The electric field of a plane electromagnetic wave is given by $$\vec{E} = E_0(\hat{x} + \hat{y})\sin(kz - \omega t)$$. Its magnetic field will be given by

We have the electric field of the plane electromagnetic wave written as

$$\vec E = E_0\,(\hat x + \hat y)\,\sin(kz-\omega t).$$

For a plane electromagnetic wave propagating in free space we use two standard facts:

1. The direction of propagation is perpendicular to both $$\vec E$$ and $$\vec B$$. 2. The magnitudes satisfy $$|\vec E| = c\,|\vec B|$$, and the vectors obey the right-hand relation

$$\hat k \times \vec E \;=\; c\,\vec B,$$

where $$\hat k$$ is the unit vector in the propagation direction and $$c$$ is the speed of light in vacuum.

In the given expression we see the argument $$kz - \omega t$$. A phase of this form tells us that the wave travels along the positive $$z$$-axis, so

$$\hat k = \hat z.$$

We therefore write

$$c\,\vec B \;=\; \hat k \times \vec E \;=\; \hat z \times \bigl[E_0(\hat x + \hat y)\sin(kz - \omega t)\bigr].$$

Because the scalar quantity $$E_0\sin(kz-\omega t)$$ does not participate in the cross product, we can place it outside:

$$c\,\vec B \;=\; E_0\sin(kz - \omega t)\; \bigl[\hat z \times (\hat x + \hat y)\bigr].$$

Now we evaluate the cross product term by term. Using the standard right-hand rule relations $$\hat z \times \hat x = \hat y$$ and $$\hat z \times \hat y = -\hat x,$$ we have

$$\hat z \times (\hat x + \hat y) = (\hat z \times \hat x) + (\hat z \times \hat y) = \hat y + (-\hat x) = -\hat x + \hat y.$$

Substituting this result back, we obtain

$$c\,\vec B = E_0\sin(kz - \omega t)\,(-\hat x + \hat y).$$

Finally, dividing both sides by $$c$$ gives the magnetic field:

$$\vec B = \frac{E_0}{c}\,(-\hat x + \hat y)\,\sin(kz - \omega t).$$

This expression exactly matches Option A.

Hence, the correct answer is Option A.