Q 12-08-202JEE MainJEE Main 2019 (9 Jan, Shift 1)Easy
A plane electromagnetic wave of frequency $50\ \text{MHz}$ travels in free space along the positive $x$-direction. At a particular point in space and time, $\vec E = 6.3\,\hat j\ \text{V/m}$. The corresponding magnetic field $\vec B$, at that point will be
Answer: (A) $2.1\times10^{-8}\,\hat k\ \text{T}$
Magnitude: $B = \dfrac Ec = \dfrac{6.3}{3\times10^8} = 2.1\times10^{-8}\ \text{T}$.
Direction: $\vec E\times\vec B$ must point along $+\hat i$. Since $\hat j\times\hat k = \hat i$, $\vec B$ is along $+\hat k$:
$$\vec B = 2.1\times10^{-8}\,\hat k\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics