A magnetic field vector in an electromagnetic wave is represented by $$\vec{B} = B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{j}$$. Its associated electric field vector is ______.

The given magnetic field of the electromagnetic (EM) wave is
$$\vec{B}=B_0\sin\!\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)\,\hat{j}$$

Step 1 — Identify the direction of propagation.
For a plane EM wave, the phase $$\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)$$ has the form $$\omega t-kx$$, which corresponds to a wave travelling along the $$+x$$-axis. Hence, the propagation vector $$\vec{k}$$ is along $$+\hat{i}$$.

Step 2 — Use the orthogonality of $$\vec{E},\;\vec{B}$$ and $$\vec{k}$$.
In a uniform EM wave, $$\vec{E}\perp\vec{B}\perp\vec{k}$$ and
$$\vec{E}\times\vec{B} = \frac{1}{\mu_0}\,\vec{S}$$ points in the propagation direction $$\vec{k}$$(i.e. $$+\hat{i}$$ here).

Step 3 — Determine the unit vector for $$\vec{E}$$.
We already have $$\vec{k}=+\hat{i}$$ and $$\vec{B}$$ along $$+\hat{j}$$. Let $$\vec{E}$$ be along $$\hat{n}$$. We need
$$\hat{n}\times\hat{j}=+\hat{i}$$.
Using the right-hand rule: $$(-\hat{k})\times\hat{j}=+\hat{i}$$. Therefore $$\hat{n}=-\hat{k}$$ and $$\vec{E}$$ must be along $$-\hat{k}$$.

Step 4 — Relate the magnitudes $$E_0$$ and $$B_0$$.
In free space, $$E_0 = cB_0$$ where $$c$$ is the speed of light. Using $$c=\nu\lambda$$, we get
$$E_0 = \nu\lambda\,B_0$$.

Step 5 — Write the complete electric field.
Keeping the same phase factor as $$\vec{B}$$, the electric field is
$$\vec{E}= -\nu\lambda B_0 \sin\!\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)\hat{k}$$.

Thus the correct option is:
Option A which is: $$-\nu\lambda B_0 \sin\!\left(2\pi\nu t-\frac{2\pi x}{\lambda}\right)\hat{k}$$.