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

The direction of propagation of the electromagnetic wave is determined by the phase term \(2\pi vt - \dfrac{2\pi x}{\lambda}\), which indicates that the wave is travelling along the \(+x\) direction (\(\hat{i}\)).

The direction of the magnetic field \(\vec{B}\) is given as \(\hat{j}\).

In an electromagnetic wave, the direction of propagation is given by the direction of the cross product \(\vec{E} \times \vec{B}\).

Let the direction of the electric field \(\vec{E}\) be \(\hat{n}\).

\(\hat{n} \times \hat{j} = \hat{i}\)

Since \((-\hat{k}) \times \hat{j} = \hat{i}\), the direction of the electric field must be \(-\hat{k}\).

The amplitude of the electric field is related to the magnetic field by \(E_0 = c B_0\), where \(c\) is the speed of the wave.

The speed of the wave is \(c = v \lambda\), where \(v\) is the frequency.

\(E_0 = v \lambda B_0\)

Therefore, the electric field vector is \(\vec{E} = -v \lambda B_0 \sin\left(2\pi vt - \dfrac{2\pi x}{\lambda}\right) \hat{k}\).

Answer: \(\vec{E}=-v\lambda B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{k}\)