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Physics Answer the following in Brief

Electromagnetic Waves and Communication System · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 45 questions.

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Answer the following in Brief

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Question 13 Marks
A transmitting antenna at the top of a tower has a height $32\ m$ and that of the receiving antenna is $50\ m$. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is $6.4 \times 10^6 m$.
Answer
Given: $h_t=32 m , h_r=50 m , R =6.4 \times 10^6 m$
To find: Maximum distance or range $(d)$
Formula: $d=\sqrt{2 R h}$
Calculation: From formula,
$ d_t=\sqrt{2 R h_t}=\sqrt{2 \times 6.4 \times 10^6 \times 32}$
$=20.238 \times 10^3 m$
$=20.238 km$
$d_r=\sqrt{2 R h_t}$
$=\sqrt{2 \times 6.4 \times 10^6 \times 50}$
$=25.298 \times 10^3 m$
$=25.298 km$
$\text { Now, } d=d_t+d_r$
$=20.238+25.298$
$=45.536\ km $
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Question 23 Marks
Height of a TV tower is $600\ m$ at a given place. Calculate its coverage range if the radius of the Earth is $6400\ km.$ What should be the height to get the double coverage area?
Answer
Given: $h=600\ m , R =6.4 \times 10^6\ m$
To find: Range $(d)$
Height to get the double coverage $\left(h^{\prime}\right)$
Formula: $d=\sqrt{2 h R}$
Calculation: From formula,
$d=\sqrt{2 \times 600 \times 6.4 \times 10^6}=87.6 \times 10^3=87.6 km$
Now, for $A^{\prime}=2 A$
$ \pi\left(d^{\prime}\right)^2=2\left(\pi d^2\right)$
$\therefore\left(d^{\prime}\right)^2=2 d^2 $
From formula,
$ h ^{\prime}=\frac{\left(d^{\prime}\right)^2}{2 R}$
$=\frac{2 d^2}{2 R}$
$=2 \times h \ldots \ldots \ldots\left(\because h =\frac{d^2}{2 R}\right)$
$=2 \times 600$
$=1200 m $
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Question 33 Marks
A TV tower has a height of $200\ m$ . How much population is covered by TV transmission if the average population density around the tower is $1000 / km ^2$ ? (Radius of the Earth $=6.4 \times 10^6 m$ )
Answer
Given: $h=200 m$,
Population density $(n)$
$=1000 / km ^2=1000 \times 10^{-6} / m ^2=10^{-3} / m ^2$
$R=6.4 \times 10^6 m $
To find: Population covered
Formulae: i. $A=\pi d^2=\pi(\sqrt{2 R h})^2=2 \pi R h$
ii. Population covered $= nA$
Calculation /From formula (i),
$A=2 \pi R h$
$=2 \times 3.142 \times 6.4 \times 10^6 \times 200$
$\approx 8 \times 10^9 m ^2$
From formula (ii),
Population covered $= nA$
$=10^{-3} \times 8 \times 10^9$
$=8 \times 10^6$
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Question 43 Marks
The amplitude of the magnetic field part of a harmonic $EM$ wave in vacuum is $B0 = 5 \times 10^{-7}\  T.$ What is the amplitude of the electric field part of the wave?
Answer
Given: $B_0=5 \times 10^{-7} T , c =3 \times 10^8$
To find: Amplitude of electric field $\left(E_0\right)$
Formula: $c =\frac{E_0}{B_0}$
Calculation /From formula,
$E_0=c \times B_0$
$=3 \times 10^8 \times 5 \times 10^{-7}$
$=150\ V / m$
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Question 53 Marks
In a EM wave the electric field oscillates sinusoidally at a frequency of $2 \times 10^{10}$ What is the wavelength of the wave?
