Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

Mobile telephony : Frequency range 800-950 kHz

Sky wave communication : Frequency upto 30 MHz

Line-of-sight communication : Frequency greater than 40 MHz

Ground wave propagation : Frequency less than a few MHz

MCQ Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 7

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Who discovered Radio activity?

0%
Rutherford
0%
Neil's Bohr
0%
Einstein
0%
Henry Becquerel

The part of the electromagnetic spectrum which is produced by bombarding a metal target with high energy electrons is :

0%
Ultraviolet
0%
X-Rays
0%
Infrared
0%
Gamma Rays

Hydrogen atom does not emit X-rays because

0%
it contains only a single electron
0%
energy levels in it are far apart
0%
its size is very small
0%
energy levels in it are very close to each other

Which one of following have highest frequency?

0%
radio waves.
0%
infrared.
0%
ultraviolet.
0%
gamma rays.

Explanation

Frequency of Radio waves =

$<3\times { 10 }^{ 9 }$ Hz

Frequency of infrared waves =

$3\times { 10 }^{ 9 }-3\times { 10 }^{ 11 }$ Hz

Frequency of ultraviolet waves =

$7.5\times { 10 }^{ 14 }-3\times { 10 }^{ 17 }$ Hz

Frequency of gamma rays =

$>3\times { 10 }^{ 20 }$ Hz

So Gamma rays have highest frequency.

Electromagnetic waves that include light belong to

0%
light spectrum.
0%
electronic spectrum.
0%
electromagnet group.
0%
electromagnetic spectrum.

${\epsilon}_0$ and

${\mu}_0$ represent the permittivity and permeability of vaccum and

$\epsilon$ and

$\mu$ represent the permittivity and permeability of medium, then refractive index of the medium is given by

0%
$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\epsilon \mu}}$
0%
$\sqrt{\dfrac{\epsilon \mu}{{\epsilon}_0{\mu}_0}}$
0%
$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\epsilon}}$
0%
$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\mu}}$

Explanation

Given,

$\epsilon_o=$ permittivity of free space and

$\mu_o=$ permeablilty of vaccum and,

$\epsilon=$ permittivity of material and

$\mu=$ permeability of meadium.

$\eta=$ refractive index of the material.

So, The relation between permittivity, permability and the refractive index of the medium is,

Refractive index of a medium

$\eta=\sqrt{\dfrac{\mu \epsilon}{{\mu}_0{\epsilon}_0}}$

In electromagnetic spectrum, the frequencies of

$\gamma$ -rays, X- rays and ultraviolet rays are denoted by

$n_1, n_2$ and

$n_3$ respectively then.

0%
$n_1 > n_2 > n_3$
0%
$n_1 < n_2 < n_3$
0%
$n_1 > n_2 < n_3$
0%
$n_1 < n_2 > n_3$

Explanation

Frequency range of

$\gamma \ rays = 5 \times 10^{20} \ to \ 3 \times { 10 }^{ 19 }\ Hz$

Frequency range of

$X- rays = 3 \times { 10 }^{ 19 } to \ { 10 }^{ 16 } \ Hz$

Frequency range of ultraviolet rays

$= 10^{16} \ to \ 8 \times 10^{14}$

So frequency of

$\gamma$ rays > frequency of

$X-$ rays > frequency of ultraviolet rays

${ n }_{ 1 } >{ n }_{ 2 } > n_3$

In electromagnetic wave, according to Maxwell, changing electric field gives.

0%
Stationary magnetic field
0%
Conduction current
0%
Eddy current
0%
Displacement current

Explanation

By Maxwell

${ I }_{ d }=\dfrac { \varepsilon dE }{ dt }$

dE is electric field

${ I }_{ d }$ is displacement current per unit area.

Hence changing electric field gives displacement current.

The working of a microwave oven involves

0%
Absorption of microwaves by matter
0%
Reception of microwaves by optical fibre
0%
Microwave amplification by stimulated emission of radiation
0%
Transmission of microwaves through a metal

The maximum frequency of transmitted radio waves above which the radio waves are no longer reflected back by ionosphere is ................. (N = maximum electron density of ionosphere, g = acceleration due to gravity)

0%
$gN$
0%
$gN^2$
0%
$g \sqrt N$
0%
$g^2N^2$

Explanation

This is called critical frequency ,at this frequency wave will not return to earth .

$f_{cr}=\sqrt{81N}=g\sqrt{N}$

Radio waves travel with a speed of.

