Electromagnetic Wave Propagation MCQ Quiz in मल्याळम - Objective Question with Answer for Electromagnetic Wave Propagation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 13, 2025
Latest Electromagnetic Wave Propagation MCQ Objective Questions
Top Electromagnetic Wave Propagation MCQ Objective Questions
Electromagnetic Wave Propagation Question 1:
A lossless line of characteristic impedance z0 is terminated in pure reactance of -jz0, Voltage Standing Wave Ratio (VSWR) is
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 1 Detailed Solution
Concept:
The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current).
\(VSWR = \frac{{{{\rm{V}}_{{\rm{max}}}}}}{{{{\rm{V}}_{{\rm{min}}}}}} = \frac{{{{\rm{I}}_{{\rm{max}}.}}}}{{{{\rm{I}}_{{\rm{min}}.}}}}\;\)
This can also be written as:
\(VSWR = \frac{{1 + {\rm{|Γ| }}}}{{1 - {\rm{|Γ|}}}}\;\)
Γ = Reflection coefficient, defined as:
\({\rm{Γ }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\;\)
ZL = Load impedance
Z0 = Characteristic Impedance
Calculation:
Given:
Z0 = Z0 Ω
ZL = -jZ0 Ω
\({\rm{Γ }} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}\;\)
\({\rm{Γ }} = \frac{{-jZ_0 - Z_0}}{{-jZ_0 + Z_0}} = \frac{-1-j}{1-j}=\frac{\sqrt2\angle -135^{\circ}}{\sqrt 2\angle-45^{\circ}}=1\angle -90^\circ\)
Γ = -j
Now:
\(VSWR = \frac{{1 + |-j|}}{{1 - |-j|}} =\frac{1+1}{1-1}=\frac{2}{0}= \infty\)
Electromagnetic Wave Propagation Question 2:
Dimensions of Poynting Vector P, are same as that of:
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 2 Detailed Solution
Poynting vector:
It states that the cross product of electric field vector (E) and magnetic field vector (H) at any point is a measure of the rate of flow of electromagnetic energy per unit area at that point that is
\( \vec S = \vec E \times \vec H\)
Where P = Poynting vector,
E = Electric field and
H = Magnetic field
Or, it can be written as,
\(S = \frac{Power}{Area}\)
The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves.
The unit of the Poynting vector is watt/m2.
Additional Information
1) The Poynting vector (i.e. energy flow per unit area per unit time) for a plane electromagnetic wave is given by
\( \vec S = \frac{1}{\mu_o}(\vec E \times \vec H)\)
\( S = \frac{1}{\mu_o}(E H\, sin\theta )\)
2) From the above, it is clear that E and H are mutually perpendicular and also they are perpendicular to the direction of propagation of the wave.
Thus, the direction of the Poynting vector is along the direction of the propagation of the electromagnetic wave.
The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves per unit volume.
∴ The dimensions of both the Poynting vector and the electromagnetic power density are the same, i.e. M1 L-1 T-2
where M = Mass, L = Length, T = Time
Electromagnetic Wave Propagation Question 3:
A non-magnetic lossy dielectric material with relative permittivity εr = 2.25 and conductivity σ = 10-4 mho/m is applied with electromagnetic wave of 2.5 MHz. What is the loss tangent?
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 3 Detailed Solution
Concept:
Loss tangent is given by, \(\tan \delta = \frac{\sigma }{{\omega \varepsilon }}\)
Where ω is the frequency is rad/sec
Loss tangents represent the dissipation of electromagnetic energy of dielectric material.
Calculation:
The given material is nonmagnetic lossy dielectric medium i.e. σ ≠ 0.
Real permittivity, εr = 2.25
Conductivity, σ = 10-4 mho/m
For nonmagnetic material, μr = 1
Frequency of EM wave = 2.5 × 106 Hz
\(\tan \delta = \frac{{{{10}^{ - 4}}}}{{2\pi\; \times \;2.5\; \times \;{{10}^6}\; \times \;\frac{1}{{36\pi }}\; \times \;{{10}^{ - 9}}\; \times \;2.25}} = 0.32\)
Electromagnetic Wave Propagation Question 4:
A solid conductor with relative permeability μr = 200, conductivity σ = 5 × 106 mho/m having outer diameter 8 mm and length 2 mm. If the total current carried by the conductor is i(t) = 2 cos (π 104 t) A. Find the skin depth.
