MCQ 511 Mark
According to Maxwell's equation, the velocity of light in any medium is expressed as.
- A
$\frac{1}{\mu_0\in_0}$
- ✓
$\frac{1}{\sqrt\mu\in}$
- C
$\sqrt\frac{\mu}{\in}$
- D
$\sqrt\frac{\mu_0}{\in}$
AnswerCorrect option: B. $\frac{1}{\sqrt\mu\in}$
Velocity of light in a medium,
$\text{c}=\frac{1}{\sqrt{\mu_0\in_0\mu_\text{r}\in_\text{r}}}=\frac{1}{\mu_0\in_0}$
View full question & answer→MCQ 521 Mark
Electromagnetic waves travel only through.
- A
- B
- C
oscillating electric and magnetic fields
- ✓
oscillating electric and magnetic fields whose directions are perpendicular to each other
AnswerCorrect option: D. oscillating electric and magnetic fields whose directions are perpendicular to each other
Electromagnetic waves travel through oscillating electric and magnetic fields whose directions are perpendicular to each other.
View full question & answer→MCQ 531 Mark
Choose the correct answer from the alternatives given. The amplitude of an electromagnetic wave in vaccum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is correct?
- A
The frequency of the wave changes only
- B
The wave length of the wave changes only
- C
The speed of the wave propagation changes only
- ✓
None of the above is correct
AnswerCorrect option: D. None of the above is correct
As we know, velocity of electromagnetic wave,
$\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}=\frac{3\times10^8\text{m}}{\text{s}}$
which is constant
So It is independent of amplitude of electromagnetic wave, frequency and wavelength of electromagnetic wave.
so none of the above is correct.
View full question & answer→MCQ 541 Mark
A plane electromagnetic wave travels in vacuum along $\hat{\text{k}}$ direction, where $\hat1\hat{\text{j}}$ and $\hat{\text{k}}$ are unit vectors along the $x, y$ and $z$ directions. The direction along which the electric and the magnetic field vectors point may be respectively.
- ✓
$\hat{\text{k}}\text{ and }\hat{\text{j}}$
- B
$\hat{\text{k}}\text{ and }\hat{-\text{j}}$
- C
$\hat{\text{j}}\text{ and }\hat{\text{i}}$
- D
$\hat{\text{k}}\text{ and }\hat{\text{i}}$
AnswerCorrect option: A. $\hat{\text{k}}\text{ and }\hat{\text{j}}$
Electromagnetic wave is a transverse wave that means the electric and magnetic field associated to it will not only be perpendicular to each other but will also be
perpendicular to the direction in which the wave travels.
So, if waves travel along $\hat{\text{k}}$ direction then the electric and the magnetic field will be along $\hat{\text{i}}$ and $\hat{\text{j}}$ directions.
View full question & answer→MCQ 551 Mark
Choose the correct answer from the alternatives given. The conduction current is the same as displacement current when the source is.
AnswerCorrect option: A. $AC$ only
For a capacitor, we have:
$Q = CV$
If $Q$ is changing, there will be a current in capacitor plates,
$\text{I}=\frac{\text{dQ}}{\text{dt}}=\frac{\text{CdV}}{\text{dt}}$when voltage across the capacitor is constant, $\frac{\text{dV}}{\text{dt}}=0$
therefore,$ I = 0$
It implies that, for a $DC ($constant$)$ voltage, the capacitor current is zero.
Hence, for a $DC$ source the conduction current and displacement current $($capacitor current$)$ are not same.
Whereas, by Maxwell's equation for a time varying voltage $(AC$ voltage$),$ both conduction and displacement currents are same.
View full question & answer→MCQ 561 Mark
The condition under which a microwave over heats up a food item containing water molecules most efficiently is:
- ✓
The frequency of the microwaves must match the resonant frequency of the water molecules.
- B
The frequency of the microwaves has no relation with natural frequency of the water molecules.
- C
Microwaves are heat waves, so always produce heating.
- D
Infrared waves produce heating in a microwave oven.
AnswerCorrect option: A. The frequency of the microwaves must match the resonant frequency of the water molecules.
When frequency of microwave matches with frequency of water molecules i.e., resonant condition. Maximum energy is transferred to water molecules as their $K.E.$ energy.
View full question & answer→MCQ 571 Mark
Which of the following type of radiations are radiated by on oscillating electric charge?
