MCQ Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 8

The correct sequence of various regions in absorption spectrum is?

0%
Ultraviolet, visible, infra-red, microwave
0%
Microwave, visible, infra-red, X-rays
0%
Infra-red, gamma rays, ultraviolet, microwave
0%
Visible, ultraviolet, microwave, infra-red

Choose the correct answer from the alternatives given.
A plane electromagnetic wave of frequency

$25 MHz$ travels in free space along

$X$ -direction. At a particular point in space and time, electric field

$\vec E=6.3\ \hat j\ V/m$ . What is

$B$ at this point.

0%
$1.2 \, \times \, 10^{-6} \, T$
0%
$1.2 \, \times \, 10^{-8} \, T$
0%
$2.1 \, \times \, 10^{-6} \, T$
0%
$2.1 \, \times \, 10^{-8} \, T$

Explanation

Given: The frequency of the electromagnetic wave is

$25\ MHz$ .

The electric field at the particular point is

$6.3\hat j\ V/m$

To find: The magnetic field at that point.

The magnetic field of the electromagnetic wave at a point is given by:

$B = \dfrac{E}{c}\\= \dfrac{6.3}{3 \times 10^8}\\ \Rightarrow2.1 \times 10^{-8} T$

So, option

$(D)$ is correct.

A microwave and an ultrasonic sound wave have the same wavelength. Their frequencies are in the ratio (approximately)

0%
$10^2$
0%
$10^4$
0%
$10^6$
0%
$10^8$

Explanation

Frequency of microwaves,

$\upsilon_m \approx 10^{11}Hz$
Frequency of ultrasonic sound waves,

$\upsilon_m \approx 10^5Hz$

$\therefore \, \dfrac{\upsilon_m}{\upsilon_u} = \dfrac{10^{11}}{10^{5}} = 10^6$

The ultra-high frequency band of radio waves in electromagnetic wave is used as in:

0%
television waves
0%
cellular phone communication
0%
commercial FM radio
0%
both (a) and (c)

Explanation

The ultra high-frequency band of radio waves in FM waves is used in cellular phone communication. It is of the order of

$10^9Hz$

Choose the correct answer from the alternatives given.
Which of the following is not true for electromagnetic waves?

0%
They transport energy.
0%
They have momentum.
0%
They travel at different speeds in air depending on their frequency.
0%
They travel at different speeds in medium depending on their frequency.

Choose the correct answer from the alternatives given.
Suppose that the electric field amplitude of an electromagnetic wave propagating along x-direction is

$E_0$ = 120 N

$C^{-1}$ and that frequency is:

$\upsilon$ = 50.0 MHz.

0%
The expression of elctric field is $\vec{E} = 120sin \left ( \dfrac{\pi}{3}x-100\pi \times10^6t \right )\hat{j}$
0%
The expression of electric field is $\vec{E} = 60sin \left ( \dfrac{\pi}{3}x-100\pi \times10^6t \right )\hat{j}$
0%
the expression of magnetic field is $\vec{B} = 40\times10^{-8}sin \left ( \dfrac{\pi}{3}x-\pi \times10^8t \right )\hat{k}$
0%
Both (a) and (c)

Explanation

$E_0{max}=120\\ \omega=2\pi f\\ \quad=2\pi\times50\times10^6\\ \quad=100\pi\times10^6\\K=\cfrac{\omega}{V}\\=\cfrac{100\pi\times10^6}{3\times10^8}\\ \quad\quad\quad=\cfrac{\pi}{3}\\ \cfrac{E_{max}}{B_{max}}=C\\ \cfrac{120}{B_{max}}=3\times10^8\\B_{max}=40\times10^{-8}$

Accordingly puttingthe equation.

