A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths a and b, then the magnitude and direction of magnetic moment of the loop ABCDEFA is:

Magnetic moment of loop ABCDA in the XY-plane: $$\vec{M}_1 = I(a \times b)\hat{k} = Iab\hat{k}$$

Magnetic moment of loop ADEFA in the XZ-plane: $$\vec{M}_2 = I(a \times b)\hat{j} = Iab\hat{j}$$

Total magnetic dipole moment vector: $$\vec{M} = \vec{M}_1 + \vec{M}_2 = Iab(\hat{j} + \hat{k})$$

Magnitude of the total magnetic moment: $$M = \sqrt{(Iab)^2 + (Iab)^2} = \sqrt{2}abI$$

Direction unit vector: $$\hat{u} = \frac{\vec{M}}{M} = \frac{Iab(\hat{j} + \hat{k})}{\sqrt{2}abI} = \left(\frac{\hat{j}}{\sqrt{2}} + \frac{\hat{k}}{\sqrt{2}}\right)$$