Q 12-08-136JEE MainJEE Main 2022 (26 Jul, Shift 2)Medium
The oscillating magnetic field in a plane electromagnetic wave is given by $B_y = 5\times10^{-6}\sin1000\pi(5x - 4\times10^8t)\ \text{T}$. The amplitude of the electric field will be
Answer: (D) $4\times10^2\ \text{V m}^{-1}$
The wave speed is $v = \dfrac{\omega}{k} = \dfrac{1000\pi\times4\times10^8}{1000\pi\times5} = 0.8\times10^8\ \text{m s}^{-1}$ (the wave is in a medium).
$$E_0 = vB_0 = 0.8\times10^8\times5\times10^{-6} = 400 = 4\times10^2\ \text{V m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics