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#### Question 1:

One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

(a) visible region.
(b) infrared region.
(c) ultraviolet region.
(d) microwave region.

#### Answer:

Given: Energy required to dissociate a carbon monoxide molecule into carbon and oxygen atoms, E = 11 eV
Energy, $E=h\nu$
Here h is Planck's constant

This frequency radiation is under the ultraviolet region.
Hence, the correct answer is option C.

#### Question 2:

A linearly polarized electromagnetic wave given as  $E={E}_{0}\stackrel{\wedge }{i}\mathrm{cos}\left(kz-\omega t\right)$ 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

#### Answer:

Given: Incident electromagnetic wave, $E={E}_{0}\stackrel{\wedge }{i}\mathrm{cos}\left(kz-\omega t\right)$
When a wave is reflected from a denser medium, then the phase changes by 180° or π radian.
Reflected electromagnetic wave is given by:

Hence, the correct answer is option B.

#### Question 3:

Light with an energy flux of 20 W/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm2. the total momentum delivered (for complete absorption) during 30 minutes is

(a) 36 × 10–5 kg m/s.
(b) 36 × 10–4 kg m/s.
(c) 108 × 104 kg m/s.
(d) 1.08 × 107 kg m/s.

#### Answer:

Given: Intensity of incident light, I = 20 W/cm2
Surface area, A = 30 cm2
Time, t = 30 minutes
Total energy falling on the surface in time t, E = IAt

Momentum of incident light:

Momentum of reflected light:
${p}_{f}=0$
Momentum delivered to the surface for complete absorption is .
Hence, the correct answer is option B.

#### Question 4:

The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

#### Answer:

Given: Power of first bulb, P1 = 100 W
Power of second bulb, P2 = 50 W
Distance, d = 3 m
Electric field intensity:
$E\propto P\phantom{\rule{0ex}{0ex}}⇒\frac{{E}_{1}}{{E}_{2}}=\frac{{P}_{1}}{{P}_{2}}=\frac{100}{50}=2\phantom{\rule{0ex}{0ex}}⇒{E}_{2}=\frac{{E}_{1}}{2}$
So, the electric field intensity produced by the radiations coming from a 50 W bulb is $\frac{{E}_{1}}{2}=\frac{E}{2}$.
Hence, the correct answer is option A.

#### Question 5:

If E and B represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along
(a) E.
(b) B.
(c) B × E.
(d) E × B.

#### Answer:

The direction of propagation of the electromagnetic wave is perpendicular to both electric field vector E and magnetic field vector B.

Here, an electromagnetic wave is propagating along the z-direction which is given by the cross product of E and B.
Hence, the correct answer is option D.

#### Question 6:

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) c2: 1
(c) 1 : 1
(d) $\sqrt{\mathrm{c}}:1$

#### Answer:

Intensity due to electric field,
Intensity due to magnetic field,
Speed of light,
Using equation (1) and (2),

$\frac{{\mathrm{I}}_{E}}{{\mathrm{I}}_{B}}=\frac{\frac{1}{2}c{\epsilon }_{0}{\mathrm{E}}^{2}}{\frac{1}{2}\frac{c{\mathrm{B}}^{2}}{{\mu }_{0}}}=\frac{{\epsilon }_{0}{\mathrm{E}}^{2}}{\frac{{\mathrm{B}}^{2}}{{\mu }_{0}}}=\left({\epsilon }_{0}{\mu }_{0}\right){\left(\frac{\mathrm{E}}{\mathrm{B}}\right)}^{2}$
$\frac{{\mathrm{I}}_{E}}{{\mathrm{I}}_{B}}=\frac{1}{{c}^{2}}{c}^{2}=\frac{1}{1}$
So, the ratio of contribution made by electric and magnetic field components to the intensity of an EM wave is 1 : 1
Hence, the correct answer is option C.

#### Question 7:

An EM wave radiates outwards from a dipole antenna, with E0 as the amplitude of its electric field vector. The electric field E0 which  transports significant energy from the source falls off as

(d) remains constant.

#### Answer:

Given: Amplitude of electric field vector = Eo
Intensity of radiation, $I=\frac{W}{At}$
Here, W is energy emitted by source t is time A is area on which radiation is incident, If energy and time are constant
$⇒I\propto \frac{1}{A}$
An antenna radiates energy in spherical shape

Hence, the correct answer is option C.