0%
$3\times 10^8$ m/sec
0%
$3\times 10^7$ m/sec
0%
$3\times 10^6$ m/sec
0%
$3\times 10^9$ m/sec

White light is passed through a dilute solution of potassium permagnate. The spectrum produced by the emergent light is

0%
Band emission spectrum
0%
Line emission spectrum
0%
Band absorption spectrum
0%
Line absorption spectrum

Explanation

when light is passed through the solution then some part of the complete electromagnetic spectrum will be absorbed by the solution.So, the emitted spectrum will have some missing E.M. waves of frequency corresponding to what has been absorbed by the solution.So, the emergent light produced will be

$Band$

$Absorption$

$Spectrum$ .

In the spectrum of visible light produced by a prism, dispersion is

0%
Uniform throughout the spectrum
0%
Maximum in the middle, decreases on either sides
0%
Maximum towards yellow
0%
Maximum towards violet

Which of the following is not one of the prime health risks associated with greater UV radiation due to the depletion of stratospheric ozone?

0%
Increased liver cancer.
0%
Increased skin cancer.
0%
Reduced immune system
0%
Damage to eyes.

The frequencies of X-rays, Gamma rays and visible light waves rays are

$a, b$ and

$c$ respectively, then:

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

Explanation

The frequency of X-rays

$a = 3\times 10^{19}$ to

$1\times 10^{16}$
The frequency of Gamma-rays

$b = 3\times 10^{22}$ to

$3\times 10^{18}$
The frequency of visible light

$c = 8\times 10^{14}$ to

$4\times 10^{14}$
So,

$b > a, b > c$

The capacity of a parallel plate air capacitor is

$2 \mu F$ and voltage between the plates is changing at the rate of

$3 {V}/{S}$ . The displacement current in the capacitor is:

0%
$2 \mu F$
0%
$3 \mu F$
0%
$5 \mu F$
0%
$6 \mu F$

Explanation

Displacement current in capacitor

$i = A\epsilon_o \dfrac{dE}{dt}$

$i = \dfrac{A\epsilon_o}{l} \dfrac{d(El)}{dt}$ where

$l$ is the separation between the two plates.

Potential difference between the two plates

$V = El$

$\implies$

$i = C \dfrac{dV}{dt}$ where capacitance of the capacitor is

$C = \dfrac{A\epsilon_o }{l}$

$\therefore$

$i = 2\mu F \times 3 V/s = 6\mu A$

IR region lies between.

0%
Radio waves and microwave regions
0%
Microwave and visible
0%
Visible and UV region
0%
UV rays and X-ray region

Which of the following has zero average value in a plane electromagnetic wave?

0%
Kinetic energy
0%
Magnetic field
0%
Electric field
0%
Both (b) and (c)

Explanation

Hint:

The average of

$sin \theta$ and

$cos \theta$ for whole cycle is is zero.

Step 1: Analyzing the average value of Kinetic energy.

Kinetic Energy is always a positive quantity, therefore its average will also be a positive quantity.

Step 2: Finding the average of electric and the magnetic field.

The equations for the electric field and the magnetic field are given as,

Find the mismatch

0%
Sky wave communication : Frequency upto 30 MHz
0%
Line-of-sight communication : Frequency greater than 40 MHz
0%
Mobile telephony : Frequency range 800-950 kHz
0%
Facsimile : Static document
0%
Ground wave propagation : Frequency less than a few MHz

The ionospheric layer acts as a reflector for the frequency range

0%
1 kHz to 10 kHz
0%
3 to 30 MHz
0%
3 to 30 kHz
0%
100 kHz to 1 MHz
0%
3 GHz to 30 GHz

Light is an electromagnetic wave. Its speed in vacuum is given by the expression

0%
$\displaystyle \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } }$
0%
$\displaystyle \sqrt { \frac { { \mu }_{ 0 } }{ { \varepsilon }_{ 0 } } }$
0%
$\displaystyle \sqrt { \frac { { \varepsilon }_{ 0 } }{ { \mu }_{ 0 } } }$
0%
$\displaystyle \frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } }$

Explanation

We know that,

$\displaystyle { \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }Wb{ A }^{ -1 }{ m }^{ -1 }$

$\displaystyle { \varepsilon }_{ 0 }=8.85\times { 10 }^{ -12 }\frac { N-{ m }^{ 2 } }{ { C }^{ 2 } }$
So, speed of light

$\displaystyle c=\frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } =3\times { 10 }^{ 8 }{ m }/{ s }$

The waves that are bent down by the ionosphere are

0%
Ground waves
0%
Surface waves
0%
Direct waves
0%
Space waves
0%
Sky waves

The wavelength of the short radiowaves, micro waves, ultraviolet waves are

${ \lambda }_{ 1 }, { \lambda }_{ 2 }$ and

${ \lambda }_{ 3 }$ respectively. Arrange them in decreasing order.

