Description

This mock test of Test: Transverse Electric Waves(TE) And Magnetic Waves(TM) for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
This contains 24 Multiple Choice Questions for Electrical Engineering (EE) Test: Transverse Electric Waves(TE) And Magnetic Waves(TM) (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Transverse Electric Waves(TE) And Magnetic Waves(TM) quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE)
students definitely take this Test: Transverse Electric Waves(TE) And Magnetic Waves(TM) exercise for a better result in the exam. You can find other Test: Transverse Electric Waves(TE) And Magnetic Waves(TM) extra questions,
long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.

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: TE_{10} 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 = V^{2}/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 η_{TE} = 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 10^{8} 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/2Z_{0}. On substituting for R = 100 and Z_{0} = 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 TM_{10}, TM_{01} and TM_{20} does not exist in any waveguide. The TM_{11}mode 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 = V^{2}/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 = V^{2}/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 η_{TM} = 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 TM_{mo} and TM_{on} are called

Solution:

Answer: d

Explanation: The modes TM_{mo} and TM_{on} 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 TM_{30} exists?

Solution:

Answer: b

Explanation: Modes in the format of TM_{mo} and TM_{on} does not exist. The given mode is in the form of TM_{mo}, 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 = 2L^{2}/λ. On substituting for L = 12 and λ = 3, we get R = 2 x 12^{2}/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.

### Lecture 41 - Transverse Electric and Magnetic Mode

Doc | 7 Pages

### MAGNETIC EFFECTS OF ELECTRIC CURRENT

Doc | 17 Pages

### Magnetic Effects of electric current

Video | 41:40 min

- Test: Transverse Electric Waves(TE) And Magnetic Waves(TM)
Test | 24 questions | 10 min

- Magnetic Effect Of Electric Current - Practice Test, Class 10 Science
Test | 20 questions | 40 min

- Test: Magnetic And Electric Properties Of Solids & Semiconductors
Test | 24 questions | 35 min

- Test: Magnetic Field Due To An Electric Current
Test | 10 questions | 10 min

- Magnetic Effect Of Electric Current - MCQ Test (22-11-2017)
Test | 30 questions | 30 min