Solved GATE Paper

GATE 2020 Electromagnetic Fields Questions and Solutions

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

Question 1

Let \(\hat{a}_r\), \(\hat{a}_\phi\) and \(\hat{a}_z\) be unit vectors along \(r\), \(\phi\) and \(z\) directions, respectively in the cylindrical coordinate system. For the electric flux density given by \(\vec{D} = (\hat{a}_r 15 + \hat{a}_\phi 2r - \hat{a}_z 3rz)\) Coulomb/m\(^2\), the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the \(z\)-axis with its base at the origin is:

  1. \(108 \pi\)
  2. \(54 \pi\)
  3. \(90 \pi\)
  4. \(180 \pi\)

Solution

The flux \(\psi\) crossing a closed surface is given by the Divergence Theorem:

Equation
\[\psi = \oiint \vec{D} \cdot d\vec{s} = \iiint (\nabla \cdot \vec{D}) dv\]

Step 1: Calculate Divergence In cylindrical coordinates:

Equation
\[\nabla \cdot \vec{D} = \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho D_\rho) + \frac{1}{\rho} \frac{\partial D_\phi}{\partial \phi} + \frac{\partial D_z}{\partial z}\]

Substituting the given components (\(D_\rho = 15, D_\phi = 2\rho, D_z = -3\rho z\)):

Equation
\[\begin{aligned} \nabla \cdot \vec{D} &= \frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho \cdot 15) + \frac{1}{\rho} \frac{\partial}{\partial \phi} (2\rho) + \frac{\partial}{\partial z} (-3\rho z) \\ &= \frac{1}{\rho} (15) + 0 - 3\rho \\ &= \frac{15}{\rho} - 3\rho \end{aligned}\]

Step 2: Integrate over Volume The differential volume element is \(dv = \rho d\rho d\phi dz\).

Equation
\[\begin{aligned} \psi &= \int_{z=0}^{5} \int_{\phi=0}^{2\pi} \int_{\rho=0}^{3} \left( \frac{15}{\rho} - 3\rho \right) \rho \, d\rho \, d\phi \, dz \\ &= \int_{z=0}^{5} dz \int_{\phi=0}^{2\pi} d\phi \int_{\rho=0}^{3} (15 - 3\rho^2) \, d\rho \\ &= [z]_0^5 \cdot [\phi]_0^{2\pi} \cdot \left[ 15\rho - \frac{3\rho^3}{3} \right]_0^3 \\ &= (5) \cdot (2\pi) \cdot \left[ 15(3) - (3)^3 \right] \\ &= 10\pi \cdot [45 - 27] \\ &= 10\pi \cdot 18 \\ &= 180\pi \text{ C} \end{aligned}\]
D
Final Answer
Correct answer: D (\(180\pi\)).
Question 02

Question 2

A conducting square loop of side length 1 m is placed at a distance of 1 m from a long straight wire carrying a current \(I = 2\) A as shown below. The mutual inductance, in nH (rounded off to 2 decimal places), between the conducting loop and the long wire is ________ .

GATE 2020 Electromagnetic Fields Q2 figure

Solution

Mutual inductance is defined as \(M = \frac{\phi}{I}\).

The magnetic field due to an infinitely long wire is:

Equation
\[\vec{B} = \frac{\mu_0 I}{2\pi \rho} \hat{a}_\phi\]

The magnetic flux \(\phi\) crossing the square loop is:

Equation
\[\phi = \iint \vec{B} \cdot d\vec{s}\]

Here, \(d\vec{s} = d\rho dz \, \hat{a}_\phi\). The limits for \(\rho\) are from 1 m to \((1+1)=2\) m. The limits for \(z\) are from 0 to 1 m.

Equation
\[\begin{aligned} \phi &= \int_{z=0}^{1} \int_{\rho=1}^{2} \frac{\mu_0 I}{2\pi \rho} \, d\rho \, dz \\ &= \frac{\mu_0 I}{2\pi} [z]_0^1 [\ln \rho]_1^2 \\ &= \frac{\mu_0 I}{2\pi} (1) (\ln 2 - \ln 1) \\ &= \frac{\mu_0 I}{2\pi} \ln 2 \end{aligned}\]

Calculating Mutual Inductance \(M\):

Equation
\[\begin{aligned} M &= \frac{\phi}{I} = \frac{\mu_0 \ln 2}{2\pi} \\ &= \frac{4\pi \times 10^{-7} \times 0.6931}{2\pi} \\ &= 2 \times 10^{-7} \times 0.6931 \\ &= 1.3862 \times 10^{-7} \text{ H} \\ &\approx 138.62 \text{ nH} \end{aligned}\]
✓
Final Answer
Answer: 138.63.
Question 03

Question 3

The static electric field inside a dielectric medium with relative permittivity \(\epsilon_r = 2.25\), expressed in cylindrical coordinate system is given by the following expression:

Equation
\[\vec{E} = \hat{a}_r 2r + \hat{a}_\phi \left( \frac{3}{r} \right) + \hat{a}_z 6\]

where \(\hat{a}_r, \hat{a}_\phi, \hat{a}_z\) are unit vectors along \(r\), \(\phi\) and \(z\) directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, \(\epsilon_0\), in SI units is given by:

  1. \(4 \epsilon_0\)
  2. \(5 \epsilon_0\)
  3. \(3 \epsilon_0\)
  4. \(9 \epsilon_0\)

Solution

First, determine the Electric Flux Density \(\vec{D}\):

Equation
\[\vec{D} = \epsilon \vec{E} = \epsilon_0 \epsilon_r \vec{E}\]

Given \(\epsilon_r = 2.25\):

Equation
\[\begin{aligned} \vec{D} &= \epsilon_0 (2.25) \left( 2r \hat{a}_r + \frac{3}{r} \hat{a}_\phi + 6 \hat{a}_z \right) \\ &= 4.5 \epsilon_0 r \hat{a}_r + \frac{6.75 \epsilon_0}{r} \hat{a}_\phi + 13.5 \epsilon_0 \hat{a}_z \end{aligned}\]

The volume charge density \(\rho_v\) is given by Maxwell's equation \(\rho_v = \nabla \cdot \vec{D}\). In cylindrical coordinates:

Equation
\[\nabla \cdot \vec{D} = \frac{1}{r} \frac{\partial}{\partial r} (r D_r) + \frac{1}{r} \frac{\partial D_\phi}{\partial \phi} + \frac{\partial D_z}{\partial z}\]

Substituting the components:

Equation
\[\begin{aligned} \rho_v &= \frac{1}{r} \frac{\partial}{\partial r} \left( r \cdot 4.5 \epsilon_0 r \right) + \frac{1}{r} \frac{\partial}{\partial \phi} \left( \frac{6.75 \epsilon_0}{r} \right) + \frac{\partial}{\partial z} (13.5 \epsilon_0) \\ &= \frac{1}{r} \frac{\partial}{\partial r} (4.5 \epsilon_0 r^2) + 0 + 0 \\ &= \frac{1}{r} (9 \epsilon_0 r) \\ &= 9 \epsilon_0 \end{aligned}\]
D
Final Answer
Correct answer: D.
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