#### Question 8:

An electromagnetic wave travels in vacuum along z-direction: $\mathbf{E}=\left({\mathrm{E}}_{1}\stackrel{\mathbf{^}}{\mathbf{i}}+{\mathrm{E}}_{2}\stackrel{\mathbf{^}}{\mathbf{j}}\right)\mathrm{cos}\left(kz-\omega t\right)$ Choose the correct options from the following:

(a) The associated magnetic field is given as $\mathbf{B}=\frac{1}{\mathrm{c}}\left({\mathrm{E}}_{1}\stackrel{^}{\mathbf{i}}+{\mathrm{E}}_{2}\stackrel{^}{\mathbf{j}}\right)\mathrm{cos}\left(kz-\omega t\right)$
(b) The associated magnetic field is given as $\mathbf{B}=\frac{1}{\mathrm{c}}\left({\mathrm{E}}_{1}\stackrel{^}{\mathbf{i}}-{\mathrm{E}}_{2}\stackrel{^}{\mathbf{j}}\right)\mathrm{cos}\left(kz-\omega t\right)$
(c) The given electromagnetic field is circularly polarised.
(d) The given electromagnetic wave is plane polarised.

#### Answer:

Given: Electric field vector, $\mathbf{E}=\left({\mathrm{E}}_{1}\stackrel{\mathbf{^}}{\mathbf{i}}+{\mathrm{E}}_{2}\stackrel{\mathbf{^}}{\mathbf{j}}\right)\mathrm{cos}\left(kz-\omega t\right)$
Relation between electric field and magnetic field:-
$\mathrm{B}=\frac{\mathrm{E}}{c}=\frac{1}{c}\left({\mathrm{E}}_{1}\stackrel{\mathbf{^}}{\mathbf{i}}+{\mathrm{E}}_{2}\stackrel{\mathbf{^}}{\mathbf{j}}\right)\mathrm{cos}\left(kz-\omega t\right)$
Em wave propagates through z-direction and E and B vectors lie in x-y plane. Here, the propagation of electromagnetic wave is perpendicular to E and B. This means electromagnetic wave is plane polarised.
Hence, the correct answer is option A and D.

Disclaimer: option a and b were same in question, so option a is corrected.

#### Question 9:

An electromagnetic wave travelling along z-axis is given as: E = E0 cos (kz − ωt.). Choose the correct options from the following;

(a) The associated magnetic field is given as $\mathbit{B}=\frac{1}{c}\stackrel{^}{\mathrm{k}}×\mathbit{E}=\frac{1}{\omega }\left(\mathbit{k}×\mathbit{E}\right)$
(b) The electromagnetic field can be written in terms of the associated magnetic field as E = c $\left(\mathbf{B}×\stackrel{^}{\mathbf{k}}\right)$
(c) $\stackrel{\mathbf{^}}{\mathbf{k}}\mathbf{.}\mathbf{E}=0,\stackrel{\mathbf{^}}{\mathbf{k}}\mathbf{.}\mathbf{B}=0.$
(d) $\stackrel{^}{\mathbf{k}}×\mathbf{E}=0,\mathbf{k}\mathbf{×}\mathbf{B}=0.$

#### Answer:

Given: Electric field, E = E0 cos (kz − ωt.).
Associated magnetic field, ${\mathrm{B}}_{0}=\frac{{\mathrm{E}}_{0}}{c}$
Also, this magnetic field is along x-axis in an electromagnetic wave
i.e., $\stackrel{\to }{\mathrm{B}}\mathit{=}\frac{\mathit{1}}{c}\mathit{\left(}\stackrel{\mathit{^}}{k}\mathit{×}\stackrel{\to }{\mathrm{E}}\mathit{\right)}\phantom{\rule{0ex}{0ex}}$

Since direction of $\stackrel{^}{k}$ is perpendicular to both $\stackrel{\to }{E}$ and $\stackrel{\to }{B}$, so

Hence, the correct answer is option a, b and c.

