NCERT Exercise Q & A

EXERCISE PROBLEMS

Q8.1: Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm , and separated by 5.0 cm . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A .
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Solution:Radius of each circular plate, \(\mathrm{r}=12 \mathrm{~cm}=0.12 \mathrm{~m}\)
Distance between the plates, \(\mathrm{d}=5 \mathrm{~cm}=0.05 \mathrm{~m}\)
Charging current, \(\mathrm{I}=0.15 \mathrm{~A}\)
Permittivity of free space, \(\varepsilon_0=8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)
(a) Capacitance between the two plates is given by the relation,
\(
\begin{aligned}
& \mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \ldots \ldots\left[\because \mathrm{~A}=\pi \mathrm{r}^2=3.14 \times(0.12)^2\right] \\
& =\frac{8.854 \times 10^{-12} \times 3.14 \times(0.12)^2}{0.05} \\
& =8.01 \times 10^{-12} \mathrm{~F} \\
& =8.01 \mathrm{pF}
\end{aligned}
\)
Charge on each plate, \(\mathrm{q}=\mathrm{CV} \Rightarrow \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}\)
\(
\therefore \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{1}{\mathrm{C}} \frac{\mathrm{dq}}{\mathrm{dt}}=\frac{1}{\mathrm{C}} \mathrm{I} \ldots\left(\because \frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{I}\right)
\)
Rate of charge of the potential difference \(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{0.15}{8.01 \times 10^{-12}}\)
\(
=1.875 \times 10^9 \mathrm{~V} \mathrm{~s}^{-1}
\)

(b) Displacement current across the plates,
\(
\mathrm{I}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{~d} \phi_{\mathrm{E}}}{\mathrm{dt}}
\)
Where \(\phi_{\mathrm{E}}\) is the electric flux passing through a closed loop between the plates.
\(\because\) The electric field E between the plates \(=\frac{\mathrm{q}}{\varepsilon_0 \mathrm{~A}}\)
\(\therefore\) If the area of the loop is \(A\) then,
\(
\begin{aligned}
& \phi_{\mathrm{E}}=\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\oint \mathrm{EdA} \ldots \cdot[\because \overrightarrow{\mathrm{E}} \perp \mathrm{dA}] \\
& \Rightarrow \phi_{\mathrm{E}}=\mathrm{EA}=\frac{\mathrm{q}}{\varepsilon_0} \Rightarrow \frac{\mathrm{~d} \phi_{\mathrm{E}}}{\mathrm{dt}}=\frac{1}{\varepsilon_0} \cdot \frac{\mathrm{dq}}{\mathrm{dt}} \\
& \therefore \mathrm{I}_{\mathrm{d}}=\varepsilon_0 \frac{1}{\varepsilon_0} \frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{I}
\end{aligned}
\)
\(\Rightarrow\) Displacement current, \(\mathrm{I}_{\mathrm{d}}=0.15 \mathrm{~A}\)
(c) Yes, Kirchhoff’s first law is very much applicable to each plate of capacitor as \(\mathrm{I}_{\mathrm{d}}=\mathrm{I}\).
So current is continuous and constant across each plate.

Q8.2: A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius \(R=6.0 \mathrm{~cm}\) has a capacitance \(C=100 \mathrm{pF}\). The capacitor is connected to a 230 V ac supply with a (angular) frequency of \(300 \mathrm{rad} \mathrm{s}^{-1}\).
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of \(\mathbf{B}\) at a point 3.0 cm from the axis between the plates.

Answer: Radius of each circular plate, \(\mathrm{R}=6.0 \mathrm{~cm}=0.06 \mathrm{~m}\)
Capacitance of a parallel plate capacitor, \(\mathrm{C}=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}\)
Supply voltage, \(\mathrm{V}=230 \mathrm{~V}\)
Angular frequency, \(\omega=300 \mathrm{rad} \mathrm{s}^{-1}\)
(a) The rms value of conduction current, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}\)
Where,
\(
\begin{aligned}
& \mathrm{X}_{\mathrm{C}}=\text { Capacitive reactance } \\
& =\frac{1}{\omega \mathrm{C}} \\
& \therefore \mathrm{I}=\mathrm{V} \times \omega \mathrm{C} \\
& =230 \times 300 \times 100 \times 10^{-12} \\
& =6.9 \times 10^{-6} \mathrm{~A} \\
& =6.9 \mu \mathrm{~A}
\end{aligned}
\)
Hence, the rms value of the conduction current is \(6.9 \mu \mathrm{~A}\).

(b) Yes, conduction current is equal to displacement current.

(c) Magnetic field is given as:
\(
\mathrm{B}=\frac{\mu_0 \mathrm{r}}{2 \pi \mathrm{R}^2} \mathrm{I}_0
\)
Where,
\(
\begin{aligned}
& \mu_0=\text { Free space permeability }=4 \pi \times 10^{-7} \mathrm{NA}^{-2} \\
& \mathrm{I}_0=\text { Maximum value of current }=\sqrt{2} \mathrm{I}
\end{aligned}
\)
\(\mathrm{r}=\) Distance between the plates from the axis \(=3.0 \mathrm{~cm}=0.03 \mathrm{~m}\)
\(
\begin{aligned}
& \therefore \mathrm{B}=\frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \pi \times(0.06)^2} \\
& =1.63 \times 10^{-11} \mathrm{~T}
\end{aligned}
\)
Hence, the magnetic field at that point is \(1.63 \times 10^{-11} \mathrm{~T}\).

