A) \[1.9\,kW/{{m}^{2}}\]
B) \[3.8\,kW/{{m}^{2}}\]
C) \[5.7\,kW/{{m}^{2}}\]
D) \[7.6\,kW/{{m}^{2}}\]
Correct Answer: A
Solution :
Given that \[Bn=(4.0\times {{10}^{-6}}T)\sin \,[(1.57\times {{10}^{7}}{{m}^{-1}})y+\omega \,t]\] This equation comparing with \[B={{B}_{0}}\sin \left[ \frac{2\pi V}{\lambda }+\omega \,t \right]\], we get \[{{B}_{o}}=4\times {{10}^{-6}}T\] Now, from \[l=\frac{CB_{0}^{2}}{2{{\mu }_{0}}}\] but \[{{C}^{2}}=\frac{1}{{{\varepsilon }_{0}}{{H}_{0}}}\] \[\therefore \] \[l=\frac{1}{2}{{C}^{3}}{{\varepsilon }_{0}}B_{0}^{2}\] \[l=1.9\,kW/{{m}^{2}}\]
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