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

(A) Comparing the given equation $B = 12 \times 10^{-8} \sin(1.20 \times 10^7 z - 3.60 \times 10^{15} t) \text{ T}$ with the standard wave equation,the amplitude of the magnetic field is $B_0 = 12 \times 10^{-8} \text{ T}$.
The average intensity $I_{\text{avg}}$ of an electromagnetic wave is given by the formula:
$I_{\text{avg}} = \frac{B_0^2}{2\mu_0} c$
Substituting the values $B_0 = 12 \times 10^{-8} \text{ T}$,$c = 3 \times 10^8 \text{ m/s}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$:
$I_{\text{avg}} = \frac{(12 \times 10^{-8})^2 \times 3 \times 10^8}{2 \times 4\pi \times 10^{-7}}$
$I_{\text{avg}} = \frac{144 \times 10^{-16} \times 3 \times 10^8}{8\pi \times 10^{-7}}$
$I_{\text{avg}} = \frac{432 \times 10^{-8}}{25.12 \times 10^{-7}} \approx 1.72 \text{ W/m}^2$.