Question 41: EM waves

Question

Due to presence of an em-wave whose electric component is given by $E=100 \sin (\omega t-k x) \mathrm{NC}^{-1}$ a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

Accepted Answer

We know,
Energy density $$ = {1 \over 2}{\varepsilon _0}{E^2}C$$
$$ \Rightarrow {	ext{Energy} \over 	ext{Volume}} = {1 \over 2}{\varepsilon _0}{E^2}C$$
$$ \Rightarrow 	ext{Energy}= {1 \over 2}{\varepsilon _0}E_c^2x\,vol$$
Given that, $${\left( {	ext{Energy}} ight)_1} = {\left( {	ext{Energy}} ight)_2}$$
$$ \Rightarrow {1 \over 2}{\varepsilon _0}E_1^2cx\pi R_1^2{L_1} = {1 \over 2}{\varepsilon _0}E_2^2cx\pi R_2^2{L_2}$$
$$ \Rightarrow E_1^2R_1^2{L_1} = E_2^2R_2^2{L_1}\,(as\,{L_1} = {L_2})$$
$$ \Rightarrow E_1^2R_1^2 = E_2^2R_2^2$$
$$ \Rightarrow {E_1}{R_1} = {E_2}{R_2}$$
$$ \Rightarrow {E_1}{R_1} = {E_2}\left( {{{{R_1}} \over 2}} ight)\,\left( {As\,{R_2} = {{{R_1}} \over 2}} ight)$$
$$ \Rightarrow {E_2} = 2{E_1} = 2 	imes 100$$
$$ \Rightarrow {E_2} = 200\,N/C$$
Hence, option (3) is correct.

Answer

$50 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$

Answer

$400 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$

Answer

$200 \sin (\omega t-k x) \mathrm{NC}^{-1}$

Answer

$25 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$

EM waves

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