Ed

1. If the electrostatic potential were given by $\phi=\phi_0\left(x^2+y^2+z^2\right)$, where $\phi_0$ is constant, then the charge density giving rise to the above potential would be
1. 0
2. $-6\phi_0\epsilon_0$
3. $-2\phi_0\epsilon_0$
4. $-\frac{6\phi_0}{\epsilon_0}$
\begin{align*} \bar E&=-\bar \nabla\phi\\ &=-\bar \nabla\left\{\phi_0\left(x^2+y^2+z^2\right)\right\}\\ &=-2\phi_0(x\hat i+y\hat j+z\hat k) \end{align*} Charge density $\sigma=\epsilon_0\bar\nabla.\bar E=-6\phi_0\epsilon_0$
2. The work done in bringing a charge $+q$ from infinity in free space, to a position at a distance $d$ in front of a semi-infinite grounded metal surface is
1. $-\frac{q^2}{4\pi\epsilon_0(d)}$
2. $-\frac{q^2}{4\pi\epsilon_0(2d)}$
3. $-\frac{q^2}{4\pi\epsilon_0(4d)}$
4. $-\frac{q^2}{4\pi\epsilon_0(6d)}$
In front of a semi-infinite grounded metal surface when a charge $q$ is held at a distance $x$, then there is an image charge $-q$ at a distance $x$ behind the metal surface. Hence force on $+q$ charge is given by $$F=-\frac{q^2}{4\pi\epsilon_0(2x)^2}$$ Hence force required to move the charge is $$F=\frac{q^2}{4\pi\epsilon_0(2x)^2}$$ \begin{align*} W&=\int_{\infty}^{d}Fdx\\ &=\frac{q^2}{4\pi\epsilon_0(4)}\int_{\infty}^d\frac{1}{x^2}dx\\ &=-\frac{q^2}{4\pi\epsilon_0(4d)} \end{align*}
3. Twelve equal charges of magnitude $q$ are kept at the corners of a regular 12-sided polygon one at each of the corner. What is the net force on a test charge $Q$ at the center of the polygon at distance $r$?
1. Zero
2. $\frac{1}{4\pi\epsilon_0}\frac{12qQ}{r^2}$
3. $\frac{1}{4\pi\epsilon_0}\frac{6qQ}{r^2}$
4. $\frac{1}{4\pi\epsilon_0}\frac{qQ}{r^2}$
As charge $Q$ is placed at symmetric position the net force acting is zero.

4. Two equal charges each of magnitude $q$ are kept $d$ distance apart along X-axis. Now, the electric field at a distance $z(z>>d)$ above the midpoint between two charges is given by:

1. $E_z=\frac{1}{4\pi\epsilon_0}\frac{2q}{z^2}$
2. $E_z=\frac{1}{4\pi\epsilon_0}\frac{q^2}{z^2}$
3. $E_z=\frac{1}{4\pi\epsilon_0}\frac{2q}{r^{3/2}}$
4. $E_z=\frac{1}{4\pi\epsilon_0}\frac{2q}{(2+r)^2}$
Electric field at a point at a distance $z$ due to single charge is $E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$. Hence electric field at the same point along z-axis is $E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta$. But $\cos\theta=\frac{z}{r}$ and $r=z^2+\frac{d^2}{4}$. Hence $E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\frac{z}{r}=\frac{1}{4\pi\epsilon_0}\frac{qz}{r^3}=\frac{1}{4\pi\epsilon_0}\frac{qz}{(z^2+\frac{d^2}{4})^{3/2}}$
For $z>>d$, $\frac{d^2}{4}$ can be neglected. Hence,$E_1=\frac{1}{4\pi\epsilon_0}\frac{q}{z^2}$. Similarly, $E_2=\frac{1}{4\pi\epsilon_0}\frac{q}{z^2}$. Hence, $E=E_1+E_2=\frac{1}{4\pi\epsilon_0}\frac{2q}{z^2}$

5. Poynting's vector represent which of the following quantities?
   1. Current density vector producing electrostatic field
   2. Power density vector producing electromagnetic field
   3. Current density vector producing electromagnetic field
   4. Power density vector producing electrostatic and magnetostatic fields
   (B) Power density vector producing electromagnetic field.
   
