A) 0.12 \[A{{m}^{2}}\]
B) 0.1 \[A{{m}^{2}}\]
C) 0.05 \[A{{m}^{2}}\]
D) 0.01 \[A{{m}^{2}}\]
Correct Answer: B
Solution :
On bending the magnet in the form of semicircle, let diameter of magnet is d. Then, length of magnet \[=\pi \frac{d}{2}\] \[\therefore \] \[31.4=\frac{\pi d}{2}\] or \[d=\frac{31.4\times 2}{3.14}=20\,\,cm\] Therefore, effective length of magnet \[2\,l=d=20\,cm=0.2\,cm\] Hence, its magnetic moment will be \[M=m\times 2\,l\] \[=0.5\times 0.2\] \[=0.1\,A{{m}^{2}}\]
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