Answer
Given: $v=2 \times 10^{10} Hz , c =3 \times 10^8 m$
To find: Wavelength $(\lambda)$
Formula: $c=v \lambda$
Calculation: From formula,
$
\lambda=\frac{c}{\lambda}=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2}
$
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Question 63 Marks
Calculate the wavelength of a microwave of frequency $8.0\ GHz.$
Answer
Given: $v=8 GHz =8 \times 10^9 Hz$ $c =3 \times 10^8 m / s$
To find: Wavelength $(\lambda)$
Formula: $c=v \lambda$
Calculation: From formula,
$\lambda=\frac{c}{\lambda}=\frac{3 \times 10^8}{8 \times 10^9}=3.75 \times 10^{-2}$
$=3.75\ cm$
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Question 73 Marks
The speed of light is $3 \times 10^8 m / s$. Calculate the frequency of red light of wavelength of $6.5 \times 10^{-7} m$.
Answer
Given: $c =3 \times 10^8 m / s , \lambda=6.5 \times 10^{-7} m$
To find: Frequency $( v )$
Formula: $c=v \lambda$
Calculation: From formula,
$
v =\frac{c}{\lambda}=\frac{3 \times 10^8}{6.5 \times 10^{-7}}=4.6 \times 10^{14} Hz
$
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Question 83 Marks
Calculate the wavelength in nm of an X-ray wave of frequency $2.0 \times 10^{18} Hz$.
Answer
Given: $c =3 \times 10^8, v =2 \times 10^{18} Hz$
To find: Wavelength $(\lambda)$
Formula: $c = v \lambda$
Calculation. From formula,
$ \lambda=\frac{c}{v}=\frac{3 \times 10^8}{2 \times 10^{18}}=1.5 \times 10^{-10}$
$=0.15\ nm$
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Question 93 Marks
Calculate the frequency in MHz of a radio wave of wavelength $250\ m$. Remember that the speed of all EM waves in vacuum is $3.0 \times 10^8\ m/s$.
Answer
Given: $\lambda=250 m , c =3 \times 10^8 m / s$
To find: Frequency $(v)$
Formula: $c=v^8$
Calculation: From formula,
$ v =\frac{c}{\lambda}=\frac{3 \times 10^8}{250}=1.2 \times 10^6 Hz$
$=1.2\ MHz$
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Question 103 Marks
Explain why is modulation needed.
Answer
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.
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Question 113 Marks
Why does amplitude modulation give noisy reception?
Answer
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

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Question 123 Marks
Explain the necessity of a carrier wave in communication.
Answer
  1. Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
  2. The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
  3. Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.
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Question 133 Marks
Why should broadcasting programs use different frequencies?
Answer
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.
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Question 143 Marks
How does the effective power radiated by an antenna vary with wavelength?
Answer
  1. To transmit a signal, an antenna or an aerial is needed.
    2. Power radiated from a linear antenna of length $I$ is, $P \propto\left(\frac{l}{\lambda}\right)^2$ where, $\lambda$ is the wavelength of the signal.
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Question 153 Marks
What is meant by noise?
Answer
  1. A random unwanted signal is called noise.
  2. The source generating the noise may be located inside or outside the system.
  3. Efforts should be made to minimize the noise level in a communication system.
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Question 163 Marks
What is modulation?
Answer
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.
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Question 173 Marks
Why high frequency carrier waves are used for transmission of audio signals?
Answer
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.
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Question 183 Marks
What is a carrier wave?
Answer
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.
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Question 193 Marks
Name the three basic units of any communication system.
Answer
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.
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Question 203 Marks
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

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Question 213 Marks
What are ultraviolet rays? Give two uses.
Answer
Production:

  1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
  2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
  3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

  1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
  2. Used in burglar alarms and security systems.
  3. Used to distinguish real and fake gems.
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Question 223 Marks
Why light waves travel in vacuum whereas sound wave cannot?
Answer
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.
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Question 233 Marks
Can we produce a pure electric or magnetic wave in space? Why?
Answer
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.
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Question 243 Marks
What are EM waves?
