# Basics of Electromagnetics

 Question 1
In an electrostatic field, the electric displacement density vector, $\vec{D}$, is given by
$\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})C/m^2$
, where $\vec{i},\vec{j},\vec{k}$ are the unit vectors along x-axis, y-axis, and z-axis, respectively. Consider a cubical region R centered at the origin with each side of length 1 m, and vertices at ($\pm 0.5 m, \pm 0.5 m, \pm 0.5 m$). The electric charge enclosed within R is _________ C (rounded off to two decimal places).
 A 0.25 B 0.5 C 0.75 D 0.85
GATE EC 2022   Electromagnetics
Question 1 Explanation:
$\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})c/m^2$
$Q_{enc.}=\int _v \rho _v\cdot dV=\int (\triangledown \cdot \vec{D})dV$
$\triangledown \cdot \vec{D}=3x^2+3y^2$
$dV=dxdydz$
$\therefore \; Q_{enc.}=\int _v 3(x^2+y^2)dxdydz =3\left [ \int_{-0.5}^{0.5} x^2 dx \int_{-0.5}^{0.5}dy \int_{-0.5}^{0.5} dz+\int_{-0.5}^{0.5}dx\int_{-0.5}^{0.5}y^2dy\int_{-0.5}^{0.5}dz \right ]$
$=3\left [ \frac{x^3}{3}|_{-0.5}^{0.5} \times 1 \times 1 + \frac{y^3}{3}|_{-0.5}^{0.5} \times 1 \times 1\right ] =0.25+0.25=0.5C$
 Question 2
In a circuit, there is a series connection of an ideal resistor and an ideal capacitor. The conduction current (in Amperes) through the resistor is $2\sin (t+\pi/2)$. The displacement current (in Amperes) through the capacitor is _________.
 A $2 \sin (t)$ B $2 \sin (t+\pi)$ C $2 \sin (t+\pi /2)$ D 0
GATE EC 2022   Electromagnetics
Question 2 Explanation:

In series connection, current pass through each element remain same. Hence, $i_c=i_d$
So, $i_d= 2\sin (t+\pi/2)$.
 Question 3
For a vector field $D=\rho\cos^{2}\:\varphi \:a_{\rho }+z^{2}\sin^{2}\:\varphi \:a_{\varphi }$ in a cylindrical coordinate system $\left ( \rho ,\varphi ,z \right )$ with unit vectors $a_{\rho },a_{\varphi }$ and $a_{z}$ , the net flux of D leaving the closed surface of the cylinder $\left ( \rho =3, 0\leq z\leq 2 \right )$ (rounded off to two decimal places) is ________________
 A 56.55 B 22.12 C 36.85 D 76.34
GATE EC 2021   Electromagnetics
Question 3 Explanation:
Method 1: $\vec{D}=\rho \cos ^{2} \phi \hat{a}_{\rho}+z^{2} \sin ^{2} \phi \hat{a}_{\phi}$
Electric flux crossing the closed surface is
\begin{aligned} \psi &=\oiint \vec{D} \cdot \overrightarrow{d S}=\iiint(\vec{\nabla} \cdot \vec{D}) d v \\ \vec{\nabla} \cdot \vec{D} &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho D_{\rho}\right)+\frac{1}{\rho} \frac{\partial D_{\phi}}{\partial \phi}+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho \rho \cos ^{2} \phi\right)+\frac{1}{\rho} \frac{\partial}{\partial \phi}\left(z^{2} \sin ^{2} \phi\right)+0 \\ &=\frac{1}{\rho}(2 \rho) \cos ^{2} \phi+\frac{1}{\rho} z^{2} 2 \sin \phi \cos \phi=2 \cos ^{2} \phi+\frac{z^{2}}{\rho} \sin 2 \phi\\ \iiint(\vec{\nabla} \cdot \vec{D}) d v &=\iiint 2 \cos ^{2} \phi(\rho d \rho d \phi d z)+\iiint\left(\frac{z^{2}}{\rho} \sin 2 \phi\right) \rho d \rho d \phi d z \\ &=2 \int_{\rho=0}^{3} \rho d \rho \int_{\phi=0}^{2 \pi}\left(\frac{1+\cos 2 \phi}{2}\right) d \phi \int_{z=0}^{2} d z+\int_{\rho=0}^{2} d \rho \int_{\phi=0}^{2 \pi} \sin 2 \phi d \phi \int_{z=0}^{2} z^{2} d z \\ &=2\left(\frac{\rho^{2}}{-2}\right)_{\rho=0}^{3} \frac{1}{2}(2 \pi)(z)_{z=0}^{2}+0 \\ &=2\left(\frac{3^{2}}{2}\right) \pi(2)=18 \pi(\text { Coulomb })=56.55(\text { Coulomb }) \end{aligned}
Method 2: Electric flux crossing the closed surface is
$\psi=\oiint \vec{D} \cdot \overrightarrow{d S}$
Electric flux crossing \rho=3 cylindrical surface is
\begin{aligned} \left.\psi\right|_{\rho=3} &=\oiint\left(\rho \cos ^{2} \phi \hat{a}_{p}\right) \cdot(\rho d \phi d z) \hat{a}_{p} \\ &=3^{2} \int_{\phi=0}^{2 \pi} \cos ^{2} \phi d \phi \int_{z=0}^{2} d z \\ &=9 \frac{1}{2}(2 \pi)(2)=18 \pi(\text { coulomb })=56.55(\text { coulomb }) \end{aligned}
 Question 4
Consider the vector field $F\:=\:a_{x}\left ( 4y-c_{1}z \right )+a_y\left ( 4x + 2z\right )+a_{z}\left ( 2y +z\right )$ in a rectangular coordinate system (x,y,z) with unit vectors $a_{x},\:a_{y}$ and $a_{z}$. If the field F is irrotational (conservative), then the constant $c_{1}$ (in integer) is _________________
 A 0 B 1 C 2 D 3
GATE EC 2021   Electromagnetics
Question 4 Explanation:
\begin{aligned} \nabla \times \vec{F}&=0\\ \left|\begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 4 y-c_{1} z & 4 x+2 z & 2 y+z \end{array}\right|&=0\\ =i(2-2)-j\left(0+c_{1}\right)+k(4-4)=0 \\ c_{1} &=0 \end{aligned}
 Question 5
Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the magnetic field $\vec{B}$ between the wires at a distance r from W1 is
 A $\frac{\mu _0 I}{6 \pi r}$ B $\frac{6 \mu _0 I}{5 \pi r}$ C $\frac{5 \mu _0 I}{6 \pi r}$ D $\frac{{\mu _0}^2 I^2}{2 \pi r^2}$
GATE EC 2019   Electromagnetics
Question 5 Explanation:

