Electromagnetic Waves

This is a long answer type question as classified in NCERT Exemplar

(i) U_E= $\frac{1}{2}{\epsilon }_o{E}^2$

U_B= $\frac{1}{2}\frac{B^2}{{\mu }_o}$

So total energy density = U_E+ U_B= $\frac{1}{2}{\epsilon }_o{E}^2$ + $\frac{1}{2}\frac{B^2}{{\mu }_o}$

E= Eo/ $\sqrt[]{2}$  and B=Bo/ $\sqrt[]{2}$

U_av= $\frac{1}{4}{\epsilon }_o{E}^2$ + $\frac{1}{4}\frac{B^2}{{\mu }_o}$

(ii) We know Eo=cBo and c = 1/ $\sqrt[]{{\mu }_0{\epsilon }_0}$

$\frac{1}{4}\frac{B^2}{{\mu }_o}=\frac{1}{4}\frac{\frac{{Eo}^2}{c^2}}{{\mu }_o}$ = 1/4 $\epsilon oEo$ ²

U_E= U_B

This is a long answer type question as classified in NCERT Exemplar

(i) $\left(i\right)\int E.dl$ = ${\int }_1^{2}E.dl+{\int }_2^{3}E.dl+{\int }_3^{4}E.dl+{\int }_4^{1}E.dl$

= E_oh[sin(kz₂-wt)-sin(kz₁-gwt)]

(ii) $\int B.ds=\int B.dscos0={\int }_{z1}^{z2}{B}_0\mathrm{s}\mathrm{i}\mathrm{n}?(kz-wt)hdz$

= $\frac{-B_{o}h}{k}\cos ?\left(kz2-wt\right)\cos ?\left(kz1-wt\right)$

(iv) $?E.dl=\frac{-d\varnothing }{dt}$ =- $?B.ds$

=Eoh[sin(kz2-wt)-sin(kz1-wt)]

=- $\frac{d}{dt}[\frac{B_{yh}}{k}\cos ?\left(kz2-wt\right)-\mathrm{c}\mathrm{o}\mathrm{s}?(kz1-wt\left)\right]$

Eo=Bow/k=Byc

Eo/Bo=c

This is a long answer type question as classified in NCERT Exemplar

E(s,t)= ${\mu }_o{I}_o\cos ?\left(2\pi vt\right)In\left(\frac{s}{a}\right)k$

Now displacement current J_d=e_{o $\frac{dE}{dt}=\frac{{\epsilon }_{o}d}{dt}\left[{\mu }_o{I}_o\cos ?\left(2\pi vt\right)In\left(\frac{s}{a}\right)k\right]$}

= $\frac{{\mu }_o{\epsilon }_o{I}_{o}vd}{dt}\left[cos2v\pi tIn\left(\frac{s}{a}\right)k\right]$

= $\frac{v^2}{c^2}2\pi I_{o}sin2\pi vtIn\left(\frac{a}{s}\right)k$

${\frac{2\pi a}{2\lambda })}^2{I}_0=({\frac{a\pi }{\lambda })}^2{I}_0$ = $\frac{2\pi I_0}{{\lambda }_2}$ ina/s sin2 $\pi vtk$

I_d= $\int J_{d}sdsd\theta ={\int }_0^1{J}_{s}sds{\int }_0^{2\pi }d\theta$

  =( $\frac{2\pi }{\lambda }$ )^{2 $I_o{\int }_0^a\frac{a}{s}sdssin\pi vt$}

= $\frac{a^2}{2}{\left(\frac{2\pi }{\lambda }\right)}^2{I}_{o}sin2\pi vt{\int }_0^{a}In\left(\frac{a}{s}\right).d({\frac{s}{a})}^2$

After solving this we get

I_d= $\frac{a^2}{4}({\frac{2\pi }{\lambda })}^2{I}_{0}sin2\pi vt$

I_d= ${\left(\frac{2\pi a}{2\lambda }\right)}^2{I}_{o}sin2\pi vt=I_{od}sin2\pi vt$

