Q2GATE 2025MSQ

Consider two coupled circuits, having selfinductances $L_{1}$ and $L_{2}$ , that carry non-zero currents $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ , respectively. The mutual inductance between the circuits is M with unity coupling coefficient. The stored magnetic energy of the coupled circuits is minimum at which of the following value(s) of $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}$ ?

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📖 Explanation

$\mathrm{E}_{\mathrm{s}}=\frac{1}{2} \mathrm{~L}_{1} \mathrm{I}_{1}^{2}+\frac{1}{2} \mathrm{~L}_{2} \mathrm{I}_{2}^{2}+\mathrm{MI}_{1} \mathrm{I}_{2}$ [Based on given options it has to be mutally adding only]

\begin{aligned} \frac{\partial \mathrm{E}_{\mathrm{s}}}{\partial \mathrm{I}_{1}}=0 & \Rightarrow \mathrm{~L}_{1} \mathrm{I}_{1}+\mathrm{MI}_{2}=0 \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=-\frac{\mathrm{M}}{\mathrm{~L}_{1}} \\ & \Rightarrow \mathrm{~L}_{1} \mathrm{I}_{1}=-\mathrm{MI}_{2} \\ \frac{\partial \mathrm{E}_{\mathrm{s}}}{\partial \mathrm{I}_{2}}= & \Rightarrow \mathrm{L}_{2} \mathrm{I}_{2}+\mathrm{MI}_{1}=0 \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=-\frac{\mathrm{L}_{2}}{\mathrm{M}} \\ & \Rightarrow \mathrm{~L}_{2} \mathrm{I}_{2}=-M I_{1} \end{aligned}

Now put both condition's together only $\mathrm{E}_{\mathrm{s}}$ is minimum,

\begin{aligned} \mathrm{E}_{\mathrm{s}} & =\frac{1}{2} \mathrm{~L}_{1} \mathrm{I}_{1}^{2}+\frac{1}{2} \mathrm{~L}_{2} \mathrm{I}_{2}^{2}+\mathrm{MI}_{1} \mathrm{I}_{2} \\ & =\frac{1}{2}\left[-\mathrm{MI}_{2}\right]_{1}+\frac{1}{2}\left[-\mathrm{MI}_{1}\right] \mathrm{I}_{2}+\mathrm{MI}_{1} \mathrm{I}_{2} \\ & =0 \mathrm{~J} \end{aligned}

Q3GATE 2024NAT

In the $(x, y, z)$ coordinate system, three point-charges $Q, Q$ and $\alpha Q$ are located in free space at $(-1,0,0),(1,0,0)$ and $(0,-1,0)$ respectively. The value of $\alpha$ for the electric field to be zero at $(0,0.5,0)$ is _____ (rounded off to 1 decimal places).

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📖 Explanation

From the figure, $\left.\bar{E}_{n e t}=\bar{E}_{1 y}+\bar{E}_{2 y}+\bar{E}_{\alpha Q}=0 \text { [as per question }\right]$ Here, $\bar{E}_{1 y}=\bar{E}_{2 y}=\bar{E}_{Q} \text { (say) }$ $\therefore \quad \bar{E}_{n e t}=\left.2 E\right|_{Q}+\left.2 E\right|_{\alpha Q}=0$ Now, $\left.\bar{E}\right|_{Q}=\frac{k Q}{\left[\sqrt{1+0.5^{2}}\right]^{2}} \sin \theta \hat{a}_{y} ; \quad k=\frac{1}{4 \pi \epsilon_{o}}$ $\left.\Rightarrow \quad \bar{E}\right|_{Q}=\frac{k Q}{1.25} \times \frac{0.5}{\sqrt{1+0.5^{2}}} \hat{a}_{y}=0.357 k Q \hat{a}_{y}$ Also, $\left.\quad \bar{E}\right|_{\alpha Q}=\frac{k(\alpha Q)}{(1.5)^{2}} \hat{a}_{y}=0.444 k(\alpha Q) \hat{a}_{y}$ Hence, $\quad \bar{E}_{n e t}=2 \times 0.357 k Q \hat{a}_{y}+0.444 k(\alpha Q) \hat{a}_{y}=0$ $\Rightarrow \quad 0.715 \mathrm{kQ}=-0.444 \mathrm{k}(\alpha Q)$ $\Rightarrow \quad \alpha=\frac{-0.715}{0.444}=-1.61$ $\therefore \quad \alpha=-1.61$

Q4GATE 2023MCQ

In the figure, the electric field $E$ and the magnetic field $B$ point to $\mathrm{x}$ and $\mathrm{z}$ directions, respectively, and have constant magnitudes. $A$ positive charge ' $q$ ' is released from rest at the origin. Which of the following statement(s) is/ are true.

