Energy stored in electromagnetic oscillations is in the form of

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

If an electromagnetic wave is propagating in the $x$-direction and the magnetic field $\vec{B}$ at any instant is in the $z$-direction,then at that instant,the electric field $\vec{E}$ will be in the ........ direction.

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 magnetic field in a plane electromagnetic wave is given by $B = 3.01 \times 10^{-7} \sin(6.28 \times 10^2 x + 2.2 \times 10^{10} t) \, T$,where $x$ is in $cm$ and $t$ is in seconds. The wavelength of the given wave is ....... $cm$.

For an $EM$ wave,the electric field intensity due to a bulb of $100\ W$ at a distance of $3\ m$ is $E$. What is the electric field intensity produced by radiations coming from a $50\ W$ bulb at the same distance?