Disclaimer: Option a is corrected, it should be $\mathbit{B}=\frac{1}{c}\stackrel{^}{\mathrm{k}}×\mathbit{E}=\frac{1}{\omega }\left(\mathbit{k}×\mathbit{E}\right)$ instead of $\mathbit{B}=\frac{1}{c}\stackrel{^}{\mathrm{k}}×\mathbit{E}=\frac{1}{\omega }\left(\stackrel{\mathbf{^}}{\mathbf{k}}×\mathbit{E}\right)$

#### Question 10:

A plane electromagnetic wave propagating along x direction can have the following pairs of E and B
(a) Ex, By.
(b) Ey, Bz.
(c) Bx, Ey.
(d) Ez, By

#### Answer:

Electric field and magnetic field vectors are perpendicular to each other and the direction of propagation of electromagnetic wave.

The following pairs of E and B can be Ey, Bz  or   Ez, By.
Hence the correct answer is option (B, D).

#### Question 11:

A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. The electromagnetic waves produced:

(a) will have frequency of 109 Hz.
(b) will have frequency of 2 × 109 Hz.
(c) will have a wavelength of 0.3 m.
(d) fall in the region of radiowaves.

#### Answer:

The frequency of em wave is equal to the frequency of oscillation of charged particle producing em wave.
Frequency, $\mathrm{\nu }={10}^{9}\mathrm{Hz}$ (lies in radio waves)
Wavelength, $\lambda \mathit{=}\frac{\mathit{c}}{\mathit{\nu }}$
Here, c = speed of light = 3 × 108 m/s

Hence, the correct answer is option (A, C, D).

#### Question 12:

The source of electromagnetic waves can be a charge

(a) moving with a constant velocity.
(b) moving in a circular orbit.
(c) at rest.
(d) falling in an electric field.

#### Answer:

An accelerated charge can emit em waves. When a charged particle is in motion inside an electric field then its velocity keeps on changing along with its acceleration.
So, a charge moving in a circular orbit or falling in an electric field has acceleration.
Hence, the correct answer is option (B, D).

#### Question 13:

An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true?

(a) Radiation pressure is I/c if the wave is totally absorbed.
(b) Radiation pressure is I/c if the wave is totally reflected.
(c) Radiation pressure is 2I/c if the wave is totally reflected.
(b) Radiation pressure is in the range I/c < p < 2I/c for real surfaces.

#### Answer:

Change in momentum per unit time is given as:

If wave is completely absorbed by surface then the momentum of reflected wave per unit area is zero.
So, radiation pressure when wave is absorbed is $\frac{∆I}{c}=\frac{I}{c}–0=\frac{I}{c}$
Now, when the wave is entirely reflected then $p=\frac{I}{c}–\left(–\frac{I}{c}\right)=\frac{2I}{c}$
If the surface is not perfectly absorbing and perfectly reflecting, then p will be more than $\frac{I}{c}$ but less than $\frac{2I}{c}$
Hence, the correct answer is option A, C and D.

#### Question 14:

Why is the orientation of the portable radio with respect to broadcasting station important?

#### Answer:

The orientation of the portable radio with respect to broadcasting station is important this is because for receiving electromagnetic waves receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave.

#### Question 15:

Why does microwave oven heats up a food item containing water molecules most efficiently?

#### Answer:

Microwave oven heats up a food item containing water molecules most efficiently because the frequency of microwave matches with the resonant frequency of water molecules. At resonant frequency, energy from the waves is transferred efficiently to the kinetic energy of water molecules.

#### Question 16:

The charge on a parallel plate capacitor varies as q = q0 cos 2πνt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor?

#### Answer:

Given: Charge on a parallel plate capacitor,
The displacement current is equal to the conduction current

Hence, the required displacement current through the capacitor is .

#### Question 17:

A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?

#### Answer:

Capacitive reactance, Xc$\frac{1}{\omega C}=\frac{1}{2\mathrm{\pi }fC}$
From the above equation, the decrease in frequency of a.c. will increase the reactance of the capacitor and hence will decrease the conduction current.
Also, in this case the displacement current is equal to conduction current, so displacement current will decrease with the decrease in frequency of a.c.

#### Question 18:

The magnetic field of a beam emerging from a filter facing a floodlight is given by B0 = 12 × 10–8 sin (1.20 × 107z – 3.60 × 1015t) T. What is the average intensity of the beam?

#### Answer:

Given magnetic feild,
General equation of magnetic field, B = Bo sin ωt
On comparing the general equation and the given equation, we get
${B}_{\mathrm{o}}=12×{10}^{-8}$
Average intensity, ${I}_{av}=\frac{1}{2}\frac{{{B}_{\mathrm{o}}}^{2}c}{{\mu }_{\mathrm{o}}}$

Hence, the required average intensity is 1.71 W/m2.

