Q 12-08-037JEE MainJEE Main 2026 (28 Jan, Shift 2)Easy
A plane electromagnetic wave is moving in free space with velocity $c = 3\times10^8$ m/s and its electric field is given as $\vec E = 54\sin(kz-\omega t)\,\hat j$ V/m, where $\hat j$ is the unit vector along y-axis. The magnetic field vector $\vec B$ of the wave is :
Answer: (A) $-1.8\times10^{-7}\sin(kz-\omega t)\,\hat i$ T
Amplitude: $B_0 = \dfrac{E_0}{c} = \dfrac{54}{3\times10^8} = 1.8\times10^{-7}$ T.
The wave travels along $+z$, and $\vec E\times\vec B$ must point along $+\hat k$. With $\vec E$ along $\hat j$: $\hat j\times(-\hat i) = \hat k$, so $\vec B$ is along $-\hat i$.
$$\vec B = -1.8\times10^{-7}\sin(kz-\omega t)\,\hat i\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics