USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
INTERNAL ASSESSMENT TEST 2
Date
:
Marks: 60
Subject & Code : Engineering Electromagnetics – 17EC36
Sec : 3rd A,B,C
Name of faculty : Mr. Muralidhar/ Mr. Shreyus
Time : 11:30 to 1:00 PM
Note: Answer FIVE full questions, selecting any ONE full question from each part.
PART 1
1
Point charges of 50 nC each are located at A (1, 0, 0), B (−1, 0, 0), C (0, 1, 0), and D (0, −1, 0) in
Marks
12
free space. Find the total force on the charge at A.
2
3
4
5
6
7
8
9
10
A uniform volume charge density of 0.2 μC/m3 is present throughout the spherical shell extending
from r = 3 cm to r = 5 cm. If ρν = 0 elsewhere, find (a) the total charge present throughout the shell,
and (b) r1 if half the total charge is located in the region 3 cm < r < r1.
PART 2
(a) State Gauss’s Law. (b) Derive the mathematical form of the Gauss’s Law. (c) An electric field
in free space is E = (5z3/ε0) az V/m. Find the total charge contained within a sphere of 3-m radius,
centered at the origin.
(a) Derive Maxwell’s First Equation. (b) State and derive Divergence Theorem. (c) Given the
1
1
field 𝑫 = 6𝜌 𝑠𝑖𝑛 2 𝜑𝒂𝝆 + 1.5𝜌 𝑐𝑜𝑠 2 𝜑𝒂𝝋 𝐶/𝑚2 , evaluate both sides of the divergence theorem
for the region bounded by ρ = 2, φ = 0, φ = π, z = 0, and z = 5.
PART 3
(a) Calculate the energy expended in moving a point charge in an electric field. (b) The value of E
at P (ρ = 2, φ = 40◦, z = 3) is given as E = 100aρ − 200aφ + 300az V/m. Determine the incremental
work required to move a 20 μC charge a distance of 6 μm: (i) in the direction of aρ; (ii) in the
direction of aφ; (iii) in the direction of az; (iv) in the direction of E; (v) in the direction of G = 2ax −
3ay + 4az.
Let G = 3xy2ax + 2zay Given an initial point P (2, 1, 1) and a final point Q (4, 3, 1), find ∫ 𝑮. 𝒅𝑳
using the path (a) straight line: y = x − 1, z = 1; (b) parabola: 6y = x2 + 2, z = 1.
PART 4
(a) Derive the relation for convection current density. (b) Given the vector current density J =
10ρ2zaρ − 4ρ cos2φ aφ mA/m2: (i) find the current density at P (ρ = 3, φ = 30◦, z = 2); (ii) determine
the total current flowing outward through the circular band ρ = 3, 0 < φ < 2π, 2 < z < 2.8.
(a) Derive Continuity Equation. (b) Current density is given in cylindrical coordinates as J =
−106z1.5az A/m2 in the region 0 ≤ ρ ≤ 20μm; for ρ ≥ 20μm, J = 0. (i) Find the total current crossing
the surface z = 0.1 m in the az direction. (ii) If the charge velocity is 2 × 106 m/s at z = 0.1 m, find
ρν there. (iii) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity
there.
PART 5
(a) Derive Poisson’s Equation. (b) Given the potential field V = (Aρ4 + Bρ−4) sin 4φ: (i) Show that
∇2V = 0. (ii) Select A and B so that V = 100 V and |E| = 500 V/m at P (ρ = 1, φ = 22.5◦, z = 2).
State and prove Uniqueness theorem.
BE III semester
6+6
4+4+4
4+4+4
4+8
6+6
4+8
4+8
4+8
12
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
SCHEME & SOLUTION
INTERNAL ASSESSMENT TEST – II
Faculty: Mr.Muralidhar/ Mr.Shreyus
Semester: III A,B,C
Subject: Engineering Electromagnetics
Sub. Code: 17EC36
Q.
No.
1
2
BE III semester
Marks
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
3
(a) Gauss’s Law: The electric flux passing through any closed surface is equal to the total charge
enclosed by that surface.
(b) Mathematical form of the Gauss’s Law:
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
(c) E = (5z3/ε0) az V/m
D = ε0E = 5z3 az C/m3
div D = 15z2 = ρv
𝑄 = ∫𝑣 𝜌𝑣 𝑑𝑣 = 15268.1 𝐶
4. (a) Maxwell’s First Equation:
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
(b) Divergence Theorem: The divergence theorem states that the total flux crossing the closed
surface is equal to the integral of the divergence of the flux density throughout the enclosed
volume.
(c) LHS = 225 ; RHS = 225
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
5. (a)
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
(b)
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
6.
7. (a)
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
(b)
(i) 180aρ - 9aφ mA/m2 ; (ii) 3.26 A
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
8. (a) Continuity Equation:
(b)
(i) -39.7 μA ; (ii) -15.8 mC/m3 ; (iii) 29.0 m/s
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
9. (a) Poisson’s Equation:
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
(b)
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
10. Uniqueness Theorem
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
BE III semester
USN
1 P E
PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of Electronics and Communication
BE III semester