A) \[1.25\text{W}\,{{\text{m}}^{-2}}\]
B) \[1.72\text{W}\,{{\text{m}}^{-2}}\]
C) \[0.2\text{W}\,{{\text{m}}^{-2}}\]
D) \[0.25\text{W}\,{{\text{m}}^{-2}}\]
Correct Answer: B
Solution :
[b]: Comparing\[B={{B}_{0}}\sin (kz-\omega t)\]T with \[B=12\times {{10}^{-8}}\sin (1.20\times {{10}^{7}}z-3.60\times {{10}^{15}}t)\]T, we get\[{{B}_{0}}=12\times {{10}^{-8}}T\] Average intensity of the beam, \[{{I}_{v}}=\frac{cB_{0}^{2}}{2{{\mu }_{0}}}=\frac{3\times {{10}^{8}}\times {{(12\times {{10}^{-8}})}^{2}}}{2\times 4\pi \times {{10}^{-7}}}=1.72W\,{{m}^{-2}}\]
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