Electromagnetics Waves
Chapter 8
ELECTROMAGNETIC WAVES
MCQ I
8.1 One requires $11 eV$ 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.
Show Answer
Answer: (c)8.2 A linearly polarized electromagnetic wave given as $\mathbf{E}=E_o \hat{\mathbf{i}} \cos (k z-\omega t)$ 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
(a) $\mathbf{E} _r=-E_o \hat{\mathbf{i}} \cos (k z-\omega t)$.
(b) $\mathbf{E} _r=E_o \hat{\mathbf{i}} \cos (k z+\omega t)$. (c) $\mathbf{E} _r=-E_o \hat{\mathbf{i}} \cos (k z+\omega t)$.
(d) $\mathbf{E} _r=E_o \hat{\mathbf{i}} \sin (k z-\omega t)$.
Show Answer
Answer: (b)8.3 Light with an energy flux of $20 W / cm^{2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $30 cm^{2}$. the total momentum delivered (for complete absorption) during 30 minutes is
(a) $36 \times 10^{-5} kg m / s$.
(b) $36 \times 10^{-4} kg m / s$.
(c) $108 \times 10^{4} kg m / s$.
(d) $1.08 \times 10^{7} kg m / s$.
Show Answer
Answer: (b)8.4 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
(a) $\frac{E}{2}$.
(b) $2 E$.
(c) $\frac{E}{\sqrt{2}}$.
(d) $\sqrt{2} E$.
Show Answer
Answer: (d)8.5 If $\mathbf{E}$ and $\mathbf{B}$ represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along
(a) E.
(b) $\mathbf{B}$.
(c) $\mathbf{B} \times \mathbf{E}$.
(d) $\mathbf{E} \times \mathbf{B}$.
Show Answer
Answer: (d)8.6 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) $c^{2}: 1$
(c) $1: 1$
(d) $\sqrt{c}: 1$
Show Answer
Answer: (c)8.7 An EM wave radiates outwards from a dipole antenna, with $E_0$ as the amplitude of its electric field vector. The electric field $E_0$ which transports significant energy from the source falls off as
(a) $\frac{1}{r^{3}}$
(b) $\frac{1}{r^{2}}$
(c) $\frac{1}{r}$
(d) remains constant.
MCQ II
Show Answer
Answer: (c)8.8 An electromognetic wave travels in vacuum along $z$ direction: $\mathbf{E}=(E_1 \hat{\mathbf{i}}+E_2 \hat{\mathbf{j}}) \cos (k z-\omega t)$. Choose the correct options from the following:
(a) The associated magnetic field is given as $\mathbf{B}=\frac{1}{c}(E_1 \hat{\mathbf{i}}-E_2 \hat{\mathbf{j}}) \cos (k z-\omega t)$.
(b) The associated magnetic field is given as $\mathbf{B}=\frac{1}{c}(E_1 \hat{\mathbf{i}}-E_2 \hat{\mathbf{j}}) \cos (k z-\omega t)$.
(c) The given electromagnetic field is circularly polarised.
(d) The given electromagnetic wave is plane polarised.
Show Answer
Answer: (a), (d)8.9 An electromagnetic wave travelling along $z$-axis is given as: $\mathbf{E}=\mathbf{E} _0 \cos (k z-\omega t$ ). Choose the correct options from the following;
(a) The associated magnetic field is given as $\mathbf{B}=\frac{1}{c} \hat{\mathbf{k}} \times \mathbf{E}=\frac{1}{\omega}(\hat{\mathbf{k}} \times \mathbf{E})$.
(b) The electromagnetic field can be written in terms of the associated magnetic field as $\mathbf{E}=c(\mathbf{B} \times \hat{\mathbf{k}})$.
(c) $\hat{\mathbf{k}} \cdot \mathbf{E}=0, \hat{\mathbf{k}} \cdot \mathbf{B}=0$.
(d) $\hat{\mathbf{k}} \times \mathbf{E}=0, \hat{\mathbf{k}} \times \mathbf{B}=0$.
Show Answer
Answer: (a), (b), (c)8.10 A plane electromagnetic wave propagating along $x$ direction can have the following pairs of $\mathbf{E}$ and $\mathbf{B}$
(a) $E_x, B_y$.
(b) $E_y, B_z$.
(c) $B_x, E_y$.
(d) $E_z, B_y$.
Show Answer
Answer: (b), (d)8.11 A charged particle oscillates about its mean equilibrium position with a frequency of $10^{9} Hz$. The electromagnetic waves produced:
(a) will have frequency of $10^{9} Hz$.