I_od= ( ${\frac{2\pi a}{2\lambda })}^2{I}_0=({\frac{a\pi }{\lambda })}^2{I}_0$

$\frac{I_{od}}{I_o}=({\frac{a\pi }{\lambda })}^2$

This is a long answer type question as classified in NCERT Exemplar

Suppose the distance between the plates is d and applied voltage is V_t= V₀2

Then electric field is E= $\frac{V}{d}$ sin(2 $\pi vt$ )

J_c= $\frac{1}{\rho }\frac{V_0}{d}sin\left(2\pi vt\right)$

  =  $J_0^{c}sin2\pi vt$

$J_0^c$   = $\frac{V_0}{\rho d}$

J_d= $\epsilon \frac{dE}{dt}$ = $\frac{V_0}{\rho d}$

   = $\epsilon \frac{2\pi v{V}_0}{d}$ cos(2 $\pi vt$ )

   = J₀^d cos 2 $\pi vt$

J₀^d= $\frac{2\pi V_{\epsilon }{V}_0}{d}$

$\frac{J_0^d}{J_0^c}$ = 2 $\pi V_{\epsilon }\rho$ = 2 $\pi 80{\epsilon }_0$ v $*$ 0.25 = 4 $\pi {\epsilon }_0$ v $*10$ =4/9

This is a long answer type question as classified in NCERT Exemplar

E= $\frac{\lambda e_s}{2\pi {\epsilon }_{0}a}j$

And B= $\frac{{\mu }_{0i}}{2\pi a}$ i= $\frac{{\mu }_{0\lambda v}}{2\pi a}$ i

Then S= $\frac{1}{{\mu }_0}$ {E $*B$ }= $\frac{1}{{\mu }_0}\{\frac{{\lambda }_j}{2\pi {\epsilon }_{0}a}*\frac{{\mu }_{0i}}{2\pi a}\mathrm{\lambda }\sqrt[]{i}\}$

 = $\frac{{\lambda }^{2}v}{4{\pi }^2{\epsilon }_{0a^2}}\left(j*i\right)=-\frac{{\lambda }^{2}v}{4{\pi }^2{\epsilon }_{0a^2}}k$

8.12 Power rating of the bulb, P = 100 W

Power of visible radiation, $P_v$ = 5% of 100 W = 5 W

Distance from the bulb, d = 1 m

Intensity of radiation at that point is given as:

I = $\frac{P_v}{4\pi d^2}$ = $\frac{5}{4\pi *1^2}$ = 0.398 W/ $m^2$

Distance from the bulb, d = 10 m

Intensity of radiation at that point is given as:

I = $\frac{P_v}{4\pi d^2}$ = $\frac{5}{4\pi *{10}^2}$ = 3.978 $*{10}^{-3}$  W/ $m^2$

8.10 Frequency of the electromagnetic wave, $\nu$  = 2.0 $*{10}^{10}$  Hz

Electric field amplitude, $E_0$  = 48 V/m

Speed of light, c = 3 $*{10}^8$ m/s

Wavelength of a wave is given by,

$\lambda =\frac{c}{\nu }$ = $\frac{3\mathrm{}*{10}^8}{2.0\mathrm{}*{10}^{10}}$ = 0.015 m

Magnetic field strength is given by

$B_0=$ $\frac{E_0}{c}$ = $\frac{48}{3\mathrm{}*{10}^8}$ = 1.6 ${10}^{-12}$ T

Energy density of the electric field is given as,

$U_E$ =  $\frac{1}{2}{?}_0{E}^2$ and energy density of magnetic field is given by,

$U_B$ = $\frac{1}{2}\frac{B^2}{{\mu }_0}$

$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}$

$E_0$  = Permittivity of free space

${\mu }_0=$  Permeability of free space

We have the relation connecting E and B as:

E = cB, where c = $\frac{1}{\sqrt[]{{?}_0{\mu }_0}}$

Hence E = $\frac{B}{\sqrt[]{{?}_0{\mu }_0}}$ $\stackrel{?}{\mathit{i}}$

$E^2$  = $\frac{B^2}{{?}_0{\mu }_0}$

${?}_0{E}^2=$ $\frac{B^2}{{\mu }_0}$

$2U_E=2U_B$  or $U_E$ = $U_B$