At $$0^oC$$ body emits :

electromagnetic radiation of all wavelengths that are emitted by it at room temperature

electromagnetic radiation of single wavelength

electromagnetic radiation of fewer wavelengths than are emitted by it at room temperature

MCQ Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 10

The spectra of radiation emitted by two distant stars are shown below.
The ratio of the surface temperature of star

$A$ to that of star

$B, T_A: T_B$ , is approximately :

0%
$2:1$
0%
$4:1$
0%
$1:2$
0%
$1:1$

Ultrasonic are not used in :

0%
SONAR
0%
Sibigraphy
0%
CUSA
0%
radio waves

$0^oC$ body emits :

0%
no radiation
0%
electromagnetic radiation of single wavelength
0%
electromagnetic radiation of all wavelengths that are emitted by it at room temperature
0%
electromagnetic radiation of fewer wavelengths than are emitted by it at room temperature

X-rays are produced due to

0%
Bombarding of electrons on solids
0%
Bombarding of $\alpha$ -particle on solids
0%
Bombarding of $\gamma$ -rays on solids
0%
Bombarding of neutron on solids

Explanation

X-rays are produced due to sudden deceleration of fast-moving electrons when they collide and interact with the target element.

Hence, option

$A$ is correct.

Write true or false for the following statements :
In order to find the force experienced by a current carrying conductor , we can use Maxwell's cork screw rule.

0%
True
0%
False

The source of electromagnetic waves can be a charge

0%
moving with a constant velocity
0%
moving in a circular orbit
0%
at rest
0%
falling in an electric field

In a dark room of photography, generally red light is used. The reason is

0%
Most of the photographic films are not red light
0%
The frequency for red light is low and hence the energy $hv$ of photons is less
0%
(a) and (b) both
0%
None of these

For the production of X-ray of wavelength

$0.1\ A^o$ the minimum potential difference will be

0%
$12.4\ kV$
0%
$24.8\ kV$
0%
$124\ kV$
0%
$248\ kV$

Explanation

As we know,

$\lambda_{min}=\dfrac{hc}{eV}=\dfrac{12375}{V} \ A^o ; \ \ \therefore V=\dfrac{12375}{\lambda \ in\ A^o}=124\ kV$

A wavelength

$0.60 \mathrm{cm}$ is produced in air and it travels at a speed of

$300 \mathrm{ms}$ . lt will be an

0%
Audible wave
0%
Infrasonic wave
0%
Ultrasonic wave
0%
None of the above

Explanation

(c)

$\quad n=\dfrac{v}{\lambda}=\dfrac{300}{0.6 \times 10^{-2}} \mathrm{Hz}=\dfrac{3}{6} \times 10^{4} \mathrm{Hz}=50,000 \mathrm{Hz}$

$\Rightarrow$ Wave is ultrasonic.

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

0%
$10^6 :1$
0%
$10^2:1$
0%
$10^4:1$
0%
$10:1$

For the production of characteristic

$K_{\gamma, X-}$ ray, the electron transition is

0%
$n=2$ to $n=1$
0%
$n=3$ to $n=2$
0%
$n=3$ to $n=1$
0%
$n=4$ to $n=1$

The waves used in telecommunication are

0%
$IR$
0%
$UV$
0%
Microwaves
0%
Cosmic rays

Explanation

$Microwaves$ are electromagnetic radiation that finds applications in communications, radar, and cooking. The frequencies of microwaves range from 1 GHz to 300 GHz.Telecommunication refers to the process of exchange of information by using electronic devices over a significant distance. Hence , in telecommunication, microwaves are used.

Which of the following waves have the maximum wavelength

0%
$X-$ rays
0%
I.R. rays
0%
UV rays
0%
Radio waves

Explanation

The cortrect order for the wavelength for the given radiations is:

$\lambda_{Radiation\ waves} > \lambda_{UV-rays} > \lambda_{J-rays} > \lambda_{X-rays}$

So, radio waves has the maximum wavelength.

