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(B) The magnetic moment $M$ of a circular loop carrying current $i$ is given by the formula $M = iA$,where $A$ is the area of the loop.For a circular

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$\sqrt{\frac{4\pi M}{i}}$

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(B) The magnetic moment $M$ of a circular loop carrying current $i$ is given by the formula $M = iA$,where $A$ is the area of the loop.For a circular loop of radius $R$,the area is $A = \pi R^2$.If the length of the wire is $L$,then the circumference of the loop is $L = 2\pi R$,which implies $R = \frac{L}{2\pi}$.Substituting the value of $R$ into the area formula,we get $A = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi}$.Now,substituting $A$ into the magnetic moment formula: $M = i \left(\frac{L^2}{4\pi}\right)$.Rearranging the equation to solve for $L$: $L^2 = \frac{4\pi M}{i}$.Therefore,the length of the wire is $L = \sqrt{\frac{4\pi M}{i}}$.

Process	Elasticity Type
$(i)$ Suspension fibre of galvanometer	$(a)$ Linear
$(ii)$ Bending of beam	$(b)$ Shear
$(iii)$ Cutting piece of paper	$(c)$ Bulk
$(iv)$ Mechanical waves in fluid	$(d)$ Shear

$A$ straight wire carrying current $i$ is turned into a circular loop. If the magnitude of the magnetic moment associated with it in $M.K.S.$ units is $M$,the length of the wire will be

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