Basics - Grad Plus
Solutions for previous GATE-EC questions of Electromagnetics (Basics)
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Solutions for previous GATE-EC questions of Electromagnetics (Uniform Plane Waves)
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Basics

Q.  [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 is related to another vector field through ,Which of the following is true? Note: C and SC 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 is the position vector of a point inside s, with the unit normal on S,the value of the integral [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 where and K is a constant. The value of n for which 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 is given by 

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 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}

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