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:

We have the electric field of the electromagnetic wave written as

$$\vec E = 2E_0 \cos kz \, \cos \omega t \; \hat i.$$

In free space the fields satisfy Faraday’s law (one of Maxwell’s equations)

$$\nabla \times \vec E = -\,\frac{\partial \vec B}{\partial t}.$$

Because the given $$\vec E$$ depends only on $$z$$ and $$t$$ and points purely along $$\hat i$$, its curl contains only a $$\hat j$$ (y-) component. Writing $$\vec E = E_x(z,t)\hat i$$ we obtain

$$\nabla \times \vec E = \left(\frac{\partial E_x}{\partial z}\right)\hat j.$$

So the magnetic field must also point along $$\hat j$$. Let us therefore set

$$\vec B = B_y(z,t)\; \hat j.$$

Substituting into Faraday’s law gives

$$\frac{\partial E_x}{\partial z}\;\hat j = -\,\frac{\partial B_y}{\partial t}\;\hat j,$$

and hence, dropping the common unit vector $$\hat j$$,

$$\frac{\partial E_x}{\partial z} = -\,\frac{\partial B_y}{\partial t}.$$

Now insert the explicit form of $$E_x$$:

$$E_x(z,t) = 2E_0 \cos kz \, \cos \omega t.$$

Differentiating with respect to $$z$$ gives

$$\frac{\partial E_x}{\partial z} = 2E_0 (-k) \sin kz \, \cos \omega t \;=\; -\,2kE_0 \sin kz \, \cos \omega t.$$

Therefore

$$-\,2kE_0 \sin kz \, \cos \omega t = -\,\frac{\partial B_y}{\partial t},$$

which simplifies to

$$2kE_0 \sin kz \, \cos \omega t = \frac{\partial B_y}{\partial t}.$$

Integrating both sides with respect to time $$t$$, we obtain

$$B_y(z,t) = 2kE_0 \sin kz \int \cos \omega t \; dt + g(z),$$

where $$g(z)$$ is an arbitrary function of $$z$$ (the “constant” of integration for the time integral). Evaluating the integral,

$$\int \cos \omega t \; dt = \frac{1}{\omega} \sin \omega t,$$

so

$$B_y(z,t) = \frac{2kE_0}{\omega} \sin kz \, \sin \omega t + g(z).$$

The physical requirement that the magnetic field be zero when $$t=0$$ (because $$\sin\omega t = 0$$ then) forces $$g(z)=0$$. Thus

$$B_y(z,t) = \frac{2kE_0}{\omega} \sin kz \, \sin \omega t.$$

For electromagnetic waves in vacuum we use the dispersion relation

$$\frac{\omega}{k} = c \quad \Longrightarrow \quad \frac{k}{\omega} = \frac{1}{c}.$$

Substituting $$k/\omega = 1/c$$, we finally get

$$B_y(z,t) = \frac{2E_0}{c} \sin kz \, \sin \omega t.$$

Restoring the unit vector $$\hat j$$, the magnetic field vector is

$$\vec B = \frac{2E_0}{c} \sin kz \, \sin \omega t \; \hat j.$$

Hence, the correct answer is Option A.