JEE Main 2022 — Electromagnetic Waves Question with Solution

The intensity is defined as energy per unit time per unit area. Therefore, $I=\frac{\mathrm{d}E}{A\mathrm{d}t}$.

Energy density is energy per unit volume. Therefore, $U_d=\frac{\mathrm{d}E}{A\mathrm{d}x}$.

Dividing $\frac{I}{U_d}=\frac{dx}{dt}=c$

$\Rightarrow U_d=\frac{I}{c}=\frac{0.22}{3\times {10}^8}=\frac{22}{3}\times {10}^{-10} \mathrm{J} {\mathrm{m}}^{-3}$

Magnetic energy density will be half of total energy density, Therefore, $U_b=\frac{11}{3}\times {10}^{-10} \mathrm{J} {\mathrm{m}}^{-3}$.

Now energy density in magnetic field is given by, $U_b=\frac{B^2}{2{\mu }_0}$. Therefore,

$\frac{B_{\mathrm{rms}}^2}{2{\mu }_0}=\frac{11}{3}\times {10}^{-10} \mathrm{J} {\mathrm{m}}^{-3}$ where $B_{\mathrm{rms}}^2=\left(\frac{B_0}{\sqrt{2}}\right)=\frac{B_0^2}{2}$

$$\begin{gathered} \Rightarrow \frac{B_0^2}{4{\mu }_0}=\frac{11}{3}\times {10}^{-10} \\ \Rightarrow B_0=\sqrt{4\times \left(4\pi \times {10}^{-7}\right)\times \frac{11}{3}\times {10}^{-10}} \\ \Rightarrow B_0=4.29\times {10}^{-8} \mathrm{T}\simeq 43\times {10}^{-9} \mathrm{T} \end{gathered}$$