Q8.3: What physical quantity is the same for X -rays of wavelength \(10^{-10} \mathrm{~m}\), red light of wavelength \(6800 Å\) and radiowaves of wavelength 500 m ?

Answer: The speed of light \(\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)\) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

Q8.4: A plane electromagnetic wave travels in vacuum along \(z\)-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz , what is its wavelength?

Answer: The electromagnetic wave travels in a vacuum along the \(z\)-direction. The electric field \((E)\) and the magnetic field \((H)\) are in the \(x-y\) plane. They are mutually perpendicular.
Frequency of the wave, \(\nu=30 \mathrm{MHz}=30 \times 10^6 \mathrm{~s}^{-1}\)
Speed of light in a vacuum, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Wavelength of a wave is given as:
\(
\begin{aligned}
\lambda & =\frac{c}{\nu} \\
& =\frac{3 \times 10^8}{30 \times 10^6}=10 \mathrm{~m}
\end{aligned}
\)

Q8.5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Answer: A radio can tune to minimum frequency, \(\nu_1=7.5 \mathrm{MHz}=7.5 \times 10^6 \mathrm{~Hz}\) Maximum frequency, \(\nu_2=12 \mathrm{MHz}=12 \times 10^6 \mathrm{~Hz}\)
Speed of light, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Corresponding wavelength for \(\nu_1\) can be calculated as:
\(
\begin{aligned}
\lambda_1 & =\frac{c}{\nu_1} \\
& =\frac{3 \times 10^8}{7.5 \times 10^6}=40 \mathrm{~m}
\end{aligned}
\)
Corresponding wavelength for \(\nu_2\) can be calculated as:
\(
\begin{aligned}
\lambda_2 & =\frac{c}{\nu_2} \\
& =\frac{3 \times 10^8}{12 \times 10^6}=25 \mathrm{~m}
\end{aligned}
\)
Thus, the wavelength band of the radio is 40 m to 25 m.

Q8.6: A charged particle oscillates about its mean equilibrium position with a frequency of \(10^9 \mathrm{~Hz}\). What is the frequency of the electromagnetic waves produced by the oscillator?

Answer: The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., \(10^9 \mathrm{~Hz}\).

Q8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is \(B_0=510 \mathrm{~nT}\). What is the amplitude of the electric field part of the wave?

Answer: Amplitude of magnetic field of an electromagnetic wave in a vacuum,
\(
B_0=510 \mathrm{nT}=510 \times 10^{-9} \mathrm{~T}
\)
Speed of light in a vacuum, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Amplitude of electric field of the electromagnetic wave is given by the relation,
\(
\begin{aligned}
& E=c B_0 \\
& =3 \times 10^8 \times 510 \times 10^{-9}=153 \mathrm{~N} / \mathrm{C}
\end{aligned}
\)
Therefore, the electric field part of the wave is 153 N/C.

Q8.8: Suppose that the electric field amplitude of an electromagnetic wave is \(E_0=120 \mathrm{~N} / \mathrm{C}\) and that its frequency is \(\nu=50.0 \mathrm{MHz}\). (a) Determine, \(B_0, \omega, k\), and \(\lambda\). (b) Find expressions for \(\mathbf{E}\) and \(\mathbf{B}\).

Answer: Electric field amplitude, \(E_0=120 \mathrm{~N} / \mathrm{C}\)
Frequency of source, \(\nu=50.0 \mathrm{MHz}=50 \times 10^6 \mathrm{~Hz}\)
Speed of light, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
(a) Magnitude of magnetic field strength is given as:
\(
\begin{aligned}
B_0 & =\frac{E_0}{c} \\
& =\frac{120}{3 \times 10^8} \\
& =4 \times 10^{-7} \mathrm{~T}=400 \mathrm{nT}
\end{aligned}
\)
Angular frequency of source is given as:
\(
\begin{aligned}
& \omega=2 \pi \nu \\
& =2 \pi \times 50 \times 10^6 \\
& =3.14 \times 10^8 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Propagation constant is given as:
\(
\begin{aligned}
&\begin{aligned}
k & =\frac{\omega}{c} \\
& =\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}
\end{aligned}\\
&\text { Wavelength of wave is given as: }\\
&\begin{aligned}
\lambda & =\frac{c}{\nu} \\
& =\frac{3 \times 10^8}{50 \times 10^6}=6.0 \mathrm{~m}
\end{aligned}
\end{aligned}
\)

(b) Suppose the wave is propagating in the positive \(x\) direction. Then, the electric field vector will be in the positive \(y\) direction and the magnetic field vector will be in the positive \(z\) direction. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as:
\(
\begin{aligned}
\vec{E} & =E_0 \sin (k x-\omega t) \hat{j} \\
& =120 \sin \left[1.05 x-3.14 \times 10^8 t\right] \hat{j}
\end{aligned}
\)
\(
\begin{aligned}
&\text { And, magnetic field vector is given as: }\\
&\begin{aligned}
& \vec{B}=B_0 \sin (k x-\omega t) \hat{k} \\
& \vec{B}=\left(4 \times 10^{-7}\right) \sin \left[1.05 x-3.14 \times 10^8 t\right] \hat{k}
\end{aligned}
\end{aligned}
\)

