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(A) The magnetic dipole moment of a current-carrying loop is given by $m = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the are

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$\frac{m}{2}$

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(A) The magnetic dipole moment of a current-carrying loop is given by $m = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the loop.For the first case,$N_1 = 1$,$R_1 = R$,and $A_1 = \pi R^2$. Thus,$m = I \pi R^2$.When the wire of length $L$ is converted into a ring of $N_2 = 2$ turns,the circumference of the new ring is $2 	imes (2 \pi R_2) = L$. Since $L = 2 \pi R$,we have $4 \pi R_2 = 2 \pi R$,which gives $R_2 = \frac{R}{2}$.The new area is $A_2 = \pi R_2^2 = \pi (\frac{R}{2})^2 = \frac{\pi R^2}{4}$.The new magnetic dipole moment $m'$ is $m' = N_2 I A_2 = 2 	imes I 	imes \frac{\pi R^2}{4} = \frac{I \pi R^2}{2}$.Since $m = I \pi R^2$,we get $m' = \frac{m}{2}$.

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