# Electromagnetic Theory Questions and Answers – Transverse Electric Waves(TE)

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This set of Electromagnetic Theory Problems focuses on “Transverse Electric Waves(TE)”.

1. In transverse electric waves, which of the following is true?
a) E is parallel to H
b) E is parallel to wave direction
c) E is transverse to wave direction
d) H is transverse to wave direction

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

2. The dominant mode in rectangular waveguide is
a) TE01
b) TE10
c) TM01
d) TM10

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

3. The number of modes in a waveguide having a V number of 10 is
a) 10
b) 25
c) 100
d) 50

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.

4. Does the waveguide with dimensions 3 cm x 5.5 cm exist?
a) Yes
b) No

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.

5. The mode which has the highest wavelength is called
a) Dominant mode
b) Evanescent mode
c) Generate mode
d) Degenerate mode

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

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
a) 628.33
b) 338.62
c) 498.76
d) 342.24

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.

7. The cut off frequency of a rectangular waveguide of dimensions 3 cm x 1.5 cm is
a) 12 GHz
b) 6 GHz
c) 4 GHz
d) 5 GHz

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.

8. The propagation constant for a lossless transmission line will be
a) Real
b) Complex
c) Real and equal to phase constant
d) Complex and equal to phase constant

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.

9. The attenuation of a 50 ohm transmission line having a resistance of 100 ohm is
a) 0.01
b) 0.1
c) 1
d) 10

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.

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
a) 2.88 GHz
b) 3.32 GHz
c) 4.5 GHz
d) 2.12 GHz

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.

11. The phase constant and frequency are related by
a) Phase constant α ω
b) Phase constant α 1/ω
c) Phase constant α 1/2ω
d) Phase constant α ω/2

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

12. Which of the following parameter is non zero for a lossless line?
a) Attenuation
b) Resistance
c) Conductance
d) Phase constant 