Solved GATE Paper

GATE 2023 Electromagnetic Fields Questions and Solutions

Instructor: Prof. Mithun Mondal Institution: BITS Pilani Subject: Electromagnetic Fields
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Question 01

Question 1

In the figure, the electric field \(E\) and the magnetic field \(B\) point to \(x\) and \(z\) 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?

GATE 2023 Electromagnetic Fields Q1 figure
  1. The charge will move in the direction of \(\mathbf{z}\) with constant velocity.
  2. The charge will always move on the \(\mathbf{y}\)-\(\mathbf{z}\) plane only.
  3. The trajectory of the charge will be a circle.
  4. The charge will progress in the direction of \(\mathbf{y}\).
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Final Answer
Marks to all candidates (per official GATE key).
Question 02

Question 2

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

GATE 2023 Electromagnetic Fields Q2 figure

Solution

The magnetic field due to an infinite sheet current is given by:

Equation
\[\bar{H} = \frac{1}{2} (\bar{K} \times \hat{a}_n) \text{ A/m}\]

Given:

Equation
\[\begin{aligned} \bar{K} &= 5\hat{a}_x \\ \hat{a}_n &= \hat{a}_z \quad (\text{Unit normal vector}) \end{aligned}\]

Calculating the cross product:

Equation
\[\begin{aligned} \bar{H} &= \frac{1}{2} (5\hat{a}_x \times \hat{a}_z) \\ &= 2.5(-\hat{a}_y) \\ &= -2.5\hat{a}_y \text{ A/m} \end{aligned}\]

The magnitude is:

Equation
\[|\bar{H}| = 2.5 \text{ A/m}\]
GATE 2023 Electromagnetic Fields Q2 figure
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Final Answer
Answer: 2.5.
Question 03

Question 3

The closed curve shown in the figure is described by \(r = 1 + \cos\theta\), where \(r = \sqrt{x^2 + y^2}\); \(x = r\cos\theta, y = r\sin\theta\). The magnitude of the line integral of the vector field \(F = -y\hat{i} + x\hat{j}\) around the closed curve is ________ (Round off to 2 decimal places).

GATE 2023 Electromagnetic Fields Q3 figure

Solution

Curve \(C\) is given by \(r = (1 + \cos\theta)\). We need to find:

Equation
\[\oint_C \bar{F} \cdot d\bar{r} = \oint_C (-y dx + x dy)\]

Using Green's Theorem: \(\oint_C (L dx + M dy) = \iint_R \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) dx dy\). Here, \(L = -y\) and \(M = x\).

Equation
\[\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1 - (-1) = 2\]

Thus, the integral becomes:

Equation
\[\begin{aligned} I &= \iint_R 2 \, dx \, dy = 2 \iint_R dA \\ &= 2 \times [\text{Area of region bounded by curve } C] \end{aligned}\]

calculating the area in polar coordinates:

Equation
\[\begin{aligned} \text{Area} &= \frac{1}{2} \int_{0}^{2\pi} r^2 d\theta \\ \implies I &= 2 \times \frac{1}{2} \int_{0}^{2\pi} (1 + \cos\theta)^2 d\theta \\ &= \int_{0}^{2\pi} (1 + \cos^2\theta + 2\cos\theta) d\theta \\ &= \int_{0}^{2\pi} d\theta + \int_{0}^{2\pi} \cos^2\theta \, d\theta + 2 \int_{0}^{2\pi} \cos\theta \, d\theta \\ &= [\theta]_0^{2\pi} + \int_{0}^{2\pi} \left(\frac{1 + \cos 2\theta}{2}\right) d\theta + 0 \\ &= 2\pi + \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_0^{2\pi} \\ &= 2\pi + \pi + 0 \\ &= 3\pi \end{aligned}\]

Numerical value:

Equation
\[3 \times 3.14159 \approx 9.42\]
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Final Answer
Answer: 9.42.
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GATE Electromagnetic Fields