Question
The magnetic field of a beam emerging from a filter facing a floodlight is given by
$${B_0} = 12 \times {10^{ - 8}}\sin \left( {1.20 \times {{10}^7}z - 3.60 \times {{10}^{15}}t} \right)T.$$
What is the average intensity of the beam?
A.
$$1.72 \times {10^2}\,W/{m^2}$$
B.
$$1.72\,W/{m^2}$$
C.
$$2.31\,W/{m^2}$$
D.
$$2.31 \times {10^2}\,W/{m^2}$$
Answer :
$$1.72\,W/{m^2}$$
Solution :
As we know that,
The standard equation of magnetic field
$$B = {B_0}\sin \left( {kx - \omega t} \right)$$
And, the given equation is
$$B = 12 \times {10^{ - 8}}\sin \left( {1.20 \times {{10}^7}z - 3.60 \times {{10}^{15}}t} \right)T.$$
On comparing this equation with standard equation (i), we have.
$${B_0} = 12 \times {10^{ - 8}}$$
So, the average intensity of the beam
$$\eqalign{
& {I_{{\text{av}}}} = \frac{1}{2}\frac{{B_0^2}}{{{\mu _0}}} \cdot c = \frac{1}{2} \times \frac{{{{\left( {12 \times {{10}^{ - 8}}} \right)}^2} \times 3 \times {{10}^8}}}{{4\pi \times {{10}^{ - 7}}}} \cr
& = 1.72\,W/{m^2} \cr} $$