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(C) The magnetic moment $M$ of a current-carrying coil is given by $M = NiA$,where $N$ is the number of turns,$i$ is the current,and $A$ is the area o

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Directly proportional to the square of the length of the wire in the coil

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(C) The magnetic moment $M$ of a current-carrying coil is given by $M = NiA$,where $N$ is the number of turns,$i$ is the current,and $A$ is the area of the coil.For a circular coil of radius $r$,the area $A = \pi r^2$.The length of the wire $l$ used to make $N$ turns is $l = N(2\pi r)$,which implies $r = \frac{l}{2\pi N}$.Substituting $A$ in the formula for $M$: $M = Ni(\pi r^2) = Ni\pi (\frac{l}{2\pi N})^2 = Ni\pi \frac{l^2}{4\pi^2 N^2} = \frac{il^2}{4\pi N}$.Assuming the number of turns $N$ is constant,we find that $M \propto l^2$.Therefore,the magnetic moment is directly proportional to the square of the length of the wire.

The magnetic moment of a circular coil carrying current is

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