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Q5GATE 2020MCQ

The vector function expressed by $F = a_x(5y - k_1z) + a_y(3z + k_2x) + a_z(k_3y - 4x)$ represents a conservative field, where $a_x, a_y, a_z$ are unit vectors along x, y and z directions, respectively. The values of constants $k_1, k_2, k_3$ are given by:

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📖 Explanation

$\bar{F}=(5y-k_{1}Z)\hat{i}+(3z+k_{2}x)\hat{j}+(k_{3}y-4x)\hat{k}$ is conservative field $\bar{F}$ is irrotational,

\begin{aligned} \triangledown \times \bar{F}&=0\\ \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 5y-k_{1}z &3z-k_{2}x &k_{3}y -4x \end{vmatrix}&=0\\ \hat{i}(k_{3}-3)-\hat{j}(-4+k_{1})+\hat{k}(k_{2}-5)&=0\\ k_{3}-3&=0\\ 4-k_{1}&=0\\ k_{2}-5&=0\\ k_{1}&=4\\ k_{2}&=5\\ k_{3}&=3 \end{aligned}

Q6GATE 2017MCQ

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

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📖 Explanation

$\vec{X}$ is going away so $\vec{\triangledown } \cdot \vec{X}\neq 0$ $\vec{Y}$ is moving circulator direction so $\vec{\triangledown } \cdot \vec{Y}\neq 0$ $\vec{Z}$ has circular rotation so $\vec{\triangledown } \cdot \vec{Z}\neq 0$

Q7GATE 2016MCQ

In cylindrical coordinate system, the potential produced by a uniform ring charge is given by $\psi = f(r, z)$ , where f is a continuous function of r and z. Let $\vec{E}$ be the resulting electric field. Then the magnitude of $\bigtriangledown \times \vec{E}$

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📖 Explanation

V is given as static field in time invariant. Hence, $\triangledown \times E=0$

Q8GATE 2016NAT

The line integral of the vector field $F = 5xz \hat{i}+ (3x^{2} + 2y) \hat{j} + x^{2}z\hat{k}$ along a path from (0,0,0) to (1,1,1) parametrized by (t, $t^{2}$ , t) is _____.

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📖 Explanation

\begin{aligned} E &=5xZ\bar{i}+(3x^2 +2y)\bar{j}+x^2z\bar{k} \\ &=\int _C \bar{F}\bar{d}r \\ &= \int _C 5xzdx +(3x^{2+}2y)dy +x^2zdz\\ x&=t, \; y=t^2,\; z=t,\; t=0 \; to \; 1 \\ dx&=dt \\ dy &=2tdt,\; dz =dt\\ &=\int_{0}^{1} 5t^2dt+(3t^{2+}2t^2)2tdt+t^3dt\\ &= \int_{0}^{1} (5t^{2+}11t^3)dt\\ &=\left[ \frac{5t^3}{3}+\frac{11t^4}{4} \right]_0^1=\frac{5}{3}+\frac{11}{4}=4.41 \end{aligned}

Q9GATE 2015MCQ

Consider a function $\vec{f}=\frac{1}{r^{2}}\hat{r}$ , where r is the distance from the origin and $\hat{r}$ is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is

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📖 Explanation

$\bar{f}=\frac{1}{r^2}\hat{r}$ From divergence theorem as we know,

\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}

Q10GATE 2015MCQ

Match the following.

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📖 Explanation

Stokes theorem $\oint \vec{A}\cdot dl=\int \int (\triangledown \times A)\cdot \hat{n}ds$ Gauss theorem $\int \int D\cdot ds=Q$ Divergence theorem $\oint \oint A\cdot \hat{n}ds=\int \int \int \triangledown \cdot \bar{A}dV$ Cauchy integral theorem $\oint _cf(z)dz=0$

Q11GATE 2012MCQ

The direction of vector A is radially outward from the origin, with $|A|=kr^{n}$ , where $r^{2}=x^{2}+y^{2}+z^{2}$ and k is a constant. The value of n for which $\bigtriangledown \cdot A= 0$ is

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📖 Explanation

\begin{aligned} \vec{A} &=kr^n\hat{i}_r \\ \triangledown \cdot \vec{A} &= \frac{1}{r^2}\frac{\partial }{\partial r}(r^2 k r^n)\\ &=\frac{1}{r^2}\frac{\partial }{\partial r}( k r^{n+2}) \\ &= (n+2)k\frac{r^{n+1}}{r^2}\\ &= (n+2)kr^{n-1}\\ \text{For}\;\; \triangledown \cdot \vec{A} &= 0\\ n+2 &=0\\ \Rightarrow \;\; n=-2 \end{aligned}

Q12GATE 2007MCQ

Divergence of the vector field $V(x,y,z)=-(x \cos xy+y)\hat{i}+(y \cos xy)\hat{j}$ $+(\sin z^{2}+x^{2}+y^{2})\hat{k}$ is

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📖 Explanation

\begin{aligned} V(x,y,z) &=-(x \cos xy+y)i +(y \cos xy)j \\ & \times \$[\sin (z^2)+(x^2)+(y^2)]\$k \\ \text{Divergence}&=\triangledown \cdot V \\ &=\frac{\partial V_x}{\partial x} +\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z}\\ &=-\cos xy+x(\sin xy)y +\cos xy \\ &-y \sin (xy)x+2z \cos z^2 \\ &= 2z \cos z^2 \end{aligned}

Q13GATE 2006MCQ

Consider the following statements with reference to the equation $\frac{\delta p}{\delta t}$ (1) This is a point form of the continuity equation. (2) Divergence of current density is equal to the decrease of charge per unit volume per unit at every point. (3) This is Max well's divergence equation (4) This represents the conservation of charge Select the correct answer.

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Q14GATE 2005MCQ

If $\vec{E}$ is the electric intensity, $\triangledown \cdot (\triangledown \times \vec{E})$ is equal to

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📖 Explanation

Divergence of a curl field is always zero. i.e. $\triangledown \cdot (\triangledown \times \vec{E})=0$

Q15GATE 2002MCQ

Given a vector field $\vec{F}$ , the divergence theorem states that

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