A current carrying wire of length L is transf | vedclass

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(A) The magnetic moment $M$ of a coil is given by $M = N I A$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the coil.For

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Must have just one turn and would be circular in shape

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(A) The magnetic moment $M$ of a coil is given by $M = N I A$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the coil.For a circular coil of $N$ turns and radius $R$,the total length of the wire is $L = N(2 \pi R)$,which implies $R = \frac{L}{2 \pi N}$.The area of the circular coil is $A = \pi R^2 = \pi \left( \frac{L}{2 \pi N} \right)^2 = \frac{L^2}{4 \pi N^2}$.Substituting this into the magnetic moment formula: $M = N I \left( \frac{L^2}{4 \pi N^2} \right) = \frac{I L^2}{4 \pi N}$.Since $I$,$L$,and $\pi$ are constants,$M$ is inversely proportional to $N$ $(M \propto \frac{1}{N})$.To maximize $M$,we must minimize $N$. Therefore,$N = 1$ gives the maximum magnetic moment.For a fixed perimeter,a circle encloses the maximum area compared to other shapes,so the coil must be circular.

$A$ current carrying wire of length $L$ is transformed into a coil of $N$ turns. To achieve maximum magnetic moment,the coil

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