Electromagnetic Waves

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

The orientation of the portable radio with respect to broadcasting station is important because 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.

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/ $m^2$