Magnetic flux density $(\vec{B})$ at r distance due to infinite line carrying current I is $|\vec{B}|=\frac{\mu_{0} I}{2 \pi \rho} .$
$\vec{B}$ at r distance due to $W_{1}$ wire
$=\left|\vec{B}_{1}\right|=\frac{\mu_{0} I}{2 \pi r}\qquad \ldots(i)$
$\vec{B}$ at 3r distance due to $W_{2}$ wire
$=\left|\vec{B}_{2}\right|=\frac{\mu_{0}(2 I)}{2 \pi(3 r)}\qquad \ldots(ii)$
From right hand thumb rule, $\vec{B}$ due to both lines add in between conductors.
\begin{aligned} {So,}\qquad |\vec{B}|&=\left|\vec{B}_{1}\right|+\left|\vec{B}_{2}\right|\\ \therefore \qquad |\vec{B}|&=\frac{\mu_{0} I}{2 \pi r}+\frac{2 \mu_{0} I}{6 \pi r}=\frac{5 \mu_{0} I}{6 \pi r} \end{aligned}
 Question 6
In the table shown, List I and List II, respectively, contain terms appearing on the left-hand side and the right-hand side of Maxwell's equations (in their standard form). Match the left-hand side with the corresponding right-hand side.
 A 1-P, 2-R, 3-Q, 4-S B 1-Q, 2-R, 3-P, 4-S C 1-Q, 2-S, 3-P, 4-R D 1-R, 2-Q, 3-S, 4-P
GATE EC 2019   Electromagnetics
Question 6 Explanation:
\begin{aligned} \nabla \cdot \vec{D} &=\rho_{v} \\ \nabla \times \vec{E} &=-\frac{\partial \vec{B}}{\partial t} \\ \nabla \cdot \vec{B} &=0 \\ \nabla \times \vec{H} &=\vec{J}+\frac{\partial \vec{D}}{\partial t} \end{aligned}
 Question 7
What is the electric flux ($\int \vec{E}\cdot d\hat{a}$) through a quarter-cylinder of height H (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q?
 A $\frac{HQ}{\varepsilon _0}$ B $\frac{HQ}{4\varepsilon _0}$ C $\frac{H\varepsilon _0}{4Q}$ D $\frac{4H}{Q\varepsilon _0}$
GATE EC 2019   Electromagnetics
Question 7 Explanation:
Electric field intensity $(\vec{E})$ at '$\rho$' distance due to infinite long line having line charge density Q is
\begin{aligned} \vec{E} &=\frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{\rho} \\ \int \vec{E} \cdot \vec{da} &=\iint \frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{p} \cdot \rho d \phi d z \hat{a}_{\rho} \\ &=\frac{Q}{2 \pi \varepsilon_{0}} \int_{<\pi / 2>} d \phi \int_{z=0}^{H} d z \\ &=\frac{Q}{2 \pi \varepsilon_{0}}\left(\frac{\pi}{2}\right) H=\frac{H Q}{4 \varepsilon_{0}} \end{aligned}
 Question 8
An electron ($q_1$) is moving in free space with velocity $10^{5}$ m/s towards a stationary electron ($q_2$) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is ___________ $\times 10^{-8}$m.
[Given, mass of electron $m=9.11 \times 10^{-31}$ kg, charge of electron $e=-1.6 \times 10^{-19}$C, and permittivity $\varepsilon_{0}=(1/36\pi)\times 10^{-9}F/m]$
 A 4 B 5 C 6 D 7
GATE EC 2017-SET-2   Electromagnetics
Question 8 Explanation:

