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 \[2l=d=20\,cm\,=0.2\,cm\] Hence, its magnetic moment will be \[M=m\times 2l\] \[=0.5\times 0.2\] \[=0.1\,A{{m}^{2}}\]
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