Hc Verma II Solutions for Class 11 Science Physics Chapter 40 Electromagnetic Waves are provided here with simple step-by-step explanations. These solutions for Electromagnetic Waves are extremely popular among Class 11 Science students for Physics Electromagnetic Waves Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma II Book of Class 11 Science Physics Chapter 40 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma II Solutions. All Hc Verma II Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 338:

#### Answer:

The natural frequency of water matches the frequency of microwave. This is the reason that food containing water gets cooked. The natural frequency of the plastic container does not match the frequency of microwave. So, the plastic container is not damaged.

#### Page No 338:

#### Answer:

The magnetic field along the axis of a solenoid carrying a high-frequency alternating current changes continuously. Due to the change in the magnetic field, e.m.f (or eddy current) is induced in the metal rod. There will be flow of charge due to the induced e.m.f. The direction of the induced e.m.f changes very frequently due to the high-frequency alternating current in the solenoid. Thus, the rod gets heated up due to the flow of charge in it.

#### Page No 338:

#### Answer:

No, an electromagnetic wave cannot be deflected by an electric field or a magnetic field. This is because according to Maxwell's theory, an electromagnetic wave does not interact with the static electric field and magnetic field. Even if we consider the particle nature of the wave, the photon is electrically neutral. So, it is not affected by the static magnetic and electric fields.

#### Page No 338:

#### Answer:

When an alternating current passes through a conductor, the changing magnetic field create a changing electric field outside it. An electromagnetic field is radiated from the surface of the conductor. There is a time-varying electric field outside the conductor. Hence, there is a time-varying electric field in the vicinity of the wire.

#### Page No 338:

#### Answer:

When an alternating-current source is connected to a capacitor, the electric field between the plates of the capacitor keeps on changing with the applied voltage. Due to the changing electric field, a magnetic field exists in between the plates of the capacitor.

#### Page No 338:

#### Answer:

An electromagnetic wave is a transverse wave; thus, it can be polarised. An unpolarised wave consists of many independent waves, whose planes of vibrations of electric and magnetic fields are randomly oriented. They are polarised by restricting the vibrations of the electric field vector or magnetic field vector in one direction only.

#### Page No 338:

#### Answer:

Let the electromagnetic wave be propagating in the z-direction. The vibrations of the electric and magnetic fields are given by:

*E*_{x}= *E*_{0} sin (*kz – ωt*)

*B*_{y}= *B*_{0} sin (*kz – ωt*)

Let the volume of the region be *V*.

The angular frequency of the vibrations of the electric and magnetic fields are same and are equal to *ω. *Therefore, their frequency, $f=\frac{\omega}{2\pi}$, is same.

The electrical energy in the region,

*U*_{E}_{ }= $\left(\frac{1}{2}{\in}_{0}{E}^{2}\right)\times V$

It can be written as:

${U}_{\mathrm{E}}=\left(\frac{1}{2}{\in}_{0}\left({{E}_{0}}^{2}{\mathrm{sin}}^{2}\right(kz-\omega t)\right)\times V\phantom{\rule{0ex}{0ex}}{U}_{E}=\left(\frac{1}{2}{\in}_{0}{{E}_{0}}^{2}\times \frac{\left(1-\mathrm{cos}2(kz-\omega t)\right)}{2}\right)\times V\phantom{\rule{0ex}{0ex}}{U}_{E}=\left(\frac{1}{4}{\in}_{0}{{E}_{0}}^{2}\times \left(1-\mathrm{cos}2(kz-\omega t)\right)\right)\times V\phantom{\rule{0ex}{0ex}}$

The magnetic energy in the region,

${U}_{\mathrm{B}}=\left(\frac{{B}^{2}}{2{\mu}_{0}}\right)\times V\phantom{\rule{0ex}{0ex}}{U}_{\mathrm{B}}=\left(\frac{{{B}_{0}}^{2}{\mathrm{sin}}^{2}(kz-\omega t)}{2{\mu}_{0}}\right)\times V\phantom{\rule{0ex}{0ex}}\Rightarrow {U}_{\mathrm{B}}=\left(\frac{{{B}_{0}}^{2}\left(1-\mathrm{cos}(2kz-2\omega t)\right)}{4{\mu}_{0}}\right)\times V$

The angular frequency of the electric and magnetic energies is same and is equal to 2*ω*.

