Question 1 |
A transparent dielectric coating is applied to glass \left(\epsilon_{r}=4, \mu_{r}=1\right) to eliminate the reflection of red light \left(\lambda_{0}=0.75 \mu \mathrm{m}\right). The minimum thickness of the dielectric coating, in \mu \mathrm{m}, that can be used is ____ (rounded off to two decimal places).
0.02 | |
0.18 | |
0.13 | |
0.52 |
Question 1 Explanation:
For no reflection, impedance must be matched.
Hence, \eta_{2} acts like a quarter wave impedance transformer.
So,
(i) \quad \eta_{2}=\sqrt{\eta_{1} \cdot \eta_{3}} \Rightarrow \epsilon_{r_{2}}=\sqrt{\epsilon_{r_{1}} \cdot \epsilon_{r_{3}}} \Rightarrow \epsilon_{r_{2}}=2
(ii) For impedance matching,
\begin{aligned} & d=(2 n+1) \frac{\lambda}{4} ; n=0,1,2 \ldots \\ & \lambda=\frac{\lambda_{0}}{\sqrt{\epsilon_{r}}}=\frac{\lambda_{0}}{\sqrt{\epsilon_{r_{2}}}} \\ Here \;\;& \lambda=\frac{0.75 \times 10^{-6}}{\sqrt{2}}=0.53 \times 10^{-6} \end{aligned}
Hence, for minimum distance, n=0
So, d=\frac{\lambda}{4}=\frac{0.53 \times 10^{-6}}{4}=0.133 \mu \mathrm{m}
Hence, \eta_{2} acts like a quarter wave impedance transformer.
So,
(i) \quad \eta_{2}=\sqrt{\eta_{1} \cdot \eta_{3}} \Rightarrow \epsilon_{r_{2}}=\sqrt{\epsilon_{r_{1}} \cdot \epsilon_{r_{3}}} \Rightarrow \epsilon_{r_{2}}=2
(ii) For impedance matching,
\begin{aligned} & d=(2 n+1) \frac{\lambda}{4} ; n=0,1,2 \ldots \\ & \lambda=\frac{\lambda_{0}}{\sqrt{\epsilon_{r}}}=\frac{\lambda_{0}}{\sqrt{\epsilon_{r_{2}}}} \\ Here \;\;& \lambda=\frac{0.75 \times 10^{-6}}{\sqrt{2}}=0.53 \times 10^{-6} \end{aligned}
Hence, for minimum distance, n=0
So, d=\frac{\lambda}{4}=\frac{0.53 \times 10^{-6}}{4}=0.133 \mu \mathrm{m}
Question 2 |
In an electrostatic field, the electric displacement density vector, \vec{D} , is given by
\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})C/m^2
, where \vec{i},\vec{j},\vec{k} are the unit vectors along x-axis, y-axis, and z-axis, respectively. Consider a cubical region R centered at the origin with each side of length 1 m, and vertices at (\pm 0.5 m, \pm 0.5 m, \pm 0.5 m ). The electric charge enclosed within R is _________ C (rounded off to two decimal places).
\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})C/m^2
, where \vec{i},\vec{j},\vec{k} are the unit vectors along x-axis, y-axis, and z-axis, respectively. Consider a cubical region R centered at the origin with each side of length 1 m, and vertices at (\pm 0.5 m, \pm 0.5 m, \pm 0.5 m ). The electric charge enclosed within R is _________ C (rounded off to two decimal places).