Q8.9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula \(E=h \nu\) (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Answer: Energy of a photon is given as:
\(
E=h \nu=\frac{h c}{\lambda}
\)
Where,
\(h=\) Planck’s constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
\(c=\) Speed of light \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
\(\lambda=\) Wavelength of radiation
\(
\begin{aligned}
\therefore E & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{\lambda}=\frac{19.8 \times 10^{-26}}{\lambda} \mathrm{~J} \\
& =\frac{19.8 \times 10^{-20}}{\lambda \times 1.6 \times 10^{-19}}=\frac{12.375 \times 10^{-7}}{\lambda} \mathrm{eV}
\end{aligned}
\)
The given table lists the photon energies for different parts of an electromagnetic spectrum for different \(\lambda\).
\(
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline \begin{array}{l}
\lambda \\
(\mathrm{m})
\end{array} & 10^3 & 1 & 10^{-3} & 10^{-6} & 10^{-8} & 10^{-10} & 10^{-12} \\
\hline E & 12.375 \times & 12.375 \times & 12.375 \times & 12.375 \times & 12.375 \times & 12.375 \times \\
(\mathrm{eV}) & 10^{-10} & 10^{-7} & 10^{-4} & 10^{-1} & 10^1 & 10^3 & 10^5 \\
\hline
\end{array}
\)
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

Q8.10: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of \(2.0 \times 10^{10} \mathrm{~Hz}\) and amplitude \(48 \mathrm{~V} \mathrm{~m}^{-1}\).
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the \(\mathbf{E}\) field equals the average energy density of the \(\mathbf{B}\) field. \(\left[c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right.\).]

Answer: Frequency of the electromagnetic wave, \(\nu=2.0 \times 10^{10} \mathrm{~Hz}\)
Electric field amplitude, \(E_0=48 \mathrm{~V} \mathrm{~m}^{-1}\)
Speed of light, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
(a) Wavelength of a wave is given as:
\(
\begin{aligned}
\lambda & =\frac{c}{\nu} \\
& =\frac{3 \times 10^8}{2 \times 10^{10}}=0.015 \mathrm{~m}
\end{aligned}
\)

(b) Magnetic field strength is given as:
\(
\begin{aligned}
B_0 & =\frac{E_0}{c} \\
& =\frac{48}{3 \times 10^8}=1.6 \times 10^{-7} \mathrm{~T}
\end{aligned}
\)

(c) Energy density of the electric field is given as:
\(
U_E=\frac{1}{2} \epsilon_0 E^2
\)
And, energy density of the magnetic field is given as:
\(
U_B=\frac{1}{2 \mu_0} B^2
\)
Where,
\(\epsilon_0=\) Permittivity of free space
\(\mu_0=\) Permeability of free space
We have the relation connecting \(E\) and \(B\) as:
\(
E=c B \dots(1)
\)
\(
c=\frac{1}{\sqrt{\epsilon_0 \mu_0}} \dots(2)
\)
Putting equation (2) in equation (1), we get
\(
E=\frac{1}{\sqrt{\epsilon_0 \mu_0}} B
\)
Squaring both sides, we get
\(
\begin{aligned}
& E^2=\frac{1}{\epsilon_0 \mu_0} B^2 \\
& \epsilon_0 E^2=\frac{B^2}{\mu_0} \\
& \frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \frac{B^2}{\mu_0} \\
& \Rightarrow U_E=U_B
\end{aligned}
\)

EXEMPLAR SECTION

VSA

Q8.14: Why is the orientation of the portable radio with respect to broadcasting station important?

Answer: The electromagnetic waves are plane polarised, so the receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave. So the receiving antenna should be parallel to electric/magnetic part of the wave. That is why the orientation of the portable radio with respect to broadcasting station is important.

Q8.15: Why does microwave oven heats up a food item containing water molecules most efficiently?

Answer: The microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules.

Q8.16: The charge on a parallel plate capacitor varies as \(q=q_0 \cos 2 \pi \nu t\). The plates are very large and close together (area \(= A\), separation \(=d\) ). Neglecting the edge effects, find the displacement current through the capacitor?

Answer: The displacement current through the capacitor is given by,
\(
I_d=I_c=\frac{d q}{d t} \dots(i)
\)
Herę we are given, \(q=q_0 \cos 2 \pi \nu t\)
Putting this value in Eq (i), we get
\(
\begin{aligned}
& I_d=I_c=-q_0 \sin 2 \pi \nu t \times 2 \pi \nu \\
& I_d=I_c=-2 \pi v q_0 \sin 2 \pi \nu t
\end{aligned}
\)

Q8.17: A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?

Answer: Capacitive reaction \(X_C=\frac{1}{2 \pi f C}\)
Hence, \(X_C \propto \frac{1}{f}\)
As frequency decreases, \(X_C\) increases and the conduction current is inversely proportional to \(X_C\left(\because I \propto \frac{1}{X_C}\right)\).
It means the displacement current decreases as the conduction current is equal to the displacement current.

Q8.18: The magnetic field of a beam emerging from a filter facing a floodlight is given by
\(
B_o=12 \times 10^{-8} \sin \left(1.20 \times 10^7 z-3.60 \times 10^{15} t\right) \mathrm{T}
\)
What is the average intensity of the beam?