(b) will have frequency of $2 \times 10^{9} Hz$.
(c) will have a wavelength of $0.3 m$.
(d) fall in the region of radiowaves.
Show Answer
Answer: (a), (c), (d)8.12 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.
Show Answer
Answer: (b), (d)8.13 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 $2 I / c$ if the wave is totally reflected.
(b) Radiation pressure is in the range $I / c<p<2 I / c$ for real surfaces.
VSA
Show Answer
Answer: (a), (c), (d)8.14 Why is the orientation of the portable radio with respect to broadcasting station important?
Show Answer
Answer: As electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric/magnetic part of the wave.8.15 Why does microwave oven heats up a food item containing water molecules most efficiently?
Show Answer
Answer: Frequency of the microwave matches the resonant frequency of water molecules.8.16 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?
Show Answer
Answer: $i_C=i_D=\frac{d q}{d t}=-2 \pi q_0 v \sin 2 \pi v t$8.17 A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?
Show Answer
Answer: On decreasing the frequency, reactance $X_c=\frac{1}{\omega C}$ will increase which will lead to decrease in conduction current. In this case $i_D=i_C$; hence displacement current will decrease.8.18 The magnetic field of a beam emerging from a filter facing a floodlight is given by
$B_0=12 \times 10^{-8} \sin (1.20 \times 10^{7} z-3.60 \times 10^{15} t) T$.
What is the average intensity of the beam?
Show Answer
Answer: $I _{a v}=\frac{1}{2} c \frac{B_0^{2}}{\mu_0}=\frac{1}{2} \times \frac{3 \times 10^{8} \times(12 \times 10^{-8})^{2}}{1.26 \times 10^{-6}}=1.71 W / m^{2}$.8.19 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.
Show Answer
Answer:
8.20 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.
SA
Show Answer
Answer: EM waves exert radiation pressure. Tails of comets are due to solar solar radiation.8.21 Show that the magnetic field $B$ at a point in between the plates of a parallel-plate capacitor during charging is $\frac{\varepsilon_0 \mu_r}{2} \frac{d E}{d t}$ (symbols having usual meaning).
Show Answer
Answer: $ \begin{gathered} B=\frac{\mu_0 2 I_D}{4 \pi r}=\frac{\mu_0 1}{4 \pi r}=\frac{\mu_0}{2 \pi r} \varepsilon_0 \frac{d \phi_E}{d t} \\ =\frac{\mu_0 \varepsilon_0}{2 \pi r} \frac{d}{d t}(E \pi r^{2}) \\ =\frac{\mu_0 \varepsilon_0 r}{2} \frac{d E}{d t} . \end{gathered} $
8.22 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.
Show Answer
Answer: (a) $\lambda_1 \to$ Microwave, $\lambda_2 \to UV$
$\lambda_3 \to X$ rays, $\quad \lambda_4 \to$ Infrared
(b) $\quad \lambda_3<\lambda_2<\lambda_4<\lambda_1$ (c) Microwave - Radar
UV - LASIK eye surgery
X-ray - Bone fracture identification (bone scanning) Infrared - Optical communication.
8.23 Show that average value of radiant flux density ’ $S$ ’ over a single period ’ $T$ is given by $S=\frac{1}{2 c \mu_0} E_0^{2}$.
Show Answer
Answer: $ \begin{aligned} S _{a v}=c^{2} \varepsilon_0|\mathbf{E} _0 \times \mathbf{B} _0| \frac{1}{T} \int_0^{T} \cos ^{2}(k x-\omega t) d t \text{ as } \mathbf{S}=c^{2} \varepsilon_0(\mathbf{E} \times \mathbf{B}) \\ \quad=c^{2} \varepsilon_0 E_0 B_0 \frac{1}{\not x} \times \frac{\not{T}}{2} \\ =c^{2} \varepsilon_0 E_0(\frac{E_0}{c}) \times \frac{1}{2}(\text{ as } c=\frac{E_0}{B_0}) \\ =\frac{1}{2} \varepsilon_0 E_0^{2} c \\ =\frac{E_0^{2}}{2 \mu_0 c} \text{ as }(c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}) \end{aligned} $8.24 You are given a $2 \mu F$ parallel plate capacitor. How would you establish an instantaneous displacement current of $1 mA$ in the space between its plates?
Show Answer
Answer: $ \quad i_D=C \frac{d V}{d t}$
$1 \times 10^{-3}=2 \times 10^{-6} \frac{d V}{d t}$
$\frac{d V}{d t}=\frac{1}{2} \times 10^{3}=5 \times 10 V / s$
Hence, applying a varying potential difference of $5 \times 10^{2} V / s$ would produce a displacement current of desired value.
