An electromagnetic wave travels in free space along the \(x\)-direction. At a particular point in space and time, \(\vec{B} = 2 \times 10^{-7} \hat{j}\) T is associated with this wave. The value of corresponding electric field \(\vec{E}\) at this point is _______ V/m.

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

Given that the wave travels along the \(x\)-direction, the direction of propagation is \(\hat{i}\).

The magnetic field is given as \(\vec{B} = 2 \times 10^{-7} \hat{j}\) T, so its direction is \(\hat{j}\).

Let the direction of the electric field be \(\hat{n}\). We must have:

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

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

The magnitude of the electric field is related to the magnetic field by \(E = cB\), where \(c = 3 \times 10^8\) m/s is the speed of light in free space.

\(E = (3 \times 10^8) \times (2 \times 10^{-7}) = 60\) V/m

Therefore, the electric field vector is \(\vec{E} = -60 \hat{k}\) V/m.

Answer: \(-60 \hat{k}\)