Question

The magnetic field of a beam emerging from a filter facing a floodlight is given by $B_0=12\times {10}^{-8}\sin (1.20\times {10}^{7}z-3.60\times {10}^{15}t)T$. What is the average intensity of the beam?

Answer 1

Given, magnetic field of the beam
$B_0=12\times {10}^{-8}\sin (1.20\times {10}^{7}z-3.60\times {10}^{15}t)T$

Amplitude of magnetic field,

$B_0=12\times {10}^{-8}T$

So, the average intensity of the beam,

$I_{av}=\frac{1}{2}\frac{c{B}_0^2}{{\mu }_0}$

$I_{av}=\frac{1}{2}\times \frac{3\times {10}^8\times (12\times {10}^{-8}{)}^2}{4\pi \times {10}^{-7}}$

$I_{av}=1.71W/m^2$

Final answer: $1.71\frac{W}{m^2}$