If the peak value of the magnetic field of an electromagnetic wave is $30 \times 10^{-9} \ T$,then the peak value of the electric field is (in $Vm^{-1}$)

If the magnetic field of a plane electromagnetic wave is given by (The speed of light $c = 3 \times 10^8 \, m/s$):
$B = 100 \times 10^{-6} \sin \left[ 2\pi \times 2 \times 10^{15} \left( t - \frac{x}{c} \right) \right]$
Then the maximum electric field associated with it is:

The magnetic field of an electromagnetic wave is given by $\vec B = 1.6 \times 10^{-6} \cos(2 \times 10^7 z + 6 \times 10^{15} t) (2\hat i + \hat j) \text{ Wb/m}^2$. The associated electric field will be

In an electromagnetic wave in free space,the root mean square value of the electric field is $E_{rms} = 6 \, V m^{-1}$. The peak value of the magnetic field is:

$A$ point source of electromagnetic radiation has an average power output of $960 \,W$. The peak value of the electric field at a distance $400 \,cm$ from the source is (in $\,Vm^{-1}$)

In an electromagnetic wave,the electric field intensity is $100 \ Vm^{-1}$ and the magnetic intensity is $H_0 = 0.265 \ Am^{-1}$. The maximum intensity of radiation is ..... $Wm^{-2}$.