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Physics 3 Marks Question

Electromagnetic Waves · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 21 questions.

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
How electromagnetic waves are produced? What is the source of the energy accompanied by any propagating electromagnetic wave?
(i) In far switches of household electrical appliances
(ii) In medical diagnosis
Answer
Accelerated charge produces electric and magnetic fields which changes with time and place both. These changing magnetic and electric fields give rise to electromagnetic waves. A charge undergoing circular motion will have accelerated motion and hence it can be a source of origin of electromagnetic waves.
Electromagnetic radiations are used : (i) In infrared radiations, (ii) In X-rays.
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Question 23 Marks
Draw a propagation diagram of a linearly polarized electromagnetic wave and write any two characteristics of electromagnetic wave.Magnitude of magnetic field associated with an electromagnetic wave in vacuum is $B_0=50 \times 10^{-8} T$. Write the magnitude of electric field associated with the wave in volt/meter
Answer
Schematic diagram of a linearly polarized electromagnetic wave :
Image
Characteristics of electromagnetic wave : (1) It is a light wave whose nature is transverse. (2) Electromagnetic waves travel with the speed of light, c in vacuum where $ C=\frac{1}{\sqrt{\mu_0 \epsilon_0}}=3 \times 10^8 m / s $ Note: Students can write other characteristics also. Sol. Given : $B_0=50 \times 10^{-8} T, C =3 \times 10^8 m / s$ For electromagnetic waves, or, $ \begin{aligned} C & =\frac{E_0}{B_0} \\ E_0 & =C B_0 \end{aligned} $ On putting the values, $ \begin{aligned} E_0 & =3 \times 10^8 \times 50 \times 10^{-8} \\ & =150 volt / metre \end{aligned} $
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Question 33 Marks
(a) Which of the following can be a source for origin of electromagnetic waves? Give reason.
(i) A charge moving with constant speed
(ii) Charge undergoing circular motion
(iii) Fixed charge.
(b) Name that part of electromagnetic spectrum with which waves of frequency
(i) $10^{20} Hz$,
(ii) $10^9 Hz$ associated.
Answer
(a) (i) There is no acceleration in a charge moving with constant speed, hence it cannot be a source of origin of electromagnetic waves. (ii) If a charge is undergoing circular motion, then its motion will be accelerated motion and it can be a source of origin of electromagnetic waves. (iii) There is no acceleration in fixed charge and hence it cannot be a source of origin of electromagnetic waves.
(b) (i) Gamma rays, (ii) Micro waves.
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Question 43 Marks
Arrange electromagnetic radiations in the ascending order of their frequencies. Write uses of any one of them-Micro waves, Radio waves, X-rays, gamma rays.
Answer
Radio waves, Micro waves, X-rays, Gamma rays.
Uses :
Micro waves : In radars, satellites, long distance communication, micro wave oven.
Radio waves : In telecast of radio and television.
X-rays : In medical diagnosis and treatment.
Gamma rays : Knowledge of nuclear structure, medical treatment etc.
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Question 53 Marks
Explain frequency and wavelength of electromagnetic spectrum.
Answer
Electromagnetic spectrum : Due to same characteristics of radio waves, micro waves, infrared, visible light, ultraviolet rays, X-rays, gamma rays etc. as of electromagnetic waves, these all are electromagnetic waves. The only difference is their wavelengths are different. It is clear that extension of wavelengths of electromagnetic waves is very wide, so these waves can be kept in a sequence depending on wavelength and it is known as electromagnetic spectrum.

Image
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Question 63 Marks
Which waves are electromagnetic waves?Which factors are responsible for their production?
Answer
It is a three-dimensional wave which is produced by oscillating electric circuit. Their electric and magnetic fields are mutually perpendicular. These waves do not require any material medium for their propagation.
Reasons for their production : (i) Mechanical energy of the charge vibrating in oscillating electric field gets converted into electromagnetic waves.
(ii) Due to oscillation of charge, kinetic energy of the charge gets converted into electromagnetic waves.
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Question 73 Marks
Explain ground wave and sky wave.
Answer
Ground waves : These waves are radio waves of smaller frequencies which travel from one point on the surface of the earth to the other point on the surface of earth travelling along the surface of earth. With their help, the radio programs which are broadcasted in a limited area are broadcasted. These waves are known as ground waves.