6. A plane wave of electric and magnetic fields $\vec E_0$ , $\vec B_0$ and frequency $\omega$ enters in a conducting bar of conductivity $\sigma$ along z-axis. Which of the following pairs of equations best represents the propagating wave? ($k\longrightarrow $ wave number)
1. $\vec E(z,t)=E_0e^{-ikz}e^{i(kz-\omega t)}\hat x$ and $\vec B(z,t)=\frac{k}{\omega} E_0e^{-ikz}e^{i(kz-\omega t+\phi)}\hat y$
2. $\vec E(z,t)=E_0e^{-kz}e^{i(kz-\omega t)}\hat x$ and $\vec B(z,t)=\frac{k}{\omega} E_0e^{-kz}e^{i(kz-\omega t+\phi)}\hat y$
3. $\vec E(z,t)=\frac{k}{\omega}E_0e^{-kz}e^{i(kz-\omega t)}\hat x$ and $\vec B(z,t)=\frac{k}{\omega} E_0e^{-ikz}e^{i(kz-\omega t+\phi)}\hat y$
4. $\vec E(z,t)=\frac{k}{\omega}E_0e^{-kz}e^{i(kz-\omega t)}\hat x$ and $\vec B(z,t)= E_0e^{-kz}e^{i(kz-\omega t+\phi)}\hat y$
In electromagnetic wave the electric and magnetic fields and propagation direction are mutually perpendicular to each other. Hence as wave travels in z-direction, electric and magnetic fields must be in x and y-directions. Let electric field is in x-direction. The wave equation for electric field in electromagnetic wave is given by $$\vec E(z,t)=E_0e^{-kz}e^{i(kz-\omega t)}\hat x$$ The term $e^{i(kz-\omega t)}$ is a plane equation and the term $e^{-kz}$ gives exponential decay of wave.
Magnetic field in plane electromagnetic wave in a conducting medium is related to the electric field by \begin{align*} \vec B&=\frac{\vec k}{\omega}\times\vec E\\ &=\frac{k}{\omega}\left|\vec E\right|\hat z\times\hat x\\ &=\frac{k}{\omega}\left|\vec E\right|\hat y\\ &=\frac{k}{\omega}E_0e^{-kz}e^{i(kz-\omega t)}\hat y \end{align*} Also in conducting media there is a phase difference of $\phi$ between $\vec E$ and $\vec B$ $$\therefore\vec B=\frac{k}{\omega}E_0e^{-kz}e^{i(kz-\omega t+\phi)}\hat y$$
7. A plane electromagnetic wave travelling in vacuum is incident normally on a non-magnetic, non-absorbing medium of refractive index $n$. The incident $(E_i)$, reflected $(E_r)$ and transmitted $(E_t)$, electric fields are given as $E_i=E_{0i}exp[i(kz-\omega t)]$, $E_r=E_{0r}exp[i(k_rz-\omega t)]$, $E_t=E_{0t}exp[i(k_tz-\omega t)]$. If $E_{0i}=2 V/m$ and $n=1.5$ then the application of appropriate boundary conditions leads to
1. $E_{0r}=-\frac{3}{5} V/m$, $E_{0t}=\frac{7}{5} V/m$
2. $E_{0r}=-\frac{1}{5} V/m$, $E_{0t}=\frac{9}{5} V/m$
3. $E_{0r}=-\frac{2}{5} V/m$, $E_{0t}=\frac{8}{5} V/m$
4. $E_{0r}=\frac{4}{5} V/m$, $E_{0t}=\frac{6}{5} V/m$
Wave is traveling in vacuum, hence $n_1=1$, for the medium $n_2=1.5$. \begin{align*} E_{0r}&=\frac{n_1-n_2}{n_1+n_2}E_{0i}\\ &=\frac{1-1.5}{1+1.5}\times 2\\ &=-\frac{1}{2.5}=-\frac{2}{5} \end{align*} \begin{align*} E_{0t}&=\frac{2n_1}{n_1+n_2}E_{0i}\\ &=\frac{2}{1.5+1}\times 2\\ &=\frac{4}{2.5}=\frac{8}{5} \end{align*}
8. The magnetic field due to the $TE_{11}$ mode in a rectangular wave guide aligned along Z-axis is given by $H_z=H_1\cos{(0.5\:\pi x)}\cos{(0.6\:\pi y)}$, where $x$ and $y$ are in cm. Then dimensions of the rectangular wave guide $a$ and $b$, respectively, are
1. 2.00 cm and 1.66 cm
2. 1.66 cm and 2.66 cm
3. 2.54 cm and 1.66 cm
4. 1.66 cm and 1.25 cm
In a rectangular wave guide the longitudinal magnetic field is given by $$H_z(x,y)=H_0\cos{(k_xx)}\cos{(k_yy)}$$ where $k_x=\frac{n\pi}{a}$ and $k_y=\frac{m\pi}{b}$. For $TE_{11}$ mode $n=m=1$.
Hence, $k_x=\frac{\pi}{a}$ and $k_y=\frac{\pi}{b}$
Hence, $a=\frac{\pi}{k_x}=\frac{1}{0.5}=2$ cm and $b=\frac{\pi}{k_y}=\frac{1}{0.6}=1.66$ cm
Hence, answer is (A) 2.00 cm and 1.66 cm.
9. A solid sphere of radius $R$ carries a uniform volume charge density $\rho$. The magnitude of electric field inside the sphere at a distance $r$ from center is
1. $\frac{r\rho}{3\epsilon_0}$
2. $\frac{R\rho}{3\epsilon_0}$
3. $\frac{R^2\rho}{r\epsilon_0}$
4. $\frac{R^3\rho}{r^2\epsilon_0}$
According to Gauss law , if the surface encloses a charge $q$ within it, then net flux through the surface is $q/\epsilon_0$. The flux is given by $$\phi_E=\int_S\vec E.d\vec A=\frac{q}{\epsilon_0}$$ If $\rho$ is the volume charge density, the charge enclosed within a distance $r$ is $q=\frac{4}{3}\pi r^3\rho$ and $$\int_S\vec E.d\vec A=E(4\pi r^2)$$ $$E(4\pi r^2)=\frac{4}{3\epsilon_0}\pi r^3\rho$$ $$E=\frac{\rho r}{3\epsilon_0}$$
10. The electric field $\vec E(\vec r,t)$ for a circularly polarized electromagnetic wave propagating along the positive $z$ direction is
    1. $E_0(\hat x+\hat y)exp[i(kz-\omega t)]$
    2. $E_0(\hat x+i\hat y)exp[i(kz-\omega t)]$
    3. $E_0(\hat x+i\hat y)exp[i(kz+\omega t)]$
    4. $E_0(\hat x+\hat y)exp[i(kz+\omega t)]$
    Plane wave traveling along the positive z-axis is given by $$\vec E(z,t)=\vec E_0 cos(kz-\omega t)$$ Plane wave traveling along the negative z-axis is given by $$\vec E(z,t)=\vec E_0 cos(kz+\omega t)$$ A beam linearly polarized along the x-axis and traveling in the positive z-direction can be represented by:$$\vec E(z,t)= E_0 cos(kz-\omega t)\hat x$$ Circularly polarized light can be represented by the expressions: \begin{align*} \vec E_{RCP}(z,t)&= E_0 \left(cos(kz-\omega t)\hat x\right.\\ &{}\left.+sin(kz-\omega t)\hat y\right) \end{align*} \begin{align*} \vec E_{LCP}(z,t)&= E_0 \left(cos(kz-\omega t)\hat x\right.\\ &\left.-sin(kz-\omega t)\hat y\right) \end{align*} Consider option (A) \begin{align*} \vec E&=E_0(\hat x+\hat y)exp[i(kz-\omega t)]\\ &=E_0(\hat x+\hat y)\left(cos(kz-\omega t)\right.\\ &{}\left.+isin(kz-\omega t)\right)\\ &=E_0\left[cos(kz-\omega t) \hat x\right.\\ &\left.{}+cos(kz-\omega t)\hat y\right]\\ &{}+iE_0\left[sin(kz-\omega t)\hat x\right.\\ &{}\left.+sin(kz-\omega t)\hat y\right] \end{align*} Hence real part of $\vec E$ is \begin{align*} Re \vec E&=E_0\left[cos(kz-\omega t) \hat x\right.\\ &{}\left.+cos(kz-\omega t)\hat y\right] \end{align*} which represents plane polarized light.
    Now consider the option (B) \begin{align*} \vec E&=E_0(\hat x+i\hat y)exp[i(kz-\omega t)]\\ &=E_0(\hat x+i\hat y)\left(cos(kz-\omega t)\right.\\ &\left. +isin(kz-\omega t)\right)\\ &=E_0\left[cos(kz-\omega t) \hat x\right.\\ &\left.-sin(kz-\omega t)\hat y\right]\\ &+iE_0\left[sin(kz-\omega t)\hat x\right.\\ &\left.+cos(kz-\omega t)\hat y\right] \end{align*} \begin{align*} Re \vec E&=E_0\left[cos(kz-\omega t) \hat x\right.\\ &\left.-sin(kz-\omega t)\hat y\right] \end{align*} Which represents a circularly polarized light, as \begin{align*} E_x^2+E_y^2&=E_0^2cos^2(kz-\omega t)\\ &+E_0^2sin^2(kz-\omega t)\\ &=E_0^2 \end{align*} is equation of circle. Hence (B) is correct option.
11. A signal of frequency 10 kHz is being digitized by an A/D converter. A possible sampling time which can be used is
1. $100\quad \mu s$
2. $40\quad \mu s$
3. $60\quad \mu s$
4. $200\quad \mu s$
The conversion time is the time required to complete a conversion of the input signal. It establishes the upper signal frequency limit that can be sampled without aliasing $$f_{max}=\frac{1}{2\times conversion~time}$$ \begin{align*} conversion~time &=\frac{1}{2\times f_{max}}\\ &=\frac{1}{2\times 10\times10^3}\\ &=5\times10^{-5}=50\mu S \end{align*} Hence answer is (B).
Supplementary information: AD converter specifications:
• The conversion time is the time required to complete a conversion of the input singnal. It establishes the upper signal frequency limit that can be sampled without aliasing $$f_{max}=\frac{1}{2\times conversion~time}$$
• Resolution: The number of bits in the converter gives the resolution $$Resolution=\frac{full-scale~signal}{2^n}$$
• Accuracy: Accuracy describes how close the measurement is to the actual value. $$Accuracy=\frac{V_{resolution}}{V_{signal}}\times 100\%$$
12. The electrostatic potential $V ( x, y)$ in free space in a region where the charge density $\rho$ is zero is given by $V ( x, y)=4e^{2x}+f(x)-3y$. Given that the x-component of the electric field, $E_x$, and $V$ are zero at the origin, $f (x)$ is
1. $3x^2-4e^{2x}+8x$
2. $3x^2-4e^{2x}+16x$
3. $4e^{2x}-8$
4. $3x^2-4e^{2x}$
$V ( x, y)$ is zero at origin is zero gives $f(0)=-4$. This requirement is satisfied by all options. So let us calculate electric field. $$\vec F=q_0\vec E$$ If test charge $q_0=1$ then \begin{align*} \vec E&=\vec F=-\vec \nabla V\\ &={\scriptstyle -\left(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z}\right)\left(4e^{2x}+f(x)-3y\right)} \end{align*} $$\vec E=-\hat i \left(8e^{2x}+\frac{\partial f(x)}{\partial x}\right)+\hat j(3)$$ Hence, x-component of electric field is $$E_x(x)=-\left(8e^{2x}+\frac{\partial f(x)}{\partial x}\right)$$ However, $E_x=0$ at origin. $$E_x(0)=-\left(8e^{0}+\frac{\partial f(0)}{\partial x}\right)=0$$ Hence, $$\frac{\partial f(0)}{\partial x}=-8$$ For $f(x)=3x^2-4e^{2x}+8x$, $$\frac{\partial f}{\partial x}=6x-8e^{2x}+8$$ $$\frac{\partial f(0)}{\partial x}=0-8+8=0$$
For $f(x)=3x^2-4e^{2x}+16x$, $$\frac{\partial f}{\partial x}=6x-8e^{2x}+16$$ $$\frac{\partial f(0)}{\partial x}=0-8+16=8$$
For $f(x)=4e^{2x}-8$, $$\frac{\partial f}{\partial x}=8e^{2x}$$ $$\frac{\partial f(0)}{\partial x}=8$$
For $f(x)=3x^2-4e^{2x}$, $$\frac{\partial f}{\partial x}=6x-8e^{2x}$$ $$\frac{\partial f(0)}{\partial x}=0-8=-8$$ Hence, Answer is (D).
13. For constant uniform electric and magnetic fields $\vec E=\vec E_0$ and $\vec B=\vec B_0$,it is possible to choose a gauge such that the scalar potential $\phi$ and vector potential $\vec A$ are given by
    1. $\phi=0$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$
    2. $\phi=-\vec E\cdot \vec r$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$
    3. $\phi=-\vec E\cdot \vec r$ and $\vec A=0$
    4. $\phi=0$ and $\vec A=-\vec E t$
    Electric field $\vec E$ and magnetic field $\vec B$ are related to vector and scalar potentials by $$\vec B=\vec\nabla\times\vec A\quad (1)$$ $$\vec E=-\vec\nabla\phi-\frac{\partial\vec A}{\partial t}\quad (2)$$ Substituting values $\phi=-\vec E\cdot \vec r$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$ in equations (1) and (2), we get $\vec E=\vec E_0$ and $\vec B=\vec B_0$ (Verify this). Hence answer is (B)$\phi=-\vec E\cdot \vec r$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$.
14. The Lagrangian of a particle of charge $e$ and mass $m$ in applied electric and magnetic fields is given by $L = \frac{1}{2}mv^2+ e\vec A\cdot \vec v - e\phi$, where $A$ and $\phi$ are the vector and scalar potentials corresponding to the magnetic and electric fields, respectively. Which of the following statements is correct?
1. The canonically conjugate momentum of the particle is given by $\vec p= m\vec v$
2. The Hamiltonian of the particle is given by $H = \frac{p^2}{2m}+ \frac{e}{m}\vec A\cdot \vec p + e\phi$
3. $L$ remains unchanged under a gauge transformation ofthe potentials
4. Under a gauge transformation of the potentials, $L$ changes by the total time derivative of a function of $r$ and $t$
The canonical momentum is given by $$p_i=\frac{\partial L}{\partial \dot x_i}=mv_i+eA_i$$ The Hamiltonian is given by \begin{align*} H&=\sum\limits_ip_ix_i-L\\ &=\frac{1}{2}mv^2+e\phi\\ &=\frac{1}{2m}\left|\vec p-e\vec A\right|^2+e\phi \end{align*} The equations of gauge transformation of the potentials are given by $${\scriptstyle\vec A(\vec x,t)\rightarrow \vec{A'}(\vec x,t)=\vec A(\vec x,t)+\vec\nabla\Lambda(\vec x,t)}$$ $${\scriptstyle\phi(\vec x,t)\rightarrow \phi'(\vec x,t)=\phi(\vec x,t)-\frac{\partial\Lambda(\vec x,t)}{\partial t}}$$ $${\scriptstyle L\rightarrow L'(\vec x,t)=L+e\left(\nabla\Lambda\cdot\vec v+\frac{\partial\Lambda}{\partial t}\right)}$$ $$L'(\vec x,t)=L+e\frac{d\Lambda}{dt}=L$$ Because, total time derivative of$\Lambda(\vec x,t)$ is zero. Hence, option (C) is correct.
15. The magnetic field due to the $TE_{11}$ mode in a rectangular wave guide aligned along Z-axis is given by $H_z=H_1\cos{(0.3\:\pi x)}\cos{(0.4\:\pi y)}$, where $x$ and $y$ are in cm.
1. The dimensions of the rectangular wave guide are
1. $a=3.33$ cm, $b=2.50$ cm
2. $a=0.40$ cm, $b=0.30$
3. $a=0.80$ cm, $b=0.60$ cm
4. $a=1.66$ cm and $1.25$ cm
2. The entire range of frequencies $f$ for which the $TE_{11}$ mode will propagate is
1. 6.0GHz < $f$ < 7.5GHz
2. 7.5GHz< $f$ < 9.0GHz
3. 7.5GHz< $f$ < 12.0GHz
4. 7.5GHz< $f$
(A) In a rectangular wave guide the longitudinal magnetic field is given by $$H_z(x,y)=H_0\cos{(k_xx)}\cos{(k_yy)}$$ where $k_x=\frac{n\pi}{a}$ and $k_y=\frac{m\pi}{b}$. For $TE_{11}$ mode $n=m=1$.
Hence, $k_x=\frac{\pi}{a}$ and $k_y=\frac{\pi}{b}$
Hence, $a=\frac{\pi}{k_x}=\frac{1}{0.3}=3.33$ cm and $b=\frac{\pi}{k_y}=\frac{1}{0.4}=2.5$ cm
(B) The cutoff frequency for $TE_{mn}$ mode is given by $$f_{c_{mn}}=\frac{c}{2n}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ The cutoff frequency for $TE_{mn}$ mode is the operating frequency below which attenuaiion occurs and above which propagation takes place. <\br> For air filled waveguide $n=1$ \begin{align*} f_{c_{11}}&={\scriptstyle\frac{3\times10^{10}}{2}\sqrt{\left(\frac{1}{3.33}\right)^2+\left(\frac{1}{2.5}\right)^2}}\\ &=7.5\:GHz \end{align*} Hence $7.5\:GHz < f$, hence, answer is (D).
16. Three charges are located on the circumference of a circle of radius $R$ as shown in the figure below. The two charges $Q$ subtend an angle $90^o$ at the centre of the circle. The charge $q$ is symmetrically placed with respect to the charges $Q$. If the electric field at the centre of the circle is zero, what is the magnitude of $Q$?