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 4 Detailed Solution
Relative permeability (μr) = 200
Conductivity (σ) = 5 × 106 mho/m
Outer diameter of conductor = 8 mm
Length of conductor = 2 mm
Total current carried by conductor, i(t) = 2 cos (π × 104t) Amp
⇒ ω = π × 104 rad/sec
⇒ f = 0.5 × 104 Hz
\(\delta = \frac{1}{{\sqrt {\pi f\;\sigma \;{\mu _0}{\mu _r}} }}\)
\(\delta = \frac{1}{{\sqrt {\pi\; \times \;0.5\; \times \;{{10}^4}\; \times \;4\pi\; \times \;{{10}^{ - 7}}\; \times \;200\; \times \;5 \;\times\; {{10}^6}} }}\)
\(\Rightarrow \delta = \frac{1}{{\sqrt {2{\pi ^2}\; \times \;{{10}^6}} }} = 0.225\;mm\)
Electromagnetic Wave Propagation Question 5:
For a dielectric-conductor interface where subscript 1 denotes the dielectric, the boundary condition that is NOT satisfied for static fields is
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 5 Detailed Solution
Analysis:
A conductor placed in a static field will exhibit or induce a charge on its surface so as to have zero fields inside it. i.e. for an ideal conductor, the electric & magnetic field inside it is zero.
The given situation can be visualized as shown:
A conductor dielectric interface satisfies the following boundary conditions:
H1t – H2t = Js
B1n = B2n
E1t = E2t = 0
D1n – D2n = ρs
For a conductor: H2t = E2t = B2n = D2n = 0
So, H1t = Js
B1n = B2n = 0
E1t = E2t = 0
D1n = ρs
Electromagnetic Wave Propagation Question 6:
Which of the following has the lowest frequency?
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 6 Detailed Solution
- Under the electromagnetic spectrum, visible light has the lowest frequency among the given option.
- The spectrum ranges from radio waves having low frequency till gamma rays, having the highest one.
Electromagnetic Wave Propagation Question 7:
The electric permittivity of free space 'ϵ0' is
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 7 Detailed Solution
Permittivity: The property of any material which opposes the formation of an electric field is called permittivity.
The permittivity of free space is denoted by ϵ0.
Permeability: The measurement of the ability of any material that allows the formation of magnetic lines of force is called permeability.
The permeability of free space is given by μ0.
Notes:
- The larger the tendency for charge distortion (also called electric polarization), the larger the value of the permittivity
- This constant is equal to approximately 8.85 × 10-12 farad per meter (F/m) in free space (a vacuum)
- In other materials, it can be much different, often substantially greater than the free-space value
Electromagnetic Wave Propagation Question 8:
The characteristic impedance η0 of a free space is
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 8 Detailed Solution
The characteristic impedance of a wave traveling in a lossy medium is given by:
\(Z = \sqrt {\frac{{j\omega μ }}{{\sigma + j\omega ϵ}}}\)
Put σ = 0 (free space and lossless medium)
\(Z = \sqrt {\frac{μ_0 }{ϵ_0}}\)
Where,
μ0 = Permeability of free space = 4π x 10-7 H / m
ϵ0 = Permittivity of free space = 8.85 x 10-12 F/m
Electromagnetic Wave Propagation Question 9:
The attenuation constant of a medium is 400/m. Find the skin depth in the medium.
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 9 Detailed Solution
CONCEPT:
- The skin effect is the tendency of an alternating electric current to become distributed within a conductor such that the current density is largest near the surface of the conductor and decreases with greater depths in the conductor.
- The electric current flows mainly at the skin of the conductor, between the outer surface and a level called the skin depth.
- Skin depth is defined as the reciprocal of attenuation constant i.e.,
\(\delta=\frac{1}{\alpha}\)
- The attenuation constant is given by:
\(\Rightarrow\alpha=\omega\sqrt{\frac{\mu\epsilon}{2}\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]}\)
- For an ideal conductor with σ ≈ ∞,
\(\Rightarrow \delta=\frac{1}{\alpha}=\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]≈\frac{σ}{\omega\epsilon}\)
CALCULATION:
Given α = 400 /m
Skin depth is defined as the reciprocal of attenuation constant i.e.,
\(\Rightarrow \delta=\frac{1}{\alpha}\)
\(\Rightarrow \delta=\frac{1}{400}=2.5~mm\)
α is the attenuation constant given by:
\(α=\omega\sqrt{\frac{\mu\epsilon}{2}\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]}\)
ω = Operating frequency
μ = Permeability of the material
σ = Conductivity of the material
For a good conductor σ >> 1. The above expression for the attenuation constant can be approximated as:
\(α=\left[\sqrt{1+\left(\frac{σ}{\omega\epsilon}\right)^2}-1\right]≈\frac{σ}{\omega\epsilon}\)
∴ The attenuation constant becomes:
\(α=\omega \sqrt{\left(\frac{\muσ}{2\omega}\right)}\)
\(α=\sqrt{\left(\frac{\omega\muσ} {2}\right)}\)
\(\therefore~α =\sqrt{\pi f\muσ}\)
Thus, the skin depth becomes:
\(\delta=\frac{1}{\sqrt{\pi f \muσ}}\)
Note: The skin depth is inversely proportional to the square root of frequency.
Electromagnetic Wave Propagation Question 10:
An unpolarised light beam strikes a glass surface at Brewster's angle. Then
Answer (Detailed Solution Below)
Electromagnetic Wave Propagation Question 10 Detailed Solution
Calculation:
According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.