AnswerAny stationary charge produce static electric field. And the field strength is given by:
$r$, is the radial distance from the point charge.
$Q,$ is the charge in Coulomb.
When electric charge oscillates electric field at any point also oscillates.And according to Maxwell's equations varying electric field produces magnetic field and an oscillating electric field produces oscillating magnetic field. This thing is used in antennas in which oscillating current of certain frequency produces oscillating electric and magnetic field which propagates through space $($electromagnetic waves$).$
View full question & answer→MCQ 581 Mark
Two opposite charged particles oscillate about their mean equilibrium position in free space, with a frequency of $10^9 \ Hz$. The wavelength of the corresponding electromagnetic wave produced is$..........$
- ✓
$0.3m$
- B
$10^9 m$
- C
$3 \times 10^{17} m$
- D
$3.3m$
AnswerCorrect option: A. $0.3m$
Electromagnetic wave $V=3 \times 10^8 m / s$
Given frequency $(f) =10^9 Hz$
$\text{V}=\text{f}\lambda$
$\lambda=\frac{\text{V}}{\text{f}}$
$=\frac{3 \times 10^8} {10^9}$
$= 0.3m$
View full question & answer→MCQ 591 Mark
Which of the following effects could not be explained by Maxwell's electromagnetic wave theory?
AnswerPhotoelectric effect was discovered by heinrich Rudoy Hertz.
Compton effect was discovered by Aethur Holl Compton.
Raman effect was discovered by Sir Chandrasekhar Venbata Ram.
So, none of these effect was discovered by Maxwell.
View full question & answer→MCQ 601 Mark
The ozone layer in the atmosphere absorbs:
AnswerCorrect option: D. $X-$rays and ultraviolet rays
View full question & answer→MCQ 611 Mark
An electromagnetic wave can be produced when the charge is
- A
moving in a circular orbit
- B
moving with a constant velocity
- C
falling in an electric field
- ✓
Both $(a)$ and $(c)$
AnswerCorrect option: D. Both $(a)$ and $(c)$
An accelerated charge is the source of electromagnetic waves $\text{(EMWs)}$. When the charge is in a circular motion, the direction of its velocity continuously changes and thus it is in accelerated motion and produces $\text{EMWs}$. A charge falling in an electric field is accelerated by the electric force and thus produces $\text{EMWs.}$
View full question & answer→MCQ 621 Mark
A $1000\Omega$ resistance and a capacitor of $100\Omega$ resistance are connected in series a $220V$ source. when the capacitor is $50\%$ charged, the value of the displacement current is.
- A
$11.\sqrt{2}\text{A}$
- ✓
$2.2A$
- C
$11A$
- D
$4.4A$
AnswerCorrect option: B. $2.2A$
Displacement current $=\text{I}_\text{D}=\text{C}\frac{\text{dV}}{\text{dT}}=\text{C}\omega\text{V}_\text{o}=\frac{\text{V}_\text{o}}{\text{X}_\text{c}}=\frac{220\text{V}}{100\Omega}=2.2\text{A}$
As we are asked amplitude of displacement current. So, we don't have to worry about charge on capacitor.
View full question & answer→MCQ 631 Mark
A plane electromagnetic wave of frequency $20\ MHz$ travels through a space along $x$ direction. If the electric field vector at a certain point in space is $6 V m ^{-1}$, what is the magnetic field vector at that point?
AnswerCorrect option: A. $2\times10^{-8}\text{T}$
Velocity of $EM$ wave $\text{v}=\frac{3\times10^8\text{m}}{\text{s}}$
Electric field vector $\text{E}=\frac{6\text{V}}{\text{m}}$
Thus magnetic field vector $\text{B}=\frac{\text{E}}{\text{v}}$
$\therefore\text{B}=\frac{6}{3\times10^8}=2\times10^{-8}\text{T}$
View full question & answer→MCQ 641 Mark
The energy contained in a small volume through which an electromagnetic wave is passing oscillates with:
- A
- B
The frequency of the wave.
- C
Half the frequency of the wave.
- ✓
Double the frequency of the wave.
AnswerCorrect option: D. Double the frequency of the wave.
The energy per unit volume of an electromagnetic wave,
$\text{u}=\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}$
The energy of the given volume can be calculated by multiplying the volume with the above expression.