8.25 Show that the radiation pressure exerted by an EM wave of intensity $I$ on a surface kept in vacuum is $I / c$.
Show Answer
Answer: Pressure
$ \begin{aligned} P & =\frac{\text{ Force }}{\text{ Area }}=\frac{F}{A}=\frac{1}{A} \frac{\Delta p}{\Delta t}(F=\frac{\Delta p}{\Delta t}=\text{ rate of change of momentum }) \\ & =\frac{1}{A} \cdot \frac{U}{\Delta t c}(\Delta p c=\Delta U=\text{ energy imparted by wave in time } \Delta t) \\ & =\frac{I}{c}(\text{ intensity } I=\frac{U}{A \Delta t}. \end{aligned} $
8.26 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?
LA
Show Answer
Answer: Intensity is reduced to one fourth. Tis is beacause the light beam spreads, as it propogates into a spherical region of area $4 \pi r^{2}$, but LASER does not spread and hence its intensity remains constant.8.27 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.
Show Answer
Answer: Electric field of an EM 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. Hence, electric field is not responsible for radiation pressure.8.28 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 $\mathbf{v}=v \hat{\mathbf{k}} _{z}$. Calculate the poynting vector $\mathbf{S}=\frac{1}{\mu_o}(\mathbf{E} \times \mathbf{B})$.
Fig. 8.1
Show Answer
Answer: $\mathbf{E}=\frac{\lambda \hat{\mathbf{e}} _{s}}{2 \pi \varepsilon_o a} \hat{\mathbf{j}}$
$\mathbf{B}=\frac{\mu_o i}{2 \pi a} \hat{\mathbf{i}}$
$=\frac{\mu_o \lambda v}{2 \pi a} \hat{i}$
$\mathbf{S}=\frac{1}{\mu_o}(\mathbf{E} \times \mathbf{B})=\frac{1}{\mu_o}(\frac{\lambda \hat{\mathbf{j}} _{s}}{2 \pi \varepsilon_o a} \hat{\mathbf{j}} \times \frac{\mu_o \lambda v}{2 \pi a} \hat{\mathbf{i}})$
$ =\frac{-\lambda^{2} v}{4 \pi^{2} \varepsilon_0 a^{2}} \hat{\mathbf{k}} $
8.29 Sea water at frequency $v=4 \times 10^{8} Hz$ has permittivity $\varepsilon \approx 80 \varepsilon_0$, permeability $\mu \approx \mu_o$ and resistivity $\rho=0.25 \Omega-m$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V(t)=V_o \sin (2 \pi v t)$. What fraction of the conduction current density is the displacement current density?
Show Answer
Answer: Let the distance between the plates be $d$. Then the electric field $E=\frac{V_o}{d} \sin (2 \pi v t)$. The conduction current density is given by the Ohm’s law $=E$.
$\Rightarrow J^{c}=\frac{1}{\rho} \frac{V_o}{d} \sin (2 \pi v t)=\frac{V_0}{\rho d} \sin (2 \pi v t)$
$ =J_o^{c} \sin 2 \pi v t $
where $J_0^{c}=\frac{V_0}{\rho d}$.
The displacement current density is given as
$ \begin{gathered} J^{d}=\varepsilon \frac{\partial E}{d t}=\varepsilon \frac{\partial}{dt}{\frac{V_o}{d} \sin (2 \pi v t)} \\ =\frac{\varepsilon 2 \pi v V_o}{d} \cos (2 \pi v t) \end{gathered} $
$=J_o^{d} \cos (2 \pi v t)$, where $J_0^{d}=\frac{2 \pi v \varepsilon V_0}{d}$
$ \begin{aligned} J_o^{d} / J_o^{c}= & \frac{2 \pi v \varepsilon V_o}{d} \cdot \frac{\rho d}{V_n}=2 \pi v \varepsilon \rho=2 \pi \times 80 \varepsilon_o v \times 0.25=4 \pi \varepsilon_o v \times 10 \\ & =\frac{10 v}{9 \times 10^{9}}=\frac{4}{9} \end{aligned} $
Exemplar Problems-Physics
8.30 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 $I(t)=I_o \sin (2 \pi v t)$ 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 $\mathbf{E}(s, t)=\mu_o I_o v \cos (2 \pi v t) In(\frac{s}{a}) \hat{\mathbf{k}}$.
(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 $I^{d}$.
(iii) Compare the conduction current $I_0$ with the dispalcement current $I_o^{d}$.
