# Electromagnetic Theory

 Question 1
An infinite surface of linear current density $K=5 \hat{a}_{x} A / m$ exists on the $x-y$ plane, as shown in the figure. The magnitude of the magnetic field intensity $(H)$ at a point $(1,1,1)$ due to the surface current in Ampere/meter is ___ (Round off to 2 decimal places).

 A 1.45 B 6.36 C 2.5 D 3.54
GATE EE 2023      Magnetostatic Fields
Question 1 Explanation:
\begin{aligned} \overrightarrow{\mathrm{a}}_{\mathrm{N}}&=\overrightarrow{\mathrm{a}}_{\mathrm{Z}}\\ \text{Now, }\quad \overrightarrow{\mathrm{H}}&=\frac{1}{2}\left(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{a}}_{\mathrm{N}}\right) \\ & =\frac{1}{2}\left(5 \hat{a}_{x} \times \hat{a}_{z}\right) \\ & =-2.5 \hat{a}_{y} A / m \end{aligned}
$\therefore$ Magnitude, $|\overrightarrow{\mathrm{H}}|=2.5 \mathrm{~A} / \mathrm{m}$
 Question 2
In the figure, the electric field $E$ and the magnetic field $B$ point to $\mathrm{x}$ and $\mathrm{z}$ directions, respectively, and have constant magnitudes. $A$ positive charge '$q$' is released from rest at the origin. Which of the following statement(s) is/ are true.

 A The charge will move in the direction of $z$ with constant velocity. B The charge will al ways move on the $y$-z plane only. C The trajectory of the charge will be a cycle. D The charge will progress in the direction of $y$.
GATE EE 2023      Coordinate Systems and Vector Calculus
Question 2 Explanation:
As per Answer key of IIT Official : MTA (Marks to All)
Given :

 Question 3
As shown in the figure below, two concentric conducting spherical shells, centered at $r=0$ and having radii $r=c$and $r=d$ are maintained at potentials such that the potential $V(r)$ at $r=c$ is $V_1$ and $V(r)$ at $r=d$ is $V_2$. Assume that $V(r)$ depends only on $r$, where $r$ is the radial distance. The expression for $V(r)$ in the region between $r=c$and $r=d$ is

 A $V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_1c+V_2d-2V_1d}{d-c}$ B $V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_2d-V_1c}{d-c}$ C $V(r)=\frac{cd(V_1-V_2)}{(d-c)r}-\frac{V_1c-V_2c}{d-c}$ D $V(r)=\frac{cd(V_2-V_1)}{(d-c)r}-\frac{V_2c-V_1c}{d-c}$
GATE EE 2022      Electrostatic Fields
Question 3 Explanation:
We have, Laplace equation
$\triangledown ^2V=0$ ...(1)
where, $\triangledown ^2V=\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )+\frac{1}{r^2 \sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta \frac{\partial V}{\partial \theta } \right )+ \frac{1}{r^2 \sin ^2\theta } \frac{\partial^2 V}{\partial \phi ^2}$
V will have only radial component.
From equation (1)
$\frac{1}{r^2}\frac{\partial }{\partial r}\left ( r^2\frac{\partial V}{\partial r} \right )=0$
Integrate both side $r^2\frac{\partial V}{\partial r} =P$
Again integrate, $V=-\frac{P}{r}+Q$
Here, P and Q are constant.
Given: at $r=c, V=V_1$
$V_1=-\frac{P}{r}+Q$ ...(2)
at $r=d, V=V_2$
$V_2=-\frac{P}{d}+Q$ ...(3)
From eq. (2) & (3)
$P=\frac{(V_2-V_1)cd}{d-c}$
$Q=V_1+\frac{d(V_2-V_1)}{d-c}$
From eq. (1),
$V=-\frac{(V_2-V_1)cd}{r(d-c)}+V_1+\frac{(V_2-V_1)cd}{(d-c)}$
$V=\frac{(V_1-V_2)cd}{r(d-c)}+\frac{(V_2d-V_1c)}{(d-c)}$
 Question 4
If the magnetic field intensity $(H)$ in a conducting region is given by the expression, $H=x^2\hat{i}+x^2y^2\hat{j}+x^2y^2z^2\hat{k}\;A/m$. The magnitude of the current density, in $A/m^2$ , at $x=1m, y=2m$ and $z=1m$, is
 A 8 B 12 C 16 D 20
GATE EE 2022      Magnetostatic Fields
Question 4 Explanation:
We have,
\begin{aligned} \vec{J}&=\triangledown \times \vec{H}\\ \vec{J}&=\begin{vmatrix} \hat{a}_x &\hat{a}_y & \hat{a}_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ x^2& x^2y^2 & x^2y^2z^2 \end{vmatrix} \\ &=\hat{a}_x(2x^2yz^2-0)-\hat{a}_y(2xy^2z^2-0)+\hat{a}_z(2xy^2-0)\\ &=2x^2yz^2\hat{a}_x-2xy^2z^2\hat{a}_y+2xy^2\hat{a}_z\\ &\text{At point (1,2,1)}\\ \vec{J}&=4\hat{a}_x-8\hat{a}_y+8\hat{a}_z\\ |\vec{J}|&=\sqrt{4^2+8^2+8^2}=12A/m^2 \end{aligned}
 Question 5
A long conducting cylinder having a radius 'b' is placed along the z axis. The current density is $J=J_ar^3\hat{z}$ for the region $r \lt b$ where $r$ is the distance in the radial direction. The magnetic field intensity ($H$) for the region inside the conductor (i.e. for $r \lt b$) is
 A $\frac{J_a}{4}r^4$ B $\frac{J_a}{3}r^3$ C $\frac{J_a}{5}r^4$ D $J_a r^3$
GATE EE 2022      Magnetostatic Fields
Question 5 Explanation:
Using Ampere?s law,
\begin{aligned} \oint \vec{H}\cdot d\vec{l}&=I_{enc}\\ \text{where, }I_{enc}&=\int \int \vec{J}\cdot d\vec{s}\\ d\vec{s}&=rdrd\phi \hat{a}_z\\ \therefore \oint \vec{H}\cdot d\vec{l}&=\int \int \vec{J}\cdot r^3\hat{a}_z\cdot rdrd\phi \hat{a}_z\\ \vec{H}\cdot 2 \pi r&=\frac{J_ar^5}{5}(2 \pi)\\ \Rightarrow \vec{H}&=\frac{J_ar^4}{5}A/m \end{aligned}

There are 5 questions to complete.