Em Waves Question 139
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?
Options:
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} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] As we know that, The standard equation of magnetic field $ B=B _{0}\sin (kx-\omega t) $ And, the given equation is $ B=12\times {{10}^{-8}}\sin (1.20\times 10^{7}z-3.60\times 10^{15}t)T $ . On comparing this equation with standard equation (i), we have. $ B _{0}=12\times {{10}^{-8}} $ So, the average intensity of the beam $ I _{av}=\frac{1}{2}\frac{B _{0}^{2}}{{\mu _{0}}}\cdot c=\frac{1}{2}\times \frac{{{(12\times {{10}^{-8}})}^{2}}\times 3\times 10^{8}}{4\pi \times {{10}^{-7}}} $
$ =1.72W/m^{2} $
BETA
