Question

The magnetic field of a beam emerging from a filter facing a flood light as given $B=12\times {10}^{-8}\sin (1.20\times {10}^{7}z-3.6\times {10}^{15}t)$
The average intensity of the beam is

• A. $1.71W{m}^{-2}$
• B. $2.4W{m}^{-2}$
• C. $3.2W{m}^{-2}$
• D. $2.9W{m}^{-2}$

Answer 1

The correct option is A $1.71W{m}^{-2}$

Given: The magnetic field of beam is $B=12\times {10}^{-8}\sin (1.20\times {10}^{7}z-3.6\times {10}^{15}t)$.

To find: The average intensity of the beam.

The standard equation of the magnetic field of a beam is:

$B=B_{0}sin(kz-\omega t)$

On comparing the equation of the magnetic field with the standard equation,
we have $B_0=12\times {10}^{-8}T$

The intensity of tehe beam is given by:
$I_{av}=\frac{1}{2}\frac{B_0^{2}c}{{\mu }_0}\Rightarrow \frac{1}{2}\times \frac{(12\times {10}^{-8}{)}^2\times 3\times {10}^8}{4\pi \times {10}^{-7}}$

$⟹1.71W{m}^{-2}$

Option $\left(A\right)$ is correct.