This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Transverse Magnetic Waves”.
1. In transverse magnetic waves, which of the following is true?
a) E is parallel to H
b) H is parallel to wave direction
c) H is transverse to wave direction
d) E is transverse to H
Explanation: In transverse magnetic waves, the magnetic field strength is transverse to the wave direction. They are also called E waves.
2. The dominant mode in the TM waves is
Explanation: The modes TM10, TM01 and TM20 does not exist in any waveguide. The TM11 mode is the dominant mode in the waveguide.
3. The modes in a waveguide having a V number of 20 is
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.
4. The v number of a waveguide having 120 modes is
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.
5. The intrinsic impedance of a TM wave will be
a) Greater than 377 ohm
b) Equal to 377 ohm
c) Lesser than 377 ohm
d) Does not exist
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.
6. The modes TMmo and TMon are called
a) Generate modes
b) Degenerate modes
c) Dominant modes
d) Evanescent modes
Explanation: The modes TMmo and TMon does not exist. These modes are said to be evanescent mode.
7. Two modes with same cut off frequency are said to be
a) Generate modes
b) Dominant modes
c) Degenerate modes
d) Regenerate modes
Explanation: Two modes with same cut off frequency are called as degenerate modes. These modes have same field distribution.
8. Does the mode TM30 exists?
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.
9. The boundary between the Fresnel and Fraunhofer zones having a length of 12 units and a wavelength of 3 units is
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.
10. The reflection coefficient, when a resonant cavity is placed between the waveguide is
Explanation: When the waveguide is shorted by conducting plates, the reflection coefficient will be unity. This will lead to the occurrence of standing waves.
11. The distance between two terminated plates is given by the
a) Guided wavelength
b) 2(guided wavelength)
c) Guided wavelength/2
d) (Guided wavelength)/4
Explanation: The distance between the terminating plates is given by Vmin = λg/2, where λg is the guided wavelength.
12. Find the guided wavelength if the distance between the two conducting plates in the waveguide is 2 cm.
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.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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