$E=120\sin(Kx-\omega t)\\ \quad=120\sin(\cfrac{\pi}{3}x-100\pi\times10^6t)\\B=40\times10^{-8}\sin(Kx-\omega t)\\ \quad=40\times10^{-8}\sin(\cfrac{\pi}{3}-100\pi\times10^{-6}t)$

The electric field of an electromagnetic wave traveling through the vacuum is given by the equation

$E=E_0\ sin (Kx-\omega t).$ The quantity that is independent of wavelength is:

0%
$k\omega$
0%
$\dfrac{k}{\omega}$
0%
$k^2\omega$
0%
$\omega$

Explanation

To find: The quantity that is independent of the wavelength.

The angular frequency

$\omega$ is given by:

$\omega \, = \, 2\pi \nu$

The frequency of a wave varies with the wavelength. So, angular frequency is dependent on wavelength.

The quantity

$k$ is defined as the wavenumber and it is given by:

$k = \dfrac{2\pi}{\lambda}$

It shows that

$k$ is dependent on wavelength.

The value of

$\dfrac{k}{\omega}$ can be obtained as:

$\dfrac {k}{\omega} \, = \, \dfrac{2\pi / \lambda}{2\pi \nu}\\\implies \, \dfrac{1}{\nu \lambda} \, = \, \dfrac{1}{c}\,\,\ \ \ \ \ \ \ \ \ \ \ \ \ (\because \, c \, = \, \nu \lambda)$
where c is the speed of electromagnetic wave in vacuum. It is a constant whose value is

$3 \, \times \, 10^8 \, ms^{-1}$ .

So, option

$(B)$ is correct.

Choose the correct answer from the alternatives given.
A radio can tune to any station in 7.5 MHz to 12 MHz band. The corresponding wavelength band is

0%
40 m to 25 m
0%
30 m to 25 m
0%
25 m to 10 m
0%
10 m to 5 m

Explanation

$Here, \, \upsilon_1 \, =\, 7.5 \, MHz, \, \upsilon_2 \, =\,12 \, MHz$

$\therefore \, \lambda_1 = \dfrac{c}{\upsilon_1} = \dfrac{3 \times 10^8}{7.5 \times 10^6} = 40m$

$and \, \lambda_2 = \dfrac{c}{\upsilon_2} = \dfrac{3 \times 10^8}{12 \times 10^6} = 25m$

Which of the following electromagnetic wave is used in high precision application like LASIK eye surgery?

0%
Microwave
0%
Ultraviolet rays
0%
Gamma rays
0%
X-rays

Which of the following electromagnetic wave play an important role in maintaining the earths warmth or average temperature through the greenhouse effect?

0%
Visible rays
0%
Infrared waves
0%
Gamma rays
0%
Ultraviolet rays

Which of the following electromagnetic waves is used in medicine to destroy cancer cells?

0%
IR-rays
0%
Visible rays
0%
Gamma rays
0%
Ultraviolet rays

Maxwell in his famous equations of electromagnetism, introduced the concept of

0%
ac current
0%
displacement current
0%
impedance
0%
reactance

Explanation

Maxwell's equations are:

$\nabla .E=\rho / \epsilon$

$\nabla.B=0$

$\nabla \times E= -\dfrac{dB}{dt}$

$\nabla \times B= \mu_0J+ \dfrac{1}{c^2} \dfrac{dE}{dt}$

so, considering the last eqn. written,

$\nabla \times B=\mu_0 J$ is the Ampere's eqn.

so, Maxwell modified the Ampere's eqn. and introduced the concept of displacement current.

So, displacement current =

$\dfrac{1}{c^2} \dfrac{dE}{dt}$

Hence the correct option is

$(B)$

Choose the correct answer from alternatives given.
The phenomena involved in the reflection of radio waves by ionosphere is similar to

0%
reflection of light by a plane mirror.
0%
total internal reflection of light in air during a mirage.
0%
dispersion of light by water molecules during the formation of a rainbow.
0%
scattering of light by the particles of air.

Explanation

The phenomenon involved in the reflection of radiowaves by ionosphere is similar to total internal reflection of light in air during a mirage (angle of incidence gt critical angle). It is same as the formation of the rainbow because it also involves the process of total internal reflection.

Choice

$(b)$ is correct.

