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

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

$(0 .05) (\overset{\land}{i} - \sqrt{3} \overset{\land}{k}) A - m^{2}$

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

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

answer is A.

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Detailed Solution

$\vec{M} = - 10 \times 10^{- 2} \cos 30 \hat{i} - 10 \times 10^{- 2} \sin 30 \hat{k}$

$\vec{M} = 0.05 (\hat{i} - \sqrt{3} \hat{k}) N - m$

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