Question 1 |

The magnetic field of a uniform plane wave in vacuum is given by

\vec{H}(x,y,z,t)=(\hat{a}_x+2\hat{a}_y+b\hat{a}_z) \cos (\omega t+3x-y-z)

The value of b is _____

\vec{H}(x,y,z,t)=(\hat{a}_x+2\hat{a}_y+b\hat{a}_z) \cos (\omega t+3x-y-z)

The value of b is _____

0 | |

1 | |

-1 | |

-2 |

Question 1 Explanation:

For uniform plane wave

\hat{a}_{H}\cdot \hat{a}_{\rho }=0

\hat{a}_{H}is unit vector in magnetic field direction \hat{a}_{\rho } is unit vector in power flow direction

\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}

\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}

\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0

-3+2+b=0

b=1

\hat{a}_{H}\cdot \hat{a}_{\rho }=0

\hat{a}_{H}is unit vector in magnetic field direction \hat{a}_{\rho } is unit vector in power flow direction

\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}

\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}

\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0

-3+2+b=0

b=1

Question 2 |

A uniform plane wave traveling in free space and having the electric field

\vec{E}=(\sqrt{2}\hat{a}_{x}-\hat{a}_{z})cos[6\sqrt{3}\pi \times 10^{8}t-2\pi (x+\sqrt{2}z)]V/m

is incident on a dielectric medium (relative permittivity \gt1, relative permeability = 1) as shown in the figure and there is no reflected wave.

The relative permittivity (correct to two decimal places) of the dielectric medium is___________.

\vec{E}=(\sqrt{2}\hat{a}_{x}-\hat{a}_{z})cos[6\sqrt{3}\pi \times 10^{8}t-2\pi (x+\sqrt{2}z)]V/m

is incident on a dielectric medium (relative permittivity \gt1, relative permeability = 1) as shown in the figure and there is no reflected wave.

The relative permittivity (correct to two decimal places) of the dielectric medium is___________.

2 | |

2.9 | |

1 | |

3 |

Question 2 Explanation:

The wave has E_{i} direction and propagation direction both in same plane (ZX).

The wave is plane of incidence (P) polarized.

\begin{aligned} \beta_{x} &=1, \quad \beta_{2}=\sqrt{2} \\ \tan \theta_{i} &=\frac{\beta_{z}}{\beta_{x}}=\sqrt{2} \end{aligned}

No reflection in P polarized wave under Brewster's angle of incidence.

\begin{aligned} \theta_{i} &=\theta_{B} \\ \tan \theta_{B} &=\sqrt{2}=\sqrt{\epsilon_{r}} \\ \epsilon_{r} &=2 \end{aligned}

The wave is plane of incidence (P) polarized.

\begin{aligned} \beta_{x} &=1, \quad \beta_{2}=\sqrt{2} \\ \tan \theta_{i} &=\frac{\beta_{z}}{\beta_{x}}=\sqrt{2} \end{aligned}

No reflection in P polarized wave under Brewster's angle of incidence.

\begin{aligned} \theta_{i} &=\theta_{B} \\ \tan \theta_{B} &=\sqrt{2}=\sqrt{\epsilon_{r}} \\ \epsilon_{r} &=2 \end{aligned}

Question 3 |

The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.1 m
so that the amplitude of the wave attenuates by 20 dB, is

0.12 | |

0.23 | |

0.46 | |

2.3 |

Question 3 Explanation:

Attenuation constant,

\begin{aligned} \alpha &=\frac{1}{\text { skin depth }}=10 \mathrm{Np} / \mathrm{m} \\ \text { 20log }_{10}\left(\frac{E_{0}}{E_{x}}\right) &=20 \mathrm{dB} \\ \frac{E_{o}}{E_{x}} &=10 \Rightarrow E_{x}=\frac{E_{0}}{10} \\ E_{x} &=E_{0} e^{-\alpha x}=E_{0} e^{-10 x}=\frac{E_{0}}{10} \\ e^{-10 x} &=\frac{1}{10} \\ x &=\frac{1}{10} \ln (10)=0.23 \mathrm{m} \end{aligned}

\begin{aligned} \alpha &=\frac{1}{\text { skin depth }}=10 \mathrm{Np} / \mathrm{m} \\ \text { 20log }_{10}\left(\frac{E_{0}}{E_{x}}\right) &=20 \mathrm{dB} \\ \frac{E_{o}}{E_{x}} &=10 \Rightarrow E_{x}=\frac{E_{0}}{10} \\ E_{x} &=E_{0} e^{-\alpha x}=E_{0} e^{-10 x}=\frac{E_{0}}{10} \\ e^{-10 x} &=\frac{1}{10} \\ x &=\frac{1}{10} \ln (10)=0.23 \mathrm{m} \end{aligned}

