GATE 2022 Electromagnetic Fields Questions and Solutions

Question 01

Question 1

A long conducting cylinder having a radius \(b\) is placed along the \(z\)-axis. The current density is \(\vec{J} = J_a r^3 \hat{z}\) for the region \(r < b\) where \(r\) is the distance in the radial direction.

The magnetic field intensity \((\vec{H})\) for the region inside the conductor (i.e., for \(r < b\)) is:

\(\frac{J_a r^4}{4}\)
\(\frac{J_a r^3}{3}\)
\(\frac{J_a r^4}{5}\)
\(J_a r^3\)

Solution

Given:

Equation

\[\vec{J} = J_a r^3 \hat{z}\]

To find the current enclosed (\(I\)):

Equation

\[\begin{aligned} I &= \int_S \vec{J} \cdot d\vec{s}; \quad \text{where } d\vec{s} = r dr d\phi \hat{z} \\ \Rightarrow I &= \int (J_a r^3 \hat{z}) \cdot (r dr d\phi \hat{z}) \\ &= J_a \int_{0}^{r} r^4 dr \int_{0}^{2\pi} d\phi \\ &= J_a \left[ \frac{r^5}{5} \right]_0^r \cdot [\phi]_0^{2\pi} \\ &= \frac{J_a (2\pi) r^5}{5} \end{aligned}\]

As per Ampere's Circuital Law:

Equation

\[\begin{aligned} \oint \vec{H} \cdot d\vec{L} &= I_{\text{enc}} \\ H(2\pi r) &= \frac{J_a (2\pi) r^5}{5} \\ H &= \frac{J_a r^4}{5} \end{aligned}\]

Final Answer

Correct answer: C.

Question 02

Question 2

If the magnetic field intensity \((\vec{H})\) in a conducting region is given by the expression \(\vec{H} = x^2\hat{i} + x^2y^2\hat{j} + x^2y^2z^2\hat{k}\) A/m, the magnitude of the current density, in A/m\(^2\), at \(x = 1\) m, \(y = 2\) m, and \(z = 1\) m is:

Solution

Given:

Equation

\[\vec{H} = x^2\hat{i} + x^2y^2\hat{j} + x^2y^2z^2\hat{k} \text{ A/m}\]

Current density is the curl of magnetic field intensity (\(\vec{J} = \nabla \times \vec{H}\)):

Equation

\[\vec{J} = \begin{vmatrix} \hat{a}_x & \hat{a}_y & \hat{a}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 & x^2y^2 & x^2y^2z^2 \end{vmatrix}\]

Equation

\[\begin{aligned} \vec{J} &= \hat{a}_x \left( \frac{\partial}{\partial y}(x^2y^2z^2) - \frac{\partial}{\partial z}(x^2y^2) \right) - \hat{a}_y \left( \frac{\partial}{\partial x}(x^2y^2z^2) - \frac{\partial}{\partial z}(x^2) \right) + \hat{a}_z \left( \frac{\partial}{\partial x}(x^2y^2) - \frac{\partial}{\partial y}(x^2) \right) \\ &= \hat{a}_x(2x^2y z^2 - 0) - \hat{a}_y(2xy^2z^2 - 0) + \hat{a}_z(2xy^2 - 0) \\ &= 2x^2y z^2 \hat{a}_x - 2xy^2z^2 \hat{a}_y + 2xy^2 \hat{a}_z \end{aligned}\]

At point \((1, 2, 1)\):

Equation

\[\begin{aligned} \vec{J}|_{1,2,1} &= 2(1)^2(2)(1)^2 \hat{a}_x - 2(1)(2)^2(1)^2 \hat{a}_y + 2(1)(2)^2 \hat{a}_z \\ &= 4\hat{a}_x - 8\hat{a}_y + 8\hat{a}_z \end{aligned}\]

The magnitude is:

Equation

\[|\vec{J}| = \sqrt{(4)^2 + (-8)^2 + (8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12\]

Final Answer

Correct answer: B.

