Solved GATE Paper

GATE 2019 Electromagnetic Fields Questions and Solutions

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

Question 1

A rectangular waveguide of width \(w\) and height \(h\) has cut-off frequencies for \(\text{TE}_{10}\) and \(\text{TE}_{11}\) modes in the ratio \(1 : 2\). The aspect ratio \(w/h\), rounded off to two decimal places, is ________ .

Solution

The general formula for cut-off frequency is:

Equation
\[f_{cmn} = \frac{c}{2} \sqrt{ \left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 }\]

(Note: In this context, width \(a=w\) and height \(b=h\)).

For \(\text{TE}_{10}\) mode (\(m=1, n=0\)):

Equation
\[f_{c10} = \frac{c}{2w} \tag{1}\]

And for \(\text{TE}_{11}\) mode (\(m=1, n=1\)):

Equation
\[f_{c11} = \frac{c}{2} \sqrt{ \left(\frac{1}{w}\right)^2 + \left(\frac{1}{h}\right)^2 } = \frac{c}{2w} \sqrt{ 1 + \left(\frac{w}{h}\right)^2 } \tag{2}\]

Given,

Equation
\[\frac{f_{c10}}{f_{c11}} = \frac{1}{2} \tag{3}\]

Substituting (1) and (2) into (3):

Equation
\[\begin{aligned} \frac{\frac{c}{2w}}{\frac{c}{2w}\sqrt{1 + \left(\frac{w}{h}\right)^2}} &= \frac{1}{2} \\[1em] \frac{1}{\sqrt{1 + \left(\frac{w}{h}\right)^2}} &= \frac{1}{2} \\[1em] \sqrt{1 + \left(\frac{w}{h}\right)^2} &= 2 \end{aligned}\]

Squaring both sides:

Equation
\[\begin{aligned} 1 + \left(\frac{w}{h}\right)^2 &= 4 \\ \left(\frac{w}{h}\right)^2 &= 3 \\ \frac{w}{h} &= \sqrt{3} \approx 1.732 \end{aligned}\]
✓
Final Answer
Answer: 1.732 (1.71 to 1.75).
Question 02

Question 2

Two identical copper wires \(W_1\) and \(W_2\) 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 \(W_1\) is:

GATE 2019 Electromagnetic Fields Q2 figure
  1. \( \frac{\mu_0 I^2}{2\pi r^2}\)
  2. \( \frac{6\mu_0 I}{5\pi r}\)
  3. \( \frac{\mu_0 I}{6\pi r}\)
  4. \( \frac{5\mu_0 I}{6\pi r}\)

Solution

GATE 2019 Electromagnetic Fields Q2 figure

The magnetic flux density (\(\vec{B}\)) at a distance \(\rho\) due to an infinite line carrying current \(I\) is given by:

Equation
\[|\vec{B}| = \frac{\mu_0 I}{2\pi \rho}\]

1. Field due to wire \(W_1\): At distance \(r\) from \(W_1\):

Equation
\[|\vec{B}_1| = \frac{\mu_0 I}{2\pi r} \tag{i}\]

2. Field due to wire \(W_2\): The point of interest is at a distance \(r\) from \(W_1\). Since the total separation is \(4r\), the distance from \(W_2\) is \(4r - r = 3r\).

Equation
\[|\vec{B}_2| = \frac{\mu_0 (2I)}{2\pi (3r)} \tag{ii}\]

Total Field: Using the right-hand thumb rule, since the currents are in opposite directions, the magnetic fields due to both wires add up in the region between the conductors.

Equation
\[\begin{aligned} |\vec{B}| &= |\vec{B}_1| + |\vec{B}_2| \\[1em] &= \frac{\mu_0 I}{2\pi r} + \frac{2\mu_0 I}{6\pi r} \\[1em] &= \frac{\mu_0 I}{2\pi r} \left( 1 + \frac{2}{3} \right) \\[1em] &= \frac{\mu_0 I}{2\pi r} \left( \frac{5}{3} \right) \\[1em] &= \frac{5\mu_0 I}{6\pi r} \end{aligned}\]
D
Final Answer
Correct answer: D.
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GATE Electromagnetic Fields