JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II ELECTROMAGNETIC THEORY AND TRANSMISSION LINES Previous Question Paper (November/December - 2020 - R15)

B.Tech | II | II-II | R15 | November/December - 2020

Answer PaperPaper Code: 124CUMaximum Marks: 75Time: 2 hours

Answer any Five Questions. All Questions Carry Equal Marks.

Main

115 marks
a

Derive the expression of electric field intensity due to uniform line charge.

9 marks

Answer

Consider an infinitely long line charge with uniform charge density ρL (C/m) placed along the z-axis. To find the electric field intensity E at a radial distance r from the line, we use Gauss's Law in integral form: ∮ D · dS = Qenclosed. Due to cylindrical symmetry, the electric flux density D has only a radial component Dr, and the field is independent of φ and z.

We choose a cylindrical Gaussian surface of radius r and length L centered on the z-axis. The surface consists of three parts: the top disk, the bottom disk, and the curved side surface. The flux through the top and bottom disks is zero because the electric field is directed radially outward, perpendicular to the surface normal vectors. The total flux is therefore only through the curved side surface:

∮ D · dS = Dr(2πrL)

The total charge enclosed by this surface is Q = ρLL. Applying Gauss's Law:

Dr(2πrL) = ρLL
Dr = ρL / (2πr)

Since D = ε0E for free space, the electric field intensity is:

E = ρL / (2πε0r) ar
b

Given V= 5x3y2z and ε=2.25ε0, find i) E at point P (-3, 1, 2) ii) ρv at P.

6 marks

Answer

Given V = 5x3y2z and ε = 2.25ε0.

i) The electric field E is the negative gradient of the potential V:

E = -∇V = -(∂V/∂x ax + ∂V/∂y ay + ∂V/∂z az)
E = -(15x2y2z ax + 10x3yz ay + 5x3y2 az)

At point P(-3, 1, 2):

E = -(15(-3)2(1)2(2) ax + 10(-3)3(1)(2) ay + 5(-3)3(1)2 az)
E = -(270 ax - 540 ay - 135 az) = -270 ax + 540 ay + 135 az V/m

ii) To find the volume charge density ρv, use Poisson's equation ∇·D = ρv, where D = εE = 2.25ε0E:

ρv = ε ∇ · E = ε (-∇2V)
∇2V = ∂2V/∂x2 + ∂2V/∂y2 + ∂2V/∂z2
∇2V = 30xy2z + 10x3z + 0 = 30xy2z + 10x3z

At P(-3, 1, 2):

∇2V = 30(-3)(1)2(2) + 10(-3)3(2) = -180 - 540 = -720
ρv = -2.25ε0(-720) = 1620ε0 C/m3
215 marks
a

Explain any one application of Gauss’s Law.

7 marks

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b

Define and explain the following: i) Electric flux density D ii) Electric field intensity E.

8 marks

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315 marks
a

Briefly discuss about magnetic scalar and vector potentials.

7 marks

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b

Explain the applications of Ampere’s Circuit Law.

8 marks

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415 marks
a

The electric field in air is given by E=ρte-ρ-t aφ V/m. Find B and J.

8 marks

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b

Show that another form of Faraday’s law is E = -∂A/∂t, where A is the magnetic vector potential.

7 marks

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515 marks
a

Show that in a good conductor, the skin depth δ is always much shorter than the wavelength.

9 marks

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b

A copper wire is 0.1mm thick. Find the frequency at which the thickness of the trace is equal to the skin depth.

6 marks

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6

State and prove Poynting theorem.

15 marks

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715 marks
a

Prove that a line of finite length terminated by its characteristic impedance Z0 is equal to a line of infinite length.

7 marks

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b

Describe the losses in transmission lines.

8 marks

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815 marks
a

Describe the applications and characteristics of λ/2 and λ/4 lossless transmission line elements.

8 marks

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b

What is stub? Why are short circuited stubs are preferred to open circuited ones? Explain.

7 marks

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This page contains the question paper of ELECTROMAGNETIC THEORY AND TRANSMISSION LINES from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech ECE II II-II, November/December - 2020, R15.

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