Show Answer
Answer: (i) Displacement curing density can be found from the relation be $\mathbf{J} _D=\varepsilon_0 \frac{d \mathbf{E}}{d t}$
$ \begin{aligned} & =\varepsilon_0 \mu_0 I_0 \frac{\partial}{\partial t} \cos (2 \pi v t) \cdot \ln (\frac{s}{a}) \hat{\mathbf{k}} \\ & =\frac{1}{c^{2}} I_0 2 \pi v^{2}(-\sin (2 \pi v t)) \ln (\frac{s}{a}) \hat{\mathbf{k}} \\ & =(\frac{v}{c})^{2} 2 \pi I_0 \sin (2 \pi v t) \ln (\frac{a}{s}) \hat{k} \\ & =\frac{2 \pi}{\lambda^{2}} I_0 \ln (\frac{a}{s}) \sin (2 \pi v t) \hat{\mathbf{k}} \end{aligned} $
(ii) $I^{d}=\int J_D s d s d \theta$
$ \begin{aligned} & =\frac{2 \pi}{\lambda^{2}} I_0 2 \pi \int _{s=0}^{a} \ln (\frac{a}{s}) \cdot s d s \sin (2 \pi v t) \\ & =(\frac{2 \pi}{\lambda})^{2} I_0 \int _{s=0}^{a} \frac{1}{2} d s^{2} \ln (\frac{a}{s}) \cdot \sin (2 \pi v t) \\ & =\frac{a^{2}}{4}(\frac{2 \pi}{\lambda})^{2} I_0 \int _{s=0}^{a} d(\frac{s}{a})^{2} \ln (\frac{a}{s})^{2} \cdot \sin (2 \pi v t) \\ & =-\frac{a^{2}}{4}(\frac{2 \pi}{\lambda})^{2} I_0 \int_0^{1} \ln \xi d \xi \cdot \sin (2 \pi v t) \\ & =+(\frac{a}{2})^{2}(\frac{2 \pi}{\lambda})^{2} I_0 \sin 2 \pi v t \quad(\therefore \text{ The integral has value }-1) \end{aligned} $
(iii) The displacement current
$ \begin{aligned} & I^{d}=(\frac{a}{2} \cdot \frac{2 \pi}{\lambda})^{2} I_0 \sin 2 \pi v t=I_0^{d} \sin 2 \pi \nu t \\ & \frac{I_0^{d}}{I_0}=(\frac{a \pi}{\lambda})^{2} . \end{aligned} $
8.31 A plane EM wave travelling in vacuum along $z$ direction is given by $\mathbf{E}=E_0 \sin (k z-\omega t) \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \sin (k z-\omega t) \hat{\mathbf{j}}$.
(i) Evaluate $\oint \mathbf{E} . \mathbf{d l}$ over the rectangular loop 1234 shown in Fig 8.2.
(ii) Evaluate $\int$ B.ds over the surface bounded by loop 1234. (iii) Use equation $\oint \mathbf{E} \cdot \mathbf{d l}=\frac{-d \phi_B}{d t}$ to prove $\frac{E_0}{B_0}=c$.
(iv) By using similar process and the equation $\oint \mathbf{B} . \mathbf{d} \mathbf{l}=\mu_0 I+\varepsilon_0 \frac{d \phi_E}{d t}$, prove that $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
Show Answer
Answer: (i) $\quad \oint E . d l=\int_1^{2} E . d l+\int_2^{3} E . d l+\int_3^{4} E . d l+\int_4^{1} E . d l$
$ \begin{aligned} & =\int_1^{2} E \cdot d l \cos 90^{\circ}+\int_2^{3} E \cdot d l \cos 0+\int_3^{4} E \cdot d l \cos 90^{\circ}+\int_4^{1} E \cdot d l \cos 180^{\circ} \\ & =\boldsymbol{{}E}_0 h[\sin (k z_2-\omega t)-\sin (k z_1-\omega t)] \end{aligned} $
(ii) For evaluating $\int$ B.ds let us consider the rectangle
1234 to be made of strips of area $d s=h d z$ each.