Therefore, their frequency, $f\text{'}=\frac{2\omega}{2\pi}=2f\phantom{\rule{0ex}{0ex}}$, will be same.

Thus, the electric and magnetic fields have same frequencies and the electrical and magnetic energies will have same frequencies.

#### Page No 338:

#### Answer:

(d) Both of them

According to Ampere-Maxwell's Law, a magnetic field is produced due to the conduction current in a conductor and the displacement current. The conduction current is actually the motion of the charge. The displacement current is due to the changing electric field. The displacement current is given by ${i}_{\mathrm{d}}={\epsilon}_{0}\frac{\mathrm{d}{\Phi}_{E}}{\mathrm{d}t}$ ($\because $ *ϕ*_{E} is the electric flux).

Thus, the magnetic field is produced by the moving charge as well as the changing electric field.

#### Page No 338:

#### Answer:

(d) deflects and gradually comes to the original position in a time that is large compared to the time constant

The compass needle deflects due to the presence of the magnetic field. Inside the capacitor, a magnetic field is produced when there is a changing electric field inside it. As the capacitor is connected across the battery, the charge on its plates at a certain time *t* is given by:

* Q = CV *(1* **$-$ **e** ^{$-$t/RC}*),

where

*Q*= charge developed on the plates of the capacitor

*R*= resistance of the resistor connected in series with the capacitor

*C*= capacitance of the capacitor

*V*= potential difference of the battery

The time constant of the capacitor is given,

*τ = RC*

The capacitor keeps on charging up to the time

*τ*. The development of charge on the plates will be gradual after

*t = RC.*The change in electric field will be up to the time the charge is developing on the plates of the capacitor. Thus, the compass needle deflects and gradually comes to the original position in a time that is large compared to the time constant.

#### Page No 338:

#### Answer:

(c) L^{2}/T^{2}

The speed of light, $C=\frac{1}{\sqrt{{\mu}_{0}{\in}_{0}}}$.

The dimensions of $\frac{1}{\sqrt{{\mu}_{0}{\in}_{0}}}$ are of velocity, i.e. L/T .

Therefore, $\frac{1}{{\in}_{0}{\mu}_{0}}$ will have dimensions L^{2}/T^{2}.

#### Page No 338:

#### Answer:

(c) an accelerating charge

A static charge produces an electrostatic field. A moving charge produces a magnetic field. Electromagnetic waves are produced by an accelerating charge.

#### Page No 338:

#### Answer:

(a) *E*_{0} *k* = *B*_{0} ω

The relation between *E*_{0} and *B*_{0}_{ }is given by $\frac{{E}_{0}}{{B}_{0}}=c$ ...(1)

Here, *c* = speed of the electromagnetic wave

The relation between *ω* (the angular frequency) and *k* (wave number):

$\frac{\omega}{k}=c$ ...(2)

Therefore, from (1) and (2), we get:

$\frac{{E}_{0}}{{B}_{0}}=$$\frac{\omega}{k}=c$

$\Rightarrow {E}_{0}k={B}_{0}\omega $

#### Page No 339:

#### Answer:

(c) An electromagnetic wave may be passing through the region.

For an electromagnetic wave,electric field ,magnetic field and direction of propagation are mutually perpendicular to each other.We can have a region in which electric and magnetic fields are applied at an angle with each other.In transmission lines Different modes exist. In transverse electric (TE) mode-no electric field exist in the direction of propagation. These are sometimes called *H modes* because there is only a magnetic field along the direction of propagation (*H* is the conventional symbol for magnetic field).

#### Page No 339:

#### Answer:

(a) Both A and B are true.

For a linearly polarised, plane electromagnetic wave

$E={E}_{0}\mathrm{sin}\omega (t-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.)\phantom{\rule{0ex}{0ex}}B={B}_{0}\mathrm{sin}\omega (t-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$c$}\right.)$

The average value of either *E* or *B* over a cycle is zero ( average of sin($\theta $) over a cycle is zero).

Also the electric energy density (*u*_{E}) and magnetic energy density (*u*_{B})are equal.