0.25 | |
0.5 | |
0.75 | |
0.85 |
Question 2 Explanation:
\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})c/m^2
Q_{enc.}=\int _v \rho _v\cdot dV=\int (\triangledown \cdot \vec{D})dV
\triangledown \cdot \vec{D}=3x^2+3y^2
dV=dxdydz
\therefore \; Q_{enc.}=\int _v 3(x^2+y^2)dxdydz =3\left [ \int_{-0.5}^{0.5} x^2 dx \int_{-0.5}^{0.5}dy \int_{-0.5}^{0.5} dz+\int_{-0.5}^{0.5}dx\int_{-0.5}^{0.5}y^2dy\int_{-0.5}^{0.5}dz \right ]
=3\left [ \frac{x^3}{3}|_{-0.5}^{0.5} \times 1 \times 1 + \frac{y^3}{3}|_{-0.5}^{0.5} \times 1 \times 1\right ] =0.25+0.25=0.5C
Q_{enc.}=\int _v \rho _v\cdot dV=\int (\triangledown \cdot \vec{D})dV
\triangledown \cdot \vec{D}=3x^2+3y^2
dV=dxdydz
\therefore \; Q_{enc.}=\int _v 3(x^2+y^2)dxdydz =3\left [ \int_{-0.5}^{0.5} x^2 dx \int_{-0.5}^{0.5}dy \int_{-0.5}^{0.5} dz+\int_{-0.5}^{0.5}dx\int_{-0.5}^{0.5}y^2dy\int_{-0.5}^{0.5}dz \right ]
=3\left [ \frac{x^3}{3}|_{-0.5}^{0.5} \times 1 \times 1 + \frac{y^3}{3}|_{-0.5}^{0.5} \times 1 \times 1\right ] =0.25+0.25=0.5C
Question 3 |
In a circuit, there is a series connection of an ideal resistor and an ideal capacitor.
The conduction current (in Amperes) through the resistor is 2\sin (t+\pi/2) .
The displacement current (in Amperes) through the capacitor is _________.
2 \sin (t) | |
2 \sin (t+\pi) | |
2 \sin (t+\pi /2) | |
0 |
Question 3 Explanation:
In series connection, current pass through each element remain same. Hence, i_c=i_d
So, i_d= 2\sin (t+\pi/2) .
Question 4 |
For a vector field D=\rho\cos^{2}\:\varphi \:a_{\rho }+z^{2}\sin^{2}\:\varphi \:a_{\varphi }
in a cylindrical coordinate system \left ( \rho ,\varphi ,z \right )
with unit vectors a_{\rho },a_{\varphi }
and a_{z}
, the net flux of D leaving the closed surface of the cylinder \left ( \rho =3, 0\leq z\leq 2 \right )
(rounded off to two decimal places) is ________________
56.55 | |
22.12 | |
36.85 | |
76.34 |
Question 4 Explanation:
Method 1: \vec{D}=\rho \cos ^{2} \phi \hat{a}_{\rho}+z^{2} \sin ^{2} \phi \hat{a}_{\phi}
Electric flux crossing the closed surface is
\begin{aligned} \psi &=\oiint \vec{D} \cdot \overrightarrow{d S}=\iiint(\vec{\nabla} \cdot \vec{D}) d v \\ \vec{\nabla} \cdot \vec{D} &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho D_{\rho}\right)+\frac{1}{\rho} \frac{\partial D_{\phi}}{\partial \phi}+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho \rho \cos ^{2} \phi\right)+\frac{1}{\rho} \frac{\partial}{\partial \phi}\left(z^{2} \sin ^{2} \phi\right)+0 \\ &=\frac{1}{\rho}(2 \rho) \cos ^{2} \phi+\frac{1}{\rho} z^{2} 2 \sin \phi \cos \phi=2 \cos ^{2} \phi+\frac{z^{2}}{\rho} \sin 2 \phi\\ \iiint(\vec{\nabla} \cdot \vec{D}) d v &=\iiint 2 \cos ^{2} \phi(\rho d \rho d \phi d z)+\iiint\left(\frac{z^{2}}{\rho} \sin 2 \phi\right) \rho d \rho d \phi d z \\ &=2 \int_{\rho=0}^{3} \rho d \rho \int_{\phi=0}^{2 \pi}\left(\frac{1+\cos 2 \phi}{2}\right) d \phi \int_{z=0}^{2} d z+\int_{\rho=0}^{2} d \rho \int_{\phi=0}^{2 \pi} \sin 2 \phi d \phi \int_{z=0}^{2} z^{2} d z \\ &=2\left(\frac{\rho^{2}}{-2}\right)_{\rho=0}^{3} \frac{1}{2}(2 \pi)(z)_{z=0}^{2}+0 \\ &=2\left(\frac{3^{2}}{2}\right) \pi(2)=18 \pi(\text { Coulomb })=56.55(\text { Coulomb }) \end{aligned}