1. $q/\sqrt{2}$
2. $\sqrt{2}q$
3. $2q$
4. $2q$

From figure, for balancing, we must have $$E_q=2E_Q\cos{(45)}$$ i.e. $\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}=2\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\frac{1}{\sqrt{2}}$. Hence, $Q=q/\sqrt{2}$.
17. Consider a hollow charged shell of inner radius $a$ and outer radius $b$. The volume charge density is $\rho(r) =\frac{k}{r^2}$ ($k$ is a constant) in the region $a < r < b$. The magnitude of the electric field produced at distance $r > a$ is
1. $\frac{k(b-a)}{\epsilon_0r^2}$ for all $r > a$
2. $\frac{k(b-a)}{\epsilon_0r^2}$ for $a < r < b$ and $\frac{kb}{\epsilon_0r^2}$ for $r > b$
3. $\frac{k(r-a)}{\epsilon_0r^2}$ for $a < r < b$ and $\frac{k(b-a)}{\epsilon_0r^2}$ for $r > b$
4. $\frac{k(r-a)}{\epsilon_0a^2}$ for $a < r < b$ and $\frac{k(b-a)}{\epsilon_0a^2}$ for $r > b$

Let us divide the region $r > a$ into two regions viz. $a\leq r\leq b$ and $r > b$.
For region $a\leq r\leq b$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^r\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^r\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(r-a) \end{align*} $$E=\frac{k(r-a)}{\epsilon_0 r^2}$$ For region $r > b$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^b\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^b\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(b-a) \end{align*} $$E=\frac{k(b-a)}{\epsilon_0 r^2}$$ Hence, answer is (C)
18. Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by $\vec E_1 = \hat i E_0 \cos{(\omega t+\phi)}$ and $\vec E_2 = \hat j E_0 \cos{(\omega t+\phi)}$ where $\phi$ is the phase difference. The intensity of the resulting wave is given by $\frac{\epsilon_0}{2}\left < E^2\right >$, where $\left < E^2\right > $ is the time average of $E^2$. The total intensity is
1. $0$
2. $\epsilon_0E_0^2$
3. $\epsilon_0E_0^2\sin^2{\phi}$
4. $\epsilon_0E_0^2\cos^2{\phi}$
Resultant electric field is given by \begin{align*} \vec E &= \hat i E_0 \cos{(\omega t+\phi)}\\ &+\hat j E_0 \cos{(\omega t+\phi)} \end{align*} $$ E^2 = 2 E_0^2 \cos^2{(\omega t+\phi)}$$ \begin{align*} \left < E^2\right > &=\frac{\int_0^TE^2\,dt}{\int_0^T\,dt}\\ &=\frac{2E_0^2}{T}\int_0^T\cos^2{(\omega t+\phi)}\;dt\\ &={\scriptstyle\frac{2E_0^2}{T}\frac{1}{2}\int_0^T\left[1-\cos2{(\omega t+\phi)}\right]\;dt}\\ &=\frac{E_0^2}{T}T=E_0^2 \end{align*} $$I=\frac{\epsilon_0}{2}\left < E^2\right > =\frac{\epsilon_0E_0^2}{2}$$
19. Four charges (two $+q$ and two $-q$) are kept fixed at the four vertices of a square of side $a$ as shown
    
    At the point $P$ which is at a distance $R$ from the centre $( R > > a)$, the potential is proportional to
1. $1/R$
2. $1/R^2$
3. $1/R^3$
4. $1/R^4$
The system forms a quadrupole. The electric field and potential energy variations for various system are as follow:
For monopole $E\propto 1/r^2$ and $V\propto 1/r$
For dipole $E\propto 1/r^3$ and $V\propto 1/r^2$
For quadrupole $E\propto 1/r^4$ and $V\propto 1/r^3$
For octapole $E\propto 1/r^5$ and $V\propto 1/r^4$
Hence, answer is (C).
20. A point charge $q$ of mass $m$ is kept at a distance $d$ below a grounded infinite conducting sheet which lies in the $xy$-plane. For what value of $d$ will the charge remains stationary?
1. $q/4\sqrt{mg\pi\epsilon_0}$
2. $q/\sqrt{mg\pi\epsilon_0}$
3. There is no finite value of $d$
4. $\sqrt{mg\pi\epsilon_0}/q$

When a charge $+q$ is placed below the plate at a distance $d$, then there is an image charge $-q$ above the plate at the same distance $d$. For the charge to remain stationary, weight $mg$ must equal to electrostatic force $\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}$. i.e. $$mg=\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}$$ $$d=\frac{q}{4\sqrt{mg\pi\epsilon_0}}$$ Hence, answer is (A).
21. An infinite solenoid with its axis of symmetry along the z-direction carries a steady current $I$. The vector potential $\vec A$ at a distance $R$ from the axis
1. is constant inside and varies as $R$ outside the solenoid
2. varies as $R$ inside and is constant outside the solenoid
3. varies as $\frac{1}{R}$ inside and as $R$ outside the solenoid
4. varies as $R$ inside and as $\frac{1}{R}$ outside the solenoid
Let $r$ is the radius of solenoid.
Magnetic filed inside the solenoid is $B=\mu_0 n I$, where $n$ is number of turns per unit length, and, $I$ is current flowing through the wire. Magnetic filed outside the solenoid is zero.

