Electromagnetic theory miscellaneous Easy Questions and Answers | Page - 35

Electromagnetic theory miscellaneous

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

1. 5 GHz
1. 10 GHz
1. 15 GHz
1. 20 GHz

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From the synopsis provided we get relation of cutoff frequency
f_c = V_cm = 3 × 10⁸
2a 2 × (0.1)

= 1.5 × 10¹⁰
= 15 GHz

Correct Option: C

From the synopsis provided we get relation of cutoff frequency
f_c = V_cm = 3 × 10⁸
2a 2 × (0.1)

= 1.5 × 10¹⁰
= 15 GHz

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—

1. 7.5 cm
1. 15 cm
1. 15 m
1. 150 m

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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.

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—

1. 1.5
1. √1.5
1. 2
1. √2

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From Snell’s law
n₁ sin θ₁ = n₂ sin θ₂
n₁ = √ε₀ (free space or air)
n₂ = √ε_rε₀
θ₁ → incident angle with normal
θ₂ → transmitted angle with normal
ε₀ sin (45°) = √ε_rε₀ sin 30°
⇒ √ε₀ = √ε_rε₀ √ε_r = √2 ⇒ ε_r = 2
√2 2

Correct Option: C

From Snell’s law
n₁ sin θ₁ = n₂ sin θ₂
n₁ = √ε₀ (free space or air)
n₂ = √ε_rε₀
θ₁ → incident angle with normal
θ₂ → transmitted angle with normal
ε₀ sin (45°) = √ε_rε₀ sin 30°
⇒ √ε₀ = √ε_rε₀ √ε_r = √2 ⇒ ε_r = 2
√2 2

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

1. Bandwidth increases
1. Bandwidth decreases
1. Gain increases
1. Gain decreases

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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.

The frequency range for satellite communication is—

1. 1 kHz to 100 kHz
1. 100 kHz to 10 kHz
1. 10 MHz to 30 MHz
1. 1 GHz to 30 GHz

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NA

Correct Option: D

NA