In an electromagnetic wave,the phase difference between the electric and magnetic field vectors $\vec{E}$ and $\vec{B}$ is:

The electric fields of two plane electromagnetic waves in vacuum are given by $\overrightarrow{E}_{1}=E_{0} \hat{j} \cos (\omega t-kx)$ and $\overrightarrow{E}_{2}=E_{0} \hat{k} \cos (\omega t-ky)$. At $t=0$,a particle of charge $q$ is at the origin with a velocity $\overrightarrow{v}=0.8 c \hat{j}$ ($c$ is the speed of light in vacuum). The instantaneous force experienced by the particle is:

$A$ radio transmitter transmits at $830 \, kHz$. At a certain distance from the transmitter,the magnetic field has an amplitude of $4.82 \times 10^{-11} \, T$. The electric field and the wavelength are respectively:

$A$ plane electromagnetic wave of frequency $100 \, MHz$ is travelling in vacuum along the $x$-direction. At a particular point in space and time,$\overrightarrow{B} = 2.0 \times 10^{-8} \hat{k} \, T$ (where $\hat{k}$ is the unit vector along the $z$-direction). What is $\overrightarrow{E}$ at this point?

$A$ 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 of $66\, Vm^{-1}$. The equations for the electric and magnetic fields as a function of $x$ and $t$ are respectively:

About $20 \%$ of the power of a $100 \ W$ bulb is converted to visible radiation. Assuming that the radiation is emitted isotropically and neglecting reflection,the average intensity of visible radiation at a distance of $5 \ m$ is $\frac{\alpha}{25 \pi} \ W/m^2$. The value of $\alpha$ is