0%
${ \lambda }_{ 1 },{ \lambda }_{ 2 },{ \lambda }_{ 3 }$
0%
${ \lambda }_{ 1 },{ \lambda }_{ 3 },{ \lambda }_{ 2 }$
0%
${ \lambda }_{ 3 },{ \lambda }_{ 2 },{ \lambda }_{ 1 }$
0%
${ \lambda }_{ 2 },{ \lambda }_{ 1 },{ \lambda }_{ 3 }$

Explanation

We know that
1. Short radio waves
Wavelength range

$=1\times { 10 }^{ -1 } m$ to

$1\times { 10 }^{ 4 } m$
Frequency range

$=3\times { 10 }^{ 9 }Hz$ to

$3\times { 10 }^{ 4 }Hz$
2. Microwaves
Wavelength range

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

$3\times { 10 }^{ -1 } m$
Frequency range

$=3\times { 10 }^{ 11 }Hz$ to

$1\times { 10 }^{ 9 }Hz$
3. Ultraviolet waves
Wavelength range

$=1\times { 10 }^{ -8 } m$ to

$4\times { 10 }^{ -8 } m$
Frequency range

$=3\times { 10 }^{ 16 }Hz$ to

$8\times { 10 }^{ 14 }Hz$
So, decreasing order is

${ \lambda }_{ 1 },{ \lambda }_{ 2 },{ \lambda }_{ 3 }$ .

In an electromagnetic wave, the electric magnetising fields are

$100V/m$ and

$0.265 A/m$ . The maximum energy flow is?

0%
$26.5W/m^2$
0%
$36.5W/m^2$
0%
$46.7W/m^2$
0%
$765W/m^2$

Explanation

Maximum rate of energy flow,

$S=E_0\times H_0$
Given,

$E_0=100V/m, H_0=0.265 A/m$

$\therefore S=100\times 0.265=26.5W/m^2$ .

The atomic number

$(Z)$ of an element whose

${ k }_{ a }$ wavelength is

$\lambda$ is

$11$ . The atomic number of an element whose

${k}_{a}$ , wavelength is

$4\lambda$ is equal to

0%
$6$
0%
$11$
0%
$44$
0%
$4$

Explanation

${ \left( Z-1 \right) }^{ 2 }\lambda =$ constant

$\therefore { (10 }^{ 2 })\lambda =4\lambda { (Z-1) }^{ 2 }$

$Z=6$

Electromagnetic waves travel only through.

0%
material medium
0%
vacuum
0%
oscillating electric and magnetic fields
0%
oscillating electric and magnetic fields whose directions are perpendicular to each other

The energy of infrared waves is greater than that of

0%
Visible light
0%
Ultraviolet waves
0%
$X$ -rays
0%
Gamma rays
0%
Micro waves

Explanation

$E \propto \dfrac { 1 }{ \lambda }$
Since, the wavelength of microwave is greater than infrared, so the energy of infrared waves is greater than that of microwaves.

A piece of blue glass heated to a high temperature and a piece of red glass at room temperature are taken inside a dimly-lit room. Then.

0%
The blue piece will look blue and the red piece will look red as usual
0%
The red piece will look brighter red and the blue piece will look ordinary blue
0%
The blue will look brighter as compared to the red piece
0%
Both the pieces will look equal red

The wavelength of

$X$ -rays is in the range

0%
$0.001\mathring { A }$ to $1\mathring { A }$
0%
$0.001 nm$ to $1 nm$
0%
$0.001 \mu m$ to $1 \mu m$
0%
$0.001 cm$ to $1 cm$

Explanation

$X$ -Rays Wavelength range

$=1\times { 10 }^{ -11 }m$ to

$3\times { 10 }^{ -8 }m$
So,

$=0.001 nm$ to

$1 nm$

If the distance that a radio wave can travel in the time of one complete cycle is

$10$ metres, then the frequency of the radio wave in cycles per second is?

