=

On solving we get,

Refer the Topic Wise Question for Waveguides Electromagnetics

Given , a=75mm

b=37.5mm

As we know, guide wavelength in rectangular waveguide,

Cut-off frequency is given by ,

And Cut-off frequency in case of air-filled waveguide ,

For mode,

For mode,

Now,

=

On solving we get,

Refer the Topic Wise Question for Waveguides Electromagnetics

Question 102 |

A waveguide of dimensions a = 15 mm and b = 7.5 mm is used as a high-pass filter. If the stop band attenuation required at 8 GHz is ~109.2 dB, what is the length of the filter? (assume conductor losses to be zero, approximate = 3.14 and 1 Np ~8.69 dB) (log

_{10}e = 0.4343)100 mm | |

869 mm | |

86.9 mm | |

54.6 mm |

Question 102 Explanation:

The Correct Answer Among All the Options is A

a=15mm, b=7.5mm

For

Propagation constant

If , the will become real and it will attenuate.

Therefore,

For m=1, n=0

For dominate mode ,

=10 =40??

So ,

Attenuation at 8GHz= -109.2dB

As we know in a waveguide, if input is then o/p after travelling z distance

20 = -109.2dB

On solving, we get ,

z=100mm

Refer the Topic Wise Question for Waveguides Electromagnetics

a=15mm, b=7.5mm

For

Propagation constant

If , the will become real and it will attenuate.

Therefore,

For m=1, n=0

For dominate mode ,

=10 =40??

So ,

Attenuation at 8GHz= -109.2dB

As we know in a waveguide, if input is then o/p after travelling z distance

20 = -109.2dB

On solving, we get ,

z=100mm

Refer the Topic Wise Question for Waveguides Electromagnetics

Question 103 |

An RF signal is applied to a 50 lossless transmission line which is terminated in a load with impedance, Z

_{L}= j50. The wavelength is 8 cm. Find the position of voltage and current maximum respectively nearest to the load measured from load end?1 cm, 3 cm | |

3 cm, 1 cm | |

3 cm, 5 cm | |

5 cm, 3 cm |

Question 103 Explanation:

The Correct Answer Among All the Options is A

, it is purely imaginary.

We know,

if load is resistive then waveform of is

if load is purely inductive then waveform of is

, this is equation to find location of maxima.

Reflection coefficient () = =

Therefore,

=1cm (it is voltage maxima)

From the above drawn waveform we have seen that current maxima occurs after distance

= 1+ =1+ = 3cm

Refer the Topic Wise Question for Transmission Lines Electromagnetics

, it is purely imaginary.

We know,

if load is resistive then waveform of is

if load is purely inductive then waveform of is

, this is equation to find location of maxima.

Reflection coefficient () = =

Therefore,

=1cm (it is voltage maxima)

From the above drawn waveform we have seen that current maxima occurs after distance

= 1+ =1+ = 3cm

Refer the Topic Wise Question for Transmission Lines Electromagnetics

Question 104 |

For a conservative vector field F below, which of the following is the scalar Potential?

F = a is an integer

F = a is an integer

Cannot be found | |

Question 104 Explanation:

The Correct Answer Among All the Options is B

for a conservative /irrotational field

=0

= 0

On solving it for ‘a’, we get

a =1

now, () +() ……………….(1)

as we know , , where V is scalar potential.

…………………(2)

On comparing (1) and (2)

= + f(y, z) {f(y,z) is a constant}

= + f(y, z)

Similarly, =

Therefore V= + +

V=

Refer the Topic Wise Question for Electrostatics Electromagnetics

for a conservative /irrotational field

=0

= 0

On solving it for ‘a’, we get

a =1

now, () +() ……………….(1)

as we know , , where V is scalar potential.

