JEE Main 2023 — Electromagnetic Induction Question with Solution

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

$u_E=\frac{1}{2}{\epsilon }_0{E}^{2}dV ...\left(1\right)$

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

$E=E_0 \sin \left(\omega t+\varphi \right) ...\left(2\right)$

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 $dV$.

$\begin{array}{rcl}{u}_E& =& \frac{1}{2}{\epsilon }_0{\left[E_0 \sin \left(\omega t+\varphi \right)\right]}^{2}dV\\ & =& \frac{1}{2}{\epsilon }_0{E_0}^2{\sin }^2\left(\omega t+\varphi \right)dV\\ & =& \frac{1}{4}{\epsilon }_0{E_0}^2\left[1-\cos \left(2\omega t+2\varphi \right)\right] ...\left(3\right)\end{array}$

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.