0%
$3\times 10^9$ Hz
0%
$3\times 10^6$ Hz
0%
$3\times 10^8$ Hz
0%
$3\times 10^7$ Hz

International broadcasting requires.

0%
Medium frequency radio waves
0%
High frequency radio waves
0%
Low frequency radio waves
0%
Longitudinal waves

The transverse nature of electromagnetic radiation is demonstrated by.

0%
Interference experiment
0%
Diffraction experiment
0%
Polarisation experiment
0%
Optical activity experiment

Unpolarized light falls first on polarizer

$\left( P \right)$ and then on analyzer

$\left( A \right)$ . If the intensity of the transmitted light from the analyser is

$\dfrac { 1 }{ 8 }$ th of the incident unpolarized light. What will be the angle between optic axes of

$P$ and

$A$ ?

0%
${ 45 }^{ o }$
0%
${ 30 }^{ o }$
0%
Zero
0%
${ 60 }^{ o }$

Explanation

Given,

$I=\dfrac { { I }_{ 0 } }{ 2 }$ ....(i)

${ I }^{ ' }=I\cos ^{ 2 }{ \theta }$

$\left( \because { I }^{ ' }=\dfrac { { I }_{ 0 } }{ 8 } \right)$

$\therefore \dfrac { { I }_{ 0 } }{ 8 } =\dfrac { { I }_{ 0 } }{ 2 } \cos ^{ 2 }{ \theta }$
From the equation (i), we have

$\dfrac { 1 }{ 4 } =\cos ^{ 2 }{ \theta } \Rightarrow \cos { \theta } ={ 1 }/{ 2 }$

$\Rightarrow \cos { \theta } =\cos { { 60 }^{ o } }$

$\Rightarrow \theta ={ 60 }^{ o }$

The sun delivers

$10^3\, Wm^{-2}$ of electromagnetic flux on the earth's surface. The total power that is incident on a roof of dimensions

$6 \,m \times 30\, m$ is

0%
$1.8 \times 10^5\, W$
0%
$7.2 \times 10^5\, W$
0%
$0.9 \times 10^5\, W$
0%
$4.5 \times 10^5\, W$

Explanation

Given, electromagnetic flux

$=10^3\, Wm^{-2}$
The area of roof

$A= 6\, m \times 30\, m$

$=180 \, m^2$
The total power incident on roof

$P=10^3 \times 180$

$=1.8\times 10^5 \, W$

Which of the following is correct for X-rays?

0%
It has a beam of electrons
0%
It is electromagnetic waves
0%
It is positively charged particle
0%
It is negatively charged particles

Explanation

X-rays are electromagnetic waves with a wavelength less than about

$10^{-9}$ m. Cathode rays comprise of a beam of electrons.

$\alpha$ -rays are made up of helium nucleus(i.e.,

$^{4}_{2}He^{2+}$ ), hence are made up of positively charged particles, whereas

$\beta$ -rays are made up of energetic electrons, hence comprise of negatively charged particles.

Radiations of intensity

$0.5\ W/m^{2}$ are striking a metal plate. The pressure on the plate is

0%
$0.166\times 10^{-8} N/m^{2}$
0%
$0.332\times 10^{-8}N/m^{2}$
0%
$0.111\times 10^{-8}N/m^{2}$
0%
$0.083\times 10^{-8}N/m^{2}$

Explanation

Intensity or power per unit area of the radiations,

$P = pv$

$\Rightarrow p = \dfrac {P}{v} = \dfrac {0.5}{3\times 10^{8}} = .166\times 10^{-8} N/m$

An electromagnetic wave passing through vacuum is described by the equations.

$E=E_0\sin(kx-\omega t)$ and

$B=B_0\sin(kx-\omega t)$ . Then.

0%
$E_0k=B_0\omega$
0%
$E_0\omega =B_0k$
0%
$E_0B_0=\omega k$
0%
$E_0=B_0$

Explanation

Equations of an electromagnetic wave

$E=E_0\sin(kx-\omega t)$ and

$B=B_0\sin(kx -\omega t)$
The relation between electric field

$(E_0)$ and magnetic field

$(B_0)$ is

$E_0=cB_0$ ........(i)
Velocity of light

$c=\displaystyle v\lambda =\displaystyle\frac{\omega}{2\pi}\lambda =\frac{\omega}{k}$ (where

$k=2\pi /\lambda$ ).
Substituting this value of

$c$ in equation (i), we get

$E_0=\displaystyle\frac{\omega}{k}B_0$
or

$E_0k=B_0\omega$ .