${u}_{\mathrm{E}}=\frac{1}{2}{\in}_{0}{E}^{2}=\frac{{B}^{2}}{2{\mu}_{0}}={u}_{\mathrm{B}}$

Energy can be found out by integrating energy density over the entire volume of full space.

As the energy of the electromagnetic wave is equally shared between electric and magnetic field so their average values will also be equal.

#### Page No 339:

#### Answer:

(a) along the electric field

As the electron is at rest initially, only the electric field will exert force on it. There will be no magnetic force on the electron in the stating. Hence, the electron will start moving along the electric field.

#### Page No 339:

#### Answer:

(c) *p* ≠ 0, *E* ≠ 0.

When an electromagnetic wave strikes a material surface, it transports the momentum, as well as the energy, to the surface. The striking electromagnetic wave exerts pressure on the surface. The total energy transferred to the surface by the electromagnetic wave is given by *E = pc. *Therefore, *p* ≠ 0, *E* ≠ 0.

#### Page No 339:

#### Answer:

(c) *k*/ω

The given quantities can be expressed as:

*k* is given by

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

*ω* is given by

$\omega =2\pi \nu \phantom{\rule{0ex}{0ex}}c=\nu \lambda \phantom{\rule{0ex}{0ex}}\Rightarrow \omega =2\pi \frac{c}{\lambda}$

*k*/ω is given by

$\frac{k}{\omega}=\frac{2\pi /\lambda}{2\pi c/\lambda}=\frac{1}{c}$

*k*ω is given by

$k\times \omega =\frac{2\pi}{\lambda}\times \frac{2\pi c}{\lambda}=\frac{4{\pi}^{2}c}{{\lambda}^{2}}$

Thus, *k*/ω is independent of the wavelength.

#### Page No 339:

#### Answer:

(a) increases

(b) decreases

Displacement current inside a capacitor,

${i}_{\mathrm{d}}={\epsilon}_{0}\frac{d{\varphi}_{\mathrm{E}}}{dt}$, where

*ϕ*_{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 electric field increases or decreases inside the capacitor.

#### Page No 339:

#### Answer:

(c) for all intensities

For any given medium, the speed (*c*) of an electromagnetic wave is given by

*c* =* νλ*,

where

*ν* = frequency of the electromagnetic wave

*λ* = wavelength of the electromagnetic wave

As the frequency and wavelength are changed, the speed of the electromagnetic wave changes. So, the speed of an electromagnetic wave is not same for all wavelengths and all frequencies in any medium. The velocity of an electromagnetic wave changes with change in medium. Also, the speed of an electromagnetic wave is same for all the intensities in any medium.

#### Page No 339:

#### Answer:

(a) Electric field

(b) Magnetic field

In a plane electromagnetic wave, the electric and the magnetic fields oscillate sinusoidally. For an electromagnetic wave propagating in the z-direction, the electric and magnetic fields are given by:

*E*_{x} = *E*_{0} sin (*kz – ωt*)

*B*_{y} = *B*_{0} sin (*kz – ωt*)

These are sinusoidal functions. Therefore, for a fixed value of z, the average value of the electric and magnetic fields are zero.

#### Page No 339:

#### Answer:

(d) double the frequency of the wave

The energy per unit volume of an electromagnetic wave,

$u=\frac{1}{2}{\in}_{0}{E}^{2}+\frac{{B}^{2}}{2{\mu}_{0}}$

The energy of the given volume can be calculated by multiplying the volume with the above expression.

$U=u\times V=\left(\frac{1}{2}{\in}_{0}{E}^{2}+\frac{{B}^{2}}{2{\mu}_{0}}\right)\times V$ ...(1)

Let the direction of propagation of the electromagnetic wave be along the z-axis. Then, the electric and magnetic fields at a particular point are given by

*E*_{x}= *E*_{0} sin (*kz – ωt*)

*B*_{y}= *B*_{0} sin (*kz – ωt*)

Substituting the values of electric and magnetic fields in (1), we get:

$U=\left(\frac{1}{2}{\in}_{0}\left({{E}_{0}}^{2}{\mathrm{sin}}^{2}\right(kz-\omega t)+\frac{{{B}_{0}}^{2}{\mathrm{sin}}^{2}(kz-\omega t)}{2{\mu}_{0}}\right)\times V\phantom{\rule{0ex}{0ex}}\Rightarrow U=\left({\in}_{0}{{E}_{0}}^{2}\frac{(1-\mathrm{cos}(2kz-2\omega t\left)\right)}{4}+\frac{{{B}_{0}}^{2}(1-\mathrm{cos}(2kz-2\omega t\left)\right)}{4{\mu}_{0}}\right)\times V$

From the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency 2*ω*. Thus, the frequency of the energy of the electromagnetic wave will also be double.

