Q 12-08-213JEE MainJEE Main 2019 (11 Jan, Shift 2)Medium
A $27$ mW laser beam has a cross-sectional area of $10\ \text{mm}^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space $\varepsilon_0 = 9\times10^{-12}$ SI units, speed of light $c = 3\times10^8$ m/s]
Answer: (D) $1.4$ kV/m
Intensity: $I = \dfrac{27\times10^{-3}}{10\times10^{-6}} = 2700\ \text{W/m}^2$.
$$I = \frac12\varepsilon_0cE_0^2 \Rightarrow E_0^2 = \frac{2\times2700}{9\times10^{-12}\times3\times10^8} = 2\times10^6$$
$$E_0 \approx 1.4\times10^3\ \text{V/m} = 1.4\ \text{kV/m}$$
Solution by Sreeraj P, M.Sc Physics