Vector potential inside the solenoid:
Now, according to Stoke's law, we have $$\oint \vec A\cdot d\vec l=\int_S\left(\nabla\times\vec A\right)\cdot d\vec S$$ But $\left(\nabla\times\vec A\right)=\vec B$ $$\oint \vec A\cdot d\vec l=\int_S\vec B\cdot d\vec S=\Phi_B=flux$$ Let us calculate flux $\Phi_B$. For this let us consider a circular path of radius $R$ centered along its axis. The flux through the circular area is $$\Phi_B=\pi R^2B=\pi R^2\mu_0 n I$$ Hence, $$\oint \vec A\cdot d\vec l=\pi R^2\mu_0 n I$$ But, $\oint \vec A\cdot d\vec l=2\pi R A_\Phi$, this is because $\vec A$ is along the direction of current which is circumferential. $$2\pi R A_\Phi=\pi R^2\mu_0 n I$$ $$A_\Phi=\mu_0 n I\frac{R}{2}$$

Vector potential outside the solenoid:
Let us consider the point $R>r$. The flux through circular area is $$\Phi_B=\pi r^2B=\pi r^2\mu_0 n I$$ $$\oint \vec A\cdot d\vec l=\pi r^2\mu_0 n I$$ But, $\oint \vec A\cdot d\vec l=2\pi R A_\Phi$ $$2\pi R A_\Phi=\pi r^2\mu_0 n I$$ $$A=\mu_0 n I\frac{r^2}{2R}$$
Hence, $\vec A$ varies as $R$ inside and as $\frac{1}{R}$ outside the solenoid
Hence, answer is (D)
22. Consider an infinite conducting sheet in the xy-plane with a time dependent cunent density $Kt\hat i$, where $K$ is a constant. The vector potential at $(x, y,z)$ is given by $\vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i$. The magnetic field $\vec B$ is
1. $\frac{\mu_0Kt}{2}\hat j$
2. $-\frac{\mu_0Kz}{2c}\hat j$
3. $-\frac{\mu_0K}{2c}(ct-z)\hat i$
4. $-\frac{\mu_0K}{2c}(ct-z)\hat j$
$$\vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i$$ \begin{align*} \vec B&=\vec\nabla\times\vec A\\ &=\begin{vmatrix} \hat i&\hat j&\hat k\\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \frac{\mu_0K}{4c}(ct-z)^2&0&0 \end{vmatrix} \end{align*} $$\vec B=-\frac{\mu_0K}{2c}(ct-z)\hat j$$ Hence, answer is (D)
23. When a charged particle emits electromagnetic radiation, the electric field $\vec E$ and the Poynting vector $\vec S = \frac{1}{\mu_0} \vec E\times\vec B$ at a large distance $r$ from the emitter vary as $\frac{1}{r^n}$ and $\frac{1}{r^m}$ respectively. Which of the following choices for $n$ and $m$ are correct?
1. $n = 1$ and $m=1$
2. $n = 2$ and $m=2$
3. $n = 1$ and $m=2$
4. $n = 2$ and $m=4$
When a charged particle emits electromagnetic radiation $E\propto\frac{1}{r}$ and $B\propto\frac{1}{r}$. As $\vec S=\frac{\vec E\times \vec B}{\mu_0}$, $S\propto\frac{1}{r^2}$
Hence, answer is (C)
24. Given a uniform magnetic field $\vec B=B_0\hat k$ (where $B_0$ is a constant), a possible choice for the magnetic vector potential $\vec A$ is
1. $B_0y\hat i$
2. $-B_0y\hat i$
3. $B_0(x\hat j+y\hat i)$
4. $B_0(x\hat i-y\hat j)$
Using, $\vec B=\vec\nabla \times\vec A$, for options
(A)$\vec B=\vec\nabla \times B_0y\hat i=-B_0\hat k$
(B)$\vec B=\vec\nabla \times (-B_0y\hat i)=B_0\hat k$
(C)$\vec B=\vec\nabla \times (B_0(x\hat j+y\hat i))=0$
(D)$\vec B=\vec\nabla \times (B_0(x\hat i-y\hat j))=2B_0\hat k$
Hence, answer is (B)
25. Consider a charge $Q$ at the origin of 3-dimensional coordinate system. The flux of the electric field through the curved surface of a cone that has a height $h$ and a circular base of radius $R$ is
1. $\frac{Q}{\epsilon_0}$
2. $\frac{Q}{2\epsilon_0}$
3. $\frac{hQ}{R\epsilon_0}$
4. $\frac{QR}{2h\epsilon_0}$
Let us consider a sphere of radius of $h$ (if $h>R$) or of radius $R$ (if $R>h$) with centre at origin. According to Gauss law the flux through entire sphere is $\frac{Q}{\epsilon_0}$. The flux through curved surfaces of cone is half of the flux through sphere i.e. $\frac{Q}{2\epsilon_0}$
Hence, answer is (B)
26. A plane electromagnetic wave is propagating in a loss-less dielectric. The electric field is given by $${\scriptstyle\vec E(x,y,z,t)=E_0(\hat x+A\hat z)\exp{\left[ik_0\left\{-ct+\left(x+\sqrt{3}z\right)\right\}\right]}}$$ where $c$ is the speed of light in vacuum, $E_0$ , $A$ and $k_0$ are constants and $\hat x$ and $\hat z$ are unit vectors along the x- and z-, axes. The relative dielectric constant of the medium, $\epsilon_r$ and the constant $A$ are
1. $\epsilon_r=4$ and $A=-\frac{1}{\sqrt{3}}$
2. $\epsilon_r=4$ and $A=+\frac{1}{\sqrt{3}}$
3. $\epsilon_r=4$ and $A=\sqrt{3}$
4. $\epsilon_r=4$ and $A=-\sqrt{3}$
General form of wave equation is $\vec E(x,y,z,t)=E_0\hat ne^{i\left(\vec k\cdot\vec r-\omega t\right)}$. Comparing with this equation, we get $$\hat n=(\hat x+A\hat z)$$ \begin{align*} &\vec k\cdot\vec r=k_0\left(x+\sqrt{3}z\right)\\ &\Rightarrow\vec k=k_0\left(\hat x+\sqrt{3}\hat z\right) \end{align*} $$\omega=k_0c$$ $$v=\frac{\omega}{k}=\frac{k_0c}{k_0^2+3k_0^2}=\frac{c}{2}$$ $$\text{Refractive index } n=\frac{c}{v}=2$$ $$\text{Refractive index } n=\sqrt{\epsilon_2}\Rightarrow\epsilon_r=4$$ \begin{align*} &\text{Since }\vec k\cdot\hat n=0\\ &\Rightarrow k_0\left(\hat x+\sqrt{3}\hat z\right)\cdot(\hat x+A\hat z)\\ &=0 \end{align*} $$k_0\left(1+\sqrt{3}A\right)=0\Rightarrow A=-\frac{1}{\sqrt{3}}$$ Hence, answer is (A)
27. A static, spherically symmetric charge distribution is given by $\rho(r) = \frac{A}{r} e^{-Kr}$ where $A$ and $K$ are positive constants. The electrostatic potential corresponding to this charge distribution varies with $r$ as
1. $re^{-Kr}$
2. $\frac{1}{r} e^{-Kr}$
3. $\frac{1}{r^2} e^{-Kr}$
4. $\frac{1}{r} (1-e^{-Kr})$
Since $\nabla^2V=-\rho/\epsilon_0$
$\nabla^2V$ must be proportional to $\frac{1}{r} e^{-Kr}$. Taking $\nabla^2=\frac{d^2}{dr^2}+\frac{2}{r}\frac{d}{dr}$, for the option (B) we get $\nabla^2\left(\frac{1}{r} e^{-Kr}\right)=\frac{K^2}{r} e^{-Kr}$
Hence, option (B) is correct.
28. The frequency dependent dielectric constant of a material is given by $$\epsilon(\omega)=1+\frac{A}{\omega_0^2-\omega^2-i\omega\gamma}$$ where $A$ is a positive constant, $\omega_0$ the resonant frequency and $\gamma$ the damping coefficient. For an electromagnetic wave of angular frequency $\omega < < \omega_0$, which of the following is true? (Assume that $\frac{\gamma}{\omega_0} < < 1$.)
1. There is negligible absorption of the wave
2. The wave propagation is highly dispersive
3. There is strong absorption of the electromagnetic wave
4. The group velocity and the phase velocity will have opposite sign
\begin{align*} &\epsilon(\omega)=1+\frac{A}{\omega_0^2-\omega^2-i\omega\gamma}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2+i\omega\gamma\right)}{\left(\omega_0^2-\omega^2+i\omega\gamma\right)\left(\omega_0^2-\omega^2-i\omega\gamma\right)}}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2+i\omega\gamma\right)}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}}\\ &={\scriptstyle1+\frac{A\left(\omega_0^2-\omega^2\right)}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}+i\frac{A\omega\gamma}{\left(\omega_0^2-\omega^2\right)^2+\omega^2\gamma^2}}\\ \end{align*} Fro $\omega < < \omega_0$, imaginary part of $\epsilon(\omega)$ is negligibly small. Hence, there is negligible absorption of the wave.

