Question

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

• A. $1.71\, W\,m^{-2}$
• B. $2.1\, W\,m^{-2}$
• C. $3.2\, W\,m^{-2}$
• D. $2.9\, W\,m^{-2}$

Answer 1

Answer: (A)

Here,
$B = 12 \times 10^{-8} \,sin \left(1.20 \times10^{7}z -3.6 \times 10^{15}\,t\right) T$
Comparing it with,
$B = B_{0} sin \left(kz - \omega t\right)$,
we have $B_{0} = 12 \times 10^{-8 }\,T$
$\therefore \quad I_{av} = \frac{1}{2} \frac{B^{2}_{0}c}{\mu_{0}}$
$= \frac{1}{2} \times \frac{\left(12 \times 10^{-8}\right)^{2} \times 3 \times 10^{8}}{4\pi \times 10^{-7}}$
$= 1.71\, W\,m^{-2}$