An electromagnetic wave radiates outwards from a dipole antenna, with

$E_0$ as the amplitude of its electric field vector. The electric field

$B_0$ which transports significant energy from the source falls off as

0%
$\dfrac{1}{r^3}$
0%
$\dfrac{1}{r^2}$
0%
$\dfrac{1}{r}$
0%
remains constant.

Explanation

An antenna that produces the Electromagnetic wave are radiated outwards. The amplitude of electric field vector

$(E_0)$ . This electric field vector transports the energy from the source through the medium.

The electric field intensity of the wave from the source at a distance is inversely proportional to the distance between the source and the point.

$E_0=\dfrac 1r$

Option

$(C)$ is correct.

Which of the following doesnt represent Maxwells equation?

0%
$\oint { \overrightarrow { E } .\bar { dA } = } \dfrac { q }{ { \varepsilon }_{ 0 } }$
0%
$\oint { \overrightarrow { B } .\bar { dA } = } 0$
0%
$\oint { \overrightarrow { E } .\bar { dl } =\dfrac { -dB }{ dt } }$
0%
$\oint { \overrightarrow { B } .\bar { dl } = } { \mu }_{ 0 }{ I }_{ C }+{ \mu }_{ 0 }{ \varepsilon }_{ 0 }\dfrac { d{ \phi }_{ E } }{ dt }$

Explanation

Following are Maxwell's equations :

Therefore, the equations

$(A), (B)$ and

$(D)$ represent the correct form of Maxwell's equations whereas the equation

$(C)$ shows the rate of change of magnetic field

$\vec B$ but actually it should be the rate of change of magnetic flux

$\phi$ . So, equation

$(C)$ is wrong.

So, option

$(C)$ is correct.

1170522_950877_ans_9978f2f8a09046779a5f72e9f0598245.jpeg

An electromagnetic waves can be produced, when charge is

0%
Moving with constant velocity
0%
Moving in a circular orbit
0%
Falling in an electric field
0%
both (b) and (c)

The charge on a parallel plate capacitor varies as

$q = {q}_{0}cos2vt$ . The plates are very large and close together (area = A, separation = d ) , the displacement current through the capacitor is

0%
${q}_{0}2v sinvt$
0%
$-{q}_{0}2v sinvt$
0%
${q}_{0}2 sinvt$
0%
${q}_{0}v sinvt$

Explanation

Given :

$q = q_0 \cos 2vt$

To find :

$I_d = ?$

Formula :

$I_d = \dfrac{dq}{dt}$

Sol : From formula,

The displacement current through the capacitor

$I_d= \dfrac{dq}{dt} = \dfrac{d}{dt} (q_0\cos 2vt) = -q_0 \sin 2vt \times 2v$

$\therefore I_d = -q_02v \sin 2vt$

$\therefore$ Correct option is (B)

A plane electromagnetic wave travels in a vacuum in the z-direction. If the frequency of the wave is 40 MHz, then its wavelength is:

0%
$5\ m$
0%
$7.5\ m$
0%
$8.5\ m$
0%
$10\ m$

Explanation

Given :

$v = 40 \times 10^6 Hz, C = 3\times 10^8m/s$

$\text{Formula: v = c}\lambda$

To find :

$\lambda = ?$

Sol : From formula

$v = c\lambda$

$\therefore = \dfrac{c}{v}$

$= \dfrac{3\times 10^8}{40 \times 10^6}$

$= 0.075 \times 10^8 \times 10^{-6}$

$= 0.075 \times 10^2$

$=7.5m$

$\therefore$ correct option is (B)

The matter-wave picture of electromagnetic wave/radiation elegantly incorporated the:

0%
Heinsenbergs uncertainty principle
0%
correspondence principle
0%
cosmic theory
0%
Hertzs observations

The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is:

0%
$c : 1$
0%
$c^2: 1$
0%
$1 : 1$
0%
$\sqrt{c} : 1$

Explanation

The intensity of an electromagnetic wave is given by:

$I=U_{av}c$

here,

$U_{av}$ is the average energy and

$c$ is the speed of light.