Answer
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.
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Question 253 Marks
Why are microwaves used in radar?
Answer
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.
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Question 263 Marks
State two characteristics of an EM wave.
Answer
i. The electric and magnetic fields, $\vec{E}$ and $\vec{B}$ are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.
ii. The cross product $(\vec{E} \times \vec{B})$ gives the direction in which the EM wave travels. $(\vec{E} \times \vec{B})$ also gives the energy carried by EM wave.
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Question 273 Marks
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
i. 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
ii. 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen known as Lamb Shift).
iii. 5890 A – 5896 A [double lines of sodium]
Answer
i. Radio waves (short wavelength or high frequency end)
ii. Radio waves (short wavelength or high frequency end)
iii. Visible region (yellow light)
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Question 283 Marks
State uses and limitations of frequency modulation.
Answer
  1. Frequency modulation (FM) is more complex as compared to amplitude modulation and, therefore is more difficult to implement.
  2. However, its main advantage is that it reproduces the original signal closely and is less susceptible to noise.
  3. This modulation is used for high quality broadcast transmission.
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Question 293 Marks
What are the two different modes of communication?
Answer
i. There are two basic modes of communication:
a. point to point communication
b. broadcast communication
ii. In point to point communication mode, communication takes place over a link between a single transmitter and a receiver e.g. telephony.
iii. In the broadcast mode, there are large number of receivers corresponding to the single transmitter e.g., Radio and Television transmission.
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Question 303 Marks
The height of a transmitting antenna is 68 m and the receiving antenna is at the top of a tower of height 34 m. Calculate the maximum distance between them for satisfactory transmission in line of sight mode. (Radius of Earth = 6400 km)
Answer
Given: $h_t=68 m, h_r=34 m$,
$R=6400 km=6.4 \times 10^6 m$
To find: Maximum distance or range (d)
Formula: $d =\sqrt{2 R h}$
Calculation:
From formula,
$\begin{aligned}
d _{ t } & =\sqrt{2 Rh _{ t }} \\
& =\sqrt{2 \times 6.4 \times 10^6 \times 68} \\
& =\sqrt{\{\operatorname{antilog}[\log 2+\log 6.4+\log 68]\} \times 10^6} \\
& =\sqrt{\{\text { antilog }[0.3010+0.8062+1.8325]\} \times 10^6} \\
& =\sqrt{\{\text { antilog }[2.9397]\} \times 10^6} \\
& =\sqrt{8.704 \times 10^2 \times 10^6} \\
& =2.951 \times 10^4 \\
& =29.51 km
\end{aligned}$
$\begin{aligned} d _{ r } & =\sqrt{2 Rh _{ r }} \\ & =\sqrt{2 \times 6.4 \times 10^6 \times 34} \\ & =\sqrt{\{\operatorname{antilog}[\log 2+\log 6.4+\log 34]\} \times 10^6} \\ & =\sqrt{\{\text { antilog }[0.3010+0.8062+1.5315]\} \times 10^6} \\ & =\sqrt{\{\operatorname{antilog}[2.6387]\} \times 10^6} \\ & =\sqrt{4.352 \times 10^2 \times 10^6} \\ = & 2.086 \times 10^4 \\ = & 20.86 km \\ d & = d _{ t }+ d _{ r }=29.51+20.86=50.37 km\end{aligned}$
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Question 313 Marks
If the height of a $TV$ transmitting antenna is $128 m,$ how much square area can be covered by the transmitted signal if the receiving antenna is at the ground level? $($Radius of the Earth $= 6400 \ km)$
Answer
Given: $h =128 m, R =6400 \ km-6400 \times 10^3 m$
To find: Area covered $(A)$
Formulae: $i. d =\sqrt{2 R h}$
$ii. A =\pi d ^2$
Calculation:
From formula $(i),$
$d =\sqrt{2 \times 6400 \times 10^3 \times 128}$
$ =\sqrt{\{\text { antilog }[\log 2+\log 6400+\log 128]\} \times 10^3}$
$ =\sqrt{\{\text { antilog }[0.3010+3.8062+2.1072]\} \times 10^3}$
$ =\sqrt{\{\text { antilog }[6.2144]\} \times 10^3}$
$ =\sqrt{1.639 \times 10^6 \times 10^3}$
$ =\sqrt{16.39 \times 10^8}$
$= 4.048 \times 10^4$
$= 40.48 \ km$
From formula $(ii).$
Area covered $=3.142 \times(40.48)^2$
$=$ antilog $[\log 3.142+2 \log 40.48]$
$=$ antilog $[0.4972+2(1.6073)]$
$=$ antilog $[3.7118]$
$=5.150 \times 10^3$
$=5150 \ km^2$
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Question 323 Marks
A radar has a power of $10 \ kW$ and is operating at a frequency of $20 \ GHz.$ It is located on the top of a hill of height $500 m.$ Calculate the maximum distance upto which it can detect object located on the surface of the Earth.
