Test: Transverse Electric Waves(TE) And Magnetic Waves(TM)


24 Questions MCQ Test Electromagnetic Theory | Test: Transverse Electric Waves(TE) And Magnetic Waves(TM)


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QUESTION: 1

In transverse electric waves, which of the following is true?

Solution:

Answer: c
Explanation: In TE waves, the electric field strength will be transverse to the wave direction. Thus the TE waves are also called H waves.

QUESTION: 2

The dominant mode in rectangular waveguide is 

Solution:

Answer: b
Explanation: TE10 is the dominant mode in the rectangular waveguide. This is because it gives the minimum cut off frequency required for transmission.

QUESTION: 3

The number of modes in a waveguide having a V number of 10 is

Solution:

Answer: d
Explanation: The number of modes is given by m = V2/2, where V is the v number. On substituting for V = 10, we get m = 100/2 = 50.

QUESTION: 4

Does the waveguide with dimensions 3 cm x 5.5 cm exist?

Solution:

Answer: b
Explanation: For a waveguide, the dimension a should be greater than b. Here a = 3 and b = 5.5, thus such waveguide does not exist.

QUESTION: 5

The mode which has the highest wavelength is called

Solution:

Answer: a
Explanation: Dominant modes are the modes having least cut off frequency. This implies they have highest cut off wavelength.

QUESTION: 6

The intrinsic impedance of a TE wave having a cut off frequency of 6 GHz at a frequency of 7.5 GHz in air is

Solution:

Answer: a
Explanation: The intrinsic impedance of a TE wave is given by ηTE = η/cos θ, where cos θ is given by √(1- (fc/f)2). On substituting for fc = 6 GHz, f = 7.5 GHz and η = 377, we get the intrinsic impedance as 628.33 units.

QUESTION: 7

The cut off frequency of a rectangular waveguide of dimensions 3 cm x 1.5 cm is

Solution:

Answer: d
Explanation: The cut off frequency in dominant mode will be fc = mc/2a. On substituting for c = 3 x 108 and a = 0.03, we get the cut off frequency as 5 GHz.

QUESTION: 8

The propagation constant for a lossless transmission line will be

Solution:

Answer: d
Explanation: The propagation constant is given by γ = α + jβ, where α and β are the attenuation and phase constants respectively. For a lossless line, the attenuation constant is zero. Thus γ = jβ. It is clear that γ is complex and equal to β.

QUESTION: 9

The attenuation of a 50 ohm transmission line having a resistance of 100 ohm is

Solution:

Answer: c
Explanation: The attenuation of a wave is given by α = R/2Z0. On substituting for R = 100 and Z0 = 50, we get α = 100/(2 x 50) = 1 unit.

QUESTION: 10

The cut off frequency of a TE wave with waveguide dimension of a= 3.5 cm in a medium of permittivity 2.2 is

Solution:

Answer: a
Explanation: The cut off frequency of a TE wave in any other medium is mc/2a√εr. On substituting for a = 0.035 and εr = 2.2, we get the cut off frequency as 2.88 GHz.

QUESTION: 11

The phase constant and frequency are related by

Solution:

Answer: a
Explanation: The phase constant is given by β = ω√LC. Thus the relation is β is directly proportional to ω.

QUESTION: 12

Which of the following parameter is non zero for a lossless line?

Solution:

Answer: d
Explanation: The attenuation constant, resistance and conductance are zero for a lossless line. Only the phase constant is non zero.

QUESTION: 13

In transverse magnetic waves, which of the following is true?

Solution:

Answer: c
Explanation: In transverse magnetic waves, the magnetic field strength is transverse to the wave direction. They are also called E waves

QUESTION: 14

The dominant mode in the TM waves is 

Solution:

Answer: d
Explanation: The modes TM10, TM01 and TM20 does not exist in any waveguide. The TM11mode is the dominant mode in the waveguide.

QUESTION: 15

The modes in a waveguide having a V number of 20 is

Solution:

Answer: b
Explanation: The number of modes in a waveguide is given by m = V2/2. On substituting for V = 20, we get m = 400/2 = 200 modes.

QUESTION: 16

The v number of a waveguide having 120 modes is

Solution:

Answer: a
Explanation: The number of modes in a waveguide is given by m = V2/2. On substituting for m = 120, we get V = √(2 x 120) = 15.5.

QUESTION: 17

The intrinsic impedance of a TM wave will be

Solution:

Answer: c
Explanation: The intrinsic impedance of the transverse magnetic wave is given by ηTM = η √(1-(fc/f)2). Here the term √(1-(fc/f)2) is always lesser than unity. Thus the intrinsic impedance of the TM wave is lesser than 377 ohms.

QUESTION: 18

The modes TMmo and TMon are called 

Solution:

Answer: d
Explanation: The modes TMmo and TMon does not exist. These modes are said to be evanescent mode.

QUESTION: 19

Two modes with same cut off frequency are said to be

Solution:

Answer: c
Explanation: Two modes with same cut off frequency are called as degenerate modes. These modes have same field distribution.

QUESTION: 20

Does the mode TM30 exists?

Solution:

Answer: b
Explanation: Modes in the format of TMmo and TMon does not exist. The given mode is in the form of TMmo, which is does not exist. It is an evanescent mode.

QUESTION: 21

The boundary between the Fresnel and Fraunhofer zones having a length of 12 units and a wavelength of 3 units is

Solution:

Answer: a
Explanation: The Fresnel- Fraunhofer boundary is related by the wavelength as R = 2L2/λ. On substituting for L = 12 and λ = 3, we get R = 2 x 122/3 = 96 units.

QUESTION: 22

The reflection coefficient, when a resonant cavity is placed between the waveguide is

Solution:

Answer: b
Explanation: When the waveguide is shorted by conducting plates, the reflection coefficient will be unity. This will lead to the occurrence of standing waves.

QUESTION: 23

The distance between two terminated plates is given by the

Solution:

Answer: c
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength.

QUESTION: 24

Find the guided wavelength if the distance between the two conducting plates in the waveguide is 2 cm.

Solution:

Answer: a
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength. On substituting for Vmin = 2cm, we get λg = 2Vmin = 2 x 0.02 = 4cm.