$\text{U}=\text{u}\times\text{V}=\Big(\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}\Big)\times\text{V}\ ....(\text{i})$
Let the direction of propagation of the electromagnetic wave be along the $z-$axls. Then, the electric and magnetic fields at a particular point are given by,
$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$
$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$
Substituting the values of electric and magnetic fields in $(1)$ we get,
$\text{U}=\Big(\frac{1}{2}\in_0\big(\text{E}_0^2\sin^2(\text{kz}-\omega\text{t})+\frac{\text{B}^2_0\sin^2(\text{kz}-\omega\text{t})}{2\mu_0}\Big)\times\text{V}$
$\text{U}=\Big(\in_0\text{E}^2_0\frac{(1-\cos(2\text{kz}-2\omega\text{t}))}{4}+\frac{\text{B}_0^2(1-\cos(2\text{kz}-2\omega\text{t}))}{4\mu_0}\Big)\times\text{V}$
From the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency $2\omega$ Thus. the frequency of the energy of the electromagnetic wave will also be double.
View full question & answer→MCQ 651 Mark
From Maxwell’s hypothesis, a charging electric field gives rise to:
AnswerA charging electric field gives rise to a magnetic field.
View full question & answer→MCQ 661 Mark
Light with an energy flux of $20 W / cm ^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30 cm^2$. the total momentum delivered (for complete absorption) during $30$ minutes is:
- A
$36 \times 10^{-5} kg m / s$
- ✓
$36 \times 10^{-4} kg m / s$.
- C
$108 \times 10^4 kg m / s$.
- D
$1.08 \times 10^7 kg m / s$
AnswerCorrect option: B. $36 \times 10^{-4} kg m / s$.
b. $36 \times 10^{-4} kg m / s$.
Solution:
Givne energy flux $\phi=20\frac{\text{W}}{\text{cm}^2}$
Area, $A=30 cm^2$
Time, $t =30 min=30 \times 60 s$
Now, total energy falling on the surface in time t is,
$\text{U}=\phi\text{At}=20\times30\times(30\times60)\text{J}$
Momentum of the incident light $=\frac{\text{U}}{\text{c}}$
$=\frac{20\times30\times(30\times60)}{3\times10^{8}}=36\times10^{-4}\text{kg-ms}^{-1}$
Momentum of the reflected light $= 0$
$\therefore$ Momentum delivered to the surface
$=36\times10^{-4}-0=36\times10^{-4}\text{kg-ms}^{-1}$
Important points
Mass of photon:
Actually rest mass of the photon is zero. But its effective mass is given as $\text{E}=\text{mc}^2=\text{hv}\Rightarrow\ \text{m}=\frac{\text{E}}{\text{C}^2}=\frac{\text{hv}}{\text{C}^2}=\frac{\text{h}}{\text{c}\lambda}$. Thsi mass is also known as kintic mass of the photon.
Momentum of the photon:
Momentum $\text{p}=\text{m}\times\text{c}=\frac{\text{E}}{\text{c}}=\frac{\text{hv}}{\text{c}}=\frac{\text{h}}{\lambda}$
Number of emitted photons:
The number of photons emitted per second from a source of monochromatic radiation of wavelengh $\lambda$ and power P is given as $\text{(n)}=\frac{\text{P}}{\text{E}}=\frac{\text{P}}{\text{hv}}=\frac{\text{P}\lambda}{\text{hc}}$; where E = energy of each photon
Inrensity of light (I):
Energy crossing per unit area normally per second is called intensity or energy flux
i.e. $\text{I}=\frac{\text{E}}{\text{At}}=\frac{\text{P}}{\text{A}}\Big(\frac{\text{E}}{\text{t}}=\text{P}=\text{radiation Power}\Big)$
At a distance r from a point source of power P intensity is given by $\text{I}=\frac{\text{P}}{4\pi\text{r}^2}\Rightarrow\ \text{I}\propto\frac{1}{\text{r}^2}$.
View full question & answer→MCQ 671 Mark
The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is.