Which of the following shows green house effect

0%
Ultraviolet rays
0%
Infrared rays
0%
$X$ - rays
0%
None of these

Radio wave diffract around building although light waves do not. The reason is that radio waves

0%
Travel with speed larger than $c$
0%
Have much larger wavelength than light
0%
Carry news
0%
Are not electromagnetic waves

Explanation

Diffraction only occurs when the wavelength of the wave is approximately equal to the obstacle size, i.e

$\lambda \approx size$ . As the size of the buildings will always remain constant we can say that diffraction will depend on the wavelength of the wave. Visible light has the wavelength in terms of

$10^{-7}m$ which is much smaller than any building size. But the wavelengths of radio waves are much closer to the size of the buildings.

Hence, the correct answer is option (B), i.e the radio waves have much larger wavelength than visible light.

Which rays are not the portion of electromagnetic spectrum

0%
$X-$ rays
0%
Microwaves
0%
$\alpha$ - rays
0%
Radio waves

Explanation

The electromagnetic spectrum consists of radio waves, micro waves, infrared waves, visible light, ultra violet rays, x rays and

$\gamma$ rays.

So, correct answer is option (C), i.e

$\alpha$ rays.

Electromagnetic radiation of highest frequency is

0%
Infrared radiations
0%
Visible radiation
0%
Radio waves
0%
$\gamma$ - rays

The electromagnetic waves travel with a velocity

0%
Equal to velocity of sound
0%
Equal to velocity of light
0%
Less than velocity of light
0%
None of these

Explanation

Velocity of

$EM$ waves

$=\dfrac {1}{\sqrt {\mu_0{ \in}_{ 0 }}}=3\times 10^8 m/s=$ velocity of light.

The frequencies of

$X-$ rays,

$\gamma$ - rays and ultraviolet rays are respectively

$a, b$ and

$c$ . Then

0%
$a < b, b > c$
0%
$a > b, b > c$
0%
$a > b, b < c$
0%
$a < b, b < c$

Explanation

The correct for the frequency of the given radiations is:

$\nu_{y-rays} > \nu_{X-rays} > \nu_{UV-rays}$

So, the correct option is

$(A)$ .

Which of the following rays has the maximum frequency

0%
Gamma rays
0%
Blue light
0%
Infrared rays
0%
Ultraviolet rays

Explanation

The correct order for the frequency of radiations is:

$\nu_{y-rays} > \nu_{UV-rays} > \nu_{Blue\ light} > \nu_{Infrared\ rays}$

Therefore the maximum frequency is for Gamma rays

Heat radiations propagate with the speed of

0%
$\alpha$ - rays
0%
$\beta$ - rays
0%
Light waves
0%
Sound waves

The wavelength

$21\ cm$ emitted by atomic hydrogen in interstellar space belongs to

0%
Radio waves
0%
Infrared waves
0%
Microwaves
0%
$\gamma$ - rays

Which one of the following have minimum wavelength

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

Explanation

These all are different types of electromagnetic radiation. Among which the cosmic rays have the minimum wavelength varying from $10^-14$ m to $10^-12$ m. [Option B]

TV waves have a wavelength range of

$1-10$ meter. Their frequency range in

$MHz$ is

0%
$30-300$
0%
$3-30$
0%
$300-3000$
0%
$3-3000$

Explanation

$v=\dfrac {C}{\lambda}\Rightarrow v_1 =\dfrac {3\times 10^8}{1}=3\times 10^8 Hz=300\ MHz$
and

$v_2=\dfrac {3\times 10^8}{10}=3\times 10^7 Hz =30\ MHz$

$\lambda_v , \lambda_r$ and

$\lambda_m$ represent the wavelength of visible light

$x-$ rays and microwaves respectively, then

0%
$\lambda_m > \lambda_x > \lambda_v$
0%
$\lambda_v > \lambda_m > \lambda_y$
0%
$\lambda_m > \lambda_v > \lambda_v$
0%
$\lambda_v > \lambda_x > \lambda_m$

Which of the following electromagnetic waves?

0%
$\alpha$ rays
0%
$\beta$ rays
0%
$\gamma$ rays
0%
Cathode rays

Explanation

$\gamma$ rays are electromagnetic waves

Which radiation has least wavelenght?

0%
$X-ray$
0%
$Y-ray$
0%
$\beta-ray$
0%
$\alpha-ray$

A parallel plate capacitor with plate are

$A$ and seperation between the plates

$d,$ is charged byb a constant current

$i,$ consider a plane surface of area

$A/2$ parallel to the plates and drawn symmetrically between the plates, the displacement current through this area, will be.