#### Page No 339:

#### Answer:

Displacement current,

${I}_{\mathrm{D}}=\frac{{\in}_{0}d{\phi}_{\mathrm{e}}}{dt}$

Electric flux,

${\varphi}_{\mathrm{e}}=EA$

$\left[{\varphi}_{e}\right]\mathit{=}\mathit{}\left[E\right]\left[A\right]\phantom{\rule{0ex}{0ex}}=\left[\frac{\mathit{1}}{\mathit{4}\pi {\mathit{\in}}_{\mathit{0}}}\frac{q}{{r}^{\mathit{2}}}\right]\left[A\right]\phantom{\rule{0ex}{0ex}}\mathrm{Also},\left[{\mathit{\in}}_{0}\right]=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^{4}{\mathrm{A}}^{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \left[{\varphi}_{\mathrm{e}}\right]=\left[{\mathrm{M}}^{1}{\mathrm{L}}^{3}{\mathrm{T}}^{-4}{\mathrm{A}}^{-2}\right]\left[\mathrm{AT}\right]\left[{\mathrm{L}}^{-2}\right]\left[{\mathrm{L}}^{2}\right]\phantom{\rule{0ex}{0ex}}=\left[{\mathrm{ML}}^{3}{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]\phantom{\rule{0ex}{0ex}}$

Displacement current,

$\left[{I}_{\mathrm{D}}\right]=\left[{\in}_{0}\right]\left[{\varphi}_{\mathrm{e}}\right]\left[{\mathrm{T}}^{-1}\right]\phantom{\rule{0ex}{0ex}}\left[{I}_{\mathrm{D}}\right]=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^{4}{\mathrm{A}}^{2}\right]\left[{\mathrm{ML}}^{3}{\mathrm{T}}^{-3}{\mathrm{A}}^{-1}\right]\left[{\mathrm{T}}^{-1}\right]\phantom{\rule{0ex}{0ex}}\left[{I}_{\mathrm{D}}\right]=\left[\mathrm{A}\right]$

[*I*_{D}]=[current]

#### Page No 339:

#### Answer:

From Coulomb's law:

Electric field strength,

$E\mathit{=}\mathit{}\frac{kq}{{x}^{\mathit{2}}}$

Electric flux,

${\varphi}_{\mathrm{E}}=EA\phantom{\rule{0ex}{0ex}}{\varphi}_{\mathrm{E}}=\frac{kqA}{{x}^{2}}$

Displacement current =* I*_{d}

${I}_{\mathrm{d}}=\left|{\in}_{0}\frac{d{\varphi}_{\mathrm{E}}}{dt}\right|\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{d}}=\left|{\in}_{0}\frac{d}{dt}\left(\frac{kqA}{{x}^{2}}\right)\right|\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{d}}=\left|{\in}_{0}kqA\frac{d}{dt}{x}^{-2}\right|\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{d}}=\left|{\mathit{\in}}_{\mathit{0}}\frac{\mathit{1}}{\mathit{4}\mathit{\pi}{\mathit{\in}}_{\mathit{0}}}\mathit{\times}\mathit{q}\mathit{\times}\mathit{A}\mathit{\times}\mathit{(}\mathit{-}\mathit{2}\mathit{)}{\mathit{x}}^{\mathit{-}\mathit{3}}\mathit{\times}\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}\right|\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{d}}=\left|\frac{qAv}{2\mathrm{\pi}{x}^{3}}\right|$

#### Page No 339:

#### Answer:

Electric field strength for a parallel plate capacitor,

$E=\frac{Q}{{\in}_{0}A}$

Electric flux linked with the area,

${\varphi}_{E}\mathit{=}EA\mathit{=}\frac{Q}{{\mathit{\in}}_{\mathit{0}}A}\mathit{\times}\frac{A}{\mathit{2}}=\frac{Q}{2{\in}_{0}}$