The Electromagnetic wave travelling in a medium has relative permeability

$1.3$ and relative premittivity

$2.14$ . The speed of EM wave in that medium will be.

0%
$13.6\times 10^6\ m/s$
0%
$1.8\times 10^6\ m/s$
0%
$3.6\times 10^8\ m/s$
0%
$1.8\times 10^8\ m/s$

Explanation

Velocity of EM wave in vacuum is given by

$\displaystyle c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$
Velocity of EM wave in medium is given by

$\displaystyle v=\frac{1}{\sqrt{\mu \varepsilon}}$

$\therefore \displaystyle\frac{c}{v}=\sqrt{\displaystyle\frac{\mu\varepsilon}{\mu_0\varepsilon_0}}=\sqrt{\displaystyle\frac{\mu_0\mu_r\varepsilon_0\cdot \varepsilon_r}{\mu_0\varepsilon}}$

$v=c/\sqrt{\mu_r\varepsilon_r}$

$=3\times 10^8/\sqrt{1.3\times 2.14}$

$=1.8\times 10^8$ m.

Radiation pressure on any surface :

0%
is dependent on wavelength of the light used
0%
is dependent on nature of surface and intensity of light used
0%
is dependent on frequency and nature of surface
0%
depends on the nature of source from which light is coming and on nature of surface on which it is falling.

Explanation

Radiation pressure is given by

$P_R = \dfrac{(1+\alpha) I}{c}$
where

$\alpha$ is the coefficient of reflection of the surface.
For completely reflecting surface

$\alpha =1$
For completely absorbing surface

$\alpha = 0$
So, radiation pressure depends on the nature of surface on which the light is falling but independent of wavelength of light falling.

Two monochromatic (wavelength

$=a/5$ ) and coherent source of electromagnetic waves are placed on the x-axis at the point

$(2a,0)$ and

$(-a,0)$ . A detector moves in a circle of radius

$R(>> 2a)$ whose Centre is at the origin. The number of maximas detected during one circular revolution by the detector are:

0%
$60$
0%
$15$
0%
$64$
0%
None

Indigo and red regions of VIBGYOR, respectively fall in the range of wavelength

0%
430-470 nm and 660-760 nm
0%
300--390 nm and 600-650 nm
0%
390-760 nm and 430-470 nm
0%
660-760 nm and 430-470 nm

Explanation

Light is the visible part of electromagnetic radiations. visible light consists of radiations having a wavelength between 390-760 nm (or 3900-7600

$\mathring { A}$ ) It can be resolved into the light of different colors -

violet (390-430 nm)

blue or indigo (430-470 nm)

blue-green or blue (470-500 nm)

green (500-580 nm)

yellow (580-600 nm)

orange (600-650 nm)

orange-red (650-660 nm)

red (660-760 nm).

So the correct answer is '430-470 nm and 660-760 nm'.

The visible part of the electromagnetic spectrum consists of radiations having a wavelength in the range of

0%
400-800 nm
0%
300-2600 nm
0%
390-760 nm
0%
650-760 nm

Explanation

Light is the visible part of electromagnetic radiations. visible light consists of radiations having a wavelength between 390-760 nm (or 3900-7600

$\mathring { A}$ ) It can be resolved into light of different colours - violet (390-430 nm), blue or indigo (430-470 nm), blue-green or blue (470-500 nm), green (500-580 nm), yellow (580-600 nm),orange (600-650 nm), orange-red (650-660 nm) and red (660-760 nm).

So the correct answer is '390-760 nm'.

Frequencies of various radiations are given as

$f_v\rightarrow$ Visible light

$f_r\rightarrow$ Radio waves

$f_{UV}\rightarrow$ Ultra Violet waves
Then which of following is true?

0%
$f_{UV} < f_v < f_r$
0%
$f_r < f_v < f_{UV}$
0%
$f_v < f_r < f_{UV}$
0%
$f_{UV} < f_r < f_v$

Explanation

Increasing order of frequency of electromagnetic spectrum is given by :
Radio waves < Micro waves < Infrared < Visible < UV < X-rays < Gamma rays.
So,

$f_r<f_v<f_{UV}$
Thus correct answer is option B.