0%
$i$
0%
$\dfrac{i}{2}$
0%
$\dfrac{i}{4}$
0%
None of these

Explanation

Suppose the charge on the capacitor at time

$t$ is

$Q,$ the electric field between the plates of the capacitor is

$E=\dfrac{Q}{\varepsilon_{0}A}.$ The flux through the area considered is

$\phi_{E}=\dfrac{Q}{\varepsilon_{0}A}\dfrac{A}{2}=\dfrac{Q}{2\varepsilon_{0}}$

$\therefore$ the displacement current

$i_{d}=\varepsilon_{0}\dfrac{d\phi_{E}}{dt}=\varepsilon_{0}\left(\dfrac{1}{2\varepsilon_{0}}\right)\dfrac{dQ}{dt}=\dfrac{i}{2}.$

Waves related to telecommunication are

0%
infrared
0%
visible light
0%
microwaves
0%
ultraviolet ray

Rank the following kinds of waves according to their wavelength ranges from those with the largest typical or average wavelength to the smallest, noting any cases of equality:

0%
gamma rays
0%
microwaves
0%
radio waves
0%
visible light
0%
x-rays

Explanation

The ranking is

$c > b> d> e> a$ . Gamma rays have the shortest wavelength.
1832362_1880816_ans_0c00bf89cb27452a96f22e241cd9e8b5.PNG

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,

0%
increase
0%
decrease
0%
same
0%
none

Explanation

The 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.

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 speed,

0%
increase
0%
decrease
0%
stay constant
0%
None of these

Explanation

The 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.

A plane electromagnetic wave with a single frequency moves in vacuum in the positive

$x$ direction. Its amplitude is uniform over the

$yz$ plane.As the wave moves, does its frequency

0%
increase
0%
decrease
0%
stay constant
0%
none

Explanation

The 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 and also frequency is the property of source so it will be remain same.

Rank the following kinds of waves according to their wavelength ranges from those with the largest typical or average wavelength to the smallest, noting any cases of equality: Rank the kinds of waves according to their speeds in vacuum from fastest to slowest.

0%
gamma rays
0%
microwaves
0%
radio waves
0%
visible light
0%
x-rays

Explanation

The ranking is

$a=b=c=d=e$ . All electromagnetic waves travel at the speed of light

$c$ in vacuum, which is assumed here.

0%
gamma rays
0%
microwaves
0%
radio waves
0%
visible light
0%
x-rays

Explanation

The ranking is

$a > e > d> b> c$ . According to

$f \lambda=c$ , as wavelength decreases, frequency increases.
1832368_1880819_ans_49653257a7ba47c8b461634e79875dce.PNG

The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is :
(c = speed of electromagnetic waves)

0%
$1:1$
0%
$1:c$
0%
$1:c^2$
0%
$c:1$

Explanation

In an electromagnetic wave, the electric field and magnetic field exist mutually perpendicular to the direction of propagation of the wave.

The contribution of both fields is equal and is in the ratio of

$1:1$

The displacement current will be

0%
$6.9\mu A$
0%
$9.6\mu A$
0%
$6.3\mu A$
0%
$5.7\mu A$

Explanation

Capacitance,

$C=100\times {10}^{-12}F={10}^{-10}F$
Voltage,

$V=230V$

Angular frequency,

$\omega=300rad/s$

Since we know that,

$I_D = \epsilon_o \dfrac{d( \phi _E)}{dt} = \epsilon_o \dfrac{d(EA)}{dt}$ (

$\because \phi_E = EA$ where

$\phi_E$ isthe electromagnetic flux and

$E$ and

$A$ are electric field and area respectively)

$\Rightarrow I_D = \epsilon_o A \dfrac{dE}{dt}$

$\Rightarrow I_D = \epsilon_o A \dfrac{d}{dt} \left(\dfrac{Q}{ \epsilon_o A} \right)$

$\left( \because E = \dfrac{ \sigma}{ \epsilon_o} = \dfrac{Q}{ \epsilon_o A} \right)$

$\Rightarrow I_D = \epsilon_o A \times \dfrac{1}{\epsilon_o A} \dfrac{dQ}{dt} = \dfrac{dQ}{dt} = I_{conduction \ current}$