Sky waves: Radio waves travel in straight lines and hence they are not able to travel long distances due spherical shape of the earth. If they are of smaller frequencies, they travel along the surface of the earth but if frequency is more, they are not able to travel to a large distance. Waves of higher frequency strike with the lower part of the ionosphere and reach larger distance on the earth. Since these waves come back from the sky and reach any place on earth, are known as sky waves.
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Question 83 Marks
How does a charge oscillating with any frequency produces electromagnetic waves? Draw a schematic diagram by showing electric and magnetic fields for an electromagnetic wave transmitting along Z-axis.
Answer
When a charge oscillates with any frequency then it produces an electric field oscillating from that place. This produces an oscillating magnetic field. Oscillating magnetic field becomes a source to produce electric field and this process continues. In this way the electric and magnetic field vectors producing each other are mutually perpendicular to each other and also to the direction of propagation of wave.
Image
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Question 93 Marks
Write some properties of electromagnetic waves.
Answer
Some of the properties of electromagnetic waves are :
(i) They are transverse in nature.
(ii) They travel with the speed of light ( $c=3 \times$ $10^8 m / s$ ) in vacuum.
(iii) They do not need any material medium for their propagation.
(iv) They are produced by accelerated or oscillating charge.
(v) These are composed of perpendicular changeable electric and magnetic fields.
(vi) Energy of electromagnetic waves is equally divided between electric and magnetic field vectors.
(vii) They follow the principle of superposition.
(viii) Maximum and minimum magnitudes of $\vec{E}$ and $\overrightarrow{ B }$ are obtained at the same time and same place.
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Question 103 Marks
What is the nature of electromagnetic waves?
Answer
It is a three-dimensional wave which is emitted by an oscillating electric circuit in which electric field and magnetic field oscillate perpendicularly. These waves do not require any material medium for their propagation and they can travel through vacuum also. Hence, they are different from mechanical waves. They travel with maximum speed in vacuum which is equal to the speed of the light. These waves are transverse in nature and the properties of the medium remain unchanged through which they travel.
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Question 113 Marks
How does Maxwell modify Ampere's circuital law? Explain.
Answer
The dispute in Ampere's circuital law arise because it was assumed that current is discontinuous between the plates of the capacitor. Maxwell gave the assumption of displacement current $\left(I_D\right)$ which states that no conduction current ( $I_c$ ) flows but only displacement current ( $I _{ D }$ ) flows.As the capacitor is charged, it gives rise to flow of displacement current between the plates of the capacitor because electric field increases between the plates of the capacitor. In this way, Ampere's circuital law $\left(\oint \overrightarrow{ B } \cdot \overrightarrow{ dI }=\mu_0 I _{ C }\right.$ ) was written as $\oint \overrightarrow{ B } \cdot \overrightarrow{ dI }=\mu_0\left( I _{ C }+ I _{ D }\right)$.
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Question 123 Marks
By defining conduction current, explain why Maxwell propounded the idea of displacement current?
Answer
Conduction current : Current flowing by the wires joining the circuit, in the circuit is known as conduction current. Let a capacitor is charged by a current I by a battery. It is clear from the figure that during charging conduction current is flowing from left to right outside the plates of the capacitor and there is no conduction current I between the plates of the capacitor.
Image
To remove this contradiction, Maxwell introduced a new concept of displacement current $\left(I_D\right)$ and its source to be the changing electric field between the plates of the capacitor. It has been introduced to show that current flowing outside the capacitor (conduction current, $I _{ C }$ ) is equal to the current flowing in the capacitor (displacement current, $I_D$ ). According to condition put by Maxwell, conduction current ( $I_C$ ) flows in the left plate of the capacitor and displacement current $\left(I_D\right)$ flows towards the right plate in the capacitor.
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Question 133 Marks
What conclusion can you draw about displacement current and conduction current?
Answer
(1) Displacement current and conduction current are discontinuous separately but both currents are continuous together at closed path.
(2) Displacement current like conduction current is a source of magnetic field.
(3) Both currents are always equal to each other.
(4) Displacement current is always produced due to change in electric field between the plates of the capacitor.
(5) Current I is between the plates of the capacitor and conduction current is in the wire joining the plates and source of emf.
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Question 143 Marks
Prove that $I _{ D }=\epsilon_0 \frac{d}{d t}\left(\phi_{ E }\right)$, when symbols have their usual meanings.
