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

Basics [Preview]

Q. $\oint\limits_C\overrightarrow A\cdot\overrightarrow{dl}=\int\limits_S\_.\overrightarrow{ds}$ [1994 1M]

Ans:- $\nabla\times\overrightarrow A$

Exp- Using Stoke’s theorem

$\oint\limits_C\overrightarrow A\cdot\overrightarrow{dl}=\int\limits_S\nabla\times\overrightarrow A\cdot\overrightarrow{ds}$

Q. If a vector field $\overrightarrow V$ is related to another vector field $\overrightarrow A$ through $\overrightarrow V=\nabla\times\overrightarrow A$ ,Which of the following is true? Note: C and S_C refer to any closed counter and any surface whose boundary is C. [2009 2M]

a) $\oint_C\overrightarrow V\cdot\overrightarrow{dl}=\int\limits_{S_C}\int\overrightarrow A\cdot\overrightarrow{ds}$

b) $\oint_C\overrightarrow A\cdot\overrightarrow{dl}=\int\limits_{S_C}\int\overrightarrow V\cdot\overrightarrow{ds}$

c) $\oint_C\nabla\times\overrightarrow V\overrightarrow{dl}=\int\limits_{S_C}\int\nabla\times\overrightarrow A\cdot\overrightarrow{ds}$

d) $\oint_C\nabla\times\overrightarrow A\cdot\overrightarrow{dl}=\int\limits_{S_C}\int\overrightarrow V\cdot\overrightarrow{ds}$

Ans-(b)

Exp- Stoke’s theorem

$\oint_C\overline A\cdot\overline{dl}=\iint_S\left(\nabla\times\overrightarrow A\right)\cdot\overline{ds}=\iint_S\overline V\cdot\overline{ds}$

Q. Consider a closed surface S surrounding a volume V. If $\overrightarrow r$ is the position vector of a point inside s, with $\widehat n$ the unit normal on S,the value of the integral $\oint\limits_S5\overrightarrow r\cdot\widehat nds$ [2011 1M]

a) 3 V

b) 5 V

c) 10 V

d) 15 V

Ans-(d)

Exp- For close surface, we can use Divergence Theorem.

$\oint_s\overrightarrow A\cdot\overrightarrow{ds}=\int_v\left(\nabla\cdot\overrightarrow A\right)dv$

$\oint\limits_S5\overrightarrow r\cdot\widehat nds=\underset V{\int\int\int}\nabla\cdot5\overrightarrow rdv$

$\begin{array}{l}\nabla\cdot r=\frac1{r^2}\frac\partial{\partial r}\left(r^2r\right)\\\\\;\;\;\;\;\;=\frac3{r^2}\cdot r^2=3\end{array}$

$5\underset V{\int\int\int}\nabla\cdot\overrightarrow rdv=5\times3V=15V$

Q. The direction of vector A is radially outward from the origin,with $\left|A\right|=Kr^n$ where $r^2=x^2+y^2+z^2$ and K is a constant. The value of n for which $\nabla\cdot A=0$ is [2012 2M]

a) -2

b) 2

c) 1

d) 0

Ans- (a)

Exp- $\left|A\right|=Kr^n$

$\overrightarrow A=Kr^n{\widehat{\;a}}_r\left(\sin ce\;it\;is\;radially\;outward\right)$

$\nabla\cdot\overrightarrow A\;in\;spherical\;coordinate\;is$

$\nabla\cdot\overrightarrow A=\frac1{r^2}\frac\partial{\partial r}\left(r^2A_r\right)+\frac1{r\;\sin\theta}\frac\partial{\partial r}\left(A_\theta\sin_\theta\right)+\frac1{r\;\sin\theta}\frac\partial{\partial\phi}A_\phi$

$\nabla\cdot\overrightarrow A=\frac1{r^2}\frac\partial{\partial r}\left(r^2Kr^2\right)+0+0$

$\nabla\cdot\overrightarrow A=\frac K{r^2}\frac\partial{\partial r}\left(r^{n+2}\right)$

For $\nabla\cdot\overline A$ to be zero, derivative must be zero in other words (rⁿ⁺²) must be constant.