Now, Conduction current is given by,

$I=\dfrac{V}{{X}_{c}}$
Capacitive reactance is given by,

${X}_{c}=\dfrac{1}{\omega C}$
Current,

$I=V\omega C=230\times300\times{10}^{-10}$

$I=6.9\times {10}^{-6}A=6.9\mu A$

Displacement current is equal to conduction current here. Hence,

$I_D = 6.9 \mu A$

A collimated beam of light of flux density

$3k Wm^{-2}$ is incident normally on a

$100 mm^2$ completely absorbing screen. If P is the pressure exerted on the screen and

$\Delta p$ is the momentum transferred to the screen during a 1000 s interval, then

0%
$P=10^{-3} N m^{-2}$
0%
$P=10^{-4} N m^{-2}$
0%
$\Delta p=10^{-4} kg ms^{-1}$
0%
$\Delta p=10^{-5} kg ms^{-1}$

Explanation

Given :

$I = 3000$

$Wm^{-2}$

$A =100 mm^2 = 10^{-4}$

$m^2$

$t = 1000$ s

Radiant pressure (completely absorbed)

$P = \dfrac{I}{c} =\dfrac{3000}{3\times 10^8} = 10^{-4}$

$N/m^2$

Momentum transferred

$\Delta p = {PA}\times{t} = {10^{-4}\times 10^{-4}}\times{1000} = 10^{-5}$

$kgm/s$

A small metallic ball is suspended in a uniform electric field by a thread. If high energy X-rays are made to fall on it , then:

0%
ball will be displaced in opposite direction to the direction of field
0%
ball will be displaced in the direction of field
0%
ball will remain in equilibrium state
0%
ball will start oscillating

The incident intensity on a horizontal surface at sea level from the sun is about

$1 kW m^{-2}$ . Assuming that 50 per cent of this intensity is reflected and 50 per cent is absorbed, determine the radiation pressure on this horizontal surface (in pascals).

0%
$8.2\times 10^{-2}$
0%
$5\times 10^{-6}$
0%
$3\times 10^{-5}$
0%
$6\times 10^{-5}$

Explanation

Given:

$I = 1000$

$Wm^{-2}$

As 50 percent of light is reflected, thus

$e=0.5$

Radiation pressure,

$P = \dfrac{(1+e)I}{c}$

$\therefore$

$P = \dfrac{(1+0.5)\times 1000}{3\times 10^8} = 5\times 10^{-6}$ Pa

A parallel beam of light is incident normally on a plane surface absorbing 40 % of the light and reflecting the rest. If the incident beam carries 60 watt of power, the force exerted by it on the surface is:

0%
$3.2 \times 10^{-8}\, N$
0%
$3.2 \times 10^{-7}\, N$
0%
$5.2 \times 10^{-7}\, N$
0%
$5.12 \times 10^{-8}\, N$

Explanation

momentum of incident ray per second

$P_1=\dfrac{E}{c}=\dfrac{60}{3\times 10^8}=2\times 10^{-7} N$

momentum of reflected ray per second

$P_2=0.6\times\dfrac{E}{c}=\dfrac{0.6\times 60}{3\times 10^8}=1.2\times 10^{-7} N$

now rate of change of momentum

$F=P_2-(-P1)=3.2\times 10^{-7} N$

Light of intensity

$= 3 W/m^{2}$ is incident on a perfectly absorbing metal surface of area

$1 m^{2}$ making an angle of

$60^0$ with the normal. If the force exerted by the photons on the surface is

$p \times 10^{-9}$ (in Newton), then the value of p is :

0%
5
0%
2
0%
1
0%
3

Explanation

We will be calculating everything per unit time.

Effective area for the light on the surface is

$= 1 \times cos 60^{0}$
Energy falling on the surface

$=$ intensity x Effective area

$= 3 \times 1 \times cos 60^{0} = \dfrac{3}{2} watt$
Momentum carried by the light per sec

$= 3/(2c) = 5\times 10^{-9}$
So,

$p= 5\times 10^{-9}$

A pulse of light of duration

$100\ ns$ is absorbed completely by a small object initially at rest. Power of the pulse is

$30\ mW$ and the speed of light

$3\times 10^8 m/s.$ The final momentum of the object is

0%
$0.3\times 10^{-17} kg\ ms^{-1}$
0%
$1.0\times 10^{-17} kg\ ms^{-1}$
0%
$3\times 10^{-17} kg\ ms^{-1}$
0%
$9.0\times 10^{-17} kg ms^{-1}$

