Electromagnetic theory miscellaneous Easy Questions and Answers | Page - 1

Electromagnetic theory miscellaneous

The ground wave coverage of the medium wave transmitter is 100 km and in the height the first reflected ray is at 800 km. The skip distance is—

1. 900 km
1. 700 km
1. 100 km
1. 800 km

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Skip distance is the minimum distance where the reflected wave returns back to earth so a wave transmit ground wave coverage of 100 km and then reflected back to the earth at 800 km from transmitter so skip distance is 700 km.

Correct Option: B

Skip distance is the minimum distance where the reflected wave returns back to earth so a wave transmit ground wave coverage of 100 km and then reflected back to the earth at 800 km from transmitter so skip distance is 700 km.

A radiowave is incident on a layer of ionosphere at an angle of 30 degree with the vertical common. If the critical frequency is 1.2 MHz the maximum usable frequency is—

1. 1.2 MHz
1. 2.4 MHz
1. 0.6 MHz
1. 1.386 MHz

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Relation between maximum usable frequency and critical frequency with incident angle is given by as follows
f_muf = f_c sec i
∴ f_muf = 1.2 × 2 × 10⁶
√3

= 1.3856 × 10⁶
= 1.385 MHz

Correct Option: D

Relation between maximum usable frequency and critical frequency with incident angle is given by as follows
f_muf = f_c sec i
∴ f_muf = 1.2 × 2 × 10⁶
√3

= 1.3856 × 10⁶
= 1.385 MHz

In a broadside array of 20 isotropic radiators equally spaced at a distance of λ/2. The beamwidth between first nulls is—

1. 51.3 degrees
1. 11.46 degrees
1. 22.9 degrees
1. 102.6 degrees

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For broadside array of 20 point sources. We have general formula for the array of n point sources have been as follow
2φ = 2 sin^-1 m λ
nd

where
m = no. of null 1^st or 2^nd
d = separation between the point sources
n = no. of point sources
for first nulls we take m = 1, n = 20 is given, so we get
2φ = 2 sin^-1 λ = 2 ×5.739
20 × λ/2

= 11.478º

Correct Option: B

For broadside array of 20 point sources. We have general formula for the array of n point sources have been as follow
2φ = 2 sin^-1 m λ
nd

where
m = no. of null 1^st or 2^nd
d = separation between the point sources
n = no. of point sources
for first nulls we take m = 1, n = 20 is given, so we get
2φ = 2 sin^-1 λ = 2 ×5.739
20 × λ/2

= 11.478º

When a wave is propagated in a good dielectric where σ/ωε << 1 the attenuation factor and the phase shift factor are given by—

1. α = σ √ μ , β = ω √με
  2 ε
1. α = σ , β = ω √με
  2
1. α = σ √ μ √ μ , β = √ ωμε
  2ε ε 2
1. α = 1, β = 45°

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Refer synopsis

Correct Option: A

Refer synopsis

For F¹ layer maximum ionic density is 2.3 × 10⁶ electrons per c.c. The critical frequency for this layer will be—

1. 1.36 MHz
1. 13.6 MHz
1. 136 MHz
1. 1.36 × 10³ MHz

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For calculation of critical frequency we have relation as follows
f_c = 9√N_max
= 9√2.3 × 10⁶× 10⁶
= 9× 10⁶√2.3
= 13.6MHz

Correct Option: B

For calculation of critical frequency we have relation as follows
f_c = 9√N_max
= 9√2.3 × 10⁶× 10⁶
= 9× 10⁶√2.3
= 13.6MHz