A monochromatic beam of light has a frequency $$\nu = \frac{3}{2\pi} \times 10^{12}$$ Hz and is propagating along the direction $$\frac{\hat{i}+\hat{j}}{\sqrt{2}}$$. It is polarized along the $$\hat{k}$$ direction. The acceptable form for the magnetic field is:

We have a plane monochromatic electromagnetic wave whose direction of propagation is given by the unit vector

$$\hat{n}= \frac{\hat{i}+\hat{j}}{\sqrt{2}}\;.$$

The electric field is said to be polarised along the $$\hat{k}$$ direction, so at any point we may write it in the standard travelling-wave form

$$\vec{E}(\vec{r},t)=E_0\,\hat{k}\,\cos\!\left(\vec{k}\!\cdot\!\vec{r}-\omega t\right).$$

First we calculate the angular frequency. The given frequency is

$$\nu=\frac{3}{2\pi}\times10^{12}\ \text{Hz}.$$

The relation between angular frequency and frequency is $$\omega=2\pi\nu.$$ Substituting,

$$\omega=2\pi\left(\frac{3}{2\pi}\times10^{12}\right)=3\times10^{12}\ \text{rad s}^{-1}.$$

For a wave travelling in vacuum, the magnitude of the wave-vector is obtained from $$k=\frac{\omega}{c},$$ where $$c=3\times10^{8}\ \text{m s}^{-1}.$$ Hence

$$k=\frac{3\times10^{12}}{3\times10^{8}}=10^{4}\ \text{m}^{-1}.$$

The full wave-vector is therefore

$$\vec{k}=k\hat{n}=10^{4}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right).$$

Next we obtain the magnetic field. For any plane electromagnetic wave in free space the magnetic field satisfies

$$\vec{B}=\frac{1}{c}\,\hat{n}\times\vec{E}.$$

We now perform the cross product. Using $$\hat{i}\times\hat{k}=-\hat{j},\quad\hat{j}\times\hat{k}=\hat{i},$$ we get

$$\hat{n}\times\hat{k}=\frac{1}{\sqrt{2}}\Big(\hat{i}\times\hat{k}+\hat{j}\times\hat{k}\Big)=\frac{1}{\sqrt{2}}\big(-\hat{j}+\hat{i}\big)=\frac{\hat{i}-\hat{j}}{\sqrt{2}}.$$

Hence the magnetic-field vector is

$$\vec{B}(\vec{r},t)=\frac{E_0}{c}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right)\cos\!\left(10^{4}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\!\cdot\!\vec{r}-3\times10^{12}t\right).$$

This expression has (i) the correct amplitude factor $$E_0/c,$$ (ii) a direction $$\frac{\hat{i}-\hat{j}}{\sqrt{2}}$$ perpendicular to both $$\hat{k}$$ and $$\hat{n},$$ and (iii) the correct phase factor $$\vec{k}\cdot\vec{r}-\omega t.$$ Comparing with the choices, it exactly matches Option A.

Hence, the correct answer is Option A.