Q 12-08-099JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
A plane electromagnetic wave of frequency $20\ \text{MHz}$ propagates in free space along $x$-direction. At a particular space and time $\vec E=6.6\hat j\ \text{V m}^{-1}$. What is $\vec B$ at this point?
Answer: (A) $2.2\times10^{-8}\hat k\ \text{T}$
$B=\dfrac Ec=\dfrac{6.6}{3\times10^8}=2.2\times10^{-8}\ \text{T}$. Propagation along $\vec E\times\vec B$: $\hat j\times\hat k=\hat i$, so $\vec B$ is along $+\hat k$.
Solution by Sreeraj P, M.Sc Physics