A) 0.1 J
B) 0.2 J
C) 0.4 J
D) 0.8 J
Correct Answer: A
Solution :
The magnetic moment of current carrying loop \[M=niA=ni\,(\pi {{r}^{2}})\] Hence the work done in rotating it through 180° \[W=MB\,(1-\cos \theta )=2MB\]\[=2(ni\pi {{r}^{2}})B\] \[=2\times (50\times 2\times 3.14\times 16\times {{10}^{-4}})\times 0.1=0.1J\]
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