Question

For a beam emerging from a filter facing a floodlight, the magnetic field is given by, $B=B_{0}sin(kz-\omega t)$ , where, $B_0=1.2\times 10^{-8}T\text{,}k=1.2\times 10^7{m}^{-1}\text{,}\omega =3.6\times 10^{15}{s}^{-1}$ . The average intensity of the beam is $\alpha$ . Find the value of $\left[10\alpha \right]$ , where $\left[\right]$ is the greatest integer function.

Answer 1

Answer: 17

Comparing given equation with,
$B=B_{0}sin(kz-\omega t)$
$\therefore I_{av}=\frac{1}{2}\frac{B_0^{2}c}{{\mu }_0}$
$=\frac{1}{2}\times \frac{{\left(12\times {\left(10\right)}^{-8}\right)}^2\times 3\times {\left(10\right)}^8}{4\pi \times {\left(10\right)}^{-7}}=\frac{270}{157}$
$=1.719W{m}^{-2}$
$\approx 1.72W{m}^{-2}=\alpha$ $\dots .$ (Rounding off to 2 decimal places)
$\left[10\alpha \right]=\left[17.2\right]=17$