Solved GATE Paper

GATE 2017 Electromagnetic Fields Questions and Solutions

Instructor: Prof. Mithun Mondal Institution: BITS Pilani Subject: Electromagnetic Fields
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Question 21

Question 21

The figures show diagrammatic representations of vector fields \(\vec{X}\), \(\vec{Y}\), and \(\vec{Z}\), respectively. Which one of the following choices is true?

GATE 2017 Electromagnetic Fields Q1 figure
  1. \(\nabla \cdot \vec{X} = 0, \nabla \times \vec{Y} \neq 0, \nabla \times \vec{Z} = 0\)
  2. \(\nabla \cdot \vec{X} \neq 0, \nabla \times \vec{Y} = 0, \nabla \times \vec{Z} \neq 0\)
  3. \(\nabla \cdot \vec{X} \neq 0, \nabla \times \vec{Y} \neq 0, \nabla \times \vec{Z} \neq 0\)
  4. \(\nabla \cdot \vec{X} = 0, \nabla \times \vec{Y} = 0, \nabla \times \vec{Z} = 0\)

Solution

GATE 2017 Electromagnetic Fields Q1 figure
Vector \(\vec{X}\)Vector \(\vec{Y}\)Vector \(\vec{Z}\)
\(\nabla \cdot \vec{X} \neq 0\)\(\nabla \cdot \vec{Y} = 0\)\(\nabla \cdot \vec{Z} \neq 0\)
\(\nabla \times \vec{X} \neq 0\)\(\nabla \times \vec{Y} \neq 0\)\(\nabla \times \vec{Z} \neq 0\)
C
Final Answer
Correct answer: C.
Question 45

Question 45

The magnitude of magnetic flux density (\(B\)) in micro Teslas (\(\mu\)T), at the center of a loop of wire wound as a regular hexagon of side length 1 m carrying a current (\(I = 1\)A) and placed in vacuum as shown in the figure is ________ .

GATE 2017 Electromagnetic Fields Q2 figure

Solution

GATE 2017 Electromagnetic Fields Q2 figure

The total magnetic field intensity \(H\) is the sum of the fields from the 6 sides:

Equation
\[H = H_1 + H_2 + H_3 + H_4 + H_5 + H_6\]

Since the figure is a regular hexagon, the field magnitude from each side is the same and points in the same direction.

Equation
\[\therefore H = 6H_1\]
Geometry of one sector
Geometry of one sector

1. Geometry calculations:

Equation
\[\begin{aligned} \tan 60^\circ &= \frac{d}{1/2} \\ \sqrt{3} &= 2d \\ d &= \frac{\sqrt{3}}{2} \text{ m} \end{aligned}\]

2. Magnetic Field Calculation: Using the formula for the magnetic field due to a finite wire:

Equation
\[H_1 = \frac{I}{4\pi d}(\sin \alpha_1 + \sin \alpha_2)\]

Here, \(\alpha_1 = \alpha_2 = 30^\circ\) (from the equilateral triangle geometry).

Equation
\[\begin{aligned} H_1 &= \frac{1}{4\pi \left(\frac{\sqrt{3}}{2}\right)} \left(\sin 30^\circ + \sin 30^\circ\right) \\ &= \frac{1}{2\sqrt{3}\pi} \left(\frac{1}{2} + \frac{1}{2}\right) \\ &= \frac{1}{2\sqrt{3}\pi} \end{aligned}\]

3. Total Magnetic Field Intensity (\(H\)):

Equation
\[H = 6H_1 = 6 \times \frac{1}{2\sqrt{3}\pi} = \frac{3}{\sqrt{3}\pi} = \frac{\sqrt{3}}{\pi} \text{ A/m}\]

4. Magnetic Flux Density (\(B\)):

Equation
\[\begin{aligned} B &= \mu_0 H \\ &= (4\pi \times 10^{-7}) \times \frac{\sqrt{3}}{\pi} \\ &= 4\sqrt{3} \times 10^{-7} \text{ T} \\ &= 4(1.732) \times 10^{-7} \text{ T} \\ &\approx 6.928 \times 10^{-7} \text{ T} \\ &= 0.6928 \, \mu\text{T} \end{aligned}\]

Result:

Equation
\[B \approx 0.693 \, \mu\text{T}\]
✓
Final Answer
Answer: 0.693.
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GATE Electromagnetic Fields