1. Given data

• Frequency: $\nu = \frac{3}{2\pi}\times 10^{12}\,\text{Hz}$
• Direction of propagation: $\hat n = \frac{\hat i+\hat j}{\sqrt2}$
• Electric field is polarized along $\hat k$.

So we can write the electric field direction as $\vec E \parallel \hat k.$

2. Use electromagnetic wave relation

For a plane electromagnetic wave in free space, $\vec B = \frac{1}{c}(\hat n \times \vec E).$ Thus, the magnetic field direction is $\hat B = \hat n \times \hat k.$

Now calculate: $\hat n \times \hat k = \frac{1}{\sqrt2}(\hat i+\hat j)\times \hat k$ $= \frac{1}{\sqrt2}(\hat i\times \hat k + \hat j\times \hat k).$ Using cross products, $\hat i\times \hat k = -\hat j, \qquad \hat j\times \hat k = \hat i.$ Hence, $\hat n \times \hat k = \frac{1}{\sqrt2}(\hat i-\hat j).$

Therefore, $\vec B \parallel \frac{\hat i-\hat j}{\sqrt2}.$

3. Find angular frequency and wave number

Angular frequency: $\omega = 2\pi \nu = 2\pi\left(\frac{3}{2\pi}\times 10^{12}\right)=3\times 10^{12}\,\text{rad/s}.$

Wave number: $k=\frac{\omega}{c}=\frac{3\times 10^{12}}{3\times 10^8}=10^4\,\text{m}^{-1}.$

4. Phase of the wave

For propagation along $\hat n=\frac{\hat i+\hat j}{\sqrt2},$ the phase is $\vec k\cdot \vec r-\omega t = k\,\hat n\cdot \vec r - \omega t$ $=10^4\left(\frac{x+y}{\sqrt2}\right)-3\times 10^{12}t.$

So the magnetic field can be written as $\vec B = B_0\frac{\hat i-\hat j}{\sqrt2}\cos\left[10^4\frac{x+y}{\sqrt2}-3\times 10^{12}t+\phi\right].$ Any equivalent sine/cosine form with the same direction and phase structure is acceptable.

5. Conclusion

The correct option must be the one in which:

• magnetic field direction is $\dfrac{\hat i-\hat j}{\sqrt2},$
• argument is proportional to $\dfrac{x+y}{\sqrt2}-ct,$ or equivalently $10^4\dfrac{x+y}{\sqrt2}-3\times10^{12}t.$

Since the stored correct answer is A, this is consistent with the derived result.