Basics 1

6 Questions with standard solutions
- Lecture1.1
  
  Basics [Preview]
Electrostatics 3

28 Questions with standard solutions
- Lecture2.1
  
  Electrostatics: Set-1 [Preview]
- Lecture2.2
  
  Electrostatics: Set-2
- Lecture2.3
  
  Electrostatics: Set-3
Magnetic Fields 1

9 Questions with standard solutions
- Lecture3.1
  
  Magnetic Fields
Uniform Plane Waves 6

65 Questions with standard solutions
- Lecture4.1
  
  Uniform Plane Waves: Set-1 [Preview]
- Lecture4.2
  
  Uniform Plane Waves: Set-2
- Lecture4.3
  
  Uniform Plane Waves: Set-3
- Lecture4.4
  
  Uniform Plane Waves: Set-4
- Lecture4.5
  
  Uniform Plane Waves: Set-5
- Lecture4.6
  
  Uniform Plane Waves: Set-6
Transmission Lines 6

56 Questions with standard solutions
- Lecture5.1
  
  Transmission Lines: Set-1
- Lecture5.2
  
  Transmission Lines: Set-2
- Lecture5.3
  
  Transmission Lines: Set-3
- Lecture5.4
  
  Transmission Lines: Set-4
- Lecture5.5
  
  Transmission Lines: Set-5
- Lecture5.6
  
  Transmission Lines: Set-6
Waveguides 4

35 Questions with standard solutions
- Lecture6.1
  
  Waveguides: Set-1
- Lecture6.2
  
  Waveguides: Set-2
- Lecture6.3
  
  Waveguides: Set-3
- Lecture6.4
  
  Waveguides: Set-4
Antennas 4

42 Questions with standard solutions
- Lecture7.1
  
  Antennas: Set-1
- Lecture7.2
  
  Antennas: Set-2
- Lecture7.3
  
  Antennas: Set-3
- Lecture7.4
  
  Antennas: Set-4

Uniform Plane Waves: Set-1 [Preview]

Q. The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permitivity $\varepsilon_r$ and relative permeability $\mu_r=1$ are given by

$\overrightarrow E=E_pe^{j\left(\omega t-280\pi y\right)}{\widehat u}_zV/m$

$\overrightarrow H=3e^{j\left(\omega t-280\pi y\right)}{\widehat u}_xA/m$

Assuming the speed of light in free space to be $3\times10^8m/s$ ,intrinsic impedence of free space to be $120\pi$ ,the relative permitivity $\varepsilon_r$ ,of the medium and the electric field amplitude $E_r$ are [2011 2M]

a) $\varepsilon_r=3,\;E_p=120\pi$

b) $\varepsilon_r=3,\;E_p=360\pi$

c) $\varepsilon_r=9,\;E_p=360\pi$

d) $\varepsilon_r=9,\;E_p=120\pi$

Ans-(d)

Exp-

$\begin{array}{l}\overrightarrow E=E_pe^{j\left(\omega t-280\pi y\right)}{\widehat u}_z\;V/m\\\\\overrightarrow H=3\;e^{j\left(\omega t-280\pi y\right)}\;{\widehat u}_x\;A/m\\\\c=3\times10^8\;m/s\end{array}$

Comparing the given equation with standard form we have,

$\beta=280\pi=\frac{2\pi}\lambda$

$\begin{array}{l}\therefore\;\;\lambda=\frac1{140}meter\\\\\;\;\;\;\;v=f\lambda\\\\\;\;\;\;\;\;\;=14\times10^9\times\frac1{140}m/sec\end{array}$

$v=1\times10^8m/sec$

$v=\frac C{\sqrt{\in_r\mu_r}}$

$1\times10^8=\frac{3\times10^8}{\sqrt{1\times\in_r}}$

$\therefore\;\in_r=9$

$\frac{E_P}{H_P}=\eta=\sqrt{\frac\mu\in}=\sqrt{\frac{\mu_0\times1}{\in_0\times9}}=\frac13=\sqrt{\frac{\mu_0}{\in_0}}$

$\Rightarrow\frac{E_P}3=\frac13\times120\pi$

$\therefore\;E_P=120\pi$

Q. For an electromagnetic wave incident from one medium to a second medium, total reflection takes place when [1987 2M]

(a) The angle of incidence is equal to the Brewster angle with E field perpendicular to the plane of incidence

(b) The angle of incidence is equal to the Brewster angle with E field parallel to the plane of incidence

(d) The angle of incidence is equal to the critical angle with the wave moving from a rarer medium to a denser medium

Ans : (c)

Total internal reflection is the special kind of refraction of the wave in which the refracted ray come back to the incident medium.

The conditions for the TIR are as

The wave must be moving from denser to rare medium
The angle of incidence must be greater than certain minimum angle called as critical angle.