Displacement current,

${I}_{\mathrm{d}}={\in}_{0}\frac{d{\varphi}_{\mathrm{E}}}{dt}={\in}_{0}\frac{d}{dt}\left(\frac{Q}{{\mathit{\in}}_{\mathit{0}}\mathit{2}}\right)\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{d}}=\frac{1}{2}\left(\frac{d\mathrm{Q}}{dt}\right)...\left(\mathrm{i}\right)$

Charge on the capacitor as a function of time during charging,

$Q=\epsilon C\left[1-{e}^{\mathit{-}t\mathit{/}RC}\right]$

Putting this in equation (i), we get:

${I}_{\mathrm{d}}=\frac{1}{2}\epsilon C\frac{d}{dt}\left(1\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)\phantom{\rule{0ex}{0ex}}{I}_{\mathrm{d}}=\frac{1}{2}\epsilon C\left(\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)\times \left(\mathit{-}\frac{1}{RC}\right)\phantom{\rule{0ex}{0ex}}\mathrm{C}=\frac{A{\mathit{\in}}_{\mathit{0}}}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow {I}_{\mathrm{d}}=\frac{\epsilon}{2R}\times {e}^{\mathit{-}\frac{td}{{\epsilon}_{\mathit{0}}AR}}$

#### Page No 339:

#### Answer:

Electric field strength for a parallel plate capacitor = $E=\frac{Q}{{\in}_{0}A}$

$\mathrm{Electric}\mathrm{flux},\varphi \mathit{=}E\mathit{.}A\mathit{=}\frac{Q}{{\mathit{\in}}_{0}A}\mathit{.}A=\frac{Q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\mathrm{Displacement}\mathrm{current},{i}_{\mathrm{d}}={\in}_{0}\frac{d{\varphi}_{\mathrm{E}}}{dt}={\in}_{0}\frac{d}{dt}\left(\frac{\mathrm{Q}}{{\in}_{0}}\right)\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{d}}=\left(\frac{dQ}{dt}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Also},Q\mathit{}\mathit{=}\mathit{}CV\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{d}}=\frac{d}{dt}\left({E}_{\mathit{0}}{{C}_{e}}^{\mathit{-}t\mathit{/}RC}\right)\phantom{\rule{0ex}{0ex}}{i}_{\mathrm{d}}={E}_{0}C\left(\mathit{-}\frac{\mathit{1}}{RC}\right){e}^{\mathit{-}t\mathit{/}RC}\phantom{\rule{0ex}{0ex}}\mathrm{Displacement}\mathrm{resistance},{R}_{d}=\frac{{E}_{\mathit{0}}}{{i}_{\mathrm{d}}}-R\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{d}=\frac{{E}_{\mathit{0}}}{\frac{{E}_{\mathit{0}}}{R}{\mathrm{e}}^{\mathit{-}t\mathit{/}RC}}-R\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{d}=R{\mathrm{e}}^{t\mathit{/}RC}-R\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{R}_{\mathit{d}}\mathit{=}R({\mathrm{e}}^{\mathit{t}\mathit{/}\mathit{R}\mathit{C}}-1)$

#### Page No 339:

#### Answer:

Given, *B = *µ_{0}*H*

For vacuum we can rewrite this equation as,

*B*_{0}* = *µ_{0}*H*_{0} ...(i)

Relation between magnetic field and electric field for vacuum is given as,

*B*_{0}* = *µ_{0${\in}_{0}$}*cE*_{0} ...(ii)

From equation (i) by (ii),

${\mathrm{\mu}}_{0}{H}_{0}={\mathrm{\mu}}_{0}{\in}_{0}c{E}_{0}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{E}_{0}}{{H}_{0}}=\frac{1}{{\in}_{0}c}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{E}_{0}}{{H}_{0}}=\frac{1}{8.85\times {10}^{-12}\times 3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{E}_{0}}{{H}_{0}}\approx 377\mathrm{\Omega}$

$\mathrm{Dimension}\text{of}\frac{1}{{\in}_{0}c}=\frac{1}{\left[{\mathrm{LT}}^{-1}\right]\left[{\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^{4}{\mathrm{A}}^{2}\right]}\phantom{\rule{0ex}{0ex}}=\frac{1}{{\mathrm{M}}^{-1}{\mathrm{L}}^{-2}{\mathrm{T}}^{3}{\mathrm{A}}^{2}}\phantom{\rule{0ex}{0ex}}={\mathrm{M}}^{4}{\mathrm{L}}^{2}{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}=\left[R\right]$

#### Page No 339:

#### Answer:

Given:

Electric field amplitude, *E*_{0} = 810 V/m

Maximum value of magnetic field = Magnetic field amplitude = *B*_{0} = ?