…………………(2)

On comparing (1) and (2)

= + f(y, z) {f(y,z) is a constant}

= + f(y, z)

Similarly, =

Therefore V= + +

V=

Refer the Topic Wise Question for Electrostatics Electromagnetics

Question 105 |

A cylindrical waveguide with radious of 3.5 cm has waves travelling in TM

_{12}mode. The value of 1^{st}zero of 2^{nd}order Bessel function is 7. Find the cut-off wavelength for this mode. cm | |

1.5 cm | |

2 cm | |

/2 cm |

Question 105 Explanation:

The Correct Answer Among All the Options is A

for circular waveguide in modes.

n= 1 (first zero);m=2(second order)

Given, , a=3.5cm

Therefore ,

Cut-off wavelength(= =>

Refer the Topic Wise Question for Waveguides Electromagnetics

for circular waveguide in modes.

n= 1 (first zero);m=2(second order)

Given, , a=3.5cm

Therefore ,

Cut-off wavelength(= =>

Refer the Topic Wise Question for Waveguides Electromagnetics

Question 106 |

Relationship between doppler frequency shifts of two radars A and B having 0.1 foot and 0.05 foot wavelengths, approaching the target at 1000 feet per second and 2000 feet per second rate respectively, will be

Doppler frequency shift of radar A will be one-fourth of doppler frequency shift of radar B | |

Doppler frequency shift of radar A will be one-half of doppler frequency shift of radar B | |

Doppler frequency shift of radar A will be double of doppler frequency shift of radar B | |

Doppler frequency shifts of radar A and radar B will be same |

Question 106 Explanation:

The Correct Answer Among All the Options is A

Given=> Wavelength of Radar A= 0.1ft, and wavelength of radar B=0.05ft.

Velocity of Radar A=1000ft/second, and velocity of radar B=2000ft/second. We have to determine the doppler shift for both the radars to check which one is greater than the other, and by how much.

Now, doppler shift of a radar is generally given by the formula: 2V/λ

Now for radar A=> 2(1000)/0.1λ=> 20,000/λ

And for radar B=> 2(2000)/0.05λ=> 80,000/λ

So, DS

So, clearly we can say that:

Doppler shift(Radar A)= (1/4)(Doppler Shift Radar B).

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Given=> Wavelength of Radar A= 0.1ft, and wavelength of radar B=0.05ft.

Velocity of Radar A=1000ft/second, and velocity of radar B=2000ft/second. We have to determine the doppler shift for both the radars to check which one is greater than the other, and by how much.

Now, doppler shift of a radar is generally given by the formula: 2V/λ

Now for radar A=> 2(1000)/0.1λ=> 20,000/λ

And for radar B=> 2(2000)/0.05λ=> 80,000/λ

So, DS

_{B}/DS_{A}=> 80,000/λ / 2000/λSo, clearly we can say that:

Doppler shift(Radar A)= (1/4)(Doppler Shift Radar B).

Refer the Topic Wise Question for RADAR and Optical Fibers Electromagnetics

Question 107 |

The temperature below which certain materials are antiferromagnetic and above which they are paramagnetic is called

Weiss temperature | |

Curie temperature | |

Neel temperature | |

None of the above |

Question 107 Explanation:

The Correct Answer Among All the Options is C

Neel temperature also called as magnetic ordering temperature is a certain temperature above which antiferro-magnetic materials become paramagnetic. This is the temperature at which the thermal energy of antiferromagnetic material gets to such a point where it becomes large enough to damage the magnetic ordering.

Refer the Topic Wise Question for Electrostatics Electromagnetics

Neel temperature also called as magnetic ordering temperature is a certain temperature above which antiferro-magnetic materials become paramagnetic. This is the temperature at which the thermal energy of antiferromagnetic material gets to such a point where it becomes large enough to damage the magnetic ordering.

Refer the Topic Wise Question for Electrostatics Electromagnetics

Question 108 |

In a specimen of ferromagnetic material with saturation magnetization as 8000 Gauss, as the flux density is increased from 0 to 2.5 T, will

Increase | |

Decrease | |

First decrease then increase | |

First increase then decrease |

Question 108 Explanation:

The Correct Answer Among All the Options is D

Generally B= µ

Now, from this we can write, µ

The B-H curve of a ferromagnetic material is shown below:

'H' represents magnetic field strength, and 'B' represents degree of magnetization.

Now, at the origin=> No magnetic field, no magnetization. Now if we apply the magnetic field, then the ferromagnetic material becomes magnetic. The B-H curve generally is an indication of how much µ

Refer the Topic Wise Question for Electrostatics Electromagnetics

Generally B= µ

_{0}µ_{r}(H+M).Now, from this we can write, µ

_{r}=(B/ µ_{0})HThe B-H curve of a ferromagnetic material is shown below:

'H' represents magnetic field strength, and 'B' represents degree of magnetization.