Answer$v \rightarrow 2 vhv - hv v _0= KG _{\max }$
$So , KG _{\max }>2 KG _{\max }$
as $h v_0$ is constant
View full question & answer→MCQ 681 Mark
Dimensions of $\frac{1}{(\mu_0\in_0)}$ is:
- A
$\frac{\text{L}}{\text{T}}$
- B
$\frac{\text{T}}{\text{L}}$
- ✓
$\frac{\text{L}^2}{\text{T}^2}$
- D
$\frac{\text{T}^2}{\text{L}^2}$
AnswerCorrect option: C. $\frac{\text{L}^2}{\text{T}^2}$
The speed of light, $\text{C}=\frac{1}{\sqrt{\mu_0\in_0}}$
The dimensions of $\frac{1}{\sqrt{\mu_0\in_0}}$ are of velocity, i.e., $\frac{\text{L}}{\text{T}}$
Therefore, $\frac{1}{\in_0\mu_0}$ will have dimensions $\frac{\text{L}^2}{\text{T}^2}$
View full question & answer→MCQ 691 Mark
$A.$ Current flow inside the capacitor due to accumulation of charges on the capacitor walls is called displacement current.
$B.$ Current due to the flow of electrons due to some potential difference is called as conduction current.
$C.$ Displacement current came into existence when Maxwell observed that if a magnetic compass is placed between the capacitors the needle gets deflected which signifies presence of magnetic fields which would possibly caused due to some changing current.
$D.$ Displacement current change the actual motion of electric charges.
Which of the above statement(s) is/ are correct?
- A
$A$ and $B$ only
- ✓
$A, B$ and $C$ only
- C
$C$ only
- D
$A, B, C$ and $D$
AnswerCorrect option: B. $A, B$ and $C$ only
Displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current came into existence when Maxwell observed that if a magnetic compass is placed between the capacitors the needle gets deflected which signifies presence of magnetic fields which would possibly caused due to some changing current. Current flow inside the capacitor due to accumulation of charges on the capacitor walls is called displacement current.However it is not an electric current of moving charges, but a 'time-varying electric field'.
View full question & answer→MCQ 701 Mark
Electromagnetic wave of intensity $1400 W / m ^2$ falls on metal surface on area $1.5 m^2$ is completely absorbed by it. Find out force exerted by beam.
- A
$14 \times 10^{-5} N$
- B
$14 \times 10^{-6} N$
- C
$7 \times 10^{-5} N$
- ✓
$7 \times 10^{-6} N$
AnswerCorrect option: D. $7 \times 10^{-6} N$
For a perfectly absorbing surface,
$\text{F}=\frac{\text{IA}}{\text{C}}$
$=\frac{(1400\text{W/m}^2\times 1.5\text{m}^2)}{(3\times 10^8\text{m/s})}$
$=7\times10^{−6}\text{N}.$
View full question & answer→MCQ 711 Mark
The ratio of contributions made by the electric field and magnetic field components to the intensity of an $EM$ wave is:
- A
$c : 1.$
- B
$c^2: 1$
- ✓
$1 : 1.$
- D
$\sqrt{\text{c}}:1.$
AnswerCorrect option: C. $1 : 1.$
The intensity of electromagnetic wave is given by,
$I = U _{ av } C$, where $U _{ av }=$ Average energy and $c =$ speed of light
Intensity in relation with electric field $\text{U}_\text{av}=\frac{1}{2}\in_0\text{E}_0^2$
Intensity relation with magnetic field $\text{U}_\text{av}=\frac{1}{2}\frac{\text{B}_0^2}{\mu_0}$
Now taking the intensity in terms of electric field,
$(\text{U}_\text{av})_\text{electric field}=\frac{1}{2}\in_0\text{E}_0^2=\frac{1}{2}\in_0(\text{cB}_0)^2\ \ (\because\ \text{E}_0=\text{cB}_0)$
But, $\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}$
$\therefore\ (\text{U}_\text{av})_\text{Electric field}=\frac{1}{2}\in_0\times\frac{1}{\mu_0\in_0}\text{B}_0^2=\frac{1}{2}\frac{\text{B}_0^2}{\mu_0}$
$=(\text{U}_\text{av})_\text{magnetic field}$
Hence the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector.
It means the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is $1:1.$
Impotant point:
Propertioe of $EM$ waves
Speed: In free, its speed $\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}=\frac{\text{E}_0}{\text{B}_0}=3\times10^8\text{m/s}$.
