In vacuum,the speed of light depends upon

$A$ point source of electromagnetic radiation has an average power output of $800 \, W$. The maximum value of the electric field at a distance $4.0 \, m$ from the source is .... $V/m$.

The electric field in an electromagnetic wave is given as $\vec{E} = 20 \sin \omega (t - \frac{x}{c}) \hat{j} \text{ N/C}$. Where $\omega$ and $c$ are the angular frequency and velocity of the electromagnetic wave,respectively. The energy contained in a volume of $5 \times 10^{-4} \text{ m}^3$ will be $..... \times 10^{-13} \text{ J}$. (Given $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$)

The efficiency of a bulb of power $60 \ W$ is $16 \%$. The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of $2 \ m$ from the bulb is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}\right)$ (in $Vm^{-1}$)

The magnetic field in a plane electromagnetic wave is given as $B = (3 \times 10^{-7} \text{ T}) \sin (3 \times 10^4 x + 9 \times 10^{12} t) \hat{j}$. The electric field of this wave is given as:

The electric field of an electromagnetic wave in free space is represented as $\vec{E} = E_0 \cos(\omega t - kz) \hat{i}$. The corresponding magnetic induction vector will be: