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JNTUH B.Tech II ELECTROMAGNETIC FIELDS Previous Question Paper (August/September - 2022 - R15)

B.Tech | II | II-I | R15 | August/September - 2022

Answer PaperPaper Code: 123BYMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

State and Explain Coulomb’s Law.

7 marks

Answer

Coulomb’s Law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.

F = (Q1Q2) / (4πε0R2) aR

In this expression, Q1 and Q2 are the point charges in Coulombs, R is the distance between them in meters, ε0 is the permittivity of free space (8.854 × 10-12 F/m), and aR is the unit vector directed from the source charge toward the test charge. The force is repulsive for like charges and attractive for unlike charges.

b

A Eight point charges of 1nC each are located at the corners of a cube in free space that is 1m on a side. Find at the centre of a face.

8 marks

Answer

For a cube of side length a = 1m with charges q = 1nC at each corner, consider the center of the top face. Let the coordinates of the corners be (0,0,1), (1,0,1), (0,1,1), (1,1,1) for the top face and (0,0,0), (1,0,0), (0,1,0), (1,1,0) for the bottom face. The center of the top face is at (0.5, 0.5, 1).

  1. The four charges on the top face corners are at distances of √(0.52 + 0.52) = 1/√2 m from the center. Due to symmetry, the horizontal components of the forces from these four charges cancel out, and their vertical components are zero as they lie in the same plane.
  2. The four charges on the bottom face corners are at a distance r from the center, where r = √(0.52 + 0.52 + 12) = √1.5 m.
  3. The electric field E from one bottom corner charge is E = q / (4πε0r2). The vertical component is Ez = E cos(θ), where cos(θ) = 1/√1.5.
  4. Summing the vertical components of the four bottom charges: Etotal = 4 × [q / (4πε0(1.5))] × (1/√1.5).
  5. Substituting q = 10-9 C and 1/(4πε0) = 9 × 109: Etotal = 4 × (9 × 109 × 10-9 / 1.5) × (1/1.2247) ≈ 19.59 V/m.
215 marks
a

Derive expression for potential and Electric field intensity due to an electric dipole.

7 marks

Answer

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b

An electric dipole, 1az nC-m is at (0, 0, 0). Find the electric field at (0, 0, 1).

8 marks

Answer

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315 marks
a

Derive an expression for the capacitance of parallel plate capacitor with composite dielectrics.

7 marks

Answer

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b

A unit vector directed from region 1 to region 2 at the planar boundary between two perfect dielectrics is given. Find the necessary field components.

8 marks

Answer

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415 marks
a

State and Explain Biot Savart’s Law.

7 marks

Answer

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b

Find H at the centre of square current loop of side ‘L’.

8 marks

Answer

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515 marks
a

State Ampere’s Law and explain how do you find H using ampere’s Law.

7 marks

Answer

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b

A current filament of 5.0A in the ay direction is parallel to the y-axis at x = 2m, z = -2m. Find H at the origin.

8 marks

Answer

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615 marks
a

Derive the expression for force between two current carrying Conductors in same direction.

7 marks

Answer

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b

A conductor of length 2.5m located at z = 0, x = 4m carries a current of 12.0A in the -ay direction. Find the uniform B in the region if the force on the conductor is 1.20 × 10-2N.

8 marks

Answer

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715 marks
a

Define Scalar magnetic potential and List out its limitations.

5 marks

Answer

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b

Determine self-inductance of a solenoid.

5 marks

Answer

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c

Calculate the self-inductances and mutual inductances between two coaxial solenoids of radius R1 and R2.

5 marks

Answer

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815 marks
a

Derive Maxwell’s Equation from Faraday’s law for time varying fields.

7 marks

Answer

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b

In a material for which σ = 5.0 S/m and εr = 1 the electric field intensity is E = 250 sin 1010t (V/m). Find the conduction and displacement current densities.

8 marks

Answer

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