Answer: The standard equation of magretic field can be expressed as \(B=B_0 \sin \omega t\). We are given equation
\(
B=12 \times 10^{-8} \sin \left(120 \times 10^7 z-3.60 \times 10^{1 / 5} t\right) \mathrm{T}
\)
On comparing this equation with standard equation, we get
\(
B_0=12 \times 10^{-8} \mathrm{~T}_1
\)
The average intensity of the beam
\(
\begin{aligned}
I_{\mathrm{av}} & =\frac{1}{2} \frac{B_0^2}{\mu_0} \cdot c=\frac{1}{2} \times \frac{\left(12 \times 10^{-8}\right)^2 \times 3 \times 10^8}{4 \pi \times 10^{-7}} \\
& =1.71 \mathrm{~W} / \mathrm{m}^2
\end{aligned}
\)

Q8.19: Poynting vectors \(\mathbf{S}\) 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 \(\mathbf{S}=\frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}\). Show the nature of \(S\) vs \(t\) graph.

Answer: Consider and electromagnetic waves, let \(\mathbf{E}\) be varying along \(y\)-axis, \(\mathbf{B}\) is along \(z\)-axis and propagation of wave be along \(x\)-axis. Then \(\mathbf{E} \times \mathbf{B}\) will tell the direction of propagation of energy flow in electromegnetic wave, along \(x\)-axis.
Let \(\mathbf{E}=E_0 \sin (\omega t-k x) \hat{\mathbf{j}}\)
\(
\mathbf{B}=B_0 \sin (\omega t-k x) \hat{\mathbf{k}}
\)
\(
\mathbf{S}=\frac{1}{\mu_0}(\mathbf{E} \times \mathbf{B})=\frac{1}{\mu_0} E_0 B_0 \sin ^2(\omega t-k x)[\hat{\mathbf{j}} \times \hat{\mathbf{k}}]
\)
\(
=\frac{E_0 B_0}{\mu_0} \sin ^2(\omega t-k x) \hat{\mathbf{i}}
\)
The variation of \(|\mathbf{S}|\) with time \(t\) will be as given in the figure below.

Q8.20: 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: The properties of an electromagnetic wave is same as other waves. Like other wave an electromagnetic wave also carries energy and momentum. Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped 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.

SA

Q8.21: Show that the magnetic field \(B\) at a point in between the plates of a parallel-plate capacitor during charging is \(\frac{\varepsilon_0 \mu_r}{2} \frac{\mathrm{~d} E}{\mathrm{~d} t}\) (symbols having usual meaning).

Answer: Let us assume \(I_d\) be the displacement current in the region between two plates of parallel plate capacitor, in the figure.

The magnetic field at a point between two plates of capacitor at a perpendicular distance \(r\) from the axis of plates is given by
\(
B=\frac{\mu_0 2 I_d}{4 \pi r}=\frac{\mu_0}{2 \pi r} I_d=\frac{\mu_0}{2 \pi r} \times \varepsilon_r \frac{d \phi_E}{d t} \quad\left[\because I_d=\frac{E_0 d \phi_E}{d t}\right]
\)
\(
\Rightarrow \quad B=\frac{\mu_0 \varepsilon_r}{2 \pi r} \frac{d}{d t}\left(E \pi r^2\right)=\frac{\mu_0 \varepsilon_r}{2 \pi r} \pi r^2 \frac{d F}{d t}
\)
\(
\Rightarrow \quad B=\frac{\mu_0 \varepsilon_r}{2} \frac{d E}{d t} \quad\left[\because \phi_E=E \pi r^2\right]
\)

Q8.22: Electromagnetic waves with wavelength
(i) \(\lambda_1\) is used in satellite communication.
(ii) \(\lambda_2\) is used to kill germs in water purifies.
(iii) \(\lambda_3\) is used to detect leakage of oil in underground pipelines.
(iv) \(\lambda_4\) 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) In satellite communications, microwave is widely used. Hence \(\lambda_1\), is the wavelength of microwave.
(ii) In water purifier, ultraviolet rays are used to kill germs. So, \(\lambda_2\) is the wavelength of UV rays.
(iii) X -rays are used to detect leakage of oil in underground pipelines. So, \(\lambda_3\) is the wavelength of X -rays.
(iv) Infrared rays are used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.

(b) Wavelength of X -rays < wavelength of UV < wavelength of infrared < wavelength of microwave.
\(
=>\lambda_3<\lambda_2<\lambda_4<\lambda_1
\)

(c)
\(
\begin{array}{|l|l|}
\hline \text { Radiation } & \text { Uses } \\
\hline \gamma \text {-rays } & \begin{array}{l}
\text { Gives informations on nuclear structure, medical } \\
\text { treatment etc. }
\end{array} \\
\hline X \text {-rays } & \begin{array}{l}
\text { Medical diagnosis and treatment study of crystal } \\
\text { structure, industrial radiograph. }
\end{array} \\
\hline U V \text {-rays } & \begin{array}{l}
\text { Preserve food, sterilizing the surgical instruments, } \\
\text { detecting the invisible writings, finger prints etc. }
\end{array} \\
\hline \text { Visible light } & \text { To see objects } \\
\hline \text { Infrared rays } & \begin{array}{l}
\text { To treat, muscular strain for taking photography } \\
\text { during the fog, haze etc. }
\end{array} \\
\hline \begin{array}{l}
\text { Micro wave } \\
\text { and radio wave }
\end{array} & \text { In radar and telecommunication. } \\
\hline
\end{array}
\)

Q8.23: Show that average value of radiant flux density ‘S’ over a single period ‘ \(T\) ‘ is given by \(\mathrm{S}=\frac{1}{2 \mathrm{c} \mu_0} E_0^2\).

