On a particular day, the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^3\right) \frac{\mathrm{W}}{\mathrm{m}^2}$ to the top of earth's atmosphere. Find the amplitude of magnetic field for the electromagnetic waves above atmosphere.

(Take, $\mu_0=4 \pi \times 10^{-7}$ SI unit)

A laser beam has intensity $2.1 \times 10^{15} \mathrm{~W} / \mathrm{m}^2$. The amplitude of magnetic field in the beam in approximately is

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 $\frac{\alpha}{25 \pi} \mathrm{~W} / \mathrm{m}^2$. The value of $\alpha$ is

An electromagnetic wave has its electric and magnetic fields given by

$$ \mathbf{E}(t)=\mathbf{E}_m \sin (k x-\omega t) ; \quad \mathbf{B}(t)=\mathbf{B}_m \sin (k x-\omega t) $$

If the direction of $\mathbf{E}_m$ and $\mathbf{B}_m$ are in the direction of $(\hat{\mathbf{i}}+\hat{\mathbf{j}})$ and $(\hat{\mathbf{i}}-\hat{\mathbf{j}})$ respectively, the unit vector that gives the direction of propagation of the wave is