Now, at the origin=> No magnetic field, no magnetization. Now if we apply the magnetic field, then the ferromagnetic material becomes magnetic. The B-H curve generally is an indication of how much µ

_{r}is changing/varying. So, definitely we can say that answer is option (d)Refer the Topic Wise Question for Electrostatics Electromagnetics

Question 109 |

An electromagnetic wave propagates through a lossless insulator with a velocity 1.5x10

^{10}cm/s. Calculate the electric and magnetic properties of the insulator if its intrinsic impedance is ohms =2.66 =1.5 | |

=1.5 = 2.66 | |

=1.2 = 2.0 | |

=2.0=1.2 |

Question 109 Explanation:

The Correct Answer Among All the Options is A

Given=> V=1.5 x 10

Intrinsic impedance (Z

We have to determine => ξ

We know that intrinsic impedance is given as:

Z

And velocity, V= C/

On putting V and C values, we get: µ

From (a) we can write: = 90ᴨ/120ᴨ

So=> µ

Multiplying (b) and (c) we get: µ

Again, from (b), putting the value of µ

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Given=> V=1.5 x 10

^{10}cm/s=1.5 x 10^{8}m/secIntrinsic impedance (Z

_{0})= 90ᴨWe have to determine => ξ

_{r}and µ_{r}We know that intrinsic impedance is given as:

Z

_{0}= 120ᴨ =>90ᴨ=120ᴨ......................... (a)And velocity, V= C/

On putting V and C values, we get: µ

_{r}ξ_{r}=4.... (b)From (a) we can write: = 90ᴨ/120ᴨ

So=> µ

_{r}/ξ_{r}= 9/16..... (c)Multiplying (b) and (c) we get: µ

_{r}=3/2Again, from (b), putting the value of µ

_{r}=3/2, we get: ξ_{r}=8/3= 2.66, and µ_{r}=3/2=1.5Refer the Topic Wise Question for Waves and Properties Electromagnetics

Question 110 |

A square waveguide carries TE

_{11}mode whose axial magnetic field is given by H_{z}= H_{0}cos(x/) cos(y/) A/m, where waveguide dimensions are in cm. What is the cut-off frequency of the mode?5.5 GHz | |

6.5 GHz | |

7.5 GHz | |

8.5 GHz |

Question 110 Explanation:

The Correct Answer Among All the Options is C

Given: TE

Axial magnetic field H

We have to determine=> Cut-off frequency of TE

General form of axial magnetic field is given as:

H

Now, TE

Comparing with the equation given in question we get: a= and b=

So, cut-off wavelength λ

So, on putting values=> λ

So, cut-off frequency f

Refer the Topic Wise Question for Waveguides Electromagnetics

Given: TE

_{11}mode(Square waveguide)Axial magnetic field H

_{z}= H_{0}cos(ᴨ/)cos(ᴨy/We have to determine=> Cut-off frequency of TE

_{11}General form of axial magnetic field is given as:

H

_{z}= H_{0}cos(mᴨx/a)cos(nᴨy/b)Now, TE

_{11}=> m=1, n=1Comparing with the equation given in question we get: a= and b=

So, cut-off wavelength λ

_{c}= 2ab/So, on putting values=> λ

_{c}=4So, cut-off frequency f

_{c}= C/λ=> 7.5GHzRefer the Topic Wise Question for Waveguides Electromagnetics

Question 111 |

The Eddy current loss is proportional to the

Frequency | |

Square of the frequency | |

Cube of the frequency | |

Square root of the frequency |

Question 111 Explanation:

The Correct Answer Among All the Options is B

Eddy current loss in a transformer is fixed, and it depends on the core material's magnetic properties. Eddy current loss is given as:

W

So definitely answer is option (b)

Refer the Topic Wise Question for Electrostatics Electromagnetics

Eddy current loss in a transformer is fixed, and it depends on the core material's magnetic properties. Eddy current loss is given as:

W

_{e}= PB^{2}_{max}.f^{2}t^{2}wattSo definitely answer is option (b)

Refer the Topic Wise Question for Electrostatics Electromagnetics

Question 112 |

From the following Relative amplitude vs Frequency plot, identify the type of noise which the sections A, B, C & D depict.

