If the amplitude of the magnetic field component of an electromagnetic wave in vacuum is $B_m = 510 \, nT$,then the amplitude of the electric field component of the wave is ......

Electromagnetic waves travel in a medium with a speed of $1.5 \times 10^8 \ m/s$. The relative permeability of the medium is $2.0$. The relative permittivity will be:

The magnetic field of a plane electromagnetic wave is given by $\vec B = B_0 \hat i \cos(kz - \omega t) + B_1 \hat j \cos(kz - \omega t)$,where $B_0 = 3 \times 10^{-5} \, T$ and $B_1 = 2 \times 10^{-6} \, T$. The rms value of the force experienced by a stationary charge $Q = 10^{-4} \, C$ at $z = 0$ is closest to:

$A$ light wave is travelling along the $y$-direction. If the corresponding $\vec E$ vector at any time is along the $x$-axis,the direction of the $\vec B$ vector at that time is along:

If $10 \%$ of the power of a $(100 \pi) \ W$ light bulb is converted to visible radiation,then the average intensity of visible radiation at a distance of $10 \ m$ is (in $W \ m^{-2}$)

$A$ plane electromagnetic wave travels in a medium of relative permeability $\mu_{r} = 1.61$ and relative permittivity $\epsilon_{r} = 6.44$. If the magnitude of the magnetic intensity $H$ is $4.5 \times 10^{-2} \; A m^{-1}$ at a point,what will be the approximate magnitude of the electric field intensity $E$ at that point? (Given: $\mu_{0} = 4 \pi \times 10^{-7} \; N A^{-2}$,$c = 3 \times 10^{8} \; m s^{-1}$)