NEET 2020 Physics Question 137 Solution

In an electromagnetic wave traveling in free space, energy is stored in both the oscillating electric and magnetic fields. The time-averaged energy densities are given by: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 E_{\text{rms}}^2 = \frac{1}{4} \varepsilon_0 E_0^2$$ $$\langle u_B \rangle = \frac{1}{2} \frac{B_{\text{rms}}^2}{\mu_0} = \frac{B_0^2}{4 \mu_0}$$

From Maxwell's equations, the ratio of electric field to magnetic field amplitude equals the speed of light in vacuum: $$E_0 = c B_0 \implies E_{\text{rms}} = c B_{\text{rms}}$$ Also, the speed of light is related to vacuum permittivity and permeability by: $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \implies c^2 = \frac{1}{\mu_0 \varepsilon_0}$$

Substitute $$E_{\text{rms}} = c B_{\text{rms}}$$ into the electric energy density expression: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 (c B_{\text{rms}})^2 = \frac{1}{2} \varepsilon_0 c^2 B_{\text{rms}}^2$$ Substituting $$c^2 = \frac{1}{\mu_0 \varepsilon_0}$$: $$\langle u_E \rangle = \frac{1}{2} \varepsilon_0 \left(\frac{1}{\mu_0 \varepsilon_0}\right) B_{\text{rms}}^2 = \frac{B_{\text{rms}}^2}{2 \mu_0} = \langle u_B \rangle$$

The intensity of an electromagnetic wave is the energy transported per unit area per unit time, given by $$I = u \cdot c$$. Since $$\langle u_E \rangle = \langle u_B \rangle$$, the intensity contributions are identical: $$I_E = \langle u_E \rangle c = \langle u_B \rangle c = I_B$$ $$\frac{I_E}{I_B} = \frac{1}{1} = 1 : 1$$