A) \[1.72\times {{10}^{2}}\,W/{{m}^{2}}\]
B) \[1.72\,W/{{m}^{2}}\]
C) \[2.31\,W/{{m}^{2}}\]
D) \[2.31\,\times {{10}^{2}}W/{{m}^{2}}\]
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 \[{{I}_{av}}=\frac{1}{2}\frac{B_{0}^{2}}{{{\mu }_{0}}}\cdot c=\frac{1}{2}\times \frac{{{(12\times {{10}^{-8}})}^{2}}\times 3\times {{10}^{8}}}{4\pi \times {{10}^{-7}}}\] \[=1.72W/{{m}^{2}}\]
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