Due to presence of an em-wave whose electric component is given by E = 100 sin (t – kx) , 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

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Solution

Q.
Due to presence of an em-wave whose electric component is given by E = 100 sin ( $ω$ t – kx) ${NC}^{- 1}$ , 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 [2025]

1 25 sin ( $ω$ t – kx) ${NC}^{- 1}$

2 200 sin ( $ω$ t – kx) ${NC}^{- 1}$

3 400 sin ( $ω$ t – kx) ${NC}^{- 1}$

4 50 sin ( $ω$ t – kx) ${NC}^{- 1}$

Ans.
(2)

Energy density of an ${EM}_{wave}$ $= \frac{1}{2} ε_{0} E_{0}^{2},$ where $E_{0}$ is the amplitude of the wave.

Since total energy is the same for both cylinders

$(\frac{1}{2} ε_{0} E_{1}^{2}) π R_{1}^{2} L_{1} = (\frac{1}{2} ε_{0} E_{2}^{2}) π R_{2}^{2} L_{2}$

$\Rightarrow E_{1}^{2} R_{1}^{2} L_{1} = E_{2}^{2} R_{2}^{2} L_{2}$

or $E_{2} = \frac{E_{1} R_{1}}{R_{2}} \sqrt{\frac{L_{1}}{L_{2}}} = \frac{100 d}{(d / 2)} \sqrt{\frac{L_{1}}{L_{1}}} = 200 N/C [∵ L_{1} = L_{2} = 200 cm]$

$\Rightarrow The amplitude of corresponding electromagnetic wave is 200 N/C.$

Hence the wave is, $E = 200 \sin (ω t - k x) {NC}^{- 1} .$

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