The electric field component of a monochromatic radiation is given by $\vec E = 2E_0\cos kz\cos\omega t\,\hat i$. Its magnetic field $\vec B$ is then given by:
Answer: (A) $\dfrac{2E_0}{c}\sin kz\sin\omega t\,\hat j$
Use Faraday's law $\nabla\times\vec E = -\dfrac{\partial\vec B}{\partial t}$. For $\vec E = E_x(z,t)\hat i$, only the $y$-component of the curl survives:
$$(\nabla\times\vec E)_y = \frac{\partial E_x}{\partial z} = -2E_0k\sin kz\cos\omega t$$
$$\frac{\partial B_y}{\partial t} = 2E_0k\sin kz\cos\omega t \;\Rightarrow\; B_y = \frac{2E_0k}{\omega}\sin kz\sin\omega t$$
With $\omega/k = c$:
$$\vec B = \frac{2E_0}{c}\sin kz\sin\omega t\,\hat j$$
(This is a standing wave: $\vec B$ is shifted by a quarter period and a quarter wavelength from $\vec E$.)
Solution by Sreeraj P, M.Sc Physics