Nature of EM Waves & Wave Equation question from NEET 2025

Answer: (c)

1. From $\cos(kx+\omega t)$: $k=5$, $\omega = 1.5\times10^{9}$, so $v = \omega/k = 3\times10^{8}$ m/s and the wave moves along $-x$.
2. $B_0 = \dfrac{E_0}{c} = \dfrac{60}{3\times10^{8}} = 2\times10^{-7}\text{ T}$
3. $\vec E\times\vec B$ must point along $-\hat x$: $\hat z\times\hat y = -\hat x$, so $\vec B$ is along $+y$ with the same phase.
4. So the answer is (c) $B_y = 2\times10^{-7}\cos(5x+1.5\times10^{9}t)$ T.