Question

In an EM wave, $E=50\sin (\omega t-kx)$. If $\mu =4{\mu }_0$ and $\epsilon ={\epsilon }_0$, then average power per unit area (in $W/m^2$ ) is

Answer 1

Answer: 1.65

Intensity, $I=\frac{1}{2}\epsilon v{E}_0^2$.
Now,
$v=\frac{1}{\sqrt{\mu \epsilon }}=\frac{1}{\sqrt{\left(4{\mu }_0\right)\left({\epsilon }_0\right)}}=\frac{c}{2}$
$I=\frac{1}{2}{\epsilon }_0\left(\frac{c}{2}\right)E_0^2=\frac{1}{2}\times 8.8\times 10^{-12}\times \left(\frac{3\times 10^8}{2}\right)(50{)}^2$
$I=1.65W/m^2$