Question 03

Question 3

Let \(R\) be a region in the first quadrant of the \(xy\) plane enclosed by a closed curve \(C\) considered in counter-clockwise direction. Which of the following expressions does not represent the area of the region \(R\)?

GATE 2022 Electromagnetic Fields Q3 figure

\(\iint_R dx dy\)
\(\oint_C x dy\)
\(\oint_C y dx\)
\(\frac{1}{2} \oint_C (x dy - y dx)\)

Solution

We evaluate the area using Green's Theorem: \(\oint P dx + Q dy = \iint (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA\). The area is \(\iint dA\).

\(\iint_R dx dy\): This is the standard definition of the Area of region \(R\).
\(\oint_C x dy\): Here \(P=0, Q=x\).
Equation
\[\iint_R \left(\frac{\partial x}{\partial x} - \frac{\partial 0}{\partial y}\right) dA = \iint_R (1) dA = \text{Area}\]
\(\oint_C y dx\): Here \(P=y, Q=0\).
Equation
\[\iint_R \left(\frac{\partial 0}{\partial x} - \frac{\partial y}{\partial y}\right) dA = \iint_R (-1) dA = -\text{Area}\]
This represents the negative area.
\(\frac{1}{2} \oint_C (x dy - y dx)\):
Equation
\[\frac{1}{2} \iint_R [1 - (-1)] dA = \frac{1}{2} \iint_R 2 dA = \text{Area}\]

Final Answer

Correct answer: C.

Question 04

Question 4

As shown in the figure below, two concentric conducting spherical shells, centred 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 \(r\). The expression for \(V(r)\) in the region between \(r = c\) and \(r = d\) is:

GATE 2022 Electromagnetic Fields Q4 figure

\(V(r) = \frac{cd(V_2 - V_1)}{(d-c)r} - \frac{V_1 c + V_2 d - 2V_1 d}{d-c}\)
\(V(r) = \frac{cd(V_1 - V_2)}{(d-c)r} + \frac{V_2 d - V_1 c}{d-c}\)
\(V(r) = \frac{cd(V_1 - V_2)}{(d-c)r} - \frac{V_1 c - V_2 c}{d-c}\)
\(V(r) = \frac{cd(V_2 - V_1)}{(d-c)r} - \frac{V_2 c - V_1 c}{d-c}\)

Solution

From Laplace's equation, \(\nabla^2 V = 0\). For a spherical coordinate system where \(V\) depends only on \(r\):

Equation

\[\begin{aligned} \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right) &= 0 \\ r^2 \frac{dV}{dr} &= A \quad (\text{Constant}) \\ \frac{dV}{dr} &= \frac{A}{r^2} \\ V(r) &= \frac{-A}{r} + B \end{aligned}\]

Applying Boundary Conditions:

At \(r = c, V = V_1 \implies V_1 = \frac{-A}{c} + B\) (Eq. 1)
At \(r = d, V = V_2 \implies V_2 = \frac{-A}{d} + B\) (Eq. 2)

Subtracting (Eq. 2) from (Eq. 1):

Equation

\[\begin{aligned} V_1 - V_2 &= -A\left(\frac{1}{c} - \frac{1}{d}\right) = -A\left(\frac{d-c}{cd}\right) \\ A &= - \frac{cd(V_1 - V_2)}{d-c} = \frac{cd(V_1 - V_2)}{c-d} \end{aligned}\]

To match the option format, we use \(A = \frac{-cd(V_1 - V_2)}{d-c}\).

Substituting \(A\) back into Eq. 1 to find \(B\):

Equation

\[\begin{aligned} B &= V_1 + \frac{A}{c} \\ &= V_1 - \frac{d(V_1 - V_2)}{d-c} \\ &= \frac{V_1(d-c) - V_1 d + V_2 d}{d-c} \\ &= \frac{V_1 d - V_1 c - V_1 d + V_2 d}{d-c} \\ &= \frac{V_2 d - V_1 c}{d-c} \end{aligned}\]

Therefore, the potential is:

Equation

\[V(r) = \frac{cd(V_1 - V_2)}{(d-c)r} + \frac{V_2 d - V_1 c}{d-c}\]

Final Answer

Correct answer: B.