#### Question 19:

Poynting vectors S is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propogation. Mathematically, it is given by $\mathbf{S}=\frac{1}{{\mu }_{0}}\mathbf{E}×\mathbf{B}$ Show the nature of S vs t graph.

#### Answer:

Let us consider an electromagnetic wave in which electric field $\stackrel{\to }{E}$ is varying along y-axis, magnetic field $\stackrel{\to }{B}$ is varying along Z-axis and the propagation of wave is along x-axis.
Let

Poynting vector, $S=\frac{1}{{\mu }_{0}}\left(\stackrel{\to }{E}×\stackrel{\to }{B}\right)$
$⇒S=\frac{1}{{\mu }_{0}}{E}_{0}{B}_{0}{\mathrm{sin}}^{2}\left(\omega t-kx\right)\stackrel{^}{i}$
Since ${\mathrm{sin}}^{2}\left(\omega t-kx\right)$ is never negative and  always points in the positive x-direction. So, the variation of |S| w.r.t. t can be shown as:

#### Question 20:

Professor C.V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.

#### Answer:

Electromagnetic waves carry energy and momentum like other waves. By exerting momentum, pressure will be exerted known as radiation pressure. The tail of comet is also due to this property. Professor CV Raman was surprised by shining beam on it which proves the propagation of em waves from vacuum.

#### Question 21:

Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is $\frac{{\in }_{0}{\mu }_{0}r}{2}\frac{dE}{dt}$ (symbols having usual meaning).

#### Answer:

Let us consider parallel plates capacitor having plates of radius r.

Magnetic field at a point in between the plates of the parallel plate capacitor, if ID is the displacement current.
$B=\frac{{\mu }_{0}}{4\mathrm{\pi }}\frac{2{I}_{\mathrm{D}}}{r}=\frac{{\mu }_{0}}{2\mathrm{\pi }}\frac{{I}_{\mathrm{D}}}{r}$
As ${I}_{\mathrm{D}}={\in }_{0}\frac{d{\varphi }_{E}}{dt}$

If E is the electric field between the plates of capacitor, then

Using equation (1) and (2)
$B=\frac{{\mu }_{0}{\in }_{0}}{2\mathrm{\pi }r}×\mathrm{\pi }{r}^{2}\frac{dE}{dt}=\frac{{\in }_{0}{\mu }_{0}r}{2}\frac{dE}{dt}$
Hence Proved.

Disclaimer : In question $\frac{{\in }_{0}{\mu }_{r}}{2}\frac{\mathrm{d}E}{\mathrm{d}t}$ is replaced by $\frac{{\in }_{0}{\mu }_{0}r}{2}\frac{dE}{dt}$

#### Question 22:

Electromagnetic waves with wavelength

(i) λ1 is used in satellite communication.
(ii) λ2 is used to kill germs in water purifies.
(iii) λ3 is used to detect leakage of oil in underground pipelines.
(iv) λ4 is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.

#### Answer:

(a)
(i) For satellite communication $\to$ Microwaves
(ii) For killing germs $\to$ Ultraviolet rays
(iii) For detecting leakage of oil $\to$x - rays
(iv) For improving visibility $\to$ Infrared

(b) Ascending order is
${\mathrm{\lambda }}_{3}<{\mathrm{\lambda }}_{2}<{\mathrm{\lambda }}_{4}<{\mathrm{\lambda }}_{1}$
Microwaves - 1 mm – 25 μm
UV rays - 400 nm – 1 nm
X-rays - 1 nm – 1 pm
Infrared - 25 μm – 2.5 μm

(c)
(i) Microwaves are used in radar system.
(ii) UV rays are used in Lasik surgery
(iii) X-rays is used for treating bones
(iv) Infrared in optical communication.

