# A single piece of wire carrying a current I is bent in the shape ABCDEF as shown where the rectangle ABCDA and AOEFA are perpendicular to each other.If the sides of the rectangles are of the length a and b then the magnitude and direction of magnetic moment of the loop  ABCDEFA   Option: 1 Option: 2 Option: 3 Option: 4

$M_1=IA\hat{k}=Iab\hat{k}$ and $M_2=IA\hat{j}=Iab\hat{j}$

$|M_{net}|=\sqrt{(M_1)^2+(M_2)^2}=\sqrt{(I^2a^2b^2)+(I^2a^2b^2)}=\sqrt{2}Iab$

And the direction of net magnetic moment is along $\vec{u}$ where $\vec{u}=\hat{j}+\hat{k}$

So $\vec{M_{net}}=\sqrt{2}Iab\hat{u}=\sqrt{2}Iab*\frac{\vec{u}}{|\vec{u}|}=\sqrt{2}Iab*\frac{(\hat{k}+\hat{j})}{\sqrt{2}}$

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