A) \[0.012\text{ }T\]
B) \[0.120\text{ }T\]
C) \[1.200\,\,T\]
D) \[2.10\,\,T\]
Correct Answer: A
Solution :
\[m=8.7\times {{10}^{-2}}A-{{m}^{2}}\] \[l=11.5\times {{10}^{-6}}kh-{{m}^{2}}\] \[T=0.67\,s\] \[\Rightarrow \] From the formula \[T=2\pi \sqrt{\frac{l}{mB}}\Rightarrow B=\frac{4{{\pi }^{2}}\times l}{m{{T}^{2}}}\] \[=\frac{4{{\pi }^{2}}\times 11.5\times {{10}^{-6}}}{8.7\times {{10}^{-2}}\times {{(0.67)}^{2}}}\] \[=\frac{453.5\times {{10}^{-6}}}{{{10}^{-2}}\times 3.9}=116.28\times {{10}^{-4}}\] \[=0.0116=0.012T\]
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