Explanation

$answer:-$ B option

using Einstein mass energy equation

$E=mc^2$

$Energy(E)=power\times time$

$E=30\times { 10 }^{ -3 }\times 100\times { 10 }^{ -9 }=m{ c }^{ 2 }=m\times { (3\times { 10 }^{ 8 }) }^{ 2 }$

$m=\dfrac { 1 }{ 3 } \times { 10 }^{ -25 }$

using

$p=mv$

$p=\dfrac { 1 }{ 3 } \times { 10 }^{ -25 }\times 3\times { 10 }^{ 8 }={ 10 }^{ -17 } kgm/s$

Sea water at frequency

$\nu \ =\ 4\ x\ { 10 }^{ 8 }$ Hz has permittivity

$\varepsilon \ \approx \ 80\ { \varepsilon }_{ 0 }$ , permeability

$\mu \ \approx \ { \mu }_{ 0 }$ and resistivity

$\rho \ =\ 0.25\ \Omega m$ . Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) =

${ V }_{ 0 }\ \sin { \ (2\pi \nu t) }$ . The of amplitude of the displacement current density to the conduction current density is

0%
$\dfrac { 2 }{ 3 }$
0%
$\dfrac { 4 }{ 9 }$
0%
$\dfrac { 9 }{ 4 }$
0%
2

Explanation

Suppose distance between the parallel plates is

$D$ and applied voltage

$V_{(t)} = V_02\pi vt$ .

thus electric field

$E = \dfrac{V_0}{d} \sin (2\pi vt)$

Now using Ohm's law

$J_c = \dfrac{1}{\phi} \dfrac{V_0}{d}\sin (2\pi vt)$

$\dfrac{V_0}{\phi d}\sin (2 \pi vt) = J_0^c \sin 2 \pi vt$

Here

$J_0^c = \dfrac{V_0}{pd}$

Now the displacement current density is given as

$Jd = \in \dfrac{\delta E}{dt} =\dfrac{\in \delta}{dt}$

$\left[\dfrac{V_0}{dt} \sin (2\pi vt)\right]$

$= \dfrac{\in 2\pi v V_0}{d} \cos (2\pi vt)$

$\Rightarrow = J^d_0 \cos (2\pi vt)$

Where

$J_0^d = \dfrac{2\pi V\in V_0}{d}$

$\Rightarrow \dfrac{J^d_0}{J^c_0} = \dfrac{2\pi v \in V_0}{d}. \dfrac{pd}{V_0} = 2\pi v \in \rho$

$= 2\pi \times 80\in_0v\times 0.25 = 4\pi \in_0v \times 10$

$= \dfrac{10v}{9\times 10^9} = \dfrac{4}{9}$

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

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

Choose the correct answer from the alternatives given.
Assume a bulb of efficiency 2.5% as a point source. The peak values of electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m is respectively

0%
$2.5 \, V \, m^1, \, 3.6 \, \times \, 10^{-8} T$
0%
$4.2 \, V \, m^1, \, 2.8 \, \times \, 10^{-8} T$
0%
$4.08 \, V \, m^1, \, 1.36 \, \times \, 10^{-8} T$
0%
$3.6 \, V \, m^1, \, 3.6 \, \times \, 10^{-8} T$

Explanation

Here intensity,

$I = \dfrac{power}{area}$

$\dfrac{100 \times 2.5 }{4\pi (3)^2 \times 100} = \dfrac{2.5}{36\pi }Wm^{-2}$

Half of this intensity belongs to electronic field and half of that to magnetic field.

$\therefore \dfrac{I}{2} \,= \, \dfrac{1}{4}\varepsilon_0E^2_0c$

$E_o = \sqrt{\dfrac{2I}{\epsilon _0c}}\,\\\Rightarrow \, \sqrt{\dfrac{2 \times \dfrac{2.5}{36\pi }}{\dfrac{1}{4\pi \times 9\times 10^9 }}\times 3\times 10^8}\\\Rightarrow 4.08 V m^{-i}$

$\therefore B_o = \dfrac{E_0}{c} \,=\, \dfrac{4.08}{3\times 10^8}\\ = 1.36 \times 10^{-8}T$

Choose the correct answer from the alternatives given.
A parallel plate capacitor with plate area

$A$ and separation between the plates

$d$ is charged by a constant current

$I$ . Consider a plane surface of area

$\dfrac{A}{2}$ parallel to the plate and drawn between the plates. The displacement current through the area is :

0%
I
0%
$\dfrac{I}{2}$
0%
$\dfrac{I}{4}$
0%
$\dfrac{I}{8}$

Explanation

Charge on capacitor plates at time

$t$ is ,

$q = It$ .