Hence, answer is (A)

29. A plane electromagnetic wave is traveling along the positive z-direction. The maximum electric field along the x-direction is 10 V/m. The approximate maximum values of the power per unit area and the magnetic induction $B$, respectively, are
1. $3.3\times 10^{-7} watts/m^2$ and 10 tesla
2. $3.3\times 10^{-7} watts/m^2$ and $3.3\times 10^{-8} $ tesla
3. $0.265 watts/m^2$ and $10$ tesla
4. $0.265 watts/m^2$ and $3.3\times 10^{-8} $ tesla

The power per unit area transported by the electromagnetic waves is given \begin{align*} I&=c\epsilon_0E_0^2\\ &=3\times 10^8\times8.854\times10^{-12}\times10^2\\ &=0.265 watts/m^2 \end{align*}

Magnetic induction \begin{align*} B&=\frac{1}{c}E_0\\ &=\frac{1}{3\times10^{-8}}\times10\\ &=3.3\times 10^{-8} \: tesla \end{align*}

Hence, answer is (D)

30. Suppose the yz-plane forms a chargeless boundary between two media of permittivities $\epsilon_{left}$ and $\epsilon_{right}$ where $\epsilon_{left}:\epsilon_{right}=1:2$. If the uniform electric field on the left is $\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$ (where $c$ is constant), then the electric field on the right $\vec E_{right}$ is
1. $c\left(2\hat i+\hat j+\hat k\right)$
2. $c\left(\hat i+2\hat j+2\hat k\right)$
3. $c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$
4. $c\left(\hat i+\frac{1}{2}\hat j+\frac{1}{2}\hat k\right)$

Using relation $$E_{left}\epsilon_{left}\cos{\theta_1}=E_{right}\epsilon_{right}\cos{\theta_2}$$ where, $\theta_1$ is the angle between $\vec E_{left}$ and the vector normal to the yz-plane and $\theta_2$ is the angle between $\vec E_{right}$ and the vector normal to the yz-plane. The vector normal to yz-plane is $\hat i$

Using $\epsilon_{left}:\epsilon_{right}=1:2$ in above equation $$E_{left}\cos{\theta_1}=2E_{right}\cos{\theta_2}\quad--(1)$$ Magnitude of $\vec E_{left}$ is given by $$E_{left}=\sqrt{c^2+c^2+c^2}=\sqrt{3}c$$ Let us find angle $\cos{\theta_1}$ \begin{align*} \cos{\theta}&={\textstyle\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}}\\ &=\frac{c}{\sqrt{3c^2}\sqrt{1}}=\frac{1}{\sqrt{3}} \end{align*} Hence, equation (1) becomes $$\sqrt{3}c\times\frac{1}{\sqrt{3}}=2E_{right}\cos{\theta_2}$$ $$E_{right}\cos{\theta_2}=\frac{c}{2}$$ One cane verify that $E_{right}=c\left(\hat i+2\hat j+2\hat k\right)$ satisfies the above condition. For this vector find $E_{right}$ and $\cos\theta_2$.

Hence, answer is (B)

31. Which of the following transformations $\left(V,\vec A\right)\rightarrow \left(V',\vec A'\right)$ of electrostatic potential $V$ and the vector potential $\vec A$ is a gauge transformation?
1. $\left(V'=V+ax,\vec A'=\vec A+at\hat k\right)$
2. $\left(V'=V+ax,\vec A'=\vec A-at\hat k\right)$
3. $\left(V'=V+ax,\vec A'=\vec A+at\hat i\right)$
4. $\left(V'=V+ax,\vec A'=\vec A-at\hat i\right)$

The general gauge transformation of electrostatic potential $V$ and the vector potential $\vec A$ is given by the equations $$ V'=V-\frac{\partial f}{\partial t}$$ $$\vec A'=\vec A+\nabla f$$ In all options, we have, $V'=V+ax$. This implies $$-\frac{\partial f}{\partial t}=ax$$ Hence, $f=-atx$

Hence, $\nabla f=-at\hat i$

Hence, we have, $\left(V'=V+ax,\vec A'=\vec A-at\hat i\right)$

Hence, answer is (D)

32. Three polarizers P₁, P₂ and P₃ are arranged as shown in figure. Optical axis of P₁ and P₃ are perpendicular to one another while the axis of P₂ makes an angle $\theta$ with that of P₁. If a beam of ordinary light of intensity I₀ is incident normally on P₁, then the intensity of light emerging from P₃ is :
    
1. 0
2. $\frac{I_0}{8}\sin^2{2\theta}$
3. $\frac{I_0}{2}$
4. $\frac{I_0}{4}\sin^2{2\theta}$
According to Malus law when a beam of polarized light of intensity $I_0$ falls on the polarizer, the intensity of transmitted light is given by $$I=I_0\cos^2\theta$$ where, $\theta$ is the angle between the polarizer axis and polarization direction of light.

When ordinary light falls on the polarizer, then the intensity of light passed through polarizer is one half the intensity of unpolarized light

Hence, for polarizer $P_1$, we have $$I_1=\frac{I_0}{2}$$ For polarizer $P_2$, according to Malus law $$I_2=I_1\cos^2\theta=\frac{I_0}{2}\cos^2\theta$$ For polarizer $P_3$, according to Malus law \begin{align*} I_3&=\frac{I_2}{2}\cos^2{(\frac{\pi}{2}\theta)}\\ &=\frac{I_0}{2}\cos^2\theta\cos^2{(\frac{\pi}{2}\theta)}\\ &=\frac{I_0}{2}\cos^2\theta\sin^2{(\theta)}\\ &=\frac{I_0}{8}\left(2\cos\theta\sin{(\theta)}\right)^2\\ &=\frac{I_0}{8}\sin^2{2\theta} \end{align*} Hence, answer is (B)

33. A hollow metallic sphere of radius $a$, which is kept at a potential $V_0$, has a charge $Q$ at its centre. The potential at a point outside the sphere, at a distance $r$ from the centre, is
1. $V_0$
2. $\frac{Q}{4\pi\epsilon_0r}+\frac{V_0a}{r}$
3. $\frac{Q}{4\pi\epsilon_0r}+\frac{V_0a^2}{r^2}$
4. $\frac{V_0a}{r}$

The potential on metallic sphere is $V_0$. Let $Q_0$ be charge on the sphere. Hence, $$V_0=\frac{Q_0}{4\pi\epsilon_0 a}$$ $$Q_0=V_04\pi\epsilon_0 a$$ Hence, potential at a distance $r$ is $$V=\frac{Q_0}{4\pi\epsilon_0 r}=\frac{V_0a}{r}$$

Hence, answer is (D)

34. The anomalous dispersion of light observed in gases is well represented by the Cauchy's formula. The refractive index of the medium in such case is given as $n=1+A+\frac{AB}{\lambda^2}$. ($\lambda$ is wavelength of light in air)

    What does the terms $A$ and $B$ respectively represent?

