Electromagnetic theory miscellaneous Easy Questions and Answers | Page - 5

Electromagnetic theory miscellaneous

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

1. N alone and not by a
1. d alone and not by N
1. The ratio (N/d)
1. The product (Nd)

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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.

In a multicavity magnetron strapping is employed primarily—

1. To prevent mode jumping
1. To increase the separation between the resonant frequencies in the π-mode and in the adjacent mode
1. To reduce the back heating of the cathode
1. To increase the output of the magnetron
1. Both A and B

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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.

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. A good conductor
1. A good dielectric
1. Neither a good conductor nor a good dielectric
1. A good magnetic material

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Material may be found either conductor or insulator by the method of loss tangent
σ = 10⁶
ωε 10 × 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
ωε 10 × 10⁹ × 10 × 8.85 × 10^-12

= 1.12 × 10⁶
so σ >> ωε

A plane wave is incident normally on a perfect conductor as shown below. Here E^r x = H'T and → P are electric fields magnetic field and Poynting vector respectively for the incident wave. The reflected wave should be—

1. E_xr = –E'_x
1. E_y = –H_y
1. → P = – → P
1. E^r_x = E_x
1. (B), (C) and (D)

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The reflected wave will travel in negative z direction. Hence P = – P and to accommodate the reversal of power flow magnetics field directions is reversed and we get
E_xr = E_xi; H_yr = – H_yi.
Hence alternatives (B), (C) and (D) are correct choices.

Correct Option: E

The reflected wave will travel in negative z direction. Hence P = – P and to accommodate the reversal of power flow magnetics field directions is reversed and we get
E_xr = E_xi; H_yr = – H_yi.
Hence alternatives (B), (C) and (D) are correct choices.

Medium wave radio signals may be received at for distance at night because—

1. Radiowaves travel faster at night
1. Ground wave attenuation is low at night
1. The sky wave is stronger at night
1. Their is no fading at night

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Propagation in medium wave frequency 525 to 1625 kHz is absorbed by D layer during day time. This vanishes at night and E layer helps in reflection at night especially at the higher range of frequency.

Correct Option: C

Propagation in medium wave frequency 525 to 1625 kHz is absorbed by D layer during day time. This vanishes at night and E layer helps in reflection at night especially at the higher range of frequency.