The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to

If electric field intensity of a uniform plane electro magnetic wave is given as

$E =-301.6 \sin ( kz -\omega t ) \hat{a}_{ x }+452.4 \sin ( kz -\omega t )$ $\hat{a}_{y} \frac{V}{m}$

Then, magnetic intensity $H$ of this wave in $Am ^{-1}$ will be

[Given: Speed of light in vacuum $c =3 \times 10^{8} ms ^{-1}$, permeability of vacuum $\mu_{0}=4 \pi \times 10^{-7} NA ^{-2}$ ]

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……$m$

Two electrons are moving with same speed $v$. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. Then after some time if the de-broglie wavelength of the two are ${\lambda _1}$ and ${\lambda _2}$  then

The magnetic field of a plane electromagnetic Wave is $\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i }\, T$ Where $c=3 \times 10^{8} \,ms ^{-1}$ is the speed of light. The corresponding electric field is