JEE Main 2025 Jan 28 Shift 1, Physics Q16: Electromagnetic Waves

JEE Main2025Jan 28, Shift 1Physics

Due to presence of an em-wave whose electric component is given by $E = 100 sin (ω t - k x)$ 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
$25 sin (ω t - k x)$ NC $^{- 1}$
B
$200 sin (ω t - k x)$ NC $^{- 1}$
C
$400 sin (ω t - k x)$ NC $^{- 1}$
D
$50 sin (ω t - k x)$ NC $^{- 1}$

Solution

Energy density of an em-wave (averaged): $u = \frac{1}{2} ϵ_{0} E_{0}^{2}$ , so the total energy in a cylinder of cross-section area $A = π R^{2}$ and length $L$ is

$U = u \cdot A \cdot L = \frac{1}{2} ϵ_{0} E_{0}^{2} π R^{2} L$ .

Setting the two energies equal (same $L$ ):

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

With $E_{1} = 100$ N/C, $R_{2} = R_{1} /2$ (half diameter):

$100 \cdot R_{1} = E_{2} \cdot \frac{R _{1}}{2} \Rightarrow E_{2} = 200$ N/C.

$∴ E = 200 sin (ω t - k x)$ NC $^{- 1}$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q16: Electromagnetic Waves

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