Electromagnetic Waves

JEE Main level, numerical answer, medium

73 of 165

Question 73 of 165

The electric field in an electromagnetic wave is given by
E=(50NC−1)sin⁡ω(t−xc)E=\left(50N{C}^{-1}\right)\sin \omega (t-\dfrac{x}{c})
The energy contained in a cylinder of volume VV is 5.5×10−12J.5.5\times {10}^{-12}J. The value of VV is _______ cm3.{cm}^{3}.
(given ϵ0=8.8×10−12C2N−1m−2{ϵ}_{0}=8.8\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2})

Show answer and solution

Answer: 500

Energy density =12ϵ0E2=\dfrac{1}{2}{\epsilon }_{0}{E}^{2}
Energy =12ϵ0E2×=\dfrac{1}{2}{\epsilon }_{0}{E}^{2}\times volume
5.5×10−12=12×8.8×10−12×502×V5.5\times {10}^{-12}=\dfrac{1}{2}\times 8.8\times {10}^{-12}\times {50}^{2}\times V
V=500cm3V=500{cm}^{3}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Electromagnetic Waves (Electromagnetic Waves and Maxwell's equations), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.