1. Coefficients of refraction and dispersion
2. Coefficients of dispersion and refraction
3. Coefficients of dispersion and polarization
4. Coefficients of refraction and scattering

(A) Coefficients of refraction and dispersion

35. A linear quadrupole is formed by joining two dipoles each of mmt $\vec P$ back-to-back. The electric potential and field at point P far away from the quadrupole is found to vary respectively with distance as :
1. $1/r^5$ and $1/r^4$
2. $1/r^2$ and $1/r^3$
3. $1/r^4$ and $1/r^3$
4. $1/r^3$ and $1/r^4$

Answer is (D)

36. In an electromagnetic field, which one of the following remains invariant under Lorentz transformation?
1. $\vec E\times\vec B$
2. $E^2-c^2B^2$
3. $B^2$
4. $E^2$

The Lorentz transformation equations of electric and magnetic field are given by

$E_x'=E_x$	$B_x'=B_x$
${\scriptstyle E_y'=\gamma\left(E_y-vB_z\right)}$	${\scriptstyle B_y'=\gamma\left(B_y+\frac{v}{c^2}E_z\right)}$
${\scriptstyle E_z'=\gamma\left(E_z+vB_y\right)}$	${\scriptstyle B_z'=\gamma\left(B_z-\frac{v}{c^2}E_y\right)}$

\begin{align*} &E^{2'}-c^2B^{2'}= E_x^{2'}+E_y^{2'}+E_z^{2'}\\ &-c^2\left(B_x^{2'}+B_y^{2'}+B_z^{2'}\right)\\ &={\textstyle E_x^2+\left(\gamma\left(E_y-vB_z\right)\right)^2}\\ &{\textstyle +\left(\gamma\left(E_z+vB_y\right)\right)^2}\\ &-c^2{\scriptstyle \left(B_x^2+\left(\gamma\left(B_y+\frac{v}{c^2}E_z\right) \right)^2+\left(\gamma\left(B_z-\frac{v}{c^2}E_y\right)\right)^2 \right)}\\ &=E^2-c^2B^2 \end{align*}

Hence, answer is (B)

37. A sphere of radius $R$ has uniform volume charge density. The electric potential at a point $r(r < R)$ is
1. due to the charge inside a sphere of radius $r$ only
2. due to the entire charge of the sphere
3. due to the charge of the spherical shell of inner and outer radii $r$ and $R$, only
4. independent of $r$

For region $r < R$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ $$E4\pi r^2=\frac{\frac{4}{3}\pi r^3\rho}{\epsilon_0}$$ $$E=\frac{\rho}{3\epsilon_0}r$$ \begin{align*} V&=-\int\limits_0^r\frac{\rho}{3\epsilon_0}r\:dr\\ &=-\frac{\rho}{6\epsilon_0}r^2 \end{align*}

Hence, answer is (A)

38. Three point charges $q$, $q$ and $-2q$ are located at $(0,-a,a)$, $(O,a,a)$ and $(0,0,-a)$, respectively. The net dipole moment of this charge distribution is
1. $4qa\hat k$
2. $2qa\hat k$
3. $-4qa\hat i$
4. $-2qa\hat j$

If the system of charges is electrically neutral as whole, then the net dipole moment of a system of charges $q_1,q_2,\dots, q_n$ with position vectors $r_1,r_2,\dots, r_n$ is given by $$\vec P=\sum\limits_{i=1}^{n}q_i\vec r_i$$ \begin{align*} \vec P&=q\vec r_1+q\vec r_2-2q\vec r_3\\ &=q\left(-a\hat j+a\hat k\right)+q\left(a\hat j+a\hat k\right)-2q\left(-a\hat k\right)\\ &=4qa\hat k \end{align*}

Hence, answer is (A)

39. The vector potential in a region is given as $\vec A(x, y, z) = -y\hat i + 2x\hat j$. The associated magnetic induction $\vec B$ is
1. $\hat i+\hat k$
2. $3\hat k$
3. $-\hat i+2\hat j$
4. $-\hat i+\hat j+\hat k$

\begin{align*} \vec B&=\vec\nabla\times\vec A\\ &=\begin{vmatrix}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ -y&2x&0\end{vmatrix}\\ &=3\hat k \end{align*}

Hence, answer is (B)

40. A proton moves with a speed of 300m/s in a circular orbit in the xy-plane in a magnetic 1 tesla along the positive z-direction. When an electric field of 1 V/m is applied along the positive y-direction, the center of the circular orbit
1. remains stationary
2. moves at 1 m/s along the negative x-direction
3. moves at 1 m/s along the positive z-direction
4. moves at 1 m/s along the positive x-direction

The addition of an electric field perpendicular to a given magnetic field simply causes the particle to drift perpendicular to both the electric and magnetic field with the fixed velocity $$\vec v_{EB}=\frac{\vec E\times\vec B}{B^2}$$ In the present case, $\vec E=1\hat j$ and $\vec B=1\hat k$ $$\vec v_{EB}=\frac{\hat j\times\hat k}{1}=\hat i$$ Hence, center of circular moves at 1 m/s along the positive x-direction

Hence, answer is (D)

41. Suppose the yz-plane forms a chargeless boundary between two media of permittivities $\epsilon_{left}$ and $\epsilon_{right}$ where $\epsilon_{left}:\epsilon_{right}=1:2$. If the uniform electric field on the left is $\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$ (where $c$ is constant), then the electric field on the right $\vec E_{right}$ is
1. $c\left(2\hat i+\hat j+\hat k\right)$
2. $c\left(\hat i+2\hat j+2\hat k\right)$
3. $c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$
4. $c\left(\hat i+\frac{1}{2}\hat j+\frac{1}{2}\hat k\right)$

Here the electric fields in two media should satisfy the condition $$ E_{left}\epsilon_{left}\cos\theta_1= E_{right}\epsilon_{right}\cos\theta_2$$ where, $\theta_1$ is the angle between $\vec E_{left}$ and unit normal to yz-plane in left medium and $\theta_2$ is the angle between $\vec E_{right}$ and unit normal to yz-plane in right medium. $$\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$$ The unit vector normal to yz-plane in left medium is $\hat n=-\hat i$. Hence, angle between $\vec E_{left}$ and $\hat n$ is given by \begin{align*} \cos\theta_1&=\frac{\left|\vec E_{left}\cdot\hat n\right|}{\left|\vec E_{left}\right|\left|\hat n\right|}\\ &=\frac{c}{c\sqrt{3}\times 1}\\ \cos\theta_1&=\frac{1}{\sqrt{3}} \end{align*} \begin{align*} E_{right}\cos\theta_2&= E_{left}\frac{\epsilon_{left}}{\epsilon_{right}}\cos\theta_1\\ &=c\sqrt{3}\times\frac{1}{2}\times\frac{1}{\sqrt{3}}\\ E_{right}\cos\theta_2&=\frac{c}{2} \end{align*} The vector which satisfies above condition is $\vec E_{right}=c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$

Hence, answer is (C)

42. Which one of the following sets of Maxwell's equations for time independent charge density $\rho$ and current density $\vec J$ is correct?
1. \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=-\frac{\partial\vec B}{\partial t}\\ \vec\nabla\times\vec B&=\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}
2. \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}
3. \begin{align*} \vec\nabla\cdot\vec E&=0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}
4. \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=\mu_0|\vec J |\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}

Maxwell equations are given by \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=-\frac{\partial\vec B}{\partial t}\\ \vec\nabla\times\vec B&=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*} For time independent case $\frac{\partial\vec B}{\partial t}=0$ and $\frac{\partial\vec E}{\partial t}=0$. Hence, Maxwell equations become \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}

Hence, answer is (B)

43. A dipole of moment $\vec p$, oscillating at frequency $\omega$, radiates spherical waves. The vector potential at large distance is $$\vec A(\vec r)=\frac{\mu_0}{4\pi}i\omega\frac{e^{ikr}}{r}\vec p$$. To order $(1/r)$ the magnetic field $\vec B$ at a point $\vec r=r\hat n$ is
1. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\left(\hat n\cdot\vec p\right)\frac{e^{ikr}}{r}$
2. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\left(\hat n\times\vec p\right)\frac{e^{ikr}}{r}$
3. $-\frac{\mu_0}{4\pi}\omega^2k\left(\hat n\cdot\vec p\right)\vec p\frac{e^{ikr}}{r}$
4. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\vec p\frac{e^{ikr}}{r}$

Since $\vec B=\vec\nabla\times\vec A$, option (B) is correct

Hence, answer is (B)

44. A small magnetic needle is kept at $(0,0)$ with its moment along the x-axis. Another small magnetic needle is at the point $(1, 1)$ and is free to rotate in the $xy$-plane. In equilibrium the angle $\theta$ between their magnetic moments is such that
1. $\tan\theta=1/3$
2. $\tan\theta=0$
3. $\tan\theta=3$
4. $\tan\theta=1$

Potential energy of two dipole system is given by \begin{align*} U&=\frac{\mu_0}{4\pi r^3}\left[\vec m_1\cdot\vec m_2-3\vec m_1\cdot\vec r)(\vec m_2\cdot\vec r) \right]\\ &={\scriptstyle\frac{\mu_0m_1m_2}{4\pi r^3}\left[\cos\theta -3(\cos{45^0})(\cos{\left(\theta-45^0\right)}) \right]}\\ \end{align*} For equilibrium potential energy is minimum i.e. $\frac{\partial U}{\partial\theta}=0$ \begin{align*} \frac{\partial U}{\partial\theta}&={\scriptstyle\frac{\mu_0m_1m_2}{4\pi r^3}\left[\sin\theta +\frac{\sqrt{3}}{2}\sin{\left(\theta-45^0\right)} \right]}\\ &=0\\ \end{align*} $$\Rightarrow \tan\theta=3$$

Hence, answer is (C)

45. A beam of unpolarized light in a medium with dielectric constant $\epsilon_1$ is reflected from a plane interface formed with another medium of dielectric constant $\epsilon_2=3\epsilon_1$. The two media have identical magnetic permeability. If the angle of incidence is $60^0$, then the reflected light
1. is plane polarized perpendicular to the plane of incidence
2. is plane polarized parallel to the plane of incidence
3. is circularly polarized
4. has the same polarization as the incident light

This is a simple method (discovered by malus in 1808) of obtaining partially plane polarized light by reflection. The percentage of polarized light is greatest in reflected beam when light beam is incident on the transparent medium with an incident angle equal to the angle of polarization. The vibrations of this plane polarized reflected light are found to be perpendicular to the plane of incidence and therefore, the reflected light is said to be plane polarized in the plane of incidence, when the incident angle is equal to the Brewster angle $\theta_B$, where $\tan\theta_B = n_2/n_1$. \begin{align*} \tan\theta_B&=\frac{n_2}{n_1}\\ &=\sqrt{\frac{\epsilon_2}{\epsilon_1}}\\ \tan\theta_B&=\sqrt{3} \end{align*} Hence, $\theta_B=60^0$. As angle of incidence is equal to Brewster angle, reflected light is plane polarized perpendicular to the plane of incidence.

