Q 12-08-176JEE MainJEE Main 2020 (8 Jan, Shift 2)Easy
A plane electromagnetic wave of frequency $25$ GHz is propagating in vacuum along the $z$-direction. At a particular point in space and time, the magnetic field is given by $\vec{B} = 5\times10^{-8}\,\hat{j}$ T. The corresponding electric field $\vec{E}$ is (speed of light $= 3\times10^8$ m s$^{-1}$)
Answer: (D) $15\,\hat{i}$ V/m
Magnitude: $E = cB = 3\times10^8\times5\times10^{-8} = 15$ V/m.
Direction: the wave travels along $\vec{E}\times\vec{B}$, which must be $+\hat{k}$. With $\vec{B}$ along $\hat{j}$, $\hat{i}\times\hat{j} = \hat{k}$, so $\vec{E}$ is along $+\hat{i}$.
$$\vec{E} = 15\,\hat{i}\ \text{V/m}$$
Solution by Sreeraj P, M.Sc Physics