Parallel Plane Waveguide MCQ Quiz in தமிழ் - Objective Question with Answer for Parallel Plane Waveguide - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The curve generated by a point on the circumference of a circle that rolls along a straight line is known as Cycloid.

Calculation:

Initially charged particle will experience electric force and will gain velocity then it will deflect in magnetic field

F ∝ (v×B) → y

The correct answer is option (1).

Concept:

Phase velocity:

\({{v}_{p}}=\frac{c}{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}\)

Where c is the free space velocity, f is the operating frequency and fc is the cut-off frequency.

1. We can readily see that the phase velocity is a non-linear function of the operating frequency.

2. We can also write phase velocity as:

\({v_p} = \frac{c}{{\cos \theta }};\)

Where θ is the angle with which the wave enters the waveguide as shown:

Calculation:

Given, f = fc

\({v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\)

\(v_p = \frac{c}{{\sqrt {1 - 1} \;}} = \frac{c}{0} = \infty \)

v_p = ∞ means that,

cos θ = 0, i.e. θ = 90°.

This situation is shown as below:

This implies that at f = fc, the wave will oscillate between the walls as shown.

Hence option (3) is the correct answer.

Concept:

Phase velocity is defined as:

\({V_p} = \frac{\omega }{\beta }\) 

β is the phase constant defined as:

\(\beta = \sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} \)

\({V_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}\) 

\({V_p} = \frac{1}{{\sqrt {\mu \epsilon - {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}\) 

\({V_p} = \frac{{\frac{1}{{\sqrt {\mu \epsilon } }}}}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu \epsilon} }}} \right)}^2}} }}\) 

Using \(c = \frac{1}{{\sqrt {\mu C} }}\) where c =  speed of light, the above expression becomes:

\({V_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{m\pi C}}{{a\omega }}} \right)}^2}} }};\) 

Also \({\omega _c} = \frac{{m\pi c}}{a}\)

\({V_p} = \frac{C}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}\) 

Using \(\sin \theta = \frac{{{\omega _c}}}{\omega }\), we get:

\({V_p} = \frac{c}{{\sqrt {1 - {{\sin }^2}\theta } }}\) 

\({V_p} = \frac{c}{{\cos \theta }};\) 

Since -1 ≤ cos θ ≤ 1

∴ V_p > c

Extra Information:

Group velocity is given by:

\({V_g} = \frac{{d\omega }}{{d\beta }}\) 

\(\beta = \sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} \) 

\(\frac{{d\beta }}{{d\omega }} = \frac{{2\omega \mu\epsilon }}{{2\sqrt {{\omega ^2}\mu\epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}\) 

\(\frac{{d\beta }}{{d\omega }} = \frac{{\sqrt {\mu\epsilon } }}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu\epsilon } }}} \right)}^2}} }}\) 

\(\frac{{d\beta }}{{d\omega }} = \frac{1}{{C\sqrt {1 - {{\left( {\frac{{{\omega _C}}}{\omega }} \right)}^2}} }}\) 

\({V_g} = c\sqrt {1 - {{\left( {\frac{{{\omega _C}}}{\omega }} \right)}^2}} \) 

V_g = c cos θ

V_g < c

Conclusion: ∴ The phase velocity is always greater than the speed of light and group velocity is always less than the speed of light.

Concept:

For parallel plate waveguide, the propagation constant is defined as:

\(\beta = \frac{{m\pi }}{a}\)

a = distance between the plates

Where the minimum value of m = 1

Cut-off frequency for the parallel plate waveguide is given by:

\({f_c} = \frac{m}{{2a}}\)

The wavelength will be:

\({\lambda _c} = \frac{{2a}}{m}\)

Explanation:

Permittivity and Relative Permittivity

Definition: Permittivity is a measure of how much electric field is 'permitted' to pass through a material. It quantifies the ability of a material to store electrical energy in an electric field. The absolute permittivity (ε) of a material is the product of the relative permittivity (εr) and the permittivity of free space (ε0), where ε0 is a constant value approximately equal to 8.854 × 10^-12 F/m (farads per meter).

Relative permittivity, also known as the dielectric constant, is a dimensionless quantity that represents the ratio of the permittivity of a material to the permittivity of free space. It indicates how much better the material is at storing electrical energy compared to a vacuum. Mathematically, it is expressed as:

εr = ε / ε0

Working Principle: When an electric field is applied to a material, the material's molecules polarize, aligning themselves with the field. This polarization reduces the overall field within the material, allowing it to store more energy. The relative permittivity reflects the extent of this polarization effect. A higher relative permittivity means the material can store more electrical energy, while a lower relative permittivity means it stores less.

Correct Option Analysis:

The correct option is:

Option 2: reduces

When the actual permittivity (ε) of a material decreases, the relative permittivity (εr) also reduces. This relationship can be understood by revisiting the formula for relative permittivity:

εr = ε / ε0

If the absolute permittivity (ε) decreases while the permittivity of free space (ε0) remains constant, the ratio εr will decrease. This means the material's ability to store electrical energy compared to a vacuum diminishes, leading to a reduction in the relative permittivity.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: increases

This option is incorrect because a reduction in actual permittivity would not lead to an increase in relative permittivity. Since relative permittivity is directly proportional to the actual permittivity, a decrease in one leads to a decrease in the other.

Option 3: becomes zero

This option is incorrect because while a reduction in actual permittivity will lower the relative permittivity, it cannot become zero unless the actual permittivity itself becomes zero. Since actual permittivity is a positive value for all materials, relative permittivity will also remain positive and will not reach zero.

Option 4: remains the same

This option is incorrect because relative permittivity is a function of actual permittivity. If actual permittivity changes, relative permittivity will also change. Therefore, it cannot remain the same if actual permittivity is reduced.

Conclusion:

Understanding the relationship between actual permittivity and relative permittivity is crucial for analyzing the dielectric properties of materials. As explained, relative permittivity reduces when actual permittivity decreases because of their direct proportionality. This fundamental concept is essential in various fields of electrical engineering and material science, where the dielectric properties of materials play a significant role in the design and application of electronic components and systems.

Electric energy density \(U_E = \dfrac{1}{2}\epsilon E^2\).

Now using \(B = \dfrac{E}{c}\) and \(c = \dfrac{1}{\sqrt{\epsilon \mu}}\), we get

\(U_E = \dfrac{1}{2}\epsilon E^2 = \dfrac{1}{2}\dfrac{B^2}{\mu} = U_B\).

Electric energy density is equal to the magnetic energy density.

CONCEPT:

For an electromagnetic wave propagating a rectangular metallic waveguide, if the perpendicular components of the Electric wave vanishes and only the transverse component exist then the corresponding wave is referred to as a transverse electric wave or TE wave. 

For a TE wave, there can be the cutoff frequency is \(\omega_{m n}=c \pi \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}} \)

The wave number K can be represented as \( K=\frac{1}{c} \sqrt{\omega^2-\omega_{m n}^2} \)

Where, m and n (where, m,n = 1, 2, 3, ...) and a and b are the dimensions of the waveguide.

EXPLANATION:

Here the dimension of the wave guide is a = b = L

We know the value of K is given by

\( \begin{aligned} &K=\frac{1}{c} \sqrt{\omega^2-\omega_{m n}^2}\\ & \Rightarrow K^2=\frac{1}{c^2}\left(\omega^2-\omega_{m n}^2\right)\\ & \Rightarrow \omega^2=c^2 K^2+\omega_{m n}^2 \\ & \left( \text {Where, } \omega_{m n}=c \pi \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}} \right) \\ & \Rightarrow \omega^2=c^2 K^2+\frac{2 c^2 \pi^2}{L^2} \quad \left( \because \omega_{11}=\sqrt{2} \frac{c \pi}{L} \right) \\ &\text{​ → Differentiating both sides w.r.t K}\\ & \Rightarrow 2 \omega \frac{d \omega}{d K}=c^2 \times 2 K \\ &\Rightarrow v_g=\frac{d \omega}{d K}=c^2 \frac{K}{\omega} \\ & \because \omega^2=c^2 K^2+2 \frac{c^2 \pi^2}{L^2}\\ & \Rightarrow \frac{\omega^2}{K^2}=c^2+\frac{2 c^2 \pi^2}{K^2 L^2} \\ & \Rightarrow \frac{\omega}{K}=\sqrt{c^2+\frac{2 c^2 \pi^2}{K^2 L^2}}\\ & \Rightarrow v_g=\frac{c^2}{\sqrt{c^2+\frac{2 c^2 \pi^2}{K^2 L^2}}} \\ &\Rightarrow v_g=\frac{c K L}{\sqrt{K^2 L^2+2 \pi^2}} \\ \end{aligned} \)

Hence the correct answer is option 4.

Concept:

For a perfectly reflective surface, the radiation pressure 𝑃 exerted by electromagnetic radiation of intensity 𝐼 is given by

𝑃 = 2 𝐼 / 𝑐

where 𝑐 is the speed of light.

Calculation:

Given:

m = 10³ kg

M = 10³⁰ kg

P = 10²⁶ W

Radiation pressure for fully reflecting Surface = \({2I \over c}\)

I = Intensity = \({Power \over Area}\) = \({P \over 4π R^2}\)

Radiation Pressure = \({2P \over 4π R^2 c}\)

Gravitational pull = \({GMm \over R^2}\)

Force on sail = Radiation Press × Area of sail

= \({2P \over 4π R^2 c}\) × A

A = 6.67×6π×10⁴

= 125.72×104

= 1.25×106

The correct answer is option (4).

Concept:

The velocity (Vp) of a plane electromagnetic wave is given as-

\({V_p} = \frac{1}{{\sqrt {μ \varepsilon } }}\)

Here, μ = μoμr

ε = εo εr

\({v_p} = \frac{1}{{\sqrt {{μ _o}{μ _r}{\varepsilon _o}{\varepsilon _r}} }}\)

\({v_p} = \frac{c}{{\sqrt {{μ _r}{\varepsilon _r}} }}\)-----(1)

\(c = \frac{1}{{\sqrt {{μ _o}{\varepsilon _o}} }} = 3 \times {10^8}\;m/sec\)

μ0 = permeability in free space 4π × 10-7 H/m

εo = permittivity in free space 8.854 × 10-12 C2/Nm2

Calculation:

Given:

μr = 4

from Equation (1)

\(v_p=\frac{3\times10^8}{\sqrt{36}}\)

vp = 0.5 x 108 m/sec

Concept:

The cutoff frequency of wave guide is given by:

\({f_c} = \frac{c}{{2\sqrt {{\mu _b}{\epsilon_b}} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

where c = 3 × 10⁸ m/s

The lowest propogating mode is TE₁₀ cutoff frequency

\({f_c} = \frac{c}{{2a}}\)

\(a = \frac{c}{{2{f_c}}}\)

Calculations:

\(a = \frac{{3 \times {{10}^8}}}{{2 \times 15 \times {{10}^5}}}m\)

a = 1 cm

The next higher mode will be either TE₂₀ or TE₀₁.

for TE₂₀:

cutoff frequency

\({f_c} = \frac{c}{{2a}} \times 2\)

\(= \frac{{2 \times {{10}^8}}}{{2 \times 0.01}} \times 0\)

= 30 GHz

But the required cutoff frequency of next mode is 20 GHz

For TE₀₁ mode

\({f_c} = \frac{c}{{2b}}\) (cutoff frequency)

\(20GHz = \frac{{3 \times {{10}^8}}}{{2 \times b}}\)

\(b = \frac{3}{4}cm = 0.75\;cm\)