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 $\pi V_{\epsilon }\rho$ 

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

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

8.15 Long distance radio broadcasts use shortwave bands because only these bands can be refracted by ionosphere.

It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus these signals are not reflected by ionospher

8.13 A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Plank’s law. It can be given by the relation,

T = 0.29 cmK

${\mathrm{}\lambda }_m=\frac{0.29}{T}$ cm K, where ${\mathrm{}\lambda }_{\mathit{m}}$

= maximum wavelength and T = t

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

Distance from the bulb, d = 10 m

Intensity of radiation at that point is given as:

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

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

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

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

Wavelength of a wave is given by,

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

Magnetic field strength is given by

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

Energy density of the electric field is given as,

$U_E$ =  $\frac{1}{2}{?}_0{E}^2$