A current i flows in a horizontal square loop | vedclass

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Question

(C) The initial magnetic moment of the square loop is $\mu_1 = iL^2$,directed perpendicular to the plane of the loop.When the loop is folded by $90^\c

Vedclass · Accepted Answer

$\frac{{|\overrightarrow {{\mu _1}} |}}{{|\overrightarrow {{\mu _2}} |}} = \sqrt 2 $

Vedclass · Answer

(C) The initial magnetic moment of the square loop is $\mu_1 = iL^2$,directed perpendicular to the plane of the loop.When the loop is folded by $90^\circ$,it consists of two perpendicular loops,each of area $A' = L 	imes (L/2) = L^2/2$.The magnetic moment of each half-loop is $M = iA' = iL^2/2 = \mu_1/2$.These two magnetic moments are perpendicular to each other,one in the horizontal plane and one in the vertical plane.The resultant magnetic moment $\mu_2$ is the vector sum of these two moments: $\mu_2 = \sqrt{M^2 + M^2} = M\sqrt{2}$.Substituting $M = \mu_1/2$,we get $\mu_2 = (\mu_1/2) 	imes \sqrt{2} = \mu_1/\sqrt{2}$.Therefore,the ratio is $\frac{|\overrightarrow {{\mu _1}} |}{|\overrightarrow {{\mu _2}} |} = \frac{\mu_1}{\mu_1/\sqrt{2}} = \sqrt{2}$.

$A$ current $i$ flows in a horizontal square loop of side $L$. Now,half of the loop is folded to be vertical. If $\overrightarrow {{\mu _1}}$ and $\overrightarrow {{\mu _2}}$ are the magnetic moments before and after folding,respectively,then:

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