Question 4 |

The permittivity of water at optical frequencies is 1.75 \varepsilon_{0}. It is found that an isotropic light source at a distance d under water forms an illuminated circular area of radius 5m, as shown in the figure. The critical angle is \theta_{c}.

The value of d (in meter) is _____________

The value of d (in meter) is _____________

1.3 | |

2.8 | |

4.3 | |

5.8 |

Question 4 Explanation:

Critical angle,

\begin{aligned} \theta_{c} &=\sin ^{-1} \sqrt{\frac{\epsilon_{2}}{\epsilon_{1}}} \\ &=\sin ^{-1} \sqrt{\frac{\epsilon_{0}}{1.75 \epsilon_{0}}}=49.1^{\circ} \end{aligned}

\begin{aligned} \tan 49.1 &=1.547=\frac{5}{d} \\ \Rightarrow \quad d &=\frac{5}{1.1547}=4.33 \mathrm{m} \end{aligned}

\begin{aligned} \theta_{c} &=\sin ^{-1} \sqrt{\frac{\epsilon_{2}}{\epsilon_{1}}} \\ &=\sin ^{-1} \sqrt{\frac{\epsilon_{0}}{1.75 \epsilon_{0}}}=49.1^{\circ} \end{aligned}

\begin{aligned} \tan 49.1 &=1.547=\frac{5}{d} \\ \Rightarrow \quad d &=\frac{5}{1.1547}=4.33 \mathrm{m} \end{aligned}

Question 5 |

The expression for an electric field in free space is E=E_{0}=(\hat{x}+\hat{y}+j2\hat{z})e^{-j(\omega t-kx+ky)}, where x, y,
z represent the spatial coordinates, t represents time, and \omega, k are constants. This electric field

does not represent a plane wave | |

represents a circular polarized plane wave propagating normal to the z-axis | |

represents an elliptically polarized plane wave propagating along x-y plane. | |

represents a linearly polarized plane wave |

Question 5 Explanation:

\vec{E} field direction \Rightarrow\left(\hat{a}_{x}+\hat{a}_{y}+j 2 \hat{a}_{z}\right)

Propagation direction \Rightarrow k \hat{a}_{x}-k \hat{a}_{y}

\vec{E} is perpendicular to propagation

\vec{E} \cdot \vec{P}=0

Component in \hat{a}_{z} has magnitude of 2.

Component in X-Y plane has magnitude of \sqrt{2}.

These two components are out of phase by 90^{\circ} and have unequal amplitudes. So, it is elliptically polarized wave.

Propagation direction \Rightarrow k \hat{a}_{x}-k \hat{a}_{y}

\vec{E} is perpendicular to propagation

\vec{E} \cdot \vec{P}=0

Component in \hat{a}_{z} has magnitude of 2.

Component in X-Y plane has magnitude of \sqrt{2}.

These two components are out of phase by 90^{\circ} and have unequal amplitudes. So, it is elliptically polarized wave.

Question 6 |

Faraday's law of electromagnetic induction is mathematically described by which one of the following equations?

\bigtriangledown \cdot \vec{B}=0 | |

\bigtriangledown \cdot \vec{D}=\rho _{v} | |

\bigtriangledown \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} | |

\bigtriangledown \times \vec{H}=\sigma \vec{E}+\frac{\partial \vec{D}}{\partial t} |

Question 6 Explanation:

Rate of change of magnetic field results in induced voltage,

\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}

\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}

Question 7 |

If a right-handed circularly polarized wave is incident normally on a plane perfect conductor, then the reflected wave will be

right-handed circularly polarized | |

left-handed circularly polarized | |

elliptically polarized with a tilt angle of 45^{\circ} | |

horizontally polarized |

Question 7 Explanation:

Left circularly polarized, due to direction change
after the reflection from conductor.

