The mean intensity of radiation on the surface of the Sun is about $10^{8}\,W/m^2.$ The $rms$ value of the corresponding magnetic field is closet to

The magnetic field in a plane electromagnetic wave is given by, $B_{y}=2 \times 10^{-7} \sin \left(\pi \times 10^{3} x+3 \pi \times 10^{11} t\right) \;T$ Calculate the wavelength.

Assume a bulb of efficiency $2.5\%$ as a point source. The peak values of electric field produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is respectively…..$V\,{m^{ – 1}}$

The electric field part of an electromagnetic wave in vacuum is

$E = 3.1\,NC^{-1}\,cos\,[\,(1.8\,rad\,m^{-1})\,y + (5.4\times 18^8\,rad\,s^{-1})\,t\,]\,\hat i$

The wavelength of this part of electromagnetic wave is……$m$

A velocity selector consists of electric field $\overrightarrow{ E }= E \hat{ k }$ and magnetic field $\overrightarrow{ B }= B \hat{ j }$ with $B =12 mT$.

The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$-axis to pass undeflected is:

(Given, , ass of electron $=9.1 \times 10^{-31} kg$ )