Answer: Radiant flux density \(S=\frac{1}{\mu_0}(\mathbf{E} \times \mathbf{B})=c^2 \varepsilon_0(\mathbf{E} \times \mathbf{B}) \quad \quad \left[\because c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right]\)
Suppose electromagnetic waves be propagating along \(x\)-axis. The electric field vector of electromagnetic wave be along \(y\)-axis and magnetic field vector be along \(z\)-axis. Therefore,
\(
\begin{aligned}
\mathbf{E}_0 & =\mathbf{E}_0 \cos (k x-\omega t) \\
\text { and } \mathbf{B} & =\mathbf{B}_0 \cos (k x-\omega t) \\
\mathbf{E} \times \mathbf{B} & =\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \\
\mathbf{S} & =\mathbf{c}^2 \varepsilon_0(\mathbf{E} \times \mathbf{B}) \\
& =c^2 \varepsilon_0\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t)
\end{aligned}
\)
Average value of the magnitude of radiant flux density over complete cycle is
\(
S_{\mathrm{av}}=c^2 \varepsilon_0\left|\mathbf{E}_0 \times \mathbf{B}_0\right| \frac{1}{T} \int_0^T \cos ^2(k x-\omega t) d t
\)
\(
=c^2 \varepsilon_0 E_0 B_0 \times \frac{1}{T} \times \frac{T}{2} \quad\left[\because \int_0^T \cos ^2(k x-\omega t) d t=\frac{T}{2}\right]
\)
\(
\Rightarrow \quad S_{\mathrm{av}}=\frac{c^2}{2} \varepsilon_0 E_0\left(\frac{E_0}{c}\right) \quad\left[\mathrm{As}, \mathrm{c}=\frac{E_0}{B_0}\right]
\)
\(
=\frac{c}{2} \varepsilon_0 E_0^2=\frac{c}{2} \times \frac{1}{c^2 \mu_0} E_0^2 \quad\left[c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \text { or } \varepsilon_0=\frac{1}{c^2 \mu_0}\right]
\)
\(\Rightarrow \quad S_{\mathrm{av}}=\frac{E_0^2}{2 \mu_0 c} \quad\). Hence proved.

Q8.24: You are given a \(2 \mu \mathrm{~F}\) parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?

Answer: The capacitance of capacitor \(C=2 \mu \mathrm{~F}\),
Displacement current \(I_d=1 \mathrm{~mA}\)
Charge in capacitor, \(q=C V\).
\(
\begin{array}{ll}
& I_d d t=C d V \quad[\because q=i t] \\
\text { or } \quad & I_d=C \frac{d V}{d t} \\
& 1 \times 10^{-3}=2 \times 10^{-6} \times \frac{d V}{d t} \\
& \frac{d V}{d t}=\frac{1}{2} \times 10^3=500 \mathrm{~V} / \mathrm{s}
\end{array}
\)
Hence by applying a varying potential difference of \(500 \mathrm{~V} / \mathrm{s}\), we would produce a displacement current of desired value.

Q8.25: 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 a surface exposed to electromagnetic radiation. The radiation is falling normally on the surface. Further, intensity of radiation is I and area of surface exposed to radiation is A.
\(E=\) Energy received by surface per second \(=I . A\)
\(N=\) Number of photons received by surface per second
\(
N=\frac{E}{E_{\text {photon }}}=\frac{E \lambda}{h c}=\frac{I A \lambda}{h c}
\)
Let the surface is perfectly absorbing, \(\Delta P_{\text {onephoton }}=\frac{h}{\lambda}\).
\(
\Rightarrow \quad F=N \times \Delta P_{\text {one photon }}=\frac{I A}{c}
\)
Also, \(\quad\) Pressure \(P=\frac{F}{A}=\frac{I}{c}\)

Q8.26: 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: As the distance is doubled, the area of spherical region \(\left(4 \pi r^2\right)\) will become four times, so the intensity becomes one fourth the initial value \(\left(\because I \propto \frac{1}{r^2}\right)\) but in case of laser it does not spread, so its intensity remain same.
Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following
(i) Unidirection
(ii) Monochromatic
(iii) Coherent light
(iv) Highly collimated
These characteristic are missing in the case of light from the bulb.

Q8.27: Even though an electric field \(\mathbf{E}\) exerts a force \(q \mathbf{E}\) 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 oscillation field. It exerts electric force on a charged particle, but this electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure though it transfer the energy. In fact, radiation pressure appears as a result of the action of the magnetic field of the wave on the electric currents induced by the electric field of the same wave.

LA

Q8.28: An infinitely long thin wire carrying a uniform linear static charge density \(\lambda\) is placed along the \(z\)-axis (Fig. 8.1). The wire is set into motion along its length with a uniform velocity \(\mathbf{v}=v \hat{\mathbf{k}}_{\mathrm{z}}\). Calculate the poynting vector \(\mathbf{S}=\frac{1}{\mu_{\mathrm{o}}}(\mathbf{E} \times \mathbf{B})\).