Question 8 |

Let the electric field vector of a plane electromagnetic wave propagating in a homogenous medium be expressed as E=\hat{x}E_{x}e^{-j(\omega t-\beta z)}, where the propagation constant \beta is a function of the angular frequency \omega. Assume that \beta (\omega)
and E_{x} are known and are real. From the information available, which one of the following CANNOT be determined?

The type of polarization of the wave. | |

The group velocity of the wave | |

The phase velocity of the wave. | |

The power flux through the z = 0 plane. |

Question 8 Explanation:

\beta(\omega) is known so V_{g} can be calculated.

v_{p}=\omega / \beta can be calculated.

Polarization can be identified.

\mu_{r} and \epsilon_{r} cannot be found, due to which power flux cannot be calculated as power flux

P=\frac{1}{2} \frac{|E|^{2}}{\eta}, \text{ where }\eta=120 \pi \times \sqrt{\frac{\mu_{r}}{\epsilon_{r}}}

v_{p}=\omega / \beta can be calculated.

Polarization can be identified.

\mu_{r} and \epsilon_{r} cannot be found, due to which power flux cannot be calculated as power flux

P=\frac{1}{2} \frac{|E|^{2}}{\eta}, \text{ where }\eta=120 \pi \times \sqrt{\frac{\mu_{r}}{\epsilon_{r}}}

Question 9 |

The electric field of a plane wave propagating in a lossless non-magnetic medium is given by the following expression

E(z,t)=a_{x}5 cos(2\pi \times 10^{9}t+\beta z)+ a_{y} 3 cos(2\pi \times 10^{9}t+\beta z-\pi/2)

The type of the polarization is

E(z,t)=a_{x}5 cos(2\pi \times 10^{9}t+\beta z)+ a_{y} 3 cos(2\pi \times 10^{9}t+\beta z-\pi/2)

The type of the polarization is

Right Hand Circular. | |

Left Hand Elliptical | |

Right Hand Elliptical | |

Linear. |

Question 9 Explanation:

\vec{E}(z, t)=\hat{a}_{x} 5 \cos \left(2 \pi \times 10^{9} t+\beta z\right)

+\hat{a}_{y} 3 \cos \left(2 \pi \times 10^{9} t+\beta z-\frac{\pi}{2}\right)

\Rightarrow Wave is travelling in -\hat{a}_{z} direction.

\Rightarrow Wave has orthogonal components with unequal amplitudes and checking the time trace.

\therefore Wave is left hand elliptically polarized.

+\hat{a}_{y} 3 \cos \left(2 \pi \times 10^{9} t+\beta z-\frac{\pi}{2}\right)

\Rightarrow Wave is travelling in -\hat{a}_{z} direction.

\Rightarrow Wave has orthogonal components with unequal amplitudes and checking the time trace.

\therefore Wave is left hand elliptically polarized.

Question 10 |

The electric field of a uniform plane electromagnetic wave is

\vec{E}=(\vec{a}_{x}+j4\vec{a}_{y})exp[j(2\pi \times 10^{7}t-0.2z)]

The polarization of the wave is

\vec{E}=(\vec{a}_{x}+j4\vec{a}_{y})exp[j(2\pi \times 10^{7}t-0.2z)]

The polarization of the wave is

right handed circular | |

left handed circular | |

right handed elliptical | |

left handed elliptical |

Question 10 Explanation:

\begin{aligned} \vec{E}(z, t)=& \hat{a}_{x} 5 \cos \left(2 \pi \times 10^{9} t+\beta z\right) \\ &+\hat{a}_{y} 3 \cos \left(2 \pi \times 10^{9} t+\beta z-\frac{\pi}{2}\right) \end{aligned}

\Rightarrow Wave is travelling in -\hat{a}_{z} direction.

\Rightarrow Wave has orthogonal components with unequal amplitudes and checking the time trace.

\therefore Wave is left hand elliptically polarized.

\Rightarrow Wave is travelling in -\hat{a}_{z} direction.

\Rightarrow Wave has orthogonal components with unequal amplitudes and checking the time trace.

\therefore Wave is left hand elliptically polarized.

There are 10 questions to complete.