In medium $\text{v}=\frac{1}{\sqrt{\mu\in}}$; where $\mu_0$ = Absolute permeability, $\in_0$ = Absolute permittivity, E = Amplitude of electric of field and magnetic field vectors.
Energy: The energy in an $EM$ waves is divided equally between the electric and magnetic fields.
Energy density of electric field $\text{u}_\text{e}=\frac{1}{2}\in_0\text{E}^2$, Energy density of magnetic field $\text{u}_\text{B}=\frac{1}{2}\frac{\text{B}^2}{\mu_0}$
It is found that $u _{ e }= u _{ B }$. Also $\text{u}_\text{av}=\text{u}_\text{e}+\text{u}_\text{B}=2\text{u}_\text{e}=2\text{u}_\text{B}=\in_0\text{E}^2=\frac{\text{B}^2}{\mu_0}$
Intensity (I): The energy crossing per unit area unit time, perpendicular to the direction of propagation of $EM$ wave is called intensity.
$\text{I}=\text{u}_\text{av}\times\text{c}=\frac{1}{2}\in_0\text{E}^2\text{c}=\frac{1}{2}\frac{\text{B}^2}{\mu_0}.\text{c}$
Momentum: EM waves also carries momentum, if a portion of $Em$ wave of energy u propagating with speed $c$, then linear momentum $=\frac{\text{Energy (u)}}{\text{Speed (c)}}$
When the incident $EM$ wave is completely absorbed by a surface, it delivers energy $u$ and momentum $u/c$ to the surface.
When a wave of energy us is totally reflected from the surface, the momentum delivered to surface is $2u/c.$
Poynting vector $(\vec{\text{S}})$: In $EM$ waves, the rate of flow of energy crossing a unit area is described by the poynting vector. Its unit is watt/m and $\vec{\text{S}}=\frac{1}{\mu_0}(\vec{\text{E}}\times\vec{\text{B}})=\text{c}^2\in_0(\vec{\text{E}}\times\vec{\text{B}})$. Because in $EM$ waves, $\vec{\text{E}}$ and $\vec{\text{B}}$ are perpendicular to each other, the magnitude of $\vec{\text{S}}$ is $|\vec{\text{S}}|=\frac{1}{\mu_0}\text{E B}\sin90^\circ=\frac{\text{EB}}{\mu_0}=\frac{\text{E}^2}{\mu_\text{C}}$.
The direction of the Poynting vector $\vec{\text{S}}$ at any point gives the wave's direction of travel and direction of energy transport the point.
Radiation Pressure: Is the momentum imparted per second per unit area on which the light falls.
For a perfectly reflecting surface $\text{P}_\text{r}=\frac{2\text{S}}{\text{c}}; S =$ Poynting vector; $c =$ speed of light
For a perfectly absorbing surface $\text{P}_\text{a}=\frac{\text{S}}{\text{c}}$.
The radiation pressure is real that's why tails of comet point away from the sun.
View full question & answer→MCQ 721 Mark
An AM radio wave is emitted by a radio antenna and travels across flat ground. Find out the direction of the magnetic field component of the wave?
- ✓
Parallel to the ground and perpendicular to the direction of propagation
- B
Perpendicular to the ground and to the direction of propagation
- C
Perpendicular to the ground and parallel to the direction of propagation
- D
Parallel to the electric field component of the wave
AnswerCorrect option: A. Parallel to the ground and perpendicular to the direction of propagation
According to Maxwell an accelerated charge produces a sinusoidal time varying magnetic field which in turn produces a time varying electric field .The two fields so produced are mutually perpendicular to each other and constitute an electromagnetic wave and propagate in space in the direction perpendicular to both the fields. An $AM$ wave is also an electromagnetic wave therefore its magnetic field component would be parallel to ground and perpendicular to the direction of propogation.
View full question & answer→MCQ 731 Mark
The matter$-$wave picture of electromagnetic wave/radiation elegantly incorporated the:
- ✓
Heinsenbergs uncertainty principle
- B
- C
- D
AnswerCorrect option: A. Heinsenbergs uncertainty principle
The matter$-$wave picture of electromagnetic wave/radiation elegantly incorporated the Heisenberg uncertainty principle.