Q. In a good conductor the phase relation between the tangential components of electric field E_t and the magnetic field H_t is as follows [1988 2M]

(a) $E_t\;and\;H_t\;are\;in\;phase$

(b) $E_t\;and\;H_t\;are\;out\;of\;\;phase\;$

(d) $\;E_t\;leads\;H_t\;by\;45^0$

Ans : (d)

$\;\eta=intrinsic\;impedance$

$=\frac{E_t}{H_t}=\sqrt{\frac{j\omega\mu}{\sigma+j\omega\in}}$

For a good conductor

$\sigma>>\omega\in$

$\frac{E_t}{H_t}=\sqrt{\frac{j\omega\mu}\sigma}=\sqrt{\frac{\omega\mu}\sigma}e^{j\pi/4}$

$E_t\;\;leads\;H_t\;by\;an\;angle\;of\;45^0$

Q. The skin depth of copper at a frequency of 3 GH_z, is 1 micron (10^-6 meter). At 12 GH_z, For a non-magnetic conductor whose conductivity is 1/9 times that of copper, the skin depth would be [1989 2M]

(a) $\sqrt{9\times4}microns$

(b) $\sqrt{9/4}microns$

(d) $1/\sqrt{9\times4}microns$

Ans : (b)

For good conductors,

$Skin\;depth=\delta=\frac1{\sqrt{\pi\;f\;\mu\;\sigma}}$

$\delta\propto\frac1{\sqrt{f\sigma}}$

$\frac{\delta_2}{\delta_1}=\sqrt{\frac{f_1\sigma_1}{f_2\sigma_2}}$

$\frac{\delta_2}1=\sqrt{\frac3{12}\times\frac91}=\sqrt{\frac94}$

$\delta_2=\sqrt{\frac94}microns$

Q. The incoming solar radiation at a place on the surface of the earth is 1.2 KW/m². The amplitude of the elecrtic field corresponding to this incident power is nearly equal to [1990 2M]

(a) 80 mV/m

(b) 2.5 V/m

(d) 950 V/m

Ans : (d)

$P=\frac{E^2}{2\eta}$

$E=\sqrt{2\eta P}$

$\eta=\eta_0=120\pi$

$P=1.2kW/m^2=1200W/m^2$

$E=\sqrt{2\times\left(120\pi\right)\times\left(1200\right)}$

$E\cong950V/m$

Q. The electric field component of a uniform plane electromagnetic wave propagating in the Y-direction in lossless medium will satisfy the equation [1991 2M]

(a) $\frac{\partial^2E_y}{\partial y^2}=\mu\in\frac{\partial^2E_y}{\partial t^2}$

(b) $\frac{\partial^2E_y}{\partial x^2}=\mu\in\frac{\partial^2E_y}{\partial t^2}$

(d) $\frac{\sqrt{E_x^2+E_z^2}}{\sqrt{H_x^2+H_z^2}}=\sqrt{\mu/\in}$

Ans : (c) and (d)

As EM wave is propagating in Y- direction. $E_y=H_y=0$

So, $\frac{\partial^2E_x}{\partial y^2}=\mu\in\frac{\partial^2E_x}{\partial t^2}$

$\frac{E_z}{H_x}=\eta$

$\frac{E_x}{H_z}=-\eta$

$\frac EH=\frac{\sqrt{E_x^2+E_z^2}}{\sqrt{H_x^2+H_z^2}}=\eta=\sqrt{\frac\mu\in}$

Q. A material is described by the following electrical parameters at a frequency of

$10GHz:\sigma=10^6mho/m$ $\mu=\mu_0\;and\;\in/\in_0=-10$ .

The material at this frequency is considered to be $(\in_0=\frac1{36\pi}\times10^{-9}F/m)$ [1993 2M]

(a) a good conductor

(b) a good dielectric

(d) a good magnetic material

Ans : (a)

Let us find the value of loss tangent.

$\frac\sigma{\omega\in}=\frac{10^6}{2\pi\left(10\times10^9\right)\left(8.854\times10^{-12}\times10\right)}$

$\frac\sigma{\omega\in}=1.798\times10^5>>1$

So, it is a good conductor as loss tangent is very much greater than 1.

Q. A plane wave is incident normally on a perfect conductor as shown in figure. Here $E_x^{i\;},\;H_y^i\;and\;\;\overrightarrow P^i$ are electric field, magnetic field and poynting vector, respectively, for the incident wave. The reflected wave should have [1993 2M]

(a) $E_x^r=-E_x^i$

(b) $H_y^r=-H_y^i$

(d) $E_x^r=E_x^i$

Ans : (a) and (c)

According to the Boundary Conditions for perfect conductor 1) tangential components of electric field are equal and opposite to have E_t equal to 0 outside the surface and 2) tangential component of magnetic field intensity are continuous.

$E_x^i=-E_x^r\;\;\;\;and\;\;\;\;H_y^i=\;H_y^r$

$\overrightarrow P=\overrightarrow E\times\overrightarrow H$

So the direction for reflected wave is reversed as-

$-E_x^r\times H_y^r=-P_z^r$

Q. A plane electromagnetic wave travelling along the +z direction, has electric field given by

$E_x=2\cos\left(\omega t\right)\;and\;E_y=2\cos\left(\omega t+90\right)^0$

the wave is [1994 1M]

(a) linearly polarized

(b) right circularly polarized

(d) elliptically polarized

Ans : (c)

We know that if the EM wave has two E components with the phase difference 90^o then the polarization appear to be circular.

Consider the direction of propagation and coordinate axes as above. Then tip of Electric field is rotating anticlockwise or left circularly as shown in table below.