$($Radius of Earth $= 6.4 \times 10^6 m)$
Answer
Given: $h=500 m, R =6.4 \times 10^6 m$
To find: Maximum distance or range $(d)$
Formula: $d =\sqrt{2 R h}$
Calculation: From formula,
$d =\sqrt{2 R h}=\sqrt{2 \times 64 \times 10^6 \times 500}$
$=8 \times 10^4$
$=80 \ km$
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Question 353 Marks
Explain how different types of waves emitted by stars and galaxies are observed?
Answer
i. Stars and galaxies emit different types of waves. Radio waves and visible light can pass through the Earth’s atmosphere and reach the ground without getting absorbed significantly. Thus, the radio telescopes and optical telescopes can be placed on the ground.
ii. All other type of waves get absorbed by the atmospheric gases and dust particles. Hence, the y-ray, X-ray, ultraviolet, infrared, and microwave telescopes are kept aboard artificial satellites and are operated remotely from the Earth.
iii. Even though the visible radiation reaches the surface of the Earth, its intensity decreases to some extent due to absorption and scattering by atmospheric gases and dust particles. Optical telescopes are therefore located at higher altitudes.
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Question 363 Marks
Identify the name and part of electromagnetic spectrum and arrange these wavelengths in ascending order of magnitude:
Electromagnetic waves with wavelength
$i. \ \lambda _1$ are used by a $\text{FM}$ radio station for broad casting.
$ii. \ \lambda _2$ are used to detect bone fracture.
$iii. \ \lambda _3$ are absorbed by the ozone layer of atmosphere.
$iv. \ \lambda _4$ are used to treat muscular strain.
Answer
$i. \ \lambda _1$ belongs to radiowaves.
$ii. \ \lambda _2$ belongs to $X-$rays.
$iii. \ \lambda _3$ belongs to ultraviolet rays.
$iv. \ \lambda _4$ belongs to infrared radiations.
Ascending order of magnitude of wavelengths:
$\lambda _3 < \lambda _3 < \lambda _4 < \lambda _1$​​​​​​​
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Question 373 Marks
The magnetic field of an $EM$ wave travelling along $X-$axis is $\vec{B}=\hat{k}\left[4 \times 10^{-4} \sin (\omega t-k x)\right]$. Here $B$ is in tesla, $t$ is in second and $x$ is in $m .$ Calculate the peak value of electric force acting on a particle of charge $5 \mu C$ travelling with a velocity of $5 \times 10^5 m / s$ along the $Y-$axis.