View full question & answer→MCQ 741 Mark
In the propagation of electromagnetic waves, the angle between the direction of propagation and plane of polarisation is
- ✓
$0^\circ$
- B
$45^\circ$
- C
$90^\circ$
- D
$180^\circ$
AnswerCorrect option: A. $0^\circ$
Plane of polarization is a confinement of the electric/ magnetic field vector to a given plane along the direction of propagation. Therefore angle between them is $0^\circ .$
View full question & answer→MCQ 751 Mark
The polarisation of electromagnetic wave is in:
- A
The directions of electric and magnetic field
- ✓
The directions of electric field
- C
The direction of magnetic field
- D
AnswerCorrect option: B. The directions of electric field
View full question & answer→MCQ 761 Mark
Mark the correct option in impure spectrum:
- A
Order of colours is reverse
- B
Order of colours is irregular
- ✓
- D
AnswerThe order of colours is are straight, hence option a is false.
Order of colours are regular.
Colours are overlapped each other in impure spectrum.
Colours are present in impure spectrum.
View full question & answer→MCQ 771 Mark
Which of the following proves that electromagnetic waves are transverse?
AnswerOnly transverse waves can be polarised.
View full question & answer→MCQ 781 Mark
Which among the following has a frequency range of $500\ kHz$ to $1000\ MHz?$
AnswerRadio Waves are generally in the frequency range $➔ 500\ kHz$ to $1000\ MHz$. Radio waves are used for long$-$distance communication, such as in television, mobiles, and radios. These devices receive radio waves and convert them to mechanical vibrations in the speaker to create sound waves.
View full question & answer→MCQ 791 Mark
What is the phase difference between electric and magnetic fields in an electromagnetic wave?
- A
$0$
- ✓
$\pi$
- C
$\frac{\pi}{2}$
- D
$\frac{\pi}{4}$
AnswerThe phase difference between electric and magnetic fields in an electromagnetic wave is$\pi$.
View full question & answer→MCQ 801 Mark
Ultraviolet rays coming from sun are absorbed by:
View full question & answer→MCQ 811 Mark
Which of the following is used to investigate the structure of solids?
AnswerCorrect option: C. $X-$Rays
$X-$Rays are used to investigate the structure of solids.
View full question & answer→MCQ 821 Mark
Which of the following has/have zero average value in a plane electromagnetic wave?
- ✓
Both magnetic and electric field
- B
- C
- D
AnswerCorrect option: A. Both magnetic and electric field
View full question & answer→MCQ 831 Mark
A plane electromagnetic wave with a single frequency moves in vacuum in the positive $x$ direction. Its amplitude is uniform over the $yz$ plane its wavelength:
AnswerThe same amount of energy passes through equal areas parallel to the $yz$ plane as the wave travels in the $+x$ direction, so the amplitude and the intensity, which is proportional to the square of the amplitude, do not change.
View full question & answer→MCQ 841 Mark
The concept of displacement current introduced by Maxwell removes asymmetry between.
- A
Gauss's law in electrostatics and Gauss's law in magnetism
- ✓
Faraday's law and Ampere' law
- C
Gauss's law and Ampere's law
- D
Gauss's law and Faraday's law
AnswerCorrect option: B. Faraday's law and Ampere' law
Maxwell added the concept of displacement current in AMpere Circuit Law which governs the conduction in the wire of conduction current. After the deviation compass between capacitor Maxwell thought of magnetic lines which would be the result of varying current known as displacement current. So by continuing the displacement in Amperes law, Maxwell was able to show the result of Amperes conduction in circuit moving electrons and also the result of faraday generation of $ME$ waves.
View full question & answer→MCQ 851 Mark
Choose the correct answer from the alternatives given.
Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor.
AnswerDisplacement current inside a capacitor is given by:-
$\text{i}_\text{d}=\in_0\frac{\phi\text{E}}{\text{dt}}$
where $\phi_\text{E}$ is the electric flux inside the capacitor
The displacement current is developed inside a capacitor when there is a change in the electric flux linked with the capacitor.
The change in electric flux can occur in both the cases either the charge increases or decreases on the capacitor. This will lead to a change in flux linked with the coil.
View full question & answer→MCQ 861 Mark
According to Maxwell’s Hypothesis, a changing electric field gives rise to?
AnswerA changing electric field gives rise to a magnetic field.
