Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Electromagnetic Fields Theory (EMFT)  >  Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Electrical Engineering (EE) MCQ

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Electrical Engineering (EE) MCQ


Test Description

24 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Transverse Electric Waves(TE) & Magnetic Waves(TM)

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) below.
Solutions of Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) | 24 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 1

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 1

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 2

The dominant mode in rectangular waveguide is 

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 2

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.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 3

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 3

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 4

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 4

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 5

The mode which has the highest wavelength is called

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 5

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - 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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 6

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 7

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 7

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 8

The propagation constant for a lossless transmission line will be

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 8

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 9

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 9

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - 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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 10

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 11

The phase constant and frequency are related by

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 11

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 12

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 12

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 13

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 13

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 14

The dominant mode in the TM waves is 

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 14

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 15

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 15

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 16

The v number of a waveguide having 120 modes is

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 16

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 17

The intrinsic impedance of a TM wave will be

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 17

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 18

The modes TMmo and TMon are called 

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 18

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 19

Two modes with same cut off frequency are said to be

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 19

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 20

Does the mode TM30 exists?

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 20

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 21

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 21

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 22

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 22

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.

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 23

The distance between two terminated plates is given by the

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 23

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

Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 24

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

Detailed Solution for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) - Question 24

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.

11 videos|48 docs|73 tests
Information about Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) Page
In this test you can find the Exam questions for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Transverse Electric Waves(TE) & Magnetic Waves(TM), EduRev gives you an ample number of Online tests for practice

Up next

Download as PDF

Up next