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In waveguides, which of the following conditions will be true?
a Explanation: In waveguides, the phase velocity will always be greater than the speed of light. This enables the wave to propagate through the waveguide. Thus V > c is the required condition.
a
See lessExplanation: In waveguides, the phase velocity will always be greater than the speed of light. This enables the wave to propagate through the waveguide. Thus V > c is the
required condition.
The expression for phase constant is given by
c Explanation: The phase constant is represented as β. It is a complex quantity representing the constant angle of the wave propagated. It is given by β = ω√(με).
c
See lessExplanation: The phase constant is represented as β. It is a complex quantity
representing the constant angle of the wave propagated. It is given by β = ω√(με).
In perfect conductors, the phase shift between the electric field and magnetic field will be
c Explanation: For perfect conductors, the electric and magnetic field E and H respectively vary by a phase of 45 degree. This is due to the polarisation phenomenon in the conductors, unlike dielectrics.
c
See lessExplanation: For perfect conductors, the electric and magnetic field E and H respectively vary by a phase of 45 degree. This is due to the polarisation phenomenon in the conductors, unlike dielectrics.
Calculate the wavelength of the wave with phase constant of 3.14 units.
b Explanation: The wavelength is the ratio of 2π to the phase constant β. On substituting for β = 3.14, we get λ = 2π/β = 2π/3.14 = 2 units.
b
See lessExplanation: The wavelength is the ratio of 2π to the phase constant β. On substituting for β = 3.14, we get λ = 2π/β = 2π/3.14 = 2 units.
Calculate the velocity of wave propagation in a conductor with frequency 5 x 108 rad/s and phase constant of 3 x 108 units.
c Explanation: The velocity of wave propagation is the ratio of the frequency to the phase constant. It is given by V = ω/β. On substituting the given values, we get V = 5/3 units.
c
See lessExplanation: The velocity of wave propagation is the ratio of the frequency to the phase constant. It is given by V = ω/β. On substituting the given values, we get V = 5/3 units.
In conductors, which two parameters are same?
b Explanation: In conductors, which are considered to be lossy, the attenuation and the phase constant are the same. It is given by α=β= √(ωμσ/2).
b
See lessExplanation: In conductors, which are considered to be lossy, the attenuation and the
phase constant are the same. It is given by α=β= √(ωμσ/2).
For metals, the conductivity will be
d Explanation: Metals are pure conductors. Examples are iron, copper etc. Their conductivity will be very high. Thus the metal conductivity will be infinity. Practically the conductivity of conductors will be maximum.
d
See lessExplanation: Metals are pure conductors. Examples are iron, copper etc. Their
conductivity will be very high. Thus the metal conductivity will be infinity. Practically the conductivity of conductors will be maximum.
In conductors, which condition will be true?
a Explanation: For conductors, the conductivity will be maximum. Thus the loss tangent is greater than unity. This is given by σ/ωε >1.
a
See lessExplanation: For conductors, the conductivity will be maximum. Thus the loss tangent is greater than unity. This is given by σ/ωε >1.
Skin depth phenomenon is found in which materials?
c Explanation: Skin depth is found in pure conductors. It the property of the conductor to allow a small amount of electromagnetic energy into its skin, but not completely. This is the reason why EM waves cannot travel inside a good conductor.
c
See lessExplanation: Skin depth is found in pure conductors. It the property of the conductor to allow a small amount of electromagnetic energy into its skin, but not completely. This is the reason why EM waves cannot travel inside a good conductor.
An example for lossless propagation is
d Explanation: There are many techniques employed to achieve zero attenuation or maximum propagation. But it is not achievable practically. Thus lossless propagation is not possible practically.
d
See lessExplanation: There are many techniques employed to achieve zero attenuation or
maximum propagation. But it is not achievable practically. Thus lossless propagation is not possible practically.