Past GATE Solved Papers Electromagnetics | Electronics & Comm. Engg.

CoursesGATEElectronics & Comm. Engineering (EC) | Topic-wise Previous Solved GATE Papers | Electromagnetics

Basics 0

6 Questions with standard solutions
- Lecture1.1
  
  Basics [Preview]
Electrostatics 2

28 Questions with standard solutions
- Lecture2.1
  
  Electrostatics: Set-1 [Preview]
- Lecture2.2
  
  Electrostatics: Set-2
- Lecture2.3
  
  Electrostatics: Set-3
Magnetic Fields 1

9 Questions with standard solutions
- Lecture3.1
  
  Magnetic Fields
Uniform Plane Waves 5

65 Questions with standard solutions
- Lecture4.1
  
  Uniform Plane Waves: Set-1 [Preview]
- Lecture4.2
  
  Uniform Plane Waves: Set-2
- Lecture4.3
  
  Uniform Plane Waves: Set-3
- Lecture4.4
  
  Uniform Plane Waves: Set-4
- Lecture4.5
  
  Uniform Plane Waves: Set-5
- Lecture4.6
  
  Uniform Plane Waves: Set-6
Transmission Lines 6

56 Questions with standard solutions
- Lecture5.1
  
  Transmission Lines: Set-1
- Lecture5.2
  
  Transmission Lines: Set-2
- Lecture5.3
  
  Transmission Lines: Set-3
- Lecture5.4
  
  Transmission Lines: Set-4
- Lecture5.5
  
  Transmission Lines: Set-5
- Lecture5.6
  
  Transmission Lines: Set-6
Waveguides 4

35 Questions with standard solutions
- Lecture6.1
  
  Waveguides: Set-1
- Lecture6.2
  
  Waveguides: Set-2
- Lecture6.3
  
  Waveguides: Set-3
- Lecture6.4
  
  Waveguides: Set-4
Antennas 4

42 Questions with standard solutions
- Lecture7.1
  
  Antennas: Set-1
- Lecture7.2
  
  Antennas: Set-2
- Lecture7.3
  
  Antennas: Set-3
- Lecture7.4
  
  Antennas: Set-4

Electrostatics: Set-1 [Preview]

Q. An electrostatic field is said to be conservative when: [1987 2M]

a) The divergence of the field is equal to zero

b) The curl of the field is equal to zero

c) The curl of the field is equal to $-\frac{\partial E}{\partial t^2}$

d) The Laplacian of the field is equal to $\mu\varepsilon\frac{\partial^2E}{\partial t^2}$

Ans- (b)

Exp- An electrostatic field is said to be conservative when the closed line integral of the field is zero.

$\oint\overrightarrow E\cdot\overrightarrow{dl}=0$

According to the Stoke’s theorem

$\oint\limits_C\overrightarrow E\cdot\overrightarrow{dl}=\int\limits_s\left(\nabla\times\overrightarrow E\right)\cdot\overrightarrow{ds}$

$\nabla\times\overrightarrow E=0$

Q. On either side of a charge-free interface between two media [1988 2M]

a) the normal components of the electric field are equal

b) the tangential component of the electric field are equal

c) the normal components of the electric flux density are equal

d) the tangential components of the electric flux density are equal

Ans- (b) and (c)

Expl:- According to Boundary conditions: Tangential components of electric field and normal components of flux density are continuous.

$\begin{array}{l}E_{t_1}=E_{t_2}\\\\D_{N_1}=D_{N_2}\end{array}$

Q. The electric field strength at a far-off point P due to a point charge,+q located at the origin, O is 100 millivolts/meter. The point charge is now enclosed by a perfectly conducting hollow metal sphere with its centre at the origin,O. The electric field strength at the point, P [1989 2M]

a) remains unchanged in its magnitude and direction

b) remains unchanged in its magnitude but reverse in direction

c) would be that due to a dipole formed by the charge,+q, at O and -q induced

d) would be zero

Ans-(a)

The charge will be reflected on the metal sphere. Now if we calculate the electric field because of the metal sphere then the same formula is used with distance from the center.

GATE EC

Q. Which of the following field equation indicate that the free magnetic charge do not exist [1990 2M]

a) $\overrightarrow H=\frac1\mu\nabla\times\overrightarrow A$

b) $\overrightarrow H=\oint\frac{I\overrightarrow{dl}\times\overrightarrow R}{4\pi R^2}$

c) $\nabla\cdot\overrightarrow H=0$

d) $\nabla\times\overrightarrow H=\overrightarrow J$

Ans-(c)

Exp- Free magnetic charge i.e monopole can not exist; always we have magnetic diapole. According to the Gauss’s law for magnetic fields.

