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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$200 \sin (ω t - k x) {NC}^{- 1}$

$25 \sin (ω t - k x) N C^{- 1}$

$50 \sin (ω t - k x) N C^{- 1}$

$400 \sin (ω t - k x) N C^{- 1}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} Energy density = \frac{1}{2} ϵ_{0} E^{2} \times c \\ Energy = \frac{1}{2} ϵ_{0} E^{2} \times c \times volume \\ (Energy)_{1} = (Energy)_{2} (Given) \end{aligned}$
$\begin{aligned} \frac{1}{2} ϵ_{0} E_{1}^{2} c π R_{1}^{2} \times L_{1} = \frac{1}{2} ϵ_{0} E_{2}^{2} c π R_{2}^{2} \times L_{2} \\ E_{1}^{2} R_{1}^{2} = E_{2}^{2} R_{2}^{2} \\ E_{1} R_{1} = E_{2} R_{2} \end{aligned}$
$\begin{aligned} 100 \times R_{1} = E_{2} \times \frac{R_{1}}{2} \\ E_{2} = 200 N / C \end{aligned}$