JNTU
JNTU B.Tech II ELECTROMAGNETIC FIELDS Previous Question Paper (September - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
State and explain Gauss Law.
Answer
Gauss's Law states that the total electric flux (Ψ) passing through any closed surface is equal to the total charge (Q) enclosed by that surface.
In this expression, D represents the electric flux density, dS is the differential surface vector, and ρv is the volume charge density. The law implies that the net electric flux emanating from a closed region is proportional to the net charge contained within it. It serves as a fundamental relationship in electromagnetics, allowing for the calculation of electric fields in systems with high degrees of symmetry.
A Point charge of 20nC is located at (4,-1,3) and a uniform Line charge of -25nC/m lies along the intersection of the planes x= -4 and z= 6. Calculate D at (3,-1,0).
Answer
To find the total electric flux density D at point P(3, -1, 0), we sum the contributions from the point charge and the line charge.
- Point Charge Contribution: The point charge Q = 20 nC is at (4, -1, 3). The vector from the charge to P is R = (3-4)ax + (-1-(-1))ay + (0-3)az = -ax - 3az. The distance is |R| = sqrt((-1)^2 + (-3)^2) = sqrt(10). The unit vector is aR = (-ax - 3az) / sqrt(10). Using D1 = Q / (4π|R|^2) * aR:D1 = 20e-9 / (4π * 10) * (-ax - 3az) / sqrt(10) = 0.159 * (-ax - 3az) / 3.162 = -0.0503ax - 0.151az nC/m^2
- Line Charge Contribution: The line charge ρL = -25 nC/m lies at the intersection of x = -4 and z = 6, which is a line parallel to the y-axis passing through (-4, y, 6). The vector from the line to P(3, -1, 0) is R = (3 - (-4))ax + (0 - 6)az = 7ax - 6az. The perpendicular distance is ρ = sqrt(7^2 + (-6)^2) = sqrt(85) ≈ 9.22 m. The unit vector is aρ = (7ax - 6az) / sqrt(85). Using D2 = ρL / (2πρ) * aρ:D2 = -25e-9 / (2π * 9.22) * (7ax - 6az) / 9.22 = -0.431 * (0.759ax - 0.651az) = -0.327ax + 0.281az nC/m^2
- Total D: D = D1 + D2 = (-0.0503 - 0.327)ax + (-0.151 + 0.281)az = -0.377ax + 0.130az nC/m^2
Derive Electric Field Intensity at a point due to surface charge.
Answer
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Given the surface charge density, ρs=2µC/m2, existing in the region ρ<0.2m, z=0, and is zero elsewhere, Find E at P(ρ=0, z=0.5).
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Derive Equation of Continuity.
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Let V= (Cos2Ø)/ρ in free space. Find the volume charge density at point P(0.5, 60o, 1).
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Derive capacitance of a two-wire line.
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Obtain boundary conditions for Dielectric-Dielectric material.
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State Biot Savart’s Law and Ampere’s circuital Law.
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Given points C(5,-2,3) and P(4,-1,2), a current element IdL = 10-4(4,-3,1)A.m at C produces a field dH at P. Specify the direction of dH by a unit vector aH and also find |dH|.
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Derive Equation for self inductance of solenoid.
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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 in the direction (-ax+az)/√2.
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Explain statically and dynamically induced electro motive forces.
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A certain material has σ=0, μr=1. If E = 800sin(106t - 0.01y)az V/m. Make use of Maxwell equation to find єR.
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Derive electromagnetic Wave Equation.
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State and explain Poynting Theorem.
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