The average energy in terms of the electric field is written as:

$U_{av}=\dfrac 12\varepsilon_oE_o^2$

Average energy in terms of magnetic field:

$U_{av}=\dfrac 12\dfrac{B_o^2}{\mu_o}$

We know that,

$E_o=cB_o$ and

$c=\dfrac{1}{\sqrt{\mu_o\varepsilon_o}}$

$\therefore U_{av\ (electric\ field)}=\dfrac 12\varepsilon_o\times\dfrac{1}{\mu_o\varepsilon_o}B_o^2$

$\Rightarrow\dfrac 12\dfrac{B_o^2}{\mu_o}$

Since the energy due to electric and magnetic field is same. so, both will have the same contribution to the intensity.

The part of the spectrum of electromagnetic radiation used to cook food is

0%
Ultraviolet rays
0%
Cosmic rays
0%
X rays
0%
Microwaves

An electromagnetic wave of frequency

$\upsilon = 3$ MHz passes from a vacuum into a dielectric medium with permittivity

$= 4$ , Then

0%
Wavelength and frequency both become half.
0%
Wavelength is doubled and frequency remains unchanged.
0%
Wavelength and frequency both remains unchanged.
0%
Wavelength is halfved and frequency remains unchanged.

Explanation

Frequency remains constant during refraction.

$v_{med}=\dfrac{1}{\sqrt{\mu_0 \epsilon_0 *4}}$ =

$\dfrac{c}{2}$

$\dfrac {\lambda_{med}}{\lambda_{air}} = \dfrac{v_{med}}{v_{air}}$

$\dfrac{\dfrac{c}{2}}{c}$ =

$\dfrac{1}{2}$

\therefore

Which of the following is not true for electromagnetic waves?

0%
They transport energy.
0%
They have momentum.
0%
They travel at different speeds in air depending on their frequency.
0%
They travel at different speeds in medium depending on their frequency.

Explanation

Electromagnetic wave have constant velocity in a particular medium and it is independant of frequency and it is equal to

$V=\sqrt{\dfrac{1}{\mu_0 \epsilon_0}}=3 \times 10^8 \ m/s$

Option C

Which of the following ray is not electromagnetic wave?

0%
X-rays
0%
$\gamma$ -rays
0%
$\beta$ -rays
0%
Heat rays

The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is

${B}_{0}$ = 510 nT. The amplitude of the electric field part of the wave is

0%
120 N ${C}^{-1}$
0%
163 N ${C}^{-1}$
0%
510 N ${C}^{-1}$
0%
153 N ${C}^{-1}$

Explanation

Given,

$B_0=510nT$

$c=3\times 10^8m/s$

The magnitude of electric field is given by

$E_0=B_0 c$

$E_0=510\times 10^{-9}\times 3\times 10^8$

$E_0=153NC^{-1}$

The correct option is D.

The rms value of the electric field of the light coming from sun is 720 N

$C^{-1}$ . The average total energy density of the electromagnetic wave is

0%
$3.3 10^{-3} J m^{-3}$
0%
$4.58 10^{-6} J m^{-3}$
0%
$6.37 10^{-9} J m^{-3}$
0%
$81.35 10^{-12} J m^{-3}$

Explanation

Given: The RMS value of the electric field of light is

$720\ N/C$ .

Energy density of electromagnetic wave is given by:

$E=\epsilon_0 E_{rms}^2$

$=8.85\times 10^{-12}\times (720)^2$

$E\approx 4.58\times 10^{-6}J/m^3$

Option

$(B)$ is corret.

Which one of the following is a property of a monochromatic, plane electromagnetic wave in free space?

0%
Electric and magnetic fields have a phase difference of ${\pi}/{2}$
0%
The energy contribution of both electric and magnetic fields are equal
0%
The direction of propagation is in the direction of $\overline { B } \times \overline { E }$
0%
The pressure exerted by the wave is the product of its speed and energy density.

