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JNTU B.Tech III ELECTROMAGNETIC THEORY AND TRANSMISSION LINES Previous Question Paper (August - 2022 - R16)

B.Tech | III | I | R16 | August - 2022

Answer PaperPaper Code: 135APMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Evaluate the electric field intensity at a point P (-5, 7, -4) in free space due to a charge of 0.2 milli coulombs placed at point R(2, -1, -2).

8 marks

Answer

The electric field intensity E at a point P due to a point charge Q is given by Coulomb's Law:

E = Q / (4πε₀|R|³) * R

Given: Q = 0.2 mC = 0.2 × 10⁻³ C, R = P - R = (-5 - 2)aₓ + (7 - (-1))aᵧ + (-4 - (-2))a₂ = -7aₓ + 8aᵧ - 2a₂. The distance vector magnitude is |R| = sqrt((-7)² + 8² + (-2)²) = sqrt(49 + 64 + 4) = sqrt(117) ≈ 10.817 m. The unit vector is aᵣ = R / |R|. Substituting values with ε₀ ≈ 8.854 × 10⁻¹² F/m:

E = (0.2 × 10⁻³) / (4π * 8.854 × 10⁻¹² * (117)¹·⁵) * (-7aₓ + 8aᵧ - 2a₂)

Calculating the magnitude factor: 1 / (4π * 8.854 × 10⁻¹² * 1265.59) ≈ 7105.8. Thus, E ≈ 0.2 × 10⁻³ * 7105.8 * (-7aₓ + 8aᵧ - 2a₂) ≈ -9.95aₓ + 11.37aᵧ - 2.84a₂ kV/m.

b

Derive the expression for point form of Gauss law.

7 marks

Answer

Gauss's Law in integral form states that the total electric flux passing through a closed surface is equal to the total charge enclosed by that surface:

∮ D · dS = Q = ∫ ρᵥ dV

Using the Divergence Theorem, the surface integral of the electric flux density D over a closed surface can be converted to a volume integral of its divergence:

∮ D · dS = ∫ (∇ · D) dV

Equating the two expressions for the total charge:

∫ (∇ · D) dV = ∫ ρᵥ dV

Since this equality holds for any arbitrary volume, the integrands must be equal, resulting in the point form of Gauss's Law:

∇ · D = ρᵥ

This equation indicates that the divergence of the electric flux density at any point is equal to the volume charge density at that point.

215 marks
a

Define Electric field intensity at a point. Derive the expression for E for a line charge.

7 marks

Answer

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b

An infinitely long uniform line charge is located at y=3, z=5 of PL = 30 nc/m. Find E at the origin.

8 marks

Answer

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315 marks
a

State Ampere's law in integral form and derive the derivative form from it. Derive the continuity equation.

8 marks

Answer

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b

A 3 meter long straight conductor carries a current of 100A in the +x direction. If the flux density is uniformly 2 Tesla and is parallel to xy plane, making an angle of 30 degrees with the x axis, find the force on the conductor.

7 marks

Answer

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415 marks
a

Starting with Ampere's law, derive Maxwell's equation in integral form based on this law and obtain the corresponding differential equation by applying Stroke's theorem.

8 marks

Answer

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b

State and prove the boundary conditions for E and H fields.

7 marks

Answer

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515 marks
a

Derive the relationship between E and H field and show that E/H = 120π.

8 marks

Answer

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b

Write briefly about instantaneous, average and complex poynting vectors.

7 marks

Answer

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615 marks
a

Discuss the wave propagation in good conductor and good dielectrics.

7 marks

Answer

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b

For a 20 MHz travelling wave with Eo=6 V/m find:

i) Average poynting vector

ii) Peak energy density.

8 marks

Answer

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715 marks
a

Derive an expression for input impedance at any point in a transmission line.

7 marks

Answer

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b

Derive the secondary constants for a low loss transmission line.

8 marks

Answer

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815 marks
a

Explain the basis for construction of Smith chart. Illustrate as to how it can be used of an Admittance chart?

8 marks

Answer

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b

Sketch input impedance versus line length for shorted and open circuited line with 0 < l < λ.

7 marks

Answer

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