A) 0.012 T
B) 0.120 T
C) 1.200 T
D) 2.10 T
Correct Answer: A
Solution :
[a] Magnetic moment, \[M=8.7\times {{10}^{-2}}\,A{{m}^{2}}\]moment of inertia, \[I=11.5\times {{10}^{-6}}\,kg{{m}^{2}}\] Time period of oscillation is \[T=\frac{6.70}{10}=0.6755\]As,\[T=2\pi \sqrt{\frac{I}{MB}}\] \[B=\frac{4{{\pi }^{2}}{{I}^{2}}}{M{{T}^{2}}}\] \[\therefore \,\,\,\,\,\,B=\frac{4\times {{(3.14)}^{2}}\times 11.5\times {{10}^{-6}}}{8.7\times {{10}^{-2}}\times {{(0.67)}^{2}}}=0.012\,T\]
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