Method 2: Electric flux crossing the closed surface is
\psi=\oiint \vec{D} \cdot \overrightarrow{d S}
Electric flux crossing \rho=3 cylindrical surface is
\begin{aligned} \left.\psi\right|_{\rho=3} &=\oiint\left(\rho \cos ^{2} \phi \hat{a}_{p}\right) \cdot(\rho d \phi d z) \hat{a}_{p} \\ &=3^{2} \int_{\phi=0}^{2 \pi} \cos ^{2} \phi d \phi \int_{z=0}^{2} d z \\ &=9 \frac{1}{2}(2 \pi)(2)=18 \pi(\text { coulomb })=56.55(\text { coulomb }) \end{aligned}
Electric flux crossing the closed surface is
\begin{aligned} \psi &=\oiint \vec{D} \cdot \overrightarrow{d S}=\iiint(\vec{\nabla} \cdot \vec{D}) d v \\ \vec{\nabla} \cdot \vec{D} &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho D_{\rho}\right)+\frac{1}{\rho} \frac{\partial D_{\phi}}{\partial \phi}+\frac{\partial D_{z}}{\partial z} \\ &=\frac{1}{\rho} \frac{\partial}{\partial p}\left(\rho \rho \cos ^{2} \phi\right)+\frac{1}{\rho} \frac{\partial}{\partial \phi}\left(z^{2} \sin ^{2} \phi\right)+0 \\ &=\frac{1}{\rho}(2 \rho) \cos ^{2} \phi+\frac{1}{\rho} z^{2} 2 \sin \phi \cos \phi=2 \cos ^{2} \phi+\frac{z^{2}}{\rho} \sin 2 \phi\\ \iiint(\vec{\nabla} \cdot \vec{D}) d v &=\iiint 2 \cos ^{2} \phi(\rho d \rho d \phi d z)+\iiint\left(\frac{z^{2}}{\rho} \sin 2 \phi\right) \rho d \rho d \phi d z \\ &=2 \int_{\rho=0}^{3} \rho d \rho \int_{\phi=0}^{2 \pi}\left(\frac{1+\cos 2 \phi}{2}\right) d \phi \int_{z=0}^{2} d z+\int_{\rho=0}^{2} d \rho \int_{\phi=0}^{2 \pi} \sin 2 \phi d \phi \int_{z=0}^{2} z^{2} d z \\ &=2\left(\frac{\rho^{2}}{-2}\right)_{\rho=0}^{3} \frac{1}{2}(2 \pi)(z)_{z=0}^{2}+0 \\ &=2\left(\frac{3^{2}}{2}\right) \pi(2)=18 \pi(\text { Coulomb })=56.55(\text { Coulomb }) \end{aligned}
Method 2: Electric flux crossing the closed surface is
\psi=\oiint \vec{D} \cdot \overrightarrow{d S}
Electric flux crossing \rho=3 cylindrical surface is
\begin{aligned} \left.\psi\right|_{\rho=3} &=\oiint\left(\rho \cos ^{2} \phi \hat{a}_{p}\right) \cdot(\rho d \phi d z) \hat{a}_{p} \\ &=3^{2} \int_{\phi=0}^{2 \pi} \cos ^{2} \phi d \phi \int_{z=0}^{2} d z \\ &=9 \frac{1}{2}(2 \pi)(2)=18 \pi(\text { coulomb })=56.55(\text { coulomb }) \end{aligned}
Question 5 |
Consider the vector field F\:=\:a_{x}\left ( 4y-c_{1}z \right )+a_y\left ( 4x + 2z\right )+a_{z}\left ( 2y +z\right ) in a rectangular coordinate system (x,y,z) with unit vectors a_{x},\:a_{y} and a_{z}. If the field F is irrotational (conservative), then the constant c_{1}
(in integer) is _________________
0 | |
1 | |
2 | |
3 |
Question 5 Explanation:
\begin{aligned} \nabla \times \vec{F}&=0\\ \left|\begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 4 y-c_{1} z & 4 x+2 z & 2 y+z \end{array}\right|&=0\\ =i(2-2)-j\left(0+c_{1}\right)+k(4-4)=0 \\ c_{1} &=0 \end{aligned}
There are 5 questions to complete.