Answer
We know that :
$ I_{D} =I$
Or $I =\frac{d q}{d t} $
By Gauss's law,
$ E=\frac{q}{\epsilon_0 A}$
Or $q=E \in_0 A $
On putting values in eqn. $(2)$
$ I=\frac{d}{d t}\left(E \epsilon_0 A\right) $
From equation $(1),$
$\therefore I_{D}=\epsilon_0 A \frac{d}{d t}(E)$
$=\epsilon_0 A \frac{d}{d t}\left(\frac{\phi_{E}}{A}\right)$
$\because \phi_{E}=EA$
$=\frac{\epsilon_0 A}{A} \frac{d \phi_{E}}{d t}$
or
$I_{D}=\epsilon_0 \frac{d}{d t}\left(\phi_{E}\right) $
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Question 153 Marks
Area of a parallel plate capacitor is A and distance between them is $d$. It is charged with a constant current of I. An element of area $\frac{A}{2}$ is kept in the middle of the plates, parallel to the plates. What is the current flowing through this area?
Answer
Let charge at any instant ' $t$ ' on plates is $q$. Electric field between the plates
$E =\frac{\sigma}{\epsilon_0}=\frac{q}{\epsilon_0 A}$
Electric flux passing through area $\frac{ A }{2}$ between the plates
$\phi_{ E }= E \times \frac{ A }{2}=\frac{q}{\epsilon_0 A} \times \frac{ A }{2}$ $=\frac{q}{2 \epsilon_0}$
$\because$ Displacement current passing through this plane
$I _d=\epsilon_0 \frac{d}{d t}\left(\phi_{ E }\right)$
$=\epsilon_0 \frac{d}{d t}\left(\frac{q}{2 \epsilon_0}\right)=\frac{1}{2} \frac{d q}{d t}$
$=\frac{ I }{2}$
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Question 163 Marks
Magnitudes of $\frac{1}{4 \pi \epsilon_0}$ and $\frac{\mu_0}{4 \pi}$ for vacuum are respectively $9 \times 10^9 N / m ^2$ and $10^{-7}$ weber/ampere respectively. Calculate the speed of electromagnetic wave in vacuum.
Answer
According to the question,
$\frac{1}{4 \pi \epsilon_0}=9 \times 10^9$
$\therefore \quad \epsilon_0=\frac{1}{4 \pi \times 9 \times 10^9}=\frac{1}{36 \pi \times 10^9}$
and $\quad \frac{\mu_0}{4 \pi}=10^{-7} \quad \therefore \mu_0=4 \pi \times 10^{-7}$
Speed of light in any medium $V=\frac{1}{\sqrt{\mu \epsilon}}$
In vacuum, for speed of electromagnetic wave, $V =$ C, $\mu=\mu_0$ and $\in=\epsilon_0$
$c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}=\frac{1}{\sqrt{4 \pi \times 10^{-7} \times \frac{1}{36 \pi \times 10^9}}}$
or $c=\sqrt{9 \times 10^{16}}=3 \times 10^8 m / s$.
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Question 173 Marks
How will establish displacement current of 6 A between the plates of a capacitor of capacitance $6 \mu F$ ?
Answer
Given : $C =6 \mu F=6 \times 10^{-6} F$
$I _d=6 A$
We know that, $I_d=\epsilon_0 A \frac{d E}{d t} $
$=\epsilon_0 A \frac{d}{d t}\left(\frac{V}{d}\right) \quad \because E =\frac{ V }{d}$
$=\frac{\epsilon_0 A}{d} \cdot \frac{d V}{d t}$
$I _d= C \cdot \frac{d V}{d t}$
$\therefore \quad \frac{d V}{d t}=\frac{ I _d}{ C }=\frac{6}{6 \times 10^{-6}}$
$=10^6 Vs ^{-1} Ans$.
A displacement current of 6 A can be established on the plates of the capacitor changing the potential difference on the plates of the capacitor at the rate of $10^6 Vs ^{-1}$.
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Question 183 Marks
The charging current on the plates of a capacitor is 0.5 A . Calculate displacement current on the plates of the capacitor.
Answer
Given $I =0.5 A$ , $I _d=?$
We know that, $I _d=\epsilon_0 \frac{d \phi_{ E }}{d t}$
But $\phi_{E}=EA $
$\therefore \quad I _d=\epsilon_0 \frac{d}{d t}( EA )$
$=\epsilon_0 A \frac{d E }{d t}$
But $E=\frac{q}{\epsilon_0 A} $
$\therefore \quad \quad I _d=\epsilon_0 A \frac{d}{d t}\left(\frac{q}{\epsilon_0 A}\right)$
$=\frac{\epsilon_0 A}{\epsilon_0 A} \frac{d q}{d t}=\frac{d q}{d t}$
$=0.5 A$ Ans. $\left(\because \frac{d q}{d t}=I=0.5 A\right)$
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Question 193 Marks
A beam of light moving along $x$-axis is represented by the electric field $E_y=600 Vm ^{-1} \sin \omega$ $\left(t-\frac{x}{c}\right)$. Calculate the maximum magnitudes of electric and magnetic fields on a charge $q=2 e$ moving along the $y$-axis with a speed of $3.0 \times 10^7 m / s$ (where $\left.e=1.6 \times 10^{-19} C \right)$.
Answer
Given : $E=600 Vm^{-1} \sin \omega\left(t-\frac{x}{c}\right) $
$E_0=600 Vm ^{-1}$
$q=2 e$
$=2 \times 1.6 \times 10^{-19} C$
Speed of light $C =3 \times 10^8 m / s$
$v=3.0 \times 10^7 m / s$
$\therefore \quad B _0=\frac{ E _0}{ C }=\frac{600}{3 \times 10^8}$
$=2 \times 10^{-6} T$
Which works along the $z$-axis.
Thus, $\quad F _0=q E _0$ $=2 e E _0$
$=2 \times 1.6 \times 10^{-19} \times 600$
$=1.92 \times 10^{-16} N$
Maximum magnetic force,
$F _{\max }=q v B_0=2 e v B_0$
$=2 \times 1.6 \times 10^{-19} \times 3 \times 10^7 \times 2 \times 10^{-6} N$
$=1.92 \times 10^{-17} N$.
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Question 203 Marks
Rate of change of electric field in the region between the plates of capacitor is $1.5 \times 10^{12} Vm ^{-1} s^{ 1 }$. If circular plate of capacitor has a radius of 55 mm , then find the magnitude of displacement current.
Answer
Given :$r=55 mm=55 \times 10^{-3} m$
$=5.5 \times 10^{-2} m$
$\therefore \quad$ Area $=\pi r^2=\frac{22}{7} \times\left(5.5 \times 10^{-2}\right)^2$
$=\frac{22}{7} \times 5.5 \times 5.5 \times 10^{-4}$
$=95.07 \times 10^{-4} m^2$
or $\frac{d E }{d t}=1.5 \times 10^{12} Vm ^{-1} s^{-1}$
Displacement current $I _d=$ ?
We know that :
Displacement current: $I _d=\epsilon_0 \frac{d}{d t}\left(\phi_{ E }\right)$
But $\phi_{ E }= EA$
$\therefore \quad I _d=\epsilon_0 \frac{d}{d t}( EA )$
$=\epsilon_0 A \frac{d E }{d t}$
On putting the values
$\begin{aligned}=8.85 \times 10^{-12} \times 95.07 & \times 10^{-4} \\ & \times 1.5 \times 10^{12}\end{aligned}$
$=8.85 \times 95.07 \times 1.5 \times 10^{-4}$
$=1262.05 \times 10^{-4}$
$=126.205 \times 10^{-3} A$
$\simeq 126 mA$
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Question 213 Marks
Magnitude of electric field of a plane electromagnetic wave is $40 V / m$ and it vibrates with a frequency of $2 \times 10^{10} s^{-1}$. Calculate the wavelength of the wave.
Answer
Given : Frequency $v=2 \times 10^{10} s^{-1}$
$c=3 \times 10^8 m / s$
$E _0=40 V / m$.
$\therefore \quad E _{\max }=\frac{ E _0}{\sqrt{2}}=\frac{\sqrt{2}}{2} E _0$
$=\frac{\sqrt{2} \times 40}{2}=20 \sqrt{2} V / m$.
Wavelength of wave $\lambda=$ ?
$\lambda=\frac{c}{v}=\frac{\text { Speed of light }}{\text { Frequency of vibration }}$
$=\frac{3 \times 10^8}{2 \times 10^{10}}=1.50 \times 10^{-2} m$
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