Answer: The electric field due to infinitely long thin wire \(\vec{E}=\frac{\lambda \hat{e}_s}{2 \pi \varepsilon_0 a} \hat{j}\)
Magnetic field due to the wire, \(\vec{B}=\frac{\mu_0 i}{2 \pi a} \hat{i}\)
Equivalent current flowing through the wire, \(i=\lambda v\)
Hence \(\vec{B}=\frac{\mu_0 \lambda v}{2 \pi a} \hat{i}\)
\(
\therefore \quad \vec{S}=\frac{1}{\mu_0}[\vec{E} \times \vec{B}]=\frac{1}{\mu_0}\left[\frac{\lambda}{2 \pi \varepsilon_0 a} \hat{j} \times \frac{\mu_0 \lambda v}{2 \pi a} \hat{i}\right]
\)

Q8.29: Sea water at frequency \(v=4 \times 10^8 \mathrm{~Hz}\) has permittivity \(\varepsilon \approx 80 \varepsilon_0\), permeability \(\mu \approx \mu_0\) and resistivity \(\rho=0.25 \Omega-\mathrm{m}\). Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source \(V(t)=V_o \sin (2 \pi \nu t)\). What fraction of the conduction current density is the displacement current density?

Answer: Let the separation between the plates of capacitor immersed in sea water plates is \(d\) and applied voltage across the plates is \(V(t)=V_0 \sin (2 \pi \nu t)\).
Thus, electric field, \(E=\frac{V(t)}{d}\)
\(
\Rightarrow \quad E=\frac{V_0}{d} \sin (2 \pi \nu t)
\)
Now using Ohm’s law, the conduction current density \(J^c=\frac{E}{\rho}=\frac{1}{\rho} \frac{V_0}{d} \sin (2 \pi \nu t)\)
\(
\Rightarrow \quad J^c=\frac{V_0}{\rho d} \sin (2 \pi \nu t)=J_0^c \sin 2 \pi \nu t
\)
Here, \(\quad J_0^c=\frac{V_0}{\rho d}\)
The displacement current density is given as
\(
J^d=\varepsilon \frac{d E}{d t}=\varepsilon \frac{d}{d t}\left[\frac{V_0}{d} \sin (2 \pi \nu t)\right]
\)
\(
= \frac{\varepsilon 2 \pi \nu V_0}{d} \cos (2 \pi \nu t)
\)
\(
\begin{aligned}
&\Rightarrow \quad J^d=J_0^d \cos (2 \pi \nu t)\\
&\text { where, } J_0^d=\frac{2 \pi \nu \varepsilon V_0}{d}\\
&\begin{aligned}
\Rightarrow \quad \frac{J_0^d}{J_0^c} & =\frac{\frac{2 \pi \nu \varepsilon V_0}{d}}{\frac{V_0}{\rho d}}=2 \pi \nu \varepsilon \rho \\
& =2 \pi \times 80 \varepsilon_0 \nu \times 0.25=4 \pi \varepsilon_0 \nu \times 10 \\
\Rightarrow \quad \frac{J_0^d}{J_0^c} & =\frac{10 \times 4 \times 10^8}{9 \times 10^9}=\frac{4}{9}
\end{aligned}
\end{aligned}
\)

Q8.30: A long straight cable of length \(l\) is placed symmetrically along \(z\)-axis and has radius \(a(\ll l)\). The cable consists of a thin wire and a co-axial conducting tube. An alternating current \(I(t)=I_o \sin (2 \pi \nu t)\) 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 \(\mathbf{E}(s, t)=\mu_0 I_0 v \cos (2 \pi \nu t) \operatorname{In}\left(\frac{s}{a}\right) \hat{\mathbf{k}}\).
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the crosssection of the cable to find the total displacement current \(I^4\).
(iii) Compare the conduction current \(I_0\) with the dispalcement current \(I_{\mathrm{0}}^{\mathrm{d}}\).

Answer: (i) Given, the induced electric field at a distance \(r\) from the wire inside the cable is
\(
\mathrm{E}(s, t)=\mu_0 I_0 v \cos (2 \pi \nu t) \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}}
\)
Now, displacement current density,
\(
J_d=\varepsilon_0 \frac{d \mathbf{E}}{d t}=\varepsilon_0 \frac{d}{d t}\left[\mu_0 I_0 v \cos (2 \pi \nu t) \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}}\right]
\)
\(
\begin{aligned}
& =\varepsilon_0 \mu_0 I_0 v \frac{d}{d t}[\cos 2 \pi \nu t] \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}} \\
& =\frac{1}{c^2} I_0 v^2 2 \pi[-\sin 2 \pi \nu t] \ln \left(\frac{s}{a}\right) \hat{\mathbf{k}} \\
& =\frac{v^2}{c^2} 2 \pi I_0 \sin 2 \pi \nu t \ln \left(\frac{a}{s}\right) \hat{\mathbf{k}} \left[\because l_4 \frac{s}{a}=-l_4 \frac{a}{s}\right]\\
& =\frac{1}{\lambda^2} 2 \pi I_0 \ln \left(\frac{a}{s}\right) \sin 2 \pi \nu t \hat{\mathbf{k}} \\
& =\frac{2 \pi I_0}{\lambda_2} \ln \frac{a}{s} \sin 2 \pi \nu t \hat{\mathbf{k}}
\end{aligned}
\)

(ii) Total displacement current, \(I^d=\int J_d 2 \pi s d s\)

