Electromagnetic theory miscellaneous

Electromagnetic theory miscellaneous

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Electromagnetic Theory

Electromagnetic theory miscellaneous
  1. →E = –uxx + –uyy + –uzz find the potential difference between (2, 0, 0) and (1, 2, 3) is given by—








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    VAB= BE.dl =B(x¯ux + y¯uy + z¯uz) (¯ux dx + ¯uy dy + ¯uz dz)
    AA

    = Bx dx + By dy + Bz dz
    AAA

    = 1x dx + 2y dy + 3z dz = 5
    200

    Correct Option: C

    VAB= BE.dl =B(x¯ux + y¯uy + z¯uz) (¯ux dx + ¯uy dy + ¯uz dz)
    AA

    = Bx dx + By dy + Bz dz
    AAA

    = 1x dx + 2y dy + 3z dz = 5
    200

  1. Medium 1 has the electrical permittivity ∈1 = 1.5 ∈0 farad/m and occupies the region to the left of the x = 0 plane. Medium 2 has the electrical permittivity ∈2 = 2.5 ∈0 and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = 2 ux – 3 uy + uz V/m, find E2 in medium 2 is—








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    We know that
    Dn1 = Dn2
    ∈r1 ∈n1 = ∈r2 · ∈n2

    Ex2 =
    ∈r1
    		 · Ex1
    ∈r2

    Ex2 =
    1.5 ∈0
    		× 2 = 1.2
    2.5 ∈0

    ∴ E2 = 1.2–ux – 3– uy + 1– uz
    Note: The field in second medium does not change w.r.t. y and z.
    Since the interface plane is x = 0. Which is y, z plane and normal to interface is x-axis when Et1 and Et2 resolved the component of Et1 in y and z direction = resolved component of Et2 in y and z direction.
    Therefore, E2 component (y and z) is same as that of E1 i.e. (–3uy + 1. uz)

    Correct Option: C

    We know that
    Dn1 = Dn2
    ∈r1 ∈n1 = ∈r2 · ∈n2

    Ex2 =
    ∈r1
    		 · Ex1
    ∈r2

    Ex2 =
    1.5 ∈0
    		× 2 = 1.2
    2.5 ∈0

    ∴ E2 = 1.2–ux – 3– uy + 1– uz
    Note: The field in second medium does not change w.r.t. y and z.
    Since the interface plane is x = 0. Which is y, z plane and normal to interface is x-axis when Et1 and Et2 resolved the component of Et1 in y and z direction = resolved component of Et2 in y and z direction.
    Therefore, E2 component (y and z) is same as that of E1 i.e. (–3uy + 1. uz)

  1. E (z, t) = 10 cos (2π × 107 t – 0.1 πz), the velocity of propagation—








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    On comparing the given equation
    E (z, t) = 10 cos (2π × 107 t – 0.1 πz) with standard equation
    E (z, t)=E0 cos (ωt – βz)
    ω = 2π × 107, β = 0.1π

    velocity, v =
    ω
    		 =
    2 π × 107
    		 = 2 × 108m/s
    β0.1 π

    Correct Option: C

    On comparing the given equation
    E (z, t) = 10 cos (2π × 107 t – 0.1 πz) with standard equation
    E (z, t)=E0 cos (ωt – βz)
    ω = 2π × 107, β = 0.1π

    velocity, v =
    ω
    		 =
    2 π × 107
    		 = 2 × 108m/s
    β0.1 π

  1. TEM wave has proper density 1.2 watt/m2 in a medium ∈r = 3, µr = 1. What is the value of E?








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    We know thatρ =
    E2max
    2η

    1.2 =
    E2max
    2 ×
    120 π
    √∈r

    or
    E2max
    1.2 × 2 × 120 π
    √3

    Correct Option: A

    We know thatρ =
    E2max
    2η

    1.2 =
    E2max
    2 ×
    120 π
    √∈r

    or
    E2max
    1.2 × 2 × 120 π
    √3

  1. A transmitting antenna radiates 251 watt isotropically. A receiving antenna locating 100 m away from the transmitting antenna has an effective aperture of 500 cm2. The total received power by the antenna is—








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    PR = PT GT GR (λ/4πd)2
    where
    GR = 1
    GT = 1 (for isotropic antenna)
    PT = 251 watt (given)
    Aeff = 500 cm2 = 500 × 10–4 m2
    d = 100 m
    Now,
    PR = 251 × 1 × 1 × 4π/λ2

    500 × 10–4
    λ2
    		= 9.99 × 10–5 = 100 µW
    (4π)2 × 100 × 100

    Correct Option: D

    PR = PT GT GR (λ/4πd)2
    where
    GR = 1
    GT = 1 (for isotropic antenna)
    PT = 251 watt (given)
    Aeff = 500 cm2 = 500 × 10–4 m2
    d = 100 m
    Now,
    PR = 251 × 1 × 1 × 4π/λ2

    500 × 10–4
    λ2
    		= 9.99 × 10–5 = 100 µW
    (4π)2 × 100 × 100