Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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}
Question 5 |
Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the magnetic field \vec{B} between the wires at a distance r from W1 is
\frac{\mu _0 I}{6 \pi r} | |
\frac{6 \mu _0 I}{5 \pi r} | |
\frac{5 \mu _0 I}{6 \pi r} | |
\frac{{\mu _0}^2 I^2}{2 \pi r^2} |
Question 5 Explanation:
Magnetic flux density (\vec{B}) at r distance due to infinite line carrying current I is |\vec{B}|=\frac{\mu_{0} I}{2 \pi \rho} .
\vec{B} at r distance due to W_{1} wire
=\left|\vec{B}_{1}\right|=\frac{\mu_{0} I}{2 \pi r}\qquad \ldots(i)
\vec{B} at 3r distance due to W_{2} wire
=\left|\vec{B}_{2}\right|=\frac{\mu_{0}(2 I)}{2 \pi(3 r)}\qquad \ldots(ii)
From right hand thumb rule, \vec{B} due to both lines add in between conductors.
\begin{aligned} {So,}\qquad |\vec{B}|&=\left|\vec{B}_{1}\right|+\left|\vec{B}_{2}\right|\\ \therefore \qquad |\vec{B}|&=\frac{\mu_{0} I}{2 \pi r}+\frac{2 \mu_{0} I}{6 \pi r}=\frac{5 \mu_{0} I}{6 \pi r} \end{aligned}
Question 6 |
In the table shown, List I and List II, respectively, contain terms appearing on the left-hand side and the right-hand side of Maxwell's equations (in their standard form). Match the left-hand side with the corresponding right-hand side.
1-P, 2-R, 3-Q, 4-S | |
1-Q, 2-R, 3-P, 4-S | |
1-Q, 2-S, 3-P, 4-R | |
1-R, 2-Q, 3-S, 4-P |
Question 6 Explanation:
\begin{aligned} \nabla \cdot \vec{D} &=\rho_{v} \\ \nabla \times \vec{E} &=-\frac{\partial \vec{B}}{\partial t} \\ \nabla \cdot \vec{B} &=0 \\ \nabla \times \vec{H} &=\vec{J}+\frac{\partial \vec{D}}{\partial t} \end{aligned}
Question 7 |
What is the electric flux (\int \vec{E}\cdot d\hat{a}) through a quarter-cylinder of height H (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q?
\frac{HQ}{\varepsilon _0} | |
\frac{HQ}{4\varepsilon _0} | |
\frac{H\varepsilon _0}{4Q} | |
\frac{4H}{Q\varepsilon _0} |
Question 7 Explanation:
Electric field intensity (\vec{E}) at '\rho' distance due to infinite long line having line charge density Q is
\begin{aligned} \vec{E} &=\frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{\rho} \\ \int \vec{E} \cdot \vec{da} &=\iint \frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{p} \cdot \rho d \phi d z \hat{a}_{\rho} \\ &=\frac{Q}{2 \pi \varepsilon_{0}} \int_{<\pi / 2>} d \phi \int_{z=0}^{H} d z \\ &=\frac{Q}{2 \pi \varepsilon_{0}}\left(\frac{\pi}{2}\right) H=\frac{H Q}{4 \varepsilon_{0}} \end{aligned}
\begin{aligned} \vec{E} &=\frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{\rho} \\ \int \vec{E} \cdot \vec{da} &=\iint \frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{p} \cdot \rho d \phi d z \hat{a}_{\rho} \\ &=\frac{Q}{2 \pi \varepsilon_{0}} \int_{<\pi / 2>} d \phi \int_{z=0}^{H} d z \\ &=\frac{Q}{2 \pi \varepsilon_{0}}\left(\frac{\pi}{2}\right) H=\frac{H Q}{4 \varepsilon_{0}} \end{aligned}
Question 8 |
An electron (q_1) is moving in free space with velocity 10^{5} m/s towards a stationary electron (q_2) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is ___________ \times 10^{-8}m.
[Given, mass of electron m=9.11 \times 10^{-31} kg, charge of electron e=-1.6 \times 10^{-19}C, and permittivity \varepsilon_{0}=(1/36\pi)\times 10^{-9}F/m]
[Given, mass of electron m=9.11 \times 10^{-31} kg, charge of electron e=-1.6 \times 10^{-19}C, and permittivity \varepsilon_{0}=(1/36\pi)\times 10^{-9}F/m]
4 | |
5 | |
6 | |
7 |
Question 8 Explanation:
r is the distance at which kinetic energy of q_{1} becomes zero [(because kinetic energy (KE) is converted into potential energy (PE)].
