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(A) The length of the wire is $L$,which forms the circumference of the circular loop.Therefore,$L = 2 \pi R$,where $R$ is the radius of the circle.Fro

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$\frac{I L^{2}}{4 \pi} \; A \cdot m^{2}$

Vedclass · Answer

(A) The length of the wire is $L$,which forms the circumference of the circular loop.Therefore,$L = 2 \pi R$,where $R$ is the radius of the circle.From this,the radius is $R = \frac{L}{2 \pi}$.The area $A$ of the circular loop is given by $A = \pi R^{2}$.Substituting the value of $R$,we get $A = \pi \left( \frac{L}{2 \pi} \right)^{2} = \pi \left( \frac{L^{2}}{4 \pi^{2}} \right) = \frac{L^{2}}{4 \pi}$.The magnetic moment $M$ of a current-carrying loop is given by $M = I A$.Substituting the value of $A$,we get $M = I \left( \frac{L^{2}}{4 \pi} \right) = \frac{I L^{2}}{4 \pi} \; A \cdot m^{2}$.

$A$ wire of length $L$ metre carrying a current of $I$ ampere is bent in the form of a circle. Its magnetic moment is:

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