\(
I^d=\int_0^a\left(\frac{2 \pi I_0}{\lambda^2} \ln \frac{a}{s} \sin 2 \pi \nu t\right) 2 \pi s d s
\)
\(
\begin{aligned}
& =\int_0^a\left[\frac{2 \pi}{\lambda^2} I_0 \int_{s=0}^a \ln \left(\frac{a}{s}\right) s d s \sin 2 \pi \nu t\right] \times 2 \pi \\
& =\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \int_0^a \ln \left(\frac{a}{s}\right) \frac{1}{2} d\left(s^2\right) \cdot \sin 2 \pi \nu t \\
& =\left(\frac{a}{2}\right)^2\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi \nu t \int_0^a \ln \left(\frac{a}{s}\right) \cdot d\left(\frac{s}{a}\right)^2 \\
& =\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi \nu t \int_0^a \ln \left(\frac{a}{s}\right)^2 \cdot d\left(\frac{s}{a}\right)^2
\end{aligned}
\)
\(
=\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi \nu t \times(1) \quad\left[\because \int_{s=0}^a \ln \left(\frac{s}{a}\right)^2 d\left(\frac{s}{a}\right)^2=1\right]
\)
\(
\begin{aligned}
& \therefore \quad I^d=\frac{a^2}{4}\left(\frac{2 \pi}{\lambda}\right)^2 I_0 \sin 2 \pi \nu t \\
& \Rightarrow \quad I^d=\left(\frac{\pi a}{\lambda}\right)^2 I_0 \sin 2 \pi \nu t
\end{aligned}
\)

(iii) The displacement current,
\(
I_d=\left(\frac{\pi a}{\lambda}\right)^2 I_0 \sin 2 \pi \nu t=I_0^d \sin 2 \pi \nu t
\)
\(
\begin{aligned}
&\text { Here, } I_0^d=\left(\frac{a \pi}{\lambda}\right)^2 I_0\\
&\Rightarrow \quad \frac{I_0^d}{I_0}=\left(\frac{a \pi}{\lambda}\right)^2
\end{aligned}
\)

Q8.31: A plane EM wave travelling in vacuum along \(z\) direction is given by
\(
\mathbf{E}=E_0 \sin (k z-\omega t) \hat{\mathbf{i}} \text { and } \mathbf{B}=B_0 \sin (k z-\omega t) \hat{\mathbf{j}} .
\)
(i) Evaluate \(\oint \mathbf{E . d l}\) over the rectangular loop 1234 shown in Fig 8.2.
(ii) Evaluate \(\int \mathbf{B} . \mathbf{d s}\) over the surface bounded by loop 1234.
(iii) Use equation \(\oint \mathbf{E} . \mathbf{d} \mathbf{l}=\frac{-\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{d} t}\) to prove \(\frac{E_0}{B_0}=\mathrm{c}\).
(iv) By using similar process and the equation \(\oint \mathbf{B} . \mathbf{d} \mathbf{l}=\mu_0 I+\varepsilon_0 \frac{\mathrm{~d} \phi_{\mathrm{E}}}{\mathrm{d} t}\), prove that \(\mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)

Answer: (i) Let electromagnetic wave is propagating along \(z\)-axis, in this case electric field vector \((\vec{E})\) be along \(x\)-axis and magnetic field vector \((\vec{B})\) along \(y\)-axis, i.e., \(\vec{E}=E_0 \hat{i}\) and \(\vec{B}=B_0 \hat{j}\).

The line integral of \(\vec{E}\) over the closed rectangular path 1234 in \(x-z\) plane of the figure is
\(
\oint \vec{E} \cdot d \vec{l}=\int_1^2 \vec{E} \cdot d \vec{l}+\int_2^3 \vec{E} \cdot d \vec{l}+\int_3^4 \vec{E} \cdot d \vec{l}+\int_4^1 \vec{E} \cdot d \vec{l}
\)
\(
=\int_1^2 E \cdot d l \cos 90^{\circ}+\int_2^3 E \cdot d l \cos 0^{\circ}+\int_3^4 E \cdot d l \cos 90^{\circ}+\int_4^1 E \cdot d l \cos 180^{\circ}
\)
\(
\oint \vec{E} \cdot d \vec{l}=E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \dots(i)
\)

(ii) Now let us evaluate \(\int \vec{B} \cdot d \vec{s}\), let us consider the rectangle 1234 to be made of strips of are \(d s=h d z\) each.

\(
\begin{aligned}
\int \vec{B} \cdot d \bar{s} & =\int B \cdot d s \cos 0=\int B \cdot d s \\
& =\int_{Z_1}^{Z_2} B_0 \sin (k z-\omega t) h d z \\
\int \vec{B} \cdot d \bar{s} & =\frac{-B_0 h}{k}\left[\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right] \dots(ii)
\end{aligned}
\)

(iii) We are given \(\oint E \cdot d l=\frac{-d \phi_B}{d t}=-\frac{d}{d t} \oint B \cdot d s\)
Substituting the values from Eqs. (i) and (ii), we get
\(
\begin{aligned}
& E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\
&=\frac{-d}{d t}\left[\frac{B_0 h}{k}\left\{\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right]\right. \\
&=\frac{B_0 h}{k} \omega\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\
& \Rightarrow \quad E_0=\frac{B_0 \omega}{k}=B_0 c \quad\left(\because \frac{\omega}{k}=c\right) \\
& \Rightarrow \quad \frac{E_0}{B_0}=c
\end{aligned}
\)

(iv) For evaluating \(\oint \vec{B} \cdot d \vec{l}\), let us consider a loop 1234 in \(y-z\) plane as shown in figure given below.

