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  • 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 m

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 \sqrt{a}abI,\ along \ (\frac{\hat{j}}{\sqrt{5}}+\frac{2\hat{k}}{\sqrt{5}})
Option: 2 abI,\ along \ (\frac{\hat{j}}{\sqrt{5}}+\frac{2\hat{k}}{\sqrt{5}})
Option: 3 abI,\ along \ (\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}})
Option: 4 \sqrt{2}abI,\ along \ (\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}})

Answers (1)

best_answer

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}}

Posted by

avinash.dongre

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