When q_{1} reaches 'r', it starts diverting.
Kinetic energy, K E=\frac{1}{2} m v^{2} and work done in moving q_{1} charge to distance 'r' is
\begin{aligned} q_{1} v_{2}&=q_{1} \frac{q_{2}}{4 \pi \varepsilon_{0} r} \\ &\qquad\left(q_{1}=q_{2}=-1.6 \times 10^{-19} \mathrm{C}\right)\\ Now, \quad \frac{1}{2} m v^{2}&=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r} \\ \end{aligned}
\Rightarrow r=\frac{\left(2 \times-1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{4 \pi \times \frac{10^{-9}}{36 \pi} \times 9.11 \times 10^{-31} \times\left(10^{5}\right)^{2}}
\simeq 5.06 \times 10^{-8} \mathrm{m}
Question 9 |
Two conducting spheres S1 and S2 of radii a and b (b>a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure.
For some charge placed on this structure, the potential and surface electric field on S1 are V_{a} and E_{a}, and that on S2 are V_{b} and E_{b}, respectively, which of the following is CORRECT?
For some charge placed on this structure, the potential and surface electric field on S1 are V_{a} and E_{a}, and that on S2 are V_{b} and E_{b}, respectively, which of the following is CORRECT?
V_{a}=V_{b} \; and \; E_{a}\lt E_{b} | |
V_{a}\gt V_{b} \; and \; E_{a} \gt E_{b} | |
V_{a}=V_{b} \; and \; E_{a} \gt E_{b} | |
V_{a}\gt V_{b} \; and \; E_{a}=E_{b} |
Question 9 Explanation:
When charge is placed on this structure equilibrium is established such that be spheres are at same potential i.e.
V_{a}=V_{b}
\begin{array}{c} V_{a}=V_{b} \\ \text { So, } \frac{Q_{a}}{4 \pi \epsilon_{o} a}=\frac{Q_{b}}{4 \pi \epsilon_{o} b} \end{array}
\frac{Q_{b}}{Q_{a}}=\frac{b}{a}
Now, surface electric fields.
\frac{E_{a}}{E_{b}}=\left[\frac{Q_{a} / 4 \pi \varepsilon_{o} a^{2}}{Q_{b} / 4 \pi \varepsilon_{o} b^{2}}\right]=\frac{Q_{a} \times b^{2}}{Q_{b} \times a^{2}}=\frac{b}{a}>1
So, E_{a}>E_{b}
Question 10 |
Consider the charge profile shown in the figure. The resultant potential distribution is best described by
A | |
B | |
C | |
D |
Question 10 Explanation:
Applying Poisson's equations
\nabla^{2} V=\frac{\partial^{2} V}{\partial x^{2}}=-\frac{\rho_{v}}{\epsilon}=K
Constant charge density
\begin{aligned} \frac{\partial V}{\partial x} &=-K x+K^{\prime} \\ V &=\frac{-K x^{2}}{2}+K^{\prime} x+K^{\prime} \end{aligned}
Towards positive x or negative side.
It is a second order parabolic increase.
Due to symmetry of + and - charges K" = 0 is expected with V = O at centre and graph passing through origin.
Beyond x \gt 0 \text{ or } x \lt b, \; E = 0 due to capacitive nature of + and - charges
V=-\int 0 \cdot \overrightarrow{d l}=\text { Constant }
This constant is same V at x=a or x=b
\nabla^{2} V=\frac{\partial^{2} V}{\partial x^{2}}=-\frac{\rho_{v}}{\epsilon}=K
Constant charge density
\begin{aligned} \frac{\partial V}{\partial x} &=-K x+K^{\prime} \\ V &=\frac{-K x^{2}}{2}+K^{\prime} x+K^{\prime} \end{aligned}
Towards positive x or negative side.
It is a second order parabolic increase.
Due to symmetry of + and - charges K" = 0 is expected with V = O at centre and graph passing through origin.
Beyond x \gt 0 \text{ or } x \lt b, \; E = 0 due to capacitive nature of + and - charges
V=-\int 0 \cdot \overrightarrow{d l}=\text { Constant }
This constant is same V at x=a or x=b
There are 10 questions to complete.