$\oint\limits_s\overrightarrow B\cdot\overrightarrow{ds}=0$

$\begin{array}{l}\nabla\cdot\overrightarrow B=0\\\\\nabla\cdot\mu\overrightarrow H=0\\\\\mu\nabla\cdot\overrightarrow H=0\\\\\nabla\cdot\overrightarrow H=0\end{array}$

Q. Given, $\mathbf{\overrightarrow v=x\cos^2y\widehat i+x^2e^z\widehat j+z\sin^2y\widehat k}$ and S the surface of a unit cube with one corner at the origin and edges parallel to the coordinate axis, the value of the integral is $\iint_C\overrightarrow v\cdot\widehat ndS\;$ …..[1993 2M]

Ans- 1

Expl:- According to Divergence Theorem for any close surfece, Now the given surface is surface of a cube which is the close surface. So we have,

$\oint\overrightarrow V\cdot\overrightarrow{ds}=\int\left(\nabla\cdot\overrightarrow V\right)\;dv$

$\nabla\cdot\overrightarrow V=\cos^2y+\sin^2y=1$

$\int_v\left(1\right)dv=\int_vdv=Volume\;of\;Cube=1$

Q. For the uniformly charged sphere of radius R and charge density, $\rho$ the ratio of a magnitude of electric field at distances R/2 and 2R from the centre i.e. is…..[1993 2M]

Ans: 2

Expl:- We know that the electric field inside the sphere is given by

$E=\frac{\rho_vr}{3\varepsilon}$

$E(at\;r=R/2)=\frac{\rho_vR}{6\varepsilon}$

Electric Field outside the sphere is given by,

$E=\frac1{4\pi\varepsilon}\times\frac Q{r^2}=\frac{\rho_vR^3}{3\times\varepsilon\times r^2}............as\;Q=\rho_v\frac43\pi R^3$

$E(at\;r=2R)=\frac{\rho_vR^3}{3\times\varepsilon\times4R^2}=\frac{\rho_vR}{12\varepsilon}$

Taking ratio we can get the answer.

Q. Match List-I with List-II and select the correct answer using the code given below the Lists: [1994 2M]

List-I	List-II
A. $\nabla\times\overrightarrow H=\overrightarrow J$	1. Continuity equation
B. $\oint\limits_C\overrightarrow E\cdot\overrightarrow{dl}=-\oint\limits_s\frac{\partial B}{\partial t}\cdot\overrightarrow{ds}$	2. Faraday’s law
C. $\nabla\cdot\overrightarrow J=-\frac{\partial\rho}{\partial t}$	3. Ampere’s Law
	4. Gauss’s Law
	5. Biot-Savart Law

Codes: A B C
(a) 3 2 1

(b) 2 1 3

(d) 1 2 3

Ans-(a)

Exp-

$\nabla\times\overrightarrow H=\overrightarrow J\rightarrow\;Ampere's\;law$

$\oint\limits_c\overrightarrow E\cdot\overrightarrow{dl}=-\oint\limits_s\frac{\partial B}{\partial t}\cdot\overrightarrow{ds}\rightarrow Faraday's\;law$

$\nabla\cdot\overrightarrow J=-\frac{\partial\rho}{\partial t}\rightarrow Continutiy\;equation$

Q. The electric field strength at distant point,P, due to a point charge,+q, located at the origin, is $100\mu V/m$ If the point charge is now enclosed by a perfectly conducting metal sheet sphere whose centre is at the origin, then the electric field strength at the point, P, outside the sphere, becomes [1995 1M]

a) zero

b) $100\mu V/m$

c) $-100\;\mu V/m$

d) $50\;\mu V/m$

Ans-(b)

Exp- The charge would appear on the surface of the sphere. And for finding the electric field at the point we need the formula for electric field due to charged sphere as the point is outside of sphere,

$E=\frac1{4\pi\varepsilon}\times\frac Q{r^2}$

where r is the distance from the center of the sphere.

EC-GATE Solutions

As r and Q are same in both cases, we can say both cases give equal electric field.

Prev Basics [Preview]

Next Electrostatics: Set-2

GATE

Basics 0

Electrostatics 2

Magnetic Fields 1

Uniform Plane Waves 5

Transmission Lines 6

Waveguides 4

Antennas 4

Electrostatics: Set-1 [Preview]

Leave A Reply Cancel reply

Login with your site account

Register a new account