Explanation

The energy contribution of both electric and magnetic fields are equal, Energy density is written as

$E=\dfrac 12\epsilon_0E^2=\dfrac{B^2}{2 \mu_0}$

And the direction of propagation of wave is written as

$\hat V= \hat E\times \hat B$

Option B

The ultra high frequency band of radiowaves in electromagnetic waves is used as in

0%
Television waves
0%
Cellular phone communication
0%
Commercial FM radio
0%
Both (b) and (c)

Explanation

Radio Waves

Radio waves are usually in the frequency range from $500\ kHz$ to $1000\ MHz$ .
TV waves range from $54\ MHz$ to $890\ MHz$ .
The FM (frequency modulated) radio band is from $88\ MHz$ to $108\ MHz$ .
Cellular phones also use radio waves to transmit voice communication in an ultra-high frequency $(UHF)$ band.

As given in last point radio waves are used in cellular phone communication.

A plane electromagnetic wave propagating along x direction can have the following pairs of

$\overrightarrow { E }$ and

$\overrightarrow { B }$ .

0%
${ E }_{ y },{ B }_{ z }$
0%
${ E }_{ z },{ B }_{ y }$
0%
${ E }_{ x },{ B }_{ y }$
0%
both (a) and (b)

Explanation

Let us consider, in electromagnetic wave

$x, y,$ and

$Z$ axis are there.

By using Right Hand Rule,

$E_y \times B_z = V_x$

By using Right Hand Rule

$E_Z \times B_Y = -V_x$

So, from above two condition there is a possible of two condition there is a possibility of two pairs in which wave can propagate in +ve direction and in second condition it can propagate in -ve x-axis.

Therefore, correct option is (D)

1502786_950890_ans_69ea3f81f7994737bf34867cbe694156.png

Which of the following electromagnetic wave play an important role in maintaining the earths warmth or average temperature through the green house effect?

0%
Visible rays
0%
Infrared rays
0%
Gamma rays
0%
Ultraviolet rays

An electromagnetic wave radiates outwards from a dipole antenna, with

${ E }_{ 0 }$ as the amplitude of its electric field vector. The electric field

${ E }_{ 0 }$ which transports significant energy from the source falls off as

0%
$\dfrac { 1 }{ { r }^{ 3 } }$
0%
$\dfrac { 1 }{ { r }^{ 2 } }$
0%
$\dfrac { 1 }{ { r }}$
0%
remains constant.

Explanation

A diode antenna radiates the electromagnetic wave outwards. The amplitude of electric field vector

$(E_0)$ which transports significant energy from the source falls intensity inversely as the distance (r) from the antenna,

i.e,

$E_{0}\propto \dfrac{1}{r}$

The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is

0%
Microwaves, Infrared, ultraviolet, gamma rays.
0%
Infrared, Microwaves, ultraviolet, gamma rays.
0%
ultraviolet, gamma rays, Infrared, Microwaves
0%
Microwaves, gamma rays, Infrared, ultraviolet

Explanation

The wavelength of the Electromagnetic waves is inversely proportional to the frequency of the wave. The frequency of the gamma rays is highest whereas the ultraviolet, Infrared and microwaves follow in decreasing order.

Therefore, the decreasing order of wavelengths will be opposite to the frequency order. So, the microwaves have the longest wavelength and the gamma rays have the smallest.

Microwave

$>$ Infrared

$>$ Ultraviolet

$>$ Gamma ray.

An electromagnetic radiation has an energy of 13.2 KeV. Then the radiation belongs to the region of

0%
Visible light
0%
Ultraviolet light
0%
Infrared
0%
X-ray

Explanation

As we already know that Energy of electromagnetic radiation depends upon the frequency of radiation.

Energy of a photon of

$EM$ radiation with frequency is

$E=h\nu$

Frequency and wave length are related by

$λ\nu =c$

$E=h\nu=\dfrac{hc}λ$

$λ=\dfrac{hc}E$

$=\dfrac{1240ev-nm}{13.2kev}$

$=\dfrac{1240ev-nm}{13.2\times 1000ev}$

$=\dfrac{1240}{13200}=0.093\ nm$

$=0.093\ nm$

$X-ray$ wavelength is in the Range of

$0.01\ nm$ to

$1\ nm$ .