SQUID helps in studying

0%
lung cancer
0%
heart beats
0%
weaker magnetic waves from brain
0%
electrical impulses from heart

Choose the correct answer from the alternatives given.
Maxwell in his famouse questions of eletromagnetism introduce the concept of:

0%
ac current
0%
displacement current
0%
impedance
0%
reatance

Explanation

Maxwell in his famous question's of electromagnetism introduce the concept of displacement current

$J=J_E+J_C\\ \quad=\epsilon_0\cfrac{dE}{dT}+\cfrac{dP}{dT}$

Study the frequency of the following electromagnetic radiations and select the correct relationship.

$1$ . Infra red

$2$ . Microwave

$3$ . Ultra violet.

0%
$3 > 2 > 1$
0%
$3 > 1 > 2$
0%
$1 > 2 > 3$
0%
$2 > 3 > 1$

Generation, propagation and detection of electromagnetic waves is the basis of

0%
lasers
0%
reactors
0%
radio and television
0%
computer

A long straight cable of lenght

$l$ is placed symmetrically along z - axis has radius

$a(< < l)$ . The cable consists of a thin wire and a coaxial conducting tube. An alternating current

$I(t) = I_0 sin (2\pi\upsilon t)$ flows down the central thin wire and return along the coaxial conducting tube. The induced electric field at a distance s from the wire inside the cable is

$E(s, t) = \mu_oI_o\upsilon cos(2\pi \upsilon t)\ ln\left (\dfrac{s}{a} \right )\hat{k}$ . The displacement current density inside the cable is

0%
$\dfrac{2\pi}{\lambda^2 }I_0 \ ln\left(\dfrac{a}{s}\right )sin(2\pi \upsilon t)\hat{k}$
0%
$\dfrac{I}{\lambda^2 }I_0 \, ln \left(\dfrac{a}{s} \right )sin(2\pi \upsilon t)\hat{k}$
0%
$\dfrac{\pi}{\lambda^2 }I_0 \, ln \left(\dfrac{a}{s} \right )sin(2\pi \upsilon t)\hat{k}$
0%
$Zero$

Explanation

Given: The induced electric field at a distance from wire is

$E(s,t)=\mu_0I_0\upsilon\ cos(2\pi\upsilon t)ln\left(\dfrac{s}{a}\right)\hat k$

Displacement current density is given by:

$\vec{J}_d=\epsilon_0\dfrac {d\bar{E}}{dt}$

$\Rightarrow\epsilon_0\mu_0I_0 v\dfrac{\partial}{\partial(t)}(cos2\pi v t)(\dfrac{s}{a})\hat{k}$

Substitute

$c$ for

$\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$ .

$\Rightarrow\dfrac{1}{c^2}I_0 2 \pi v^2(-sin(2\pi v t)\ ln(\dfrac{s}{a})\hat{k}$

$\Rightarrow(\dfrac{v}{c})^2 \, 2\pi I_0 sin (2\pi \,v \,t)\ ln(\dfrac{a}{s})\hat{k}$

$\implies\dfrac{2\pi}{\lambda^2}I_0\ ln(\dfrac{a}{s})sin(2\pi v t) \hat {k}$

Option

$(A)$ is correct.

Choose the correct answer from the alternatives given.
Which among the following does not represent Maxwell's equation?

0%
$\phi \, \overset{-}{E}.\overset{-}{dA} \, = \, \dfrac{q}{\varepsilon_0}$
0%
$\phi \, \overset{-}{B}.\overset{-}{dA} \, = 0$
0%
$\phi \, \overset{-}{E}.\overset{-}{-dl} \, = \, \dfrac{-dB}{dt}$
0%
$\phi \, \overset{-}{E}.\overset{-}{dl} \, = \, \mu_0I_C \, + \, \mu_0\varepsilon_0 \, \dfrac{d \phi E}{dt}$

Explanation

Maxwell's equation are as follows
(i)

$\phi \overline{E}.\overline{dA} = \dfrac{q}{\varepsilon _0}$ (Gauss's law of electricity)
(ii)

$\phi \overline{B}.\overline{dA} = 0$ (Gauss's law of magnetism)
(iii)

$\phi \overline{E}.\overline{dl} = -\dfrac{d\phi E}{dt}$ (Saraday's law)
(iv)

$\phi \overline{B}.\overline{dl} =\mu_0I_C + \mu_0 \varepsilon_0 \dfrac{d\phi E}{dt}$ (Ampere - Maxwell law)

CBSE Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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