Find the direction of vibration of Electric field if vibration of magnetic field is in positive $x-$ axis and propagation of em wave is along positive $y-$ axis.

The electric field of a plane electromagnetic wave is given by

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is

A plane electromagnetic wave travelling along the $X$-direction has a wavelength of $3\ mm$ . The variation in the electric field occurs in the $Y$-direction with an amplitude $66\  Vm^{-1}$. The equations for the electric and magnetic fields as a function of $x$ and $t$ are respectively :-

An electromagnetic wave with frequency $\omega $ and wavelength $\lambda $ travels in the $+ y$ direction . Its magnetic field is along $+\, x-$ axis. The vector equation for the associated electric field ( of amplitude $E_0$) is

A plane electromagnetic wave is travelling in the positive $X-$axis. At the instant shown electric field at the extremely narrow dashed rectangle is in the $-ve$  $z$ direction and its magnitude is increasing. Which diagram  correctly shows the direction and relative magnitudes of magnetic field at the edges of rectangle :-

A plane electromagnetic wave of frequency $500\, MHz$ is travelling in vacuum along $y-$direction. At a particular point in space and

time, $\overrightarrow{ B }=8.0 \times 10^{-8} \hat{ z } \;T$. The value of electric field at this point is

(speed of light $\left.=3 \times 10^{8}\, ms ^{-1}\right)$

$\hat{ x }, \hat{ y }, \hat{ z }$ are unit vectors along $x , y$ and $z$ direction.