Hence answer is X-rays.

One requires 2 IeV of energy to dissociate a carbon molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dis-acceleration lines is

0%
visible region
0%
interacted region
0%
ultraviolet region
0%
microwave region

Explanation

$E = 21 eV = 21 \times 1.6 \times 10^{-19}$

$\alpha = \dfrac{21 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}} = 5.07 \times 10^{15} Hz$

This is belongs ultraviolet region

A. wavelength of microwave is greater than that of ultraviolet rays.
B. The wave length of infrared rays is lesser than that of ultraviolet rays.
C. wavelength of microwave is lesser than that of ultraviolet rays
D. Gamma rays has shortest wavelength in the electromagnetic spectrum.
Choose the correct option from the given options.

0%
A and B are true
0%
B and C are true
0%
C and D are true
0%
A and D are true

Explanation

The wavelength and the frequency of the Electromagnetic waves are inversely proportional to each other.

$A$ . The wavelength of Microwave is greater than UV rays.

The frequency of the microwave is less than that of the Ultraviolet rays, the wavelength of the Microwaves will be longer than the Ultraviolter rays.

$D$ . Gamma rays have the shortest wavelength in the EM spectrum.

The frequency of the gamma rays is the highest in the electromagnetic spectrum. So, it will have the shortest wavelength.

The wavelength of the infrared is higher than the Ultraviolet rays and also the wavelength of the microwave is higher than Ultraviolet rays. So, options

$B$ and

$C$ are not true.

Option

$(D)$ is correct.

Photons of an electromagneitc radiation has an energy

$11keV$ each. To which region of electromagnetic spectrum does it belong?

0%
$X-ray\ region$
0%
$Ultra\ violet\ region$
0%
$infrared\ region$
0%
$visible\ region$

Radio waves of constant amplitude can be generated with:

0%
Filter
0%
Rectifier
0%
FET
0%
Oscillator

The operating voltage in coolidge tube is

${10}^{5}V$ . The speed of X-Rays produced is ________

$m{s}^{-1}$

0%
$2\times { 10 }^{ 8 }\quad$
0%
${ 10 }^{ 6 }$
0%
${ 10 }^{ 5 }$
0%
$3\times { 10 }^{ 8 }$

Explanation

X-ray is an electromagnetic waves which moves with a speed equal to

$3\times 10^8 \ m/s$

A beam of ultraviolet light of all wavelengths passes through hydrogen gas at room temperature, in the

$x-$ direction. Assume that all photons emitted due to electron transitions inside the gas emerge in the

$y-$ direction. Let

$A$ and

$B$ denote the lights emerging from the gas in the

$x-$ and

$y-$ directions respectively. Then,

0%
some of the incident wavelengths will be absent in $A$
0%
only those wavelengths will be present in $B$ which are absent in $A$
0%
$B$ will contain some visible light
0%
$B$ will contain some infrared light

Explanation

$A, C, D$

If a source of power

$4$ kW produces

$10^{26}$ photons/second, the radiation belongs to a part of the spectrum called:

0%
X-rays
0%
Ultraviolet rays
0%
Microwaves
0%
$\gamma$ -rays

Explanation

$4000 \ J$ of energy is produced from

$10^{26}$ photons in one second.

Energy of each photon

$E = \dfrac{4000}{10^{26}} = 4\times 10^{-23} \ J$

Frequency of each photon

$\nu = \dfrac{E}{h} = \dfrac{4\times 10^{-23}}{6.63\times 10^{-34}} = 6\times 10^{10} \ Hz$

So the radiation belongs to micro waves.

Correct answer is option C.

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is:

0%
$\dfrac{E}{c}$
0%
$\dfrac{2E}{c}$
0%
$Ec$
0%
$\dfrac{E}{c^2}$

Explanation

Hint:-Momentum transfered to surface is equal to change in momentum of radiation.