Hence, answer is (A)

46. An electric dipole of moment $\vec P$ and length $l$ aligned along the $z$-axis is used to generate electromagnetic waves. Initially it was operated at frequency 10 MHz and its power along the equatorial plane at certain distance $d (d >> l)$ was measured as $P_1$. Later, the same dipole was operated at frequency 40 MHz and power $P_2$ was measured at the same point. How do you compare the two powers ?
1. $P_2= 256 P_1$
2. $P_2= 64 P_1$
3. $P_2= 128 P_1$
4. $P_2= 16 P_1$

The average total power radiated by oscillating electric dipole is given by $$ P=\frac{\mu_0\omega^4p_0^2}{12\pi c}$$ Thus, $$ P\propto \omega^4$$ $$ P_1\propto (10)^4$$ $$ P_2\propto (40)^4$$ $$\frac{P_2}{P_1}=\left(\frac{40}{10}\right)^4$$ $$P_2= 256 P_1$$

Hence, answer is (A)

47. Consider a rectangular wave guide with transverse dimensions 2 m $\times$ 1 m driven with an angular frequency $\omega=10^9\: rad/s$. Which transverse electric (TE) modes will propagate in this wave guide?
1. $TE_{10}$, $TE_{01}$, and $TE_{20}$
2. $TE_{10}$, $TE_{11}$, and $TE_{20}$
3. $TE_{01}$, $TE_{10}$, and $TE_{11}$
4. $TE_{01}$, $TE_{10}$, and $TE_{22}$

Cut off frequency for rectangular waveguide is given by $$f_{c_{mn}}=\frac{v}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ where, $v=\frac{1}{\sqrt{\epsilon_0\mu_0}}\frac{1}{\sqrt{\epsilon_r}}=\frac{c}{\sqrt{\epsilon_r}}$ is velocity propagation in dielectric.

The cutoff frequency for $TE_{mn}$ mode is the operating frequency below which attenuation occurs and above which propagation takes place.

For air filled wave guide $v=c$ $$f_{c_{mn}}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ Now, $\omega=10^9\: rad/s$, hence, $f=\frac{\omega}{2\pi}=1.6\times10^8\:1/sec$ \begin{align*} f_{c_{22}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{2}{2}\right)^2+\left(\frac{2}{1}\right)^2}\\ &=3.35\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{20}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{2}{2}\right)^2+\left(\frac{0}{1}\right)^2}\\ &=1.5\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{11}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{1}\right)^2}\\ &=1.68\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{10}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{0}{1}\right)^2}\\ &=0.75\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{01}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{0}{2}\right)^2+\left(\frac{1}{1}\right)^2}\\ &=1.5\times10^8\:1/sec \end{align*} As, $f < f_{c_{22}}$ and $f < f_{c_{11}}$, these modes will attenuate and as $f < f_{c_{22}}$, $f > f_{c_{10}}$, $f > f_{c_{01}}$, these modes will propagate.

Hence, answer is (A)

48. The electric and magnetic fields in the charge free region $z > 0$ are given by $$\vec E(\vec r,t)=E_0e^{-k_1z}\cos{(k_2x-\omega t)}\hat j$$ $${\scriptstyle\vec B(\vec r,t)=\frac{E_0}{\omega}e^{-k_1z}\left[k_1\sin{(k_2x-\omega t)}\hat i+k_2\cos{(k_2x-\omega t)}\hat k\right]}$$ where $\omega$, $k_1$, and $k_2$ are positive constants. The average energy flow in $x$-direction is
1. $\frac{E_0^2k_2}{2\mu_0\omega}e^{-2k_1z}$
2. $\frac{E_0^2k_2}{\mu_0\omega}e^{-2k_1z}$
3. $\frac{E_0^2k_1}{2\mu_0\omega}e^{-2k_1z}$
4. $\frac{1}{2}c\epsilon_0E_0^2e^{-2k_1z}$

According to Poynting theorem the energy flux is given by $$\vec S=\frac{1}{\mu_0}\vec E\times \vec B$$ Hence, \begin{align*} S_x\hat i&=\frac{1}{\mu_0}\left(E_0e^{-k_1z}\cos{(k_2x-\omega t)}\right)\hat j\times\\ &~~~~\left(\frac{E_0}{\omega}e^{-k_1z}k_2\cos{(k_2x-\omega t)} \right)\hat k\\ &=\frac{E_0^2k_2}{\mu_0\omega}e^{-2k_1z}\cos^2{(k_2x-\omega t)}\hat i\\ \end{align*} $$\left < S_x \right > =\frac{E_0^2k_2}{2\mu_0\omega}e^{-2k_1z}$$

Hence, answer is (A)

49. A uniform magnetic field in the positive $z$-direction passes through a circular wire loop of radius 1 cm and resistance 1 $\omega$ lying in $xy$-plane. The field strength is reduced from 10 tesla to 9 tesla in 1 s. The charge transferred across any point in the wire is approximately
1. $3.1\times10^{-4}$ coulomb
2. $3.4\times10^{-4}$ coulomb
3. $4.2\times10^{-4}$ coulomb
4. $5.2\times10^{-4}$ coulomb

Magnetic flux through circular wire is \begin{align*} \phi&=\int B\:dA\\ &=B\pi r^2\\ &=B\times3.14\times1\times10^{-4}\\ &=3.14\times10^{-4}B\\ \end{align*} Change in magnetic flux per unit time is \begin{align*} \frac{d\phi}{dt}&=3.14\times10^{-4}(B_2-B_1)\\ &=3.14\times10^{-4}(10-9)\\ &=3.14\times10^{-4} \end{align*} Faraday's law of induction can now be used to determine the induced emf: $$\epsilon=-\frac{d\phi}{dt}=-3.14\times10^{-4}\:volts$$ Current $$I=\frac{\epsilon}{R}=-3.14\times10^{-4}\:Amp$$ Hence, charge transferred in one second at point is $3.14\times10^{-4}$.

Hence, answer is (A)

50. A dielectric sphere of radius $R$ carries polarization $\vec P = kr^2\hat r$, where $r$ is the distance from the centre and $k$ is a constant. In the spherical polar coordinate system, $\hat r$, $\hat \theta$ and $\hat \phi$ are the unit vectors.
1. The bound volume charge density inside the sphere at a distance $r$ from the centre is
1. $-4kR$
2. $-4kr$
3. $-4kr^2$
4. $-4kr^3$
2. The electric field inside the sphere at a distanced $d$ from the centre is
1. $\frac{-kd^2}{\epsilon_0}\hat r$
2. $\frac{-kR^2}{\epsilon_0}\hat r$
3. $\frac{-kd^2}{\epsilon_0}\hat\theta$
4. $\frac{-kR^2}{\epsilon_0}\hat\theta$

1. The unit vector $\hat n$ on the surface of the sphere is equal to the radial unit vector. The bound surface charge is equal to $\sigma_b=\vec P\cdot\hat n|_{r=R}$

   The bound volume charge is equal to \begin{align*} \rho_b&=-\vec\nabla\cdot\vec P\\ &=-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\:kr^2\right)\\ &=-4kr \end{align*}

   Hence, answer is (B)

2. The electric field inside the sphere at a distanced $d$ from the centre is \begin{align*} \vec E_{volume}&=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi d^3\rho_b}{d^2}\hat r\\ &=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi d^3(-4kd)}{d^2}\hat r\\ &=-\frac{4kd^2}{3\epsilon_0}\hat r \end{align*}

   Hence, answer is ()

51. The $x$- and $z$-components of a static magnetic field in a region are $B_x=B_0(x^2-y^2)$ and $B_z=0$, respectively. Which of the following solutions for its $y$-component is consistent with the Maxwell equations?
1. $B_y=B_0xy$
2. $B_y=-2B_0xy$
3. $B_y=B_0(x^2-y^2)$
4. $B_y=B_0(\frac{1}{3}x^3-xy^2)$

