Electromagnetic theory miscellaneous


Electromagnetic theory miscellaneous

Electromagnetic Theory

  1. In the infinite plane y = 6 m there exists a uniform surface charge density of
    1
    600

    μ c/m2. The associated electric field strength is—









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    The associated electric field strength


    1
    =
    1
    × 10−6 ×
    1
    0600π2 × 8.85 × 10−12

    = 30 kV/m

    Correct Option: C

    The associated electric field strength


    1
    =
    1
    × 10−6 ×
    1
    0600π2 × 8.85 × 10−12

    = 30 kV/m


  1. The electric field strength at a distance point P due to a point charge +q located on the origins is 100 µ V/m if the point charge is now enclosed by a perfectly conducting metal sheet sphere whose centre is at origin then the electric field strength at the point P outside the sphere becomes—









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    Charge +q induces negative charge inside the sphere and positive charge on the outside of sphere. If the sphere is enclosed by a Gaussian sphere the field strength E will be given by Gauss's law

    E =
    1
    Q
    4πε0R2

    where R is radius of sphere and Q is total charge. This is same although the charge is present at the origin of sphere. Therefore + 100 µ V/m.

    Correct Option: B

    Charge +q induces negative charge inside the sphere and positive charge on the outside of sphere. If the sphere is enclosed by a Gaussian sphere the field strength E will be given by Gauss's law

    E =
    1
    Q
    4πε0R2

    where R is radius of sphere and Q is total charge. This is same although the charge is present at the origin of sphere. Therefore + 100 µ V/m.



  1. For a short wave radio link between two stations via the ionosphere. The ratio of the maximum usable frequency to the critical frequency—









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    Since we know that fmuf = fc sec i
    so,

    fmuf
    = sec i > 1
    fc

    Correct Option: B

    Since we know that fmuf = fc sec i
    so,

    fmuf
    = sec i > 1
    fc


  1. A beam antenna consists of a dipole one reflector and one director. The velocity factor is 0.90 and antenna is to be cut for a frequency of 150 MHz. The length of the dipole will be—









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    By calculation as follows

    λ =
    C
    velocity factor × f

    λ =
    3 ×108
    0.9 × 150 × 106

    = 1.8
    Now effective length of dipole =
    1
    1.8
    = 0.9
    22

    Correct Option: C

    By calculation as follows

    λ =
    C
    velocity factor × f

    λ =
    3 ×108
    0.9 × 150 × 106

    = 1.8
    Now effective length of dipole =
    1
    1.8
    = 0.9
    22



  1. Input impedance for the given line is—









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    Graph shows clearly that shorted stub of given length is of capacitive type.

    Correct Option: B

    Graph shows clearly that shorted stub of given length is of capacitive type.