Q63 / 75 · JEE Main 2025 (28 Jan Shift 1) · Physics

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

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

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

Hence, the correct answer is option: 'B'.

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