(i) Thermal Noise (ii) Power line pick up

(iii) Power supply (EPC) switching noise (iv) 1/f noise

(i) Thermal Noise (ii) Power line pick up

(iii) Power supply (EPC) switching noise (iv) 1/f noise

A-i, B-ii, D-iv | |

A-ii, B-i, C-iv, D-iii | |

A-iv, B-ii, C-iii, D-i | |

B-iv, C-ii, D-i |

Question 112 Explanation:

The Correct Answer Among All the Options is C

Here, from the graph we have to determine the type of noise at each region.

First region A=> power is minus, so it is extremely low frequency, which is why it can said as Pink noise or '1/f' noise. The PSD of this noise is inversely proportional to the frequency of the signal.

Next region B=> This is power line pick up noise. Power line noise generally interferes with wireless communication and broadcasting. It can also disrupt television, radio and other important communication systems.

Next region C=> It is the noise that mainly arises in switched mode power supply systems. It is the power supply switching noise.

Last is region D=> It is thermal noise. This noise is also called as Johnson nyquist noise, and it arises due to the thermal distress (charge carriers).

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Here, from the graph we have to determine the type of noise at each region.

First region A=> power is minus, so it is extremely low frequency, which is why it can said as Pink noise or '1/f' noise. The PSD of this noise is inversely proportional to the frequency of the signal.

Next region B=> This is power line pick up noise. Power line noise generally interferes with wireless communication and broadcasting. It can also disrupt television, radio and other important communication systems.

Next region C=> It is the noise that mainly arises in switched mode power supply systems. It is the power supply switching noise.

Last is region D=> It is thermal noise. This noise is also called as Johnson nyquist noise, and it arises due to the thermal distress (charge carriers).

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Question 113 |

The electric field of a linearly polarized electromagnetic wave is given by is incident upon a linearly polarized antenna whose electric field polarization is expressed as Find the polarization loss factor.

1/2 | |

3/2 | |

2/3 | |

1/4 |

Question 113 Explanation:

The Correct Answer Among All the Options is A

Here we have to find out polarization loss factor. Now, Polarization loss is generally given by the formula:

|E(incident). E(transmitted)|

Now, from the data given in the question, we can put the values in the formula as:

|a

On solving this we get= 1/2

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Here we have to find out polarization loss factor. Now, Polarization loss is generally given by the formula:

|E(incident). E(transmitted)|

^{2}/ |E(incident)|^{2}. |E(transmitted)|^{2}Now, from the data given in the question, we can put the values in the formula as:

|a

_{x}(a_{x}+a_{y})|^{2}/|a_{x}|^{2}|a_{x}+a_{y}|^{2}On solving this we get= 1/2

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Question 114 |

A lossless T-junction two way power divider has a source impedance, input transmission line impendence and o/p port load impendence of 50. Find the output characterization impedances so that the input power is divided in a 2:1 ratio.

z _{1}= 150, z_{2} = 75 | |

z _{1} = 50, z_{2}= 100 | |

z _{1}= 60, z_{2}= 120 | |

z _{1} = 30, z_{2}= 60 |

Question 114 Explanation:

The Correct Answer Among All the Options is A

The T junction power divider is a network that generally has 3 ports, and is used for power dividing or power combining.

The equivalent electrical circuit is shown below:

Here, Yin= jB+(1/Z1)+(1/Z2)=(1/Zc)

Now, input power(at the matched divider) can be given as: P

So, based on this the other output powers can be calculated as follows:

P

So, now, Z

Z

Also, Z

Refer the Topic Wise Question for Transmission Lines Electromagnetics

The T junction power divider is a network that generally has 3 ports, and is used for power dividing or power combining.

The equivalent electrical circuit is shown below:

Here, Yin= jB+(1/Z1)+(1/Z2)=(1/Zc)

Now, input power(at the matched divider) can be given as: P

_{in}= (1/2)(|V|^{2}/Z_{1})=P_{in}/3So, based on this the other output powers can be calculated as follows:

P

_{1}= (1/3)P_{in}and P_{2}= (2/3)P_{in}[Since input power is divided in the ratio 2:1]So, now, Z

_{1}and Z_{2}can be calculated as:Z

_{1}= 150Ω[Since Z_{1}=3Z_{c}=3(50) is given in the question]Also, Z

_{2}= 75Ω[Since Z_{2}= (3/2)Z_{c}=(3/2)(50)]Refer the Topic Wise Question for Transmission Lines Electromagnetics

Question 115 |

A gain-standard horn is known to have a gain G = 10. It is being used to measure the gain of a large directional antenna by the comparison method. When the antenna being measured is connected to the receiver it is found to be necessary to insert an attenuator adjusted to attenuate by 23 dB in order to have the same receiver output that was observed with the horn connected. What is the gain of the large antenna?