$\vec\nabla\cdot\vec B=0$ gives $$2B_0x+\frac{\partial B_y}{\partial y}=0$$ $$\frac{\partial B_y}{\partial y}=-2B_0x$$ $$B_y=-2B_0xy$$

Hence, answer is (B)

52. A magnetic field $\vec B$ is $B\hat z$ in the region $x > 0$ and zero elsewhere. A rectangular loop, in the $xy$-plane, of sides $l$ (along the $x$-direction) and $h$ (along the $y$-direction) is inserted into the $x > 0$ region from the $x < 0$ region at a constant velocity $\vec v =v \hat x$. Which of the following values of $l$ and $h$ will generate the largest EMF?
1. $l=8$, $h=3$
2. $l=4$, $h=6$
3. $l=6$, $h=4$
4. $l=12$, $h=2$

Let us consider an area element $\vec A=hx\hat z$ passing through magnetic field $\vec B=B\hat z$. Magnetic flux passing through area element is $$\Phi=\vec B\cdot\vec A=Bhx$$ Rate of change of magnetic flux $$\frac{d\Phi}{dt}=Bh\frac{dx}{dt}=Bhv$$ $$\epsilon=-\frac{d\Phi}{dt}=-Bhv$$ Hence, EMF $\epsilon$ will be largest for largest value of $h$

Hence, answer is (B)

53. Two parallel plate capacitors, separated by distances $x$ and $1.1x$ respectively, have a dielectric material of dielectric constant 3.0 inserted between the plates, and are connected to a battery of voltage $V$. The difference in charge on the second capacitor compared to the first is
1. +66%
2. +20%
3. -3.3%
4. -10%

Capacitance of a parallel plate capacitor is inversely proportional to the separation between the plates i.e. $C\propto\frac{1}{d}$ $$C_1\propto\frac{1}{x}$$ $$C_2\propto\frac{1}{1.1x}$$ Let us take area of plates such that $$C_1=\frac{1}{x}$$ $$C_2=\frac{1}{1.1x}$$ $$C_2-C_1=\frac{1}{1.1x}-\frac{1}{x}=-\frac{0.1}{1.1x}$$ $$\% difference=\frac{100\times\left(-\frac{0.1}{1.1x}\right)}{\frac{1}{1.1x}}$$ $$\% difference=-10\%$$

Hence, answer is (D)

54. Which of the following defines a conservative force field?
1. $\vec\nabla\cdot\vec F=0$
2. $\vec\nabla\times\vec F=0$
3. $\oint\vec F\cdot\vec dr=0$
4. $\frac{d\vec F}{dt}=0$

A force field $\vec F$, is called a conservative force or conservative vector field if it satisfies following three equivalent conditions: $$\vec\nabla\times\vec F=0$$ $$\oint\vec F\cdot\vec dr=0$$ $$\vec F=-\vec\nabla\Phi$$ where $\Phi$ is scalar potential.

Hence, answer is (B and C)

55. $\nabla\left(\frac{1}{|\vec r|}\right)$ is given by
1. $\frac{1}{r}\hat r$
2. $\frac{1}{r^3}(\hat i+\hat j+\hat k)$
3. $-\frac{\vec r}{r^3}$
4. $r(\hat i-\hat j-\hat k)$

\begin{align*} \nabla\left(\frac{1}{|\vec r|}\right)&={\scriptstyle \left(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial z}\right)\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)}\\ &={\scriptstyle -\left(\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\right)\left(x\hat i+y\hat j+z\hat k\right)}\\ &=-\frac{\vec r}{r^3}\\ \end{align*}

Hence, answer is (C)

56. A plane polarized EM wave of frequency $\omega$ is incident at an angle $\theta$ in a rectangular waveguide of resonant frequency $\omega_{mn}$. Then energy carried by the wave propagating inside the cavity will propagate with the group velocity of :
1. $\frac{c}{\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}}$
2. $c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)}$
3. $\frac{c}{\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)}}$
4. $c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}$

$$v_g=c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}$$

Hence, answer is (D)

57. The electric field of an electromagnetic wave propagating in the free space is given by : $\vec E(r,t)=E_0\hat z\cos{\left[200\sqrt{3}\pi x-200\pi y-\omega t\right]}$. Then the wave vector $\vec k$ is given by
1. $200\frac{\sqrt{3}}{2}\pi\hat x-200\pi\hat y$
2. $400\pi\left[\frac{\sqrt{3}}{2}\hat x-\frac{1}{2}\hat y\right]$
3. $200\sqrt{3}\pi\hat x$
4. $-200\pi\hat y$

General format of electromagnetic wave equation is $$\vec E(r,t)=E_0\hat z\cos{(\vec k\cdot\vec r-\omega t)}$$ $$\vec k\cdot\vec r= 200\sqrt{3}\pi x-200\pi y$$ $$\Rightarrow \vec k=400\pi\left[\frac{\sqrt{3}}{2}\hat x-\frac{1}{2}\hat y\right]$$

Hence, answer is (B)

58. The Ampere's law in the free space takes the form:
1. $\vec\nabla\times\vec B=\mu_0\vec J$
2. $\vec\nabla\times\vec B=\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$
3. $\vec\nabla\times\vec B=\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$
4. $\vec\nabla\times\vec B=\mu_0\vec J-\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$

$\vec\nabla\times\vec B=\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$

Hence, answer is (B)

59. An electric charge $+Q$ is placed at the center of a cube of sides 10 cm. The electric flux emanating from each of the face of the cube is :
1. $\frac{Q}{\epsilon_0}$
2. $\frac{Q}{10\epsilon_0}$
3. $\frac{Q}{6\epsilon_0}$
4. $\frac{10Q}{\epsilon_0}$

According to Gauss’ Theorem, the total electric flux through all the six faces of the cube is, $$\Phi = \frac{Q}{\epsilon_0}$$ Each face of the cube comprises $1/6$ of the total surface of the cube. Therefore, the flux passing through each face is, $$\Phi = \frac{Q}{6\epsilon_0}$$

Hence, answer is (C)

60. A field at certain point in the space is expressed as the potential function $V=3x^2z-xy^3+z$. Then the potential $V$ at point $(2,-1,1)$ is :
1. 15 V
2. 13 V
3. 0 V
4. 8 V

$$V=32^2\times1-2\times(-1)^3+1=15 V$$

Hence, answer is (A)

61. Which of the following equations signifies the conservative nature of the electric field $\vec E$?
1. $\vec\nabla\cdot\vec E(\vec r)=\frac{\rho(\vec r)}{\epsilon_0}$
2. $\vec\nabla\times\vec E(\vec r)=\vec 0$
3. $\vec\nabla\times\vec E(\vec r,t)=\frac{-\partial\vec B(\vec r,t)}{\partial t}$
4. $\epsilon_0\mu_0\frac{\partial\vec E(\vec r,t)}{\partial t}=\vec\nabla\times\vec B(\vec r,t)-\mu_0\vec J(\vec r,t)$

A field is said to be conservative if it can be expressed as a gradient of scalar potential $V$. The equation $\vec\nabla\times\vec E(\vec r)=\vec 0$ implies that $\vec E(\vec r)$ can be expressed as $\vec E(\vec r)=-\vec\nabla V$.

Hence, answer is (B)

62. Plane electromagnetic wave is propagating through a perfect dielectric material of refractive index $\frac{3}{2}$. The phase difference between the fields $\vec E$ and $\vec B$ associated with the wave passing through the material is
1. Zero
2. $\pi$
3. $\frac{3}{2}\pi$
4. any non-zero value between $-\pi$ and $\pi$

For a perfect dielectric the phase difference between $\vec E$ and $\vec B$ is zero.

Hence, answer is (A)

63. An electromagnetic wave is propagating in a dielectric medium of permittivity $\epsilon$ and permeability $\mu$ having an electric field vector $\vec E$ associated with the wave. The associated magnetic field $\vec H$ is
1. Parallel to $\vec E$ with magnitude $E\sqrt{\mu/\epsilon}$
2. Parallel to $\vec E$ with magnitude $E\sqrt{\epsilon/\mu}$
3. Perpendicular to $\vec E$ with magnitude $E\sqrt{\mu/\epsilon}$
4. Perpendicular to $\vec E$ with magnitude $E\sqrt{\epsilon/\mu}$

Perpendicular to $\vec E$ with magnitude $E\sqrt{\epsilon/\mu}$

Hence, answer is (D)

64. Power radiated by a point charge moving with constant acceleration of magnitude $\alpha$ is proportional to
1. $\alpha$
2. $\alpha^2$
3. $\alpha^{-1}$
4. $\alpha^{-2}$

The power radiated by a point charge is given by Larmor formula as $$P=\frac{\mu_0q^2\alpha^2}{6\pi c} $$

Hence, answer is (B)