Q 12-08-199JEE MainJEE Main 2019 (10 Apr, Shift 1)Medium
The electric field of a plane electromagnetic wave is given by $\vec E = E_0\,\hat i\cos(kz)\cos(\omega t)$. The corresponding magnetic field $\vec B$ is then given by:
Answer: (C) $\vec B = \dfrac{E_0}{c}\,\hat j\sin(kz)\sin(\omega t)$
Use Faraday's law $\nabla\times\vec E = -\dfrac{\partial\vec B}{\partial t}$. For $\vec E = E_x(z,t)\,\hat i$:
$$\nabla\times\vec E = \frac{\partial E_x}{\partial z}\hat j = -E_0k\sin(kz)\cos(\omega t)\,\hat j$$
So $\dfrac{\partial\vec B}{\partial t} = E_0k\sin(kz)\cos(\omega t)\,\hat j$, and integrating in time:
$$\vec B = \frac{E_0k}{\omega}\sin(kz)\sin(\omega t)\,\hat j = \frac{E_0}{c}\,\hat j\sin(kz)\sin(\omega t)$$
Solution by Sreeraj P, M.Sc Physics