View full question & answer→MCQ 871 Mark
In the propagation of light waves, the angle between the plane of vibration $\&$ plane of polarisation is:
- A
$0^\circ$
- ✓
$90^\circ$
- C
$45^\circ$
- D
$80^\circ$
AnswerCorrect option: B. $90^\circ$
The plane of polarisations is that plane in which there is no vibration. While a plane including the direction of light propagation and the direction of electric field is called the plane of vibration. The angle between them is $90^\circ .$
View full question & answer→MCQ 881 Mark
Who first proposed the light as an electromagnetic wave?
AnswerIn $1864$, Maxwell predicted the existence of electromagnetic waves, the existence of which had not been confirmed before that time, and out of his prediction came the concept of light being a wave, or more specifically, a type of electromagnetic wave. It is a wide intuition that Albert Einstein proposed the dual nature theory but is not correct.
View full question & answer→MCQ 891 Mark
A magnetic field can be produced by:
- A
- B
A changing electric field.
- C
- ✓
AnswerAccording to Ampere$-$Maxwell's Law, a magnetic field is produced due to the conduction current in a conductor and the displacement current. The conduction current is actually the motion of the charge. The displacement current is due to the changing electric field. The displacement current is given by,
$\text{i}_\text{d}=\in_0\frac{\text{d}\phi_\text{E}}{\text{dt}}$ $\big(\because\phi_\text{E}$ is the electric flux$\big)$
Thus, the magnetic field is produced by the moving charge as well as the electric field.
View full question & answer→MCQ 901 Mark
The wavelength of the wave is.
- A
$3.5m$
- B
$5.5m$
- ✓
$7.5m$
- D
$9.5m$
AnswerCorrect option: C. $7.5m$
Using $\text{v}=\text{v}\lambda$ where $v$ is the speed of $EM$ wave.
As $\text{V}=\text{c}=3\times\frac{10\text{m}}{\text{s}}$
$\Rightarrow3\times10^8=40\times10^6\lambda$
$\Rightarrow\lambda=7.5\text{m}$
View full question & answer→MCQ 911 Mark
A linearly polarized electromagnetic wave given as $\text{E}=\text{E}_0\hat{\text{i}}\cos(\text{kz}-\omega\text{t})$ is incident normally on a perfectly reflecting infinite wall at $z = a$. Assuming that the material of the wall is optically inactive, the reflected wave will be given as:
- A
$\text{E}_\text{r}=-\text{E}_0\hat{\text{i}}\cos(\text{kz}-\omega\text{t})$
- ✓
$\text{E}_\text{r}=\text{E}_0\hat{\text{i}}\cos(\text{kz}+\omega\text{t})$
- C
$\text{E}_\text{r}=-\text{E}_0\hat{\text{i}}\cos(\text{kz}+\omega\text{t})$
- D
$\text{E}_\text{r}=\text{E}_0\hat{\text{i}}\sin(\text{kz}-\omega\text{t})$
AnswerCorrect option: B. $\text{E}_\text{r}=\text{E}_0\hat{\text{i}}\cos(\text{kz}+\omega\text{t})$
Key concept: When a wave is reflected from a denser medium or perfectly reflecting wall made with optically inactive material, then the type of wave doesn't change but only its phase changes by $180^\circ $ or $\hat{\text{I}}€$ radian.
View full question & answer→MCQ 921 Mark
In case of the electromagnetic waves the angle between the electric and magnetic field vectors is.
AnswerCorrect option: B. $\frac{\pi}{2}$
Electromagnetic waves are formed when an electric field couples with a magnetic field. The magnetic and electric fields of an electromagnetic wave are perpendicular to each other and to the direction of the wave, as shown in figure.
View full question & answer→MCQ 931 Mark
In an electromagnetic wave, the direction of the magnetic induction $B$ is
- A
parallel to the electric field $E$
- ✓
perpendicular to the electric field $\overrightarrow{\text{E}}$
- C
antiparallel to the pointing vector $\overrightarrow{\text{S}}$
- D
AnswerCorrect option: B. perpendicular to the electric field $\overrightarrow{\text{E}}$
The electric field is always perpendicular to the magnetic field, and both fields are directed at right$-$angles to the direction of propagation of the wave. In fact, the wave propagates in the direction $\overrightarrow{\text{E}}\times\overrightarrow{\text{B}}$ Electromagnetic waves are clearly a type of transverse wave.
