Electromagnetic theory miscellaneous Easy Questions and Answers | Page - 14

Electromagnetic theory miscellaneous

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—

1. Zero
1. 100 µ V/m
1. – 100 µ V/m
1. 50 µ V/m

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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
4πε₀ R²

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
4πε₀ R²

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.

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—

1. 30 i V/m
1. 30 j V/m
1. 30 k V/m
1. 60 i V/m

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The associated electric field strength

1 = 1 × 10⁻⁶ × 1
2ε₀ 600π 2 × 8.85 × 10⁻¹²

= 30 kV/m

Correct Option: C

The associated electric field strength

1 = 1 × 10⁻⁶ × 1
2ε₀ 600π 2 × 8.85 × 10⁻¹²

= 30 kV/m

A transverse electromagnetic wave with circular polarization is received by a dipole antenna due to polarization mismatch. The power transfer efficiency from the wave to the antenna is reduced to about—

1. 50%
1. 35.5%
1. 25%
1. 0%

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Because when wave is left polarized than antenna is right circularly polarized.

Correct Option: D

Because when wave is left polarized than antenna is right circularly polarized.

Radiation resistance of a monopole of height h = l/2 is given by—

1. ≃ 400 (hλ)²
1. ≃ 400
  λ 2
  h
1. ≃ 400
  h 2
  λ
1. ≃ 200
  λ 2
  h

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For monopole radiation resistance is

R_r = 400 dl ²
λ

Correct Option: C

For monopole radiation resistance is

R_r = 400 dl ²
λ

If wave of critical frequency 30 MHz is departing an angle of 60º then the MUF is given to be—

1. 10 MHz
1. 15 MHz
1. 30 MHz
1. 60 MHz

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We have relation of maximum usable frequency
f_muf = f_c sec i
=
f_c =
30 MHz = 60 MHz
cos i cos 60

Correct Option: D

We have relation of maximum usable frequency
f_muf = f_c sec i
=
f_c =
30 MHz = 60 MHz
cos i cos 60