Due to presence of an em-wave whose electric component is given by...

Due to presence of an em-wave whose electric component is given by

\mathrm{E}=100 \sin (\omega \mathrm{t}-\mathrm{kx}) \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

Options

A $400 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$
B $200 \sin (\omega t-k x) \mathrm{NC}^{-1}$
C $50 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$
D $25 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$

Correct answer

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

Step-by-step solution

Energy density of an

E M_{\text{wave }}=\frac{1}{2} \varepsilon E_0^2

, whare

E_0

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}

\Rightarrow

The amplitude of corresponding

E M

wave is 200 N/C or the wave is

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

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