Electric Field Vector of Electromagnetic Wave

An electromagnetic wave propagating in space satisfies the relation between its electric field vector $\vec{E}$, magnetic field vector $\vec{B}$, and the direction of wave propagation $\hat{c}$:

$\hat{E} \times \hat{B} = \hat{c}$

1. Identify the direction of wave propagation ($\hat{c}$): The given magnetic field is: $\vec{B} = B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{j}$ The phase term $\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) = (\omega t - kx)$ indicates that the wave propagates along the positive $x$-axis. Thus, the unit vector in the direction of propagation is: $\hat{c} = \hat{i}$

2. Determine the direction of the electric field vector ($\hat{E}$): Given that the magnetic field acts along the positive $y$-axis ($\hat{B} = \hat{j}$): $\hat{E} \times \hat{j} = \hat{i}$ Using the standard cross-product relation for orthogonal unit vectors: $(-\hat{k}) \times \hat{j} = \hat{i}$ Therefore, the electric field vector points along the negative $z$-axis ($-\hat{k}$).

3. Determine the magnitude of the electric field amplitude ($E_0$): The relation between the amplitudes of the electric and magnetic fields in an electromagnetic wave is: $E_0 = c B_0$ Where $c$ is the wave speed, given in terms of frequency $\nu$ and wavelength $\lambda$ as $c = \nu \lambda$. Substituting $c$: $E_0 = \nu \lambda B_0$

4. Construct the electric field vector ($\vec{E}$): Combining the magnitude, phase, and direction: $\vec{E} = -\nu \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}$

Thus, the correct option is A.