Step 1: WRITE INITIAL AND FINAL MOMENTUM OF RADIATION

$P_i = \dfrac{E}{c}$ and

$P_f = \dfrac{-E}{c}$ too.

Step 2: CALCULATE MOMENTUM TRANSFERED TO SURFACE

Momentum transferred to the surface

$\Rightarrow P_{final} – P_{initial}.$

So,

Net Momentum

$= \dfrac{E}{c}-(-\dfrac{E}{c}) \Rightarrow \dfrac{2E}{c}$ .

The intensity of sun on earth is

$1400 W/m^2$ . Assuming earth to be a black body. Calculate radiation pressure?

0%
$4.66 \times 10^6 N/m^2$
0%
$1.4 \times 10^{-6} N/m^2$
0%
$4.66 \times 10^{-6} N/m^2$
0%
$7 \times 10^{-6} N/m^2$

A parallel plate capacitor with circular plates of radius

$R$ is being charged as shown. At the instant shown, the displacement current in the region between the plates enclosed between

$\cfrac{R}{2}$ and

$R$ is given by

0%
$\cfrac{3}{4}i$
0%
$\cfrac{1}{4}i$
0%
$3i$
0%
$\cfrac{4}{3}i$

Explanation

Displacement current is given by

$I={{\varepsilon }_{0}}\dfrac{d{{\phi }_{e}}}{dt}$

$=\,{{\varepsilon }_{0}}A\dfrac{dE}{dt}\,\,\,\,(\because \,\phi =A.E)$

$=A\dfrac{d}{dt}\left(\dfrac{q}{{{A}^{'}}}\right)$

$=\dfrac{A}{{{A}^{'}}}i\,\,(\because \dfrac{dq}{dt}=i)$

$=\dfrac{\pi {{\left(\dfrac{R}{2}\right)}^{2}}}{\pi {{R}^{2}}}$

$=\dfrac{1}{4}i$

The electric field associated with an e.m. wave in vacuum is given by

$\vec {E} = 40\cos (kz - 6\times 10^{8}t)\hat {i}$ , where

$E, z$ and

$t$ in

$volt/m$ , meter and seconds respectively. The value of wave vector

$k$ is

0%
$6m^{-1}$
0%
$3m^{-1}$
0%
$2m^{-1}$
0%
$0.5m^{-1}$

Explanation

Given: The electric field associated with an electromagnetic wave in vacuum is given by

$\vec E =40 \cos(kz−6\times 10^8t)\hat i$ , where E, z and t are in volt per meter, meter and second respectively.

To find the value of wave vector k

Solution:

We know electromagnetic wave eqution is

$E=E_0\cos(kz-\omega t)$

And given equation is

$\vec E =40 \cos(kz−6\times 10^8t)\hat i$

By comparing these two, we get

$\omega=6\times10^8$ and

$E_0=40\hat i$

we also know,

Speed of electromagnetic wave,

$v=\dfrac \omega k$

where v is the speed of the light

Hence,

$k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$

is the required value

The electric field part of an electromagnetic wave in a medium is represented by

$Ex=0$

$Ey= 2.5 N/C ((2*10^6 rad/m)t - (*10^-2 rad/s) x)$

$Ez=0$ the wave is

0%
Moving along X direction with frequency 10^6 Hz and wavelength 100m
0%
Moving along X direction with frequency 10^6 Hz and wavelength 200m
0%
Moving along - X direction with frequency 10^6 Hz and wavelength 100m
0%
Moving along y direction with frequency 2*10^6 Hz and lwavelength 100m

The x -ray beam coming from an x - ray tube will be

0%
Monochromatic
0%
Having all wavelengths smaller than a certain maximum wavelength
0%
Having all wavelengths larger than a certain minimum wavelength
0%
Having all wavelengths lying between a minimum and a maximum wavelength

