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JEE Main 202310 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2023 · 10 April · Morning Shift · Physics · Electromagnetic Induction

The energy of an electromagnetic wave contained in a small volume...

The energy of an electromagnetic wave contained in a small volume oscillates with

Options

  1. ADouble the frequency of the wave
  2. BThe frequency of the wave
  3. CHalf the frequency of the wave
  4. DZero frequency

Correct answer

A. Double the frequency of the wave

Step-by-step solution

The formula to calculate the energy confined within the electric component of an electromagnetic wave can be written as

uE=12ε0E2dV...(1){u}_{E} = \dfrac{1}{2}{ε}_{0}{E}^{2}dV . . . \left(1\right)

Let's assume that the equation of the electric component is given by

E=E0sin⁡(\omegat+ϕ)...(2)E = {E}_{0}\sin \left(\omegat + ϕ\right) . . . \left(2\right)

Substitute the expression for the electric field component from equation (2) into equation (1) and simplify to obtain the expression for the energy confined within the small volume dVdV.

{uE=12ε0[E0sin⁡(\omegat+ϕ)]2dV=12ε0E02sin⁡2(\omegat+ϕ)dV=14ε0E02[1−cos⁡(2\omegat+2ϕ)]...(3)\begin{cases}{u}_{E} & = & \dfrac{1}{2}{ε}_{0}{\left[{E}_{0}\sin \left(\omegat + ϕ\right)\right]}^{2}dV \\ & = & \dfrac{1}{2}{ε}_{0}{{E}_{0}}^{2}\sin^{2}\left(\omegat + ϕ\right)dV \\ & = & \dfrac{1}{4}{ε}_{0}{{E}_{0}}^{2}\left[1 - \cos \left(2\omegat + 2ϕ\right)\right] . . . \left(3\right)\end{cases}

From equation (3), it can be concluded that the frequency of oscillation of the energy confined within a small volume is double the frequency of the wave.

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