r is the distance at which kinetic energy of $q_{1}$ becomes zero [(because kinetic energy (KE) is converted into potential energy (PE)].
When $q_{1}$ reaches 'r', it starts diverting.
Kinetic energy, $K E=\frac{1}{2} m v^{2}$ and work done in moving $q_{1}$ charge to distance 'r' is
\begin{aligned} q_{1} v_{2}&=q_{1} \frac{q_{2}}{4 \pi \varepsilon_{0} r} \\ &\qquad\left(q_{1}=q_{2}=-1.6 \times 10^{-19} \mathrm{C}\right)\\ Now, \quad \frac{1}{2} m v^{2}&=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r} \\ \end{aligned}
$\Rightarrow r=\frac{\left(2 \times-1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{4 \pi \times \frac{10^{-9}}{36 \pi} \times 9.11 \times 10^{-31} \times\left(10^{5}\right)^{2}}$
$\simeq 5.06 \times 10^{-8} \mathrm{m}$
 Question 9
Two conducting spheres S1 and S2 of radii a and b (b>a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure.

For some charge placed on this structure, the potential and surface electric field on S1 are $V_{a}$ and $E_{a}$, and that on S2 are $V_{b}$ and $E_{b}$, respectively, which of the following is CORRECT?
 A $V_{a}=V_{b} \; and \; E_{a}\lt E_{b}$ B $V_{a}\gt V_{b} \; and \; E_{a} \gt E_{b}$ C $V_{a}=V_{b} \; and \; E_{a} \gt E_{b}$ D $V_{a}\gt V_{b} \; and \; E_{a}=E_{b}$
GATE EC 2017-SET-2   Electromagnetics
Question 9 Explanation:

When charge is placed on this structure equilibrium is established such that be spheres are at same potential i.e.
$V_{a}=V_{b}$
$\begin{array}{c} V_{a}=V_{b} \\ \text { So, } \frac{Q_{a}}{4 \pi \epsilon_{o} a}=\frac{Q_{b}}{4 \pi \epsilon_{o} b} \end{array}$
$\frac{Q_{b}}{Q_{a}}=\frac{b}{a}$
Now, surface electric fields.
$\frac{E_{a}}{E_{b}}=\left[\frac{Q_{a} / 4 \pi \varepsilon_{o} a^{2}}{Q_{b} / 4 \pi \varepsilon_{o} b^{2}}\right]=\frac{Q_{a} \times b^{2}}{Q_{b} \times a^{2}}=\frac{b}{a}>1$
$So, E_{a}>E_{b}$
 Question 10
Consider the charge profile shown in the figure. The resultant potential distribution is best described by

 A A B B C C D D
GATE EC 2016-SET-3   Electromagnetics
Question 10 Explanation:
Applying Poisson's equations
$\nabla^{2} V=\frac{\partial^{2} V}{\partial x^{2}}=-\frac{\rho_{v}}{\epsilon}=K$
Constant charge density
\begin{aligned} \frac{\partial V}{\partial x} &=-K x+K^{\prime} \\ V &=\frac{-K x^{2}}{2}+K^{\prime} x+K^{\prime} \end{aligned}
Towards positive x or negative side.
It is a second order parabolic increase.
Due to symmetry of + and - charges K" = 0 is expected with V = O at centre and graph passing through origin.
Beyond $x \gt 0 \text{ or } x \lt b, \; E = 0$ due to capacitive nature of + and - charges
$V=-\int 0 \cdot \overrightarrow{d l}=\text { Constant }$
This constant is same V at x=a or x=b
There are 10 questions to complete.