Electric field between the plates at this instant is

$E = \dfrac{Q}{A\varepsilon _0}\, =\, \dfrac{It}{A\varepsilon _0}$

The Electric flux through the given area is:

$\phi=A.E$

The Electric flux through the area

$\dfrac{A}{2}$ is:

$\phi _E \, = \, \left(\dfrac{A}{2}\right)E$

$\phi_E=\dfrac{A}{2}\dfrac{It}{A\varepsilon _0}=\dfrac{It}{2\varepsilon_0}$

The displacement current through the area is given as:

$I_D \, = \, \varepsilon_0 \dfrac{d \phi _E}{dt}$

$\,\,\,\implies\varepsilon_0 \dfrac{d}{dt} \left(\dfrac{It}{2 \varepsilon_0} \right) = \dfrac{I}{2}$

Option

$(B)$ is correct.

$1.5\ kW$ (kilo-watt) laser beam of wavelength

$6400 \mathring A$ is used to levitate a thin aluminium disc of same area as the cross section of the beam. The laser light is reflected by the aluminium disk without any absorption. The mass of the disc is close to:

0%
$10^{-9}\ kg$
0%
$10^{-3}\ kg$
0%
$10^{-4}\ kg$
0%
$10^{-6}\ kg$

Explanation

Let the power of the beam be P.

Force acting on the disc is given by:

$F=\dfrac{2P}{c}=\dfrac{2\times1.5\times10^3}{3\times10^8} = 10^{-5}\ N$

The above force balances the weight of the disc. Hence,

$F=mg$

$10^{-5} = 10 g$

$m = 10^{-6}\ kg$

$v_{g}, v_{x}$ , and

$v_{m}$ are speeds of gamma rays,

$X$ rays, and microwaves respectively in vacuum, then:

0%
$V_{g} < V_{x} < V_{m}$
0%
$V_{g}>V_{x}>V_{m}$
0%
$V_{g} < V_{x}> V_{m}$
0%
$V_{g}=V_{x}=V_{m}$

Explanation

The gamma rays,

$X-$ rays and the microwaves are the Electromagnetic waves and the speed of all the electromagnetic waves remains the same in vacuum irrespective of other parameters. All will travel at the speed of light.

$V_{g}=V_{x}=V_{m}$

The speed of Electromagnetic waves is always the same in vacuum. So, option

$(D)$ is correct.

A plane electromagnetic wave travels in free space along X-direction. If the value of

$\overrightarrow { B }$ (in tesla) at a particular point in space and time is

$1.2 \times {10}^{-8} \hat {k}$ , the value of

$\overrightarrow { E }$ (in V

${m}^{-1}$ ) at that point is,

0%
$1.2\ \hat {j}$
0%
$3.6\ \hat {k}$
0%
$1.2\ \hat {k}$
0%
$3.6\ \hat {j}$

Explanation

Given: The magnetic field of the plane electromagnetic wave is

$1.2×10^{-8} \hat k\ T$

The direction of propagation of the electromagnetic wave is along X-direction.

The magnitude of

$\vec E$ is given by:

$E\, = \, B\cdot c\\ \ \ \ \ = (1.2 \, \times \, 10^{-8} T)(3 \, \times \, 10^{-8} m \, s^{-1})\\ \ \ \ \ = 3.6 \, V/m$

Since the magnetic field is along

$Z-$ direction and the wave propagates along

$X -$ direction. Therefore

$\vec E$ should be in a direction perpendicular to both

$X$ and

$Z$ axes.

Using vector algebra should be along X-direction.

Since

$(+\hat j) \times (\hat k )= \hat i$

$\vec E$ is along the

$Y-$ direction.

Thus,

$\vec E= 3.6\hat j\ Vm^{-1}$

CBSE Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 10 - 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 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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