\(
\begin{aligned}
& \oint \vec{B} \cdot d \vec{l}=\int_1^2 \vec{B} \cdot d \vec{l}+\int_2^3 \vec{B} \cdot d \vec{l}+\int_3^4 \vec{B} \cdot \vec{d} l+\int_4^1 \vec{B} \cdot d \vec{l} \\
& =\int_1^2 B \cdot d l \cos 0^{\circ}+\int_2^3 B \cdot d l \cos 90^{\circ} \\
& +\int_3^4 B \cdot d l \cos 180^{\circ}+\int_4^1 B \cdot d l \cos 90^{\circ}
\end{aligned}
\)
\(
\oint \vec{B} \cdot d \vec{l}=B_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \dots(iii)
\)
Now to evaluate \(\phi_E=\int \vec{B} \cdot d \vec{s}\), let us consider the rectangle 1234 to be made of strips of area \(h d s\) each.

\(
\begin{aligned}
& \phi_E=\int \vec{E} \cdot d \vec{s}=\int E d s \cos 0=\int E d s \\
&=\int_{z_1}^z E_0 \sin \left(k z_1-\omega t\right) h d z \\
& \oint \vec{E} \cdot d \vec{s}=-\frac{E_0 h}{k}\left[\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right] \\
& \therefore \quad \frac{d \phi_E}{d t}=\frac{E_0 h \omega}{k}\left[\sin \left(k z_1-\omega t\right)-\sin \left(k z_2-\omega t\right)\right] \dots(iv)
\end{aligned}
\)
Let \(\oint B \cdot d l=\mu_0\left(I+\frac{\varepsilon_o d \phi_E}{d t}\right)\) where \(I=\) conduction current \(=0\) in vacuum
\(
\therefore \oint B \cdot d l=\mu_0 \varepsilon \frac{d \phi_E}{d t}
\)
Using relations obtained in Eqs. (iii) and (iv) and simplifying, we get
\(
\begin{aligned}
& B_0=E_0 \frac{\omega \mu_0 \varepsilon_0}{k} \\
& \Rightarrow \frac{E_0}{B_0} \frac{\omega}{k}=\frac{1}{\mu_0 \varepsilon_0}
\end{aligned}
\)
But \(\frac{E_0}{B_0}=c\) and \(\omega=c k \Rightarrow c \cdot c=\frac{1}{\mu_0 \varepsilon_0}\), therefore \(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)

Q8.32: A plane EM wave travelling along \(z\) direction is described by \(\mathbf{E}=E_0 \sin (k z-\omega t) \hat{\mathbf{i}}\) and \(\mathbf{B}=B_0 \sin (k z-\omega t) \hat{\mathbf{j}}\). Show that
(i) The average energy density of the wave is given by
\(
u_{\mathrm{av}}=\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \frac{\mathrm{~B}_0^2}{\mu_0} .
\)
(ii) The time averaged intensity of the wave is given by
\(
I_{\mathrm{av}}=\frac{1}{2} \mathrm{c} \varepsilon_0 E_0^2 .
\)

Answer: (i) The electromagnetic wave carry energy which is due to electric field vector and magnetic field vector. In electromagnetic wave, \(E\) and \(B\) vary from point to point and from moment to moment. Let \(E\) and \(B\) be their time averages.
The energy density due to electric field \(E\) is
\(
u_E=\frac{1}{2} \varepsilon_0 E^2
\)
The energy density due to magnetic field \(B\) is
\(
u_B=\frac{1}{2} \frac{B^2}{\mu_0}
\)
Total average energy density of electromagnetic wave
\(
u_{a v}=u_E+u_B=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \frac{B^2}{\mu_0}
\)
Let the EM wave be propagating along z-direction. The electric field vector and magnetic field vector be represented by
\(
\begin{aligned}
& E=E_0 \sin (k z-\omega t) \\
& B=B_0 \sin (k z-\omega t)
\end{aligned}
\)
The time average value of \(E^2\) over complete cycle \(=\frac{E_0^2}{2}\) and time average value of \(B^2\) over complete cycle \(=\frac{B_0^2}{2}\)
\(
\begin{aligned}
u_{\mathrm{av}} & =\frac{1}{2} \frac{\varepsilon_0 E_0^2}{2}+\frac{1}{2} \mu_0\left(\frac{B_0^2}{2}\right) \\
& =\frac{1}{4} \varepsilon_0 E_0^2+\frac{B_0^2}{4 \mu_0}
\end{aligned}
\)

(ii) We know that \(E_0=c B_0\) and \(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
\(
\therefore \quad \frac{1}{4} \frac{B_0^2}{\mu_0}=\frac{1}{4} \frac{E_0^2 / c^2}{\mu_0}=\frac{E_0^2}{4 \mu_0} \times \mu_0 \varepsilon_0=\frac{1}{4} \varepsilon_0 E_0^2
\)
\(
\therefore \quad U_B=U_E
\)
Hence,
\(
u_{\mathrm{av}}=\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \frac{B_0^2}{\mu_0}
\)
\(
\begin{aligned}
& =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \varepsilon_0 E_0^2 \\
& =\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \frac{B_0^2}{\mu_0}
\end{aligned}
\)
Time average intensity of the wave \(I_{\mathrm{av}}=U_{\mathrm{av}} C=\frac{1}{2} \varepsilon_0 E_0^2 \mathrm{C}=\frac{1}{2} \varepsilon_0 E_0^2\)