The electric field for a plane electromagnetic wave travelling in the $+y$ direction is  shown.  Consider a point where $\vec E$ is in $+z$ direction. The $\vec B$ field is

The electric field intensity produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is $E$. The electric field intensity produced by the radiation coming from $60\, W$ at the same distance is $\sqrt{\frac{x}{5}} E$. Where the value of $x=……… .$

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 $\left.=1.6 \times 10^{-19} C \right)$

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

An em wave is propagating in a medium with a velocity $\vec v =v\hat i.$ The instantaneous oscillating electric field of this em wave is along $+y$ axis. Then the direction of oscillating magnetic field of the em wave will be along