The photon energy in units of $eV$ for electromagnetic wave of wavelength $2\,cm$ is

A plane electromagnetic wave is propagating along the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}},$ with its polarization along the direction $\hat{\mathrm{k}}$. The correct form of the magnetic field of the wave would be (here $\mathrm{B}_{0}$ is an appropriate constant)

A point source of $e. m.$ radiation has an average power output of $800\,W$ . The maximum value of electric field at a distance $4.0\,m$ from the source is…$V/m$

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.

An electron is constrained to move along the $y-$axis with a speed of $0.1\, c$ (c is the speed of light) in the presence of electromagnetic wave, whose electric field is $\overrightarrow{ E }=30 \hat{ j } \sin \left(1.5 \times 10^{7} t -5 \times 10^{-2} x \right)\, V / m$ The maximum magnetic force experienced by the electron will be: (given $c=3 \times 10^{8}\, ms ^{-1}$ and electron charge $\left.=1.6 \times 10^{-19} C \right)$