This could be possible only if n = -2.

Q. A vector $\overrightarrow P$ is given by $\overrightarrow P=x^3y{\overrightarrow a}_x-x^2y^2{\overrightarrow a}_y-x^2yz{\overrightarrow a}_z$

Which of the following statements is True? [2015 1M,set-2]

a) $\overrightarrow P$ is solenoidal,but not irrotational

b) $\overrightarrow P$ is irrotational,but not solensoidal

c) $\overrightarrow P$ is neither solenoidal nor irrotational

d) $\overrightarrow P$ is both solenoidal and irrotational

Ans-(a)

Exp- $\overrightarrow P=x^3y{\overrightarrow a}_x-x^2y^2{\overrightarrow a}_yx^2yz{\overrightarrow a}_z$

For solenoidal, $\nabla\cdot\overrightarrow P=0$

$\begin{array}{l}\Rightarrow\nabla\cdot\overrightarrow P=\frac{\partial P_x}{\partial x}+\frac{\partial P_y}{\partial y}+\frac{\partial P_z}\partial\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;=3x^2y-2x^2y-x^2y\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;=0\end{array}$

$\Rightarrow\overrightarrow P$ is solenoidal

For irrotational, $\nabla\times\overrightarrow P=0$

$\Rightarrow\nabla\times\overrightarrow P=\begin{vmatrix}{\overrightarrow a}_x&{\overrightarrow a}_y&{\overrightarrow a}_z\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\x^3y&-x^2y^2&-x^2yz\end{vmatrix}$

$\begin{array}{l}={\overrightarrow a}_x\left(-x^2z\right)+{\overrightarrow a}_y\left(2xyz\right)+{\overrightarrow a}_z\left(-2x^2-x^3\right)\\\\\neq0\\\\\Rightarrow\overrightarrow P\;is\;not\;irrotational\end{array}$

Q. If the vector function

$\overrightarrow F={\widehat a}_x\left(3y-K_1z\right)+{\widehat a}_y\left(k_2x-2z\right)-{\widehat a}_z\left(k_3y+z\right)$

is irrotational,then the values of the constants $K_1,K_2\;and\;K_3$ respectively,are[2017 2M Set-2]

a) $0.3,-2.5,0.5$

b) $0.0,3.0,2.0$

c) $0.3,0.33,0.5$

d) $4.0,3.0,2.0$

Ans-(b)

Exp- $\overrightarrow F={\widehat a}_x\left(3y-k_1z\right)+{\widehat a}_y\left(k_2x-2z\right)-{\widehat a}_z\left(k_3y+z\right)$

$\nabla\times\overrightarrow F=0\;\left(irrotational\right)$

$\nabla\times\overrightarrow F=\begin{vmatrix}{\widehat a}_x&{\widehat a}_y&{\widehat a}_z\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\3y-k_1z&k_2x-2z&-\left(k_3y+z\right)\end{vmatrix}$

$={\widehat a}_x\left[\frac\partial{\partial y}\left[-\left(k_3y+z\right)\right]-\frac\partial{\partial z}\left(k_2x-2z\right)\right]$

$-{\widehat a}_y\left[\frac\partial{\partial x}\left[-\left(k_3y+z\right)\right]-\frac\partial{\partial z}\left(3y-k_1z\right)\right]$

$+{\widehat a}_z\left[\frac\partial{\partial x}\left(k_2x-2z\right)-\frac\partial{\partial y}\left(3y-k_1z\right)\right]$

${\widehat a}_x\left[-k_3+2\right]-{\widehat a}_y\left[k_1\right]+{\widehat a}_z\left[k_2-3\right]=0$

$\begin{array}{l}\Rightarrow k_3=2,\;k_1=0,\;k_2=3\\\\or\;\;k_1=0,\;k_2=3,k_3=2\end{array}$

Next Electrostatics: Set-1 [Preview]

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

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

Basics 0

Electrostatics 2

Magnetic Fields 1

Uniform Plane Waves 5

Transmission Lines 6

Waveguides 4

Antennas 4

Basics [Preview]

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