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(B) The magnetic moment $m$ 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 o

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$\frac{2 \pi BR^3}{\mu_0}$

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(B) The magnetic moment $m$ 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 a single circular loop $(n=1)$ of radius $R$,the magnetic induction $B$ at the centre is given by $B = \frac{\mu_0 I}{2R}$.From this,we can express the current $I$ as $I = \frac{B 	imes 2R}{\mu_0}$.The area of the loop is $A = \pi R^2$.Substituting these into the formula for magnetic moment:$m = I 	imes A = \left( \frac{B 	imes 2R}{\mu_0} ight) 	imes (\pi R^2) = \frac{2 \pi B R^3}{\mu_0}$.

Magnetic induction produced at the centre of a circular loop of radius $R$ carrying a current is $B$. The magnetic moment of the loop is $(\mu_0 = \text{permeability of free space})$

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