$ \int \mathbf{B \cdot ds} = \int B d s \cos 0 = \int B d s = \int_{Z_1} ^{Z_2} B_0 \sin (k z-\omega t) h d z $
$=\frac{-B_o h}{k}[\cos (k Z_2-\omega t)-\cos (k Z_1-\omega t)]$
(iii) $\oint \mathbf{E} . \mathbf{d} \mathbf{l}=\frac{-d \phi_B}{d t}$
Using the relations obtained in Equations (1) and (2) and simplifiying, we get
$ \begin{aligned} E_0 h & {[\sin (k z_2-\omega t)-\sin (k z_1-\omega t)]=\frac{B_o h}{k} \omega[\sin (k z_2-\omega t)-\sin (k z_1-\omega t)] } \\ E_0 & =B_0 \frac{\omega}{k} \\ \frac{E_0}{B_0} & =c \end{aligned} $
(iv) For evaluating $\oint \mathbf{B} . d 1$, let us consider the loop 1234 in yz plane as shown in Fig.
y
$B_0=E_0 \frac{\omega}{k} \cdot \mu \varepsilon_0$
$\frac{E_0}{B_0} \frac{\omega}{k}=\frac{1}{\mu_0 \varepsilon_0}$ But $E_0 / B_0=c$, and $\omega=c k$
or $c . c=\frac{1}{\mu_0 \varepsilon_0}$ Therefore, $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
$\therefore \oint \mathbf{B} \cdot \mathbf{d l}=\mu_0 \varepsilon_0 \frac{d \phi_E}{d t}$ we get
$ B_0=E_0 \frac{\omega}{k} \cdot \mu_0 \varepsilon_0 $
$ \overline{\overline{\varepsilon_0}} $
be made of strips of area $h d z$ each.
In $\oint \mathbf{B} . \mathbf{d l}=\mu_0(I+\varepsilon_0 \frac{d \phi_E}{d t}), I=$ conduction current $ \hspace {49mm} =0 \text{ in vacuum.} $
Using relations obtained in Equations (3) and (4) and ssimplifying,
$B_0 = E_0 \frac{ \omega }{K} \cdot \mu_0 \varepsilon_0 $
$ \frac{E_0}{B_0} \frac { \omega } {k} = \frac{1}{ \mu_0 \varepsilon_0} \ \text{But} \ E_0 / B_0 = c, \ \text{and} \ \omega = ck $
or $ c.c =\frac{1}{\mu_0 \varepsilon_0} \ \text{Therefore,} \ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
8.32 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 _{av}=\frac{1}{4} \varepsilon_0 E_0^{2}+\frac{1}{4} \frac{B_0^{2}}{\mu_0} . $
(ii) The time averaged intensity of the wave is given by
$ I _{av}=\frac{1}{2} c \varepsilon_0 E_0^{2} $
Show Answer
Answer: (a) E - field contribution is $u_E=\frac{1}{2} \varepsilon_0 E^{2}$
B - field contribution is $u_B=\frac{1}{2} \frac{B^{2}}{\mu_0}$
Total energy density $u=u_E+u_B=\frac{1}{2} \varepsilon_0 E^{2}+\frac{1}{2} \frac{B^{2}}{\mu_0}$
The values of $E^{2}$ and $B^{2}$ vary from point to point and from moment to moment. Hence, the effective values of $E^{2}$ and $B^{2}$ are their time averages.
$(E^{2}) _{a v}=E_0^{2}[\sin ^{2}(k z-\omega t)] _{a v}$
$(B^{2}) _{a v}=(B^{2}) _{a v}=B_0^{2}[\sin ^{2}(k z-\omega t)] _{a v}$
The graph of $\sin ^{2} \theta$ and $\cos ^{2} \theta$ are identical in shape but shifted by $\pi /$ 2 , so the average values of $\sin ^{2} \theta$ and $\cos ^{2} \theta$ are also equal over any integral multiple of $\pi$.
and also $\sin ^{2} \theta+\cos ^{2} \theta=1$
So by symmetry the average of $\sin ^{2} \theta=$ average of $\cos ^{2} \theta=\frac{1}{2}$
$\therefore(E^{2}) _{a v}=\frac{1}{2} E_0^{2}$ and $(B^{2}) _{a v}=\frac{1}{2} B_0^{2}$
Substuting in Equation (1),
$u=\frac{1}{4} \varepsilon_0 E^{2}+\frac{1}{4} \frac{B_0^{2}}{\mu}$
(b) We know $\frac{E_0}{B_0}=$ cand $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \therefore \frac{1}{4} \frac{B_0^{2}}{\mu_0}=\frac{E_0^{2} / c^{2}}{4 \mu_0}=\frac{E_0^{2}}{4 \mu_0} \mu_0 \varepsilon_0=\frac{1}{4} \varepsilon_0 E_0^{2}$.
Therefore, $u _{a v}=\frac{1}{4} \varepsilon_0 E_0^{2}+\frac{1}{4} \varepsilon_0 E_0^{2}=\frac{1}{2} \varepsilon_0 E_0^{2}$, and $I _{a v}=u _{a v} c=\frac{1}{2} \varepsilon_0 E_0^{2}$.
