Electromagnetic theory miscellaneous

Electromagnetic theory miscellaneous

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Electromagnetic Theory

Electromagnetic theory miscellaneous
  1. A rectangular waveguide has dimensions 1 cm × 0.5 cm its cut-off frequency is—








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    From the synopsis provided we get relation of cutoff frequency

    fc =
    Vcm
    		 =
    3 × 108
    2a2 × (0.1)

    = 1.5 × 1010
    = 15 GHz

    Correct Option: C

    From the synopsis provided we get relation of cutoff frequency

    fc =
    Vcm
    		 =
    3 × 108
    2a2 × (0.1)

    = 1.5 × 1010
    = 15 GHz

  1. For an 8 feet (2.4 m) parabolic dish antenna operating at 4 GHz the minimum distance required for the far field measurement is closest to—








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    For measurement of far field or secondary we apply following relation

    r ≥
    2d2
    λ

    d = antenna aperture, r = radius Substituting the value
    r ≥
    2 × (2.4)2
    		 = 153.6 m
    0.075

    So the closest answer is 150 m.

    Correct Option: D

    For measurement of far field or secondary we apply following relation

    r ≥
    2d2
    λ

    d = antenna aperture, r = radius Substituting the value
    r ≥
    2 × (2.4)2
    		 = 153.6 m
    0.075

    So the closest answer is 150 m.

  1. A uniform place wave in air impinges at 45º angle on a lossless dielectric material with dielectric constant ∈r. The transmitted wave propagate in a 30º direction with respect to the normal the wave of ∈r is—








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    From Snell’s law
    n1 sin θ1 = n2 sin θ2
    n1 = √ε0 (free space or air)
    n2 = √εrε0
    θ1 → incident angle with normal
    θ2 → transmitted angle with normal
    ε0 sin (45°) = √εrε0 sin 30°

    ⇒    
    √ε0
    		 =
    √εrε0
    		√εr = √2 ⇒ εr = 2
    √2 2

    Correct Option: C

    From Snell’s law
    n1 sin θ1 = n2 sin θ2
    n1 = √ε0 (free space or air)
    n2 = √εrε0
    θ1 → incident angle with normal
    θ2 → transmitted angle with normal
    ε0 sin (45°) = √εrε0 sin 30°

    ⇒    
    √ε0
    		 =
    √εrε0
    		√εr = √2 ⇒ εr = 2
    √2 2

  1. If the diameter of a λ/2 dipole antenna is increased from λ/100 to λ/50 then its—








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    See from synopsis provided that smaller diameter result in narrower bandwidth and larger diameter gives wider bandwidth.

    Correct Option: A

    See from synopsis provided that smaller diameter result in narrower bandwidth and larger diameter gives wider bandwidth.

  1. The frequency range for satellite communication is—








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    NA

    Correct Option: D

    NA