Displacement current goes through the gap between the plates of a capaitor when the charge of the capacitor:
(i) does not change
(ii) change with time
(iii) increases
(iv) decreases

0%
( i ), ( iii ) and ( iv )
0%
( i ), and ( ii ) and ( iv )
0%
( ii ), ( iii ) and ( iv )
0%
( i ), ( ii ) and ( iii )

Explanation

Displacement current inside a capacitor

$I_D=\epsilon_0\dfrac {d\phi_E}{dt}$

where

$\phi_E$ is the electric flux inside the capacitor

Up to the time the electric flux changes, there will be a displacement current. This is possible when the charge on a capacitor changes. Therefore, the displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor or the electric field increases or decreases inside the capacitor.

The velocity of light is

$3.0 \times {10^8}m{s^{ - 1}}$ . Which value is closest to the wavelength in nanometres of a quantum of light with frequency of

$8 \times {10^{15}}{s^{ - 1}}$ ?

0%
$3 \times {10^7}$
0%
$2 \times {10^{ - 25}}$
0%
$5 \times {10^{ - 18}}$
0%
$3.7 \times {10^1}$

Explanation

Given,

$velocity = 3 \times 10^{8} m/s$

$frequency = 8 \times 10^{15} s^{-1}$

As we know ,

$f = \frac{c}{\lambda}$

putting the values , in above equation , we get ;

$8 \times 10^{15} = \frac{3 \times 10^{8}}{\lambda}$

After calculating , we get

$\lambda = 3 \times 10^{7}$ (approx)

Hence , option A is correct .

A parallel plate condenser consists of two circular plates each of radius

$2cm$ separated by a distance of

$0.1mm$ . A time varying potential difference of

$5\times {10}^{13}v/s$ is applied across the plates of the condenser. The displacement current is:

0%
$5.50A$
0%
$5.56\times {10}^{2}A$
0%
$5.56\times {10}^{3}A$
0%
$2.28\times {10}^{4}A$

Explanation

Given: A parallel plate condenser consists of two circular plates each of radius 2cm separated by a distance of 0.1mm. A time varying potential difference of

$5\times 10^{13}v/s$ is applied across the plates of the condenser.

To find the displacement current

Solution:

Radius of each circular plate,

$r = 2 cm = 0.02 m$

Distance between the plates,

$d =0.1mm=0.1\times10^{-3}m$

the change in potential difference between the plates,

$\dfrac {dV}{dt}=5\times10^{13}vs$

Permittivity of free space,

$\varepsilon_0= 8.85 \times 10^{−12} C^2 N^{−1} m^{−2}$

Hence area of the each plate,

$A=\pi r^2=\pi (0.02)^2=1.26\times 10^{-3}m^2......(i)$

Capacitance between the two plates is given by the relation,

$C=\dfrac {\varepsilon_0A}{d}\\\implies C=\dfrac {8.85\times10^{-12}\times1.26\times10^{-3}}{0.1\times10^{-3}}\\\implies C=11.15\times10^{-11}F$

Now we know

displacement current,

$I_d=\dfrac {dq}{dt}$

where

$q$ is the charge on each plate and it is equal to

$q=CV$

$I_d=C\dfrac {dV}{dt}$

By substituting the values, we get

$I_d=11.15\times10^{-11}\times5\times10^{13}\\\implies I_d=57.55\times10^2\\\implies I_d=5.7\times10^3A$

is the displacement current.

A plane electromagnetic wave of frequency

$28MHz$ travels in free space along the positive x-direction. At a particular point in space and time, electric field is

$9.3V/m$ along positive y-direction. The magnetic field (in T) at that point is

0%
$3.1\times10^{-8}$ along positive z-direction
0%
$3.1\times10^{-8}$ along negative z-direction
0%
$3.2\times10^{7}$ along positive z-direction
0%
$3.2\times10^{7}$ along negative z-direction

Explanation

$B = \dfrac{E}{C} = \dfrac{{9.3}}{{3 \times {{10}^8}}} = 3.1 \times {10^-8}$

CBSE Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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