We know:

Speed of a wave =$\frac{E}{B}$

For electromagnetic waves, speed = speed of light

*B*_{0} = µ_{0} ε_{0} c*E*_{0}

Putting the values in the above relation, we get:

${B}_{0}=4\mathrm{\pi}\times {10}^{-7}\times 8.85\times {10}^{-12}\times 3\times {10}^{8}\times 810\phantom{\rule{0ex}{0ex}}{B}_{0}=4\times 3.14\times 8.85\times 3\times 81\times {10}^{-10}\phantom{\rule{0ex}{0ex}}{B}_{0}=27010.9\times {10}^{-10}\phantom{\rule{0ex}{0ex}}{B}_{0}=2.7\times {10}^{-6}\mathrm{T}=2.7\mathrm{\mu T}$

#### Page No 339:

#### Answer:

Maximum value of a magnetic field, B_{0} = 200 $\mathrm{\mu}$T

The speed of an electromagnetic wave is c.

So, maximum value of electric field,

${E}_{0}=c{B}_{0}$

${E}_{0}=\mathrm{c}\times {B}_{0}=200\times {10}^{-6}\times 3\times {10}^{8}\phantom{\rule{0ex}{0ex}}{E}_{0}=6\times {10}^{4}{\mathrm{NC}}^{-1}$

(b) Average energy density of a magnetic field,

${U}_{av}=\frac{1}{2{\mathrm{\mu}}_{0}}{{B}_{0}}^{2}=\frac{(200\times {10}^{-6}{)}^{2}}{2\times 4\mathrm{\pi}\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}{U}_{av}=\frac{4\times {10}^{-8}}{8\mathrm{\pi}\times {10}^{-7}}=\frac{1}{20\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}{U}_{av}=0.0159\approx 0.016\mathrm{J}/{\mathrm{m}}^{3}$

For an electromagnetic wave, energy is shared equally between the electric and magnetic fields.

Hence, energy density of the electric field will be equal to the energy density of the magnetic field.

#### Page No 339:

#### Answer:

Given:

Intensity, *I* = 2.5 × 10^{14} W/m^{2}

We know:

$I=\frac{1}{2}{\in}_{0}{{E}_{0}}^{2}c$

$\Rightarrow {{E}_{0}}^{2}=\frac{2I}{{\mathit{\in}}_{0}c}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=\sqrt{\frac{2I}{{\mathit{\in}}_{0}c}}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=\sqrt{\frac{2\times 2.5\times {10}^{14}}{8.85\times {10}^{-12}\times 3\times {10}^{8}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=0.4339\times {10}^{9}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=4.33\times {10}^{8}\mathrm{N}/\mathrm{C}$

Maximum value of magnetic field,

${B}_{0}=\frac{{E}_{0}}{c}\phantom{\rule{0ex}{0ex}}{B}_{0}=\frac{4.33\times {10}^{8}}{3\times {10}^{8}}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{0}=1.43\mathrm{T}$

#### Page No 339:

#### Answer:

Given:

$\mathit{}I=1380\mathrm{W}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}I=\frac{1}{2}{\in}_{0}{{E}_{0}}^{2}c\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=\sqrt{\frac{2I}{{\in}_{0}c}}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=\sqrt{\frac{2\times 1380}{8.83\times {10}^{-12}\times 3\times {10}^{8}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=103.95\times {10}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=10.195\times {10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{0}=1.02\times {10}^{3}\mathrm{N}/\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since}{E}_{0}={B}_{\mathit{0}}c,\phantom{\rule{0ex}{0ex}}{B}_{0}=\frac{{E}_{0}}{c}=\frac{1.02\times {10}^{2}}{3\times {10}^{28}}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{0}=3.398\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{0}=3.4\times {10}^{-6}\mathrm{T}$

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