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:

We are told that the electric field of the electromagnetic wave is

$$\vec E = E_0\,\hat i \,\cos(kz)\,\cos(\omega t).$$

The wave is in free space, so Maxwell’s equations in vacuum apply. The relation that connects the curl of the electric field to the time‐rate of change of the magnetic field is Faraday’s law:

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

We therefore begin by calculating the curl of the given electric field. Because the only component of $$\vec E$$ is along $$\hat i$$ and it depends only on $$z$$ and $$t$$, we write

$$\vec E = (E_0\cos kz\cos\omega t)\,\hat i,$$ $$E_x = E_0\cos kz\cos\omega t,\qquad E_y = 0,\qquad E_z = 0.$$

The curl in Cartesian coordinates is

$$\nabla\times\vec E \;=\; \begin{vmatrix} \hat i & \hat j & \hat k\\[4pt] \displaystyle\frac{\partial}{\partial x} & \displaystyle\frac{\partial}{\partial y} & \displaystyle\frac{\partial}{\partial z}\\[6pt] E_x & E_y & E_z \end{vmatrix}.$$

Since $$E_x$$ depends only on $$z$$ (and $$t$$) and there are no $$y$$ or $$z$$ components, every derivative is zero except $$\partial E_x/\partial z$$. Hence only the $$\hat j$$ component survives:

$$\nabla\times\vec E = \hat j\left(\frac{\partial E_x}{\partial z} - 0\right).$$

We now differentiate $$E_x$$ with respect to $$z$$:

$$\frac{\partial E_x}{\partial z} = \frac{\partial}{\partial z}\Bigl[E_0\cos(kz)\cos(\omega t)\Bigr] = -E_0 k\,\sin(kz)\,\cos(\omega t).$$

Therefore,

$$\nabla\times\vec E = -E_0 k\,\sin(kz)\,\cos(\omega t)\;\hat j.$$

Invoking Faraday’s law,

$$-\,\frac{\partial\vec B}{\partial t} = -E_0 k\,\sin(kz)\,\cos(\omega t)\;\hat j.$$

The minus signs on both sides cancel, giving

$$\frac{\partial\vec B}{\partial t} = E_0 k\,\sin(kz)\,\cos(\omega t)\;\hat j.$$

To obtain $$\vec B$$ we integrate with respect to time:

$$\vec B = \int \frac{\partial\vec B}{\partial t}\,dt = E_0 k\,\sin(kz)\int \cos(\omega t)\,dt\;\hat j.$$

The time integral is elementary:

$$\int \cos(\omega t)\,dt = \frac{1}{\omega}\,\sin(\omega t) + C,$$

where the constant of integration $$C$$ can be taken as zero because it corresponds to adding a constant (static) magnetic field, which is not part of the electromagnetic wave. Hence,

$$\vec B = \frac{E_0 k}{\omega}\,\sin(kz)\,\sin(\omega t)\;\hat j.$$

For a wave in vacuum the dispersion relation is $$\omega = ck,$$ so $$k/\omega = 1/c$$. Substituting this, we find

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

This matches exactly Option C:

$$\vec B = \frac{E_0}{C}\,\hat j\,\sin(kz)\,\sin(\omega t) \quad(\text{with }C=c).$$

Hence, the correct answer is Option C.