Q 12-08-043JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
A plane electromagnetic wave of frequency $20\ \text{MHz}$ travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $E_y = 9.3\ \text{V m}^{-1}$. Then, the magnetic field vector of the wave at that point is
Answer: (B) $B_z = 3.1\times10^{-8}\ \text{T}$
In an EM wave $B = \dfrac{E}{c}$:
$$B = \frac{9.3}{3\times10^8} = 3.1\times10^{-8}\ \text{T}$$
$\vec E\times\vec B$ must point along $+x$; with $\vec E$ along $\hat j$, $\vec B$ is along $\hat k$ ($\hat j\times\hat k = \hat i$). So $B_z = 3.1\times10^{-8}\ \text{T}$.
Solution by Sreeraj P, M.Sc Physics