Question 25 of 165
Due to presence of an em-wave whose electric component is given by , 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
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Answer: (B)
Energy density of an , whare is the amplitude of the wave.
Since total energy is same for both cylinders
$\begin{aligned}
& \left(\frac{1}{2} \varepsilon E_1^2\right) \pi R_1^2 L_1=\left(\frac{1}{2} \varepsilon E_2^2\right) \pi R_2^2 L_2 \\
& \Rightarrow E_1^2 R_1^2 L_1=E_2^2 R_2^2 L_2 \\
& \text { or } E_2=\frac{E_1 R_1}{R_2} \sqrt{\frac{L_1}{L_2}}=\frac{100 \mathrm{~d}}{(\mathrm{~d} / 2)} \sqrt{\frac{L_1}{L 1}}=200 \mathrm{~N} / \mathrm{C} \\
& \qquad \quad\left[\because L_1=L_2=200 \mathrm{~cm}\right]
\end{aligned}$
The amplitude of corresponding wave is 200 N/C
or the wave is
Since total energy is same for both cylinders
$\begin{aligned}
& \left(\frac{1}{2} \varepsilon E_1^2\right) \pi R_1^2 L_1=\left(\frac{1}{2} \varepsilon E_2^2\right) \pi R_2^2 L_2 \\
& \Rightarrow E_1^2 R_1^2 L_1=E_2^2 R_2^2 L_2 \\
& \text { or } E_2=\frac{E_1 R_1}{R_2} \sqrt{\frac{L_1}{L_2}}=\frac{100 \mathrm{~d}}{(\mathrm{~d} / 2)} \sqrt{\frac{L_1}{L 1}}=200 \mathrm{~N} / \mathrm{C} \\
& \qquad \quad\left[\because L_1=L_2=200 \mathrm{~cm}\right]
\end{aligned}$
The amplitude of corresponding wave is 200 N/C
or the wave is