Figure shows a square current carrying loop $A B C D$ of side 10 cm and current $i = 10 A$ . The magnetic moment $\vec{M}$ of the loop is

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$(0.05) (\hat{i} - \sqrt{3} \hat{k}) A - m^{2}$

$(0.05) (\sqrt{3} \hat{i} + \hat{k}) A - m^{2}$

$(\hat{i} + \hat{k}) A - m^{2}$

$(0.05) (\hat{j} + \hat{k}) A - m^{2}$

answer is A.

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Detailed Solution

$\vec{M} = i A (\cos 60^{0} \hat{i} - \sin 60^{0} \hat{j}) = 10 \times (10 \times 10 \times 10^{- 4}) (\frac{\hat{i}}{2} - \frac{\sqrt{3} \hat{j}}{2}) = 0.05 (\hat{i} - \sqrt{3} \hat{j}) A - m^{2}$

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