In a plane electromagnetic wave,the electric field oscillates sinusoidally at a frequency of $2 \times 10^{10} \, Hz$ and an amplitude of $48 \, V/m$. Find the amplitude of the magnetic field.

If the rms value of the electric field of electromagnetic waves at a distance of $3 \ m$ from a point source is $3 \ N C^{-1}$,then the power of the source is (in $W$)

In a plane electromagnetic wave travelling in free space,the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \text{ Hz}$ and amplitude $48 \text{ V m}^{-1}$. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^8 \text{ m s}^{-1}$)

In an electromagnetic wave,at an instant and at a particular position,the electric field is along the negative $z$-axis and the magnetic field is along the positive $x$-axis. Then the direction of propagation of the electromagnetic wave is

$A$ plane electromagnetic wave of frequency $35 \ MHz$ travels in free space along the $X$-direction. At a particular point (in space and time) $\overrightarrow{E} = 9.6 \ \hat{j} \ V/m$. The value of the magnetic field at this point is:

The speed of electromagnetic waves is the same for all ........