View full question & answer→MCQ 941 Mark
- ✓
continuous when electric field is changing in the circuit
- B
continuous when magnetic field is changing in the circuit
- C
continuous in both types of fields
- D
continuous through wires and resistance only
AnswerCorrect option: A. continuous when electric field is changing in the circuit
View full question & answer→MCQ 951 Mark
Electromagnetic waves are produced by:
AnswerA static charge produces an electrostatic field. A moving charge produces a magnetic field. Electromagnetic waves are produced by an accelerating charge.
View full question & answer→MCQ 961 Mark
Pick out the correct increasing order of energy of electromagnetic waves from the following:
- A
$E _{\text {infrared }}< E _{\text {micro }}< E _{\text {visible }}< E _{\text {ultraviolet }}< E _{\text {gamma }}$
- ✓
$E _{\text {micro }}< E _{\text {infrared }}< E _{\text {visible }}< E _{\text {ultraviolet }}< E _{\text {gamma }}$
- C
$E _{\text {micro }}< E _{\text {infrared }}< E _{\text {visible }}< E _{\text {gamma }}< E _{\text {ultraviolet }}$
- D
$E _{\text {micro }}$
AnswerCorrect option: B. $E _{\text {micro }}< E _{\text {infrared }}< E _{\text {visible }}< E _{\text {ultraviolet }}< E _{\text {gamma }}$
The energy of electromagnetic waves is directly proportional to the frequency of the electromagnetic waves. So the order of frequency is given as:
$v_{\text {micro }} < v_{\text {infrared }}$ Since $E=h v$
$\rightarrow E \propto v$
The order of energy is as follows:
$E _{\text {micro }} < E _{\text {infrared }} < E _{\text {visible }} < E _{\text {ultraviolet }} < E _{\text {gamma }}$
View full question & answer→MCQ 971 Mark
Unpolarized light falls first on polarizer (P) and then on analyzer (A). If the intensity of the transmitted light from the analyser is $\frac{1}{8}\text{th}$ of the incident unpolarized light. What will be the angle between optic axes of P and A?
AnswerExplanation:
Given,
$\text{I}=\frac{\text{I}_\text{0}}{2}....(\text{i})$
$\text{I}=\text{I}\cos^2\theta$ $\Big(\because\text{I}=\frac{\text{I}_0}{8}\Big)$
$\therefore\frac{\text{I}_0}{8}=\frac{\text{I}_0}{2}\cos^2\theta$
From the equation (i), we have
$\frac{1}{4}=\cos^2\theta$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\Rightarrow\cos\theta=\cos60^\circ$
$\Rightarrow\theta=60^\circ$
View full question & answer→MCQ 981 Mark
Radiations of intensity $\frac{0.5\text{W}}{\text{m}^2}$ are striking a metal plate. The pressure on the plate is.
- ✓
$\frac{0.166\times10^{-8}\text{N}}{\text{m}^2}$
- B
$\frac{0.332\times10{-8}\text{N}}{\text{m}^2}$
- C
$\frac{0.111\times10{-8}\text{N}}{\text{m}^2}$
- D
$\frac{0.083\times10{-8}\text{N}}{\text{m}^2}$
AnswerCorrect option: A. $\frac{0.166\times10^{-8}\text{N}}{\text{m}^2}$
Intensity or power per unit area of the radiations,
$P = pv$
$\Rightarrow\text{P}=\frac{\text{P}}{\text{v}}=\frac{0.5}{3\times10^8}=\frac{0.166\times10^{-8}\text{N}}{\text{m}^2}$
View full question & answer→MCQ 991 Mark
A parallel plate capacitor consists of two circular plates each of radius $12\ cm$ and separated by $5.0\ mm$. The capacitor is being charged by an external source. The charging current is constant and is equal to $0.15A$.The displacement current is:
- A
$15A$
- B
$1.5A$
- ✓
$0.15A$
- D
$0.015A$
AnswerCorrect option: C. $0.15A$
According to Maxwell's hypothesis, a displacement current will flow through a capacitor when the potential difference across its plates is varying.
Thus a varying electric field will exist between the plates and this displacement current is same in magnitude to the current flowing in outer circuit.
Here, the current in the outer circuit is $0.15 A$. Thus $0.15A$ will be the displacement current.
View full question & answer→MCQ 1001 Mark
Infrared radiation are detected by:
AnswerInfrared radiation is detected by photometer.
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