#### Question 23:

Show that average value of radiant flux density ‘S’ over a single period ‘T’ is given by

#### Answer:

Let us consider electromagnetic waves be propagating along x-axis if electric field is along y-axis and magnetic field vector is along z-axis. Then
E = E0 cos (kx – ωt)
B = B0 cos (kx ωt)

Radiant flux density, $S=\frac{1}{{\mathrm{\mu }}_{0}}\left(\mathbit{E}×\mathbit{B}\right)$
= c0(E × B)
S = c2 0(E0 × B0) cos2(kx – ωt)
Average value of radiant flux density,
${S}_{\mathrm{av}}={c}^{2}{\in }_{0}|{E}_{0}×{B}_{0}|\frac{1}{T}{\int }_{0}^{\mathrm{T}}{\mathrm{cos}}^{2}\left(kx-\mathrm{\omega }t\right)dt$
$⇒{S}_{\mathrm{av}}=\frac{{c}^{2}}{2}{\in }_{0}{E}_{0}\left(\frac{{E}_{0}}{c}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{\mathrm{av}}={\in }_{0}\frac{{E}_{0}^{2}c}{2}=\frac{c}{2}×\frac{1}{{c}^{2}{\mathrm{\mu }}_{0}}{E}_{0}^{2}\phantom{\rule{0ex}{0ex}}⇒{S}_{\mathrm{av}}=\frac{{E}_{0}^{2}}{2{\mathrm{\mu }}_{0}c}$

#### Question 24:

You are given a 2μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1mA in the space between its plates?

#### Answer:

Given: Displacement current, ID = 1 mA
Capacitance  C = 2μF
Charge, Q = cv and charge, Q = IDdt

Hence, by applying a varying potential difference of 500 v/s, a displacement current of desired value could be produced.

#### Question 25:

Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is I/c.

#### Answer:

Let us consider intensity falls on the surface of area A be I.
Energy received by surface per second = I × A
Let N be the number of photons per second.

$⇒N=\frac{E}{{E}_{\mathrm{photon}}}=\frac{E}{\frac{hc}{\mathrm{\lambda }}}\phantom{\rule{0ex}{0ex}}⇒N=\frac{IA\mathrm{\lambda }}{hc}$
If surface is perfectly absorbing:
$∆P=\frac{h}{\mathrm{\lambda }}$
Force, F = N × $∆P=\frac{IA}{c}$
Pressure = $\frac{\mathrm{Force}}{\mathrm{Area}}=\frac{IA}{cA}=\frac{I}{c}$
This is the required relation.

#### Question 26:

What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant.

What geomatrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?

#### Answer:

Given: Distance from the bulb is doubled.
Intensity of light,
This means intensity of light reduces to one-fourth. But the laser beam does not spread as it travels across the length of a room, so its intensity remains constant.

Some geometrical characteristics of laser beam which are missing in case of normal light from the bulb are:
(i) Unidirectional
(ii) Monochromatic
(iii) Coherent light
(iv) Highly collimated.

#### Question 27:

Even though an electric field E exerts a force qE on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.

#### Answer:

The electric field of an electromagnetic wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero since its direction changes every half cycle.
Therefore, electric field is not responsible for radiation pressure.

#### Question 28:

An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis (Fig. 8.1). The wire is set into
motion along its length with a uniform velocity $\mathbf{v}=\mathrm{\nu }{\stackrel{\mathbf{^}}{\mathbf{k}}}_{\mathrm{z}}$ . Calculate the poynting vector $\mathbf{S}=\frac{1}{{\mu }_{0}}\left(\mathbf{E}×\mathbf{B}\right)$

#### Answer:

Let us consider a cylindrical Gaussian surface in such a way that the axis of cylinder lies on wire. Electric field intensity due to long straight wire at a distance r and charge density

Electric field due to infinitely long thin wire, $\stackrel{\to }{E}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\in }_{0}r}\stackrel{^}{j}$
Magnetic field due to the wire, $\stackrel{\to }{B}=\frac{{\mathrm{\mu }}_{0}I}{2\mathrm{\pi }r}\stackrel{^}{i}$
Equivalent current flowing through the wire:-
$I=\frac{Q}{t}=\frac{\mathrm{\lambda }l}{t}=\mathrm{\lambda }\nu \phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{B}=\frac{{\mathrm{\mu }}_{0}\mathrm{\lambda }\nu }{2\mathrm{\pi }r}\stackrel{^}{i}$
Poynting vector, $S=\frac{1}{{\mathrm{\mu }}_{0}}\left[\stackrel{\to }{E}×\stackrel{\to }{B}\right]$
$=\frac{1}{{\mathrm{\mu }}_{0}}\left[\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\in }_{0}r}\stackrel{^}{j}×\frac{{\mathrm{\mu }}_{0}\mathrm{\lambda }\nu }{2\mathrm{\pi }r}\stackrel{^}{i}\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\lambda }}^{2}\nu }{4{\mathrm{\pi }}^{2}{r}^{2}{\in }_{0}}\left(\stackrel{^}{j}×\stackrel{^}{i}\right)\phantom{\rule{0ex}{0ex}}=\frac{-{\mathrm{\lambda }}^{2}\nu }{4{\mathrm{\pi }}^{2}{r}^{2}{\in }_{0}}\stackrel{^}{k}$

Hence, this is the required value of poynting vector.

