Electromagnetic theory miscellaneous


Electromagnetic theory miscellaneous

Electromagnetic Theory

  1. A material is described by following electrical parameters as a frequency of 10 GHz, σ = 106 mho/m µ = µ0 and ε/ε0 = 10. The material at this frequencies is considered to be—
    σ0 =
    1
    ×10−9 F/m
    36 π









  1. View Hint View Answer Discuss in Forum

    Material may be found either conductor or insulator by the method of loss tangent

    σ
    =
    106
    ωε10 × 109 × 10 × 8.85 × 10-12

    = 1.12 × 106
    so σ >> ωε

    Correct Option: A

    Material may be found either conductor or insulator by the method of loss tangent

    σ
    =
    106
    ωε10 × 109 × 10 × 8.85 × 10-12

    = 1.12 × 106
    so σ >> ωε


  1. In a multicavity magnetron strapping is employed primarily—











  1. View Hint View Answer Discuss in Forum

    To prevent mode jumping and increase separation between the resonant frequencies in the π mode and in the adjacent modes.

    Correct Option: E

    To prevent mode jumping and increase separation between the resonant frequencies in the π mode and in the adjacent modes.



  1. The beamwidth between first null of uniform linear array of N equally spaced (element spacing = a) equally excited antennas is determined by—









  1. View Hint View Answer Discuss in Forum

    Beamwidth between first nulls is given by relation as follows

    φ = sin-1
    λ
    nd

    where n = no. of sources, separated by d distance apart.

    Correct Option: D

    Beamwidth between first nulls is given by relation as follows

    φ = sin-1
    λ
    nd

    where n = no. of sources, separated by d distance apart.


  1. An open wire transmission line has primary constant
    R = 20 Ω/m     G = 0.3 × 103 s/km
    L = 0.25 1 m H/km     C = 0.1 µ F/m
    at a frequency of 1 kHz its characteristic impedance is—









  1. View Hint View Answer Discuss in Forum

    We have characteristic impedance as follows

    Z0 =
    R + jωL
    G + jωc

    R = 20 Ω/ km,
    G = 0.3 × 103 s/ km
    L = 0.25 mH/ km,
    C = 0.1 µF and ω = 103 Hz so we get
    Z0 =
    20 + j2π(103 × 0.25 × 10−3)
    0.3 × 10−3 + j2π(103 × 0.1 × 10−6)

    =
    20 + j2π( 0.25 )
    0.30 −3 + j2π(0.1 × 10−6)

    = 169.75 ∠– 25º

    Correct Option: A

    We have characteristic impedance as follows

    Z0 =
    R + jωL
    G + jωc

    R = 20 Ω/ km,
    G = 0.3 × 103 s/ km
    L = 0.25 mH/ km,
    C = 0.1 µF and ω = 103 Hz so we get
    Z0 =
    20 + j2π(103 × 0.25 × 10−3)
    0.3 × 10−3 + j2π(103 × 0.1 × 10−6)

    =
    20 + j2π( 0.25 )
    0.30 −3 + j2π(0.1 × 10−6)

    = 169.75 ∠– 25º



  1. Length of dipole can be calculated from the relationship—









  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: B

    NA