The figure shows a square current-carrying lo | vedclass

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(A) The area of the square loop is $A = (10\,cm)^2 = (0.1\,m)^2 = 0.01\,m^2$.The magnitude of the magnetic moment is $M = iA = 10\,A 	imes 0.01\,m^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The area of the square loop is $A = (10\,cm)^2 = (0.1\,m)^2 = 0.01\,m^2$.The magnitude of the magnetic moment is $M = iA = 10\,A 	imes 0.01\,m^2 = 0.1\,A\cdot m^2$.According to the right-hand rule,the direction of the area vector $\vec{A}$ is perpendicular to the plane of the loop.The loop lies in a plane such that its normal makes an angle of $60^{\circ}$ with the $x$-axis and $30^{\circ}$ with the $-z$-axis (based on the geometry provided in the figure).Thus,the magnetic moment vector is $\vec{M} = M \cos(60^{\circ}) \hat{i} - M \cos(30^{\circ}) \hat{k}$.Substituting the values: $\vec{M} = 0.1 	imes (1/2) \hat{i} - 0.1 	imes (\sqrt{3}/2) \hat{k}$.$\vec{M} = 0.05 \hat{i} - 0.05\sqrt{3} \hat{k} = 0.05 (\hat{i} - \sqrt{3} \hat{k})\,A\cdot m^2$.

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

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