#### Question 29:

Sea water at frequency ν = 4 × 108 Hz has permittivity ε ≈ 80 εo, permeability μ ≈ μo and resistivity ρ = 0.25 Ω–m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) = Vo sin (2π νt). What fraction of the conduction current density is the displacement current density?

#### Answer:

Given: Frequency,
Permittivity,
Permeability, $\mathrm{\mu }={\mathrm{\mu }}_{0}$
Resistivity, ρ = 0.25Ω m
Let d be the distance between the plates of the parallel plate capacitor.
Electric field between the plates, $E=\frac{v}{d}\left(t\right)$
$⇒E=\frac{{v}_{0}}{d}\mathrm{sin}\left(2\mathrm{\pi }\nu t\right)$
Using ohm's law, conduction current density is given by
${J}^{\mathrm{C}}=\frac{E}{\mathrm{\rho }}=\frac{{v}_{0}}{\mathrm{\rho }d}\mathrm{sin}\left(2\mathrm{\pi }\nu t\right)\phantom{\rule{0ex}{0ex}}⇒{J}^{\mathrm{C}}={J}_{0}^{\mathrm{C}}\mathrm{sin}\left(2\mathrm{\pi }\nu t\right)$
Displacement current density:-
${J}^{d}=\in \frac{dE}{dt}=\in \frac{d}{dt}\left[\frac{{V}_{0}}{d}\mathrm{sin}\left(2\pi vt\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{2\pi v\in {v}_{0}}{d}\mathrm{cos}\left(2\pi vt\right)\phantom{\rule{0ex}{0ex}}⇒{J}^{d}={J}_{0}^{d}\mathrm{cos}\left(2\pi vt\right)$
This means ${J}_{0}^{C}=\frac{{v}_{0}}{\mathrm{\rho }d}$ and ${J}_{0}^{\mathrm{d}}=\frac{2\mathrm{\pi }v\in {v}_{0}}{d}$
$⇒\frac{{J}_{0}^{\mathrm{d}}}{{J}_{0}^{\mathrm{d}}}=\frac{2\pi v\in {\nu }_{0}×\rho d}{d×{v}_{0}}=2\mathrm{\pi }\nu \in \mathrm{\rho }\phantom{\rule{0ex}{0ex}}⇒\frac{{J}_{0}^{\mathrm{d}}}{{J}_{0}^{\mathrm{c}}}=\frac{2\mathrm{\pi }\left(4×{10}^{8}\right)\left(80{\in }_{0}\right)\left(0.25\right)}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{{J}_{0}^{\mathrm{d}}}{{J}_{0}^{\mathrm{c}}}=4\mathrm{\pi }{\in }_{0}\left(4×{10}^{9}\right)=\frac{4×{10}^{9}}{9×{10}^{9}}\phantom{\rule{0ex}{0ex}}⇒\frac{{J}_{0}^{\mathrm{d}}}{{J}_{0}^{\mathrm{c}}}=\frac{4}{9}$

#### Question 30:

A long straight cable of length l is placed symmetrically along z-axis and has radius a(<<l). The cable consists of a thin wire and
a co-axial conducting tube. An alternating current I(t) = Io sin (2πνt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current Id.
(iii) Compare the conduction current I0 with the dispalcement current ${I}_{0}^{\mathrm{d}}$

#### Answer:

Given→ E(s, t) = µ0 Icos
= Displacement current density, J${\in }_{0}\frac{dE}{dt}$

(ii) Total displacement current, ${I}_{\mathrm{d}}={\int }_{}{J}_{\mathrm{d}}sds\mathrm{cos}\theta$

(iii) Displacement current = $\begin{array}{rcl}{I}_{\mathrm{d}}& =& {\left(\frac{2\mathrm{\pi }a}{2\mathrm{\lambda }}\right)}^{2}{I}_{0}\mathrm{sin}2\mathrm{\pi }vt\end{array}$