13 dB | |

23 dB | |

33 dB | |

230 dB |

Question 115 Explanation:

The Correct Answer Among All the Options is C

Given=> Gain(G) of horn antenna=10

Si, in dB, gain G= 10log10=10dB

Also=> We have a large directional antenna.

Now, gain of large directional antenna - Attenuator=10dB

Let large directional antenna gain = 'X' dB

So, we can write: X-23dB=10dB=>X=33dB

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Given=> Gain(G) of horn antenna=10

Si, in dB, gain G= 10log10=10dB

Also=> We have a large directional antenna.

Now, gain of large directional antenna - Attenuator=10dB

Let large directional antenna gain = 'X' dB

So, we can write: X-23dB=10dB=>X=33dB

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Question 116 |

A paraboloidal-reflector antenna is designed for operation at 3 GHz. Its largest aperture dimension is 20 feet. It is desired to build a scale model of this antenna with the largest aperture dimension scaled to 18 inches. At what frequency must this model be operated in order to have the same pattern as the full-size antenna?

10 GHz | |

20 GHz | |

40 GHz | |

4 GHz |

Question 116 Explanation:

The Correct Answer Among All the Options is C

Given=> Frequency(Paraboloid reflector antenna)=3GHz

and D=20 feet

We know that for a paraboloid reflector antenna gain is given as: G= 6(λf)

Now 18 inches=1.5 feet.

We know that Gain 6(Df)

From this we can say that Gain (Df)

So=> (D

D

Therefore=> (20 x 3)

On solving we get: f

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Given=> Frequency(Paraboloid reflector antenna)=3GHz

and D=20 feet

We know that for a paraboloid reflector antenna gain is given as: G= 6(λf)

^{2}Now 18 inches=1.5 feet.

We know that Gain 6(Df)

^{2}From this we can say that Gain (Df)

^{2}So=> (D

_{1}f_{1})^{2}= (D_{2}f_{2})^{2}[D_{2}f_{2}=> parameter of scale model of original antenna]D

_{2}=20feet and D_{1}=18 inch=1.5 feetTherefore=> (20 x 3)

^{2}= (1.5 x f_{2})^{2}On solving we get: f

_{2}=40GHzRefer the Topic Wise Question for Waves and Properties Electromagnetics

Question 117 |

An antenna has a radiation resistance of 72 , a loss resistance of 8 Ω and power gain of 16. Calculate its directivity.

15.8 | |

16.8 | |

17.8 | |

.18.7 |

Question 117 Explanation:

The Correct Answer Among All the Options is C

Given=> R

We have to determine=> Directivity

Now, gain of antenna=Directivity x (ո)

And ո= R

So, D= Gain/ո => 17.8

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Given=> R

_{r}=72Ω, R_{Loss}=8Ω and P_{gain}=16We have to determine=> Directivity

Now, gain of antenna=Directivity x (ո)

And ո= R

_{rad}/ (R_{rad}+R_{loss})So, D= Gain/ո => 17.8

Refer the Topic Wise Question for Waves and Properties Electromagnetics

Question 118 |

The current density at the surface of a thick metal plate is 100 A/m

^{2}. What is the skin depth if the current density at a depth of 0.0059 cm is 0.272 A/m^{2}?5m | |

10 m | |

15m | |

20m |

Question 118 Explanation:

The Correct Answer Among All the Options is B

Given=> Current density=> 100A/m

We have to find=> Skin depth(δ)

Now, δ= (1/σ)

Current density reduces in metal(from surface) at the rate equal to=>

So, we can write: 100=0.27

Taking Log on both sides:

- .0059 x 10

On solving this we will get δ=10 µm

Refer the Topic Wise Question for Electrostatics Electromagnetics

Given=> Current density=> 100A/m

^{2}We have to find=> Skin depth(δ)

Now, δ= (1/σ)

Current density reduces in metal(from surface) at the rate equal to=>

So, we can write: 100=0.27

Taking Log on both sides:

- .0059 x 10

^{-2}/ δ = ln[0.272/1000]On solving this we will get δ=10 µm

Refer the Topic Wise Question for Electrostatics Electromagnetics

There are 118 questions to complete.