Answer
Expression for $EM$ wave travelling along
$X-$axis, $\vec{B}=\hat{k}\left[4 \times 10^{-4} \sin (\omega t-k x)\right]$
Here, $B_0=4 \times 10^{-4}$
Given: $q =5 \mu C =5 \times 10^{-6} C$
$v =5 \times 10^5 m / s$ along $Y-$axis
$\therefore E_0=c B_0=3 \times 10^8 \times 4 \times 10^{-4}$
$=12 \times 10^4 N / C$
Maximum electric force $=q E_0$
$=5 \times 10^{-6} \times 12 \times 10^4$
$=0.6 N$
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Question 383 Marks
For an $EM$ wave propagating along X direction, the magnetic field oscillates along the $Z-$direction at a frequency of $3 \times 10^{10} Hz$ and has amplitude of $10^{-9} T$.
$i.$ What is the wavelength of the wave?
$ii.$ Write the expression representing the corresponding electric field.
Answer
Given: $v=3 \times 10^{10} Hz, B=10^{-9} T$
$i.$ For wavelength of the wave:
$\lambda=\frac{ c }{ v }=\frac{3 \times 10^8}{3 \times 10^{10}}=10^{-2} m$
$ii.$ Since $B$ acts along $Z-$axis, $E$ acts along $Y -$axis.
Expression representing the oscillating electric field is
$E_y=E_0 \sin (k x-\omega t)$
$E_y=E_0 \sin \left[\left(\frac{2 \pi}{\lambda}\right) x-(2 \pi v) t\right]$
$E_y=E_0 \sin 2 \pi\left[\frac{x}{\lambda}-v t\right]$
$E_y=E_0 \sin 2 \pi\left[\frac{x}{10^{-2}}-3 \times 10^{10} t \right]$
$E_y=E_0 \sin 2 \pi\left[100 x -3 \times 10^{10} t \right] V / m$
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Question 393 Marks
A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer
Since the electromagnetic waves are transverse in nature, the electric and magnetic field vectors are mutually perpendicular to each other as well as perpendicular to the direction of propagation of wave.
As the wave is travelling along Z-direction,
$\vec{E}$ and $\vec{B}$ are in XY plane.
For $v=30 MHz =30 \times 10^6 Hz$
Wavelength, $\lambda=\frac{c}{v}=\frac{3 \times 10^8}{30 \times 10^6}=10 m$
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Question 403 Marks
How can energy be transported in the form of EM waves?
Answer
  1. Maxwell proposed that an oscillating electric charge radiates energy in the form of EM wave.
  2. EM waves are periodic changes in electric and magnetic fields, which propagate through space.
  3. Thus, energy can be transported in the form of EM waves.
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Question 423 Marks
Varying electric and magnetic fields regenerate each other. Explain.
Answer
  1. According to Maxwell’s theory, accelerated charges radiate EM waves.
  2. Consider a charge oscillating with some frequency. This produces an oscillating electric field in space, which produces an oscillating magnetic field which in turn is a source of oscillating electric field.
  3. Thus, varying electric and magnetic fields regenerate each other.
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Question 433 Marks
In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write respectively against (i), (ii), and (iii).
 Name of the Physicist Work
 i. H. Hertz a. Existence of EM waves
 ii. J. Maxwell b. Properties of EM waves
 iii. G. Marconi c. Wireless communication
  d. Displacement current
Answer
(i – a, b), (ii – d), (iii – c)
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Question 443 Marks
Explain the origin of displacement current?
Answer
  1. Maxwell pointed a major flaw in the Ampere’s law for time dependant fields.
  2. He noticed that the magnetic field can be generated not only by electric current but also by changing electric field.
  3. Therefore, he added one more term to the equation describing Ampere’s law. This term is called the displacement current.
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Question 453 Marks
Describe Faraday’s law along with Lenz’s law.
Answer
i. Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.
i. Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.
iii. According to Faraday's law with Lenz's law,
$\int \overrightarrow{ E } \cdot \overrightarrow{ d l}=-\frac{ d \phi_{ m }}{ dt }$
where, $\varnothing_{ m }$ is the magnetic flux and the integral is over a closed loop.
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Answer the following in Brief - Physics STD 11 Science Questions - Vidyadip