Hence, the comparison between conduction current and displacement current is ${\left(\frac{a\mathrm{\pi }}{\mathrm{\lambda }}\right)}^{2}$

#### Question 31:

A plane EM wave travelling in vacuum along z direction is given by E = E0 sin

(i) Evaluate âˆ®E.dl over the rectangular loop 1234 shown in Fig 8.2.
(ii) Evaluate ∫B.ds over the surface bounded by loop 1234.
(iii) Use equation
(iv) By using similar process and the equation

#### Answer:

As shown in the diagram

(i) $\oint \stackrel{\to }{\mathit{\text{E}}}.d\stackrel{\to }{l}=\underset{1}{\overset{2}{\int }}\stackrel{\to }{\mathit{\text{E}}}.d\stackrel{\to }{l}+\underset{2}{\overset{3}{\int }}\stackrel{\to }{\mathit{\text{E}}}.d\stackrel{\to }{l}+\underset{3}{\overset{4}{\int }}\stackrel{\to }{\mathit{\text{E}}}.d\stackrel{\to }{l}+\underset{4}{\overset{1}{\int }}\stackrel{\to }{\mathit{\text{E}}}.d\stackrel{\to }{l}$
But angle between electric field
$\stackrel{\to }{E}$ = E0 sin(kz$\omega$t)i and side 1-2 and side 3-4 is 90º.
But angle between $\stackrel{\to }{E}$ and side 2-3 and 4-1 is 0º and 180º respectively
So,
⇒ $\oint \stackrel{\to }{\mathit{\text{E}}}.d\stackrel{\to }{l}=$Esin(kz$\omega$t)h – Esin(kz1$\omega$t)h
= E
0 h [sin(kz2$\omega$t) – sin(kz1$\omega$t)]    .....(1)
(ii) Let us take a small element of area ds = h × dz as shown in the diagram

Given,

It is given that:
$\oint E.dl=\frac{d\varphi }{dt}$
From equation (3) and (4)

(iv) Now consider the loop in y-z plane as shown in the diagram:

$\oint \stackrel{\to }{B}.d\stackrel{\to }{l}=\underset{1}{\overset{2}{\int }}\stackrel{\to }{B}.d\stackrel{\to }{l}+\underset{2}{\overset{3}{\int }}\stackrel{\to }{B}.d\stackrel{\to }{l}+\underset{3}{\overset{4}{\int }}\stackrel{\to }{B}.d\stackrel{\to }{l}+\underset{4}{\overset{1}{\int }}\stackrel{\to }{B}.d\stackrel{\to }{l}$
Here $\stackrel{\to }{B}$ = Bsin(kz$\omega$t)$\stackrel{^}{j}$
Angle between $\stackrel{\to }{B}$ and side 1-2 is 0°.
Angle between $\stackrel{\to }{B}$ and side 2-3 is 90º.
Angle between $\stackrel{\to }{B}$ and side 3-4 is 180º
Angel between $\stackrel{\to }{B}$ and side 4-1 is 90º
So,

Now, flux due to electric field

It is given that, conduction current,

#### Question 32:

A plane EM wave travelling along z direction is described by  show that

(i) The average energy density of the wave is given by
(ii) The time averaged intensity of the wave is given by

#### Answer:

(a) Energy contributed due to electric field,
${u}_{E}=\frac{1}{2}{\epsilon }_{0}{E}^{2}$
Energy contributed due to magnetic field,
${u}_{\text{(B)}}=\frac{1}{2}\frac{{B}^{2}}{{\mu }_{0}}$
So, total energy density:
u = u(B) + u(E)

By symmetry of graph of

The average value of
So,

Putting the average values of
E2 and B2 in the equation (1),
We get:
$\begin{array}{l}{u}_{\text{av}}=\frac{1}{2}\left({\epsilon }_{0}\frac{{E}_{0}^{2}}{2}\right)=\frac{1}{2}\frac{\left(\frac{{B}_{0}^{2}}{2}\right)}{{\mu }_{0}}\\ =\frac{1}{4}{\epsilon }_{0}{E}_{0}^{2}+\frac{1}{4}\frac{{B}_{0}^{2}}{{\mu }_{0}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\text{(2)}\end{array}$
(b) we have $\frac{{E}_{0}}{{B}_{0}}=c⇒{B}_{0}=\frac{{E}_{0}}{c}$

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